Solutions Manual For Linear Algebra And Its Applications (4th Edition)

INSTRUCTOR

SOLUTIONS

MANUAL

INSTRUCTOR’S

SOLUTIONS MANUAL

THOMAS POLASKI

Winthrop University

JUDITH MCDONALD

Washington State University

LINEAR ALGEBRA

AND ITS APPLICATIONS

FOURTH EDITION

David C. Lay

University of Maryland

The author and publisher of this book have used their best efforts in preparing this book. These efforts include the

development, research, and testing of the theories and programs to determine their effectiveness. The author and

publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation

contained in this book. The author and publisher shall not be liable in any event for incidental or consequential

damages in connection with, or arising out of, the furnishing, performance, or use of these programs.

Reproduced by Pearson Addison-Wesley from electronic files supplied by the author.

Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116.

form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written

permission of the publisher. Printed in the United States of America.

ISBN-13: 978-0-321-38888-9

ISBN-10: 0-321-38888-7

1 2 3 4 5 6 BB 15 14 13 12 11

iii

_____________________________________________________

Contents

CHAPTER 1 Linear Equations in Linear Algebra 1

CHAPTER 2 Matrix Algebra 87

CHAPTER 3 Determinants 167

CHAPTER 4 Vector Spaces 197

CHAPTER 5 Eigenvalues and Eigenvectors 273

CHAPTER 6 Orthogonality and Least Squares 357

CHAPTER 7 Symmetric Matrices and Quadratic Forms 405

CHAPTER 8 The Geometry of Vector Spaces 453

1

1.1 SOLUTIONS

Notes:

The key exercises are 7 (or 11 or 12), 19–22, and 25. For brevity, the symbols R1, R2,…, stand

for row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section.

1.

12

57

27 5

xx

+=

−− =−

157

275

ª

º

«

»

−−−

¬

¼

Replace R2 by R2 + (2)R1 and obtain:

12

2

57

39

xx

x

+=

=

157

039

ª

º

«

»

¬

¼

Scale R2 by 1/3:

12

2

57

3

xx

x

+=

=

157

013

ª

º

«

»

¬

¼

Replace R1 by R1 + (–5)R2:

1

2

8

3

x

=−

=

10 8

01 3

−

ª

º

«

»

¬

¼

The solution is (x

1

, x

2

) = (–8, 3), or simply (–8, 3).

2.

12

36 3

57 10

xx

+=−

+=

36 3

5710

−

ª

º

«

»

¬

¼

Scale R1 by 1/3 and obtain:

12

21

57 10

xx

+=−

+=

12 1

5710

−

ª

º

«

»

¬

¼

Replace R2 by R2 + (–5)R1:

12

2

21

315

xx

x

+=−

−=

121

0315

−

ª

º

«

»

−

¬

¼

Scale R2 by –1/3:

12

2

21

5

xx

x

+=−

=−

12 1

01 5

−

ª

º

«

»

−

¬

¼

Replace R1 by R1 + (–2)R2:

1

2

9

5

x

=

=−

10 9

01 5

ª

º

«

»

−

¬

¼

The solution is (x

1

, x

2

) = (9, –5), or simply (9, –5).

2 CHAPTER 1 • Linear Equations in Linear Algebra

3. The point of intersection satisfies the system of two linear equations:

12

24

1

xx

+=

−=

124

111

ªº

«»

−

¬¼

Replace R2 by R2 + (–1)R1 and obtain:

12

2

24

33

xx

x

+=

−=−

124

033

ª

º

«

»

−−

¬

¼

Scale R2 by –1/3:

12

2

24

1

xx

x

+=

=

124

011

ª

º

«

»

¬

¼

Replace R1 by R1 + (–2)R2:

1

2

1

x

=

102

011

ª

º

«

»

¬

¼

The point of intersection is (x

1

, x

2

) = (2, 1).

4. The point of intersection satisfies the system of two linear equations:

12

213

32 1

xx

+=−

−=

1213

32 1

−

ªº

«»

−

¬¼

Replace R2 by R2 + (–3)R1 and obtain:

12

2

213

840

xx

x

+=−

−=

1213

0840

−

ª

º

«

»

−

¬

¼

Scale R2 by –1/8:

12

2

213

5

xx

x

+=−

=−

12 13

01 5

−

ª

º

«

»

−

¬

¼

Replace R1 by R1 + (–2)R2:

1

2

3

5

x

=−

10 3

01 5

−

ª

º

«

»

−

¬

¼

The point of intersection is (x

1

, x

2

) = (–3, –5).

5. The system is already in “triangular” form. The fourth equation is x

4

= –5, and the other equations do

not contain the variable x

4

. The next two steps should be to use the variable x

3

in the third equation to

eliminate that variable from the first two equations. In matrix notation, that means to replace R2 by

its sum with –4 times R3, and then replace R1 by its sum with 3 times R3.

6. One more step will put the system in triangular form. Replace R4 by its sum with –4 times R3, which

produces

164 0 1

02704

00123

000714

−−

ªº

«»

−

«»

−

«»

−

¬¼

. After that, the next step is to scale the fourth row by –1/7.

7. Ordinarily, the next step would be to interchange R3 and R4, to put a 1 in the third row and third

column. But in this case, the third row of the augmented matrix corresponds to the equation 0 x

1

+ 0

x

2

+ 0 x

3

= 1, or simply, 0 = 1. A system containing this condition has no solution. Further row

operations are unnecessary once an equation such as 0 = 1 is evident. The solution set is empty.

1.1 • Solutions 3

8. The standard row operations are:

15400 15400 15400 15400

01010 01010 01000 01000

~~~

00300 00300 00300 00100

00020 00010 00010 00010

15000 10000

01000 01000

~~

00100 00100

00010 00010

−−−−

ªºªºªºªº

«»«»«»«»

¬¼¬¼¬¼¬¼

−

ªºªº

«»«»

«»«

¬¼¬¼

»

The solution set contains one solution: (0, 0, 0, 0).

9. The system has already been reduced to triangular form. Begin by replacing R3 by R3 + (3)R4:

11005 11005

01207 01207

~

00 132 00 1014

000 14 00014

−−−−

ªºªº

«»«»

−− −−

«»«»

−

«»«»

¬¼¬¼

Next, replace R2 by R2 + (2)R3. Finally, replace R1 by R1 + R2:

11005 100016

010021 010021

~~

001014 001014

00014 00014

−−

ªºªº

«»«»

¬¼¬¼

The solution set contains one solution: (16, 21, 14, 4).

10. The system has already been reduced to triangular form. Use the 1 in the fourth row to change the 3

and –2 above it to zeros. That is, replace R2 by R2 + (-3)R4 and replace R1 by R1 + (2)R4. For the

final step, replace R1 by R1 + (-3)R2.

130 2 7 1300 11 1000 47

010 3 6 0100 12 0100 12

~~

001 0 2 0010 2 0010 2

000 1 2 0001 2 0001 2

−− − −

ªºªºªº

«»«»«»

−−−

¬¼¬¼¬¼

The solution set contains one solution: (–47, 12, 2, –2).

11. First, swap R1 and R2. Then replace R3 by R3 + (–2)R1. Finally, replace R3 by R3 + (1)R2.

015 4 143 2 1 4 3 2 143 2

143 2~0 15 4~0 1 5 4~0 15 4

271 2 271 2 0 1 5 2 000 2

−− −−

ªºªºª ºªº

«»«»« »«»

−− −−

«»«»« »«»

−−−−−

¬¼¬¼¬ ¼¬¼

The system is inconsistent, because the last row would require that 0 = –2 if there were a solution.

The solution set is empty.

4 CHAPTER 1 • Linear Equations in Linear Algebra

12. Replace R2 by R2 + (–2)R1 and replace R3 by R3 + (2)R1. Finally, replace R3 by R3 + (3)R2.

1543 1543 1543

2732~0354~0354

217109157 0005

−− − − − −

ªºªºªº

«»«»«»

−− − −

«»«»«»

−−−−

¬¼¬¼¬¼

The system is inconsistent, because the last row would require that 0 = 5 if there were a solution.

The solution set is empty.

13.

10 3 8 10 3 8 10 3 8 10 3 8

22 9 7~0215 9~01 5 2~01 5 2

01 5 2 01 5 2 0215 9 00 5 5

−−−−

ªºªºªºªº

«»«»«»«»

−−−

«»«»«»«»

−−−−

¬¼¬¼¬¼¬¼

10 3 8 100 5

~0 1 5 2~0 1 0 3

00 1 1 001 1

−

ªºªº

«»«»

−

«»«»

−−

¬¼¬¼

. The solution is (5, 3, –1).

14.

20 6 8 10 3 4 10 3 4 10 3 4

01 2 3~01 2 3~01 2 3~01 2 3

36 2 4 36 2 4 06 7 8 00 5 10

−− −− −− − −

ªºªºªºªº

«»«»«»«»

−− −− −−

¬¼¬¼¬¼¬¼

10 3 4 10 3 4 100 2

~0 1 2 3~0 1 0 1~0 1 0 1.

00 1 2 00 1 2 001 2

−− −−

ªºªºªº

«»«»«»

−−

«»«»«»

¬¼¬¼¬¼

The solution is (2, –1, 2).

15. First, replace R3 by R3 + (1)R1, then replace R4 by R4 + (1)R2, and finally replace R4 by R4 + (–

1)R3.

16005 16005 16005 16005

01410 01410 01410 01410

~~~

16 153 00 158 00 158 00 158

01540 01540 00150 00008

−−−−

ªºªºªºªº

«»«»«»«»

−−−−

«»«»«»«»

−

«»«»«»«»

−− −

¬¼¬¼¬¼¬¼

The system is inconsistent, because the last row would require that 0 = –8 if there were a solution.

16. First replace R4 by R4 + (3/2)R1 and replace R4 by R4 + (–2/3)R2. (One could also scale R1 and R2

before adding to R4, but the arithmetic is rather easy keeping R1 and R2 unchanged.) Finally, replace

R4 by R4 + (–1)R3.

200 4 10 200 4 10 200 4 10 200 4 10

033 0 0 033 0 0 033 0 0 033 0 0

~~~

001 4 1 001 4 1 001 4 1 001 4 1

323 1 5 023 5 10 001 5 10 000 9 9

−− −− −− −−

ªºªºªºªº

«»«»«»«»

−−−−

«»«»«»«»

−−− − − − −

¬¼¬¼¬¼¬¼

The system is now in triangular form and has a solution. In fact, using the argument from Example 2,

one can see that the solution is unique.

1.1 • Solutions 5

17. Row reduce the augmented matrix corresponding to the given system of three equations:

231 231 231

650~043~043

257 088 002

−−−

ªºªºªº

«»«»«»

−−

«»«»«»

−−

¬¼¬¼¬¼

The third equation, 0 = 2, shows that the system is inconsistent, so the three lines have no point in

common.

18. Row reduce the augmented matrix corresponding to the given system of three equations:

24 4 4 2 4 4 4 24 4 4

01 2 2~0 1 2 2~01 2 2

23 0 0 0 1 4 4 00 6 6

ªºª ºªº

«»« »«»

−− −− −−

«»« »«»

−−− −−

¬¼¬ ¼¬¼

The system is consistent, and using the argument from Example 2, there is only one solution. So the

three planes have only one point in common.

19.

141 4

~

368 063 4

hh

h

ªºª º

«»« »

−−

¬¼¬ ¼

Write c for 6 – 3h. If c = 0, that is, if h = 2, then the system has no

solution, because 0 cannot equal –4. Otherwise, when h ≠ 2, the system has a solution.

20.

151 5

~

286 08216

hh

h

−−

ªºª º

«»« »

−−−

¬¼¬ ¼

Write c for

82.h−−

If c = 0, that is, if h = –4, then the system

has no solution, because 0 cannot equal 16. Otherwise, when h ≠ –4, the system has a solution.

21.

14 2 1 4 2

~

360120hh

−−

ªºª º

«»« »

−−

¬¼¬ ¼

Write c for

12h−

. Then the second equation cx

2

= 0 has a solution

for every value of c. So the system is consistent for all h.

22.

412

412 ~

263 003

2

h

−

ªº

−

ªº

«»

−− −+

¬¼

The system is consistent if and only if

32

h

−+

= 0, that is, if

and only if h = 6.

23. a. True. See the remarks following the box titled Elementary Row Operations.

b. False. A 5 × 6 matrix has five rows.

c. False. The description applies to a single solution. The solution set consists of all possible

solutions. Only in special cases does the solution set consist of exactly one solution. Mark a

statement True only if the statement is always true.

d. True. See the box before Example 2.

24. a. False. The definition of row equivalent requires that there exist a sequence of row operations that

transforms one matrix into the other.

b. True. See the box preceding the subsection titled Existence and Uniqueness Questions.

c. False. The definition of equivalent systems is in the second paragraph after equation (2).

d. True. By definition, a consistent system has at least one solution.

6 CHAPTER 1 • Linear Equations in Linear Algebra

25.

147 147 147

035 ~035 ~035

259 035 2 000 2

gg g

hh h

kkgkgh

−− −

ªºª ºª º

«»« »« »

−− −

«»« »« »

−− − +++

¬¼¬ ¼¬ ¼

Let b denote the number k + 2g + h. Then the third equation represented by the augmented matrix

above is 0 = b. This equation is possible if and only if b is zero. So the original system has a solution

if and only if k + 2g + h = 0.

26. Row reduce the augmented matrix for the given system:

24 12 /2 1 2 /2

~~

02 (/2)

ff f

cd g cd g d cgcf

ªºª ºª º

«»« »« »

−−

¬¼¬ ¼¬ ¼

This shows that d – 2c must be nonzero, since f and g are arbitary. Otherwise, for some choices of f

and g the second row would correspond to an equation of the form 0 = b, where b is nonzero. Thus

d 2c.

27. Row reduce the augmented matrix for the given system. Scale the first row by 1/a, which is possible

since a is nonzero. Then replace R2 by R2 + (–c)R1.

1/ / 1 / /

~~

0(/)(/)

abf bafa ba fa

cd g c d g dcba gcfa

ªºª ºª º

«»« »« »

−−

¬¼¬ ¼¬ ¼

The quantity d – c(b/a) must be nonzero, in order for the system to be consistent when the quantity

g – c( f /a) is nonzero (which can certainly happen). The condition that d – c(b/a) ≠ 0 can also be

written as ad – bc ≠ 0, or ad ≠ bc.

28. A basic principle of this section is that row operations do not affect the solution set of a linear

system. Begin with a simple augmented matrix for which the solution is obviously (3, –2, –1), and

then perform any elementary row operations to produce other augmented matrices. Here are three

examples. The fact that they are all row equivalent proves that they all have the solution set (3, –2, –

1).

100 3 100 3 100 3

010 2~210 4~2104

001 1 001 1 2015

ªºªºªº

«»«»«»

−

«»«»«»

−−

¬¼¬¼¬¼

29. Swap R1 and R3; swap R1 and R3.

30. Multiply R3 by –1/5; multiply R3 by –5.

31. Replace R3 by R3 + (–4)R1; replace R3 by R3 + (4)R1.

32. Replace R3 by R3 + (–4)R2; replace R3 by R3 + (4)R2.

33. The first equation was given. The others are:

21 3213

(2040)/4,or4 60TT T TTT=+++ −−=

342 342

(4030)/4,or4 70TTT TTT=+++ −−=

413 413

(10 30) / 4, or 4 40TTT TTT=+++ −−=

1.1 • Solutions 7

Rearranging,

12 4

123

234

134

430

460

470

440

TT T

TTT

−−=

−+−=

−−+=

34. Begin by interchanging R1 and R4, then create zeros in the first column:

410130 101440 101440

14 1060 14 1060 04 0 420

~~

014170 014170 014170

10 1440 4 10 130 0 1415190

−− −− −−

ªºªºªº

«»«»«»

−− −− −

«»«»«»

−− −− − −

«»«»«»

−− −− −−

¬¼¬¼¬¼

Scale R1 by –1 and R2 by 1/4, create zeros in the second column, and replace R4 by R4 + R3:

10 14 40 10 14 40 1014 40

0101 5 0101 5 0101 5

~~~

014170 004275 004275

01415190 00414195 00012270

−− −− −−

ªºªºªº

«»«»«»

−−−

«»«»«»

−− − −

«»«»«»

−− −

¬¼¬¼¬¼

Scale R4 by 1/12, use R4 to create zeros in column 4, and then scale R3 by 1/4:

10 1 4 40 10 10 50 10 10 50

010 1 5 010027.5 010027.5

~~~

004 2 75 0040 120 0010 30

000 122.5 000122.5 000122.5

−−

ªºªºªº

«»«»«»

−

«»«»«»

−

«»«»«»

¬¼¬¼¬¼

The last step is to replace R1 by R1 + (–1)R3:

100020.0

010027.5

~.

001030.0

000122.5

ªº

«»

¬¼

The solution is (20, 27.5, 30, 22.5).

Notes:

The Study Guide includes a “Mathematical Note” about statements, “If … , then … .”

This early in the course, students typically use single row operations to reduce a matrix. As a result,

even the small grid for Exercise 34 leads to about 80 multiplications or additions (not counting operations

with zero). This exercise should give students an appreciation for matrix programs such as MATLAB.

Exercise 14 in Section 1.10 returns to this problem and states the solution in case students have not

already solved the system of equations. Exercise 31 in Section 2.5 uses this same type of problem in

connection with an LU factorization.

For instructors who wish to use technology in the course, the Study Guide provides boxed MATLAB

notes at the ends of many sections. Parallel notes for Maple, Mathematica, and the TI-83+/84+/89

calculators appear in separate appendices at the end of the Study Guide. The MATLAB box for Section

1.1 describes how to access the data that is available for all numerical exercises in the text. This feature

has the ability to save students time if they regularly have their matrix program at hand when studying

linear algebra. The MATLAB box also explains the basic commands replace, swap, and scale.

These commands are included in the text data sets, available from the text web site,

www.pearsonhighered.com/lay.

8 CHAPTER 1 • Linear Equations in Linear Algebra

1.2 SOLUTIONS

Notes:

The key exercises are 1–20 and 23–28. (Students should work at least four or five from Exercises

7–14, in preparation for Section 1.5.)

1. Reduced echelon form: a and b. Echelon form: d. Not echelon: c.

2. Reduced echelon form: a. Echelon form: b and d. Not echelon: c.

3.

124 8 12 4 8 12 4 8

246 8~00 2 8~00 1 4

36912 00 3 12 00 3 12

ªºª ºª º

«»« »« »

−−

«»« »« »

−− −−

¬¼¬ ¼¬ ¼

1248 120 8

~0 0 1 4~0 0 1 4

0000 000 0

−

ªºª º

«»« »

¬¼¬ ¼

. Pivot cols 1 and 3.

124 8

246 8

36912

ª

º

«

»

«

»

«

»

¬

¼

4.

1245 1 2 4 5 1 2 4 5 1 2 4 5

2454~0 0 3 6~0 3 12 18~0 1 4 6

4542 0 3 12 18 0 0 3 6 0 0 3 6

ªºª ºª ºª º

«»« »« »« »

−− −− −

«»« »« »« »

−− − − − −−

¬¼¬ ¼¬ ¼¬ ¼

1245 124 5 100 1

~0 1 4 6~0 1 0 2~0 1 0 2

0012 001 2 001 2

ªºª ºª º

«»« »« »

−−

«»« »« »

¬¼¬ ¼¬ ¼

.

Pivot cols

1, 2, and 3

1245

2454

4542

ªº

«»

¬¼

5. **0

,,

00000

ªºªºªº

«»«»«»

¬¼¬¼¬¼

 



6.

**0

0,00,00

00 0000

ª

ºª ºª º

«

»« »« »

«

»« »« »

«

»« »« »

¬

¼¬ ¼¬ ¼

 



7. 1347 13 4 7 1347 130 5

~~~

397 6 00 5 15 00 13 00 1 3

−

ªºªºªºªº

«»«»«»«»

−−

¬¼¬¼¬¼¬¼

Corresponding system of equations:

12

3

35

3

xx

x

+=−

=

The basic variables (corresponding to the pivot positions) are x

1

and x

3

. The remaining variable x

2

is

free. Solve for the basic variables in terms of the free variable. The general solution is

12

2

3

53

is free

3

xx

x

=−−



°

®

°=

¯

Note:

Exercise 7 is paired with Exercise 10.

1.2 • Solutions 9

8. 1305130513051004

~~~

3709020601030103

−− −− −−

ªºªºªºªº

«»«»«»«»

−−−

¬¼¬¼¬¼¬¼

Corresponding system of equations:

1

2

4

3

x

=

The basic variables (corresponding to the pivot positions) are x

1

and x

2

. The remaining variable x

3

is

free. Solve for the basic variables in terms of the free variable. In this particular problem, the basic

variables do not depend on the value of the free variable.

General solution:

1

2

3

4

3

is free

x

=



°=

®

°

¯

Note:

A common error in Exercise 8 is to assume that x

3

is zero. To avoid this, identify the basic

variables first. Any remaining variables are free. (This type of computation will arise in Chapter 5.)

9. 0123 1346 1023

~~

1346 0 12 3 0123

−−−−

ªºªºªº

«»«»«»

−− − −

¬¼¬¼¬¼

Corresponding system:

13

23

xx

−=

Basic variables: x

1

, x

2

; free variable: x

3

. General solution:

13

23

3

32

is free

xx

x

=+



°=+

®

°

¯

10. 12 14 12 14 1202

~~

2456 00714 0012

−− −− −

ªºªºªº

«»«»«»

−− − −

¬¼¬¼¬¼

Corresponding system:

12

3

22

2

xx

x

−=

=−

Basic variables: x

1

, x

3

; free variable: x

2

. General solution:

12

2

3

22

is free

2

xx

x

=+



°

®

°=−

¯

11.

3240 3240 123430

96120~0000~0 0 0 0

6480 0000 0 0 0 0

−− −

ªºª ºª º

«»« »« »

−

«»« »« »

−

¬¼¬ ¼¬ ¼

Corresponding system:

123

24

0

33

00

xx x−+=

=

10 CHAPTER 1 • Linear Equations in Linear Algebra

Basic variable: x

1

; free variables x

2

, x

3

. General solution:

123

2

3

24

33

is free

xx

x

=−

°

®

°

¯

12. Since the bottom row of the matrix is equivalent to the equation 0 = 1, the system has no solutions.

13.

130102 130092 100035

010041010041010041

~~

000194 000194 000194

000000 000000 000000

−−− − −

ªºªºªº

«»«»«»

−−−

«»«»«»

¬¼¬¼¬¼

Corresponding system:

15

25

45

35

41

94

00

xx

−=

+=

=

Basic variables: x

1

, x

2

, x

4

; free variables: x

3

, x

5

. General solution:

15

25

3

45

5

53

14

is free

49

is free

xx

x

xx

x

=+



°=+

°

®

°=−

°

¯

Note:

The Study Guide discusses the common mistake x

3

= 0.

14.

10 5 0 83 10 5 003

01 4 1 06 01 4 106

~

00 0 0 10 00 0 010

00 0 0 00 00 0 000

−− −

ªºªº

«»«»

−−

«»«»

¬¼¬¼

Corresponding system:

13

234

5

53

46

0

00

xx

xxx

x

−=

+−=

=

Basic variables: x

1

, x

2

, x

5

; free variables: x

3

, x

4

. General solution:

13

234

3

4

5

35

64

is free

0

xx

xxx

x

=+



°=−+

°

®

°

=

°

¯

15. a. The system is consistent. There are many solutions because x

3

is a free variable.

b. The system is consistent. There are many solutions because x

1

is a free variable.

16. a. The system is inconsistent. (The rightmost column of the augmented matrix is a pivot column).

1.2 • Solutions 11

b. The system is consistent. There are many solutions because x

2

is a free variable.

17. 114 11 4

~

23 01 8hh

−−

ªºª º

«»« »

−+

¬¼¬ ¼

The system has a solution for all values of h since the augmented

column cannot be a pivot column.

18. 13 1 1 3 1

~

62 036 2hhh

−−

ªºª º

«»« »

−+−−

¬¼¬ ¼

If 3h + 6 is zero, that is, if h = –2, then the system has a

solution, because 0 equals 0. When

2,h≠−

the system has a solution since the augmented column

cannot be a pivot column. Thus the system has a solution for all values of h.

19.

121 2

~

48 084 8

hh

khk

ªºª º

«»« »

−−

¬¼¬ ¼

a. When h = 2 and 8,k≠ the augmented column is a pivot column, and the system is inconsistent.

b. When

2,h≠

the system is consistent and has a unique solution. There are no free variables.

c. When h = 2 and k = 8, the system is consistent and has many solutions.

20.

131 1 3 1

~

2062hk h k

−−

ªºª º

«»« »

+−

¬¼¬ ¼

a. When h = –6 and 2

,

k≠ the system is inconsistent, because the augmented column is a pivot

column.

b. When 6,h≠− the system is consistent and has a unique solution. There are no free variables.

c. When h = –6 and k = 2, the system is consistent and has many solutions.

21. a. False. See Theorem 1.

b. False. See the second paragraph of the section.

c. True. Basic variables are defined after equation (4).

d. True. This statement is at the beginning of Parametric Descriptions of Solution Sets.

e. False. The row shown corresponds to the equation 5x

4

= 0, which does not by itself lead to a

contradiction. So the system might be consistent or it might be inconsistent.

22. a. True. See Theorem 1.

b. False. See Theorem 2.

c. False. See the beginning of the subsection Pivot Positions. The pivot positions in a matrix are

determined completely by the positions of the leading entries in the nonzero rows of any echelon

form obtained from the matrix.

d. True. See the paragraph just before Example 4.

e. False. The existence of at least one solution is not related to the presence or absence of free

variables. If the system is inconsistent, the solution set is empty. See the solution of Practice

Problem 2.

12 CHAPTER 1 • Linear Equations in Linear Algebra

23. Since there are four pivots (one in each row), the augmented matrix must reduce to the form

1

2

3

4

1000

0100 and so

0010

0001

xa

a

xb

b

xc

c

xd

d

=

ªº

«» =

«»

«» =

¬¼

No matter what the values of a, b, c, and d, the solution exists and is unique.

24. The system is consistent because there is not a pivot in column 5, which means that there is not a row

of the form [0 0 0 0 1]. Since the matrix is the augmented matrix for a system, Theorem 2 shows

that the system has a solution.

25. If the coefficient matrix has a pivot position in every row, then there is a pivot position in the bottom

row, and there is no room for a pivot in the augmented column. So, the system is consistent, by

Theorem 2.

26. Since the coefficient matrix has three pivot columns, there is a pivot in each row of the coefficient

matrix. Thus the augmented matrix will not have a row of the form [0 0 0 0 0 1], and the

system is consistent.

27. “If a linear system is consistent, then the solution is unique if and only if every column in the

coefficient matrix is a pivot column; otherwise there are infinitely many solutions. ”

This statement is true because the free variables correspond to nonpivot columns of the coefficient

matrix. The columns are all pivot columns if and only if there are no free variables. And there are no

free variables if and only if the solution is unique, by Theorem 2.

28. Every column in the augmented matrix except the rightmost column is a pivot column, and the

rightmost column is not a pivot column.

29. An underdetermined system always has more variables than equations. There cannot be more basic

variables than there are equations, so there must be at least one free variable. Such a variable may be

assigned infinitely many different values. If the system is consistent, each different value of a free

variable will produce a different solution, and the system will not have a unique solution. If the

system is inconsistent, it will not have any solution.

30. Example:

123

4

22 25

xx x

++=

31. Yes, a system of linear equations with more equations than unknowns can be consistent.

Example (in which x

1

= x

2

= 1):

12

2

0

32 5

xx

+=

−=

+=

32. According to the numerical note in Section 1.2, when n = 20 the reduction to echelon form takes

about 2(20)

3

/3 5,333 flops, while further reduction to reduced echelon form needs at most (20)

2

=

400 flops. Of the total flops, the “backward phase” is about 400/5733 = .07 or about 7%. When n =

200, the estimates are 2(200)

3

/3 5,333,333 flops for the reduction to echelon form and (200)

2

=

40,000 flops for the backward phase. The fraction associated with the backward phase is about

(4×10

4

) /(5.3×10

6

) = .007, or about .7%.

1.2 • Solutions 13

33. For a quadratic polynomial p(t) = a

0

+ a

1

t + a

2

t

2

to exactly fit the data (1, 6), (2, 15), and (3, 28), the

coefficients a

0

, a

1

, a

2

must satisfy the systems of equations given in the text. Row reduce the

augmented matrix:

111 6 111 6 1116 1116

12415~0 13 9~0 139~0 139

13928 02822 0024 0012

ªºªºªºªº

«»«»«»«»

¬¼¬¼¬¼¬¼

1104 1001

~0 1 0 3~0 1 0 3

0012 0012

ªºªº

«»«»

¬¼¬¼

The polynomial is p(t) = 1 + 3t + 2t

2

.

34. [M] The system of equations to be solved is:

2345

01 2 3 4 5

2345

01 2 3 4 5

2345

01 2 3 4 5

2345

01 2 3 4 5

2345

01 2 3 4 5

23

01 2 3

00 0 0 00

22 2 2 22.90

4 4 4 4 4 14.8

6 6 6 6 6 39.6

8 8 8 8 8 74.3

10 10 10

aa a a a a

aa a a

+⋅+⋅+⋅+⋅+⋅=

+⋅+⋅+⋅+

45

10 10 119aa⋅+⋅=

The unknowns are a

0

, a

1

, …, a

5

. Use technology to compute the reduced echelon of the augmented

matrix:

23 4 5

10 0 0 0 0 0 10 0 0 0 0 0

12 4 8 16 32 2.9 02 4 8 16 32 2.9

1 4 16 64 256 1024 14.8 00 8 48 224 960 9

~

16362161296 777639.6 0 0 24 192 1248 7680 30.9

1 8 64 512 4096 32768 74.3 0 0 48 480 4032 32640 62.7

008096099209984010

11010 10 10 10 119

ªº

«»

¬¼4.5

ª

º

«

»

«

»

«

»

«

»

«

»

«

»

«

»

«

»

¬

¼

100 0 0 0 0 100 0 0 0 0

024 8 16 32 2.9 024 8 16 32 2.9

008 48 224 960 9 00848 224 960 9

~~

0 0 0 48 576 4800 3.9 0 0 0 48 576 4800 3.9

000192268826880 8.7 000 0 384 7680 6.9

0 0 0 480 7680 90240 14.5 0 0 0 0 1920 42240 24.5

ªºªº

«»«»

−

«»«

−

«»«

¬¼¬¼

»

14 CHAPTER 1 • Linear Equations in Linear Algebra

100 0 0 0 0 100 0 0 0 0

024 8 16 32 2.9 024 8 16 32 2.9

00848224 960 9 00848224 960 9

~~

000485764800 3.9 000485764800 3.9

000 03847680 6.9 000 03847680 6.9

000 0 03840 10 000 0 0 1.0026

ªºª º

«»« »

−−

«»« »

¬¼¬ ¼

100 0 00 0 100000 0

024 8 160 2.8167 010000 1.7125

008482240 6.5000 001000 1.1948

~~~

000485760 8.6000 0 0 0 1 0 0 .6615

000 03840 26.900 000010 .0701

000 0 01.002604 000001 .0026

ªºªº

«»«»

−

«»«»

−

«»«»

−−

«»«»

¬¼¬¼

"

Thus p(t) = 1.7125t – 1.1948t

2

+ .6615t

3

– .0701t

4

+ .0026t

5

, and p(7.5) = 64.6 hundred lb.

Notes:

In Exercise 34, if the coefficients are retained to higher accuracy than shown here, then p(7.5) =

64.8. If a polynomial of lower degree is used, the resulting system of equations is overdetermined. The

augmented matrix for such a system is the same as the one used to find p, except that at least column 6 is

missing. When the augmented matrix is row reduced, the sixth row of the augmented matrix will be

entirely zero except for a nonzero entry in the augmented column, indicating that no solution exists.

Exercise 34 requires 25 row operations. It should give students an appreciation for higher-level

commands such as gauss and bgauss, discussed in Section 1.4 of the Study Guide. The command

ref (reduced echelon form) is available, but I recommend postponing that command until Chapter 2.

The Study Guide includes a “Mathematical Note” about the phrase, “If and only if,” used in Theorem

2.

1.3 SOLUTIONS

Notes:

The key exercises are 11–16, 19–22, 25, and 26. A discussion of Exercise 25 will help students

understand the notation [a

1

a

2

a

3

], {a

1

, a

2

, a

3

}, and Span{a

1

, a

2

, a

3

}.

1.

131(3) 4

212(1)1

−−−+−−

ªºªºª ºªº

+= + = =

«»«»« »«»

−+−

¬¼¬¼¬ ¼¬¼

uv

.

Using the definitions carefully,

131(2)(3)165

2(2)

212(2)(1)224

−−−−−−+

ªº ªºªºª ºª ºªº

−=+−=+ = =

«» «»«»« »« »«»

−−−+

¬¼ ¬¼¬¼¬ ¼¬ ¼¬¼

uv

, or, more quickly,

13165

22

21224

−−−+

ªº ªºª ºªº

−=−==

«» «»« »«»

−+

¬¼ ¬¼¬ ¼¬¼

uv

. The intermediate step is often not written.

1.3 • Solutions 15

2.

32325

21211

+

ªº ª º ª º ªº

+= + = =

«» « » « » «»

−−

¬¼ ¬ ¼ ¬ ¼ ¬¼

uv

.

Using the definitions carefully,

323(2)(2)341

2(2)

212(2)(1)224

−−−

ªº ª º ªº ª º ª º ª º

−=+−=+ = =

«» « » «» « » « » « »

−−−+

¬¼ ¬ ¼ ¬¼ ¬ ¼ ¬ ¼ ¬ ¼

uv

, or, more quickly,

32341

22

21224

−−

ªº ª º ª º ª º

−=−==

«» « » « » « »

−+

¬¼ ¬ ¼ ¬ ¼ ¬ ¼

uv

. The intermediate step is often not written.

3. 4.

5.

12

352

203

898

xx

ªº ªºªº

«» «»«»

−+=−

«» «»«»

−

¬¼ ¬¼¬¼

,

12

1

12

35 2

203

89 8

xx

x

xx

ªºª ºªº

«»« »«»

−+=−

«»« »«»

−

¬¼¬ ¼¬¼

,

12

1

12

35 2

23

89 8

xx

x

xx

+

ª

ºªº

«

»«»

−=−

«

»«»

«

»«»

−

¬

¼¬¼

12

1

12

35 2

23

89 8

xx

x

xx

+=

−=−

−=

Usually the intermediate steps are not displayed.

6.

123

37 20

23 10

xxx

−

ªº ªº ªºªº

++ =

«» «» «»«»

−

¬¼ ¬¼ ¬¼¬¼

,

3

12

312

2

37 0

23 0

x

xx

xxx

−

ªºªºªº

ª

º

++ =

«»«»«»

«

»

−

¬

¼

¬¼¬¼¬¼ ,

123

37 2 0

23 0

xxx

+−

ªº

=

«»

−++ ¬¼

¬¼

223

123

3720

23 0

xxx

xx x

+−=

−++=

Usually the intermediate steps are not displayed.

7. See the figure below. Since the grid can be extended in every direction, the figure suggests that every

vector in R

2

can be written as a linear combination of u and v.

To write a vector a as a linear combination of u and v, imagine walking from the origin to a along

the grid "streets" and keep track of how many "blocks" you travel in the u-direction and how many in

the v-direction.

a. To reach a from the origin, you might travel 1 unit in the u-direction and –2 units in the v-

direction (that is, 2 units in the negative v-direction). Hence a = u – 2v.

b. To reach b from the origin, travel 2 units in the u-direction and –2 units in the v-direction. So

b = 2u – 2v. Or, use the fact that b is 1 unit in the u-direction from a, so that

16 CHAPTER 1 • Linear Equations in Linear Algebra

b = a + u = (u – 2v) + u = 2u – 2v

c. The vector c is –1.5 units from b in the v-direction, so

c = b – 1.5v = (2u – 2v) – 1.5v = 2u – 3.5v

d. The “map” suggests that you can reach d if you travel 3 units in the u-direction and –4 units in

the v-direction. If you prefer to stay on the paths displayed on the map, you might travel from the

origin to –3v, then move 3 units in the u-direction, and finally move –1 unit in the v-direction. So

d = –3v + 3u – v = 3u – 4v

Another solution is

d = b – 2v + u = (2u – 2v) – 2v + u = 3u – 4v

Figure for Exercises 7 and 8

8. See the figure above. Since the grid can be extended in every direction, the figure suggests that every

vector in R

2

can be written as a linear combination of u and v.

w. To reach w from the origin, travel –1 units in the u-direction (that is, 1 unit in the negative

u-direction) and travel 2 units in the v-direction. Thus, w = (–1)u + 2v, or w = 2v – u.

x. To reach x from the origin, travel 2 units in the v-direction and –2 units in the u-direction. Thus,

x = –2u + 2v. Or, use the fact that x is –1 units in the u-direction from w, so that

x = w – u = (–u + 2v) – u = –2u + 2v

y. The vector y is 1.5 units from x in the v-direction, so

y = x + 1.5v = (–2u + 2v) + 1.5v = –2u + 3.5v

z. The map suggests that you can reach z if you travel 4 units in the v-direction and –3 units in the

u-direction. So z = 4v – 3u = –3u + 4v. If you prefer to stay on the paths displayed on the “map,”

you might travel from the origin to –2u, then 4 units in the v-direction, and finally move –1 unit

in the u-direction. So

z = –2u + 4v – u = –3u + 4v

9.

23

123

50

46 0

380

xx

xxx

+=

+−=

−+−=

,

23

123

50

46 0

38 0

xx

xxx

+

ªº

ª

º

«»

«

»

+−=

«»

«

»

«»

«

»

−+−

¬

¼

¬¼

23

12 3

050

46 0

380

xx

xx x

ªºªºª ºªº

«»«»« »«»

++−=

«»«»« »«»

−−

¬¼¬¼¬ ¼¬¼

,

123

01 50

46 10

13 80

xxx

ª

ºªºªºªº

«

»«»«»«»

++−=

«

»«»«»«»

«

»«»«»«»

−−

¬

¼¬¼¬¼¬¼

Usually, the intermediate calculations are not displayed.

w

x

v

u

a

c

d

2v

b

z

y

–2v–u

–v

0

1.3 • Solutions 17

Note:

The Study Guide says, “Check with your instructor whether you need to “show work” on a

problem such as Exercise 9.”

10.

123

32 4 3

27 5 1

54 32

xx x

−+=

−− +=

+−=

,

123

32 4 3

27 5 1

543 2

xx x

xxx

−+

ªº

ª

º

«»

«

»

−− +=

«»

«

»

«»

«

»

+−

¬

¼¬¼

123

3243

27 51

5432

xxx

−ªºªºª º ªº

«»

«»« » «»

−+−+=

«»

«»« » «»

«»«»« » «»

−

¬¼¬ ¼ ¬¼¬¼

,

123

3243

2751

5432

xx x

−

ª

ºªºªºªº

«

»«»«»«»

−+−+=

«

»«»«»«»

«

»«»«»«»

−

¬

¼¬¼¬¼¬¼

Usually, the intermediate calculations are not displayed.

11. The question

Is b a linear combination of a

1

, a

2

, and a

3

?

is equivalent to the question

Does the vector equation x

1

a

1

+ x

2

a

2

+ x

3

a

3

= b have a solution?

The equation

123

10 52

21 61

02 86

xxx

ªº ªº ªºªº

«» «» «»«»

−++−=−

«» «» «»«»

¬¼ ¬¼ ¬¼¬¼

↑↑↑↑

aa ab

(*)

has the same solution set as the linear system whose augmented matrix is

10 5 2

21 6 1

02 8 6

M

ªº

«»

=−−−

«»

¬¼

Row reduce M until the pivot positions are visible:

1052 1052

~0 1 4 3~0 1 4 3

0286 0000

M

ªºªº

«»«»

¬¼¬¼

The linear system corresponding to M has a solution, so the vector equation (*) has a solution, and

therefore b is a linear combination of a

1

, a

2

, and a

3

.

18 CHAPTER 1 • Linear Equations in Linear Algebra

12. The equation

12 3

123

12611

0375

1259

xx x

−−

ªº ª º ª º ª º

«» « » « » « »

++=−

«» « » « » « »

−

¬¼ ¬ ¼ ¬ ¼ ¬ ¼

↑↑↑↑

aaab

(*)

has the same solution set as the linear system whose augmented matrix is

12611

0375

1259

M

−−

ªº

«»

=−

«»

−

¬¼

Row reduce M until the pivot positions are visible:

12611

~0 3 7 5

00112

M

−−

ªº

«»

−

«»

−

¬¼

The linear system corresponding to M has a solution, so the vector equation (*) has a solution, and

therefore b is a linear combination of a

1

, a

2

, and a

3

.

13. Denote the columns of A by a

1

, a

2

, a

3

. To determine if b is a linear combination of these columns,

use the boxed fact in the subsection Linear Combinations. Row reduce the augmented matrix

[a

1

a

2

a

3

b] until you reach echelon form:

[a

1

a

2

a

3

b] =

142 3 142 3

0357~0357

2843 0003

−−

ª

ºª º

«

»« »

−−

«

»« »

«

»« »

−−−

¬

¼¬ ¼

The system for this augmented matrix is inconsistent, so b is not a linear combination of the columns

of A.

14. Row reduce the augmented matrix [a

1

a

2

a

3

b] until you reach echelon form:

[a

1

a

2

a

3

b] =

10 5 2 1052 1052

21 6 1~0143~0143

02 8 6 0286 0000

ªºªºªº

«»«»«»

−−−

«»«»«»

¬¼¬¼¬¼

.

The linear system corresponding to this matrix has a solution, so b is a linear combination of the

columns of A.

15. [a

1

a

2

b] =

153 15 3 15 3 15 3

385~07 14~01 2~01 2

12 0 3 3 0 3 3 00 3hh h h

−− − −

ªºª ºª ºª º

«»« »« »« »

−− − − −

«»« »« »« »

−−+−+−

¬¼¬ ¼¬ ¼¬ ¼

. The vector b

is in Span{a

1

, a

2

} when h – 3 is zero, that is, when h = 3.

1.3 • Solutions 19

16. [v

1

v

2

y] =

12 12 12

013~01 3~01 3

275 0352 0042

hh h

hh

−− −

ªºª ºª º

«»« »« »

−− −

«»« »« »

−− −++

¬¼¬ ¼¬ ¼

. The vector y is in

Span{v

1

, v

2

} when 4 + 2h is zero, that is, when h = –2.

17. Noninteger weights are acceptable, of course, but some simple choices are 0 v

1

+ 0 v

2

= 0, and

1

v

1

+ 0 v

2

=

3

1

2

ªº

«»

¬¼

, 0 v

1

+ 1 v

2

=

4

0

1

−

ª

º

«

»

«

»

«

»

¬

¼

, 1 v

1

+ 1 v

2

=

1

3

−

ª

º

«

»

«

»

«

»

¬

¼

, 1 v

1

– 1 v

2

=

7

1

ªº

«»

¬¼

18. Some likely choices are 0 v

1

+ 0 v

2

= 0, and

1

v

1

+ 0 v

2

=

1

2

ªº

«»

−

¬¼

, 0 v

1

+ 1 v

2

=

2

3

0

−

ªº

«»

¬¼

, 1 v

1

+ 1 v

2

=

1

4

2

−

ª

º

«

»

«

»

«

»

−

¬

¼

, 1 v

1

– 1 v

2

=

3

2

ªº

«»

−

«»

−

¬¼

19. By inspection, v

2

= (3/2)v

1

. Any linear combination of v

1

and v

2

is actually just a multiple of v

1

. For

instance,

av

1

+ bv

2

= av

1

+ b(3/2)v

1

= (a + 3b/2)v

1

So Span{v

1

, v

2

} is the set of points on the line through v

1

and 0.

Note:

Exercises 19 and 20 prepare the way for ideas in Sections 1.4 and 1.7.

20. Span{v

1

, v

2

} is a plane in R

3

through the origin, because neither vector in this problem is a multiple

of the other.

21. Let y =

h

k

ªº

«»

¬¼

. Then [u v y] =

22 22

~

11 02 /2

hh

kkh

ªºª º

«»« »

−+

¬¼¬ ¼

. This augmented matrix

corresponds to a consistent system for all h and k. So y is in Span{u, v} for all h and k.

22. Construct any 3×4 matrix in echelon form that corresponds to an inconsistent system. Perform

sufficient row operations on the matrix to eliminate all zero entries in the first three columns.

23. a. False. The alternative notation for a (column) vector is (–4, 3), using parentheses and commas.

b. False. Plot the points to verify this. Or, see the statement preceding Example 3. If

5

2

−

ªº

«»

¬¼

were on

the line through

2

5

−

ªº

«»

¬¼

and the origin, then

5

2

−

ª

º

«

»

¬

¼

would have to be a multiple of

2

5

−

ªº

«»

¬¼

, which is

not the case.

c. True. See the line displayed just before Example 4.

d. True. See the box that discusses the matrix in (5).

20 CHAPTER 1 • Linear Equations in Linear Algebra

e. False. The statement is often true, but Span{u, v} is not a plane when v is a multiple of u, or

when u is the zero vector.

24. a. False. Span{u, v} can be a plane.

b. True. See the beginning of the subsection Vectors in R

n

.

c. True. See the comment following the definition of Span{v

1

, …, v

p

}.

d. False. (u – v) + v = u – v + v = u.

e. False. Setting all the weights equal to zero results in a legitimate linear combination of a set of

vectors.

25. a. There are only three vectors in the set {a

1

, a

2

, a

3

}, and b is not one of them.

b. There are infinitely many vectors in W = Span{a

1

, a

2

, a

3

}. To determine if b is in W, use the

method of Exercise 13.

[a

1

a

2

a

3

b] =

10 4 4 10 44 10 44

03 2 1~03 21~03 21

26 3 4 06 54 00 12

−−−

ª

ºª ºª º

«

»« »« »

−−−

«

»« »« »

«

»« »« »

−− − −

¬

¼¬ ¼¬ ¼

The system for this augmented matrix is consistent, so b is in W.

c. a

1

= 1a

1

+ 0a

2

+ 0a

3

. See the discussion in the text following the definition of Span{v

1

, …, v

p

}.

26. a. [a

1

a

2

a

3

b] =

20610 1035 1035 1035

1853~1853~0888~0888

1217 1217 0222 0004

−−

−−−−

ª

ºª ºª ºª º

«

»« »« »« »

«

»« »« »« »

«

»« »« »« »

¬

¼¬ ¼¬ ¼¬ ¼

No, b is not a linear combination of the columns of A, that is, b is not in W.

b. The second column of A is in W because a

2

= 0 a

1

+ 1 a

2

+ 0 a

3

.

27. a. 5v

1

is the output of 5 days’ operation of mine #1.

b. The total output is x

1

v

1

+ x

2

v

2

, so x

1

and x

2

should satisfy

11 2 2

240

2824

xx

ª

º

+=

«

»

¬

¼

vv

.

c. [M] Reduce the augmented matrix

30 40 240 1 0 1.73

~

600 380 2824 0 1 4.70

ª

ºª º

«

»« »

¬

¼¬ ¼

.

Operate mine #1 for 1.73 days and mine #2 for 4.70 days. (This is an approximate solution.)

28. a. The amount of heat produced when the steam plant burns x

1

tons of anthracite and x

2

tons of

bituminous coal is 27.6x

1

+ 30.2x

2

million Btu.

b. The total output produced by x

1

tons of anthracite and x

2

tons of bituminous coal is given by the

vector

12

27.6 30.2

3100 6400

250 360

xx

ªºªº

«»«»

+

«»«»

¬¼¬¼

.

c. [M] The appropriate values for x

1

and x

2

satisfy

12

27.6 30.2 162

3100 6400 23,610

250 360 1,623

xx

ª

ºªºª º

«

»«»« »

+=

«

»«»« »

«

»«»« »

¬

¼¬¼¬ ¼

.

To solve, row reduce the augmented matrix:

1.3 • Solutions 21

27.6 30.2 162 1.000 0 3.900

3100 6400 23610 ~ 0 1.000 1.800

250 360 1623 0 0 0

ªºªº

«»«»

¬¼¬¼

The steam plant burned 3.9 tons of anthracite coal and 1.8 tons of bituminous coal.

29. The total mass is 4 + 2 + 3 + 5 = 14. So v = (4v

1

+2v

2

+ 3v

3

+ 5v

4

)/14. That is,

2441 8812517141.214

11

42 22 30 56 84030 177 2.429

14 14

4320 16660871.143

§·−−++

ªº ªº ªº ªº ª ºª ºª º

¨¸

«» «» «» «» « »« »« »

=−+++−=−++−=−≈−

¨¸

«» «» «» «» « »« »« »

¨¸

«» «» «» «» « »« »« »

−+−+

¬¼ ¬¼ ¬¼ ¬¼ ¬ ¼¬ ¼¬ ¼

©¹

v

30. Let m be the total mass of the system. By definition,

1

11 1

1()

k

kk k

m

mm

mmm

=++=++vv vv v""

The second expression displays v as a linear combination of v

1

, …, v

k

, which shows that v is in

Span{v

1

, …, v

k

}.

31. a. The center of mass is

08210/3

1111

114 2

3

§·

ªº ªº ªº ª º

⋅+⋅+⋅=

¨¸

«» «» «» « »

¬¼ ¬¼ ¬¼ ¬ ¼

©¹

.

b. The total mass of the new system is 9 grams. The three masses added, w

1

, w

2

, and w

3

, satisfy the

equation

() ( ) ( )

123

0822

1111

1142

9www

§·

ª

ºªºªºªº

+⋅++⋅++⋅=

¨¸

«

»«»«»«»

¬

¼¬¼¬¼¬¼

©¹

which can be rearranged to

() ( )

()

123

08218

111

11418

www

ªº ªº ªº ª º

+⋅++⋅++⋅=

«» «» «» « »

¬¼ ¬¼ ¬¼ ¬ ¼

and

123

0828

11412

www

ªº ªº ªº ª º

⋅+⋅+⋅=

«» «» «» « »

¬¼ ¬¼ ¬¼ ¬ ¼

The condition w

1

+ w

2

+ w

3

= 6 and the vector equation above combine to produce a system of

three equations whose augmented matrix is shown below, along with a sequence of row

operations:

111 6 1116 1116

082 8~0828~0828

11412 00 36 00 12

ªºªºªº

«»«»«»

¬¼¬¼¬¼

1104 1003.5 1003.5

~0 8 0 4~0 8 0 4~0 1 0 .5

0012 001 2 001 2

ªºª ºª º

«»« »« »

¬¼¬ ¼¬ ¼

Answer: Add 3.5 g at (0, 1), add .5 g at (8, 1), and add 2 g at (2, 4).

22 CHAPTER 1 • Linear Equations in Linear Algebra

Extra problem: Ignore the mass of the plate, and distribute 6 gm at the three vertices to make the center

of mass at (2, 2). Answer: Place 3 g at (0, 1), 1 g at (8, 1), and 2 g at (2, 4).

32. See the parallelograms drawn on the figure from the text that accompanies this exercise. Here c

1

, c

2

,

c

3

, and c

4

are suitable scalars. The darker parallelogram shows that b is a linear combination of v

1

and v

2

, that is

c

1

v

1

+ c

2

v

2

+ 0 v

3

= b

The larger parallelogram shows that b is a linear combination of v

1

and v

3

, that is,

c

4

v

1

+ 0 v

2

+ c

3

v

3

= b

So the equation x

1

v

1

+ x

2

v

2

+ x

3

v

3

= b has at least two solutions, not just one solution. (In fact, the

equation has infinitely many solutions.)

33. a. For j = 1,…, n, the jth entry of (u + v) + w is (u

j

+ v

j

) + w

j

. By associativity of addition in R, this

entry equals u

j

+ (v

j

+ w

j

), which is the jth entry of u + (v + w). By definition of equality of

vectors, (u + v) + w = u + (v + w).

b. For any scalar c, the jth entry of c(u + v) is c(u

j

+ v

j

), and the jth entry of cu + cv is cu

j

+ cv

j

(by

definition of scalar multiplication and vector addition). These entries are equal, by a distributive

law in R. So c(u + v) = cu + cv.

34. a. For j = 1,…, n, u

j

+ (–1)u

j

= (–1)u

j

+ u

j

= 0, by properties of R. By vector equality,

u + (–1)u = (–1)u + u = 0.

b. For scalars c and d, the jth entries of c(du) and (cd )u are c(du

j

) and (cd )u

j

, respectively. These

entries in R are equal, so the vectors c(du) and (cd)u are equal.

Note:

When an exercise in this section involves a vector equation, the corresponding technology data (in

the data files on the web) is usually presented as a set of (column) vectors. To use MATLAB or other

technology, a student must first construct an augmented matrix from these vectors. The MATLAB note in

the Study Guide describes how to do this. The appendices in the Study Guide give corresponding

information about Maple, Mathematica, and the TI calculators.

c

2

v

2

c

3

v

3

0

v

3

c

4

v

1

c

1

v

1

v

1

v

2

b

1.4 • Solutions 23

1.4 SOLUTIONS

Notes:

Key exercises are 1–20, 27, 28, 31 and 32. Exercises 29, 30, 33, and 34 are harder. Exercise 34

anticipates the Invertible Matrix Theorem but is not used in the proof of that theorem.

1. The matrix-vector Ax is not defined because the number of columns (2) in the 3×2 matrix A does not

match the number of entries (3) in the vector x.

2. The matrix-vector Ax is not defined because the number of columns (1) in the 3×1 matrix A does not

match the number of entries (2) in the vector x.

3.

12 12 264

2

31 (2)3 31 6 3 9

3

16 16 21816

A

−

ªº ªºªºªºªºªº

−

ªº

«» «»«»«»«»«»

=−=−−+=+=

«»

«» «»«»«»«»«»

¬¼

«» «»«»«»«»«»

−

¬¼ ¬¼¬¼¬¼¬¼¬¼

x

, and

12 1(2)23 4

2

31 (3)(2)13 9

3

16 1(2)63 16

A

⋅− +⋅

ªº ª ºªº

−

ªº

«» « »«»

=−=−⋅−+⋅=

«»

«» « »«»

¬¼

«» « »«»

⋅− +⋅

¬¼ ¬ ¼¬¼

x

4.

1

13 4 1 3 4 164 3

21 2 1

32 1 3 2 1 341 8

1

A

ªº

−−+−

ªºªºªºªºªºªº

«»

==⋅+⋅+⋅==

«»«»«»«»«»«»

«» ++

¬¼¬¼¬¼¬¼¬¼¬¼

«»

¬¼

x

, and

1

13 4 1132(4)1 3

2

32 1 312211 8

1

A

ªº

−⋅+⋅+−⋅

ªºª ºªº

«»

== =

«»« »«»

«» ⋅+⋅+⋅

¬¼¬ ¼¬¼

«»

¬¼

x

5. On the left side of the matrix equation, use the entries in the vector x as the weights in a linear

combination of the columns of the matrix A:

12 314

2111

23111

−−

ªº ªº ªº ªºªº

⋅−⋅+⋅−⋅=

«» «» «» «»«»

−− −

¬¼ ¬¼ ¬¼ ¬¼¬¼

6. On the left side of the matrix equation, use the entries in the vector x as the weights in a linear

combination of the columns of the matrix A:

2321

321

35

8549

2111

−−

ªº ªºª º

«» «»« »

−⋅ +⋅=

«» «»« »

−−

«» «»« »

−

¬¼ ¬¼¬ ¼

7. The left side of the equation is a linear combination of three vectors. Write the matrix A whose

columns are those three vectors, and create a variable vector x with three entries:

24 CHAPTER 1 • Linear Equations in Linear Algebra

457 457

138 138

750 750

412 412

A

ªº−−

ªºªºªº ª º

«»

«»«»«» « »

−−−−

«»

«»«»«» « »

==

«»

«»«»«» « »

−−

«»

«»«»«» « »

−−

«»

¬¼¬¼¬¼ ¬ ¼

¬¼

and

1

2

3

x

ª

º

«

»

=

«

»

«

»

¬

¼

x

. Thus the equation Ax = b is

1

2

3

457 6

138 8

750 0

412 7

x

−

ªºªº

ªº

«»«»

−− −

«»

«»«»

=

«»

«»«»

−«»

«»«»

¬¼

−−

¬¼¬¼

For your information: The unique solution of this equation is (5, 7, 3). Finding the solution by hand

would be time-consuming.

Note:

The skill of writing a vector equation as a matrix equation will be important for both theory and

application throughout the text. See also Exercises 27 and 28.

8. The left side of the equation is a linear combination of four vectors. Write the matrix A whose

columns are those four vectors, and create a variable vector with four entries:

2140 2140

4532 4532

Aªº−− −−

ªºªºªºªº ª º

==

«»

«»«»«»«» « »

−−

¬¼¬¼¬¼¬¼ ¬ ¼

¬¼

, and

1

2

3

4

z

ª

º

«

»

«

»

=

«

»

«

»

«

»

¬

¼

z

. Then the equation Az = b

is

1

2

3

4

2140 5

4532 12

z

ªº

«»

−−

ªºªº

«»

=

«»«»

«»

−

¬¼¬¼

«»

¬¼

.

For your information: One solution is (8, 7, 1, 3). The general solution is z

1

= 37/6 + (17/6)z

3

–

(1/3)z

4

, z

2

= 22/3 +(5/3)z

3

– (2/3)z

4

, with z

3

and z

4

free.

9. The system has the same solution set as the vector equation

123

51 38

02 40

xx x

−

ªº ªº ª º ªº

++ =

«» «» « » «»

¬¼ ¬¼ ¬ ¼ ¬¼

and this equation has the same solution set as the matrix equation

1

2

3

51 3 8

02 4 0

x

ªº

−

ªºªº

«»

=

«»«»

«»

¬¼¬¼

«»

¬¼

10. The system has the same solution set as the vector equation

12

418

532

311

xx

−

ªº ª º ªº

«» « » «»

+=

«» « » «»

−

¬¼ ¬ ¼ ¬¼

and this equation has the same solution set as the matrix equation

1.4 • Solutions 25

1

2

41 8

53 2

31 1

x

−

ªº ªº

ªº

«» «»

=

«»

«» «»

¬¼

«» «»

−

¬¼ ¬¼

11. To solve Ax = b, row reduce the augmented matrix [a

1

a

2

a

3

b] for the corresponding linear

system:

1342 1342 1342 1302 10011

1524~0266~0133~0103~010 3

37612 0266 0010 0010 001 0

−− −− −− − −

ªºªºªºªºªº

«»«»«»«»«»

−− −

¬¼¬¼¬¼¬¼¬¼

The solution is

1

2

3

11

3

0

x

=−



°=

®

°=

¯

. As a vector, the solution is x =

1

2

3

11

3

0

x

−

ªº

ª

º

«»

«

»

=

«»

«

»

«»

«

»

¬

¼¬¼

.

12. To solve Ax = b, row reduce the augmented matrix [a

1

a

2

a

3

b] for the corresponding linear

system:

1211 1211 1211 1211

3422~0215~0888~0111

5233 0888 0215 0215

−−−−

ªºªºªºªº

«»«»«»«»

−− − − − −

«»«»«»«»

−−− − −

¬¼¬¼¬¼¬¼

12 11 1204 100 4

~0 1 1 1~0 1 0 4~0 1 0 4

00 13 0013 001 3

−−

ªºªºªº

«»«»«»

−

«»«»«»

¬¼¬¼¬¼

The solution is

1

2

3

4

3

x

=−



°=

®

°=

¯

. As a vector, the solution is x =

1

2

3

4

3

x

−

ªº

ª

º

«»

«

»

=

«»

«

»

«»

«

»

¬

¼¬¼

.

13. The vector u is in the plane spanned by the columns of A if and only if u is a linear combination of

the columns of A. This happens if and only if the equation Ax = u has a solution. (See the box

preceding Example 3 in Section 1.4.) To study this equation, reduce the augmented matrix [A u]

350 114 11 4 114

264~264~0812~0812

114 350 0812 000

−

ªºªºª ºªº

«»«»« »«»

−−

«»«»« »«»

−−−

¬¼¬¼¬ ¼¬¼

The equation Ax = u has a solution, so u is in the plane spanned by the columns of A.

For your information: The unique solution of Ax = u is (5/2, 3/2).

14. Reduce the augmented matrix [A u] to echelon form:

25 1 4 12 0 4 12 0 4 12 0 4

01 1 1~01 1 1~01 1 1~01 1 1

12 0 4 25 1 4 0 1 1 4 00 0 3

−

ªºªºªºªº

«»«»«»«»

−− −− −− −−

«»«»«»«»

−−−−

¬¼¬¼¬¼¬¼

26 CHAPTER 1 • Linear Equations in Linear Algebra

The equation Ax = u has no solution, so u is not in the subset spanned by the columns of A.

15. The augmented matrix for Ax = b is

1

2

31

93

b

−

ªº

«»

−

¬¼

, which is row equivalent to

1

21

31

00 3

b

bb

−

ªº

«»

+

¬¼

.

This shows that the equation Ax = b is not consistent when 3b

1

+ b

2

is nonzero. The set of b for

which the equation is consistent is a line through the origin–the set of all points (b

1

, b

2

) satisfying b

2

= –3b

1

.

16. Row reduce the augmented matrix [A b]:

1

2

3

121

220, .

413

b

Ab

b

−−

ª

ºªº

«

»

«»

=−=

«

»

«»

«

»«»

−

¬¼

¬

¼

b

11

221

331

121 12 1

220 ~022 2

413 077 4

bb

bbb

−− −−

ªºªº

«»«»

−−−+

«»«»

−−

¬¼¬¼

11

21 21

31 21 1 23

12 1 12 1

~0 2 2 2 0 2 2 2

000 4(72)( 2) 0003(72)

bb

bb bb

bb bb b bb

−− −−

ªºªº

«»«»

−− +=−− +

«»«»

−++ ++

¬¼¬¼

The equation Ax = b is consistent if and only if 3b

1

+ (7/2)b

2

+ b

3

= 0, or 6b

1

+ 7b

2

+ 2b

3

= 0. The set

of such b is a plane through the origin in R

3

.

17. Row reduction shows that only three rows of A contain a pivot position:

1303 1303 1303 1303

11110214 0214 0214

~~~

0428 0428 0000 0005

2031 0637 0005 0000

A

ªºªºªºªº

«»«»«»«»

−−− − − −

«»«»«»«»

=«»«»«»«»

−− −−

«»«»«»«»

−−−

¬¼¬¼¬¼¬¼

Because not every row of A contains a pivot position, Theorem 4 in Section 1.4 shows that the

equation Ax = b does not have a solution for each b in R

4

.

18. Row reduction shows that only three rows of B contain a pivot position:

14 1 2 14 1 2 14 1 2 1 4 1 2

013 4 013 4 013 4 0 13 4

~~~

026 7 026 7 00015 0 0015

295 7 013 11 000 7 0 00 0

B

ªºª ºªºª º

«»« »«»« »

−−−−

«»« »«»« »

=«»« »«»« »

«»« »«»« »

−− −

¬¼¬ ¼¬¼¬ ¼

Because not every row of B contains a pivot position, Theorem 4 in Section 1.4 shows that not all

vectors in R

4

can be written as a linear combination of the columns of B. The columns of B certainly

do not span R

3

, because each column of B is in R

4

, not R

3

. (This question was asked to alert students

to a fairly common misconception among students who are just learning about spanning.)

19. The work in Exercise 17 shows that statement (d) in Theorem 4 is false. So all four statements in

Theorem 4 are false. Thus, not all vectors in R

4

can be written as a linear combination of the columns

of A. Also, the columns of A do not span R

4

.

1.4 • Solutions 27

20. The work in Exercise 18 shows that statement (d) in Theorem 4 is false. So all four statements in

Theorem 4 are false. Thus, the equation Bx = y does not have a solution for each y in R

4

, and the

columns of B do not span R

4

.

21. Row reduce the matrix [v

1

v

2

v

3

] to determine whether it has a pivot in each row.

10 1 10 1 10 1 101

010 010 010 010

~~~.

100 00 1 00 1 001

011011001000

ªºªºªºªº

«»«»«»«»

−−−

«»«»«»«»

−

«»«»«»«»

−−−

¬¼¬¼¬¼¬¼

The matrix [v

1

v

2

v

3

] does not have a pivot in each row, so the columns of the matrix do not span

R

4

, by Theorem 4. That is, {v

1

, v

2

, v

3

} does not span R

4

.

Note:

Some students may realize that row operations are not needed, and thereby discover the principle

covered in Exercises 31 and 32.

22. Row reduce the matrix [v

1

v

2

v

3

] to determine whether it has a pivot in each row.

004 396

032~032

396 004

−−

ªºªº

«»«»

−− −−

«»«»

−−

¬¼¬¼

The matrix [v

1

v

2

v

3

] has a pivot in each row, so the columns of the matrix span R

3

, by Theorem 4.

That is, {v

1

, v

2

, v

3

} spans R

3

.

23. a. False. See the paragraph following equation (3). The text calls Ax = b a matrix equation.

b. True. See the box before Example 3.

c. False. See the warning following Theorem 4.

d. True. See Example 4.

e. True. See parts (c) and (a) in Theorem 4.

f. True. In Theorem 4, statement (a) is false if and only if statement (d) is also false.

24. a. True. This statement is in Theorem 3. However, the statement is true without any "proof"

because, by definition, Ax is simply a notation for x

1

a

1

+ ⋅ ⋅ ⋅ + x

n

a

n

, where a

1

, …, a

n

are the

columns of A.

b. True. See the box before Example 3.

c. True. See Example 2.

d. False. In Theorem 4, statement (d) is true if and only if statement (a) is true.

e. True. See Theorem 3.

f. False. In Theorem 4, statement (c) is false if and only if statement (a) is also false.

25. By definition, the matrix-vector product on the left is a linear combination of the columns of the

matrix, in this case using weights –3, –1, and 2. So c

1

= –3, c

2

= –1, and c

3

= 2.

26. The equation in x

1

and x

2

involves the vectors u, v, and w, and it may be viewed as

[]

1

2

.

x

ªº

=

«»

¬¼

uv w

By definition of a matrix-vector product, x

1

u + x

2

v = w. The stated fact that

2u – 3v – w = 0 can be rewritten as 2u – 3v = w. So, a solution is x

1

= 2, x

2

= –3.

28 CHAPTER 1 • Linear Equations in Linear Algebra

27. The matrix equation can be written as c

1

v

1

+ c

2

v

2

+ c

3

v

3

+ c

4

v

4

+ c

5

v

5

= v

6

, where

c

1

= –3, c

2

= 1, c

3

= 2, c

4

= –1, c

5

= 2, and

123 45 6

35 4 9 7 11

,, , , ,

58 1 2 4 11

−−

ªº ªº ªº ªº ªº ª º

======

«» «» «» «» «» « »

−−−

¬¼ ¬¼ ¬¼ ¬¼ ¬¼ ¬ ¼

vvv v v v

28. Place the vectors q

1

, q

2

, and q

3

into the columns of a matrix, say, Q and place the weights x

1

, x

2

, and

x

3

into a vector, say, x. Then the vector equation becomes

Qx = v, where Q = [q

1

q

2

q

3

] and

1

2

3

x

ª

º

«

»

=

«

»

«

»

¬

¼

x

Note: If your answer is the equation Ax = b, you need to specify what A and b are.

29. Start with any 3×3 matrix B in echelon form that has three pivot positions. Perform a row operation

(a row interchange or a row replacement) that creates a matrix A that is not in echelon form. Then A

has the desired property. The justification is given by row reducing A to B, in order to display the

pivot positions. Since A has a pivot position in every row, the columns of A span R

3

, by Theorem 4.

30. Start with any nonzero 3×3 matrix B in echelon form that has fewer than three pivot positions.

Perform a row operation that creates a matrix A that is not in echelon form. Then A has the desired

property. Since A does not have a pivot position in every row, the columns of A do not span R

3

, by

Theorem 4.

31. A 3×2 matrix has three rows and two columns. With only two columns, A can have at most two pivot

columns, and so A has at most two pivot positions, which is not enough to fill all three rows. By

Theorem 4, the equation Ax = b cannot be consistent for all b in R

3

. Generally, if A is an m×n matrix

with m > n, then A can have at most n pivot positions, which is not enough to fill all m rows. Thus,

the equation Ax = b cannot be consistent for all b in R

3

.

32. A set of three vectors in R

4

cannot span R

4

. Reason: the matrix A whose columns are these three

vectors has four rows. To have a pivot in each row, A would have to have at least four columns (one

for each pivot), which is not the case. Since A does not have a pivot in every row, its columns do not

span R

4

, by Theorem 4. In general, a set of n vectors in R

m

cannot span R

m

when n is less than m.

33. If the equation Ax = b has a unique solution, then the associated system of equations does not have

any free variables. If every variable is a basic variable, then each column of A is a pivot column. So

the reduced echelon form of A must be

100

010

001

000

ª

º

«

»

«

»

«

»

«

»

«

»

¬

¼

.

Note:

Exercises 33 and 36 are difficult in the context of this section because the focus in Section 1.4 is on

existence of solutions, not uniqueness. However, these exercises serve to review ideas from Section 1.2,

and they anticipate ideas that will come later.

34. Given Au

1

= v

1

and Au

2

= v

2

, you are asked to show that the equation Ax = w has a solution, where

w = v

1

+ v

2

. Observe that w = Au

1

+ Au

2

and use Theorem 5(a) with u

1

and u

2

in place of u and v,

respectively. That is, w = Au

1

+ Au

2

= A(u

1

+ u

2

). So the vector x = u

1

+ u

2

is a solution of w = Ax.

1.4 • Solutions 29

35. Suppose that y and z satisfy Ay = z. Then 5z = 5Ay. By Theorem 5(b), 5Ay = A(5y). So 5z = A(5y),

which shows that 5y is a solution of Ax = 5z. Thus, the equation Ax = 5z is consistent.

36. If the equation Ax = b has a unique solution, then the associated system of equations does not have

any free variables. If every variable is a basic variable, then each column of A is a pivot column. So

the reduced echelon form of A must be

1000

0100

0010

0001

ª

º

«

»

«

»

«

»

«

»

¬

¼

. Now it is clear that A has a pivot position in

each row. By Theorem 4, the columns of A span R

4

.

37. [M]

7258 7 2 5 8 7 2 5 8

5349 011/73/723/7 011/7 3/7 23/7

~~

610 2 7 0 58/716/7 1/7 0 050/11 189/11

79215 0 11 3 23 0 0 0 0

−− −

ªºª ºª º

«»« »« »

−− − − − − −

«»« »« »

−−

«»« »« »

−−

«»« »« »

¬¼¬ ¼¬ ¼

or, approximately

7258

0 1.57 .429 3.29

0 0 4.55 17.2

0000

−

ªº

«»

−−

«»

−

«»

¬¼

, to three significant figures. The original matrix does

not have a pivot in every row, so its columns do not span R

4

, by Theorem 4.

38. [M]

4518 4 5 1 8 4 5 1 8

3742 013/413/4 4 013/413/4 4

~~

5614 0 1/4 1/4 6 0 0 082/13

91107 049/449/4110 0 0 0

−− − − − −

ªºª ºª º

«»« »« »

−− − − − − − −

«»« »« »

−− − −

«»« »« »

−

¬¼¬ ¼¬ ¼

With pivots only in the first three rows, the original matrix has columns that do not span R

4

, by

Theorem 4.

39. [M]

10 7 1 4 6 10 7 1 4 6

846103 0 8/5 26/534/5 9/5

~

711 5 1 8 061/10 43/10 9/5 19/5

31101212 011/1097/1054/551/5

−−

ªºª º

«»« »

−−−− −−−

«»« »

−−−− − −

«»« »

−

¬¼¬ ¼

10 7 1 4 6 10 7 1 4 6

08/5 26/5 34/5 9/5 08/5 26/5 34/5 9/5

~~

00193/8193/849/1600193/8193/849/16

0049/849/8183/1600 0 04715/386

−−

ªºª º

«»« »

−− − −− −

«»« »

−− −−

«»« »

¬¼¬ ¼

The original matrix has a pivot in every row, so its columns span R

4

, by Theorem 4.

40. [M]

511 6 712 5 11 6 7 12

7346 9 062/5 62/519/539/5

~

11 5 6 9 3 0 96 / 5 96 / 5 32 / 5 147 / 5

34 72 7 053/5 53/5 11/5 71/5

−− −−

ªºª º

«»« »

−−− − −−

«»« »

−− − −

«»« »

−− −−

¬¼¬ ¼

30 CHAPTER 1 • Linear Equations in Linear Algebra

511 6 7 12 511 6 7 12

062/5 62/5 19/5 39/5 062/5 62/5 19/5 39/5

~~

00 0 16/31537/3100 0 16/31537/31

00 0 65/62467/62 00 0 0 1367/62

−− −−

ªºªº

«»«»

−− −−

«»«»

−−

«»«»

¬¼¬¼

The original matrix has a pivot in every row, so its columns span R

4

, by Theorem 4.

41. [M] Examine the calculations in Exercise 39. Notice that the fourth column of the original matrix,

say A, is not a pivot column. Let A

o

be the matrix formed by deleting column 4 of A, let B be the

echelon form obtained from A, and let B

o

be the matrix obtained by deleting column 4 of B. The

sequence of row operations that reduces A to B also reduces A

o

to B

o

. Since B

o

is in echelon form, it

shows that A

o

has a pivot position in each row. Therefore, the columns of A

o

span R

4

.

It is possible to delete column 3 of A instead of column 4. In this case, the fourth column of A

becomes a pivot column of A

o

, as you can see by looking at what happens when column 3 of B is

deleted. For later work, it is desirable to delete a nonpivot column.

Note:

Exercises 41 and 42 help to prepare for later work on the column space of a matrix. (See Section

2.9 or 4.6.) The Study Guide points out that these exercises depend on the following idea, not explicitly

mentioned in the text: when a row operation is performed on a matrix A, the calculations for each new

entry depend only on the other entries in the same column. If a column of A is removed, forming a new

matrix, the absence of this column has no affect on any row-operation calculations for entries in the other

columns of A. (The absence of a column might affect the particular choice of row operations performed

for some purpose, but that is not being considered here.)

42. [M] Examine the calculations in Exercise 40. The third column of the original matrix, say A, is not a

pivot column. Let A

o

be the matrix formed by deleting column 3 of A, let B be the echelon form

obtained from A, and let B

o

be the matrix obtained by deleting column 3 of B. The sequence of row

operations that reduces A to B also reduces A

o

to B

o

. Since B

o

is in echelon form, it shows that A

o

has

a pivot position in each row. Therefore, the columns of A

o

span R

4

.

It is possible to delete column 2 of A instead of column 3. (See the remark for Exercise 41.)

However, only one column can be deleted. If two or more columns were deleted from A, the

resulting matrix would have fewer than four columns, so it would have fewer than four pivot

positions. In such a case, not every row could contain a pivot position, and the columns of the matrix

would not span R

4

, by Theorem 4.

Notes:

At the end of Section 1.4, the Study Guide gives students a method for learning and mastering

linear algebra concepts. Specific directions are given for constructing a review sheet that connects the

basic definition of “span” with related ideas: equivalent descriptions, theorems, geometric interpretations,

special cases, algorithms, and typical computations. I require my students to prepare such a sheet that

reflects their choices of material connected with “span”, and I make comments on their sheets to help

them refine their review. Later, the students use these sheets when studying for exams.

The MATLAB box for Section 1.4 introduces two useful commands gauss and bgauss that

allow a student to speed up row reduction while still visualizing all the steps involved. The command

B = gauss(A,1) causes MATLAB to find the left-most nonzero entry in row 1 of matrix A, and use

that entry as a pivot to create zeros in the entries below, using row replacement operations. The result is a

matrix that a student might write next to A as the first stage of row reduction, since there is no need to

write a new matrix after each separate row replacement. I use the gauss command frequently in lectures

to obtain an echelon form that provides data for solving various problems. For instance, if a matrix has 5

rows, and if row swaps are not needed, the following commands produce an echelon form of A:

B = gauss(A,1), B = gauss(B,2), B = gauss(B,3), B = gauss(B,4)

1.5 • Solutions 31

If an interchange is required, I can insert a command such as B = swap(B,2,5) . The command

bgauss uses the left-most nonzero entry in a row to produce zeros above that entry. This command,

together with scale, can change an echelon form into reduced echelon form.

The use of gauss and bgauss creates an environment in which students use their computer

program the same way they work a problem by hand on an exam. Unless you are able to conduct your

exams in a computer laboratory, it may be unwise to give students too early the power to obtain reduced

echelon forms with one command—they may have difficulty performing row reduction by hand during an

exam. Instructors whose students use a graphic calculator in class each day do not face this problem. In

such a case, you may wish to introduce rref earlier in the course than Chapter 4 (or Section 2.8), which

is where I finally allow students to use that command.

1.5 SOLUTIONS

Notes

: The geometry helps students understand Span{u, v}, in preparation for later discussions of sub-

spaces. The parametric vector form of a solution set will be used throughout the text. Figure 6 will appear

again in Sections 2.9 and 4.8.

For solving homogeneous systems, the text recommends working with the augmented matrix, al-

though no calculations take place in the augmented column. See the Study Guide comments on Exercise 7

that illustrate two common student errors.

All students need the practice of Exercises 1–14. (Assign all odd, all even, or a mixture. If you do not

assign Exercise 7, be sure to assign both 8 and 10.) Otherwise, a few students may be unable later to find

a basis for a null space or an eigenspace. Exercises 28–36 are important. Exercises 35 and 36 help

students later understand how solutions of Ax = 0 encode linear dependence relations among the columns

of A. Exercises 37–40 are more challenging. Exercise 37 will help students avoid the standard mistake of

forgetting that Theorem 6 applies only to a consistent equation Ax = b.

1. Reduce the augmented matrix to echelon form and circle the pivot positions. If a column of the

coefficient matrix is not a pivot column, the corresponding variable is free and the system of

equations has a nontrivial solution. Otherwise, the system has only the trivial solution.

2580 2 580 2 580

2710~01290~01290

4270 01290 0 000

−−−

ªºªºªº

«»«»«»

−− − −

«»«»«»

−

¬¼¬¼¬¼

The variable x

3

is free, so the system has a nontrivial solution.

2.

1230 1230

2340~0720

2490 0030

−−

ªºªº

«»«»

−−− −

«»«»

−

¬¼¬¼

There is no free variable; the system has only the trivial solution.

3.

34 80 3 4 80

~

25 40 07/328/30

−− − −

ªºª º

«»« »

−

¬¼¬ ¼

. The variable x

3

is free; the system has nontrivial

solutions. An alert student will realize that row operations are unnecessary. With only two equations,

there can be at most two basic variables. One variable must be free. Refer to Exercise 29 in Section

1.2.

32 CHAPTER 1 • Linear Equations in Linear Algebra

4.

5320 5 3 20

~

3420 029/516/50

−−

ªºª º

«»« »

−− −

¬¼¬ ¼

. The variable x

3

is free; the system has nontrivial

solutions. As in Exercise 3, row operations are unnecessary.

5.

2240 2240 1120 1010

4480~0000~0110~0110

0330 0330 0000 0000

ªºªºªºªº

«»«»«»«»

−−−

«»«»«»«»

−− −−

¬¼¬¼¬¼¬¼

13

23

0

00

xx

+=

=

. The variable x

3

is free, x

1

= –x

3

, and x

2

= –x

3

.

In parametric vector form, the general solution is

13

233

33

1

xx

xxx

xx

−−

ªºª º

ª

º

«»« »

«

»

==−=−

«»« »

«

»

«»« »

«

»

¬

¼¬¼¬ ¼

x

.

6.

12 3 0 1 2 3 0 1 2 3 0 1 0 1 0

21 3 0~0 3 3 0~0 1 10~0 1 10

11 0 0 0 3 3 0 0 0 0 0 0 0 0 0

−−−−

ªºªºªºªº

«»«»«»«»

−− − −

«»«»«»«»

−−

¬¼¬¼¬¼¬¼

13

23

0

00

xx

−=

=

. The variable x

3

is free, x

1

= x

3

, and x

2

= x

3

.

In parametric vector form, the general solution is

13

233

33

1

xx

xxx

xx

ªºªº

ª

º

«»«»

«

»

===

«»«»

«

»

«»«»

«

»

¬

¼¬¼¬¼

x

.

7.

13 37 0 10 9 80

~

01 450 01 4 50

−−

ªºª º

«»« »

−−

¬¼¬ ¼

.

134

234

980

450

xxx

+−=

−+=

The basic variables are x

1

and x

2

, with x

3

and x

4

free. Next, x

1

= –9x

3

+ 8x

4

, and x

2

= 4x

3

– 5x

4

. The

general solution is

134 4

3

234 4

3

34

33 3

44 4

98 8

998

45 5

445

010

001

xxx x

x

xxx x

xxx

xx x

−+−−

ªºª º ª º

ªº ªºªº

«»« » « »

«» «»«»

−− −

«»« » « »«» «»«»

== = + = +

«»« » « »

«» «»«»

«»« » « »

«» «»«»

«»« » « »

¬¼ ¬¼¬¼¬¼¬ ¼ ¬ ¼

x

8.

13850 10270

~

01240 01240

−− −−

ªºªº

«»«»

−−

¬¼¬¼

.

134

234

270

240

xxx

−−=

+−=

The basic variables are x

1

and x

2

, with x

3

and x

4

free. Next, x

1

= 2x

3

+ 7x

4

and x

2

= –2x

3

+ 4x

4

. The

general solution in parametric vector form is

1.5 • Solutions 33

134 4

3

234 4

3

34

33 3

44 4

27 7

227

24 4

224

010

001

xxx x

x

xxx x

xxx

xx x

+

ªºª º ª º

ªº ªºªº

«»« » « »

«» «»«»

−+−−

«»« » « »

«» «»«»

== = + = +

«»« » « »

«» «»«»

«»« » « »

«» «»«»

¬¼ ¬¼¬¼

¬¼¬ ¼ ¬ ¼

x

9.

3660 1220 1220 1200

~~~

2420 2420 0020 0010

−−−−

ªºªºªºªº

«»«»«»«»

−− −−

¬¼¬¼¬¼¬¼

12

3

20

0

xx

x

−=

=

.

The solution is x

1

= 2x

2

, x

3

= 0, with x

2

free. In parametric vector form,

22

222

3

22 2

1

00

xx

xxx

x

ªºªº ªº

«»

«» «»

===

«»

«» «»

«»«» «»

¬¼ ¬¼¬¼

x

.

10.

14040140401404010040

~~~

28080280800100001000

−− −

ªºªºªºªº

«»«»«»«»

−−

¬¼¬¼¬¼¬¼

14

2

40

0

xx

x

+=

=

.

The basic variables are x

1

and x

2

, so x

3

and x

4

are free. (Note that x

3

is not zero.) Also, x

1

= –4x

4

. The

general solution is

14 4

2

34

333

44 4

44

004

00

000

010

001

xx x

xxx

xx x

−− −

ªºª º ª º

ª

ºªºªº

«»« » « »

«

»«»«»

«»« » « »

«

»«»«»

== =+ = +

«»« » « »

«

»«»«»

«»« » « »

«

»«»«»

¬

¼¬¼¬¼

¬¼¬ ¼ ¬ ¼

x

11.

1420350 1420070 1400050

0010010 0010010 0010010

~~

0000140 0000140 0000140

0000000 0000000 0000000

−− − −− −

ªºªºªº

«»«»«»

−−−

«»«»«»

−−−

«»«»«»

¬¼¬¼¬¼

12 6

36

56

450

0

40

00

xx x

xx

−+=

−=

=

.

The basic variables are x

1

, x

3

, and x

5

. The remaining variables are free. In particular, x

4

is free (and

not zero as some may assume). The solution is x

1

= 4x

2

– 5x

6

, x

3

= x

6

, x

5

= 4x

6

, with x

2

, x

4

, and x

6

free.

In parametric vector form,

34 CHAPTER 1 • Linear Equations in Linear Algebra

126 6

2

22 2

36 6

24

44 4

56 6

66 6

45 5

40 40

0

010

00 0 0

0

00

44

00 0

xxx x

x

xx x

xx xxx

xx x

−−

ªºª º ªº

ªºªº ªº

«»« » «»

«»«» «»

«»« » «»

«»«» «»

«»« » «»

«»«» «»

== = ++ = +

«»« » «»

«»«» «»

«»« » «»

«»«» «»

«»« » «»

«»«» «»

«»« » «»

«»«» «»

«»« » «»

¬¼¬¼ ¬¼

¬¼¬ ¼ ¬¼

x6

5

0

1

10

04

01

x

−

ª

ºªº

«

»«»

«

»«»

«

»«»

+

«

»«»

«

»«»

«

»«»

«

»«»

«

»«»

¬

¼¬¼

Note

: The Study Guide discusses two mistakes that students often make on this type of problem.

12.

1236500123650012302900

000146000014000001400

~~

000001000000100000010

000000000000000000000

−− −− −

ªºªºªº

«»«»«»

−

«»«»«»

¬¼¬¼¬¼

123 5

45

6

23 29 0

40

0

00

xxx x

xx

x

−++=

+=

=

.

The basic variables are x

1

, x

4

, and x

6

; the free variables are x

2

, x

3

, and x

5

. The general solution is

x

1

= 2x

2

– 3x

3

– 29x

5

, x

4

= – 4x

5

, and x

6

= 0. In parametric vector form, the solution is

123 5 2 3 5

222

333

2

455

555

6

2329 2 3 29 2

00 1

000

40040

00 0

00000

xxx x x x x

xxx

x

xxx

x

−− − −

ªºªºªºªºªºªº

«»«»«»«»«»«»

== =+ + =

«»«»« » «»«»«»

−−

«»«»« » «»«»«»

«»«»« » «»«»«

«»« » «»«»«

«»¬¼¬ ¼ ¬¼¬¼¬¼

¬¼

x

35

329

00

10

04

01

00

xx

−−

ª

ºªº

«

»«»

«

»«»

«

»«»

++

«

»«»

−

«

»«»

»

«» « »

»

«» « »

»

«» « »

¬

¼¬¼

13. To write the general solution in parametric vector form, pull out the constant terms that do not

involve the free variable:

13 3

23 33 3

33 3

54 5 4 5 4

27 2 7 2 7 .

001

xx x

xx xx x

xx x

+

ªºª º ª º

ªº ªº ªº

«»« » « »

«» «» «»

==−− =−+−=−+−=+

«»« » « »

«» «» «»

«»« » « »

«» «» «»

¬¼ ¬¼ ¬¼

¬¼¬ ¼ ¬ ¼

↑↑

xpq

pq

Geometrically, the solution set is the line through

5

2

0

ª

º

«

»

−

«

»

«

»

¬

¼

in the direction of

4

7

1

ª

º

«

»

−

«

»

«

»

¬

¼

.

1.5 • Solutions 35

14. To write the general solution in parametric vector form, pull out the constant terms that do not

involve the free variable:

14 4

24 4

44

34 4

44 4

55

005

32 2

332

25 5

225

001

xx x

xx xxx

xx x

ªºª º ª º

ªº ªº ª º

«»« » « »

«» «» « »

−− −

«»« » « »

«» «» « »

== =+ =+ =+

«»« » « »

«» «» « »

+

«»« » « »

«» «» « »

¬¼ ¬¼ ¬ ¼

¬¼¬ ¼ ¬ ¼

↑↑

xpq

pq

The solution set is the line through p in the direction of q.

15. Solve x

1

+ 5x

2

– 3x

3

= –2 for the basic variable: x

1

= –2 – 5x

2

+ 3x

3

, with x

2

and x

3

free. In vector

form, the solution is

123 23

22 2 23

33 3

25 3 2 5 3 2 5 3

00010

00 0 0 1

xxx xx

xx x xx

xx x

−− +−− − −

ªºª º ª ºªºª º ªº ªº ªº

«»« » « »

«»« » «» «» «»

== =+ + =+ +

«»« » « »

«»« » «» «» «»

«»« » « »«»« » «» «» «»

¬¼¬ ¼ ¬¼ ¬¼ ¬¼¬¼¬ ¼ ¬ ¼

x

The solution of x

1

+ 5x

2

– 3x

3

= 0 is x

1

= – 5x

2

+ 3x

3

, with x

2

and x

3

free. In vector form,

12323

22 2 23

33 3

53 5 3 5 3

010

001

xxxxx

xx x xx

xx x

−+−−

ªºª º ª ºªº ªºªº

«»« » « »

«» «»«»

== = + = +

«»« » « »

«» «»«»

«»« » « »«» «»«»

¬¼ ¬¼¬¼¬¼¬ ¼ ¬ ¼

x

= x

2

u + x

3

v

The solution set of the homogeneous equation is the plane through the origin in R

3

spanned by

u and v. The solution set of the nonhomogeneous equation is parallel to this plane and passes through

the point p =

2

0

−

ªº

«»

¬¼

.

16. Solve x

1

– 2x

2

+ 3x

3

= 4 for the basic variable: x

1

= 4 + 2x

2

– 3x

3

, with x

2

and x

3

free. In vector form,

the solution is

123 23

22 2 23

33 3

42 3 4 2 3 4 2 3

00010

00 0 0 1

xxx xx

xx x xx

xx x

+−− −

ªºª º ª ºªº ª º ªº ªº ª º

«»« » « »

«» « » «» «» « »

== =+ + =+ +

«»« » « »

«» « » «» «» « »

«»« » « »«» « » «» «» « »

¬¼ ¬ ¼ ¬¼ ¬¼ ¬ ¼¬¼¬ ¼ ¬ ¼

x

The solution of x

1

– 2x

2

+ 3x

3

= 0 is x

1

= 2x

2

– 3x

3

, with x

2

and x

3

free. In vector form,

1232 3

22 2 23

33 3

23 2 3 2 3

010

001

xxxx x

xx x xx

xx x

−− −

ªºª º ª ºªº ªº ªº

«»« » « »

«» «» «»

== = + = +

«»« » « »

«» «» «»

«»« » « »«» «» «»

¬¼ ¬¼ ¬¼¬¼¬ ¼ ¬ ¼

x

= x

2

u + x

3

v

The solution set of the homogeneous equation is the plane through the origin in R

3

spanned by u and

v. The solution set of the nonhomogeneous equation is parallel to this plane and passes through the

point p =

4

0

ªº

«»

¬¼

.

36 CHAPTER 1 • Linear Equations in Linear Algebra

17. Row reduce the augmented matrix for the system:

224 8 2248 2248 1124 1018

44816~0000~03312~0114~0114

03312 03312 0000 0000 0000

ªºªºªºªºªº

«»«»«»«»«»

−−−− −− − −

«»«»«»«»«»

−− −−

¬¼¬¼¬¼¬¼¬¼

13

23

8

4

00

xx

+=

+=−

=

. Thus x

1

= 8 – x

3

, x

2

= –4 – x

3

, and x

3

is free. In parametric vector form,

13 3

23 3 3

33 3

88 81

44 41

001

xx x

xx x x

xx x

−− −

ªºª º ª º

ª

ºªºªº

«»« » « »

«

»«»«»

==−− =−+−=−+−

«»« » « »

«

»«»«»

«»« » « »

«

»«»«»

¬

¼¬¼¬¼¬¼¬ ¼ ¬ ¼

x

The solution set is the line through

8

4

0

ªº

«»

−

«»

¬¼

, parallel to the line that is the solution set of the

homogeneous system in Exercise 5.

18. Row reduce the augmented matrix for the system:

12 3 5 1 2 3 5 1 2 3 5 1 0 1 7

21 313~0 3 3 3~0 1 1 1~0 1 1 1

11 0 8 0 3 3 3 0 0 0 0 0 0 0 0

−−−−

ªºªºªºªº

«»«»«»«»

−− −−−−

«»«»«»«»

−− −−

¬¼¬¼¬¼¬¼

13

23

7

1

00

xx

−=

−=−

=

. Thus x

1

= 7 + x

3

, x

2

= –1 + x

3

, and x

3

is free. In parametric vector form,

13 3

23 3 3

33 3

77 71

11 11

001

xx x

xx x x

xx x

+

ªºª º ªº

ª

ºªºªº

«»« » «»

«

»«»«»

==−+=−+=−+

«»« » «»

«

»«»«»

«»« » «»

«

»«»«»

¬

¼¬¼¬¼¬¼¬ ¼ ¬¼

x

The solution set is the line through

7

1

0

ªº

«»

−

«»

¬¼

, parallel to the line that is the solution set of the

homogeneous system in Exercise 6.

19. The line through a parallel to b can be written as x = a + t b, where t represents a parameter:

x =

1

2

25

03

xt

x

−−

ªºª º ª º

=+

«»«» «»

¬¼ ¬¼

¬¼

, or

1

2

25

3

xt

=−−



®=

¯

20. The line through a parallel to b can be written as x = a + tb, where t represents a parameter:

x =

1

2

37

26

xt

x

−

ªºªº ªº

=+

«»«» «»

−

¬¼ ¬¼

¬¼

, or

1

2

37

26

xt

=−



®=−+

¯

1.5 • Solutions 37

21. The line through p and q is parallel to q – p. So, given

34

and

31

ª

ºªº

==

«

»«»

−

¬

¼¬¼

pq

, form

43 1

1(3) 4

−

ªºªº

−==

«»«»

−−

¬¼¬¼

qp

, and write the line as x = p + t(q – p) =

31

34

t

ª

ºªº

+

«

»«»

−

¬

¼¬¼

.

22. The line through p and q is parallel to q – p. So, given

30

and

23

−

ª

ºªº

==

«

»«»

−

¬

¼¬¼

pq

, form

0(3) 3

32 5

−−

ªºªº

−==

«»«»

−− −

¬¼¬¼

qp

, and write the line as x = p + t(q – p) =

33

25

t

−

ª

ºªº

+

«

»«»

−

¬

¼¬¼

Note

: Exercises 21 and 22 prepare for Exercise 26 in Section 1.8.

23. a. True. See the first paragraph of the subsection titled Homogeneous Linear Systems.

b. False. The equation Ax = 0 gives an implicit description of its solution set. See the subsection

entitled Parametric Vector Form.

c. False. The equation Ax = 0 always has the trivial solution. The box before Example 1 uses the

word nontrivial instead of trivial.

d. False. The line goes through p parallel to v. See the paragraph that precedes Fig. 5.

e. False. The solution set could be empty! The statement (from Theorem 6) is true only when there

exists a vector p such that Ap = b.

24. a. False. The trivial solution is always a solution to a homogeneous system of linear equations.

b. False. A nontrivial solution of Ax = 0 is any nonzero x that satisfies the equation. See the

sentence before Example 2.

c. True. See the paragraph following Example 3.

d. True. If the zero vector is a solution, then b = Ax = A0 = 0.

e. True. See Theorem 6.

25. Suppose p satisfies Ax = b. Then Ap = b. Theorem 6 says that the solution set of Ax = b equals the

set S ={w : w = p + v

h

for some v

h

such that Av

h

= 0}. There are two things to prove: (a) every vector

in S satisfies Ax = b, (b) every vector that satisfies Ax = b is in S.

a. Let w have the form w = p + v

h

, where Av

h

= 0. Then

Aw = A(p + v

h

) = Ap + Av

h

. By Theorem 5(a) in section 1.4

= b + 0 = b

So every vector of the form p + v

h

satisfies Ax = b.

b. Now let w be any solution of Ax = b, and set v

h

= w − p. Then

Av

h

= A(w – p) = Aw – Ap = b – b = 0

So v

h

satisfies Ax = 0. Thus every solution of Ax = b has the form w = p + v

h

.

26. When A is the 3×3 zero matrix, every x in R

3

satisfies Ax = 0. So the solution set is all vectors in R

3

.

27. (Geometric argument using Theorem 6.) Since Ax = b is consistent, its solution set is obtained by

translating the solution set of Ax = 0, by Theorem 6. So the solution set of Ax = b is a single vector if

and only if the solution set of Ax = 0 is a single vector, and that happens if and only if Ax = 0 has

only the trivial solution.

38 CHAPTER 1 • Linear Equations in Linear Algebra

(Proof using free variables.) If Ax = b has a solution, then the solution is unique if and only if there

are no free variables in the corresponding system of equations, that is, if and only if every column of

A is a pivot column. This happens if and only if the equation Ax = 0 has only the trivial solution.

28. a. When A is a 3×3 matrix with three pivot positions, the equation Ax = 0 has no free variables and

hence has no nontrivial solution.

b. With three pivot positions, A has a pivot position in each of its three rows. By Theorem 4 in

Section 1.4, the equation Ax = b has a solution for every possible b. The term "possible" in the

exercise means that the only vectors considered in this case are those in R

3

, because A has three

rows.

29. a. When A is a 4×4 matrix with three pivot positions, the equation Ax = 0 has three basic variables

andone free variable. So Ax = 0 has a nontrivial solution.

b. With only three pivot positions, A cannot have a pivot in every row, so by Theorem 4 in Section

1.4, the equation Ax = b cannot have a solution for every possible b (in R

4

).

30. a. When A is a 2×5 matrix with two pivot positions, the equation Ax = 0 has two basic variables and

three free variables. So Ax = 0 has a nontrivial solution.

b. With two pivot positions and only two rows, A has a pivot position in every row. By Theorem 4

in Section 1.4, the equation Ax = b has a solution for every possible b (in R

2

).

31. a. When A is a 3×2 matrix with two pivot positions, each column is a pivot column. So the equation

Ax = 0 has no free variables and hence no nontrivial solution.

b. With two pivot positions and three rows, A cannot have a pivot in every row. So the equation Ax

= b cannot have a solution for every possible b (in R

3

), by Theorem 4 in Section 1.4.

32. No. If the solution set of Ax = b contained the origin, then 0 would satisfy A0= b, which is not true

since b is not the zero vector.

33. Look for A = [a

1

a

2

a

3

] such that 1 a

1

+ 1 a

2

+ 1 a

3

= 0. That is, construct A so that each row sum

(the sum of the entries in a row) is zero.

34. Look for A = [a

1

a

2

a

3

] such that 2 a

1

– 1 a

2

+ 1 a

3

= 0. That is, construct A so that subtracting the

third column from the second column is twice the first column.

35. Look at

12

13

721

26

xx

−−

ªº ªº

«» «»

+

«» «»

−−

¬¼ ¬¼

and notice that the second column is 3 times the first. So suitable values

for x

1

and x

2

would be 3 and –1 respectively. (Another pair would be 6 and –2, etc.) Thus

3

1

ª

º

=

«

»

−

¬

¼

x

satisfies Ax = 0.

36. Inspect how the columns a

1

and a

2

of A are related. The second column is –2/3 times the first. Put

another way, 2a

1

+ 3a

2

= 0. Thus

2

3

ªº

«»

¬¼

satisfies Ax = 0.

Note

: Exercises 35 and 36 set the stage for the concept of linear dependence.

1.6 • Solutions 39

37. Since the solution set of Ax = 0 contains the point (4,1), the vector x = (4,1) satisfies Ax = 0. Write

this equation as a vector equation, using a

1

and a

2

for the columns of A:

4

a

1

+ 1 a

2

= 0

Then a

2

= –4a

1

. So choose any nonzero vector for the first column of A and multiply that column by

– 4 to get the second column of A. For example, set

14

A

−

ª

º

=

«

»

−

¬

¼

.

Finally, the only way the solution set of Ax = b could not be parallel to the line through (1,4) and the

origin is for the solution set of Ax = b to be empty. This does not contradict Theorem 6, because that

theorem applies only to the case when the equation Ax = b has a nonempty solution set. For b, take

any vector that is not a multiple of the columns of A.

Note

: In the Study Guide, a “Checkpoint” for Section 1.5 will help students with Exercise 37.

38. If w satisfies Ax = 0, then Aw = 0. For any scalar c, Theorem 5(b) in Section 1.4 shows that

A(cw)=cAw = c0 = 0.

39. Suppose Av = 0 and Aw = 0. Then, since A(v + w) = Av + Aw by Theorem 5(a) in Section 1.4,

A(v + w) = Av + Aw = 0 + 0 = 0.

Now, let c and d be scalars. Using both parts of Theorem 5,

A(cv + dw) = A(cv) + A(dw) = cAv + dAw = c0 + d0 = 0.

40. No. If Ax = b has no solution, then A cannot have a pivot in each row. Since A is 3×3, it has at most

two pivot positions. So the equation Ax = y for any y has at most two basic variables and at least one

free variable. Thus, the solution set for Ax = y is either empty or has infinitely many elements.

Note

: The MATLAB box in the Study Guide introduces the zeros command, in order to augment a

matrix with a column of zeros.

1.6 SOLUTIONS

1. Fill in the exchange table one column at a time. The entries in a column describe where a sector's

output goes. The decimal fractions in each column sum to 1.

Distribution of

Output From:

Goods Services Purchased by:

output input

.2 .7 Goods

.8 .3 Services

↓↓

→

Denote the total annual output (in dollars) of the sectors by p

G

and p

S

. From the first row, the total

input to the Goods sector is .2 p

G

+ .7 p

S

. The Goods sector must pay for that. So the equilibrium

prices must satisfy

income expenses

=.2 .7

GGS

ppp+

40 CHAPTER 1 • Linear Equations in Linear Algebra

From the second row, the input (that is, the expense) of the Services sector is .8 p

G

+ .3 p

S

.

The equilibrium equation for the Services sector is

income expenses

=.8 .3

SGS

ppp+

Move all variables to the left side and combine like terms:

.8 .7 0

GS

pp

−=

−+=

Row reduce the augmented matrix:

.8 .7 0 .8 .7 0 1 .875 0

~~

.8 .7 0 0 0 0 0 0 0

−−−

ªºªºªº

«»«»«»

−

¬¼¬¼¬¼

The general solution is p

G

= .875 p

S

, with p

S

free. One equilibrium solution is p

S

= 1000 and p

G

=

875. If one uses fractions instead of decimals in the calculations, the general solution would be

written p

G

= (7/8) p

S

, and a natural choice of prices might be p

S

= 80 and p

G

= 70. Only the ratio of

the prices is important: p

G

= .875 p

S

. The economic equilibrium is unaffected by a proportional

change in prices.

2. Take some other value for p

S

, say 200 million dollars. The other equilibrium prices are then

p

C

= 188 million, p

E

= 170 million. Any constant nonnegative multiple of these prices is a set of

equilibrium prices, because the solution set of the system of equations consists of all multiples of one

vector. Changing the unit of measurement to another currency such as Japanese yen has the same

effect as multiplying all equilibrium prices by a constant. The ratios of the prices remain the same,

no matter what currency is used.

3. a. Fill in the exchange table one column at a time. The entries in a column describe where a sector’s

output goes. The decimal fractions in each column sum to 1.

Distribution of Output From :

Purchased by :

Fuels and Power Manufacturing Services

output input

.10 .10 .20 Fuels and Power

.80 .10 .40 Manufacturing

.10 .80 .40 Services

↓↓↓

→

b. Denote the total annual output (in dollars) of the sectors by p

F

, p

M

, and p

S

. From the first row of

the table, the total input to the Fuels & Power sector is .1p

F

+ .1p

M

+ .2p

S

. So the equilibrium

prices must satisfy

income expenses

=.1 .1 .2

FFMS

pppp++

From the second and third rows of the table, the income/expense requirements for the

Manufacturing sector and the Services sector are, respectively,

.8 .1 .4

.1 .8 .4

MFMS

SFM S

pppp

ppp p

=++

=+ +

1.6 • Solutions 41

Move all variables to the left side and combine like terms:

.9 – .1 – .2 0

–.8 .9 – .4 0

–.1 – .8 .6 0

FM S

FMS

pp p

=

+=

.9 .1 .2 0

.8 .9 .4 0

.1 .8 .6 0

−−

ª

º

«

»

−−

«

»

«

»

−−

¬

¼

c. [M] You can obtain the reduced echelon form with a matrix program.

.9 .1 .2 0 1 0 .301 0 The number of decimal

.8 .9 .4 0 ~ 0 1 .712 0 places displayed is

.1 .8 .6 0 0 0 0 0 somewhat arbitrary.

−− −

ªºªº

«»«»

−− −

«»«»

−−

¬¼¬¼

The general solution is p

F

= .301 p

S

, p

M

= .712 p

S

, with p

S

free. If p

S

is assigned the value of 100,

then p

F

= 30.1 and p

M

= 71.2. Note that only the ratios of the prices are determined. This makes

sense, for if they were converted from, say, dollars to yen or Euros, the inputs and outputs of each

sector would still balance. The economic equilibrium is not affected by a proportional change in

prices.

4. a. Fill in the exchange table one column at a time. The entries in each column must sum to 1.

Distribution of Output From :

Purchased by :

Mining Lumber Energy Transportation

output input

.30 .15 .20 .20 Mining

.10 .15 .15 .10 Lumber

.60 .50 .45 .50 Energy

0 .20 .20 .20 Transportation

↓↓↓ ↓

→

b. [M] Denote the total annual output of the sectors by p

M

, p

L

, p

E

, and p

T

, respectively. From the first

row of the table, the total input to Agriculture is .30p

M

+ .15p

L

+ .20p

E

+ .20 p

T

. So the

equilibrium prices must satisfy

income expenses

.30 .15 .20 .20

MMLET

ppppp

=++ +

From the second, third, and fourth rows of the table, the equilibrium equations are

T

.10 .15 .15 .10

.60 .50 .45 .50

.20 .20 .20

LMLET

EMLET

TLE

ppppp

pppp

=+++

=++

Move all variables to the left side and combine like terms:

.70 .15 .20 .20 0

.10 .85 .15 .10 0

.60 .50 .55 .50 0

.20 .20 .80 0

MLET

LET

pppp

ppp

−−−=

−+−−=

−−+−=

−− +=

Reduce the augmented matrix to reduced echelon form:

42 CHAPTER 1 • Linear Equations in Linear Algebra

.70 .15 .20 .20 0 1 0 0 1.37 0

.10 .85 .15 .10 0 0 1 0 .84 0

~

.60 .50 .55 .50 0 0 0 1 3.16 0

0 .20 .20 .80 0 0 0 0 0 0

−−− −

ªºªº

«»«»

−−− −

«»«»

−− − −

«»«»

−−

¬¼¬¼

Solve for the basic variables in terms of the free variable p

T

, and obtain p

M

= 1.37p

T

, p

L

= .84p

T

,

and p

E

= 3.16p

T

. The data probably justifies at most two significant figures, so take p

T

= 100 and

round off the other prices to p

M

= 137, p

L

= 84, and p

E

= 316.

5. a. Fill in the exchange table one column at a time. The entries in each column must sum to 1.

Distribution of Output From :

Purchased by :

Agriculture Manufacturing Services Transportation

output input

.20 .35 .10 .20 Agriculture

.20 .10 .20 .30 Manufacturing

.30 .35 .50 .20 Services

.30 .20 .20 .30 Transportation

↓↓↓↓

→

b. [M] Denote the total annual output of the sectors by p

A

, p

M

, p

S

, and p

T

, respectively. The

equilibrium equations are

T

.20 .35 .10 .20

.20 .10 .20 .30

.30 .35 .50 .20

.30 .20 .20 .30

A

AM ST

MAMST

SAMST

TAMS

ppppp

=+++

Move all variables to the left side and combine like terms:

.80 .35 .10 .20 0

.20 .90 .20 .30 0

.30 .35 .50 .20 0

.30 .20 .20 .70 0

AM ST

pp pp

−−−=

−+−−=

−− +−=

−− −+=

Reduce the augmented matrix to reduced echelon form:

.80 .35 .10 .20 0 1 0 0 .799 0

.20 .90 .20 .30 0 0 1 0 .836 0

~

.30 .35 .50 .20 0 0 0 1 1.465 0

.30 .20 .20 .70 0 0 0 0 0 0

−−− −

ªºªº

«»«»

−−− −

«»«»

−− − −

«»«»

−−−

¬¼¬¼

Solve for the basic variables in terms of the free variable p

T

, and obtain p

A

= .799p

T

, p

M

= .836p

T

,

and p

S

= 1.465p

T

. Take p

T

= $10.00 and round off the other prices to p

A

= $7.99, p

M

= $8.36, and

p

S

= $14.65 per unit.

c. Construct the new exchange table one column at a time. The entries in each column must sum to 1.

1.6 • Solutions 43

Distribution of Output From :

Purchased by :

Agriculture Manufacturing Services Transportation

output input

.20 .35 .10 .20 Agriculture

.10 .10 .20 .30 Manufacturing

.40 .35 .50 .20 Services

.30 .20 .20 .30 Transportation

↓↓↓↓

→

d. [M] The new equilibrium equations are

T

.20 .35 .10 .20

.10 .10 .20 .30

.40 .35 .50 .20

.30 .20 .20 .30

A

AM ST

MAMST

SAMST

TAMS

ppppp

=+++

Move all variables to the left side and combine like terms:

.80 .35 .10 .20 0

.10 .90 .20 .30 0

.40 .35 .50 .20 0

.30 .20 .20 .70 0

AM ST

pp pp

−−−=

−+−−=

−− +−=

−− −+=

Reduce the augmented matrix to reduced echelon form:

.80 .35 .10 .20 0 1 0 0 .781 0

.10 .90 .20 .30 0 0 1 0 .767 0

~

.40 .35 .50 .20 0 0 0 1 1.562 0

.30 .20 .20 .70 0 0 0 0 0 0

−−− −

ªºªº

«»«»

−−− −

«»«»

−− − −

«»«»

−−−

¬¼¬¼

Solve for the basic variables in terms of the free variable p

T

, and obtain p

A

= .781p

T

, p

M

= .767p

T

,

and

p

S

= 1.562p

T

. Take p

T

= $10.00 and round off the other prices to p

A

= $7.81, p

M

= $7.67, and

p

S

= $15.62 per unit. The campaign has caused unit prices for the Agriculture and

Manufacturing sectirs to go down slightly, while increasing the unit price for the Services sector

to increase by $.10 per unit. The campaign has benefited the Services sector the most.

6. The following vectors list the numbers of atoms of aluminum (Al), oxygen (O), and carbon (C):

23 2

20 1 0

aluminum

Al O : 3 , C: 0 , Al: 0 , CO : 2 oxygen

01 0 1

carbon

ªº ªº ªº ªº

«» «» «» «»

¬¼ ¬¼ ¬¼ ¬¼

The coefficients in the equation x

1

⋅Al

2

O

3

+ x

2

⋅C → x

3

⋅Al + x

4

⋅CO

2

satisfy

44 CHAPTER 1 • Linear Equations in Linear Algebra

1234

2010

3002

0101

xx xx

ªº ªº ªº ªº

«» «» «» «»

+=+

«» «» «» «»

¬¼ ¬¼ ¬¼ ¬¼

Move the right terms to the left side (changing the sign of each entry in the third and fourth vectors)

and row reduce the augmented matrix of the homogeneous system:

20 1 00 10 1/2 00 10 1/2 00

30 0 20~30 0 20~00 3/2 20

01 0 10 01 0 10 01 0 10

−− −

ªºª ºª º

«»« »« »

−−−

«»« »« »

−−−

¬¼¬ ¼¬ ¼

10 1/2 00 10 1/2 00 100 2/30

~0 1 0 1 0~0 1 0 1 0~0 1 0 1 0

00 3/2 20 00 1 4/30 001 4/30

−− −

ªºª ºªº

«»« »«»

−−−

«»« »«»

−−−

¬¼¬ ¼¬¼

The general solution is x

1

= (2/3)x

4

, x

2

= x

4

, x

3

= (4/3)x

4

, with x

4

free. Take x

4

= 3. Then x

1

= 2,

x

2

= 3, and x

3

= 4. The balanced equation is

2Al

2

O

3

+ 3C → 4Al + 3CO

2

7. The following vectors list the numbers of atoms of sodium (Na), hydrogen (H), carbon (C), and

oxygen (O):

33657 3657 2 2

1 0 300sodium

1 8 520hydrogen

NaHCO : , H C H O : , Na C H O : , H O : , CO :

1 6 601carbon

3 7 712oxygen

ªº ªº ªº ªº ªº

«» «» «» «» «»

¬¼ ¬¼ ¬¼ ¬¼ ¬¼

The order of the various atoms is not important. The list here was selected by writing the elements in

the order in which they first appear in the chemical equation, reading left to right:

x

1

· NaHCO

3

+ x

2

· H

3

C

6

H

5

O

7

→ x

3

· Na

3

C

6

H

5

O

7

+ x

4

· H

2

O + x

5

· CO

2

.

The coefficients x

1

, …, x

5

satisfy the vector equation

12 3 4 5

10300

18520

16601

37712

xx x x x

ªº ªº ªº ªº ªº

«» «» «» «» «»

+=++

«» «» «» «» «»

¬¼ ¬¼ ¬¼ ¬¼ ¬¼

Move all the terms to the left side (changing the sign of each entry in the third, fourth, and fifth

vectors) and reduce the augmented matrix:

10 3 0 0 0 1000 1 0

18 5 2 00 0100 1/30

~

16 6 0 10 0010 1/30

37 7 1 2 0 0001 1 0

−−

ªºªº

«»«»

−− −

«»«»

−− −

«»«»

−−− −

¬¼¬¼

The general solution is x

1

= x

5

, x

2

= (1/3)x

5

, x

3

= (1/3)x

5

, x

4

= x

5

, and x

5

is free. Take x

5

= 3. Then x

1

=

x

4

= 3, and x

2

= x

3

= 1. The balanced equation is

3NaHCO

3

+ H

3

C

6

H

5

O

7

→ Na

3

C

6

H

5

O

7

+ 3H

2

O + 3CO

2

1.6 • Solutions 45

8. The following vectors list the numbers of atoms of hydrogen (H), oxygen (O), calcium (Ca), and

carbon (C):

332 2

30200

13102

HO: , CaCO : , H : , Ca: , CO:

01010

01001

hydrogen

oxygen

Ocalcium

carbon

ªº ªº ªº ªº ªº

«» «» «» «» «»

¬¼ ¬¼ ¬¼ ¬¼ ¬¼

The coefficients in the chemical equation

x

1

⋅H

3

O + x

2

⋅CaCO

3

→ x

3

⋅H

2

O + x

4

⋅Ca + x

5

⋅CO

2

satisfy the vector equation

12 345

30200

13102

01010

01001

xx x x x

ªº ªº ªº ªº ªº

«» «» «» «» «»

+=++

«» «» «» «» «»

¬¼ ¬¼ ¬¼ ¬¼ ¬¼

Move the terms to the left side (changing the sign of each entry in the last three vectors) and reduce

the augmented matrix:

30 2 0 00 1000 20

13 1 0 20 0100 10

~

01 0 1 00 0010 30

01 0 0 10 0001 10

−−

ªºªº

«»«»

−− −

«»«»

−−

«»«»

−−

¬¼¬¼

The general solution is x

1

= 2x

5

, x

2

= x

5

, x

3

= 3x

5

, x

4

= x

5

, and x

5

is free. Take x

5

= 1. Then x

1

= 2, and

x

2

= x

4

= 1, and x

3

= 3. The balanced equation is

2H

3

O + CaCO

3

→ 3H

2

O + Ca + CO

2

9.The following vectors list the numbers of atoms of boron (B), sulfur (S), hydrogen (H), and oxygen

(O):

23 233 2

20 10boron

30 01sulfur

BS: , HO: , HBO: , HS:

02 32hydrogen

01 30oxygen

ªº ªº ªº ªº

«» «» «» «»

¬¼ ¬¼ ¬¼ ¬¼

The coefficients in the equation x

1

⋅B

2

S

3

+ x

2

⋅H

2

O → x

3

⋅H

3

BO

3

+ x

4

⋅H

2

S satisfy

1234

2010

3001

0232

0130

xx xx

ªº ªº ªº ªº

«» «» «» «»

+=+

«» «» «» «»

¬¼ ¬¼ ¬¼ ¬¼

Move the terms to the left side (changing the sign of each entry in the third and fourth vectors) and

row reduce the augmented matrix of the homogeneous system:

20 1 00 100 1/30

30 0 10 010 2 0

~

02 3 20 001 2/30

01 3 00 000 0 0

−−

ªºª º

«»« »

−−

«»« »

−− −

«»«»

−

¬¼¬¼

46 CHAPTER 1 • Linear Equations in Linear Algebra

The general solution is x

1

= (1/3) x

4

, x

2

= 2x

4

, x

3

= (2/3) x

4

, with x

4

free. Take x

4

= 3. Then x

1

= 1,

x

2

= 6, and x

3

= 2. The balanced equation is

B

2

S

3

+ 6H

2

O → 2H

3

BO

3

+ 3H

2

S

10. [M] Set up vectors that list the atoms per molecule. Using the order lead (Pb), nitrogen (N),

chromium (Cr), manganese (Mn), and oxygen (O), the vector equation to be solved is

12 3456

103000lead

600001nitrogen

010200chromium

020010manganese

084321oxygen

xx xx xx

ªº ªº ªº ªº ªº ªº

«» «» «» «» «» «»

+=+++

«» «» «» «» «» «»

¬¼ ¬¼ ¬¼ ¬¼ ¬¼ ¬¼

The general solution is x

1

= (1/6)x

6

, x

2

= (22/45)x

6

, x

3

= (1/18)x

6

, x

4

= (11/45)x

6

, x

5

= (44/45)x

6

, and x

6

is free. Take x

6

= 90. Then x

1

= 15, x

2

= 44, x

3

= 5, x

4

= 22, and x

5

= 88. The balanced equation is

15PbN

6

+ 44CrMn

2

O

8

→ 5Pb

3

O

4

+ 22Cr

2

O

3

+ 88MnO

2

+ 90NO

11. [M] Set up vectors that list the atoms per molecule. Using the order manganese (Mn), sulfur (S),

arsenic (As), chromium (Cr), oxygen (O), and hydrogen (H), the vector equation to be solved is

12 3 456 7

1001000

1010030

0200100

0100 00 1 0

035440121

0021302

xx x x xx x

ªº ª º ªº ªº ªº ª º ªº

«» « » «» «» «» « » «»

++=+++

«» « » «» «» «» « » «

¬¼ ¬ ¼ ¬¼ ¬¼ ¬¼ ¬ ¼ ¬¼

manganese

sulfur

arsenic

chromium

oxygen

hydrogen

»

In rational format, the general solution is x

1

= (16/327)x

7

, x

2

= (13/327)x

7

, x

3

= (374/327)x

7

,

x

4

= (16/327)x

7

, x

5

= (26/327)x

7

, x

6

= (130/327)x

7

, and x

7

is free. Take x

7

= 327 to make the other

variables whole numbers. The balanced equation is

16MnS + 13As

2

Cr

10

O

35

+ 374H

2

SO

4

→ 16HMnO

4

+ 26AsH

3

+ 130CrS

3

O

12

+ 327H

2

O

Note that some students may use decimal calculation and simply "round off" the fractions that relate

x

1

, ..., x

6

to x

7

. The equations they construct may balance most of the elements but miss an atom or

two. Here is a solution submitted by two of my students:

5MnS + 4As

2

Cr

10

O

35

+ 115H

2

SO

4

→ 5HMnO

4

+ 8AsH

3

+ 40CrS

3

O

12

+ 100H

2

O

Everything balances except the hydrogen. The right side is short 1 hydrogen atom. Perhaps the

students thought that it escaped!

12. Write the equations for each intersection:

14 2

23

34

Intersection Flow in Flow out

A

B100

C80

xx x

xx

+=

=+

+=

Rearrange the equations:

1.6 • Solutions 47

12 4

23

34

0

100

80

xx x

xx

−+=

−=

−=−

Reduce the augmented matrix:

1101 0 100020

0110100~~010120

001180 001180

−

ªºªº

«»«»

−⋅⋅⋅−

«»«»

−− −−

¬¼¬¼

The general solution (written in the style of Section 1.2) is

1

24

34

4

20

= 80 +

is free

x

xx

x

=



°=+

°

®−

°

¯

Since x

3

cannot be negative, the minimum value of x

4

is 80.

13. Write the equations for each intersection:

21

35 24

65

46

13

Intersection Flow in Flow out

A3080

B

C100 40

D4090

E60 20

xx

xx xx

xx

+=+

+= +

+=+

+= +

+=+

Rearrange the equations:

12

2345

56

46

13

50

0

60

50

40

xx

xxxx

xx

−=−

−+−=

−=

−=−

Reduce the augmented matrix:

11000050 10100040

011110 0 01100010

~~

00001160 00010150

000 10150 00001160

10 1000 40 000000 0

−−−−

ªºªº

«»«»

−− −

«»«»

⋅⋅⋅

−−

«»«»

−−

«»«»

−−

¬¼¬¼

48 CHAPTER 1 • Linear Equations in Linear Algebra

a. The general solution is

13

23

3

46

56

6

40

10

is free

50

60

is free

xx

x

xx

x

=−



°

=+

°

®

=+

°

=+

°

¯

b. To find minimum flows, note that since x

1

cannot be negative, x

3

> 40. This implies that

x

2

> 50. Also, since x

6

cannot be negative, x

4

> 50 and x

5

> 60. The minimum flows are

x

2

= 50, x

3

= 40, x

4

= 50, x

5

= 60 (when x

1

= 0 and x

6

= 0).

14. Write the equations for each intersection:

15

12 4

32

45 3

Intersection Flow in Flow out

A80

B100

C90

D90

xx

xx x

xx

xx x

=+

++ =

=+

+=+

Rearrange the equations:

15

12 4

23

345

80

100

90

xx

xx x

xx

xxx

+=

+−=−

−=−

−−=−

Reduce the augmented matrix:

10 0 0 1 80 100 0 1 80

11 0 1 0 100 010 1 1 180

~~

01 1 0 0 90 001 1 1 90

00 1 1 1 90 000 0 0 0

ªºªº

«»«»

−− −−−

«»«»

⋅⋅⋅

«»«»

−− −−−

«»«»

−− −

¬¼¬¼

a. The general solution is

15

245

345

4

5

80

180

= 90

is free

xx

xxx

x

=−



°

=+−

°

+−

®

°

¯

b. If x

5

= 0, then the general solution is

1

24

34

4

80

180

= 90

is free

x

xx

x

=



°

=−

°

®

−

°

¯

c. Since x

2

cannot be negative, the minimum value of x

4

when x

5

= 0 is 180.

1.6 • Solutions 49

15. Write the equations for each intersection.

61

12

23

34

45

56

Intersection Flow in Flow out

A60

B70

C100

D90

E80

F80

xx

+=

=+

+=

=+

+=

=+

Rearrange the equations:

16

12

23

34

45

56

60

70

100

90

80

xx

−=

−=−

−=

−=−

−=

Reduce the augmented matrix:

10000 1 60 10000 160

110000 70 01000110

011000100 00100190

~~

001100 90 000101 0

000110 80 00001180

000011 80 000000 0

−−

ªºªº

«»«»

−−−

«»«»

−− −

⋅⋅⋅

«»«»

−−

«»«»

−− −

«»«»

−

«»«»

¬¼¬¼

The general solution is

16

26

36

46

56

6

60

10

90

80

is free

xx

x

=+



°

=−+

°

=+

°

®

=

°

=+

°

¯

.

Since x

2

cannot be negative, the minimum value of x

6

is 10.

Note: The MATLAB box in the Study Guide discusses rational calculations, needed for

balancing the chemical equations in Exercises 10 and 11. As usual, the appendices cover this

material for Maple, Mathematica, and the TI calculators.

50 CHAPTER 1 • Linear Equations in Linear Algebra

1.7 SOLUTIONS

Note:

Key exercises are 9–20 and 23–30. Exercise 30 states a result that could be a theorem in the text.

There is a danger, however, that students will memorize the result without understanding the proof, and

then later mix up the words row and column. Exercises 37 and 38 anticipate the discussion in Section 1.9

of one-to-one transformations. Exercise 44 is fairly difficult for my students.

1. Use an augmented matrix to study the solution set of x

1

u + x

2

v + x

3

w = 0 (*), where u, v, and w are

the three given vectors. Since

5790 5790

0240~0240

0680 0040

ªºªº

«»«»

−−

¬¼¬¼

, there are no free variables. So

the homogeneous equation (*) has only the trivial solution. The vectors are linearly independent.

2. Use an augmented matrix to study the solution set of x

1

u + x

2

v + x

3

w = 0 (*), where u, v, and w are

the three given vectors. Since

0010 20 30

2030~087/20

3810 00 10

−

ªºª º

«»« »

−−

«»« »

−−

¬¼¬ ¼

, there are no free

variables. So the homogeneous equation (*) has only the trivial solution. The vectors are linearly

independent.

3. Use the method of Example 3 (or the box following the example). By comparing entries of the

vectors, one sees that the second vector is –2 times the first vector. Thus, the two vectors are linearly

dependent.

4. From the first entries in the vectors, it seems that the second vector of the pair

13

,

39

−−

ªºªº

«»«»

−

¬¼¬¼

may be 3

times the first vector. But there is a sign problem with the second entries. So neither of the vectors is

a multiple of the other. The vectors are linearly independent.

5. Use the method of Example 2. Row reduce the augmented matrix for Ax = 0:

0390 1420 1420 1420 1420

2170 2170 0930 0930 0930

~~~~

1450 14 50 0070 0070 0070

1420 0390 0 390 0080 0000

−−−−−−−−−

ªºªºªºªºªº

«»«»«»«»«»

−−−−−

«»«»«»«»«»

−− −− − − −

«»«»«»«»«»

−− − −

¬¼¬¼¬¼¬¼¬¼

There are no free variables. The equation Ax = 0 has only the trivial solution and so the columns of A

are linearly independent.

6. Use the method of Example 2. Row reduce the augmented matrix for Ax = 0:

43 00 11 50 11 50 11 50 11 50

01 50 01 50 01 50 01 50 01 50

~~~~

11 50 43 00 01200 00150 00150

21100 21100 01 00 00 50 00 00

−− − − − −

ªºªºªºªºªº

«»«»«»«»«»

−−−−−

«»«»«»«»«»

−−− −−−

«»«»«»«»«»

−−− −

¬¼¬¼¬¼¬¼¬¼

1.7 • Solutions 51

There are no free variables. The equation Ax = 0 has only the trivial solution and so the columns of A

are linearly independent.

7. Study the equation Ax = 0. Some people may start with the method of Example 2:

14300 14300 14300

27510~01110~01110

45750 011550 00660

−−−

ªºªºªº

«»«»«»

−− − −

«»«»«»

−− − −

¬¼¬¼¬¼

But this is a waste of time. There are only 3 rows, so there are at most three pivot positions. Hence, at

least one of the four variables must be free. So the equation Ax = 0 has a nontrivial solution and the

columns of A are linearly dependent.

8. Same situation as with Exercise 7. The (unnecessary) row operations are

12 320 12320 12 320

24620~00060~01 130

01130 01130 00 060

−−−

ªºªºªº

«»«»«»

−− −

«»«»«»

−−

¬¼¬¼¬¼

Again, because there are at most three pivot positions yet there are four variables, the equation

Ax = 0 has a nontrivial solution and the columns of A are linearly dependent.

9. a. The vector v

3

is in Span{v

1

, v

2

} if and only if the equation x

1

v

1

+ x

2

v

2

= v

3

has a solution. To find

out, row reduce [v

1

v

2

v

3

], considered as an augmented matrix:

135 13 5

397~00 8

26 00 10hh

−−

ªºª º

«»« »

−−

«»« »

−−

¬¼¬ ¼

At this point, the equation 0 = 8 shows that the original vector equation has no solution. So v

3

is

in Span{v

1

, v

2

} for no value of h.

b. For {v

1

, v

2

, v

3

} to be linearly independent, the equation x

1

v

1

+ x

2

v

2

+ x

3

v

3

= 0 must have only the

trivial solution. Row reduce the augmented matrix [v

1

v

2

v

3

0]

1350 13 5 0 1350

3970~00 8 0~0080

26 0 00 100 0000hh

−− −

ªºª ºªº

«»« »«»

−−

«»« »«»

−−

¬¼¬ ¼¬¼

For every value of h, x

2

is a free variable, and so the homogeneous equation has a nontrivial

solution. Thus {v

1

, v

2

, v

3

} is a linearly dependent set for all h.

10. a. The vector v

3

is in Span{v

1

, v

2

} if and only if the equation x

1

v

1

+ x

2

v

2

= v

3

has a solution. To find

out, row reduce [v

1

v

2

v

3

], considered as an augmented matrix:

132 13 2

395~00 1

515 0 0 10hh

−−

ªºª º

«»« »

−−

«»« »

−+

¬¼¬ ¼

At this point, the equation 0 = 1 shows that the original vector equation has no solution. So v

3

is

in Span{v

1

, v

2

} for no value of h.

52 CHAPTER 1 • Linear Equations in Linear Algebra

b. For {v

1

, v

2

, v

3

} to be linearly independent, the equation x

1

v

1

+ x

2

v

2

+ x

3

v

3

= 0 must have only the

trivial solution. Row reduce the augmented matrix [v

1

v

2

v

3

0]

1320 13 2 0 1320

3950~00 1 0~0010

515 0 0 0 10 0 0 0 0 0hh

−− −

ªºª ºªº

«»« »«»

−−

«»« »«»

−+

¬¼¬ ¼¬¼

For every value of h, x

2

is a free variable, and so the homogeneous equation has a nontrivial

solution. Thus {v

1

, v

2

, v

3

} is a linearly dependent set for all h.

11. To study the linear dependence of three vectors, say v

1

, v

2

, v

3

, row reduce the augmented matrix

[v

1

v

2

v

3

0]:

2420 24 20 24 20

2620~02 00~02 00

47 0 01 40 00 40hh h

−− −

ªºª ºª º

«»« »« »

−− − −

«»« »« »

−++

¬¼¬ ¼¬ ¼

The equation x

1

v

1

+ x

2

v

2

+ x

3

v

3

= 0 has a nontrivial solution if and only if h + 4 = 0 (which

corresponds to x

3

being a free variable). Thus, the vectors are linearly dependent if and only if h = –4.

12. To study the linear dependence of three vectors, say v

1

, v

2

, v

3

, row reduce the augmented matrix

[v

1

v

2

v

3

0]:

3690 36 90

64 0~01 00

1330 00 180

h

−−

ªºª º

«»« »

−−

«»« »

−+

¬¼¬ ¼

The equation x

1

v

1

+ x

2

v

2

+ x

3

v

3

= 0 has a nontrivial solution if and only if h + 18 = 0 (which

corresponds to x

3

being a free variable). Thus, the vectors are linearly dependent if and only if h = –

18.

13. To study the linear dependence of three vectors, say v

1

, v

2

, v

3

, row reduce the augmented matrix

[v

1

v

2

v

3

0]:

1230 12 3 0

59 0~01 150

3690 00 0 0

hh

−−

ªºªº

«»«»

−−

«»«»

−−

¬¼¬¼

The equation x

1

v

1

+ x

2

v

2

+ x

3

v

3

= 0 has a free variable and hence a nontrivial solution no matter what

the value of h. So the vectors are linearly dependent for all values of h.

14. To study the linear dependence of three vectors, say v

1

, v

2

, v

3

, row reduce the augmented matrix

[v

1

v

2

v

3

0]:

1320 13 20 13 2 0

2710~0 1 50~01 5 0

46 0 06 80 00 380hh h

−− −

ªºª ºª º

«»« »« »

−

«»« »« »

−−++

¬¼¬ ¼¬ ¼

The equation x

1

v

1

+ x

2

v

2

+ x

3

v

3

= 0 has a nontrivial solution if and only if h + 38 = 0 (which

corresponds to x

3

being a free variable). Thus, the vectors are linearly dependent if and only

if h = –38.

1.7 • Solutions 53

15. The set is linearly dependent, by Theorem 8, because there are four vectors in the set but only two

entries in each vector.

16. The set is linearly dependent because the second vector is –3/2 times the first vector.

17. The set is linearly dependent, by Theorem 9, because the list of vectors contains a zero vector.

18. The set is linearly dependent, by Theorem 8, because there are four vectors in the set but only two

entries in each vector.

19. The set is linearly independent because neither vector is a multiple of the other vector. [Two of the

entries in the first vector are – 4 times the corresponding entry in the second vector. But this multiple

does not work for the third entries.]

20. The set is linearly dependent, by Theorem 9, because the list of vectors contains a zero vector.

21. a. False. A homogeneous system always has the trivial solution. See the box before Example 2.

b. False. See the warning after Theorem 7.

c. True. See Fig. 3, after Theorem 8.

d. True. See the remark following Example 4.

22. a. True. See Theorem 7.

b. True. See Example 4.

c. False. For instance, the set consisting of

12

2 and –4

36

ª

ºªº

«

»«»

−

«

»«»

«

»«»

¬

¼¬¼

is linearly dependent. See the warning

after Theorem 8.

d. False. See Example 3(a).

23.

*0 00

,,

00 00 00

ªºªºªº

«»«»«»

¬¼¬¼¬¼



24.

**

0*

00

ª

º

«

»

«

»

«

»

¬

¼



25.

*0

000

and

00 00

ª

ºªº

«

»«»

«

»«»

«

»«»

«

»«»

«

»«»

¬

¼¬¼





26.

**

0*

00

000

ªº

«»

¬¼



. The columns must be linearly independent, by Theorem 7, because the first column is

not zero, the second column is not a multiple of the first, and the third column is not a linear

combination of the preceding two columns (because a

3

is not in Span{a

1

, a

2

}).

27. All four columns of the 6×4 matrix A must be pivot columns. Otherwise, the equation Ax = 0 would

have a free variable, in which case the columns of A would be linearly dependent.

28. If the columns of a 4×6 matrix A span R

4

, then A has a pivot in each row, by Theorem 4. Since each

pivot position is in a different column, A has four pivot columns.

54 CHAPTER 1 • Linear Equations in Linear Algebra

29. A: any 3×2 matrix with one column a multiple of the other.

B: any 3×2 matrix with two nonzero columns such that neither column is a multiple of the other. In

this case the columns are linearly independent and so the equation Bx = 0 has only the trivial

solution.

30. a. n

b. The columns of A are linearly independent if and only if the equation Ax = 0 has only the trivial

solution. This happens if and only if Ax = 0 has no free variables, which in turn happens if and

only if every variable is a basic variable, that is, if and only if every column of A is a pivot

column.

31. Think of A = [a

1

a

2

a

3

]. The text points out that a

3

= a

1

+ a

2

. Rewrite this as a

1

+ a

2

– a

3

= 0. As a

matrix equation, Ax = 0 for x = (1, 1, –1).

32. Think of A = [a

1

a

2

a

3

]. The text points out that a

1

– 3a

2

= a

3

. Rewrite this as a

1

– 3a

2

– a

3

= 0. As

a matrix equation, Ax = 0 for x = (1, –3, –1).

33. True, by Theorem 7. (The Study Guide adds another justification.)

34. False. The vector v

1

could be the zero vector.

35. True, by Theorem 9.

36. False. Counterexample: Take v

1

and v

2

to be multiples of one vector. Take v

3

to be not a multiple of

that vector. For example,

12 3

121

1, 2, 0

120

ªº ª º ª º

«» « » « »

== =

«» « » « »

¬¼ ¬ ¼ ¬ ¼

vv v

37. True. A linear dependence relation among v

1

, v

2

, v

3

may be extended to a linear dependence relation

among v

1

, v

2

, v

3

, v

4

by placing a zero weight on v

4

.

38. True. If the equation x

1

v

1

+ x

2

v

2

+ x

3

v

3

= 0 had a nontrivial solution (with at least one of x

1

, x

2

, x

3

nonzero), then so would the equation x

1

v

1

+ x

2

v

2

+ x

3

v

3

+ 0¸v

4

= 0. But that cannot happen because

{v

1

, v

2

, v

3

, v

4

} is linearly independent. So {v

1

, v

2

, v

3

} must be linearly independent. This problem can

also be solved using Exercise 37, if you know that the statement there is true.

39. If for all b the equation Ax = b has at most one solution, then take b = 0, and conclude that the

equation Ax = 0 has at most one solution. Then the trivial solution is the only solution, and so the

columns of A are linearly independent.

40. An m×n matrix with n pivot columns has a pivot in each column. So the equation Ax = b has no free

variables. If there is a solution, it must be unique.

1.7 • Solutions 55

41. [M]

3410 74 3 4 10 7 4

5371115 029/329/3 2/325/3

~

435 21025/325/322/319/3

8723 415 011/311/344/377/3

A

−−− −

ªºªº

«»«»

−−−− −

«»«»

=

«»«»

−−

«»«»

−−−

¬¼¬¼

3410 7 43410 7 4

029/329/3 2/3 25/3 029/329/3 2/3 25/3

~~

000196/29392/29000196/29392/29

000418/29836/29000 0 0

−−−−

ªºªº

«»«»

−−

«»«»

−−

«»«»

−

¬¼¬¼

Use the pivot columns of A to form

34 7

5311

43 2

87 4

B

−

ª

º

«

»

−−−

«

»

=

«

»

«

»

−

¬

¼

. Other choices are possible.

42. [M]

12 10 6 8 4 14 12 10 6 8 4 14

76457 9 0 1/61/21/314/35/6

~~

9999918 0 0 2 2 16 2

43108 1 0 0 0 0 36 0

8756 111 0 0 0 0 0 0

−− − −

ªºªº

«»«»

−− −− −−−

«»«»

⋅⋅⋅

−− −−−

«»«»

−−− − −

«»«»

−−

¬¼¬¼

Use the pivot columns of A to form

12 10 6 4

7647

9999

4318

875 1

B

−

ª

º

«

»

−− −

«

»

«

»

=−

«

»

−−−−

«

»

«

»

−

¬

¼

. Other choices are possible.

43. [M] Make v any one of the columns of A that is not in B and row reduce the augmented matrix

[B v]. The calculations will show that the equation Bx = v is consistent, which means that v is a

linear combination of the columns of B. Thus, each column of A that is not a column of B is in the set

spanned by the columns of B.

44. [M] Calculations made as for Exercise 43 will show that each column of A that is not a column of B

is in the set spanned by the columns of B. Reason: The original matrix A has only four pivot

columns. If one or more columns of A are removed, the resulting matrix will have at most four pivot

columns. (Use exactly the same row operations on the new matrix that were used to reduce A to

echelon form.) If v is a column of A that is not in B, then row reduction of the augmented matrix

[B v] will display at most four pivot columns. Since B itself was constructed to have four pivot

columns, adjoining v cannot produce a fifth pivot column. Thus the first four columns of [B v] are

the pivot columns. This implies that the equation Bx = v has a solution.

Note:

At the end of Section 1.7, the Study Guide has another note to students about “Mastering Linear

Algebra Concepts.” The note describes how to organize a review sheet that will help students form a

mental image of linear independence. The note also lists typical misuses of terminology, in which an

adjective is applied to an inappropriate noun. (This is a major problem for my students.) I require my

students to prepare a review sheet as described in the Study Guide, and I try to make helpful comments on

their sheets. I am convinced, through personal observation and student surveys, that the students who

56 CHAPTER 1 • Linear Equations in Linear Algebra

prepare many of these review sheets consistently perform better than other students. Hopefully, these

students will remember important concepts for some time beyond the final exam.

1.8 SOLUTIONS

Notes:

The key exercises are 17–20, 25 and 31. Exercise 20 is worth assigning even if you normally

assign only odd exercises. Exercise 25 (and 26) can be used to make a few comments about computer

graphics, even if you do not plan to cover Section 2.6. For Exercise 31, the Study Guide encourages

students not to look at the proof before trying hard to construct it. Then the Guide explains how to create

the proof.

Exercises 19 and 20 provide a natural segue into Section 1.9. I arrange to discuss the homework on

these exercises when I am ready to begin Section 1.9. The definition of the standard matrix in Section 1.9

follows naturally from the homework, and so I’ve covered the first page of Section 1.9 before students

realize we are working on new material.

The text does not provide much practice determining whether a transformation is linear, because the

time needed to develop this skill would have to be taken away from some other topic. If you want your

students to be able to do this, you may need to supplement Exercises 23, 24, 32 and 33.

If you skip the concepts of one-to-one and “onto” in Section 1.9, you can use the result of Exercise 31

to show that the coordinate mapping from a vector space onto R

n

(in Section 4.4) preserves linear

independence and dependence of sets of vectors. (See Example 6 in Section 4.4.)

1. T(u) = Au =

20 1 2

02 3 6

ªºªºªº

=

«»«»«»

−−

¬¼¬¼¬¼

, T(v) =

20 2

02 2

aa

bb

ª

ºª º ª º

=

«

»« » « »

¬

¼¬ ¼ ¬ ¼

2. T(u) = Au =

13 0 0 3 1

013 0 6 2

00139 3

ªºªºªº

«»«»«»

=

«»«»«»

−−

¬¼¬¼¬¼

, T(v) =

13 0 0 3

013 0 3

0013 3

aa

bb

cc

ª

ºª º ª º

«

»« » « »

=

«

»« » « »

«

»« » « »

¬

¼¬ ¼ ¬ ¼

3.

[]

1032 1032 1032

3163~0133~0133

2211 0253 0013

A

−− −− −−

ªºªºªº

«»«»«»

=−−−−−

«»«»«»

−−− − −−

¬¼¬¼¬¼

b

10 3 2 1007 7

~0 1 3 3~0 1 0 6 6,

00 1 3 0011 1

−−

ªºªºªº

«»«»«»

−− =

«»«»«»

¬¼¬¼¬¼

x

unique solution

4.

[]

1236 1236 1236

0134~0134~0134

2565 0107 0033

A

−− −− −−

ªºªºªº

«»«»«»

=−− −− −−

«»«»«»

−− − −

¬¼¬¼¬¼

b

1236 10017 17

~0 1 3 4~0 1 0 7 7

0011 001 1 1

−− − −

ªºªºªº

«»«»«»

−− − =−

«»«»«»

−−−

¬¼¬¼¬¼

x

, unique solution

1.8 • Solutions 57

5.

[]

1572 1572 1033

~~

37 52 0 12 1 0121

A

−−− −−−

ªºªºªº

=

«»«»«»

−−

¬¼¬¼¬¼

b

Note that a solution is not

3

1

ªº

«»

¬¼

. To avoid this common error, write the equations:

13

23

33

21

xx

+=

and solve for the basic variables:

13

23

3

33

12

is free

xx

x

=−



°=−

®

°

¯

The general solution is

13

23 3

33

33 3 3

12 1 2

01

xx

xxx

xx

−−

ªºª º

ª

ºªº

«»« »

«

»«»

==−=+−

«»« »

«

»«»

«»« »

«

»«»

¬

¼¬¼

¬¼¬ ¼

x

. For a particular solution, one might

choose x

3

= 0 and

3

1

0

ª

º

«

»

=

«

»

«

»

¬

¼

x

.

6.

[]

1321 1321 1321 10810

3886 0123 0123 0123

~~~

0123 0123 0000 0000

10810 0369 0000 0000

A

−−−

ªºªºªºªº

«»«»«»«»

−

«»«»«»«»

=

«»«»«»«»

¬¼¬¼¬¼¬¼

b

Write the

equations:

13

23

810

23

xx

+=

and solve for the basic variables:

13

23

3

10 8

32

is free

xx

x

=−



°=−

®

°

¯

The general solution is

13

23 3

33

10 8 10 8

32 3 2

01

xx

xx x

xx

−−

ªºª º

ª

ºªº

«»« »

«

»«»

==−=+−

«»« »

«

»«»

«»« »

«

»«»

¬

¼¬¼¬¼¬ ¼

x

. For a particular solution, one might

choose x

3

= 0 and

10

3

0

ª

º

«

»

=

«

»

«

»

¬

¼

x

.

7. The value of a is 5. The domain of T is R

5

, because a 6×5 matrix has 5 columns and for Ax to be

defined, x must be in R

5

. The value of b is 6. The codomain of T is R

6

, because Ax is a linear

combination of the columns of A, and each column of A is in R

6

.

8. The matrix A must have 7 rows and 5 columns. For the domain of T to be R

5

, A must have five

columns so that Ax is defined for x in R

5

. For the codomain of T to be R

7

, the columns of A must

have seven entries (in which case A must have seven rows), because Ax is a linear combination of the

columns of A.

9. Solve Ax = 0:

13550 13550 13550

01350~01350~01350

24440 02660 00040

−− −− −−

ªºªºªº

«»«»«»

−−−

«»«»«»

−− − −

¬¼¬¼¬¼

58 CHAPTER 1 • Linear Equations in Linear Algebra

10 400

~0 1 3 0 0

00 010

−

ªº

«»

−

«»

¬¼

13

23

4

40

30

0

xx

x

−=

=

,

13

23

3

4

3

is free

= 0

xx

x

=



°

=

°

®

°

¯

x =

13

23

3

33

4

44

33

1

00

xx

xxx

xx

x

ªºªº ªº

«»«» «»

==

«»«» «»

¬¼ ¬¼

¬¼

10. Solve Ax = 0.

3210 60 10 2 40 102 40

10 2 40 3210 60 024 60

~~

01 2 30 012 30 012 30

1410 80 1410 80 048120

−−−

ªºªºªº

«»«»«»

−−

«»«»«»

¬¼¬¼¬¼

102 40 102 40

012 30 012 30

~~

024 60 000 00

048120 000 00

−−

ªºªº

«»«»

¬¼¬¼

134

234

240

230

xxx

+−=

++=

134

234

3

4

24

23

is free

xxx

x

=−+



°

=−−

°

®

°

¯

4

3

4

3

34

3

4

224

3

223

010

001

x

xxx

x

−−

ªº

ª

ºªºªº

«»

«

»«»«»

−

−−−

«»

«

»«»«»

=+=+

«»

«

»«»«»

«»

«

»«»«»

¬

¼¬¼¬¼

¬¼

x

11. Is the system represented by [A b] consistent? Yes, as the following calculation shows.

13551 13551 13551

01351~01351~01351

24440 02662 00040

−−− −−− −−−

ªºªºªº

«»«»«»

−−−

«»«»«»

−− − −

¬¼¬¼¬¼

The system is consistent, so b is in the range of the transformation

A

xx6

.

12. Is the system represented by [A b] consistent?

3210 6 1 10 2 4 3 102 4 3

10 2 4 3 3210 6 1 024 6 10

~~

01 2 3 1 012 3 1 012 3 1

1410 8 4 1410 8 4 04812 1

−− − −

ªºªºªº

«»«»«»

−−−−

«»«»«»

−− −

«»«»«»

¬¼¬¼¬¼

102 4 3 1 0 2 4 3

012 3 1 0 1 2 3 1

~~

024 6 10 0 0 0 0 8

04812 1 0 0 0 0 5

−−

ªºªº

«»«»

−−

«»«»

−−

«»«»

¬¼¬¼

1.8 • Solutions 59

The system is inconsistent, so b is not in the range of the transformation

A

xx6

.

13. 14.

A reflection through the origin. A scaling by the factor 2.

The transformation in Exercise 13 may also be described as a rotation of π radians about the origin or

a rotation of –π radians about the origin.

15. 16.

A reflection through the line x

2

= x

1

. A scaling by a factor of 2 and a projection onto the x

2

axis.

17. T(2u) = 2T(u) =

48

212

ªº ªº

=

«» «»

¬¼ ¬¼

, T(3v) = 3T(v) =

13

339

−−

ª

ºªº

=

«

»«»

¬

¼¬¼

, and

T(2u + 3v) = 2T(u) + 3T(v) =

835

2911

−

ªº ª º ªº

+=

«» « » «»

¬¼ ¬ ¼ ¬¼

.

18. Draw a line through w parallel to v, and draw a line through w parallel to u. See the left part of the

figure below. From this, estimate that w = u + 2v. Since T is linear, T(w) = T(u) + 2T(v). Locate T(u)

and 2T(v) as in the right part of the figure and form the associated parallelogram to locate T(w).

x

1

x

2

x

1

x

2

uw

v2v

T(v)

2T(v)

T(u)

T(w)

60 CHAPTER 1 • Linear Equations in Linear Algebra

19. All we know are the images of e

1

and e

2

and the fact that T is linear. The key idea is to write

x =

12

510

5353

301

.

=−=−

−

ªº ªº ªº

«» «» «»

¬¼ ¬¼ ¬¼

ee

Then, from the linearity of T, write

T(x) = T(5e

1

– 3e

2

) = 5T(e

1

) – 3T(e

2

) = 5y

1

– 3y

2

=

2113

53 .

567

−

−=

ª

ºªºªº

«

»«»«»

¬

¼¬¼¬¼

To find the image of

1

2

x

ª

º

«

»

¬

¼, observe that

1

121122

2

10

01

xxx xx

x

ªº ªº ªº

== + =+

«» «» «»

¬¼ ¬¼

¬¼

xee

. Then

T(x) = T(x

1

e

1

+ x

2

e

2

) = x

1

T(e

1

) + x

2

T(e

2

) =

12

221

56

xx

xx xx

−−

ª

ºªº ª º

+=

«

»

«» « »

+

¬¼ ¬ ¼

¬

¼

20. Use the basic definition of Ax to construct A. Write

[]

1

11 2 2 1 2

2

37 37

() ,

52 52

x

Txx A

x

−−

ªº

ª

ºªº

=+ = = =

«»

«

»«»

−−

¬

¼¬¼

¬¼

xvvvv x

21. a. True. Functions from R

n

to R

m

are defined before Fig. 2. A linear transformation is a function

with certain properties.

b. False. The domain is R

5

. See the paragraph before Example 1.

c. False. The range is the set of all linear combinations of the columns of A. See the paragraph

before Example 1.

d. False. See the paragraph after the definition of a linear transformation.

e. True. See the paragraph following the box that contains equation (4).

22. a. True. See the subsection on Matrix Transformations.

b. True. See the subsection on Linear Transformations.

c. False. The question is an existence question. See the remark about Example 1(d), following the

solution of Example 1.

d. True. See the discussion following the definition of a linear transformation.

e. True. T(0) = 0

.

See the box after the definition of a linear transformation.

23. a. When b = 0, f (x) = mx. In this case, for all x,y in R and all scalars c and d,

f (cx + dy) = m(cx + dy) = mcx + mdy = c(mx) + d(my) = cf (x) + d f (y)

This shows that f is linear.

b. When f (x) = mx + b, with b nonzero, f(0) = m(0) = b = b ≠ 0. This shows that f is not linear,

because every linear transformation maps the zero vector in its domain into the zero vector in the

codomain. (In this case, both zero vectors are just the number 0.) Another argument, for instance,

would be to calculate f (2x) = m(2x) + b and 2f (x) = 2mx + 2b. If b is nonzero, then f (2x) is not

equal to 2f (x) and so f is not a linear transformation.

c. In calculus, f is called a “linear function” because the graph of f is a line.

24. Let T(x) = Ax + b for x in R

n

. If b is not zero, T(0) = A0 + b = b

≠

0. Actually, T fails both

properties

of a linear transformation. For instance, T(2x) = A(2x) + b = 2Ax + b, which is not the same as 2T(x)

= 2(Ax + b) = 2Ax + 2b. Also,

T(x + y) = A(x + y) + b = Ax + Ay + b

1.8 • Solutions 61

which is not the same as

T(x) + T(y) = Ax + b + Ay + b

25. Any point x on the line through p in the direction of v satisfies the parametric equation

x = p + tv for some value of t. By linearity, the image T(x) satisfies the parametric equation

T(x) = T(p + tv) = T(p) + tT(v)

(*)

If T(v) = 0, then T(x) = T(p) for all values of t, and the image of the original line is just a single

point. Otherwise, (*) is the parametric equation of a line through T(p) in the direction of T(v).

26. a. From the figure following Exercise 22 in Section 1.5, the line through p and q is in the direction

of q – p, and so the equation of the line is x = p + t(q – p) = p + tq – tp = (1 – t)p + tq.

b. Consider x = (1 – t)p + tq for t such that 0 < t < 1. Then, by linearity of T,

T(x) = T((1 – t)p + tq) = (1 – t)T(p) + tT(q) 0 < t < 1

(*)

If T(p) and T(q) are distinct, then (*) is the equation for the line segment between T(p) and T(q),

as shown in part (a) Otherwise, the set of images is just the single point T(p), because

(1 – t)T(p) + tT(q) =(1 – t)T(p) + tT(p) = T(p)

27. Any point x on the plane P satisfies the parametric equation x = su + tv for some values of s and t.

By linearity, the image T(x) satisfies the parametric equation

T(x) = sT(u) + tT(v) (s, t in R)

The set of images is just Span{T(u), T(v)}. If T(u) and T(v) are linearly independent, Span{T(u),

T(v)} is a plane through T(u), T(v), and 0. If T(u) and T(v) are linearly dependent and not both zero,

then Span{T(u), T(v)} is a line through 0. If T(u) = T(v) = 0, then Span{T(u), T(v)} is {0}.

28. Consider a point x in the parallelogram determined by u and v, say x = au + bv for 0 < a < 1,

0 < b < 1. By linearity of T, the image of x is

T(x) = T(au + bv) = aT(u) + bT(v), for 0 < a < 1, 0 < b < 1

This image point lies in the parallelogram determined by T(u) and T(v). Special “degenerate” cases

arise when T(u) and T(v) are linearly dependent. If one of the images is not zero, then the

“parallelogram” is actually the line segment from 0 to T(u) + T(v). If both T(u) and T(v) are zero,

then the parallelogram is just {0}. Another possibility is that even u and v are linearly dependent, in

which case the original parallelogram is degenerate (either a line segment or the zero vector). In this

case, the set of images must be degenerate, too.

62 CHAPTER 1 • Linear Equations in Linear Algebra

29.

30. Given any x in R

n

, there are constants c

1

, …, c

p

such that x = c

1

v

1

+ c

p

v

p

, because v

1

, …, v

p

span

R

n

. Then, from property (5) of a linear transformation,

T(x) = c

1

T(v

1

) + + c

p

T(v

p

) = c

1

0 + + c

p

0 = 0

31. (The Study Guide has a more detailed discussion of the proof.) Suppose that {v

1

, v

2

, v

3

} is linearly

dependent. Then there exist scalars c

1

, c

2

, c

3

, not all zero, such that

c

1

v

1

+ c

2

v

2

+ c

3

v

3

= 0

Then T(c

1

v

1

+ c

2

v

2

+ c

3

v

3

) = T(0) = 0. Since T is linear,

c

1

T(v

1

) + c

2

T(v

2

) + c

3

T(v

3

) = 0

Since not all the weights are zero, {T(v

1

), T(v

2

), T(v

3

)} is a linearly dependent set.

32. Take any vector (x

1

, x

2

) with x

2

≠

0, and use a negative scalar. For instance, T(0, 1) = (–2, –4), but

T(–1 (0, 1)) = T(0, –1) = (–2, 4)

≠

(–1) T(0, 1).

33. One possibility is to show that T does not map the zero vector into the zero vector, something that

every linear transformation does do. T(0, 0) = (0, –3, 0).

34. Take u and v in R

3

and let c and d be scalars. Then

cu + dv = (cu

1

+ dv

1

, cu

2

+ dv

2

, cu

3

+ dv

3

). The transformation T is linear because

T(cu + dv) = (cu

1

+ dv

1

, cu

2

+ dv

2

, – (cu

3

+ dv

3

)) = (cu

1

+ dv

1

, cu

2

+ dv

2

, cu

3

– dv

3

)

= (cu

1

, cu

2

, –cu

3

) + (dv

1

, dv

2

, –dv

3

) = c(u

1

, u

2

, –u

3

) + d(v

1

, v

2

, –v

3

)

= cT(u) + dT(v)

35. Take u and v in R

3

and let c and d be scalars. Then

cu + dv = (cu

1

+ dv

1

, cu

2

+ dv

2

, cu

3

+ dv

3

). The transformation T is linear because

T(cu + dv) = (cu

1

+ dv

1

, 0, cu

3

+ dv

3

) = (cu

1

, 0, cu

3

) + (dv

1

, 0, dv

3

)

= c(u

1

, 0, u

3

) + d(v

1

, 0, v

3

)

= cT(u) + dT(v)

36. Suppose that {u, v} is a linearly independent set in R

n

and yet T(u) and T(v) are linearly dependent.

Then there exist weights c

1

, c

2

, not both zero, such that c

1

T(u) + c

2

T(v) = 0 . Because T is linear,

T(c

1

u + c

2

v) = 0. That is, the vector x = c

1

u + c

2

v satisfies T(x) = 0. Furthermore, x cannot be the

zero vector, since that would mean that a nontrivial linear combination of u and v is zero, which is

impossible because u and v are linearly independent. Thus, the equation T(x) = 0 has a nontrivial

solution.

1.8 • Solutions 63

37. [M]

23 5 5 0 1010 0

77 0 0 0 0110 0

~,

34 1 3 0 0001 0

93 6 4 0 0000 0

−

ªºªº

«»«»

−

«»«»

−

«»«»

−−−

¬¼¬¼

13

23

3

4

–

is free

0

xx

x

=



°=

°

®

°

°=

¯

3

1

0

x

−

ª

º

«

»

−

«

»

=

«

»

«

»

¬

¼

x

38. [M]

34700 10010

58740 01010

~

68640 00110

97200 00000

−

ªºªº

«»«»

−

«»«»

−

«»«»

−−

¬¼¬¼

,

14

24

34

4

is free

xx

x

=−



°=−

°

®=−

°

¯

4

1

x

−

ª

º

«

»

−

«

»

=

«

»

−

«

»

¬

¼

x

39. [M]

23 5 5 8 1010 1

77 0 0 7 01102

~,

34 1 3 5 00010

93 6 4 3 00000

−

ªºªº

«»«»

−

«»«»

−

«»«»

−−−−

¬¼¬¼

yes, b is in the range of the transformation,

because the augmented matrix shows a consistent system. In fact,

the general solution is

13

23

3

4

1–

2–

is free

0

xx

x

=



°=

°

®

°

°=

¯

; when x

3

= 0 a solution is

1

2

0

ª

º

«

»

«

»

=

«

»

«

»

¬

¼

x

.

40. [M]

34704 10011

58744 01012

~

68644 00111

97207 00000

−

ªºªº

«»«»

−−

«»«»

−−

«»«»

−− −

¬¼¬¼

, yes, b is in the range of the transformation,

because the augmented matrix shows a consistent system. In fact,

the general solution is

14

24

34

4

1

2

1

is free

xx

x

=−



°=−

°

®=−

°

¯

; when x

4

= 0 a solution is

1

2

1

0

ª

º

«

»

«

»

=

«

»

«

»

¬

¼

x

.

Notes:

At the end of Section 1.8, the Study Guide provides a list of equations, figures, examples, and

connections with concepts that will strengthen a student’s understanding of linear transformations. I

encourage my students to continue the construction of review sheets similar to those for “span” and

“linear independence,” but I refrain from collecting these sheets. At some point the students have to

assume the responsibility for mastering this material.

If your students are using MATLAB or another matrix program, you might insert the definition of

matrix multiplication after this section, and then assign a project that uses random matrices to explore

properties of matrix multiplication. See Exercises 34–36 in Section 2.1. Meanwhile, in class you can

continue with your plans for finishing Chapter 1. When you get to Section 2.1, you won’t have much to

do. The Study Guide’s MATLAB note for Section 2.1 contains the matrix notation students will need for

a project on matrix multiplication. The appendices in the Study Guide have the corresponding material for

Mathematica, Maple, and the TI-83+/84+/89 calculators.

.

64 CHAPTER 1 • Linear Equations in Line

a

r

1.9 SOLUTIONS

Notes

: This section is optional if yo

u

instructors will want to cover at least

T

illustrate a fast way to solve Exercises

basis.

The purpose of introducing one-to-

4.4) and to acquaint math majors with t

h

digression, and some instructors prefer

use the result of Exercise 31 in Section

R

n

(in Section 4.4) preserves linear ind

e

Section 4.4.) The notions of one-to-one

but can be omitted there if desired

Exercises 25–28 and 31–36 offer

important links to earlier material.

1. A = [T(e

1

) T(e

2

)] =

35

12

30

10

−

ªº

«»

¬¼

2. A = [T(e

1

) T(e

2

) T(e

3

)] =

12

49

−

ª

«

¬

3. T(e

1

) = e

1

– 3e

2

=

1

3

ªº

«»

−

¬¼

, T(e

2

) = e

2

,

A

4. T(e

1

) = e

1

, T(e

2

) = e

2

+ 2e

1

=

2

1

ªº

«»

¬¼

,

A

5. T(e

1

) = e

2

, T(e

2

) = –e

1

. A =

[

2

−e

e

6. T(e

1

) = e

2

, T(e

2

) = –e

1

. A =

[

2

−e

e

7. Follow what happens to e

1

and e

2

.

S

circle in the plane, it rotates throug

h

point on the unit circle that lies in t

h

on the line

21

xx= (that is, yx= i

n

The point (–1,–1) is on the ine

2

x

=

from the origin is 2. So the rotati

o

(–1/ 2, –1/ 2)

. Then this image r

e

axis to

(–1/ 2,1/ 2)

. Similarly, e

2

the unit circle that lies in the secon

d

ar Algebra

r

son Education, Inc. Publishing as Addison-Wesley.

u plan to treat linear transformations only lightl

y

T

heorem 10 and a few geometric examples. Exercis

e

17–22 without explicitly computing the images of

one and onto is to prepare for the term isomorphis

m

h

ese terms. Mastery of these concepts would require

to omit these topics (and Exercises 25–40). In this c

a

1.8 to show that the coordinate mapping from a vect

o

e

pendence and dependence of sets of vectors. (See

E

and onto appear in the Invertible Matrix Theorem (

S

fairly easy writing practice. Exercises 31, 32, and

3

8

º

»

−

¼

A

=

10

31

ªº

«»

−

¬¼

A

=

12

01

ªº

«»

¬¼

]

1

01

10

=

−

ªº

«»

¬¼

e

]

1

01

10

=

−

ªº

«»

¬¼

e

S

ince e

1

is on the unit

h

–3 /4

radians into a

h

e third quadrant and

n

more familiar notation).

1

x

=

, but its distance

o

nal image of e

1

is

e

flects in the horizontal

2

rotates into a point on

d

quadrant and on the

y

, but many

e

s 15 and 16

the standard

m

(in Section

a substantial

a

se, you can

o

r space onto

E

xample 6 in

S

ection 2.3),

35 provide

1.9 • Solutions 65

line

21

xx=−, namely,

(1 / 2 , –1 / 2 )

. Then this image reflects in the horizontal axis to

(1 / 2 ,1 / 2 )

. When the two calculations described above are written in vertical vector notation, the

transformation’s standard matrix [T(e

1

) T(e

2

)] is easily seen:

12

1/ 2 1/ 2 1/ 2 1/ 2

,

1/ 2 1/ 2 1/ 2 1/ 2

ªºªº ªºªº

−−

→→ →→

«»«» «»«»

−−

«»«» «»«»

¬¼¬¼ ¬¼¬¼

ee

,

1/ 2 1/ 2

A

ª

º

−

=

«

»

«

»

¬

¼

8. The horizontal shear maps e

1

into e

1

, and then the reflection in the line x

2

= –x

1

maps e

1

into –e

2

. (See

Table 1.) The horizontal shear maps e

2

into e

2

into e

2

+ 2e

1

. To find the image of e

2

+ 2e

1

when it is

reflected in the line x

2

= –x

1

, use the fact that such a reflection is a linear transformation. So, the

image of e

2

+ 2e

1

is the same linear combination of the images of e

2

and e

1

, namely,

–e

1

+ 2(–e

2

) = – e

1

– 2e

2

. To summarize,

11 2 2 2 1 1 2

01

and 2 2 , so 12

A

−

ª

º

→→− → +→− − =

«

»

−−

¬

¼

ee e e e e ee

9.

11 2 2 2 1

,

01

and so 10

A

→→ →− →−

−

ª

º

−=

«

»

−

¬

¼

ee e e e e

10.

[]

11 2 2 2 1 2 1

01

and , so 10

A

−

ª

º

→→ →−→− =−=

«

»

¬

¼

eee e e e e e

11. The transformation T described maps

11 1

→→−

ee e

and maps

222

.

→− →−

eee

A rotation through

π

radians also maps e

1

into –e

1

and maps e

2

into –e

2

. Since a linear transformation is completely

determined by what it does to the columns of the identity matrix, the rotation transformation has the

same effect as T on every vector in

2

.



12. The transformation T in Exercise 10 maps

11 2

→→

eee

and maps

221

→− →−

eee

. A rotation about

the origin through

/

2

π

radians also maps e

1

into e

2

and maps e

2

into –e

1

. Since a linear

transformation is completely determined by what it does to the columns of the identity matrix, the

rotation transformation has the same effect as T on every vector in

2

.



13. Since (2, 1)=2 e

1

+ e

2

, the image of (2, 1) under T is 2T(e

1

) + T(e

2

), by linearity of T. On the figure in

the exercise, locate 2T(e

1

) and use it with T(e

2

) to form the parallelogram shown below.

14. Since

[]

12 11221 2

() 2

TA xx== = + =−xxaaxaaaa

, when x = (1, –2), the image of x is located by

forming the parallelogram shown below.

66 CHAPTER 1 • Linear Equations in Linear Algebra

15. By inspection,

112

213

323

240 24

101

013 3

xxx

−−

ªºª ºªº

«»« »

«»

−=−

«»« »

«»

«»« »«»

−−+

¬¼¬¼¬¼

16. By inspection,

12

1

12

2

32 32

14 4

01

xx

xxx

xx

−−

ªº ª º

ªº

«» « »

=+

«»

«» « »

¬¼

«» « »

¬¼ ¬ ¼

17. To express T(x) as Ax , write T(x) and x as column vectors, and then fill in the entries in A by

inspection, as done in Exercises 15 and 16. Note that since T(x) and x have four entries, A must be a

4×4 matrix.

T(x) =

12 1 1

22

24 3 3

24 4 4

2120 0

0000 0

2020 1

010 1

xx x x

xx

A

xx x x

+

ªº ªº ªº

ªºª º

«» «» «»

«»« »

«» «» «»

«»« »

==

«» «» «»

«»« »

+

«» «» «»

«»« »

−−

¬¼¬ ¼

¬¼ ¬¼ ¬¼

18. As in Exercise 17, write T(x) and x as column vectors. Since x has 2 entries, A has 2 columns. Since

T(x) has 4 entries, A has 4 rows.

12

11

12 2 2

1

414

000

313

10

xx

A

xx x x

x

+

ªº

ªº ª º

«»

«» « »

ªº ªº

«»

«» « »

==

«» «»

«»

«» « »

−−

¬¼ ¬¼

«»

«» « »

¬¼ ¬ ¼

¬¼

19. Since T(x) has 2 entries, A has 2 rows. Since x has 3 entries, A has 3 columns.

11

12 3

22

23

33

54 154

6016

xx

xx x Ax x

xx xx

ªº ªº

−+−

ªº

ªºª º

«» «»

==

«»

«»« »

«» «»

−−

¬¼¬ ¼

¬¼

«» «»

¬¼ ¬¼

20. Since T(x) has 1 entry, A has 1 row. Since x has 4 entries, A has 4 columns.

11

22

134

33

44

[3 4 2 ] [ ] [3 0 4 2]

xx

xxx A xx

xx

ªº ªº

«» «»

+−==−

«» «»

¬¼ ¬¼

1.9 • Solutions 67

21. T(x) =

12 1 1

12 2 2

11

45 45

xx x x

A

xx x x

+

ªºªº ªºªº ª º

==

«»«» «»

«» « »

+¬¼ ¬ ¼

¬¼¬¼ ¬¼

. To solve T(x) = 3

8

ª

º

«

»

¬

¼

, row reduce the augmented

matrix: 113 11 3 10 7 7

~~,

458 01 4 01 4 4

ªºª ºª ºªº

=

«»« »« »«»

−−−

¬¼¬ ¼¬ ¼¬¼

x

.

22. T(x) =

12

11

12

22

12

221

331

23 23

xx xx

xx A xx

xx

−−

ªºªºªº

ªº ªº

«»«»«»

−+= =−

«» «»

«»«»«»

¬¼ ¬¼

«»«»«»

−−

¬¼¬¼¬¼

. To solve T(x) =

0

1

4

ª

º

«

»

−

«

»

«

»

−

¬

¼

, row reduce the

augmented matrix:

210 2 10 210 210 202 101

311~0121~012~012~012~012

234 0 24 024 000 000 000

−−−−

ªºª ºªºªºªºªº

«»« »«»«»«»«»

−− −−

«»« »«»«»«»«»

−− −− −−

¬¼¬ ¼¬¼¬¼¬¼¬¼

,

1.

2

ªº

=«»

¬¼

x

23. a. True. See Theorem 10.

b. True. See Example 3.

c. False. See the paragraph before Table 1.

d. False. See the definition of onto. Any function from R

n

to R

m

maps each vector onto another

vector.

e. False. See Example 5.

24. a. False. See Theorem 12.

b. True. See Theorem 10.

c. True. See Theorem 10.

d. False. See the definition of one-to-one. Any function from R

n

to R

m

maps a vector onto a single

(unique) vector.

e. False. See Table 3.

25. A row interchange and a row replacement on the standard matrix A of the transformation T in

Exercise 17 produce

120 0

010 1

000 3

000 0

ª

º

«

»

−

«

»

«

»

«

»

¬

¼

. This matrix shows that A has only three pivot positions, so

the equation Ax = 0 has a nontrivial solution. By Theorem 11, the transformation T is not one-to-one.

Also, since A does not have a pivot in each row, the columns of A do not span R

4

. By Theorem 12, T

does not map R

4

onto R

4

.

26. The standard matrix A of the transformation T in Exercise 2 is 2×3. Its columns are linearly

dependent because A has more columns than rows. So T is not one-to-one, by Theorem 12. Also, A is

row equivalent to 12 3

017 20

−

ªº

«»

−

¬¼

, which shows that the rows of A span R

2

. By Theorem 12, T maps

R

3

onto R

2

.

68 CHAPTER 1 • Linear Equations in Linear Algebra

27. The standard matrix A of the transformation T in Exercise 19 is 154

016

−

ª

º

«

»

−

¬

¼

. The columns of A

are linearly dependent because A has more columns than rows. So T is not one-to-one, by Theorem

12. Also, A has a pivot in each row, so the rows of A span R

2

. By Theorem 12, T maps R

3

onto R

2

.

28. The standard matrix A of the transformation T in Exercise 14 has linearly independent columns,

because the figure in that exercise shows that a

1

and a

2

are not multiples. So T is one-to-one, by

Theorem 12. Also, A must have a pivot in each column because the equation Ax = 0 has no free

variables. Thus, the echelon form of A is

*

0

.

ª

º

«

»

¬

¼



Since A has a pivot in each row, the columns of A

span R

2

. So T maps R

2

onto R

2

. An alternate argument for the second part is to observe directly from

the figure in Exercise 14 that a

1

and a

2

span R

2

. This is more or less evident, based on experience

with grids such as those in Figure 8 and Exercise 7 of Section 1.3.

29. By Theorem 12, the columns of the standard matrix A must be linearly independent and hence the

equation Ax = 0 has no free variables. So each column of A must be a pivot column:

**

0*

~.

00

000

A

ªº

«»

¬¼



Note that T cannot be onto because of the shape of A.

30. By Theorem 12, the columns of the standard matrix A must span 

3

. By Theorem 4, the matrix must

have a pivot in each row. There are four possibilities for the echelon form:

*** *** *** 0 **

0**,0**,00*,00*

00 * 000 000 000

ªºªºªºªº

«»«»«»«»

¬¼¬¼¬¼¬¼

  



Note that T cannot be one-to-one because of the shape of A.

31. “T is one-to-one if and only if A has n pivot columns.” By Theorem 12(b), T is one-to-one if and only

if the columns of A are linearly independent. And from the statement in Exercise 30 in Section 1.7,

the columns of A are linearly independent if and only if A has n pivot columns.

32. The transformation T maps R

n

onto R

m

if and only if the columns of A span R

m

, by Theorem 12. This

happens if and only if A has a pivot position in each row, by Theorem 4 in Section 1.4. Since A has m

rows, this happens if and only if A has m pivot columns. Thus, “T maps R

n

onto R

m

if and only A has

m pivot columns.”

33. Define :

nm

T→

 by T(x) = Bx for some m×n matrix B, and let A be the standard matrix for T.

By definition, A = [T(e

1

) ⋅ ⋅ ⋅ T(e

n

)], where e

j

is the jth column of I

n

. However, by matrix-vector

multiplication, T(e

j

) = Be

j

= b

j

, the jth column of B. So A = [b

1

⋅ ⋅ ⋅ b

n

] = B.

34. Take u and v in R

p

and let c and d be scalars. Then

T(S(cu + dv)) = T(c⋅S(u) + d

⋅

S(v)) because S is linear

= c

⋅

T(S(u)) + d⋅T(S(v)) because T is linear

This calculation shows that the mapping x → T(S(x)) is linear. See equation (4) in Section 1.8.

1.10 • Solutions 69

35. If :

nm

T→

 maps

n



onto

m



, then its standard matrix A has a pivot in each row, by Theorem

12 and by Theorem 4 in Section 1.4. So A must have at least as many columns as rows. That is, m <

n. When T is one-to-one, A must have a pivot in each column, by Theorem 12, so m > n.

36. The transformation T maps R

n

onto R

m

if and only if for each y in R

m

there exists an x in R

n

such that

y = T(x).

37. [M]

5656 1001

8338 0101

~~

29512 0011

32712 0000

−−−

ªºªº

«»«»

−−

«»«»

⋅⋅⋅

«»«»

−−

«»«»

−−

¬¼¬¼

. There is no pivot in the fourth column of

the standard matrix A, so the equation Ax = 0 has a nontrivial solution. By Theorem 11, the

transformation T is not one-to-one. (For a shorter argument, use the result of Exercise 31.)

38. [M]

7599 1000

5644 0100

~~

4807 0010

6665 0001

−

ªºªº

«»«»

−

«»«»

⋅⋅⋅

«»«»

−−

¬¼¬¼

. Yes. There is a pivot in every column of the

standard matrix A, so the equation Ax = 0 has only the trivial solution. By Theorem 11, the trans-

formation T is one-to-one. (For a shorter argument, use the result of Exercise 31.)

39. [M]

47375 10050

685128 01010

~~

710 8 914 0 0 1 2 0

35426 00001

56673 00000

−

ªºªº

«»«»

−−

«»«»

⋅⋅⋅

−−− −

«»«»

−−

«»«»

−−−

¬¼¬¼

. There is not a pivot in every row,

so the columns of the standard matrix do not span R

5

. By Theorem 12, the transformation T does not

map R

5

onto R

5

.

40. [M]

943 5 6 1 10000

14 15 7 5 4 0 1 0 0 0

~~

861259 00100

56498 00010

13 14 15 3 11 0 0 0 0 1

−

ªºªº

«»«»

−−

«»«»

⋅⋅⋅

−− −−

«»«»

−−−

«»«»

¬¼¬¼

. There is a pivot in every row, so the

columns of the standard matrix span R

5

. By Theorem 12, the transformation T maps R

5

onto R

5

.

1.10 SOLUTIONS

1. a. If x

1

is the number of servings of Cheerios and x

2

is the number of servings of 100% Natural

Cereal, then x

1

and x

2

should satisfy

12

nutrients nutrients quantities

per serving per serving of of nutrients

of Cheerios 100% Natural required

xx

ªºª ºª º

+=

«»« »« »

¬¼¬ ¼¬ ¼

70 CHAPTER 1 • Linear Equations in Linear Algebra

That is,

12

110 130 295

439

20 18 48

258

xx

ªº ªºªº

«» «»«»

+=

«» «»«»

¬¼ ¬¼¬¼

b. The equivalent matrix equation is

1

2

110 130 295

43 9

20 18 48

25 8

x

ª

ºªº

«

»«»

ªº

«

»«»

=

«»

«

»«»

¬¼

«

»«»

«

»«»

¬

¼¬¼

. To solve this, row reduce the

augmented matrix for this equation.

110 130 295 2 5 8 1 2.5 4

439 439 439

~~

20 18 48 20 18 48 10 9 24

258110130295110130295

ªºªºªº

«»«»«»

¬¼¬¼¬¼

12.5 4 12.54 101.5

077011011

~~~

01616000000

0145145 000 000

ªºªºªº

«»«»«»

−−

«»«»«»

−−

«»«»«»

−−

«»«»«»

¬¼¬¼¬¼

The desired nutrients are provided by 1.5 servings of Cheerios together with 1 serving of 100%

Natural Cereal.

2. Set up nutrient vectors for one serving of Shredded Wheat (SW) and Kellogg's Crispix (Crp):

Nutrients: SW Crp

calories 160 110

protein 5 2

fiber 6 .1

fat 1 .4

ªºªº

«»«»

¬¼¬¼

.

a. Let

[]

160 110

52 3

SW Crp ,

6.1 2

1.4

B

ªº

«»

ª

º

«»

== =

«

»

«»

¬

¼

«»

¬¼

u

.

Then Bu lists the amounts of calories, protein, carbohydrate, and fat in a mixture of three servings

of Shredded Wheat and two servings of Crispix.

b. [M] Let u

1

and u

2

be the number of servings of Shredded Wheat and Crispix, respectively. Can

these numbers satisfy the equation

1

2

120

3.2

2.46

.64

Bu

u

ª

º

«

»

ªº

«

»

=

«»

«

»

¬¼

«

»

¬

¼

? To find out, row reduce the augmented

matrix

1.10 • Solutions 71

160 110 130 1 .4 .64 1 .4 .64 1 .4 .64 1 0 .4

523.20 0 004627.601.601.6

~~~~

6.12.4602.31.3802.31.38000000

1.4.64 0 4627.6 0 0 0 00 0 000

ªºªºªºªºªº

«»«»«»«»«»

−− −−

«»«»«»«»«»

¬¼¬¼¬¼¬¼¬¼

Since the system is consistent, it is possible for a mixture of the two creals to provide the desired

nutrients. The mixture is .4 servings of Shredded Wheat and .6 servings of Crispix.

3. a. [M] Let x

1

, x

2

, and x

3

be the number of servings ofAnnies’s Mac and Cheese, broccoli, and

chicken, respectively, needed for the lunch. The values of x

1

, x

2

, and x

3

should satisfy

123

nutrients nutrients nutrients quantities

per serving per serving per serving of nutrients

of Mac and Cheese of broccoli of chicken required

xxx

ªºªºªºªº

++=

«»«»«»«»

¬¼¬¼¬¼¬¼

From the given data,

123

270 51 70 400

10 5.4 15 30

25.2 010

xxx

ªº ªº ªºªº

«» «» «»«»

++=

«» «» «»«»

¬¼ ¬¼ ¬¼¬¼

To solve, row reduce the corresponding augmented matrix:

270 51 70 400 1 0 0 .99

10 5.4 15 30 ~ ... ~ 0 1 0 1.54

25.2 0 10 001 .79

ªºªº

«»«»

¬¼¬¼

.99 servings of Mac and Cheese

1.54 servings of broccoli

.74 servings of chicken

ªºª º

«»« »

≈=

«»« »

¬¼¬ ¼

x

b. [M] Changing from Annie’s Mac and Cheese to Annie’s Whole Wheat Shells and White Cheddar

changes the vector equation to

123

260 51 70 400

95.41530

55.2 010

xxx

ªº ªº ªºªº

«» «» «»«»

++=

«» «» «»«»

¬¼ ¬¼ ¬¼¬¼

To solve, row reduce the corresponding augmented matrix:

260 51 70 400 1 0 0 1.09

95.415 30~...~010 .88

55.2 0 10 00 11.03

ªºªº

«»«»

¬¼¬¼

1.09 servings of Shells

.88 servings of broccoli

1.03 servings of chicken

ªºª º

«»« »

≈=

«»« »

¬¼¬ ¼

x

Notice that the number of servings of broccoli has decreased as was desired.

72 CHAPTER 1 • Linear Equations in Linear Algebra

4. Here are the data, assembled from Table 1 and Exercise 4:

Mg of Nutrients/Unit Nutrients

Required

soy soy

Nutrient (milligrams)

milk flour whey prot.

protein 36 51 13 80 33

carboh. 52 34 74 0 45

fat 0 7 1.1 3.4 3

calcium 1.26 .19 .8 .18 .8

a. Let x

1

, x

2

, x

3

, x

4

represent the number of units of nonfat milk, soy flour, whey, and isolated soy

protein, respectively. These amounts must satisfy the following matrix equation

1

2

3

4

36 51 13 80 33

52 34 74 0 45

071.13.4 3

1.26 .19 .8 .18 .8

x

ªº

ªºªº

«»

«»«»

«»«»«»

=

«»

«»«»

«»

«»«»

«»

¬¼¬¼¬¼

b. [M]

36 51 13 80 33 000 .64

1

52 34 74 0 45 0 0 0 .54

1

~~

071.13.43 00 0.09

1

1.26 .19 .8 .18 .8 0 0 0 .21

1

ªºªº

«»«»

⋅⋅⋅

«»«»

−

«»«»

−

¬¼¬¼

The “solution” is x

1

= .64, x

2

= .54, x

3

= –.09, x

4

= –.21. This solution is not feasible, because the

mixture cannot include negative amounts of whey and isolated soy protein. Although the

coefficients of these two ingredients are fairly small, they cannot be ignored. The mixture of .64

units of nonfat milk and .54 units of soy flour provide 50.6 g of protein, 51.6 g of carbohydrate,

3.8 g of fat, and .9 g of calcium. Some of these nutrients are nowhere close to the desired

amounts.

5. Loop 1: The resistance vector is

1

22

1

3

4

Total of RI voltage drops for current

11

Voltage drop for is negative; flows in opposite direction

5

Current does not flow in loop 1

0

Current does not flow in loop 1

0

I

II

I

ªº

«»

−

«»

=«»

«»

¬¼

r

Loop 2: The resistance vector is

11

2

33

4

5Voltage drop for is negative; flows in opposite direction

10 Total of RI voltage drops for current

Voltage drop for is negative; flows in opposite direction

1

Current d

0

II

I

II

I

ªº

−

«»

=«»

−

«»

¬¼

r

oes not flow in loop 2

Also, r

3

=

0

1

9

2

ªº

«»

−

«»

−

¬¼

, r

4

=

0

2

10

ª

º

«

»

«

»

«

»

−

«

»

¬

¼

, and R = [r

1

r

2

r

3

r

4

] =

500

11

510 0

1

09

12

00 10

2

ª

º

−

«

»

−−

«

»

«

»

−−

«

»

−

¬

¼

.

1.10 • Solutions 73

Notice that each off-diagonal entry of R is negative (or zero). This happens because the loop current

directions are all chosen in the same direction on the figure. (For each loop j, this choice forces the

currents in other loops adjacent to loop j to flow in the direction opposite to current I

j

.)

Next, set v =

50

40

30

ªº

«»

−

«»

−

¬¼

. The voltages in loops 2 and 4 are negative because the battery orientation in

each loop is opposite to the direction chosen for positive current flow. Thus, the equation Ri = v

becomes

1

2

3

4

500 50

11

510 0 40

1

09 30

12

00 10 30

2

I

ªº

ªºªº

−«»

«»«»

−−

−«»

«»«»

=

«»

«»«»

−−

«»

«»«»

−

¬¼¬¼

¬¼

. [M]: The solution is i =

1

2

3

4

3.68

1.90

2.57

2.49

I

ªº

ª

º

«»

«

»

−

«»

«

»

=

«»

«

»

«»

«

»

−

¬

¼

¬¼

.

6. Loop 1: The resistance vector is

1

22

1

3

4

6Total of RI voltage drops for current

1Voltage drop for is negative; flows in opposite direction

0Current does not flow in loop 1

I

II

I

ªº

«»

−

«»

=«»

«»

¬¼

r

Loop 2: The resistance vector is

11

2

33

4

Voltage drop for is negative; flows in opposite direction

1

9Total of RI voltage drops for current

4Voltage drop for is negative; flows in opposite direction

Current do

0

II

I

II

I

ªº

−

«»

=«»

−

«»

¬¼

r

es not flow in loop 2

Also, r

3

=

0

4

7

2

ªº

«»

−

«»

−

¬¼

, r

4

=

0

2

7

ª

º

«

»

«

»

«

»

−

«

»

¬

¼

, and R = [r

1

r

2

r

3

r

4

]=

6100

1940

0472

0027

−

ª

º

«

»

−−

«

»

«

»

−−

«

»

−

¬

¼

. Set v =

30

20

40

10

ªº

«»

¬¼

. Then Ri =

v becomes

1

2

3

4

6100 30

1940 20

0472 40

0027 10

I

−ªº

ªºªº

«»

«»«»

−−

«»

«»«»

=

«»

«»«»

−−

«»

«»«»

−

¬¼¬¼

¬¼

. [M]: The solution is i =

1

2

3

4

6.36

8.14

11.73

4.78

I

ªº

ª

º

«»

«

»

«»

«

»

=

«»

«

»

«»

«

»

¬

¼

¬¼

.

74 CHAPTER 1 • Linear Equations in Linear Algebra

7. Loop 1: The resistance vector is

1

22

1

3

4

Total of RI voltage drops for current

12

7Voltage drop for is negative; flows in opposite direction

0Current does not flow in loop 1

Voltage drop for is negative; flows

4

I

II

I

II

ªº

«»

−

«»

=«»

«»

−¬¼

r

in opposite direction

Loop 2: The resistance vector is

1

2

33

4

Voltage drop for is negative; flows in opposite direction

7

15 Total of RI voltage drops for current

6Voltage drop for is negative; flows in opposite direction

0Current d

II

I

II

I

ªº

−

«»

=«»

−

«»

¬¼

r

oes not flow in loop 2

Also, r

3

=

0

6

14

5

ªº

«»

−

«»

−

«»

¬¼

, r

4

=

4

0

5

13

−

ª

º

«

»

«

»

«

»

−

«

»

«

»

¬

¼

, and R = [r

1

r

2

r

3

r

4

] =

12704

715 6 0

06145

40513

−−

ª

º

«

»

−−

«

»

«

»

−−

«

»

−−

«

»

¬

¼

.

Notice that each off-diagonal entry of R is negative (or zero). This happens because the loop current

directions are all chosen in the same direction on the figure. (For each loop j, this choice forces the

currents in other loops adjacent to loop j to flow in the direction opposite to current I

j

.)

Next, set v

40

30

20

10

ªº

«»

=«»

«»

−

«»

¬¼

. Note the negative voltage in loop 4. The current direction chosen in loop 4 is

opposed by the orientation of the voltage source in that loop. Thus Ri = v becomes

1

2

3

4

12 7 0 4 40

715 6 0 30

06145 20

40513 10

I

−−

ªº

ªºªº

«»

«»«»

−−

«»«»«»

=

«»

«»«»

−−

«»

«»«»

−− −

«»«»

«»

¬¼¬¼¬¼

. [M]: The solution is i =

1

2

3

4

11.43

10.55

8.04

5.84

I

ªº

ª

º

«»

«

»

«»

«

»

=

«»

«

»

«»

«

»

«

»

«»

¬

¼¬¼

.

8. Loop 1: The resistance vector is

1

22

13

44

Total of RI voltage drops for current

9

1Voltage drop for is negative; flows in opposite direction

0Current does not flow in loop 1

Voltage drop for is negative;

1

4

I

II

I

II

ªº

«»

−

«»

=«»

−

«»

−¬¼

r

55

flows in opposite direction

Voltage drop for is negative; flows in opposite directionII

Loop 2: The resistance vector is

1.10 • Solutions 75

11

2

233

Voltage drop for is negative; flows in opposite direction

1

7Total of RI voltage drops for current

2Voltage drop for is negative; flows in opposite direction

0Current

3

II

I

II

ªº

−

«»

=−

«»

−

¬¼

r

4

55

does not flow in loop 2

Voltage drop for is negative; flows in opposite direction

I

II

Also, r

3

=

0

2

10

3

ªº

«»

−

«»

−

«»

−

¬¼

, r

4

=

1

0

3

7

2

−

ª

º

«

»

«

»

«

»

−

«

»

«

»

«

»

−

¬

¼

, r

5

=

4

3

2

12

−

ªº

«»

−

«»

−

«»

−

«»

¬¼

, and R =

91014

17 20 3

021033

10372

433212

−−−

ª

º

«

»

−−−

«

»

«

»

−−−

«

»

−−−

«

»

«

»

−−−−

¬

¼

. Set v =

50

30

20

40

0

ªº

«»

−

«»

−

«»

¬¼

. Note

the negative voltages for loops where the chosen current direction is opposed by the orientation of

the voltage source in that loop. Thus Ri = v becomes:

1

2

3

4

5

91014 50

17 20 3 30

021033 20

10372 40

433212 0

I

−−−

ªº

ªºªº

«»

«»«»

−−− −

«»

«»« »

«»

«»« »

=

−−−

«»

«»« »

−−− −

«»

«»« »

«»

«»« »

−−−−

¬¼¬ ¼

¬¼

. [M] The solution is

1

2

3

4

5

4.00

4.38

.90

5.80

.96

I

ªº

ª

º

«»

«

»

−

«»

«

»

«»

«

»

=−

«»

«

»

−

«»

«

»

«»

«

»

−

¬

¼

¬¼

.

9. The population movement problems in this section assume that the total population is constant, with

no migration or immigration. The statement that “about 7% of the city’s population moves to the

suburbs” means also that the rest of the city’s population (93%) remain in the city. This determines

the entries in the first column of the migration matrix (which concerns movement from the city).

From:

City Suburbs To:

.93 City

.07 Suburbs

ªº

«»

¬¼

Likewise, if 5% of the suburban population moves to the city, then the other 95% remain in the

suburbs. This determines the second column of the migration matrix:, M = .93 .05

.07 .95

ª

º

«

»

¬

¼. The

difference equation is x

k+1

= Mx

k

for k = 0, 1, 2, …. Also, x

0

= 800,000

500,000

ª

º

«

»

¬

¼

The population in 2011 (when k = 1) is x

1

= Mx

0

= .93 .05 800,000 769,000

.07 .95 500,000 531,000

ª

ºª º ª º

=

«

»« » « »

¬

¼¬ ¼ ¬ ¼

The population in 2012 (when k = 2) is x

2

= Mx

1

= .93 .05 769,000 741,720

.07 .95 531,000 558,280

ª

ºª º ª º

=

«

»« » « »

¬

¼¬ ¼ ¬ ¼

76 CHAPTER 1 • Linear Equations in Linear Algebra

10. The data in the first sentence implies that the migration matrix has the form:

From:

City Suburbs To:

.04 City

.06 Suburbs

ªº

«»

¬¼

The remaining entries are determined by the fact that the numbers in each column must sum to 1.

(For instance, if 6% of the city people move to the suburbs, then the rest, or 94%, remain in the city.)

So the migration matrix is M = .94 .04

.06 .96

ªº

«»

¬¼

. The initial population is x

0

= 10,000,000

800,000

ª

º

«

»

¬

¼.

The population in 2011 (when k = 1) is x

1

= Mx

0

= .94 .04 10,000,000 9,432,000

.06 .96 800,000 1,368,000

ª

ºª º ª º

=

«

»« » « »

¬

¼¬ ¼ ¬ ¼

The population in 2012 (when k = 2) is x

2

= Mx

1

= .94 .04 9,432,000 8,920,800

.06 .96 1,368,000 1,879,200

ª

ºª º ª º

=

«

»« » « »

¬

¼¬ ¼ ¬ ¼

11. The problem concerns two groups of people–those living in California and those living outside

California (and in the United States). It is reasonable, but not essential, to consider the people living

inside California first. That is, the first entry in a column or row of a vector will concern the people

living in California. With this choice, the migration matrix has the form:

From:

Calif. Outside To:

Calif.

Outside

ªº

«»

¬¼

a. For the first column of the migration matrix M, compute

{

}

{}

Calif. persons

who moved 516,100 .016372

Total Calif. pop. 31,524,000

==

The other entry in the first column is 1 – .016372 = .983628. The exercise requests that 5 decimal

places be used. So this number should be rounded to .98363. Whatever number of decimal places

is used, it is important that the two entries sum to 1. So, for the first fraction, use .01637.

For the second column of M, compute

{

}

{}

outside persons

who moved 381,262 .00167

Total outside pop. 228,680,000

==

. The other

entry is 1 – .00167 = .99833. Thus, the migration matrix is

From:

Calif. Outside To:

.98363 .00167 Calif.

.01637 .99833 Outside

ªº

«»

¬¼

b. [M] The initial vector is x

0

= (31.524, 228.680), with data in millions of persons. Since x

0

describes the population in 1994, and x

1

describes the population in 1995, the vector x

6

describes

the projected population for the year 2000, assuming that the migration rates remain constant and

1.10 • Solutions 77

there are no deaths, births, or migration. Here are the vectors x

0

through x

6

with the first 5 figures

displayed. Numbers are in millions of persons:

x

0

=31.524 31.390 31.258 31.129 31.002 30.877 30.755

,,,,,,

228.68 228.82 228.95 229.08 229.20 229.33 229.45

ªºªºªºªºªºªºªº

«»«»«»«»«»«»«»

¬¼¬¼¬¼¬¼¬¼¬¼¬¼

= x

6

.

12. Set M =

0

.97 .05 .10 295

.00 .90 .05 and 55

.03 .05 .85 150

ªºªº

«»«»

=

«»«»

¬¼¬¼

x

. Then x

1

=

.97 .05 .10 295 304

.00 .90 .05 55 57

.03 .05 .85 150 139

ª

ºª º ª º

«

»« » « »

≈

«

»« » « »

«

»« » « »

¬

¼¬ ¼ ¬ ¼

, and

x

2

=

.97 .05 .10 304 312

.00 .90 .05 57 58

.03 .05 .85 139 130

ªºªºªº

«»«»«»

≈

«»«»«»

¬¼¬¼¬¼

. The entries in x

2

give the approximate distribution of cars on

Wednesday, two days after Monday.

13. [M] The order of entries in a column of a migration matrix must match the order of the columns. For

instance, if the first column concerns the population in the city, then the first entry in each column

must be the fraction of the population that moves to (or remains in) the city. In this case, the data in

the exercise leads to M = .95 .03

.05 .97

ªº

«»

¬¼

and x

0

= 600,000

400,000

ª

º

«

»

¬

¼

a. Some of the population vectors are

5101520

523,293 472,737 439,417 417,456

,,,

476,707 527,263 560,583 582,544

ªºªºªºªº

====

«»«»«»«»

¬¼¬¼¬¼¬¼

xx x x

The data here shows that the city population is declining and the suburban population is

increasing, but the changes in population each year seem to grow smaller.

b. When x

0

= 350,000

650,000

ªº

«»

¬¼

, the situation is different. Now

5101520

358,523 364,140 367,843 370,283

,,,

641, 477 635,860 632,157 629,717

ªºªºªºªº

====

«»«»«»«»

¬¼¬¼¬¼¬¼

xx x x

The city population is increasing slowly and the suburban population is decreasing. No other

conclusions are expected. (This example will be analyzed in greater detail later in the text.)

14. Here are Figs. (a) and (b) for Exercise 13, followed by the figure for Exercise 34 in Section 1.1:

10˚

40˚

20˚ 20˚

30˚ 30˚

12

43

0˚

20˚ 20˚

12

43

10˚

40˚

0˚ 0˚

10˚ 10˚

12

43

(b) Section 1.1

(a)

For Fig. (a), the equations are

124

21 3

342

413

4020

4200

4020

40 20

TTT

TT T

TTT

=+ + +

=+ ++

=+++

=+ + +

To solve the system, rearrange the equations and row reduce the augmented matrix. Interchanging

rows 1 and 4 speeds up the calculations. The first five steps are shown in detail.

410120 101420 101420 101420

14 1020 14 10 20 04 0 4 0 0 10 1 0

~~~

014120 014 120 014120 014120

10 1420 4 10 120 0 1415100 0 1415100

−− −− −− −−

−− −− − −

−− − − − − − −

−− −− −− −−

ªºªºªºªº

«»«»«»«»

¬¼¬¼¬¼¬¼

10 1 4 20 10 1 4 20 100010

01 0 1 0 010 1 0 010010

~~~

00 4 2 20 004 2 20 001010

00 414100 00012120 000110

~

−− −−

−−

⋅⋅⋅

−−

−

ªºªºªº

«»«»«»

¬¼¬¼¬¼

For Fig (b), the equations are

124

21 3

342

413

4100

4 0 40

44010

410 10

TTT

TT T

TTT

=+++

=++ +

=+++

Rearrange the equations and row reduce the augmented matrix:

410110 100010

14 1040 010017.5

014150 001020

10 1420 000112.5

~~

−−

ªºªº

«»«»

⋅⋅⋅

«»«»

¬¼¬¼

a. Here are the solution temperatures for the three problems studied:

Fig. (a) in Exercise 14 of Section 1.10: (10, 10, 10, 10)

Fig. (b) in Exercise 14 of Section 1.10: (10, 17.5, 20, 12.5)

Figure for Exercises 34 in Section 1.1 (20, 27.5, 30, 22.5)

When the solutions are arranged this way, it is evident that the third solution is the sum of the first

two solutions. What might not be so evident is that list of boundary temperatures of the third

problem is the sum of the lists of boundary temperatures of the first two problems. (The

temperatures are listed clockwise, starting at the left of T

1

.)

Fig. (a): ( 0, 20, 20, 0, 0, 20, 20, 0)

Fig. (b): (10, 0, 0, 40, 40, 10, 10, 10)

Fig. from Section 1.1: (10, 20, 20, 40, 40, 30, 30, 10)

b. When the boundary temperatures in Fig. (a) are multiplied by 3, the new interior temperatures are

also multiplied by 3.

c. The correspondence from the list of eight boundary temperatures to the list of four interior

temperatures is a linear transformation. A verification of this statement is not expected. However,

it can be shown that the solutions of the steady-state temperature problem here satisfy a

superposition principle. The system of equations that approximate the interior temperatures can

Chapter 1 • Supplementary Exercises 79

be written in the form Ax = b, where A is determined by the arrangement of the four interior

points on the plate and b is a vector in R

4

determined by the boundary temperatures.

Note:

The MATLAB box in the Study Guide for Section 1.10 discusses scientific notation and shows

how to generate a matrix whose columns list the vectors x

0

, x

1

, x

2

, …, determined by an equation

x

k+1

= Mx

k

for k = 0 , 1, ….

Chapter 1 SUPPLEMENTARY EXERCISES

1. a. False. (The word “reduced” is missing.) Counterexample:

12 1 2 12

,,

34 0 2 0 1

AB C

ªº ª º ªº

== =

«» « » «»

−

¬¼ ¬ ¼ ¬¼

The matrix A is row equivalent to matrices B and C, both in echelon form.

b. False. Counterexample: Let A be any n×n matrix with fewer than n pivot columns. Then the

equation Ax = 0 has infinitely many solutions. (Theorem 2 in Section 1.2 says that a system has

either zero, one, or infinitely many solutions, but it does not say that a system with infinitely

many solutions exists. Some counterexample is needed.)

c. True. If a linear system has more than one solution, it is a consistent system and has a free

variable. By the Existence and Uniqueness Theorem in Section 1.2, the system has infinitely

many solutions.

d. False. Counterexample: The following system has no free variables and no solution:

12

2

12

1

5

2

xx

x

xx

+=

=

+=

e. True. See the box after the definition of elementary row operations, in Section 1.1. If [A b] is

transformed into [C d] by elementary row operations, then the two augmented matrices are row

equivalent.

f. True. Theorem 6 in Section 1.5 essentially says that when Ax = b is consistent, the solution sets

of the nonhomogeneous equation and the homogeneous equation are translates of each other. In

this case, the two equations have the same number of solutions.

g. False. For the columns of A to span R

m

, the equation Ax = b must be consistent for all b in R

m

,

not for just one vector b in R

m

.

h. False. Any matrix can be transformed by elementary row operations into reduced echelon form,

but not every matrix equation Ax = b is consistent.

i. True. If A is row equivalent to B, then A can be transformed by elementary row operations first

into B and then further transformed into the reduced echelon form U of B. Since the reduced

echelon form of A is unique, it must be U.

j. False. Every equation Ax = 0 has the trivial solution whether or not some variables are free.

k. True, by Theorem 4 in Section 1.4. If the equation Ax = b is consistent for every b in R

m

, then A

must have a pivot position in every one of its m rows. If A has m pivot positions, then A has m

pivot columns, each containing one pivot position.

l. False. The word “unique” should be deleted. Let A be any matrix with m pivot columns but more

than m columns altogether. Then the equation Ax = b is consistent and has m basic variables and

at least one free variable. Thus the equation does not does not have a unique solution.

m. True. If A has n pivot positions, it has a pivot in each of its n columns and in each of its n rows.

The reduced echelon form has a 1 in each pivot position, so the reduced echelon form is the n×n

identity matrix.

n. True. Both matrices A and B can be row reduced to the 3×3 identity matrix, as discussed in the

previous question. Since the row operations that transform B into I

3

are reversible, A can be

transformed first into I

3

and then into B.

o. True. The reason is essentially the same as that given for question f.

p. True. If the columns of A span R

m

, then the reduced echelon form of A is a matrix U with a pivot

in each row, by Theorem 4 in Section 1.4. Since B is row equivalent to A, B can be transformed

by row operations first into A and then further transformed into U. Since U has a pivot in each

row, so does B. By Theorem 4, the columns of B span R

m

.

q. False. See Example 5 in Section 1.7.

r. True. Any set of three vectors in R

2

would have to be linearly dependent, by Theorem 8 in

Section 1.7.

s. False. If a set {v

1

, v

2

, v

3

, v

4

} were to span R

5

, then the matrix A = [v

1

v

2

v

3

v

4

] would have

a pivot position in each of its five rows, which is impossible since A has only four columns.

t. True. The vector –u is a linear combination of u and v, namely, –u = (–1)u + 0v.

u. False. If u and v are multiples, then Span{u, v} is a line, and w need not be on that line.

v. False. Let u and v be any linearly independent pair of vectors and let w = 2v. Then w = 0u + 2v,

so w is a linear combination of u and v. However, u cannot be a linear combination of v and w

because if it were, u would be a multiple of v. That is not possible since {u, v} is linearly

independent.

w. False. The statement would be true if the condition v

1

is not zero were present. See Theorem 7 in

Section 1.7. However, if v

1

= 0, then {v

1

, v

2

, v

3

} is linearly dependent, no matter what else might

be true about v

2

and v

3

.

x. True. “Function” is another word used for “transformation” (as mentioned in the definition of

“transformation” in Section 1.8), and a linear transformation is a special type of transformation.

y. True. For the transformation x 6 Ax to map R

5

onto R

6

, the matrix A would have to have a pivot

in every row and hence have six pivot columns. This is impossible because A has only five

columns.

z. False. For the transformation x 6 Ax to be one-to-one, A must have a pivot in each column.

Since A has n columns and m pivots, m might be less than n.

2. If a ≠ 0, then x = b/a; the solution is unique. If a = 0, and b ≠ 0, the solution set is empty, because

0x = 0 ≠ b. If a = 0 and b = 0, the equation 0x = 0 has infinitely many solutions.

3. a. Any consistent linear system whose echelon form is

*** * ** 0 **

0** or 00* or 00*

0000 0000 0000

ªºªºªº

«»«»«»

¬¼¬¼¬¼

 



b. Any consistent linear system whose coefficient matrix has reduced echelon form I

3

.

c. Any inconsistent linear system of three equations in three variables.

4. Since there are three pivots (one in each row), the augmented matrix must reduce to the form

Chapter 1 • Supplementary Exercises 81

***

0**

00 *

ªº

«»

¬¼



. A solution of Ax = b exists for all b because there is a pivot in each row of A. Each

solution is unique because there are no free variables.

5. a.

13 1 3

~

4801284

kk

hhk

ªºª º

«»« »

−−

¬¼¬ ¼

. If h = 12 and k v 2, the second row of the augmented matrix

indicates an inconsistent system of the form 0x

2

= b, with b nonzero. If h = 12, and k = 2, there is

only one nonzero equation, and the system has infinitely many solutions. Finally, if h v 12, the

coefficient matrix has two pivots and the system has a unique solution.

b.

212 1

~

62031

hh

kkh

−−

ªºª º

«»« »

−+

¬¼¬ ¼

. If k + 3h = 0, the system is inconsistent. Otherwise, the

coefficient matrix has two pivots and the system has a unique solution.

6. a. Set

12 3

427

,,

8310

−

ªº ª º ª º

== =

«» « » « »

−

¬¼ ¬ ¼ ¬ ¼

vv v

, and

5

3

−

ª

º

=

«

»

−

¬

¼

b

. “Determine if b is a linear combination of v

1

,

v

2

, v

3

.” Or, “Determine if b is in Span{v

1

, v

2

, v

3

}.” To do this, compute

4275 4275

~

83103 0147

−− − −

ªºªº

«»«»

−− −

¬¼¬¼

. The system is consistent, so b is in Span{v

1

, v

2

, v

3

}.

b. Set A =

427 5

,

8310 3

−−

ªºªº

=

«»«»

−−

¬¼¬¼

b

. “Determine if b is a linear combination of the columns of A.”

c. Define T(x) = Ax. “Determine if b is in the range of T.”

7. a. Set

123

242

5, 1, 1

753

−−

ªº ªº ªº

«» «» «»

=−==

«» «» «»

−−

¬¼ ¬¼ ¬¼

vv v

and

1

2

3

b

ª

º

«

»

=

«

»

«

»

¬

¼

b

. “Determine if v

1

, v

2

, v

3

span R

3

.” To do this,

row reduce [v

1

v

2

v

3

]:

242 242 242

511~094~094

753 094 000

−− −− −−

ªºªºªº

«»«»«»

−−−−−

«»«»«»

−−

¬¼¬¼¬¼

. The matrix does not have a pivot in each row,

so its columns do not span R

3

, by Theorem 4 in Section 1.4.

b. Set A =

242

511

753

−−

ªº

«»

−

«»

−−

¬¼

. “Determine if the columns of A span R

3

.”

c. Define T(x) = Ax. “Determine if T maps R

3

onto R

3

.”

8. a.

** ** 0 *

,,

0*00 00

ªºªºªº

«»«»«»

¬¼¬¼¬¼

 



b.

**

0*

00

ª

º

«

»

«

»

«

»

¬

¼



9. The first line is the line spanned by

1

2

ªº

«»

¬¼

. The second line is spanned by

2

1

ª

º

«

»

¬

¼. So the problem is to

write

5

6

ªº

«»

¬¼

as the sum of a multiple of

1

2

ª

º

«

»

¬

¼ and a multiple of

2

1

ª

º

«

»

¬

¼. That is, find x

1

and x

2

such that

12

215

126

xx

ªº ªº ªº

+=

«» «» «»

¬¼ ¬¼ ¬¼

. Reduce the augmented matrix for this equation:

215 126 1 2 6 12 6 104/3

~~ ~ ~

126 2 15 0 3 7 0 17/3 0 17/3

ªºªºª ºª ºª º

«»«»« »« »« »

−−

¬¼¬¼¬ ¼¬ ¼¬ ¼

Thus,

47

33

521

612

ªº ªº ªº

=+

«» «» «»

¬¼ ¬¼ ¬¼

or

58/3 7/3

64/314/3

ªº ª º ª º

=+

«» « » « »

¬¼ ¬ ¼ ¬ ¼

.

10. The line through a

1

and the origin and the line through a

2

and the origin determine a “grid” on the

x

1

x

2

-plane as shown below. Every point in R

2

can be described uniquely in terms of this grid. Thus, b

can be reached from the origin by traveling a certain number of units in the a

1

-direction and a certain

number of units in the a

2

-direction.

11. A solution set is a line when the system has one free variable. If the coefficient matrix is 2×3, then

two of the columns should be pivot columns. For instance, take

12*

03*

ª

º

«

»

¬

¼. Put anything in column

3. The resulting matrix will be in echelon form. Make one row replacement operation on the second

row to create a matrix not in echelon form, such as

121 12 1

~

031 152

ª

ºª º

«

»« »

¬

¼¬ ¼

12. A solution set is a plane where there are two free variables. If the coefficient matrix is 2×3, then only

one column can be a pivot column. The echelon form will have all zeros in the second row. Use a

row replacement to create a matrix not in echelon form. For instance, let A =

123

ªº

«»

¬¼

.

13. The reduced echelon form of A looks like

10 *

01*

000

E

ª

º

«

»

=

«

»

«

»

¬

¼

. Since E is row equivalent to A, the

equation Ex = 0 has the same solutions as Ax = 0. Thus

10 * 3 0

01* 2 0

000 1 0

ª

ºª º ª º

«

»« » « »

−=

«

»« » « »

«

»« » « »

¬

¼¬ ¼ ¬ ¼

.

x

1

x

2

a

2

a

1

b

Chapter 1 • Supplementary Exercises 83

By inspection,

10 3

01 2

00 0

E

−

ªº

«»

=

«»

¬¼

.

14. Row reduce the augmented matrix for

12

10

20

a

xx

aa

ª

ºªºªº

+=

«

»«»«»

+

¬

¼¬¼¬¼

(*).

2

10

10 1 0

~

20 0(2 )(1 )0

02 0

a

aa

aa aa

aa

ªº

ªº ª º

=

«»

«» « »

+−+

+−

¬¼ ¬ ¼

¬¼

The equation (*) has a nontrivial solution only when (2 – a)(1 + a) = 0. So the vectors are linearly

independent for all a except a = 2 and a = –1.

15. a. If the three vectors are linearly independent, then a, c, and f must all be nonzero. (The converse is

true, too.) Let A be the matrix whose columns are the three linearly independent vectors. Then A

must have three pivot columns. (See Exercise 30 in Section 1.7, or realize that the equation

Ax = 0 has only the trivial solution and so there can be no free variables in the system of

equations.) Since A is 3×3, the pivot positions are exactly where a, c, and f are located.

b. The numbers a, …, f can have any values. Here's why. Denote the columns by v

1

, v

2

, and v

3

.

Observe that v

1

is not the zero vector. Next, v

2

is not a multiple of v

1

because the third entry of v

2

is nonzero. Finally, v

3

is not a linear combination of v

1

and v

2

because the fourth entry of v

3

is

nonzero. By Theorem 7 in Section 1.7, {v

1

, v

2

, v

3

} is linearly independent.

16. Denote the columns from right to left by v

1

, …, v

4

. The “first” vector v

1

is nonzero, v

2

is not a

multiple of v

1

(because the third entry of v

2

is nonzero), and v

3

is not a linear combination of v

1

and

v

2

(because the second entry of v

3

is nonzero). Finally, by looking at first entries in the vectors, v

4

cannot be a linear combination of v

1

, v

2

, and v

3

. By Theorem 7 in Section 1.7, the columns are

linearly independent.

17. Here are two arguments. The first is a “direct” proof. The second is called a “proof by contradiction.”

i. Since {v

1

, v

2

, v

3

} is a linearly independent set, v

1

≠

0. Also, Theorem 7 shows that v

2

cannot be a

multiple of v

1

, and v

3

cannot be a linear combination of v

1

and v

2

. By hypothesis, v

4

is not a linear

combination of v

1

, v

2

, and v

3

. Thus, by Theorem 7, {v

1

, v

2

, v

3

, v

4

} cannot be a linearly dependent

set and so must be linearly independent.

ii. Suppose that {v

1

, v

2

, v

3

, v

4

} is linearly dependent. Then by Theorem 7, one of the vectors in the

set is a linear combination of the preceding vectors. This vector cannot be v

4

because v

4

is not in

Span{v

1

, v

2

, v

3

}. Also, none of the vectors in {v

1

, v

2

, v

3

} is a linear combinations of the preceding

vectors, by Theorem 7. So the linear dependence of {v

1

, v

2

, v

3

, v

4

} is impossible. Thus {v

1

, v

2

, v

3

,

v

4

} is linearly independent.

18. Suppose that c

1

and c

2

are constants such that

c

1

v

1

+ c

2

(v

1

+ v

2

) = 0 (*)

Then (c

1

+ c

2

)v

1

+ c

2

v

2

= 0. Since v

1

and v

2

are linearly independent, both c

1

+ c

2

= 0 and c

2

= 0. It

follows that both c

1

and c

2

in (*) must be zero, which shows that {v

1

, v

1

+ v

2

} is linearly independent.

19. Let M be the line through the origin that is parallel to the line through v

1

, v

2

, and v

3

. Then v

2

– v

1

and

v

3

– v

1

are both on M. So one of these two vectors is a multiple of the other, say v

2

– v

1

= k(v

3

– v

1

).

This equation produces a linear dependence relation (k – 1)v

1

+ v

2

– kv

3

= 0.

A second solution: A parametric equation of the line is x = v

1

+ t(v

2

– v

1

). Since v

3

is on the line,

there is some t

0

such that v

3

= v

1

+ t

0

(v

2

– v

1

) = (1 – t

0

)v

1

+ t

0

v

2

. So v

3

is a linear combination of v

1

and v

2

, and {v

1

, v

2

, v

3

} is linearly dependent.

20. If T(u) = v, then since T is linear,

T(–u) = T((–1)u) = (–1)T(u) = –v.

21. Either compute T(e

1

), T(e

2

), and T(e

3

) to make the columns of A, or write the vectors vertically in the

definition of T and fill in the entries of A by inspection:

11

22

33

??? 100

?? , 010

??? 001

xx

AAxxA

xx

ªºªºªºª º

«»«»«»« »

==−=−

«»«»«»« »

¬¼¬¼¬¼¬ ¼

x

22. By Theorem 12 in Section 1.9, the columns of A span R

3

. By Theorem 4 in Section 1.4, A has a pivot

in each of its three rows. Since A has three columns, each column must be a pivot column. So the

equation Ax = 0 has no free variables, and the columns of A are linearly independent. By Theorem 12

in Section 1.9, the transformation x 6 Ax is one-to-one.

23.

45 4 3 5

implies that

30 3 4 0

ab ab

ba ab

=

−−=

ªºªºªº

«»«»«» +=

¬¼¬¼¬¼ . Solve:

435 4 3 5 43 5 4016/5 104/5

~~~~

340 025/415/4 013/5 013/5 013/5

−− −

ªºª ºª ºª ºª º

«»« »« »« »« »

−−−−

¬¼¬ ¼¬ ¼¬ ¼¬ ¼

Thus a = 4/5 and b = –3/5.

24. The matrix equation displayed gives the information

2425

ab−= and

420.ab+=

Solve for a and

b:

2425 12 5 101/5

2425

~~ ~

42 0 010 45 0 1 2/5 0 1 2/5

ª

ºª ºª º

ªº

−−

−

«

»« »« »

«»

−− −

«

»« »« »

¬¼

¬

¼¬ ¼¬ ¼

So

1/ 5, 2/ 5.ab==−

25. a. The vector lists the number of three-, two-, and one-bedroom apartments provided when x

1

floors

of plan A are constructed.

b.

123

345

743

889

xxx

ªº ªº ªº

«» «» «»

++

«» «» «»

¬¼ ¬¼ ¬¼

c. [M] Solve

123

34566

74374

889136

xx x

ªº ªº ªº ª º

«» «» «» « »

++=

«» «» «» « »

¬¼ ¬¼ ¬¼ ¬ ¼

13

23

345 66 10 1/2 2 (1 / 2) 2

743 74~ 0113/815 (13 / 8) 15

889136 00 0 0 00

xx

−−=

ªºªº

«»«»

⋅⋅⋅ + =

«»«»

«»«» =

¬¼¬¼

The general solution is

Chapter 1 • Supplementary Exercises 85

13

233

33

2(1/2) 2 1/2

15 (13/ 8) 15 13/8

01

xx

xxx

xx

+

ªºª ºªº ª º

«»« »«» « »

==−=+−

«»« »«» « »

¬¼¬ ¼¬¼ ¬ ¼

x

However, the only feasible solutions must have whole numbers of floors for each plan. Thus, x

3

must be a multiple of 8, to avoid fractions. One solution, for x

3

= 0, is to use 2 floors of plan A

and 15 floors of plan B. Another solution, for x

3

= 8, is to use 6 floors of plan A , 2 floors of plan

B, and 8 floors of plan C. These are the only feasible solutions. A larger positive multiple of 8 for

x

3

makes x

2

negative. A negative value for x

3

, of course, is not feasible either.

r

2.1 SOLUTIONS

Notes:

The definition here of a matri

x

calculations. (The dual fact about the r

o

vectors here are usually written as c

o

reinforce the definition of AB.

Exercises 23 and 24 are used in the

23–25 are mentioned in a footnote in S

e

can provide a transition to Section 2.2.

O

Exercises 27 and 28 are optional, b

u

also appear in Exercises 31–34 of Secti

o

Section 7.1. Exercises 29–33 provide goo

1. 201 4

2(2)

452 8

1

A

−−

ªºª

−=−=

«»«

−−

¬¼¬

751 4

2143 8

BA

−−

ªºª

−=+

«»«

−− −

¬¼¬

The product AC is not defined beca

u

rows of C. 12 35

21 14

CD ªºªº

=«»«»

−−

¬¼¬¼

computation, the row-column rule i

2. 201 75

33

452 14

AB

−−

ªºª

+= +

«»«

−−

¬¼¬

The expression 2C – 3E is not defi

n

357 5 1

3

141 4 3

1

DB

−

ªºª ºª

==

«»« »«

−−−−

¬¼¬ ¼¬

The product EC is not defined beca

u

rows of C.

r

son Education, Inc. Publishing as Addison-Wesley.

x

product AB gives the proper view of AB for near

l

o

ws of A and the rows of AB is seldom needed, ma

i

o

lumns.) I assign Exercise 13 and most of Exercis

e

proof of the Invertible Matrix Theorem, in Section 2

.

e

ction 2.2. A class discussion of the solutions of Exe

r

O

r, these exercises could be assigned after starting Se

c

u

t they are mentioned in Example 4 of Section 2.4. O

u

o

n 4.6 and in the spectral decomposition of a symmet

r

d training for mathematics majors.

02

1

04

º

»

−¼. Next, use B – 2A = B + (–2A):

02 353

10 4 7 6 7

−

ºª º

=

»« »

−− −

¼¬ ¼

u

se the number of columns of A does not match the n

u

13 2( 1) 15 2 4 1 13

23 1(1) 25 14 7 6

⋅+−⋅+⋅

ªºªº

==

«»«»

−⋅ +−−⋅+⋅−−

¬¼¬¼

. For men

s probably easier to use than the definition.

1221015132315

2

343512297177

+−−+−

ºª ºª

==

»« »«

−+−− − − −

¼¬ ¼¬

n

ed because 2C has 2 columns and –3E has only 1 col

u

3

751 3(5)5(4) 315(3) 26 3

1

741 1(5)4(4) 114(3) 3 1

⋅+⋅−+−⋅+−−

ºª

=

»«

⋅+⋅−−+−−⋅+−−−

¼¬

u

se the number of columns of E does not match the n

u

87

l

y all matrix

i

nly because

e

s 17–22 to

.

3. Exercises

r

cises 23–25

c

tion 2.2.

u

ter products

r

ic matrix, in

u

mber of

tal

2

º

»

¼

u

mn.

512

113

−º

»

−¼

u

mber of

88 CHAPTER 2 • Matrix Algebra

3.

2

30 2 5 320(5) 15

303 3 2 033(2) 35

IA

−−−−

ªºª ºª ºª º

−=−==

«»« »« »« »

−−−−−

¬¼¬ ¼¬ ¼¬ ¼

22

25 615

(3 ) 3( ) 3 32 9 6

IA IA

−−

ªºª º

== =

«»« »

−−

¬¼¬ ¼

, or

2

302 5 3203(5)0 6 15

(3 ) 033 2 03303(2) 9 6

IA

−⋅+−+−

ªºª ºª ºª º

== =

«»« »« »« »

−+⋅+−−

¬¼¬ ¼¬ ¼¬ ¼

4.

3

513 500 013

5436050426

312 005 313

AI

−−

ªºªºªº

«»«»«»

−=−−− =−−−

«»«»«»

−−−

¬¼¬¼¬¼

33

513 25515

(5 ) 5( ) 5 5 4 3 6 20 15 30

312 15510

IA IA A

−−

ªºª º

«»« »

===−−=−−

«»« »

−−

¬¼¬ ¼

, or

3

500 5 1 3

(5 ) 0 5 0 4 3 6

005 3 1 2

IA

−

ªºª º

«»« »

=−−

«»« »

−

¬¼¬ ¼

55 0 0 5( 1) 0 0 53 0 0 25 5 15

0 5( 4) 0 0 5 3 0 0 5( 6) 0 20 15 30

005(3) 0051 0052 15 5 10

⋅++ −++ ⋅++ −

ªºª º

«»« »

=+−++⋅++−+=−−

«»« »

++ −++⋅++⋅−

¬¼¬ ¼

5. a.

12

13 10 13 11

42

24 0, 24 8

23

53 26 53 19

AA

−−−

ªºªº ªºªº

−

ªº ªº

«»«» «»«»

====

«» «»

«»«» «»«»

−

¬¼ ¬¼

«»«» «»«»

−−−

¬¼¬¼ ¬¼¬¼

bb

[]

12

10 11

08

26 19

AB A A

−

ªº

«»

==

«»

−

¬¼

bb

b.

13 14 3( 2) 1( 2) 33 10 11

42

24 24 4(2) 2(2) 43 0 8

23

53 54 3(2) 5(2) 33 26 19

−−⋅+− − − +⋅ −

ªº ª ºª º

−

ªº

«» « »« »

=⋅+−−+⋅=

«»

«» « »« »

−

¬¼

«» « »« »

−⋅−−−−⋅−

¬¼ ¬ ¼¬ ¼

6. a.

12

43 5 43 22

14

35 12, 35 22

32

01 3 01 2

AA

−− −

ªºªº ªºªº

ªº ªº

«»«» «»«»

=−==−=−

«» «»

«»«» «»«»

−

¬¼ ¬¼

«»«» «»«»

−

¬¼¬¼ ¬¼¬¼

bb

[]

12

522

12 22

32

AB A A

−

ªº

«»

==−

«»

−

¬¼

bb

2.1 • Solutions 89

b.

43 4133443(2) 522

14

35 3153345(2) 1222

32

01 0113041(2) 3 2

−⋅−⋅⋅−−−

ªº ª ºª º

ªº

«» « »« »

−=−⋅ +⋅−⋅+−=−

«»

«» « »« »

−

¬¼

«» « »« »

⋅+⋅⋅+−−

¬¼ ¬ ¼¬ ¼

7. Since A has 3 columns, B must match with 3 rows. Otherwise, AB is undefined. Since AB has 7

columns, so does B. Thus, B is 3×7.

8. The number of rows of B matches the number of rows of BC, so B has 5 rows.

9.

23 19 7183

,

11 3 4 9

k

AB kk

−+

ªºªºª º

==

«»«»« »

−− −−+

¬¼¬¼¬ ¼

while

1 9 2 3 7 12

31169

BA kkk

−

ª

ºª º ª º

==

«

»« » « »

−− −−−+

¬

¼¬ ¼ ¬ ¼

.

Then AB = BA if and only if 18 + 3k = 12 and –4 = –6 – k, which happens if and only if k = –2.

10.

3611 2121 3635 2121

,

1234 7 7 122 1 7 7

AB AC

−− − − − − − − −

ªºªºª ºªºªºª º

=== =

«»«»« »«»«»« »

−−

¬¼¬¼¬ ¼¬¼¬¼¬ ¼

11.

123500 5 6 6

245030 101210

356002 151512

AD

ªºªºª º

«»«»« »

==

«»«»« »

¬¼¬¼¬ ¼

500123 51015

030245 61215

002356 61012

DA

ªºªºª º

«»«»« »

==

«»«»« »

¬¼¬¼¬ ¼

Right-multiplication (that is, multiplication on the right) by the diagonal matrix D multiplies each

column of A by the corresponding diagonal entry of D. Left-multiplication by D multiplies each row

of A by the corresponding diagonal entry of D. To make AB = BA, one can take B to be a multiple of

I

3

. For instance, if B = 4I

3

, then AB and BA are both the same as 4A.

12. Consider B = [b

1

b

2

]. To make AB = 0, one needs Ab

1

= 0 and Ab

2

= 0. By inspection of A, a

suitable

b

1

is

2,

1

ªº

«»

¬¼

or any multiple of

2

1

ªº

«»

¬¼

. Example:

26

.

13

B

ª

º

=

«

»

¬

¼

13. Use the definition of AB written in reverse order: [Ab

1

⋅ ⋅ ⋅ Ab

p

] = A[b

1

⋅ ⋅ ⋅ b

p

]. Thus

[Qr

1

⋅ ⋅ ⋅ Qr

p

] = QR, when R = [r

1

⋅ ⋅ ⋅ r

p

].

14. By definition, UQ = U[q

1

⋅ ⋅ ⋅ q

4

] = [Uq

1

⋅ ⋅ ⋅ Uq

4

]. From Example 6 of Section 1.8, the vectorUq

1

lists the total costs (material, labor, and overhead) corresponding to the amounts of products B andC

specified in the vector q

1

. That is, the first column of UQ lists the total costs for materials, labor, and

overhead used to manufacture products B and C during the first quarter of the year. Columns 2, 3,and

4 of UQ list the total amounts spent to manufacture B and C during the 2

nd

, 3

rd

, and 4

th

quarters,

respectively.

90 CHAPTER 2 • Matrix Algebra

15. a. False. See the definition of AB.

b. False. The roles of A and B should be reversed in the second half of the statement. See the box

after Example 3.

c. True. See Theorem 2(b), read right to left.

d. True. See Theorem 3(b), read right to left.

e. False. The phrase “in the same order” should be “in the reverse order.” See the box after Theorem

3.

16. a. True. See the box after Example 4.

b. False. AB must be a 3×3 matrix, but the formula given here implies that it is a 3×1 matrix. The

plus signs should just be spaces (between columns). This is a common mistake.

c. True. Apply Theorem 3(d) to A

2

=AA

d. False. The left-to-right order of (ABC)

T

, is C

T

B

T

A

T

. The order cannot be changed in general.

e. True. This general statement follows from Theorem 3(b).

17. Since

[]

12

311 ,

117AB A A

−−

ªº

==

«»

¬¼

bb

the first column of B satisfies the equation

1.

6

A

−

ªº

=«»

¬¼

x Row

reduction:

[]

1

133 103

~~

351 012

AA

−−

ª

ºª º

«

»« »

−

¬

¼¬ ¼

b. So b

1

=

3.

2

ª

º

«

»

¬

¼

Similarly,

[]

2

1311 101

~~

3517 014

AA

−−

ªºªº

«»«»

−

¬¼¬¼

b and b

2

=

1.

4

ª

º

«

»

¬

¼

Note:

An alternative solution of Exercise 17 is to row reduce [A Ab

1

Ab

2

] with one sequence of row

operations. This observation can prepare the way for the inversion algorithm in Section 2.2.

18. The third column of AB is also all zeros because Ab

3

= A0 = 0

19. (A solution is in the text). Write B = [b

1

b

2

b

3

]. By definition, the third column of AB is Ab

3

. By

hypothesis, b

3

= b

1

+ b

2

. So Ab

3

= A(b

1

+ b

2

) = Ab

1

+ Ab

2

, by a property of matrix-vector

multiplication. Thus, the third column of AB is the sum of the first two columns of AB.

20. The first two columns of AB are Ab

1

and Ab

2

. They are equal since b

1

and b

2

are equal.

21. Let b

p

be the last column of B. By hypothesis, the last column of AB is zero. Thus, Ab

p

= 0.

However, b

p

is not the zero vector, because B has no column of zeros. Thus, the equation Ab

p

= 0 is a

linear dependence relation among the columns of A, and so the columns of A are linearly dependent.

Note:

The text answer for Exercise 21 is, “The columns of A are linearly dependent. Why?” The Study

Guide supplies the argument above, in case a student needs help.

22. If the columns of B are linearly dependent, then there exists a nonzero vector x such that Bx = 0.

From this, A(Bx) = A0 and (AB)x = 0 (by associativity). Since x is nonzero, the columns of AB must

be linearly dependent.

23. If x satisfies Ax = 0, then CAx = C0 = 0 and so I

n

x = 0 and x = 0. This shows that the equation Ax = 0

has no free variables. So every variable is a basic variable and every column of A is a pivot column.

(A variation of this argument could be made using linear independence and Exercise 30 in Section

1.7.) Since each pivot is in a different row, A must have at least as many rows as columns.

2.1 • Solutions 91

24. Write I

3

=[e

1

e

2

e

3

] and D = [d

1

d

2

d

3

]. By definition of AD, the equation AD = I

3

is equivalent

|to the three equations Ad

1

= e

1

, Ad

2

= e

2

, and Ad

3

= e

3

. Each of these equations has at least one

solution because the columns of A span R

3

. (See Theorem 4 in Section 1.4.) Select one solution of

each equation and use them for the columns of D. Then AD = I

3

.

25. By Exercise 23, the equation CA = I

n

implies that (number of rows in A) > (number of columns), that

is, m > n. By Exercise 24, the equation AD = I

m

implies that (number of rows in A) < (number of

columns), that is, m < n. Thus m = n. To prove the second statement, observe that CAD = (CA)D =

I

n

D = D, and also CAD = C(AD) = CI

m

= C. Thus C = D. A shorter calculation is

C =C I

n

= C(AD) = (CA)D = I

n

D= D

26. Take any b in R

m

. By hypothesis, ADb = I

m

b = b. Rewrite this equation as A(Db) = b. Thus, the

vector x = Db satisfies Ax = b. This proves that the equation Ax = b has a solution for each b in R

m

.

By Theorem 4 in Section 1.4, A has a pivot position in each row. Since each pivot is in a different

column, A must have at least as many columns as rows.

27. The product u

T

v is a 1×1 matrix, which usually is identified with a real number and is written

without the matrix brackets.

[]

32 5 3 2 5

T

a

babc

c

ªº

«»

=−−=−+−

«»

¬¼

uv ,

[]

3

2325

5

T

abc a b c

−

ªº

«»

==−+−

«»

−

¬¼

vu

[]

3333

2222

5555

T

abc

abc a b c

abc

−−−−

ªº ª º

«» « »

==

«» « »

−−−−

¬¼ ¬ ¼

uv

[]

32 5

32 5 3 2 5

32 5

T

aaa a

bbb b

ccc c

−−

ªº ª º

«» « »

=−−=−−

«» « »

−−

¬¼ ¬ ¼

vu

28. Since the inner product u

T

v is a real number, it equals its transpose. That is,

u

T

v = (u

T

v)

T

= v

T

(u

T

)

T

= v

T

u, by Theorem 3(d) regarding the transpose of a product of matrices and

by Theorem 3(a). The outer product uv

T

is an n×n matrix. By Theorem 3, (uv

T

)

T

= (v

T

)

T

u

T

= vu

T

.

29. The (i, j)-entry of A(B + C) equals the (i, j)-entry of AB + AC, because

111

()

nnn

ik kj kj ik kj ik kj

kkk

ab c ab ac

===

+= +

¦¦¦

The (i, j)-entry of (B + C)A equals the (i, j)-entry of BA + CA, because

111

()

nnn

ik ik kj ik kj ik kj

kkk

bca ba ca

===

+= +

¦¦¦

30. The (i, j))-entries of r(AB), (rA)B, and A(rB) are all equal, because

11 1

() ()

nn n

ik kj ik kj ik kj

kk k

rab rab arb

== =

==

¦¦ ¦

31. Use the definition of the product I

m

A and the fact that I

m

x = x for x in R

m

.

92 CHAPTER 2 • Matrix Algebra

I

m

A = I

m

[a

1

⋅ ⋅ ⋅ a

n

] = [I

m

a

1

⋅ ⋅ ⋅ I

m

a

n

] = [a

1

⋅ ⋅ ⋅ a

n

] = A

32. Let e

j

and a

j

denote the jth columns of I

n

and A, respectively. By definition, the jth column of AI

n

is

Ae

j

, which is simply a

j

because e

j

has 1 in the jth position and zeros elsewhere. Thus corresponding

columns of AI

n

and A are equal. Hence AI

n

= A.

33. The (i, j)-entry of (AB)

T

is the ( j, i)-entry of AB, which is

11ji jnni

ab ab+⋅⋅⋅ +

The entries in row i of B

T

are b

1i

, … , b

ni

, because they come from column i of B. Likewise, the

entries in column j of A

T

are a

j1

, …, a

jn

, because they come from row j of A. Thus the (i, j)-entry in

B

T

A

T

is

11ji jnni

ab ab++"

, as above.

34. Use Theorem 3(d), treating x as an n×1 matrix: (ABx)

T

= x

T

(AB)

T

= x

T

B

T

A

T

.

35. [M] The answer here depends on the choice of matrix program. For MATLAB, use the help

command to read about zeros, ones, eye, and diag. For other programs see the

appendices in the Study Guide. (The TI calculators have fewer single commands that produce

special matrices.)

36. [M] The answer depends on the choice of matrix program. In MATLAB, the command

rand(5,6) creates a 5×6 matrix with random entries uniformly distributed between 0 and 1. The

command

round(19*(rand(4,4)–.5))

creates a random 4×4 matrix with integer entries between –9 and 9. The same result is produced by

the command randomint in the Laydata4 Toolbox on text website. For other matrix programs

see the appendices in the Study Guide.

37. [M] The equality AB = BA is very likely to be false for 4×4 matrices selected at random.

38. [M] (A + I)(A – I) – (A

2

– I) = 0 for all 5×5 matrices. However, (A + B)(A – B) – A

2

– B

2

is the zero

matrix only in the special cases when AB = BA. In general,

(A + B)(A – B) = A(A – B) + B(A – B) = AA – AB + BA – BB.

39. [M] The equality (A

T

+B

T

)=(A+B)

T

and (AB)

T

=B

T

A

T

should always be true, whereas (AB)

T

= A

T

B

T

is

very likely to be false for 4×4 matrices selected at random.

40. [M] The matrix S “shifts” the entries in a vector (a, b, c, d, e) to yield (b, c, d, e, 0). The entries in S

2

result from applying S to the columns of S, and similarly for S

3

, and so on. This explains the patterns

of entries in the powers of S:

2.2 • Solutions 93

234

00100 00010 00001

00010 00001 00000

,,

00001 00000 00000

00000 00000 00000

SSS

ªºªºªº

«»«»«»

== =

«»«»«»

¬¼¬¼¬¼

S

5

is the 5×5 zero matrix. S

6

is also the 5×5 zero matrix.

41. [M]

510

.3339 .3349 .3312 .333341 .333344 .333315

.3349 .3351 .3300 , .333344 .333350 .333306

.3312 .3300 .3388 .333315 .333306 .333379

AA

ªºª º

«»« »

==

«»« »

¬¼¬ ¼

The entries in A

20

all agree with .3333333333 to 8 or 9 decimal places. The entries in A

30

all agree

with .33333333333333 to at least 14 decimal places. The matrices appear to approach the matrix

1/3 1/3 1/3

ªº

«»

¬¼

. Further exploration of this behavior appears in Sections 4.9 and 5.2.

Note:

The MATLAB box in the Study Guide introduces basic matrix notation and operations,

including the commands that create special matrices needed in Exercises 35, 36 and elsewhere. The

Study Guide appendices treat the corresponding information for the other matrix programs.

2.2 SOLUTIONS

Notes:

The text includes the matrix inversion algorithm at the end of the section because this topic is

popular. Students like it because it is a simple mechanical procedure. However, I no longer cover it in my

classes because technology is readily available to invert a matrix whenever needed, and class time is

better spent on more useful topics such as partitioned matrices. The final subsection is independent of the

inversion algorithm and is needed for Exercises 35 and 36.

Key Exercises: 8, 11–24, 35. (Actually, Exercise 8 is only helpful for some exercises in this section.

Section 2.3 has a stronger result.) Exercises 23 and 24 are used in the proof of the Invertible Matrix

Theorem (IMT) in Section 2.3, along with Exercises 23 and 24 in Section 2.1. I recommend letting

students work on two or more of these four exercises before proceeding to Section 2.3. In this way

students participate in the proof of the IMT rather than simply watch an instructor carry out the proof.

Also, this activity will help students understand why the theorem is true.

1.

1

86 46 2 3

1

54 58 5/24

32 30

−

−−

ªº ª ºª º

==

«» « »« »

−−

−

¬¼ ¬ ¼¬ ¼

2.

1

32 5 2 52

1

85 8 3 8 3

15 16

−

−−

ªº ª ºª º

==

«» « »« »

−−

−

¬¼ ¬ ¼¬ ¼

3.

1

73 33 33 1 1

11

or

63 67 67 27/3

21 ( 18) 3

−

−− −−

ªº ªºªºª º

==−

«» «»«»« »

−− −−

−−−

¬¼ ¬¼¬¼¬ ¼

94 CHAPTER 2 • Matrix Algebra

4.

()

1

24 64 64 3/2 1

11

or

46 42 42 11/2

12 16 4

−

−−−−

ªº ªºªºª º

==

«» «»«»« »

−−−−

−−−

¬¼ ¬¼¬¼¬ ¼

5. The system is equivalent to Ax = b, where

86 2

and =

54 1

A

ª

ºªº

=

«

»«»

−

¬

¼¬¼

b, and the solution is

x = A

–1

b =

232 7

.

5/2 4 1 9

−

ªºªºªº

=

«»«»«»

−−−

¬¼¬¼¬¼

Thus x

1

= 7 and x

2

= –9.

6. The system is equivalent to Ax = b, where

73 9

and

63 4

A

−

ª

ºªº

==

«

»«»

−−

¬

¼¬¼

b, and the solution is x = A

–1

b.

To compute this by hand, the arithmetic is simplified by keeping the fraction 1/det(A) in front of the

matrix for A

–1

. (The Study Guide comments on this in its discussion of Exercise 7.) From Exercise 3,

x = A

–1

b =

339 15 5

11

674 26 26/3

33

−−− −

ªºªºªºªº

−=−=

«»«»«»«»

−

¬¼¬¼¬¼¬¼

. Thus x

1

= −5 and x

2

= 26/3.

7. a.

1

12 12 2 12 2 6 1

11

or

512 51 51 2.5.5

112 2 5 2

−

−− −

ªº ª ºª ºª º

==

«» « »« »« »

−− −

⋅−⋅

¬¼ ¬ ¼¬ ¼¬ ¼

x = A

–1

b

1

=

12 2 1 18 9

11

513 8 4

22

−− − −

ªºªºªºªº

==

«»«»«»«»

−

¬¼¬¼¬¼¬¼

. Similar calculations give

111

234

11 613

,,

525

AAA

−−−

ªº ªº ªº

===

«» «» «»

−− −

¬¼ ¬¼ ¬¼

bbb

.

b. [A b

1

b

2

b

3

b

4

] =

12 1 123

512 3 5 6 5

−

ªº

«»

−

¬¼

12 1 1 2 3 12 1 1 2 3

~~

02 8 10 4 10 01 4 5 2 5

−−

ªºªº

«»«»

−−− −−−

¬¼¬¼

10 911 613

~01 4 5 2 5

−

ªº

«»

−−−

¬¼

The solutions are

911 6 13

,,, and ,

452 5

−

ª

ºª ºª º ª º

«

»« »« » « »

−− −

¬

¼¬ ¼¬ ¼ ¬ ¼

the same as in part (a).

Note:

The Study Guide also discusses the number of arithmetic calculations for this Exercise 7, stating

that when A is large, the method used in (b) is much faster than using A

–1

.

8. Left-multiply each side of A = PBP

–1

by P

–1

:

P

–1

A = P

–1

PBP

–1

, P

–1

A = IBP

–1

, P

–1

A = BP

–1

Then right-multiply each side of the result by P:

P

–1

AP = BP

–1

P, P

–1

AP = BI, P

–1

AP = B

Parentheses are routinely suppressed because of the associative property of matrix multiplication.

9. a. True, by definition of invertible.

2.2 • Solutions 95

b. False. See Theorem 6(b).

c. False. If

11

00

Aªº

=«»

¬¼

, then ab – cd = 1 – 0 ≠ 0, but Theorem 4 shows that this matrix is not

invertible, because ad – bc = 0.

d. True. This follows from Theorem 5, which also says that the solution of Ax = b is unique, for

each b.

e. True, by the box just before Example 6.

10. a. False. The last part of Theorem 7 is misstated here.

b. True, by Theorem 6(a).

c. False. The product matrix is invertible, but the product of inverses should be in the reverse order.

See Theorem 6(b).

d. True. See the subsection “Another View of Matrix Inversion”.

e. True, by Theorem 7.

11. (The proof can be modeled after the proof of Theorem 5.) The n×p matrix B is given (but is

arbitrary). Since A is invertible, the matrix A

–1

B satisfies AX = B, because A(A

–1

B) = A A

–1

B = IB =

B. To show this solution is unique, let X be any solution of AX = B. Then, left-multiplication of each

side by A

–1

shows that X must be A

–1

B:

A

–1

(AX) = A

–1

B, IX = A

–1

B, and X = A

–1

B.

12. Left-multiply each side of the equation AD = I by A

–1

to obtain

A

–1

AD = A

–1

I, ID = A

–1

, and D = A

–1

.

Parentheses are routinely suppressed because of the associative property of matrix multiplication.

13. Left-multiply each side of the equation AB = AC by A

–1

to obtain

A

–1

AB = A

–1

AC, IB = IC, and B = C.

This conclusion does not always follow when A is singular. Exercise 10 of Section 2.1 provides a

counterexample.

14. Right-multiply each side of the equation (B – C)D = 0 by D

–1

to obtain

(B – C)DD

–1

= 0D

–1

, (B – C)I = 0, B – C = 0, and B = C.

15. If you assign this exercise, consider giving the following Hint: Use elementary matrices and imitate

the proof of Theorem 7. The solution in the Instructor’s Edition follows this hint. Here is another

solution, based on the idea at the end of Section 2.2.

Write B = [b

1

⋅ ⋅ ⋅ b

p

] and X = [u

1

⋅ ⋅ ⋅ u

p

]. By definition of matrix multiplication,

AX = [Au

1

⋅ ⋅ ⋅ Au

p

]. Thus, the equation AX = B is equivalent to the p systems:

Au

1

= b

1

, … Au

p

= b

p

Since A is the coefficient matrix in each system, these systems may be solved simultaneously,

placing the augmented columns of these systems next to A to form [A b

1

⋅ ⋅ ⋅ b

p

] = [A B]. Since A

is invertible, the solutions u

1

, …, u

p

are uniquely determined, and [A b

1

⋅ ⋅ ⋅ b

p

] must row reduce to

[I u

1

⋅ ⋅ ⋅ u

p

] = [I X]. By Exercise 11, X is the unique solution A

–1

B of AX = B.

96 CHAPTER 2 • Matrix Algebra

16. Let C = AB. Then CB

–1

= ABB

–1

, so CB

–1

= AI = A. This shows that A is the product of invertible

matrices and hence is invertible, by Theorem 6.

Note:

The Study Guide warns against using the formula (AB)

–1

= B

–1

A

–1

here, because this formula can

be used only when both A and B are already known to be invertible.

17. The box following Theorem 6 suggests what the inverse of ABC should be, namely, C

–1

B

–1

A

–1

. To

verify that this is correct, compute:

(ABC) C

–1

B

–1

A

–1

= ABCC

–1

B

–1

A

–1

= ABIB

–1

A

–1

= ABB

–1

A

–1

= AIA

–1

= AA

–1

= I

and

C

–1

B

–1

A

–1

(ABC) = C

–1

B

–1

A

–1

ABC = C

–1

B

–1

IBC = C

–1

B

–1

BC = C

–1

IC = C

–1

C = I

18. Right-multiply each side of AB = BC by B

–1

:

ABB

–1

= BCB

–1

, AI = BCB

–1

, A = BCB

–1

.

19. Unlike Exercise 18, this exercise asks two things, “Does a solution exist and what is it?” First, find

what the solution must be, if it exists. That is, suppose X satisfies the equation C

–1

(A + X)B

–1

= I.

Left-multiply each side by C, and then right-multiply each side by B:

CC

–1

(A + X)B

–1

= CI, I(A + X)B

–1

= C, (A + X)B

–1

B = CB, (A + X)I = CB

Expand the left side and then subtract A from both sides:

AI + XI = CB, A + X = CB, X = CB – A

If a solution exists, it must be CB – A. To show that CB – A really is a solution, substitute it for X:

C

–1

[A + (CB – A)]B

–1

= C

–1

[CB]B

–1

= C

–1

CBB

–1

= II = I.

Note:

The Study Guide suggests that students ask their instructor about how many details to include in

their proofs. After some practice with algebra, an expression such as CC

–1

(A + X)B

–1

could be simplified

directly to (A + X)B

–1

without first replacing CC

–1

by I. However, you may wish this detail to be included

in the homework for this section.

20. a. Left-multiply both sides of (A – AX)

–1

= X

–1

B by X to see that B is invertible because it is the

product of invertible matrices.

b. Invert both sides of the original equation and use Theorem 6 about the inverse of a product

(which applies because X

–1

and B are invertible):

A – AX = (X

–1

B)

–1

= B

–1

(X

–1

)

–1

= B

–1

X

Then A = AX + B

–1

X = (A + B

–1

)X. The product (A + B

–1

)X is invertible because A is invertible.

Since X is known to be invertible, so is the other factor, A + B

–1

, by Exercise 16 or by an

argument similar to part (a). Finally,

(A + B

–1

)

–1

A = (A + B

–1

)

–1

(A + B

–1

)X = X

Note:

This exercise is difficult. The algebra is not trivial, and at this point in the course, most students

will not recognize the need to verify that a matrix is invertible.

21. Suppose A is invertible. By Theorem 5, the equation Ax = 0 has only one solution, namely, the zero

solution. This means that the columns of A are linearly independent, by a remark in Section 1.7.

22. Suppose A is invertible. By Theorem 5, the equation Ax = b has a solution (in fact, a unique solution)

for each b. By Theorem 4 in Section 1.4, the columns of A span R

n

.

2.2 • Solutions 97

23. Suppose A is n×n and the equation Ax = 0 has only the trivial solution. Then there are no free

variables in this equation, and so A has n pivot columns. Since A is square and the n pivot positions

must be in different rows, the pivots in an echelon form of A must be on the main diagonal. Hence A

is row equivalent to the n×n identity matrix.

24. If the equation Ax = b has a solution for each b in R

n

, then A has a pivot position in each row, by

Theorem 4 in Section 1.4. Since A is square, the pivots must be on the diagonal of A. It follows that A

is row equivalent to I

n

. By Theorem 7, A is invertible.

25. Suppose

ab

Acd

=

ªº

«»

¬¼

and ad – bc = 0. If a = b = 0, then examine

1

2

00 0

0

x

cd

ªº

ª

ºªº

=

«»

«

»«»

¬

¼¬¼

¬¼

This has the

solution x

1

=

d

c

ªº

«»

−

¬¼

. This solution is nonzero, except when a = b = c = d. In that case, however, A is

the zero matrix, and Ax = 0 for every vector x. Finally, if a and b are not both zero, set x

2

=

b

a

−

ªº

«»

¬¼

.

Then

2

0

ab b abba

Acd a cbda

−−+

===

−+

ªºªºª ºªº

«»«»« »«»

¬¼¬¼¬ ¼¬¼

x

, because –cb + da = 0. Thus, x

2

is a nontrivial solution

of Ax = 0. So, in all cases, the equation Ax = 0 has more than one solution. This is impossible when A

is invertible (by Theorem 5), so A is not invertible.

26.

0

dbabdabc

cacd cbad

−−

ªºªºª º

=

«»«»« »

−−+

¬¼¬¼¬ ¼

. Divide both sides by ad – bc to get CA = I.

0

ab d b adbc

cd c a cbda

−−

ªºª ºª º

=

«»« »« »

−−+

¬¼¬ ¼¬ ¼

.

Divide both sides by ad – bc. The right side is I. The left side is AC, because

11

ab d b ab d b

cd c a cd c a

ad bc ad bc

−−

ªºª ºªº ª º

=

«»« »«» « »

−−

¬¼¬ ¼¬¼ ¬ ¼

= AC

27. a. Interchange A and B in equation (1) after Example 6 in Section 2.1: row

i

(BA) = row

i

(B)⋅A. Then

replace B by the identity matrix: row

i

(A) = row

i

(IA) = row

i

(I)⋅A.

b. Using part (a), when rows 1 and 2 of A are interchanged, write the result as

22 2

11 1

33 3

row ( ) row ( ) row ( )

AIAI

AIAIAEA

AIAI

⋅

ªºª ºªº

«»« »«»

=⋅==

«»« »«»

⋅

¬¼¬ ¼¬¼

(*)

Here, E is obtained by interchanging rows 1 and 2 of I. The second equality in (*) is a

consequence of the fact that row

i

(EA) = row

i

(E)¸A.

c. Using part (a), when row 3 of A is multiplied by 5, write the result as

11 1

22 2

33 3

row ( ) row ( ) row ( )

5row( ) 5row() 5row()

AIAI

AIAIAEA

AIAI

⋅

ªºª ºªº

«»« »«»

=⋅==

«»« »«»

⋅⋅⋅⋅

¬¼¬ ¼¬¼

Here, E is obtained by multiplying row 3 of I by 5.

98 CHAPTER 2 • Matrix Algebra

28. When row 2 of A is replaced by row

2

(A) – 3⋅row

1

(A), write the result as

11

212 1

33

row ( ) row ( )

row ( ) 3 row ( ) row ( ) 3 row ( )

row ( ) row ( )

AIA

AAIAIA

AIA

⋅

ªºªº

«»«»

−⋅ =⋅−⋅ ⋅

«»«»

⋅

¬¼¬¼

11

21 21

33

row ( ) row ( )

[row ( ) 3 row ( )] row ( ) 3 row ( )

row ( ) row ( )

IA I

IIAIIAEA

IA I

⋅

ªºªº

«»«»

=−⋅ ⋅ =−⋅ =

«»«»

⋅

¬¼¬¼

Here, E is obtained by replacing row

2

(I) by row

2

(I) – 3⋅row

1

(I).

29.

1310131010 3110 31

[] ~ ~ ~

490103410341014/31/3

AI

−− − −

ªºªºªºªº

=«»«»«»«»

−−−−

¬¼¬¼¬¼¬¼

A

–1

=

31

4/3 1/3

−

ªº

«»

−

¬¼

30.

36 1 0 121/30 1 2 1/3 0

[] ~ ~

47 0 1 47 0 1 0 1 4/3 1

AI ªºª ºª º

=«»« »« »

−−

¬¼¬ ¼¬ ¼

1

12 1/3 0 1 0 7/3 2 7/3 2

~~ .

014/3 1 0 1 4/3 1 4/3 1

A

−

−−

ªºª ºªº

=

«»« »«»

−−−

¬¼¬ ¼¬¼

31.

102100 102 100

[]314010~012310

234001038201

AI

−−

ªºª º

«»« »

=−−

«»« »

−−−

¬¼¬ ¼

10 2 100 100 831

~0 1 2 3 1 0~0 1 0 10 4 1

00 2731 002731

−

ªºªº

«»«»

−

«»«»

¬¼¬¼

1

100 8 3 1 8 3 1

~0 1 0 10 4 1 . 10 4 1

0017/23/21/2 7/23/21/2

A

−

ªºª º

«»« »

=

«»« »

¬¼¬ ¼

32.

121100 121100

[]473010~011410

264001 022201

AI

−−

ªºªº

«»«»

=−− −

«»«»

−− −

¬¼¬¼

12 1 1 0 0

~0 1 1 4 1 0

00 010 2 1

−

ªº

«»

−

«»

¬¼

. The matrix A is not invertible.

2.2 • Solutions 99

33. Let B =

100 0

110 0

011

00 11

ªº

«»

−

«»

−

«»

−

¬¼

"

#%%#

"

, and for j = 1, …, n, let a

j

, b

j

, and e

j

denote the jth columns of A, B,

and I, respectively. Note that for j = 1, …, n – 1, a

j

– a

j+1

= e

j

(because a

j

and a

j+1

have the same

entries except for the jth row), b

j

= e

j

– e

j+1

and a

n

= b

n

= e

n

.

To show that AB = I, it suffices to show that Ab

j

= e

j

for each j. For j = 1, …, n – 1,

Ab

j

= A(e

j

– e

j+1

) = Ae

j

– Ae

j+1

= a

j

– a

j+1

= e

j

and Ab

n

= Ae

n

= a

n

= e

n

. Next, observe that a

j

= e

j

+ ⋅ ⋅ ⋅ + e

n

for each j. Thus,

Ba

j

= B(e

j

+ ⋅ ⋅ ⋅ + e

n

) = b

j

+ ⋅ ⋅ ⋅ + b

n

= (e

j

– e

j+1

) + (e

j+1

– e

j+2

) + ⋅ ⋅ ⋅ + (e

n–1

– e

n

) + e

n

= e

j

This proves that BA = I. Combined with the first part, this proves that B = A

–1

.

Note:

Students who do this problem and then do the corresponding exercise in Section 2.4 will appreciate

the Invertible Matrix Theorem, partitioned matrix notation, and the power of a proof by induction.

34. Let

A =

100 0 1 0 0 0

220 0 1 1/2 0

, and

33 3 0 0 1/21/3

00 1/(1)1/

B

nn n n nn

ªºª º

«»« »

−

«»« »

=−

«»« »

−−

¬¼¬ ¼

""

#%# #%%#

"

and for j = 1, …, n, let a

j

, b

j

, and e

j

denote the jth columns of A, B, and I, respectively. Note that for

j = 1, …, n–1, a

j

= je

j

+(j+1)e

j+1

⋅ ⋅ ⋅ +n e

n

, a

n

= n e

n

, b

j

=

1

1()

jj

j

+

−ee , and

1.

nn

n

=be

To show that AB = I, it suffices to show that Ab

j

= e

j

for each j. For j = 1, …, n–1,

Ab

j

= A

1

1()

jj

j

+

§·

−

¨¸

©¹

ee =

1

1()

jj

j

+

−aa

=

()

jj1 j 1 jj

11

(1e ) – ((1)e ) .

nn

jj n j n j

jj

++

ªº

++ ++ + ++ = =

¬¼

ee eee!!

Also, Ab

n

= 11

nnn

Ann

§·

==

¨¸

©¹

eae.

Moreover,

()

jj1 1

1B (1)

jnj j n

BjBj nBj j n

++

=++ ++ =++ ++ae e eb b b!!

=

jj+1 j+1j+2 1 j

()( )( ) .

nnn−

−+−++ −+=ee e e e e e e!

which proves that BA = I. Combined with the first part, this proves that B = A

–1

.

Note:

If you assign Exercise 34, you may wish to supply a hint using the notation from Exercise 33:

Express each column of A in terms of the columns e

1

, …, e

n

of the identity matrix. Do the same for B.

100 CHAPTER 2 • Matrix Algebra

35. Row reduce [A e

3

]:

1730 1730 1730 1030 1003

215 6 0~0 1 00~01 00~010 0~010 0

1321 0411 0011 0011 0011

−− −

ªºªºªºªºªº

«»«»«»«»«»

−− − − −

¬¼¬¼¬¼¬¼¬¼

Answer: The third column of A

–1

is

3

0.

1

ªº

«»

−

¬¼

36. [M] Write B = [A F], where F consists of the last two columns of I

3

, and row reduce:

B =

25 9 27 0 0

536 185 537 1 0

154 52 143 0 1

−−−

ªº

«»

¬¼

100 .1126 .1559

~0 1 0 .5611 1.0077

001 .0828 .1915

−

ªº

«»

−

«»

−

¬¼

The last two columns of A

–1

are

.1126 .1559

~.56111.0077

.0828 .1915

−

ªº

«»

−

«»

−

¬¼

37. There are many possibilities for C, but C =

11 1

11 0

−

ª

º

«

»

−

¬

¼

is the only one whose entries are 1, –1,

and 0. With only three possibilities for each entry, the construction of C can be done by trial and

error. This is probably faster than setting up a system of 4 equations in 6 unknowns. The fact that A

cannot be invertible follows from Exercise 25 in Section 2.1, because A is not square.

38. Write AD = A[d

1

d

2

] = [Ad

1

Ad

2

]. The structure of A shows that D =

11

01

00

ª

º

«

»

«

»

«

»

«

»

¬

¼

and D =

10

11

01

ªº

«»

¬¼

are

two possibilities. There are 9 possible answers. However, there is no 4×2 matrix C such that CA =

I

4

. If this were true, then CAx would equal x for all x in R

4

. This cannot happen because the columns

of A are linearly dependent and so Ax = 0 for some nonzero vector x. For such an x,

CAx = C(0) = 0. An alternate justification would be to cite Exercise 23 or 25 in Section 2.1.

39. y = Df =

.011 .003 .001 40 .62

.003 .009 .003 50 .66

.001 .003 .011 30 .52

ªºªºªº

«»«»«»

=

«»«»«»

¬¼¬¼¬¼

. The deflections are .62 in., .66 in., and .52 in. at points

1, 2, and 3, respectively.

40. [M] The stiffness matrix is D

–1

. Use an “inverse” command to produce

2.3 • Solutions 101

D

–1

=

310

100 14 1

3013

−

ªº

«»

−−

«»

−

¬¼

To find the forces (in pounds) required to produce a deflection of .04 cm at point 3, most students

will use technology to solve Df = (0, 0, .04) and obtain (0, –4/3, 4).

Here is another method, based on the idea suggested in Exercise 42. The first column of D

–1

lists the

forces required to produce a deflection of 1 in. at point 1 (with zero deflection at the other points).

Since the transformation y 6 D

–1

y is linear, the forces required to produce a deflection of .04 cm at

point 3 is given by .04 times the third column of D

–1

, namely (.04)(100/3) times (0, –1, 3), or (0, –

4/3, 4) pounds.

41. To determine the forces that produce deflections of .07, .12, .16, and .12 cm at the four points on the

beam, use technology to solve Df = y, where y = (.07, .12, .16, .12). The forces at the four points are

.95, 6.19, 11.43, and 3.81 newtons, respectively.

42. [M] To determine the forces that produce a deflection of .22 cm at the second point on the beam, use

technology to solve Df = y, where y = (0, .22, 0, 0). The forces at the four points are –10.476,

31.429,

–10.476, and 0 newtons, respectively (to three significant digits). These forces are .22 times the

entries in the second column of D

–1

. Reason: The transformation

1

D−

yy6

is linear, so the forces

required to produce a deflection of .22 cm at the second point are .22 times the forces required to

produce a deflection of 1 cm at the second point. These forces are listed in the second column of D

–1

.

Another possible discussion: The solution of Dx = (0, 1, 0, 0) is the second column of D

–1

.

Multiply both sides of this equation by .22 to obtain D(.22x) = (0, .22, 0, 0). So .22x is the solution

of Df = (0, .22, 0, 0). (The argument uses linearity, but students may not mention this.)

Note:

The Study Guide suggests using gauss, swap, bgauss, and scale to reduce [A I]

because I prefer to postpone the use of ref (or rref) until later. If you wish to introduce ref now,

see the Study Guide’s technology notes for Sections 2.8 or 4.3. (Recall that Sections 2.8 and 2.9 are only

covered when an instructor plans to skip Chapter 4 and get quickly to eigenvalues.)

2.3 SOLUTIONS

Notes:

This section ties together most of the concepts studied thus far. With strong encouragement from

an instructor, most students can use this opportunity to review and reflect upon what they have learned,

and form a solid foundation for future work. Students who fail to do this now usually struggle throughout

the rest of the course. Section 2.3 can be used in at least three different ways.

(1) Stop after Example 1 and assign exercises only from among the Practice Problems and Exercises

1to 28. I do this when teaching “Course 3” described in the text's “Notes to the Instructor. ” If you did not

cover Theorem 12 in Section 1.9, omit statements (f) and (i) from the Invertible Matrix Theorem.

(2) Include the subsection “Invertible Linear Transformations” in Section 2.3, if you covered Section

1.9. I do this when teaching “Course 1” because our mathematics and computer science majors take this

class. Exercises 29–40 support this material.

(3) Skip the linear transformation material here, but discusses the condition number and the

Numerical Notes. Assign exercises from among 1–28 and 41–45, and perhaps add a computer project on

102 CHAPTER 2 • Matrix Algebra

the condition number. (See the projects on our web site.) I do this when teaching “Course 2” for our

engineers.

The abbreviation IMT (here and in the Study Guide) denotes the Invertible Matrix Theorem (Theorem

8).

1. The columns of the matrix

57

36

ªº

«»

−−

¬¼

are not multiples, so they are linearly independent. By (e) in

the IMT, the matrix is invertible. Also, the matrix is invertible by Theorem 4 in Section 2.2 because

the determinant is nonzero.

2. The fact that the columns of

42

63

−

ªº

«»

−

¬¼

are multiples of each other is one way to show that this matrix

is not invertible. Another is to check the determinant. In this case it is easily seen to be zero. By

Theorem 4 in Section 2.2, the matrix is not invertible.

3. Row reduction to echelon form is trivial because there is really no need for arithmetic calculations:

300 300 500

340~040~040

853 053 003

ªºªºªº

«»«»«»

−− − −

«»«»«»

−−−

¬¼¬¼¬¼

The 3×3 matrix has 3 pivot positions and hence is

invertible, by (c) of the IMT. [Another explanation could be given using the transposed matrix. But

see the note below that follows the solution of Exercise 14.]

4. The matrix

514

000

149

−

ªº

«»

¬¼

cannot row reduce to the identity matrix since it already contains a row of

zeros. Hence the matrix is not invertible (or singular) by (b) in the IMT.

5. The matrix

30 3

20 4

40 7

−

ªº

«»

−

¬¼

obviously has linearly dependent columns (because one column is zero),

and so the matrix is not invertible (or singular) by (e) in the IMT.

6.

136 13 6 136 13 6

043~04 3~043~04 3

360 0318 016 0021

−− − − −− − −

ªºªºªºªº

«»«»«»«»

−−− −

¬¼¬¼¬¼¬¼

The matrix is invertible because it is row equivalent to the identity matrix.

7.

1301 1301 1301

3583 0480 0480

~~

2632 0030 0030

0121 0121 0001

−− −− −−

ªºªºªº

«»«»«»

−− −

«»«»«»

−−

«»«»«»

−−

«»«»«»

¬¼¬¼¬¼

The 4×4 matrix has four pivot positions and so is invertible by (c) of the IMT.

2.3 • Solutions 103

8. The 4×4 matrix

3474

0146

0028

0001

ªº

«»

¬¼

is invertible because it has four pivot positions, by (c) of the IMT.

9. [M] Using technology,

4037 1000

6999 0100

~

751019 0010

124 1 0001

−−

ªºªº

«»«»

−

«»«»

−

«»«»

−−

¬¼¬¼

.

The 4×4 matrix is invertible because it has four pivot positions, by (c) of the IMT.

10. [M]

531 7 9 5 3 1 7 9

642 8 8 0 .4 .8 .4 18.8

~

75310 9 0 .81.6 .2 3.6

964 9 5 0 .62.2 21.6 21.2

85211 4 0 .2 .4 .2 10.4

ªºª º

«»« »

−−−

«»« »

−

«»« »

−− − −

«»« »

−−

¬¼¬ ¼

531 7 9 531 7 9

0 .4 .8 .4 18.8 0 .4 .8 .4 18.8

~~

000 1 34 001 21 7

001 21 7 000 1 34

000 0 1 000 0 1

ªºªº

«»«»

−− −−

«»«»

−

«»«»

−

«»«»

−−

¬¼¬¼

The 5×5 matrix is invertible because it has five pivot positions, by (c) of the IMT.

11. a. True, by the IMT. If statement (d) of the IMT is true, then so is statement (b).

b. True. If statement (h) of the IMT is true, then so is statement (e).

c. False. Statement (g) of the IMT is true only for invertible matrices.

d. True, by the IMT. If the equation Ax = 0 has a nontrivial solution, then statement (d) of the IMT

is false. In this case, all the lettered statements in the IMT are false, including statement (c),

which means that A must have fewer than n pivot positions.

e. True, by the IMT. If A

T

is not invertible, then statement (1) of the IMT is false, and hence

statement (a) must also be false.

12. a. True. If statement (k) of the IMT is true, then so is statement ( j). Use the first box after the IMT.

b. False. Notice that (i) if the IMT uses the work onto rather than the word into.

c. True. If statement (e) of the IMT is true, then so is statement (h).

d. False. Since (g) if the IMT is true, so is (f).

e. False, by the IMT. The fact that there is a b in

n

such that the equation Ax = b is consistent, does

not imply that statement (g) of the IMT is true, and hence there could be more than one solution.

Note:

The solutions below for Exercises 13–30 refer mostly to the IMT. In many cases, however, part or

all of an acceptable solution could also be based on various results that were used to establish the IMT.

104 CHAPTER 2 • Matrix Algebra

13. If a square upper triangular n×n matrix has nonzero diagonal entries, then because it is already in

echelon form, the matrix is row equivalent to I

n

and hence is invertible, by the IMT. Conversely, if

the matrix is invertible, it has n pivots on the diagonal and hence the diagonal entries are nonzero.

14. If A is lower triangular with nonzero entries on the diagonal, then these n diagonal entries can be

used as pivots to produce zeros below the diagonal. Thus A has n pivots and so is invertible, by the

IMT. If one of the diagonal entries in A is zero, A will have fewer than n pivots and hence be

singular.

Notes:

For Exercise 14, another correct analysis of the case when A has nonzero diagonal entries is to

apply the IMT (or Exercise 13) to A

T

. Then use Theorem 6 in Section 2.2 to conclude that since A

T

is

invertible so is its transpose, A. You might mention this idea in class, but I recommend that you not spend

much time discussing A

T

and problems related to it, in order to keep from making this section too lengthy.

(The transpose is treated infrequently in the text until Chapter 6.)

If you do plan to ask a test question that involves A

T

and the IMT, then you should give the students

some extra homework that develops skill using A

T

. For instance, in Exercise 14 replace “columns” by

“rows.” Also, you could ask students to explain why an n×n matrix with linearly independent columns

must also have linearly independent rows.

15. Part (h) of the IMT shows that a 4×4 matrix cannot be invertible when its columns do not span R

4

.

16. If A is invertible, so is A

T

, by (l) of the IMT. By (e) of the IMT applied to A

T

, the columns of A

T

are

linearly independent.

17. If A has two identical columns then its columns are linearly dependent. Part (e) of the IMT shows that

A cannot be invertible.

18. If A contains two identical rows, then it cannot be row reduced to the identity because subtracting

one row from the other creates a row of zeros. By (b) of the IMT, such a matrix cannot be invertible.

19. By (e) of the IMT, D is invertible. Thus the equation Dx = b has a solution for each b in R

7

, by (g) of

the IMT. Even better, the equation Dx = b has a unique solution for each b in R

7

, by Theorem 5 in

Section 2.2. (See the paragraph following the proof of the IMT.)

20. By (g) of the IMT, A is invertible. Hence, each equation Ax = b has a unique solution, by Theorem 5

in Section 2.2. This fact was pointed out in the paragraph following the proof of the IMT.

21. The matrix C cannot be invertible, by Theorem 5 in Section 2.2 or by the box following the IMT. So

(h) of the IMT is false and the columns of C do not span R

n

.

22. By the box following the IMT, E and F are invertible and are inverses. So FE = I = EF, and so E and

F commute.

23. Statement (g) of the IMT is false for F, so statement (d) is false, too. That is, the equation Fx = 0 has

a nontrivial solution.

24. Statement (b) of the IMT is false for G, so statements (e) and (h) are also false. That is, the columns

of G are linearly dependent and the columns do not span R

n

.

25. Suppose that A is square and AB = I. Then A is invertible, by the (k) of the IMT. Left-multiplying

each side of the equation AB = I by A

–1

, one has

A

–1

AB = A

–1

I, IB = A

–1

, and B = A

–1

.

2.3 • Solutions 105

By Theorem 6 in Section 2.2, the matrix B (which is A

–1

) is invertible, and its inverse is (A

–1

)

–1

,

which is A.

26. If the columns of A are linearly independent, then since A is square, A is invertible, by the IMT. So

A

2

, which is the product of invertible matrices, is invertible. By the IMT, the columns of A

2

span R

n

.

27. Let W be the inverse of AB. Then ABW = I and A(BW) = I. Since A is square, A is invertible, by (k) of

the IMT.

Note:

The Study Guide for Exercise 27 emphasizes here that the equation A(BW) = I, by itself, does not

show that A is invertible. Students are referred to Exercise 38 in Section 2.2 for a counterexample.

Although there is an overall assumption that matrices in this section are square, I insist that my students

mention this fact when using the IMT. Even so, at the end of the course, I still sometimes find a student

who thinks that an equation AB = I implies that A is invertible.

28. Let W be the inverse of AB. Then WAB = I and (WA)B = I. By (j) of the IMT applied to B in place of

A, the matrix B is invertible.

29. Since the transformation

A

xx6

is one-to-one, statement (f) of the IMT is true. Then (i) is also true

and the transformation Axx6 does map R

n

onto R

n

. Also, A is invertible, which implies that the

transformation Axx6 is invertible, by Theorem 9.

30. Since the transformation Axx6 is not one-to-one, statement (f) of the IMT is false. Then (i) is also

false and the transformation Axx6 does not map R

n

onto R

n

. Also, A is not invertible, which

implies that the transformation Axx6 is not invertible, by Theorem 9.

31. Since the equation Ax = b has a solution for each b, the matrix A has a pivot in each row (Theorem 4

in Section 1.4). Since A is square, A has a pivot in each column, and so there are no free variables in

the equation Ax = b, which shows that the solution is unique.

Note:

The preceding argument shows that the (square) shape of A plays a crucial role. A less revealing

proof is to use the “pivot in each row” and the IMT to conclude that A is invertible. Then Theorem 5 in

Section 2.2 shows that the solution of Ax = b is unique.

32. If Ax = 0 has only the trivial solution, then A must have a pivot in each of its n columns. Since A is

square (and this is the key point), there must be a pivot in each row of A. By Theorem 4 in Section

1.4, the equation Ax = b has a solution for each b in R

n

.

Another argument: Statement (d) of the IMT is true, so A is invertible. By Theorem 5 in Section

2.2, the equation Ax = b has a (unique) solution for each b in R

n

.

33. (Solution in Study Guide) The standard matrix of T is

59

,

47

A

−

ª

º

=

«

»

−

¬

¼

which is invertible because

det A ≠ 0. By Theorem 9, the transformation T is invertible and the standard matrix of T

–1

is A

–1

.

From the formula for a 2×2 inverse,

1

79

.

45

A

−

ª

º

=

«

»

¬

¼

So

()

1

12 1 2 1 2

2

79

(, ) 7 9,4 5

45

x

Txx x x x x

x

−

ªº

==++

«»«»

¬¼¬¼

106 CHAPTER 2 • Matrix Algebra

34. The standard matrix of T is

28

,

27

A

−

ªº

=«»

−

¬¼

which is invertible because det A = -2 ≠ 0. By Theorem

9, T is invertible, and

1()

T

−

x = Bx, where

1

78

1

22

2

BA

−

ª

º

==−

«

»

¬

¼

. Thus

1

12 1 2 1 2

2

78

17

(, ) 4,

22

x

Txx x x xx

x

−

ªº

ªº §·

=−=−− −−

¨¸

«»

«» ©¹

¬¼

35. (Solution in Study Guide) To show that T is one-to-one, suppose that T(u) = T(v) for some vectors u

and v in R

n

. Then S(T(u)) = S(T(v)), where S is the inverse of T. By Equation (1), u = S(T(u)) and

S(T(v)) = v, so u = v. Thus T is one-to-one. To show that T is onto, suppose y represents an arbitrary

vector in R

n

and define x = S(y). Then, using Equation (2), T(x) = T(S(y)) = y, which shows that T

maps R

n

onto R

n

.

Second proof: By Theorem 9, the standard matrix A of T is invertible. By the IMT, the columns of A

are linearly independent and span R

n

. By Theorem 12 in Section 1.9, T is one-to-one and maps R

n

onto R

n

.

36. Let A be the standard matrix of T. By hypothesis, T is not a one-to-one mapping. So, by Theorem 12

in Section 1.9, the standard matrix A of T has linearly dependent columns. Since A is square, the

columns of A do not span R

n

. By Theorem 12, again, T cannot map R

n

onto R

n

.

37. Let A and B be the standard matrices of T and U, respectively. Then AB is the standard matrix of the

mapping (())TUxx6, because of the way matrix multiplication is defined (in Section 2.1). By

hypothesis, this mapping is the identity mapping, so AB = I. Since A and B are square, they are

invertible, by the IMT, and B = A

–1

. Thus, BA = I. This means that the mapping (())UTxx6is the

identity mapping, i.e., U(T(x)) = x for all x in R

n

.

38. Given any v in R

n

, we may write v = T(x) for some x, because T is an onto mapping. Then, the

assumed properties of S and U show that S(v) = S(T(x)) = x and U(v) = U(T(x)) = x. So S(v) and U(v)

are equal for each v. That is, S and U are the same function from R

n

into R

n

.

39. If T maps R

n

onto R

n

, then the columns of its standard matrix A span R

n

, by Theorem 12 in Section

1.9. By the IMT, A is invertible. Hence, by Theorem 9 in Section 2.3, T is invertible, and A

–1

is the

standard matrix of T

–1

. Since A

–1

is also invertible, by the IMT, its columns are linearly independent

and span R

n

. Applying Theorem 12 in Section 1.9 to the transformation T

–1

, we conclude that T

–1

is a

one-to-one mapping of R

n

onto R

n

.

40. Given u, v in

n

, let x = S(u) and y = S(v). Then T(x)=T(S(u)) = u and T(y) = T(S(v)) = v, by

equation (2). Hence

()(()())

(( )) Because islinear

By equation (1)

() ()

SSTT

ST T

SS

+= +

=+

uv x y

xy

uv

So, S preserves sums. For any scalar r,

2.3 • Solutions 107

() (()) (()) Becauseislinear

Byequation (1)

()

Sr SrT STr T

r

rS

==

=

ux x

x

u

So S preserves scalar multiples. Thus S is a linear transformation.

41. [M] a. The exact solution of (3) is x

1

= 3.94 and x

2

= .49. The exact solution of (4) is x

1

= 2.90 and

x

2

= 2.00.

b. When the solution of (4) is used as an approximation for the solution in (3) , the error in using the

value of 2.90 for x

1

is about 26%, and the error in using 2.0 for x

2

is about 308%.

c. The condition number of the coefficient matrix is 3363. The percentage change in the solution

from (3) to (4) is about 7700 times the percentage change in the right side of the equation. This is

the same order of magnitude as the condition number. The condition number gives a rough

measure of how sensitive the solution of Ax = b can be to changes in b. Further information about

the condition number is given at the end of Chapter 6 and in Chapter 7.

Note:

See the Study Guide’s MATLAB box, or a technology appendix, for information on condition

number. Only the TI-83+ and TI-89 lack a command for this.

42. [M] MATLAB gives cond(A) 10, which is approximately 10

1

. If you make several trials with

MATLAB, which records 16 digits accurately, you should find that x and x

1

agree to at least 14 or 15

significant digits. So about 1 significant digit is lost. Here is the result of one experiment. The

vectors were all computed to the maximum 16 decimal places but are here displayed with only four

decimal places:

.9501

.2311

rand(4,1) .6068

.4860

ª

º

«

»

«

»

==

«

»

«

»

¬

¼

x

, b = Ax =

1.4219

6.2149

20.7973

1.4535

−

ª

º

«

»

«

»

«

»

«

»

¬

¼

. The MATLAB solution is x

1

= A\b =

.9501

.2311

.6068

.4860

ªº

«»

¬¼

.

However, x – x

1

=

.2220

0

.1665

−

ªº

«»

−

«»

−

¬¼

×10

–15

. The computed solution x

1

is accurate to about

14 decimal places.

43. [M] MATLAB gives cond(A) = 69,000. Since this has magnitude between 10

4

and 10

5

, the

estimated accuracy of a solution of Ax = b should be to about four or five decimal places less than

the 16 decimal places that MATLAB usually computes accurately. That is, one should expect the

solution to be accurate to only about 11 or 12 decimal places. Here is the result of one experiment.

The vectors were all computed to the maximum 16 decimal places but are here displayed with only

four decimal places:

x = rand(5,1) =

.8214

.4447

.6154

.7919

.9218

ª

º

«

»

«

»

«

»

«

»

«

»

«

»

¬

¼

, b = Ax =

19.8965

6.8991

26.0354

0.7861

22.4242

ª

º

«

»

«

»

«

»

«

»

«

»

«

»

¬

¼

. The MATLAB solution is x

1

= A\b =

.8214

.4447

.6154

.7919

.9218

ªº

«»

¬¼

.

108 CHAPTER 2 • Matrix Algebra

However, x – x

1

=

11

1679

.3578

10

.1775

.0084

.0002

−

ªº

«»

×

−

«»

−

«»

¬¼

. The computed solution x

1

is accurate to about 11 decimal

places.

44. [M] Solve Ax = (0, 0, 0, 0, 1). MATLAB shows that

5

cond( ) 4.8 10 .A≈×

Since MATLAB

computes numbers accurately to 16 decimal places, the entries in the computed value of x should be

accurate to at least 11 digits. The exact solution is (630, –12600, 56700, –88200, 44100).

45. [M] Some versions of MATLAB issue a warning when asked to invert a Hilbert matrix of order 12

or larger using floating-point arithmetic. The product AA

–1

should have several off-diagonal entries

that are far from being zero. If not, try a larger matrix.

Note:

All matrix programs supported by the Study Guide have data for Exercise 45, but only MATLAB

and Maple have a single command to create a Hilbert matrix.

Notes:

The Study Guide for Section 2.3 organizes the statements of the Invertible Matrix Theorem in a

table that imbeds these ideas in a broader discussion of rectangular matrices. The statements are arranged

in three columns: statements that are logically equivalent for any m×n matrix and are related to existence

concepts, those that are equivalent only for any n×n matrix, and those that are equivalent for any n×p

matrix and are related to uniqueness concepts. Four statements are included that are not in the text’s

official list of statements, to give more symmetry to the three columns. You may or may not wish to

comment on them.

I believe that students cannot fully understand the concepts in the IMT if they do not know the correct

wording of each statement. (Of course, this knowledge is not sufficient for understanding.) The Study

Guide’s Section 2.3 has an example of the type of question I often put on an exam at this point in the

course. The section concludes with a discussion of reviewing and reflecting, as important steps to a

mastery of linear algebra.

2.4 SOLUTIONS

Notes:

Partitioned matrices arise in theoretical discussions in essentially every field that makes use of

matrices. The Study Guide mentions some examples (with references).

Every student should be exposed to some of the ideas in this section. If time is short, you might omit

Example 4 and Theorem 10, and replace Example 5 by a problem similar to one in Exercises 1–10. (A

sample replacement is given at the end of these solutions.) Then select homework from Exercises 1–13,

15, and 21–24.

The exercises just mentioned provide a good environment for practicing matrix manipulation. Also,

students will be reminded that an equation of the form AB = I does not by itself make A or B invertible.

(The matrices must be square and the IMT is required.)

2.4 • Solutions 109

1. Apply the row-column rule as if the matrix entries were numbers, but for each product always write

the entry of the left block-matrix on the left.

000IABIACIBD A B

EICD EAICEBID EACEBD

++

ªºª ºª ºª º

==

«»« »« »« »

++ ++

¬¼¬ ¼¬ ¼¬ ¼

2. Apply the row-column rule as if the matrix entries were numbers, but for each product always write

the entry of the left block-matrix on the left.

000

EPQEPREQSEPEQ

FRS PFR QFS FRFS

++

ªºªºª ºª º

==

«»«»« »« »

++

¬¼¬¼¬ ¼¬ ¼

3. Apply the row-column rule as if the matrix entries were numbers, but for each product always write

the entry of the left block-matrix on the left.

000

IAB AIC BID CD

ICDIACIBDAB

++

ªºª ºª ºª º

==

«»« »« »« »

++

¬¼¬ ¼¬ ¼¬ ¼

4. Apply the row-column rule as if the matrix entries were numbers, but for each product always write

the entry of the left block-matrix on the left.

000

IWXIWYIXZ W X

EIY Z EWIY EXIZ EWY EXZ

++

ªºªºª ºª º

==

«»«»« »« »

−−+−+−+−+

¬¼¬¼¬ ¼¬ ¼

5. Compute the left side of the equation:

00

0000

ABI AIBX A BY

CXYCIXCY

++

ªºª ºª º

=

«»« »« »

++

¬¼¬ ¼¬ ¼

Set this equal to the right side of the equation:

00

so that

00 00

ABX BY I ABX BY I

CZ CZ

++==

ªºªº

=

«»«» ==

¬¼¬¼

Since the (2, 1) blocks are equal, Z = C. Since the (1, 2) blocks are equal, BY = I. To proceed further,

assume that B and Y are square. Then the equation BY =I implies that B is invertible, by the IMT, and

Y = B

–1

. (See the boxed remark that follows the IMT.) Finally, from the equality of the (1, 1) blocks,

BX = –A, B

–1

BX = B

–1

(–A), and X = –B

–1

A.

The order of the factors for X is crucial.

Note:

For simplicity, statements (j) and (k) in the Invertible Matrix Theorem involve square matrices

C and D. Actually, if A is n×n and if C is any matrix such that AC is the n×n identity matrix, then C must

be n×n, too. (For AC to be defined, C must have n rows, and the equation AC = I implies that C has n

columns.) Similarly, DA = I implies that D is n×n. Rather than discuss this in class, I expect that in

Exercises 5–8, when students see an equation such as BY = I, they will decide that both B and Y should be

square in order to use the IMT.

6. Compute the left side of the equation:

00 000 0

0

XA XABXCXA

YZBC YAZBY ZC YAZBZC

++

ªºªºª ºª º

==

«»«»« »« »

++ +

¬¼¬¼¬ ¼¬ ¼

Set this equal to the right side of the equation:

110 CHAPTER 2 • Matrix Algebra

00 00

so that

00

XA IXA I

YA ZB ZC I YA ZB ZC I

==

ªºªº

=

«»«»

++= =

¬¼¬¼

To use the equality of the (1, 1) blocks, assume that A and X are square. By the IMT, the equation

XA =I implies that A is invertible and X = A

–1

. (See the boxed remark that follows the IMT.)

Similarly, if C and Z are assumed to be square, then the equation ZC = I implies that C is invertible,

by the IMT, and Z = C

–1

. Finally, use the (2, 1) blocks and right-multiplication by A

–1

:

YA = –ZB = –C

–1

B, YAA

–1

= (–C

–1

B)A

–1

, and Y = –C

–1

BA

–1

The order of the factors for Y is crucial.

7. Compute the left side of the equation:

00 00 00

00

000

AZ

XXA B XZ I

YI YA IB YZ II

BI

ªº

++ ++

ªº ª º

«»

=

«» « »

«» ++ ++

¬¼ ¬ ¼

«»

¬¼

Set this equal to the right side of the equation:

00

so that

00

XA XZ I XA I XZ

YA B YZ I I YA B YZ I I

==

ªºªº

=

«»«»

++ +=+=

¬¼¬¼

To use the equality of the (1, 1) blocks, assume that A and X are square. By the IMT, the equation XA

=I implies that A is invertible and X = A

–1

. (See the boxed remark that follows the IMT) Also, X is

invertible. Since XZ = 0, X

–1

XZ = X

–1

0 = 0, so Z must be 0. Finally, from the equality of the (2, 1)

blocks, YA = –B. Right-multiplication by A

–1

shows that YAA

–1

= –BA

–1

and Y = –BA

–1

. The order of

the factors for Y is crucial.

8. Compute the left side of the equation:

00

000 00000

ABXY Z AXB AYB AZBI

IIXIYIZII

+++

ªºª ºª º

=

«»« »« »

++ +

¬¼¬ ¼¬ ¼

Set this equal to the right side of the equation:

00

00 00

AX AY AZ B I

II

+

ªºªº

=

«»«»

¬¼¬¼

To use the equality of the (1, 1) blocks, assume that A and X are square. By the IMT, the equation XA

=I implies that A is invertible and X = A

–1

. (See the boxed remark that follows the IMT. Since AY =

0, from the equality of the (1, 2) blocks, left-multiplication by A

–1

gives A

–1

AY = A

–1

0 = 0, so Y = 0.

Finally, from the (1, 3) blocks, AZ = –B. Left-multiplication by A

–1

gives A

–1

AZ = A

–1

(–B), and Z = –

A

–1

B. The order of the factors for Z is crucial.

Note:

The Study Guide tells students, “Problems such as 5–10 make good exam questions. Remember to

mention the IMT when appropriate, and remember that matrix multiplication is generally not

commutative.” When a problem statement includes a condition that a matrix is square, I expect my

students to mention this fact when they apply the IMT.

2.4 • Solutions 111

9. Compute the left side of the equation:

11 12 11 21 31 12 22 32

21 21 22 21 11 21 31 21 12 22 32

31 31 32 31 11 21 31 31 12 22 32

00 00 00

000

IBBIBBBIBBB

AI BB ABIB BABIB B

AIBBABBIBABBIB

++ ++

ªºªºª º

«»«»« »

=++ ++

«»«»« »

++ ++

¬¼¬¼¬ ¼

Set this equal to the right side of the equation:

11 12 11 12

21 11 21 21 12 22 22

31 11 31 31 12 32 32

0

BBCC

AB B AB B C

ªºªº

«»«»

++=

«»«»

++

¬¼¬¼

so that

11 11 12 12

21 11 21 21 12 22 22

31 11 31 31 12 32 32

0

BC BC

AB B AB B C

==

+= +=

Since the (2,1) blocks are equal,

21 11 21 21 11 21

0and .AB B AB B+= =−

Since B

11

is invertible, right

multiplication by

11

11 21 21 11

gives .BABB

−−

=−

Likewise since the (3,1) blocks are equal,

31 11 31 31 11 31

0and .AB B AB B+= =−

Since B

11

is invertible, right multiplication by

11

11 31 31 11

gives .BABB

−−

=−

Finally, from the (2,2) entries,

11

21 12 22 22 21 21 11 22 21 11 12 22

. Since , .AB B C A BB C BB B B

−−

+= =−=−+

10. Since the two matrices are inverses,

00 00 00

0000

00

II I

AI PI I

BDIQRI I

ªºªºªº

«»«»«»

=

«»«»«»

¬¼¬¼¬¼

Compute the left side of the equation:

00 00 0 0 00 0 0000

00 000000

000

II IIPQIIRII

AI PI AIIP Q A II R A I I

BDIQRI BIDPIQB DIIRB D II

++ ++ ++

ªºªºª º

«»«»« »

=++ ++ ++

«»«»« »

++ ++ ++

¬¼¬¼¬ ¼

Set this equal to the right side of the equation:

00 00

00 0

00

II

AP I I

BDPQ DR I I

ªºªº

«»«»

+=

«»«»

++ +

¬¼¬¼

so that

00 00

000

00

II

AP I I

BDPQ DR I I

===

+= = =

++= += =

Since the (2,1) blocks are equal, 0andAP P A+= =−. Likewise since the (3, 2) blocks are equal,

0and .DR R D+= =− Finally, from the (3,1) entries, 0and .

B

DP Q Q B DP++= =−−

Since , ( )PAQBDA BDA=−=−− − =−+.

11. a. True. See the subsection Addition and Scalar Multiplication.

b. False. See the paragraph before Example 3.

112 CHAPTER 2 • Matrix Algebra

12. a. False. The both AB and BA are defined.

b. False. The R

T

and Q

T

also need to be switched.

13. You are asked to establish an if and only if statement. First, supose that A is invertible,

and let

1

DE

AFG

−

ªº

=«»

¬¼

. Then

00

BDEBDBEI

CFG CFCG I

ªºª ºª ºªº

==

«»« »« »«»

¬¼¬ ¼¬ ¼¬¼

Since B is square, the equation BD = I implies that B is invertible, by the IMT. Similarly, CG = I

implies that C is invertible. Also, the equation BE = 0 imples that

1

E

B

−

=0 = 0. Similarly F = 0.

Thus

11

1

00

BDEB

ACEG C

−

ªº

ªºª º

===

«»

«»« »

¬¼¬ ¼

«»

¬¼

(*)

This proves that A is invertible only if B and C are invertible. For the “if ” part of the statement,

suppose that B and C are invertible. Then (*) provides a likely candidate for

1

A

−

which can be used

to show that A is invertible. Compute:

11

00 0 0

00

BB BB I

CI

CCC

−−

ªºª º

ªº ªº

==

«»« »

«» «»

¬¼ ¬¼

«»« »

¬¼¬ ¼

Since A is square, this calculation and the IMT imply that A is invertible. (Don’t forget this final

sentence. Without it, the argument is incomplete.) Instead of that sentence, you could add the

equation:

11

00 0 0

00

BBBB I

CI

CCC

−−

ªº ª º

ªº ªº

==

«» « »

«» «»

¬¼ ¬¼

«» « »

¬¼ ¬ ¼

14. You are asked to establish an if and only if statement. First suppose that A is invertible. Example 5

shows that A

11

and A

22

are invertible. This proves that A is invertible only if A

11

A

22

are invertible. For

the if part of this statement, suppose that A

11

and A

22

are invertible. Then the formula in Example 5

provides a likely candidate for

1

A

−

which can be used to show that A is invertible . Compute:

111 1

111

11 11 12 11 11 12 22 12 22

11 12 11 11 12 22

111 1

1

22 22 11 12 22 22 22

22 11

11 1

11 11 12 22 12 22

11

12 22 12 22

0()

0000()

0

()

0

AA A A A A A A A

AAAAAA

AAA AAAAA

A

IAAAAAA

I

IAAAA

I

−−− −

−−−

−

−− −

−

−− −

−−

ª

º

+−+

ªº

ªº−

«

»

=

«»

+−+

«

»

«»

¬¼

¬

¼

ªº

−+

=«»

¬¼

ªº

−+

=«

¬

0

I

ªº

=

»«»

¬¼

¼

Since A is square, this calculation and the IMT imply that A is invertible.

15. The column-row expansions of G

k

and G

k+1

are:

11

...

col ( ) row ( ) col ( ) row ( )

T

kkk

TT

kk kkkk

GXX

XX X X

=

=++

2.4 • Solutions 113

and

111

11 11 111111

11 11 1

11 1

...

col ( )row ( ) col ( ) row ( ) col ( ) row ( )

...

col ( ) row ( ) col ( ) row ( ) col ( ) row ( )

col ( ) row ( )

T

kkk

TTT

kk kkkkkkkk

TT T

kk kkkkkkkk

T

kkk kk

GXX

XX X X X X

GX X

+++

++ + +++++

++ +

=

=+++

=+

since the first k columns of X

k+1

are identical to the first k columns of X

k

. Thus to update G

k

to

produce G

k+1

, the matrix col

k+1

(X

k+1

) row

k+1

()

T

k

X

should be added to G

k

.

16. Compute the right side of the equation:

11 11

11

11 11

11

0

00

00 0

AAY

A

IA IY IY

XA XA Y S

XA S

XI S I I

ª

º

ªº

ªºª ºªº ªº

==

«

»

«»«»« »«» «» +

¬¼¬ ¼¬¼ ¬¼

¬¼

¬

¼

Set this equal to the left side of the equation:

11 11

11 11 11 12 11 12

11 21

11 11 21 22 11 22

so that

AA

AAYAA AYA

XA A

XA XA Y S A A XA Y S A

==

ªºªº

=

«»«» =

++=

¬¼¬¼

Since the (1, 2) blocks are equal,

11 12.AY A=

Since A

11

is invertible, left multiplication by

1

11

A

−

gives

Y =

1

11 12

.AA

−

Likewise since the (2,1) blocks are equal, X A

11

= A

21

. Since A

11

is invertible, right

multiplication by

1

11

A

−

gives that

1

21 11

.XAA

−

=

One can check that the matrix S as given in the exercise

satisfies the equation

11 22

XA Y S A+=

with the calculated values of X and Y given above.

17. Suppose that A and A

11

are invertible. First note that

000

0

II I

XI XI I

ªºª ºªº

=

«»« »«»

−

¬¼¬ ¼¬¼

and

0

00 0

IYI Y I

II I

−

ªºª ºªº

=

«»« »«»

¬¼¬ ¼¬¼

Since the matrices

0 and 0

IIY

XI I

ªºªº

«»«»

¬¼¬¼

are square, they are both invertible by the IMT. Equation (7) may be left multipled by

1

0I

XI

−

ªº

«»

¬¼

and right multipled by

1

0

IY

I

−

ªº

«»

¬¼

to find

11

00

AI IY

A

SXI I

−−

ªºªºªº

=

«»«»«»

¬¼¬¼¬¼

Thus by Theorem 6, the matrix

11

0

A

S

ªº

«»

¬¼

is invertible as the product of invertible matrices. Finally,

Exercise 13 above may be used to show that S is invertible.

114 CHAPTER 2 • Matrix Algebra

18. Since

[]

0

,WX=x

0

0000

[ ]

TTT

T

TTT

XXXX

WW X

X

ªº ª º

==

«» « »

¬¼ ¬ ¼

x

xxxx

By applying the formula for S from Exercise 15, S may be computed:

1

00 0 0

1

00

()

(())

TTT T

TTT

m

T

SXXXX

IXXXX

M

−

=−

=

xx x x

xx

19. The matrix equation (8) in the text is equivalent to

( ) and

n

AsI B C−+= +=xu0 xuy

Rewrite the first equation as

() .

n

AsI B−=−xu

When

n

AsI−

is invertible,

11

()()()

nn

AsI B AsI B

−−

=−−=−−xuu

Substitute this formula for x into the second equation above:

11

(( ) ) sothat ( )

nmn

CAsI B ICAsIB

−−

−− += −− =uuy u uy,

Thus

1

(( )).

mn

ICAsI B

−

=−−yu

If

1

() ( ) ,

mn

Ws I CA sI B

−

=−−

then ().Ws=yu

The matrix W(s) is

the Schur complement of the matrix

n

AsI−

in the system matrix in equation (8)

20. The matrix in question is

n

m

ABCsI B

CI

−−

ªº

«»

−

¬¼

By applying the formula for S from Exercise 16, S may be computed:

1

()( )

()

mn

SI CABCsI B

ICABCsI B

−

=−− − −

=+ −−

21. a.

2

10 00

1010 10

2121 01

22 0(1)

A

++

ªº

ªºªº ªº

== =

«»

«»«» «»

−−

−+−

¬¼¬¼ ¬¼

¬¼

b.

2

00 000 0

0

0( )

AA A I

M

IAIA I

AA A

ªº

++

ªºªº ªº

== =

«»

«»«» «»

−−

−+−

¬¼¬¼ ¬¼

«»

¬¼

22. Let C be any nonzero 2×2 matrix. Define

2

00

0

I

MI

CI

ª

º

«

»

=

«

»

«

»

−

¬

¼

. Then

22 2 2

2

22 2 2

00 00 00 00

00000 000

00 000

II I I

MI I I I

CICICCI I

ªºªºª ºªº

«»«»« »«»

===

«»«»« »«»

−−−

¬¼¬¼¬ ¼¬¼

2.4 • Solutions 115

23. The product of two 1×1 “lower triangular” matrices is “lower triangular.” Suppose that for n = k, the

product of two k×k lower triangular matrices is lower triangular, and consider any (k+1)× (k+1)

matrices A

1

and B

1

. Partition these matrices as

11

,

TT

ab

AB

ªº ª º

==

«» « »

¬¼ ¬ ¼

00

vw

where A and B are k×k matrices, v and w are in R

k

, and a and b are scalars. Since A

1

and B

1

are lower

triangular, so are A and B. Then

11

TTT

TT T

T

ab a B

ab ab

AB AB bAAB

bA AB

ªº

ªºª º ªº

++

== =

«»

«»« » «»

+

++

«»

¬¼¬ ¼ ¬¼

¬¼

0w 0 0

00 0

vw vw

vwv0

Since A and B are k×k, AB is lower triangular. The form of A

1

B

1

shows that it, too, is lower

triangular. Thus the statement about lower triangular matrices is true for n = k +1 if it is true for n =

k. By the principle of induction, the statement is true for all n > 1.

Note:

Exercise 23 is good for mathematics and computer science students. The solution of Exercise 23 in

the Study Guide shows students how to use the principle of induction. The Study Guide also has an

appendix on “The Principle of Induction,” at the end of Section 2.4. The text presents more applications

of induction in Section 3.2 and in the Supplementary Exercises for Chapter 3.

24. Let

100 0 1 0 0 0

110 0 1 1 0 0

,

111 0 0 1 1 0

111 1 0 11

nn

AB

ªºª º

«»« »

−

«»« »

==

−

«»« »

−

¬¼¬ ¼

""

#% #%%

""

.

By direct computation A

2

B

2

= I

2

. Assume that for n = k, the matrix A

k

B

k

is I

k

, and write

11

and

TT

kk

AB

++

ªº ª º

==

«» « »

¬¼ ¬ ¼

00

vw

where v and w are in R

k

, v

T

= [1 1 ⋅ ⋅ ⋅ 1], and w

T

= [–1 0 ⋅ ⋅ ⋅ 0]. Then

11 1

1

11 1

TTT

TT T

k

kk k

T

kk k

kkk

B

AB I

AAB

++ +

ªº

ªºª º ª º

++

== ==

«»

«»« » « »

++

«»

¬¼¬ ¼ ¬ ¼

¬¼

0w 0 0

00 0

vw 0

vwv0

The (2,1)-entry is 0 because v equals the first column of A

k

., and A

k

w is –1 times the first column of

A

k

. By the principle of induction, A

n

B

n

= I

n

for all n > 2. Since A

n

and B

n

are square, the IMT shows

that these matrices are invertible, and

1

.

nn

BA

−

=

Note:

An induction proof can also be given using partitions with the form shown below. The details are

slightly more complicated.

11

and

11

kk

TT

AB

++

ªº ª º

==

«» « »

¬¼ ¬ ¼

00

vw

missing

2.4 • Solutions 117

F o r [ j=10, j<=19, j++,

A [[ i,j ]] = B [[ i-4, j-9 ]] ] ]; Colon suppresses

output.

c. To create

0

T

A

B

A

ªº

=«»

¬¼

with MATLAB, build B out of four blocks:

B = [ A z e r o s ( 2 0 , 2 0 ) ; z e r o s ( 3 0 , 3 0 ) A ’ ] ;

Another method: first enter B = A ; and then enlarge B with the command

B ( 2 1 : 5 0 , 3 1 : 5 0 ) = A ’ ;

This places A

T

in the (2, 2) block of the larger B and fills in the (1, 2) and (2, 1) blocks with zeros.

For Maple:

B : = m a t r i x ( 5 0 , 5 0 , 0 ) :

c o p y i n t o ( A , B , 1 , 1 ) :

c o p y i n t o ( t r a n s p o s e ( A ) , B , 2 1 , 3 1 ) :

For Mathematica:

B = B l o c k M a t r i x [ { { A , Z e r o M a t r i x [ 2 0 , 2 0 ] } , Z e r o M a t r i x [ 3 0 , 3 0 ] ,

Transpose[A]}} ]

27. a. [M] Construct A from four blocks, say C

11

, C

12

, C

21

, and C

22

, for example with C

11

a 30×30

matrix and C

22

a 20×20 matrix.

MATLAB: C11 = A(1:30, 1:30) + B(1:30, 1:30)

C12 = A(1:30, 31:50) + B(1:30, 31:50)

C21 = A(31:50, 1:30)+ B(31:50, 1:30)

C22 = A(31:50, 31:50) + B(31:50, 31:50)

C = [C11 C12; C21 C22]

The commands in Maple and Mathematica are analogous, but with different syntax. The first

commands are:

Maple:

C11 := submatrix(A, 1..30, 1..30) + submatrix(B, 1..30,

1..30)

Mathematica:

C11 := Take[ A, {1,30), {1,30} ] + Take[B, {1,30), {1,30}

]

b. The algebra needed comes from block matrix multiplication:

11 12 11 12 11 11 12 21 11 12 12 22

21 22 21 22 21 11 22 21 21 12 22 22

AABB ABAB ABAB

AB

AABB ABABABAB

++

ªºªºª º

==

«»«»« »

++

¬¼¬¼¬ ¼

Partition both A and B, for example with 30×30 (1, 1) blocks and 20×20 (2, 2) blocks. The four

necessary submatrix computations use syntax analogous to that shown for (a).

c. The algebra needed comes from the block matrix equation

11 11

21 22 2 2

0A

AA

ª

ºª º ª º

=

«

»« » « »

¬

¼¬ ¼ ¬ ¼

xb

, where x

1

and b

1

are in R

20

and x

2

and b

2

are in R

30

. Then A

1 1

x

1

= b

1

, which can be solved to produce x

1

.

Once x

1

is found, rewrite the equation A

21

x

1

+ A

22

x

2

= b

2

as A

22

x

2

= c, where c = b

2

– A

21

x

1

, and

solve A

22

x

2

= c for x

2

.

Notes:

The following may be used in place of Example 5:

2.4 • Solutions 118

Example 5: Use equation (*) to find formulas for X, Y, and Z in terms of A, B, and C. Mention any

assumptions you make in order to produce the formulas.

00 0XI I

YZAB CI

ªºªºªº

=

«»«»«»

¬¼¬¼¬¼

(*)

Solution:

This matrix equation provides four equations that can be used to find X, Y, and Z:

X + 0 = I, 0 = 0

YI + ZA = C, Y0 + ZB = I (Note the order of the factors.)

The first equation says that X = I. To solve the fourth equation, ZB = I, assume that B and Z are

square. In this case, the equation ZB = I implies that B and Z are invertible, by the IMT. (Actually, it

suffices to assume either that B is square or that Z is square.) Then, right-multiply each side of ZB = I

to get ZBB

–1

= IB

–1

and Z = B

–1

. Finally, the third equation is Y + ZA = C. So, Y + B

–1

A = C, and Y =

C – B

–1

A.

The following counterexample shows that Z need not be square for the equation (*) above to be true.

100 0

1000 0 1000

0100

0100 0 0100

1125

1213 1 65 10

1113

3410 1 360 1

1124

ªº

ªº ªº

«»

«» «»

«»

«» «»

«»

=

«» «»

«»

−−

«» «»

«»

−

«» «»

¬¼ ¬¼

«»

−

¬¼

Note that Z is not determined by A, B, and C, when B is not square. For instance, another Z that

works in this counterexample is

350

120

Zªº

=«»

−−

¬¼

.

2.5 SOLUTIONS

Notes:

Modern algorithms in numerical linear algebra are often described using matrix factorizations.

For practical work, this section is more important than Sections 4.7 and 5.4, even though matrix

factorizations are explained nicely in terms of change of bases. Computational exercises in this section

emphasize the use of the LU factorization to solve linear systems. The LU factorization is performed

using the algorithm explained in the paragraphs before Example 2, and performed in Example 2. The text

discusses how to build L when no interchanges are needed to reduce the given matrix to U. An appendix

in the Study Guide discusses how to build L in permuted unit lower triangular form when row

interchanges are needed. Other factorizations are introduced in Exercises 22–26.

1.

100 37 2 7

110, 02 1, 5.First,solve .

251 001 2

LU L

−− −

ªºª ºªº

«»« »«»

=−=−− ==

«»« »«»

−−

¬¼¬ ¼¬¼

byb

1007 1007

[] 1105~0102

2512 05116

L

−−

ªºªº

«»«»

=−−

«»«»

−−

¬¼¬¼

b

. The only arithmetic is in column 4

2.5 • Solutions 119

100 7 7

0 1 0 2 , so 2.

001 6 6

−−

ªºªº

«»«»

∼−=−

«»«»

¬¼¬¼

y

Next, solve Ux = y, using back-substitution (with matrix notation).

3727 3727 37019

[]02120212020 8

0016 0016 001 6

U

−−− −−− − −

ªºªºªº

«»«»«»

=−−− ∼ −−− ∼ −−

«»«»«»

−−−

¬¼¬¼¬¼

y

37019 3009 1003

~0 1 0 4 0 1 0 4 0 1 0 4

001 6 0016 0016

−−

ªºªºªº

«»«»«»

∼ ∼

«»«»«»

−−−

¬¼¬¼¬¼

, So x =

3

4.

6

ª

º

«

»

«

»

«

»

−

¬

¼

To confirm this result, row reduce the matrix [A b]:

3727 3727 3727

[]351502120212

64020104160016

A

−−− −−− −−−

ªºªºªº

«»«»«»

=− ∼ −−− ∼ −−−

«»«»«»

−−

¬¼¬¼¬¼

b

From this point the row reduction follows that of [U y] above, yielding the same result.

2.

100 2 6 4 2

210, 0 4 8, 4

011 0 0 2 6

LU

−

ªºª ºªº

«»« »«»

=−=−=−

«»« »«»

−

¬¼¬ ¼¬¼

b

. First, solve Ly = b:

100 2 1002

[] 21040100 ,

011 6 0016

L

ªºªº

«»«»

=−− ∼

«»«»

¬¼¬¼

b

so

2

0.

6

ª

º

«

»

=

«

»

«

»

¬

¼

y

Next solve Ux = y, using back-substitution (with matrix notation):

2642 26014

[]048004024

0026 0013

U

−−

ªºªº

«»«»

=− ∼ −

«»«»

−−

¬¼¬¼

y

200 22 100 11

010 6 010 6,

001 3 001 3

−−

ªºªº

«»«»

∼ −∼ −

«»«»

−−

¬¼¬¼

so

11

6.

3

−

ª

º

«

»

=−

«

»

«

»

−

¬

¼

x

To confirm this result, row reduce the matrix [A b]:

2642 2642

[]48040480

0466 0026

A

−−

ªºªº

«»«»

=−− ∼ −

«»«»

−−

¬¼¬¼

b

From this point the row reduction follows that of [U y] above, yielding the same result.

3.

100 2 42 6

210, 036, 0

311 001 6

LU

−

ªºªºªº

«»«»«»

=−=−=

«»«»«»

−

¬¼¬¼¬¼

b

. First, solve Ly = b:

120 CHAPTER 2 • Matrix Algebra

1006 100 6 1006

[] 21000101201012 ,

3116 01112 0010

L

ªºª ºªº

«»« »«»

=−∼ ∼

«»« »«»

−−−

¬¼¬ ¼¬¼

b

so

6

12 .

0

ª

º

«

»

=

«

»

«

»

¬

¼

y

Next solve Ux = y, using back-substitution (with matrix notation):

2426 2406 20010 1005

[]03612 03012 010 4 0104

0010 0010 001 0 0010

U

−− − −

ªºªºªºªº

«»«»«»«»

=−∼−∼ −∼ −

«»«»«»«»

¬¼¬¼¬¼¬¼

y

,

so

5

4.

0

−

ªº

«»

=−

«»

¬¼

x

4.

100 112 0

110, 021, 5

351 006 7

LU

−

ªºª ºªº

«»« »«»

==−− =−

«»« »«»

−−

¬¼¬ ¼¬¼

b

. First, solve Ly = b:

1000 1000 100 0

[]11050105010 5 ,

3517051700118

L

ªºªºªº

«»«»«»

=−∼ −∼ −

«»«»«»

−− −

¬¼¬¼¬¼

b

so

0

5 .

18

ªº

«»

=−

«»

−

¬¼

y

Next solve Ux = y, using back-substitution (with matrix notation):

112 0 1120 1106

[]021 5 0215 0202

00618 0013 0013

U

−−−−

ªºªºªº

«»«»«»

=−− −∼ −−−∼ − −

«»«»«»

−−

¬¼¬¼¬¼

y

1106 1005

01010101 ,

0013 0013

−− −

ªºªº

«»«»

∼∼

«»«»

¬¼¬¼

so x =

5

1

3

−

ª

º

«

»

«

»

«

»

¬

¼

.

5.

10 00 1 2 2 3 1

31 00 0 3 6 0 6

,, .

10 10 0 0 2 4 0

34 21 0 0 0 1 3

LU

−−−

ªºª ºªº

«»« »«»

−

«»« »«»

== =

«»« »«»

−

«»« »«»

−−

¬¼¬ ¼¬¼

b First solve Ly = b:

10 00 1 10 00 1

31 006 01 003

[] 10 100 00 10 1

34 21 3 04 21 6

L

ªºªº

«»«»

=∼

«»«»

−

«»«»

−− −

¬¼¬¼

b

10 00 1 1000 1

01 00 3 0100 3

,

00 10 1 0010 1

00 21 6 0001 4

ªºªº

«»«»

∼∼

«»«»

−− −

¬¼¬¼

so

1

3.

1

4

ª

º

«

»

«

»

=

«

»

«

»

−

¬

¼

y

Next solve Ux = y, using back-substitution (with matrix notation):

2.5 • Solutions 121

1223 1 122011 1220 11

03603 0360 3 0360 3

[]

00241002017 001017/2

00014 0001 4 0001 4

U

−−− −− − −− −

ªºªºªº

«»«»«»

−−−

«»«»«»

=∼∼

«»«»«»

−− −

¬¼¬¼¬¼

y

1200 6 1200 6 1000 38

0300 48 0100 16 0100 16

001017/2 001017/2 001017/2

0001 4 0001 4 0001 4

−−

ªºªºªº

«»«»«»

−−

«»«»«»

∼∼∼∼

«»«»«»

−−−

¬¼¬¼¬¼

, so x =

38

16

17 / 2

4−

ªº

«»

¬¼

.

6.

1000 1320 1

2100 03012 2

, , .

3310 0020 1

5411 0001 2

LU

ªºªºªº

«»«»«»

−−

«»«»«»

===

«»«»«»

−−−

«»«»«»

−−

¬¼¬¼¬¼

b First, solve Ly = b:

1000 1 1000 1 1000 1

21002 01000 01000

[] 3310 1 03104 00104

54112 04117 00117

L

ªºªºªº

«»«»«»

−−

«»«»«»

=∼∼

«»«»«»

−−−− −

«»«»«»

−− − −

¬¼¬¼¬¼

b

1000 1

0100 0

,

0010 4

0001 3

ªº

«»

∼

«»

−

«»

¬¼

so

1

0.

4

3

ª

º

«

»

«

»

=

«

»

−

«

»

¬

¼

y Next solve Ux = y, using back-substitution (with

matrix notation):

13 2 0 1 13 20 1 1300 3

03 012 0 03 00 36 0300 36

[]

00 2 0 4 00 20 4 0010 2

00 0 1 3 00 01 3 0001 3

U

−

ªºªºªº

«»«»«»

−−

«»«»«»

=∼∼

«»«»«»

−− − −

«»«»«»

¬¼¬¼¬¼

y

1000 33

0100 12

0010 2

0001 3

ªº

«»

−

«»

∼

«»

¬¼

, so x

33

12

2

3

ª

º

«

»

−

«

»

=

«

»

«

»

¬

¼

.

122 CHAPTER 2 • Matrix Algebra

7. Place the first pivot column of

25

34

ªº

«»

−−

¬¼

into L, after dividing the column by 2 (the pivot), then add

3/2 times row 1 to row 2, yielding U.

252 5

~

34 07/2

AU

ªºªº

==

«»«»

−−

¬¼¬¼

2

3[7/2]

ªº

«»

−

¬¼

27/2÷÷

110

,

3/2 1 3/2 1

L

ªºªº

=

«»«»

−−

¬¼¬¼

8. Row reduce A to echelon form using only row replacement operations. Then follow the algorithm in

Example 2 to find L.

64 64

12 5 0 3

AU

ªºª º

=∼ =

«»« »

−

¬¼¬ ¼

[]

6 3

12

ªº

«»

−

¬¼

63

110

,

21 21

L

÷÷−

ªºªº

=

«»«»

¬¼¬¼

9.

31 2 312 312

90 4 032~032

9914 068 004

AU

ªºªºªº

«»«»«»

=−−∼ =

«»«»«»

¬¼¬¼¬¼

[]

3

93

94

6

ªº

«»

−ªº

«» «»

«»

¬¼ ¬¼

÷3 ÷ 3 ÷ 4

2.5 • Solutions 123

110 0

31 , 310

32 1 32 1

L

ªºªº

«»«»

−=−

«»«»

¬¼¬¼

10.

50 4 50 4 504

10 2 5 ~ 0 2 3 ~ 0 2 3

10 10 16 0 10 24 0 0 9

AU

−−−

ªºªºªº

«»«»«»

=−=

«»«»«»

¬¼¬¼¬¼

[]

5

10 2

10 9

10

529

1100

21 , 210

251 251

L

−

ªº

«»

ªº

«»

¬¼

÷−÷÷

ªºªº

«»«»

−=−

«»«»

−−

¬¼¬¼

11.

372 372 372

6194 050 050

323 055 005

AU

ªºªºªº

«»«»«»

=∼∼=

«»«»«»

−−

¬¼¬¼¬¼

3

65

3

5[5]

355

1100

21 , 210

11 1 111

L

ªº

«»

ªº

«»

−

¬¼

÷÷÷

ªºªº

«»«»

=

«»«»

−−

¬¼¬¼

12. Row reduce A to echelon form using only row replacement operations. Then follow the algorithm in

Example 2 to find L. Use the last column of I

3

to make L unit lower triangular.

124 CHAPTER 2 • Matrix Algebra

232232232

4139 0 7 5 0 7 5

65401410000

2

47

614

27

1100

21 , 210

321 321

AU

L

ªºªºªº

«»«»«»

=∼∼=

«»«»«»

−

¬¼¬¼¬¼

ªº

«»

ªº

«»

−

¬¼

÷÷

ªºªº

«»«»

=

«»«»

−−

¬¼¬¼

13.

13 53 1 353 1353

1584 0 231 0231 No more pivots!

42 57 010155 0000

2475 0 231 0000

U

−− −− −−

ªºªºªº

«»«»«»

−− − −

«»«»«»

∼∼=

«»«»«»

−− −

«»«»«»

−− −−

¬¼¬¼¬¼

4

1

12

410

Use the last two columns of to make unit lower triangular.

22IL

ªº

«»

−−

ªº

«»

−

«»

−

¬¼

1–2

11000

11 1100

,

451 4510

2101 2101

L

÷÷

ªºªº

«»«»

−−

«»«»

=

«»«»

−− −−

«»«»

¬¼¬¼

14.

1315 1315 1315

520 6 31 0 51 6 0516

2114 0516 0000

17 17 010212 0000

AU

ªºªºªº

«»«»«»

=∼∼=

«»«»«»

−−−−

«»«»«»

−

¬¼¬¼¬¼

2.5 • Solutions 125

4

1

5

2

Use the last two columns of to make unit lower triangular.

10

1IL

ªº

«»ªº

«»«»

«»

−«»

«»«»

−¬¼

¬¼

15

11000

51 5100

,

211 2110

1201 120 1

L

÷÷

ªºªº

«»«»

=

«»«»

−−

«»«»

−−

¬¼¬¼

15.

20 5 2 205 2 2052

63 13 3 032 3 0323

49 1617 09613 0004

AU

ªºªºªº

«»«»«»

=−−−∼ ∼ =

«»«»«»

¬¼¬¼¬¼

[]

2

63

44

9

234

1100

31 , 310

231 23 1

L

ªº

«»

−ªº

«»

¬¼

÷÷÷

ªºªº

«»«»

−=−

«»«»

¬¼¬¼

16.

23 4 234 234

48 7 021 021

~~

6514 042 000

6912 000 000

8 6 19 0 6 3 0 0 0

AU

−−−

ªºªºªº

«»«»«»

−−

«»«»«»

==

−

«»«»«»

−−

«»«»«»

−

¬¼¬¼¬¼

5

2

42

64

60

86Use the last three columns of to make unit lower triangular.IL

ªº

«»

−ªº

«»«»

−

«»«»

«»

¬¼¬¼

126 CHAPTER 2 • Matrix Algebra

22

110000

21 21000

,

321 32100

300 1 30010

43001 43001

L

÷÷

ªºªº

«»«»

−−

«»«»

=

«»«»

−−

«»«»

¬¼¬¼

17.

100 2 6 4

210, 0 4 8

011 0 0 2

LU

−

ªºª º

«»« »

=−=−

«»« »

−

¬¼¬ ¼

To find L

–1

, use the method of Section 2.2; that is, row

reduce [L I ]:

1

100 100 100 1 00

[] 210010 010210[ ],

011001 001 2 11

LI I L

−

ªºªº

«»«»

=− ∼ =

«»«»

−−

¬¼¬¼

so

1

100

210

211

L

−

ªº

«»

=«»

«»

−−

¬¼

. Likewise to find U

–1

, row reduce []UI :

264100 260102

[]048010040014

002001002001

UI

−−

ªºªº

«»«»

=−∼−

«»«»

−−

¬¼¬¼

1

2001 3/2 4 1001/2 3/4 2

0100 1/4 1 010 0 1/4 1[ ],

0010 0 1/2 001 0 0 1/2

IU

−

−− −−

ªºªº

«»«»

∼−−∼−−=

«»«»

−−

¬¼¬¼

1

1/ 2 3/ 4 2

so 0 1 / 4 1 . Thus

001/2

U

−

−−

ªº

«»

=−−

«»

−

¬¼

111

1/2 3/ 4 2 1 0 0 3 5/4 2

01/4 1210 3/23/4 1

001/2211 11/21/2

AUL

−−−

−− −

ª

ºª º ª º

«

»« » « »

== −− =−

«

»« » « »

«

»« » « »

−−− −

¬

¼¬ ¼ ¬ ¼

18.

1

100 2 42

210, 036Tofind,rowreduce[]:

311 001

LU LLI

−

ªºªº

«»«»

=−=−

«»«»

−

¬¼¬¼

[]

100100 100 100

210010~010210

311001 011301

LI

ªºªº

«»«»

=−

«»«»

−−−

¬¼¬¼

2.5 • Solutions 127

1

100 100

~0 1 0 2 1 0 ,

001 111

IL

−

ªº

«»

ªº

=¬¼

«»

−

¬¼

1

100

so 2 1 0 .

111

L

−

ª

º

«

»

=

«

»

«

»

−

¬

¼

1

Likewise to find ,U

−

[]

row reduce :UI

[]

242100 240102 2401 02

036010~030016~01001/32

001001 001001 0010 01

UI

−−−−−

ªºªºªº

«»«»«»

=−−− −

«»«»«»

¬¼¬¼¬¼

2001 4/36 1001/2 2/33

~0 1 0 0 1/3 2~0 1 0 0 1/3 2

0010 01 001 0 0 1

−−

ªºª º

«»« »

−−

«»« »

¬¼¬ ¼

1

[],IU

−

=

1

1/ 2 2/3 3

so 0 1 / 3 2 . Thus

001

U

−

ªº

«»

=−

«»

¬¼

111

1/ 2 2/3 3 1 0 0 23/6 7/3 3

01/32210 8/35/32

001111 111

AUL

−−−

−−

ª

ºª º ª º

«

»« » « »

== −=−

«

»« » « »

«

»« » « »

−−

¬

¼¬ ¼ ¬ ¼

19. Let A be a lower-triangular n × n matrix with nonzero entries on the diagonal, and consider the

augmented matrix [A I].

a. The (1, 1)-entry can be scaled to 1 and the entries below it can be changed to 0 by adding

multiples of row 1 to the rows below. This affects only the first column of A and the first column

of I. So the (2, 2)-entry in the new matrix is still nonzero and now is the only nonzero entry of

row 2 in the first n columns (because A was lower triangular). The (2, 2)-entry can be scaled to

1, the entries below it can be changed to 0 by adding multiples of row 2 to the rows below. This

affects only columns 2 and n + 2 of the augmented matrix. Now the (3, 3) entry in A is the only

nonzero entry of the third row in the first n columns, so it can be scaled to 1 and then used as a

pivot to zero out entries below it. Continuing in this way, A is eventually reduced to I, by scaling

each row with a pivot and then using only row operations that add multiples of the pivot row to

rows below.

b. The row operations just described only add rows to rows below, so the I on the right in [A I]

changes into a lower triangular matrix. By Theorem 7 in Section 2.2, that matrix is A

–1

.

20. Let A

= LU be an LU factorization for A. Since L is unit lower triangular, it is invertible by Exercise

19. Thus by the Invertible Matrix Theroem, L may be row reduced to I. But L is unit lower triangular,

so it can be row reduced to I by adding suitable multiples of a row to the rows below it, beginning

with the top row. Note that all of the described row operations done to L are row-replacement

operations. If elementary matrices E

1

, E

2

, … E

p

implement these row-replacement operations, then

21 21

... ( ... )

pp

EEEAEEELUIUU===

This shows that A may be row reduced to U using only row-replacement operations.

128 CHAPTER 2 • Matrix Algebra

21. (Solution in Study Guide.) Suppose A = BC, with B invertible. Then there exist elementary matrices

E

1

, …, E

p

corresponding to row operations that reduce B to I, in the sense that E

p

… E

1

B = I.

Applying the same sequence of row operations to A amounts to left-multiplying A by the product E

p

… E

1

. By associativity of matrix multiplication.

11

... ...

pp

EEAEEBCICC===

so the same sequence of row operations reduces A to C.

22. First find an LU factorization for A. Row reduce A to echelon form using only row replacement

operations:

2423 2423 2423

6958 0311 0311

~~

2739 0316 0005

4221 0627 0005

6334 09313 00010

A

−− −− −−

ªºªºªº

«»«»«»

−− − −

«»«»«»

=−− −−

«»«»«»

−−− − −

«»«»«»

−−−

¬¼¬¼¬¼

2423

0311

~0005

0000

U

−−

ªº

«»

−

«»

=

«»

¬¼

then follow the algorithm in Example 2 to find L. Use the last two columns of I

5

to make L unit lower

triangular.

2

63

5

23

5

46

10

69

235

110000

31 31000

,

111 11100

2211 22110

33201 33201

L

ªº

«»

ªº

«»

«»−ªº

«»

«» «»

«»

−

«» «»

«»

«» «»

−−

¬¼

¬¼¬¼

÷÷÷

ªºªº

«»«»

=

−−

«»«»

−−

«»«»

−− −−

¬¼¬¼

Now notice that the bottom two rows of U contain only zeros. If one uses the row-column method to

find LU, the entries in the final two columns of L will not be used, since these entries will be

multiplied by zeros from the bottom two rows of U. So let B be the first three columns of L and let C

be the top three rows of U. That is,

2.5 • Solutions 129

100

2423

310

,0311

111

0005

221

332

BC

ªº

«»

−−

ªº

«»

==−

−«»

«»

−¬¼

«»

−−

¬¼

Then B and C have the desired sizes and BC = LU = A. We can generalize this process to the case

where A is m

× n, A = LU, and U has only three non-zero rows: let B be the first three columns of L

and let C be the top three rows of U.

23. a. Express each row of D as the transpose of a column vector. Then use the multiplication rule for

partitioned matrices to write

[]

1

2

14

1234 1 22 3 4

3

4

T

TTT

T

ACD

ªº

«»

== = + + +

«»

¬¼

d

cc c c cd cd cd cd

d

which is the sum of four outer products.

b. Since A has 400 × 100 = 40000 entries, C has 400 × 4 = 1600 entries and D has 4 × 100 = 400

entries, to store C and D together requires only 2000 entries, which is 5% of the amount of entries

needed to store A directly.

24. Since Q is square and Q

T

Q = I, Q is invertible by the Invertible Matrix Theorem and Q

–1

= Q

T

. Thus

A is the product of invertible matrices and hence is invertible. Thus by Theorem 5, the equation

Ax = b has a unique solution for all b. From Ax = b, we have QRx = b, Q

T

QRx = Q

T

b, Rx = Q

T

b, and

finally x = R

–1

Q

T

b. A good algorithm for finding x is to compute Q

T

b and then row reduce the matrix

[ R Q

T

b ]. See Exercise 11 in Section 2.2 for details on why this process works. The reduction is fast

in this case because R is a triangular matrix.

25. A = UDV

T

. Since U and V

T

are square, the equations U

T

U = I and V

T

V = I imply that U and V

T

are invertible, by the IMT, and hence U

–1

= U

T

and (V

T

)

–1

= V. Since the diagonal entries

1,,

n

σσ

!

in D are nonzero, D is invertible, with the inverse of D being the diagonal matrix with

11

1

,,

n

σσ

−−

!

on

the diagonal. Thus A is a product of invertible matrices. By Theorem 6, A is invertible and A

–1

=

(UDV

T

)

–1

= (V

T

)

–1

D

–1

U

–1

= VD

–1

U

T

.

26. If A = PDP

–1

, where P is an invertible 3 × 3 matrix and D is the diagonal matrix

200

030

001

D

ªº

«»

=«»

«»

¬¼

then

211 11 121

()()()APDPPDP PDPPDPPDIDPPDP

−− −− − −

====

and since

2

200200 400

030030 090

001001 001

D

ªºªºªº

«»«»«»

==

«»«»«»

¬¼¬¼¬¼

,

21

400

090

001

AP P

−

ªº

«»

=«»

«»

¬¼

130 CHAPTER 2 • Matrix Algebra

Likewise, A

3

= PD

3

P

–1

, so

3

331 1

3

200 800

03 0 0270

001

AP P P P

−−

ªº

«»

==

«»

¬¼

«»

¬¼

In general, A

k

= PD

k

P

–1

, so

1

200

03 0

001

k

kk

AP P

−

ªº

«»

=«»

«»

¬¼

27. First consider using a series circuit with resistance R

1

followed by a shunt circuit with resistance R

2

for the network. The transfer matrix for this network is

1

22122

10 11

1/ 1 1/ ( )/01

RR

R

RRRR

−−

ªº ª º

ªº

=

«» « »«»

−−+

¬¼¬¼ ¬ ¼

For an input of 12 volts and 6 amps to produce an output of 9 volts and 4 amps, the transfer matrix

must satisfy

11

2122 122

112 612 9

1/ ( )/ ( 12 6 6 )/64

RR

RRRR RRR

−−

ªºªº

ªº ªº

==

«»«»«» «»

−+−++

¬¼ ¬¼¬¼¬¼

Equate the top entries and obtain

1

12

ohm.R=

Substitute this value in the bottom entry and solve to

obtain

9

22

ohms.R=

The ladder network is

Next consider using a shunt circuit with resistance R

1

followed by a series circuit with resistance R

2

for the network. The transfer matrix for this network is

121 2

2

11

10( )/1

1/ 1 1/ 101

R

RR RR

RR

+−− ªºª º

ªº

=

«»« »«»

−−

¬¼¬ ¼¬ ¼

For an input of 12 volts and 6 amps to produce an output of 9 volts and 4 amps, the transfer matrix

must satisfy

121 2 1212

11

()/ (12 1 2 ) / 612 9

1/ 1 12 / 664

RRR R R RR R

RR

+−+−

ªºª º

ªº ªº

==

«»« »«» «»

−−+

¬¼ ¬¼¬¼¬ ¼

Equate the bottom entries and obtain R

1

= 6 ohms. Substitute this value in the top entry and solve to

obtain

3

24

ohms.R=

The ladder network is

a.

i

2

i

1

i

2

i

3

v

3

v

2

v

1

1/2 ohm 9/ 2

ohms

2.5 • Solutions 131

28. The three shunt circuits have transfer matrices

312

10

10 10

,,and

1/ 1

1/ 1 1/ 1 R

RR

ªºªºªº

«»

«»«»

−

−−

¬¼¬¼

¬¼

respectively. To find the transfer matrix for the series of circuits, multiply these matrices

312 321

10 10

1/ 1 (1 / 1 / 1 / ) 11/ 1 1/ 1RRR RRR

ªº ª ºªºªº

=

«» « »

«»«»

−−+ +−−

¬¼¬¼¬¼ ¬ ¼

Thus the resulting network is itself a shunt circuit with resistance

123

12 13 23

RRR

R

RRRRR++

29. a. The first circuit is a series circuit with resistance R

1

ohms, so its transfer matrix is

1

01

R

−

ªº

«»

¬¼

.

The second circuit is a shunt circuit with resistance R

2

ohms, so its transfer matrix is

2

10

.

1/ 1R

ªº

«»

−

¬¼

The third circuit is a series circuit with resistance R

3

ohms so its transfer matrix is

3

1

01

R

−

ªº

«»

¬¼

.The transfer matrix of the network is the product of these matrices, in right-to-left

order:

32 13 132

31

2212

10 1/ /

11

1/ 1 1/ 1 /

01 01

R

RRRRRR

RR

RRRR

+−−−

−−

ªº ª º

ªº ªº

==

«» « »

«» «»

−−+

¬¼ ¬¼

¬¼ ¬ ¼

b. To find a ladder network with a structure like that in part (a) and with the given transfer matrix A,

we must find resistances R

1

, R

2

, and R

3

such that

32 13 132

212

1/ /

312

1/ 1 /

1/3 5/3

R

RRRRRR

ARRR

+−−−

−ªº

ªº

==

«»

−+

−

¬¼

From the (2, 1) entries, R

2

= 3 ohms. The (1, 1) entries now give

32

1/3,RR+=

which may be

solved to obtain R

3

= 6 ohms. Likewise the (2, 2) entries gives

12

1/5/3,RR+=

which also may

be solved to obtain R

1

= 2 ohms. Thus the matrix A may be factored as

31

2

10

11

1/ 1

01 01

R

AR

−−

ªº

ªº ªº

=«»

«» «»

−

¬¼ ¬¼

¬¼

16 1 012

01 1/3101

−−

ªºª ºªº

=«»« »«»

−

¬¼¬ ¼¬¼

30. Answers may vary. For example,

31216101012

1/3 5/3 0 1 1/6 1 1/6 1 0 1

−− −

ªºªºªºªºªº

=

«»«»«»«»«»

−−−

¬¼¬¼¬¼¬¼¬¼

b. i

2

i

1

i

2

i

3

3/4 ohm

v

3

v

2

v

1

6

ohms

132 CHAPTER 2 • Matrix Algebra

The network corresponding to this factorization consists of a series circuit, followed by two shunts,

followed by another series circuit. The resistances would be R

1

=2, R

2

=6, R

3

=6, R

4

=6.

Note:

The Study Guide’s MATLAB box for Section 2.5 suggests that for most LU factorizations in this

section, students can use the gauss command repeatedly to produce U, and use paper and mental

arithmetic to write down the columns of L as the row reduction to U proceeds. This is because for

Exercises 7–16 the pivots are integers and other entries are simple fractions. However, for Exercises 31

and 32 this is not reasonable, and students are expected to solve an elementary programming problem.

(The Study Guide provides no hints.)

31. [M] Store the matrix A in a temporary matrix B and create L initially as the 8×8 identity matrix. The

following sequence of MATLAB commands fills in the entries of L below the diagonal, one column

at a time, until the first seven columns are filled. (The eighth column is the final column of the

identity matrix.)

L(2:8, 1) = B(2:8, 1)/B(1, 1)

B = gauss(B, 1)

L(3:8, 2) = B(3:8, 2)/B(2, 2)

B = gauss(B, 2)

#

L(8:8, 7) = B(8:8, 7)/B(7, 7)

U = gauss(B,7)

Of course, some students may realize that a loop will speed up the process. The for..end syntax

is illustrated in the MATLAB box for Section 5.6. Here is a MATLAB program that includes the

initial setup of B and L:

B = A

L = eye(8)

for j=1:7

L(j+1:8, j) = B(j+1:8, j)/B(j, j)

B = gauss(B, j)

end

U = B

a. To four decimal places, the results of the LU decomposition are

10 0 0 0 0 00

.25 1 0 0 0 0 0 0

.25 .0667 1 0 0 0 0 0

0.2667.28571 0 0 00

00.2679.08331 0 00

00 0.2917.29211 00

00 0 0.2697.086110

00 0 0 0.2948.29311

L

ªº

«»

−

«»

−−

«»

−−

«»

=«»

−−

«»

−−

«»

−−

«»

−−

«»

¬¼

2.5 • Solutions 133

41 1 0 0 0 0 0

03.75 .25 1 0 0 0 0

003.73331.0667 1 0 0 0

00 0 3.4286 .2857 1 0 0

00 0 0 3.70831.0833 1 0

00 0 0 0 3.3919 .2921 1

00 0 0 0 0 3.70521.0861

00 0 0 0 0 0 3.3868

U

−−

ªº

«»

−−

«»

−−

«»

−−

«»

=«»

−−

«»

−−

«»

−

«»

¬¼

b. The result of solving Ly = b and then Ux = y is

x = (27.1292, 19.2344, 29.2823, 19.8086, 30.1914, 20.7177, 30.7656, 22.8708)

c.

1

.2953 .0866 .0945 .0509 .0318 .0227 .0010 .0082

.0866 .2953 .0509 .0945 .0227 .0318 .0082 .0100

.0945 .0509 .3271 .1093 .1045 .0591 .0318 .0227

.0509 .0945 .1093 .3271 .0591 .1045 .0227 .0318

.0318 .0227 .1045 .0591 .3271 .1093 .0945 .

A−

=

0509

.0227 .0318 .0591 .1045 .1093 .3271 .0509 .0945

.0010 .0082 .0318 .0227 .0945 .0509 .2953 .0866

.0082 .0100 .0227 .0318 .0509 .0945 .0866 .2953

ªº

«»

¬¼

32. [M]

3100

1310

0131

0013

A

−

ªº

«»

−−

«»

=«»

−−

«»

−

¬¼

. The commands shown for Exercise 31, but modified for 4×4 matrices,

produce

1

3

8

21

10 00

100

010

00 1

L

ªº

«»

−

«»

=«»

−

«»

−

«»

¬¼

8

3

21

8

55

21

3100

010

00 1

00 0

U

−

ªº

«»

−

«»

=«»

−

«»

¬¼

2.4 • Solutions 134

b. Let s

k+1

be the solution of Ls

k+1

= t

k

for k = 0, 1, 2, …. Then t

k+1

is the solution of Ut

k+1

= s

k+1

for k = 0, 1, 2, …. The results are

1122

10.0000 7.0000 7.0000 5.0000

18.3333 11.0000 13.3333 8.0000

,, ,,

21.8750 11.0000 16.0000 8.0000

18.3333 7.0000 13.0952 5.0000

ªºªºªºªº

«»«»«»«»

====

«»«»«»«»

¬¼¬¼¬¼¬¼

stst

3344

5.0000 3.6000 3.6000 2.6000

9.6667 5.8000 7.0000 4.2000

,,,.

11.6250 5.8000 8.4250 4.2000

9.4286 3.6000 6.8095 2.6000

ªºªºªºªº

«»«»«»«»

====

«»«»«»«»

¬¼¬¼¬¼¬¼

stst

2.6 SOLUTIONS

Notes:

This section is independent of Section 1.10. The material here makes a good backdrop for the

series expansion of (I–C)

–1

because this formula is actually used in some practical economic work.

Exercise 8 gives an interpretation to entries of an inverse matrix that could be stated without the economic

context.

1. The answer to this exercise will depend upon the order in which the student chooses to list the

sectors. The important fact to remember is that each column is the unit consumption vector for the

appropriate sector. If we order the sectors manufacturing, agriculture, and services, then the

consumption matrix is

.10 .60 .60

.30 .20 0

.30 .10 .10

C

ªº

«»

=«»

«»

¬¼

The intermediate demands created by the production vector x are given by Cx. Thus in this case the

intermediate demand is

.10 .60 .60 0 60

.30 .20 .00 100 20

.30 .10 .10 0 10

C

ªºªºªº

«»«»«»

==

«»«»«»

¬¼¬¼¬¼

x

2. Solve the equation x = Cx + d for d:

111 2 3

221 2

331 2 3

.10 .60 .60 .9 .6 .6 0

.30 .20 .00 .3 .8 20

.30 .10 .10 .3 .1 .9 0

xxxxx

Cx x x x

xxxxx

−−

ªº ªºª ºªº ªº

«» «»« »«» «»

=−=−=−+=

«» «»« »«» «»

−−+

¬¼ ¬¼¬¼ ¬¼¬ ¼

dx x

This system of equations has the augmented matrix

.90 .60 .60 0 1 0 0 37.03

.30 .80 .00 20 ~ 0 1 0 38.89

.30 .10 .90 0 0 0 1 16.67

−−

ªºªº

«»«»

−

«»«»

−−

¬¼¬¼

, so x =

37.03

38.89

16.67

ª

º

«

»

«

»

«

»

¬

¼

.

2.6 • Solutions 135

3. Solving as in Exercise 2:

111 2 3

221 2

331 2 3

.10 .60 .60 .9 .6 .6 20

.30 .20 .00 .3 .8 0

.30 .10 .10 .3 .1 .9 0

xxxxx

xxxx

xxxxx

−−

ªº ªºª ºªº ªº

«» «»« »«» «»

=−=−=−+=

«» «»« »«» «»

−−+

¬¼ ¬¼¬¼ ¬¼¬ ¼

dx xC

This system of equations has the augmented matrix

.90 .60 .60 20 1 0 0 44.44

.30 .80 .00 0 ~ 0 1 0 16.67

.30 .10 .90 0 0 0 1 16.67

−−

ªºªº

«»«»

−

«»«»

−−

¬¼¬¼

, so x =

44.44

16.67

ª

º

«

»

«

»

«

»

¬

¼

.

4. Solving as in Exercise 2:

111 2 3

221 2

331 2 3

.10 .60 .60 .9 .6 .6 20

.30 .20 .00 .3 .8 20

.30 .10 .10 .3 .1 .9 0

xxxxx

Cx x x x

xxxxx

−−

ªº ªºª ºªº ªº

«» «»« »«» «»

=−=−=−+=

«» «»« »«» «»

−−+

¬¼ ¬¼¬¼ ¬¼¬ ¼

dx x

This system of equations has the augmented matrix

.90 .60 .60 20 1 0 0 81.48

.30 .80 .00 20 ~ 0 1 0 55.56

.30 .10 .90 0 0 0 1 33.33

−−

ªºªº

«»«»

−

«»«»

−−

¬¼¬¼

, so x =

81.48

55.56

33.33

ª

º

«

»

«

»

«

»

¬

¼

.

Note:

Exercises 2–4 may be used by students to discover the linearity of the Leontief model.

5.

1

1.5 50 1.6150 110

() .6 .8 30 1.2 2 30 120

IC

−

ªºªºªºªºªº

=−===

«»«»«»«»«»

−

¬¼¬¼¬¼¬¼¬¼

xd

6.

1

.8 .5 16 .9 .5 16 48.57

1

() .6 .9 12 .6 .8 12 45.71

(.72 .30)

IC

−

ªºªº ªºªºªº

=−== =

«»«» «»«»«»

−−

¬¼¬¼ ¬¼¬¼¬¼

xd

7. a. From Exercise 5,

11.6 1

()1.2 2

IC

−

ª

º

−=

«

»

¬

¼

so

1

11

1.6 1 1 1.6

() 1.2 2 0 1.2

IC

−ªºªºªº

=−==

«»«»«»

¬¼¬¼¬¼

xd

which is the first column of

1

().IC

−

b.

1

22

1.6 1 51 111.6

() 1.2 2 30 121.2

IC

−ªºªºªº

=−==

«»«»«»

¬¼¬¼¬¼

xd

c. From Exercise 5, the production x corressponding to

50 110

is .

30 120

ª

ºªº

==

«

»«»

¬

¼¬¼

dx

Note that

21

.=+ddd

Thus

136 CHAPTER 2 • Matrix Algebra

1

22

1

11

1

()

()( )

() ()

IC

IC IC

−

−−

=−

=−+

=−+−

=+

xd

dd

xx

8. a. Given () and() ,IC IC−=−=xd x d∆∆

()( )()()IC IC IC−+=−+−=+xx x xdd∆∆∆

Thus +xx∆ is the production level corresponding to a demand of .+dd∆

b. Since

1

()IC

−

=−xd∆∆

and d∆ is the first column of I, x∆ will be the first column of

1

()IC

−

−.

9. In this case

.8 .2 .0

.3 .9 .3

.1 .0 .8

IC

−

ªº

«»

−=−−

«»

−

¬¼

Row reduce

[]IC−d to find

.8 .2 .0 40.0 1 0 0 82.8

.3 .9 .3 60.0 ~ 0 1 0 131.0

.1 .0 .8 80.0 0 0 1 110.3

−

ªºªº

«»«»

−−

«»«»

−

¬¼¬¼

So x = (82.8, 131.0, 110.3).

10. From Exercise 8, the (i, j) entry in (I – C)

–1

corresponds to the effect on production of sector i when

the final demand for the output of sector j increases by one unit. Since these entries are all positive,

an increase in the final demand for any sector will cause the production of all sectors to increase.

Thus an increase in the demand for any sector will lead to an increase in the demand for all sectors.

11. (Solution in study Guide) Following the hint in the text, compute p

T

x in two ways. First, take the

transpose of both sides of the price equation, p = C

T

p + v, to obtain

(v)()

TT TTTTT T

CC C=+= +=+pp pvpv

and right-multiply by x to get

()

TTT T T

CC=+= +px p v x p x v x

Another way to compute p

T

x starts with the production equation x = Cx + d. Left multiply by p

T

to

get

()

TT T T

CC=+=+px p x d p x pd

The two expression for p

T

x show that

TTTT

CC+= +pxvxpxpd

so v

T

x = p

T

d. The Study Guide also provides a slightly different solution.

2.6 • Solutions 137

12. Since

21

1

... (...)

mm

m m

DICC C ICIC CICD

+

=+ + + + =+ + + + =+

1

m

D+

may be found iteratively by

1

.

mm

DICD

+=+

13. [M] The matrix I – C is

0.8412 0.0064 0.0025 0.0304 0.0014 0.0083 0.1594

0.0057 0.7355 0.0436 0.0099 0.0083 0.0201 0.3413

0.0264 0.1506 0.6443 0.0139 0.0142 0.0070 0.0236

0.3299 0.0565 0.0495 0.6364 0.0204 0.0483 0.0649

0.0089

−−−− −−

−−−−−−

−− −− − −

−−− − −−

−−0.0081 0.0333 0.0295 0.6588 0.0237 0.0020

0.1190 0.0901 0.0996 0.1260 0.1722 0.7632 0.3369

0.0063 0.0126 0.0196 0.0098 0.0064 0.0132 0.9988

ªº

«»

−− − −

«»

−−−−− −

«»

−− − −− −

¬¼

so the augmented matrix []IC−d may be row reduced to find

0.8412 0.0064 0.0025 0.0304 0.0014 0.0083 0.1594 74000

0.0057 0.7355 0.0436 0.0099 0.0083 0.0201 0.3413 56000

0.0264 0.1506 0.6443 0.0139 0.0142 0.0070 0.0236 10500

0.3299 0.0565 0.0495 0.6364 0.0204 0.0483

−−−− −−

−−−−−−

−− −− − −

−−− − − 0.0649 25000

0.0089 0.0081 0.0333 0.0295 0.6588 0.0237 0.0020 17500

0.1190 0.0901 0.0996 0.1260 0.1722 0.7632 0.3369 196000

0.0063 0.0126 0.0196 0.0098 0.0064 0.0132 0.9988 5000

ª º

« »

−

« »

−−−− −−

« »

−−−−− −

« »

−− − −− −

¬ ¼

1000000 99576

0100000 97703

0010000 51231

~0001000131570

0000100 49488

0000010329554

0000001 13835

ªº

«»

¬¼

so x = (99576, 97703, 51231, 131570, 49488, 329554, 13835). Since the entries in d seem to be

accurate to the nearest thousand, a more realistic answer would be x = (100000, 98000, 51000,

132000, 49000, 330000, 14000).

14. [M] The augmented matrix []IC−d in this case may be row reduced to find

0.8412 0.0064 0.0025 0.0304 0.0014 0.0083 0.1594 99640

0.0057 0.7355 0.0436 0.0099 0.0083 0.0201 0.3413 75548

0.0264 0.1506 0.6443 0.0139 0.0142 0.0070 0.0236 14444

0.3299 0.0565 0.0495 0.6364 0.0204 0.0483

−−−− −−

−−−−−−

−− −− − −

−−− − − 0.0649 33501

0.0089 0.0081 0.0333 0.0295 0.6588 0.0237 0.0020 23527

0.1190 0.0901 0.0996 0.1260 0.1722 0.7632 0.3369 263985

0.0063 0.0126 0.0196 0.0098 0.0064 0.0132 0.9988 6526

ª º

« »

−

« »

−−−− −−

« »

−−−−− −

« »

−− − −− −

¬ ¼

138 CHAPTER 2 • Matrix Algebra

1000000134034

0100000131687

0010000 69472

~0001000176912

0000100 66596

0000010443773

0000001 18431

ªº

«»

¬¼

so x = (134034, 131687, 69472, 176912, 66596, 443773, 18431). To the nearest thousand, x =

(134000, 132000, 69000, 177000, 67000, 444000, 18000).

15. [M] Here are the iterations rounded to the nearest tenth:

(0)

(1)

(2)

(3)

(74000.0, 56000.0, 10500.0, 25000.0, 17500.0, 196000.0, 5000.0)

(89344.2, 77730.5, 26708.1, 72334.7, 30325.6, 265158.2, 9327.8)

(94681.2, 87714.5, 37577.3, 100520.5, 38598.0, 296563.8, 11480.0)

(97091.

=

x

(4)

(5)

(6)

9, 92573.1, 43867.8, 115457.0, 43491.0, 312319.0, 12598.8)

(98291.6, 95033.2, 47314.5, 123202.5, 46247.0, 320502.4, 13185.5)

(98907.2, 96305.3, 49160.6, 127213.7, 47756.4, 324796.1, 13493.8)

(99226.6, 96969.

=

x

(7)

(8)

(9)

6, 50139.6, 129296.7, 48569.3, 327053.8, 13655.9)

(99393.1, 97317.8, 50656.4, 130381.6, 49002.8, 328240.9, 13741.1)

(99480.0, 97500.7, 50928.7, 130948.0, 49232.5, 328864.7, 13785.9)

(99525.5, 97596.8, 51071.

=

x

(10)

(11)

(12)

9, 131244.1, 49353.8, 329192.3, 13809.4)

(99549.4, 97647.2, 51147.2, 131399.2, 49417.7, 329364.4, 13821.7)

(99561.9, 97673.7, 51186.8, 131480.4, 49451.3, 329454.7, 13828.2)

(99568.4, 97687.6, 51207.5, 131

=

x

x523.0, 49469.0, 329502.1, 13831.6)

so x

(12)

is the first vector whose entries are accurate to the nearest thousand. The calculation of x

(12)

takes about 1260 flops, while the row reduction above takes about 550 flops. If C is larger than

20 20,×

then fewer flops are required to compute x

(12)

by iteration than by row reduction. The

advantage of the iterative method increases with the size of C. The matrix C also becomes more

sparse for larger models, so fewer iterations are needed for good accuracy.

2.7 SOLUTIONS

Notes:

The content of this section seems to have universal appeal with students. It also provides practice

with composition of linear transformations. The case study for Chapter 2 concerns computer graphics –

see this case study (available as a project on the website) for more examples of computer graphics in

action. The Study Guide encourages the student to examine the book by Foley referenced in the text. This

section could form the beginning of an independent study on computer graphics with an interested

student.

2.7 • Solutions 139

1. Refer to Example 5. The representation in homogenous coordinates can be written as a partitioned

matrix of the form ,

1

T

A

ªº

«»

¬¼

0

where A is the matrix of the linear transformation. Since in this case

1.25

,

01

Aªº

=«»

¬¼

the representation of the transformation with respect to homogenous coordinates is

1.250

010

001

ªº

«»

¬¼

Note:

The Study Guide shows the student why the action of 1

T

A

ª

º

«

»

¬

¼

0

on the vector

x

1

ªº

«»

¬¼

corresponds to

the action of A on x.

2. The matrix of the transformation is

10

01

A

−

ª

º

=

«

»

¬

¼

, so the transformed data matrix is

10425 4 2 5

01023 0 2 3

AD

−−− −

ªºª ºª º

==

«»« »« »

¬¼¬ ¼¬ ¼

3. Following Examples 4–6,

010102 011

100011 102

001001 001

−−−

ªºªºª º

«»«»« »

=

«»«»« »

¬¼¬¼¬ ¼

4.

1/ 2 0 0 1 0 1 1/ 2 0 1/ 2

03/2001 4 03/2 6

001001 00 1

−−

ªºªºª º

«»«»« »

=

«»«»« »

¬¼¬¼¬ ¼

5.

2/2 2/2 0 2/2 2/2 0

100

2/2 2/2 0 0 1 0 2/2 2/2 0

001001001

ªºªº

−ªº

«»«»

«»

−=−

«»«»

«»

«»«»

«»

¬¼

«»«»

¬¼¬¼

6.

2/2 2/2 0 2/2 2/2 0

100

0102/2 2/20 2/2 2/20

001 0 01 0 01

ªºªº

−−

ªº

«»«»

«»

−=− −

«»«»

«»

«»«»

«»

¬¼

«»«»

¬¼¬¼

7. A 60° rotation about the origin is given in homogeneous coordinates by the matrix

1/ 2 3 / 2 0

3/2 1/2 0

001

ªº

−

«»

¬¼

. To rotate about the point (6, 8), first translate by (–6, –8), then rotate about

140 CHAPTER 2 • Matrix Algebra

the origin, then translate back by (6, 8) (see the Practice Problem in this section). A 60° rotation

about (6, 8) is thus is given in homogeneous coordinates by the matrix

1/ 2 3 / 2 0 1/ 2 3 / 2 3 4 3

106 10 6

018 3/2 1/2001 8 3/2 1/2433

001 0 0100 1 0 0 1

ªºª º

−−+

−

ªº ªº

«»« »

«» «»

−=−

«»« »

«» «»

«»« »

«» «»

¬¼ ¬¼

«»« »

¬¼¬ ¼

8. A 45° rotation about the origin is given in homogeneous coordinates by the matrix

2/2 2/2 0

001

ªº

−

«»

¬¼

. To rotate about the point (3, 7), first translate by (–3, –7), then rotate about

the origin, then translate back by (3, 7) (see the Practice Problem in this section). A 45° rotation

about (3, 7) is thus is given in homogeneous coordinates by the matrix

2/2 2/2 0 2/2 2/2 3 2 2

103 10 3

017 2/2 2/2001 7 2/2 2/2752

001 0 0100 1 0 0 1

ªºª º

−−+

−

ªº ªº

«»« »

«» «»

−=−

«»« »

«» «»

«»« »

«» «»

¬¼ ¬¼

«»« »

¬¼¬ ¼

9. To produce each entry in BD two multiplications are necessary. Since BD is a

2100×

matrix, it will

take

22100 400×× =

multiplications to compute BD. By the same reasoning it will take

22100×× =

400 multiplications to compute A(BD). Thus to compute A(BD) from the beginning will take

400 + 400 = 800 multiplications.

To compute the 22× matrix AB it will take

2228××=

multiplications, and to compute

(AB)D it will take

22100 400×× =

multiplications. Thus to compute (AB)D from the beginning will

take 8 + 400 = 408 multiplications.

For computer graphics calculations that require applying multiple transformations to data

matrices, it is thus more efficient to compute the product of the transformation matrices before

applying the result to the data matrix.

10. Let the transformation matrices in homogeneous coordinates for the dilation, rotation, and translation

be called respectively D, and R, and T. Then for some value of s,

ϕ

, h, and k,

00 cos sin 0 10

00,sincos0,01

001 0 0 1 001

sh

DsR T k

ϕϕ

−

ªºª ºªº

«»« »«»

== =

«»« »«»

¬¼¬ ¼¬¼

Compute the products of these matrices:

cos sin 0 cos sin 0

sin cos 0 , sin cos 0

001 001

ss ss

DR s s RD s s

ϕϕ ϕϕ

−−

ªºªº

«»«»

==

«»«»

¬¼¬¼

00

0,0

00 1 001

ssh sh

DT s sk TD s k

ªºªº

«»«»

==

«»«»

¬¼¬¼

2.7 • Solutions 141

cos sin cos sin cos sin

sin cos sin cos , sin cos

00 1 001

hk h

R

Thk TRk

ϕϕ ϕϕ ϕϕ

ϕϕ ϕ ϕ ϕϕ

−− −

ªºª º

«»« »

=+=

«»« »

¬¼¬ ¼

Since DR = RD, DT ≠ TD and RT ≠ TR, D and R commute, D and T do not commute and R and T do

not commute.

11. To simplify A

2

A

1

completely, the following trigonometric identities will be needed:

1.

sin

cos

tan cos cos sin

ϕ

ϕϕ ϕ ϕ

−=−=−

2.

22

sin 1sin cos

1

cos cos cos cos

sec tan sin sin cos

ϕϕϕ

ϕϕ ϕ ϕ

ϕϕϕ ϕ ϕ

−

−=−===

Using these identities,

21

sec tan 0 1 0 0

010sincos0

001001

AA

ϕϕ

−

ªºªº

«»«»

=

«»«»

¬¼¬¼

sec tan sin tan cos 0

sin cos 0

001

ϕϕϕ ϕϕ

ϕϕ

−−

ªº

«»

=

«»

¬¼

cos sin 0

sin cos 0

001

ϕϕ

−

ªº

«»

=

«»

¬¼

which is the transformation matrix in homogeneous coordinates for a rotation in

2

.

12. To simplify this product completely, the following trigonometric identity will be needed:

1cos sin

tan / 2 sin 1 cos

ϕϕ

ϕ

ϕϕ

−

==

+

This identity has two important consequences:

1cos

1(tan/2)(sin)1 sin cos

sin

ϕ

ϕϕ ϕϕ

ϕ

−

−=−=

sin

(cos )( tan / 2) tan / 2 (cos 1) tan / 2 (cos 1) sin

1cos

ϕ

ϕϕ ϕ ϕ ϕ ϕ ϕ

ϕ

−−=−+=−+=−

+

The product may be computed and simplified using these results:

1tan/201001tan/20

010sin10010

001001001

ϕϕ

ϕ

−−

ªºªºªº

«»«»«»

¬¼¬¼¬¼

1(tan/2)(sin) tan/2 0 1 tan/2 0

sin 10010

001001

ϕϕ ϕ ϕ

ϕ

−−−

ªºª º

«»« »

=

«»« »

¬¼¬ ¼

142 CHAPTER 2 • Matrix Algebra

cos tan / 2 0 1 tan / 2 0

sin 1 0 0 1 0

001001

ϕϕ ϕ

ϕ

−−

ªºªº

«»«»

=

«»«»

¬¼¬¼

cos (cos )( tan / 2) tan / 2 0

sin (sin )(tan / 2) 1 0

001

ϕϕϕ ϕ

ϕϕϕ

−−

ªº

«»

=−+

«»

¬¼

cos sin 0

sin cos 0

001

ϕϕ

−

ªº

«»

=

«»

¬¼

which is the transformation matrix in homogeneous coordinates for a rotation in

2

.

13. Consider first applying the linear transformation on

2

whose matrix is A, then applying a translation

by the vector p to the result. The matrix representation in homogeneous coordinates of the linear

transformation is ,

1

T

A

ªº

«»

¬¼

0

while the matrix representation in homogeneous coordinates of the

translation is .

1

T

I

ªº

«»

¬¼

p

0

Applying these transformations in order leads to a transformation whose

matrix representation in homogeneous coordinates is

11 1

TT T

IA A

ªºªºªº

=

«»«»«»

¬¼¬¼¬¼

p0 p

00 0

which is the desired matrix.

14. The matrix for the transformation in Exercise 7 was found to be

1/ 2 3 / 2 3 4 3

3/2 1/2 4 3 3

00 1

ªº

−+

«»

−

«»

¬¼

This matrix is of the form ,

1

T

A

ªº

«»

¬¼

p

0

where

1/ 2 3 / 2 343

,

3/2 1/2 433

A

ªº

−+

==

«»

−

«»

¬¼

p

By Exercise 13, this matrix may be written as

11

TT

IA

ªºªº

«»«»

¬¼¬¼

p0

00

that is, the composition of a linear transformation on

2

and a translation. The matrix A is the matrix

of a rotation of 60 about the origin in

2

. Thus the transformation in Exercise 7 is the composition of

a rotation about the origin and a translation by

343

.

433

ª

º

+

=

«

»

−

«

»

¬

¼

p

2.7 • Solutions 143

15. Since

1111

24824

(,,, )(, , ,),XYZH=−−

the corresponding point in

3

has coordinates

1

11

8

24

11 1

24 24 24

(, ,) , , , , (12, 6, 3)

XY Z

xyz HHH

§·

−−

§·

== =−−

¨¸

©¹

16. The homogeneous coordinates (1, –2, −3, 4) represent the point

(1 / 4, 2 / 4, 3 / 4) (1 / 4, 1 / 2, 3 / 4)−−=−−

while the homogeneous coordinates (10, –20,− 30, 40) represent the point

(10 / 40, 20 / 40, 30 / 40) (1 / 4, 1 / 2, 3 / 4)−−=−−

so the two sets of homogeneous coordinates represent the same point in

3

.

17. Follow Example 7a by first constructing that

33×

matrix for this rotation. The vector e

1

is not

changed by this rotation. The vector e

2

is rotated 60° toward the positive z-axis, ending up at the

point (0, cos 60°, sin 60°) =

(0, 1/ 2, 3 / 2).

The vector e

3

is rotated 60° toward the negative y-axis,

stopping at the point

(0, cos 150°, sin 150°) =

(0, 3 / 2, 1/ 2).−

The matrix A for this rotation is thus

10 0

01/2 3/2

03/2 1/2

A

ªº

«»

=−

«»

¬¼

so in homogeneous coordinates the transformation is represented by the matrix

10 00

01/2 3/20

103/2 1/20

00 01

T

A

ªº

«»

ªº −

«»

=

«»

¬¼

«»

¬¼

0

18. First construct the

33×

matrix for the rotation. The vector e

1

is rotated 30° toward the negative y-

axis, ending up at the point (cos(–30)°, sin (–30)°, 0) =

(3/2,1/2,0).−

The vector e

2

is rotated 30°

toward the positive x-axis, ending up at the point (cos 60°, sin 60°, 0) =

(1 / 2, 3 / 2, 0).

The vector e

3

is not changed by the rotation. The matrix A for the rotation is thus

3/2 1/2 0

1/ 2 3 / 2 0

001

A

ªº

«»

=−

«»

¬¼

so in homogeneous coordinates the rotation is represented by the matrix

3/2 1/2 0 0

1/2 3 / 2 0 0

10010

0001

T

A

ªº

«»

ªº

−

«»

=

«»

¬¼

«»

¬¼

0

Following Example 7b, in homogeneous coordinates the translation by the vector (5, –2, 1) is

represented by the matrix

144 CHAPTER 2 • Matrix Algebra

100 5

010 2

001 1

000 1

ªº

«»

−

«»

¬¼

Thus the complete transformation is represented in homogeneous coordinates by the matrix

100 5 3/2 1/200 3/2 1/20 5

010 2 1/ 2 3 / 2 0 0 1/ 2 3 / 2 0 2

001 1 0010 0011

000 1 0001 0001

ª

ºª º

ªº

«

»« »

«»

−−−−

«

»« »

«» =

«

»« »

«»

«

»« »

«»

«

»« »

¬¼

¬

¼¬ ¼

19. Referring to the material preceding Example 8 in the text, we find that the matrix P that performs a

perspective projection with center of projection (0, 0, 10) is

10 00

01 00

00 00

00 .11

ªº

«»

−

«»

¬¼

The homogeneous coordinates of the vertices of the triangle may be written as (4.2, 1.2, 4, 1), (6, 4,

2, 1), and (2, 2, 6, 1), so the data matrix for S is

4.2 6 2

1.2 4 2

426

111

ªº

«»

¬¼

and the data matrix for the transformed triangle is

10 004.262 4.2 6 2

01 001.242 1.2 4 2

00 00 426 0 0 0

00 .11 111 .6.8.4

ªºªºªº

«»«»«»

=

«»«»«»

−

«»«»«»

¬¼¬¼¬¼

Finally, the columns of this matrix may be converted from homogeneous coordinates by dividing by

the final coordinate:

(4.2, 1.2, 0, .6) (4.2 / .6, 1.2 / .6, 0 / .6) (7, 2, 0)

(6, 4, 0, .8) (6/.8, 4/.8, 0/.8) = (7.5, 5, 0)

(2, 2, 0, .4) (2 / .4, 2 / .4, 0 / .4) (5, 5, 0)

→=

→

→=

So the coordinates of the vertices of the transformed triangle are (7, 2, 0), (7.5, 5, 0), and (5, 5, 0).

20. As in the previous exercise, the matrix P that performs the perspective projection is

10 00

01 00

00 00

00 .11

ªº

«»

−

«»

¬¼

 s 3OLUTIONS 

The homogeneous coordinates of the vertices of the triangle may be written as (7, 3, –5, 1), (12, 8, 2,

1), and (1, 2, 1, 1), so the data matrix for S is

712 1

382

521

111

ªº

«»

−

«»

¬¼

and the data matrix for the transformed triangle is

10 00 712 1 712 1

01 00 3 82 3 82

00 00 5 21 0 0 0

00 .11 1 11 1.5 .8.9

ªºªºªº

«»«»«»

=

«»«»«»

−

«»«»«»

−

¬¼¬¼¬¼

Finally, the columns of this matrix may be converted from homogeneous coordinates by dividing by

the final coordinate:

(7, 3, 0, 1.5) (7 / 1.5,3 / 1.5,0 / 1.5) (4.67, 2,0)

(12, 8, 0, .8) (12 / .8, 8 / .8, 0 / .8) (15,10, 0)

(1, 2, 0, .9) (1 / .9, 2 / .9, 0 / .9) (1.11, 2.22, 0)

→=

So the coordinates of the vertices of the transformed triangle are (4.67, 2, 0), (15, 10, 0),

and (1.11, 2.22, 0).

21. [M] Solve the given equation for the vector (R, G, B), giving

1

.61 .29 .15 2.2586 1.0395 .3473

.35 .59 .063 1.3495 2.3441 .0696

.04 .12 .787 .0910 .3046 1.2777

RXX

GY Y

BZ Z

−

−−

ªº ª ºªºª ºª º

«» « »«»« »« »

==−

«» « »«»« »« »

−

¬¼ ¬ ¼¬¼¬ ¼¬ ¼

22. [M] Solve the given equation for the vector (R, G, B), giving

1

.299 .587 .114 1.0031 .9548 .6179

.596 .275 .321 .9968 .2707 .6448

.212 .528 .311 1.0085 1.1105 1.6996

RYY

GI I

BQ Q

−

ªº ª ºªº ª ºª º

«» « »«» « »« »

=−− =−−

«» « »«» « »« »

−−

¬¼ ¬ ¼¬¼ ¬ ¼¬ ¼

2.8 SOLUTIONS

Notes

: Cover this section only if you plan to skip most or all of Chapter 4. This section and the next

cover everything you need from Sections 4.1–4.6 to discuss the topics in Section 4.9 and Chapters 5–7

(except for the general inner product spaces in Sections 6.7 and 6.8). Students may use Section 4.2 for

review, particularly the Table near the end of the section. (The final subsection on linear transformations

should be omitted.) Example 6 and the associated exercises are critical for work with eigenspaces in

Chapters 5 and 7. Exercises 31–36 review the Invertible Matrix Theorem. New statements will be added

to this theorem in Section 2.9.

Key Exercises: 5–20 and 23–26.

146 CHAPTER 2 • Matrix Algebra

1. The set is closed under sums but not under multiplication

by a negative scalar. A counterexample to the subspace

condition is shown at the right.

Note

: Most students prefer to give a geometric counterexample, but some may choose an algebraic calcu-

lation. The four exercises here should help students develop an understanding of subspaces, but they may

be insufficient if you want students to be able to analyze an unfamiliar set on an exam. Developing that

skill seems more appropriate for classes covering Sections 4.1–4.6.

2.The set is closed under scalar multiples but not sums.

For example, the sum of the vectors in H shown

here is not in H.

3. No. The set is not closed under sums or scalar multiples. See the diagram.

4. No. The set is closed under sums, but not under multiplication by a

negative scalar.

u

(–1)u

2.8 • Solutions 147

5. The vector w is in the subspace generated by v

1

and v

2

if and only if the vector equation x

1

v

1

+ x

2

v

2

= w is consistent. The row operations below show that w is in the subspace generated by v

1

and v

2

.

12

123 12 3 123

[]~333~036~036

4710 012 000

−− −− −−

ªºªºªº

«»«»«»

−−

«»«»«»

−−−

¬¼¬¼¬¼

vvw

6. The vector u is in the subspace generated by {v

1

, v

2

, v

3

} if and only if the vector equation x

1

v

1

+ x

2

v

2

+ x

3

v

3

= u is consistent. The row operations below show that u is not in the subspace generated by

{v

1

, v

2

, v

3

}.

123

145 1 14 5 1 145 1

3437 081210 012 1

[]~ ~ ~

2561 03 4 1 081210

3752 0510 5 034 1

−−−

ª

ºª ºª º

«

»« »« »

−−−− − −

«

»« »« »

«

»« »« »

−−− −

«

»« »« »

−− −−

¬

¼¬ ¼¬ ¼

vv vu

14 5 1 14 5 1 145 1

01 2 1 01 2 1 012 1

00 4 2 00 1 1 001 1

00 2 2 00 4 2 000 6

−−−

ªºªºªº

«»«»«»

−

«»«»«»

−− − −

«»«»«»

−−− −

¬¼¬¼¬¼



Note

: For a quiz, you could use w = (9, –7, 11, 12), which is in Span{v

1

, v

2

, v

3

}.

7. a. There are three vectors: v

1

, v

2

, and v

3

in the set {v

1

, v

2

, v

3

}.

b. There are infinitely many vectors in Span{v

1

, v

2

, v

3

} = Col A.

c. Deciding whether p is in Col A requires calculation:

234 6 23 46 23 46

[]~88610~041014~041014

677 11 02 57 00 00

A

−− − − − −

ªºªºªº

«»«»«»

−−−− −−

«»«»«»

−− −

¬¼¬¼¬¼

p

The equation Ax = p has a solution, so p is in Col A.

8.

2206 2206 2206

[]0351~0351~0351

63517 0351 0000

A

−− − −− − −− −

ªºªºªº

«»«»«»

=−−−

«»«»«»

−−

¬¼¬¼¬¼

p

Yes, the augmented matrix [A p] corresponds to a consistent system, so p is in Col A.

9. To determine whether p is in Nul A, simply compute Ap. Using A and p as in Exercise 7,

Ap =

234 6 2

88610 62.

67711 29

−− −

ªºªºªº

«»«»«»

−−=−

«»«»«»

−−

¬¼¬¼¬¼

Since Ap ≠ 0, p is not in Nul A.

148 CHAPTER 2 • Matrix Algebra

10. To determine whether u is in Nul A, simply compute Au. Using A as in Exercise 7 and u = (–5, 5, 3),

Au =

2205 0

0355 0.

6353 0

−− −

ªºªºªº

«»«»«»

−=

«»«»«»

¬¼¬¼¬¼

Yes, u is in Nul A.

11. p = 4 and q = 3. Nul A is a subspace of R

4

because solutions of Ax = 0 must have 4 entries, to match

the columns of A. Col A is a subspace of R

3

because each column vector has 3 entries.

12. p = 3 and q = 5. Nul A is a subspace of R

3

because solutions of Ax = 0 must have 3 entries, to match

the columns of A. Col A is a subspace of R

5

because each column vector has 5 entries.

13. To produce a vector in Col A, select any column of A. For Nul A, solve the equation Ax = 0. (Include

an augmented column of zeros, to avoid errors.)

32150 32150 32150

94170~02480~02480

92510 048160 00000

−−−

ªºªºªº

«»«»«»

−− − −

«»«»«»

−−−

¬¼¬¼¬¼

321 50 10 1 10

~0 1 2 4 0~0 1 2 4 0,

000 00 00 0 00

−−

ªºª º

«»« »

−−

«»« »

¬¼¬ ¼

134

234

0

240

00

xxx

−+=

+−=

=

The general solution is x

1

= x

3

– x

4

, and x

2

= –2x

3

+ 4x

4

, with x

3

and x

4

free. The general solution in

parametric vector form is not needed. All that is required here is one nonzero vector. So choose any

values for x

3

and x

4

(not both zero). For instance, set x

3

= 1 and x

4

= 0 to obtain the vector (1, –2, 1,

0) in Nul A.

Note

: Section 2.8 of Study Guide introduces the ref command (or rref, depending on the technol-

ogy), which produces the reduced echelon form of a matrix. This will greatly speed up homework for

students who have a matrix program available.

14. To produce a vector in Col A, select any column of A. For Nul A, solve the equation Ax = 0:

12 3 0 10 1/30

1230 12 30

015/30 015/30

4570 0350

00 0 0

~~ ~

00 0 0

5100 09150

00 0 0 00 0 0

27110 0 350

00 0 0 00 0 0

3340 0350

ªº

−

ª

º

ªºªº

«»

«

»

«»«»

−− «»

«

»

«»«»

«»

«

»

«»«»

−−

«»

«

»

«»«»

«»

«

»

«»«»

«»

«

»

«»«»

−−

¬¼¬¼

¬

¼

¬¼

The general solution is x

1

= (1/3)x

3

and x

2

= (–5/3) x

3

, with x

3

free. The general solution in parametric

vector form is not needed. All that is required here is one nonzero vector. So choose any nonzero

value of x

3

. For instance, set x

3

= 3 to obtain the vector (1, –5, 3) in Nul A.

15. Yes. Let A be the matrix whose columns are the vectors given. Then A is invertible because its

determinant is nonzero, and so its columns form a basis for R

2

, by the Invertible Matrix Theorem (or

by Example 5). (Other reasons for the invertibility of A could be given.)

16. No. One vector is a multiple of the other, so they are linearly dependent and hence cannot be a basis

for any subspace.

2.8 • Solutions 149

17. Yes. Place the three vectors into a 3×3 matrix A and determine whether A is invertible:

05 6 24 2

00 3~ 05 6

24 2 00 3

A

−

ªºªº

«»«»

=«»«»

«»«»

−

¬¼¬¼

The matrix A has three pivots, so A is invertible by the IMT and its columns form a basis for R

3

(as

pointed out in Example 5).

18. No. Place the three vectors into a 3×3 matrix A and determine whether A is invertible:

135 13 5 13 5

111~044~044

324 01111 000

A

ªºªºªº

«»«»«»

=−−−−−

«»«»«»

−−

¬¼¬¼¬¼

The matrix A has two pivots, so A is not invertible by the IMT and its columns do not form a basis

for R

3

(as pointed out in Example 5).

19. No. The vectors cannot be a basis for R

3

because they only span a plan in R

3

. Or, point out that the

columns of the matrix

36

82

15

ª

º

«

»

−

«

»

«

»

−

¬

¼

cannot possibly span R

3

because the matrix cannot have a pivot in

every row. So the columns are not a basis for R

3

.

Note:

The Study Guide warns students not to say that the two vectors here are a basis for R

2

.

20. No. The vectors are linearly dependent because there are more vectors in the set than entries in each

vector. (Theorem 8 in Section 1.7.) So the vectors cannot be a basis for any subspace.

21. a. False. See the definition at the beginning of the section. The critical phrases “for each” are

missing.

b. True. See the paragraph before Example 4.

c. False. See Theorem 12. The null space is a subspace of R

n

, not R

m

.

d. True. See Example 5.

e. True. See the first part of the solution of Example 8.

22. a. False. See the definition at the beginning of the section. The condition about the zero vector is

only one of the conditions for a subspace.

b. False. See the warning that follows Theorem 13.

c. True. See Example 3.

d. False. Since y need not be in H, it is not gauranteed by the definition of a subspace that x+y will

be in H.

e. False. See the paragraph after Example 4.

150 CHAPTER 2 • Matrix Algebra

23. (Solution in Study Guide)

459 2 126 5

65112~0 15 6

348 3 000 0

A

−−

ªºªº

«»«»

=−

«»«»

−

¬¼¬¼

. The echelon form identifies

columns 1 and 2 as the pivot columns. A basis for Col A uses columns 1 and 2 of A:

45

6, 5

34

ªºªº

«»«»

¬¼¬¼

. This

is not the only choice, but it is the “standard” choice. A wrong choice is to select columns 1 and 2 of

the echelon form. These columns have zero in the third entry and could not possibly generate the

columns displayed in A.

For Nul A, obtain the reduced (and augmented) echelon form for Ax = 0:

10 4 70

01 5 60

00 0 00

−

ªº

«»

−

«»

¬¼

. This corresponds to:

134

234

470

560

00

xxx

−+=

+−=

=

.

Solve for the basic variables and write the solution of Ax = 0 in parametric vector form:

134

234

34

33

44

47 47

56 5 6

10

01

xxx

xx

−−

ªºª º ªº ª º

«»« » «» «»

−+−

«»« » «» «»

==+

«»« » «» «»

«» «»

«»« » ¬¼ ¬¼

¬¼¬ ¼

. Basis for Nul A:

47

56

,

10

01

−

ª

ºª º

«

»« »

−

«

»« »

«

»« »

«

»« »

«

»« »

¬

¼¬ ¼

Notes

: (1) A basis is a set of vectors. For simplicity, the answers here and in the text list the vectors

without enclosing the list inside set brackets. This style is also easier for students. I am careful,

however, to distinguish between a matrix and the set or list whose elements are the columns of the

matrix.

(2) Recall from Chapter 1 that students are encouraged to use the augmented matrix when solving Ax

= 0, to avoid the common error of misinterpreting the reduced echelon form of A as itself the augmented

matrix for a nonhomogeneous system.

(3) Because the concept of a basis is just being introduced, I insist that my students write the

parametric vector form of the solution of Ax = 0. They see how the basis vectors span the solution space

and are obviously linearly independent. A shortcut, which some instructors might introduce later in the

course, is only to solve for the basic variables and to produce each basis vector one at a time. Namely, set

all free variables equal to zero except for one free variable, and set that variable equal to a suitable

nonzero number.

24.

3690 1254

2472~0036

3666 0000

A

−−

ªºªº

«»«»

=−

«»«»

−−

¬¼¬¼

. Basis for Col A:

39

2,7

36

ª

ºªº

«

»«»

«

»«»

«

»«»

¬

¼¬¼

.

For Nul A, obtain the reduced (and augmented) echelon form for Ax = 0:

12060

00120

00000

−−

ªº

«»

¬¼

. This corresponds to:

12 4

34

260

20

00

xx x

xx

−−=

+=

=

.

Solve for the basic variables and write the solution of Ax = 0 in parametric vector form:

2.8 • Solutions 151

124

22

24

34

44

26 26

10

202

01

xxx

xx

−−

ªºª º ªº ª º

«»« » «» « »

==+

«»« » «» « »

−−

«»« » «» « »

¬¼ ¬ ¼

¬¼¬ ¼

. Basis for Nul A:

26

10

,

02

01

−

ª

ºª º

«

»« »

«

»« »

«

»« »

−

«

»« »

¬

¼¬ ¼

25.

148 3 7 1480 5

127 3 4 0250 1

~

229 5 5 0001 4

369 5 2 0000 0

A

−−

ªºªº

«»«»

−−

«»«»

=«»«»

−

«»«»

−−

«»«»

¬¼¬¼

. Basis for Col A:

14 3

12 3

,,

22 5

36 5

−

ª

ºª ºª º

«

»« »« »

−

«

»« »« »

«

»« »« »

−

«

»« »« »

−

«

»« »« »

¬

¼¬ ¼¬ ¼

.

For Nul A, obtain the reduced (and augmented) echelon form for Ax = 0:

10 20 7 0

012.50 .50

[]~

00 01 40

00 00 00

A

−

ªº

«»

−

«»

¬¼

0

.

135

23 5

45

270

2.5 .5 0

40

00

xxx

xx x

xx

−+=

+−=

+=

=

.

135

235

3533

45

55

27 27

2.5 .5 2.5 .5

Thesolution of 0in parametric vector form : .

10

404

01

xxx

Axxxx

xx

−−

ªºª º

ª

ºªº

«»« »

«

»«»

−+−

«»« »

«

»«»

«»« »

«

»«»

===+

«»« »

«

»«»

−−

«»« »

«

»«»

«»« »

«

»«»

¬

¼¬¼

¬¼¬ ¼

x

uv

Basis for Nul A: {u, v}.

Note

: The solution above illustrates how students could write a solution on an exam, when time is

precious, namely, describe the basis by giving names to appropriate vectors found in the calculations.

26.

313 18 31 306

313 02 02 604

~

039 14 00 012

639 26 00 000

A

−− − − −

ªºªº

«»«»

−

«»«»

=

«»«»

−− −

«»«»

−

¬¼¬¼

. Basis for Col A:

311

310

,,

031

632

−−

ªºª ºª º

«»« »« »

−

«»« »« »

−

¬¼¬ ¼¬ ¼

.

For Nul A,

[]

10004/30

0130 20

~0001 20

0000 00

A

ªº

«»

−

«»

−

«»

¬¼

0

.

15

23 5

45

4/3 0

320

20

00

xx

xx x

xx

+=

+−=

−=

=

The solution of Ax = 0 in parametric vector form:

152 CHAPTER 2 • Matrix Algebra

15

235

3335

45

55

4/3 04/3

32 32

10

202

01

xx

xxx

xxxx

xx

−−

ªºª º ªº ª º

«»« » «» « »

−+−

«»« » «» « »

==+

«»« » «» « »

¬¼ ¬ ¼

¬¼¬ ¼

u v

. Basis for Nul A: {u, v}.

27. Construct a nonzero 3×3 matrix A and construct b to be almost any convenient linear combination of

the columns of A.

28. The easiest construction is to write a 3×3 matrix in echelon form that has only 2 pivots, and let b be

any vector in R

3

whose third entry is nonzero.

29. (Solution in Study Guide) A simple construction is to write any nonzero 3×3 matrix whose columns

are obviously linearly dependent, and then make b a vector of weights from a linear dependence

relation among the columns. For instance, if the first two columns of A are equal, then b could be (1,

–1, 0).

30. Since Col A is the set of all linear combinations of a

1

, … , a

p

, the set {a

1

, … , a

p

} spans Col A.

Because {a

1

, … , a

p

} is also linearly independent, it is a basis for Col A. (There is no need to discuss

pivot columns and Theorem 13, though a proof could be given using this information.)

31. If Col F ≠ R

5

, then the columns of F do not span R

5

. Since F is square, the IMT shows that F is not

invertible and the equation Fx = 0 has a nontrivial solution. That is, Nul F contains a nonzero vector.

Another way to describe this is to write Nul F ≠ {0}.

32. If Col B = R

7

, then the columns of B span R

7

. Since B is square, the IMT shows that B is invertible

and the equation Bx = b has a solution for each b in R

7

. Also, each solution is unique, by Theorem 5

in Section 2.2.

33. If Nul C = {0}, then the equation Cx = 0 has only the trivial solution. Since C is square, the IMT

shows that C is invertible and the equation Cx = b has a solution for each b in R

6

. Also, each solution

is unique, by Theorem 5 in Section 2.2.

34. If the columns of A form a basis, they are linearly independent. This means that A cannot have more

columns than rows. Since the columns also span R

m

, A must have a pivot in each row, which means

that A cannot have more rows than columns. As a result, A must be a square matrix.

35. If Nul B contains nonzero vectors, then the equation Bx = 0 has nontrivial solutions. Since B is

square, the IMT shows that B is not invertible and the columns of B do not span R

5

. So Col B is a

subspace of R

5

, but Col B ≠ R

5

.

36. If the columns of C are linearly independent, then the equation Cx = 0 has only the trivial (zero)

solution. That is, Nul C = {0}.

37. [M] Use the command that produces the reduced echelon form in one step (ref or rref depending

on the program). See the Section 2.8 in the Study Guide for details. By Theorem 13, the pivot

columns of A form a basis for Col A.

2.8 • Solutions 153

3501 3 102.54.53.5

794911011.52.51.5

~

5725 7 00 0 0 0

3734 0 00 0 0 0

A

−− −

ªºªº

«»«»

−−− −

«»«»

=

«»«»

−− −

«»«»

−−

«»«»

¬¼¬¼

Basis for Col A:

35

79

,

57

37

−

ªºªº

«»«»

−

«»«»

−

«»«»

−

«»«»

¬¼¬¼

For Nul A, obtain the solution of Ax = 0 in parametric vector form:

1345

2345

2.5 4.5 3.5 0

1.5 2.5 1.5 0

xxxx

+−+=

Solution:

1345

2345

34 5

2.5 4.5 3.5

1.5 2.5 1.5

,,and are free

xxxx

xx x

=−+−



°=−+−

®

°

¯

1345

2345

345

33

44

55

2.5 4.5 3.5 2.5 4.5 3.5

1.5 2.5 1.5 1.5 2.5 1.5

100

010

001

xxxx

xxx

xx

−+−−−

ªºª º

ª

ºªºªº

«»« »

«

»«»«»

−+−−−

«»« »

«

»«»«»

«»« »

«

»«»«»

== = + +

«»« »

«

»«»«»

«»« »

«

»«»«»

«»« »

«

»«»«»

¬

¼¬¼¬¼

¬¼¬ ¼

x = x

3

u + x

4

v + x

5

w

By the argument in Example 6, a basis for Nul A is {u, v, w}.

38. [M]

53268 10100

41387 01100

~

514519 00010

75285 00001

A

−−

ªºªº

«»«»

−− −

«»«»

=«»«»

«»«»

−−−

¬¼¬¼

.

The pivot columns of A form a basis for Col A:

53 68

41 87

,, ,

51 519

75 85

−−

ª

ºª ºª ºª º

«

»« »« »« »

−−

«

»« »« »« »

«

»« »« »« »

«

»« »« »« »

−−

¬

¼¬ ¼¬ ¼¬ ¼

.

For Nul A, solve Ax = 0:

13

23

4

5

0

xx

x

+=

−=

=

Solution:

13

23

3

4

5

is free

0

xx

x

=−



°=

°

=

®

°=

°

¯

154 CHAPTER 2 • Matrix Algebra

1

2

33

4

5

1

0

x

xx

x

−

ªº ªº

«» «»

==

«» «»

¬¼

x = x

3

u

By the method of Example 6, a basis for Nul A is {u}.

Note

: The Study Guide for Section 2.8 gives directions for students to construct a review sheet for the

concept of a subspace and the two main types of subspaces, Col A and Nul A, and a review sheet for the

concept of a basis. I encourage you to consider making this an assignment for your class.

2.9 SOLUTIONS

Notes

: This section contains the ideas from Sections 4.4–4.6 that are needed for later work in Chapters

5–7. If you have time, you can enrich the geometric content of “coordinate systems” by discussing crystal

lattices (Example 3 and Exercises 35 and 36 in Section 4.4.) Some students might profit from reading

Examples 1–3 from Section 4.4 and Examples 2, 4, and 5 from Section 4.6. Section 4.5 is probably not a

good reference for students who have not considered general vector spaces.

Coordinate vectors are important mainly to give an intuitive and geometric feeling for the

isomorphism between a k-dimensional subspace and R

k

. If you plan to omit Sections 5.4, 5.6, 5.7 and 7.2,

you can safely omit Exercises 1–8 here.

Exercises 1–16 may be assigned after students have read as far as Example 2. Exercises 19 and 20 use

the Rank Theorem, but they can also be assigned before the Rank Theorem is discussed.

The Rank Theorem in this section omits the nontrivial fact about Row A which is included in the

Rank Theorem of Section 4.6, but that is used only in Section 7.4. The row space itself can be introduced

in Section 6.2, for use in Chapter 6 and Section 7.4.

Exercises 9–16 include important review of techniques taught in Section 2.8 (and in Sections 1.2 and

2.5). They make good test questions because they require little arithmetic. My students need the practice

here. Nearly every time I teach the course and start Chapter 5, I find that at least one or two students

cannot find a basis for a two-dimensional eigenspace!

1. If [x]

B

= 3

2

ªº

«»

¬¼

, then x is formed from b

1

and b

2

using

weights 3 and 2:

x = 3b

1

+ 2b

2

= 127

32

111

ª

ºªºªº

+=

«

»«»«»

−

¬

¼¬¼¬¼

2. If [x]

B

= 1

2

−

ªº

«»

¬¼

, then x is formed from b

1

and b

2

using weights –1 and 2:

x = (–1)b

1

+ 2b

2

= 339

(1) 2

123

−

ªº ªºªº

−+=

«» «»«»

¬¼ ¬¼¬¼

b

2

2b

2

b

1

2b

1

3b

1

x

1

x

2

r

3. To find c

1

and c

2

that satisfy x = c

1

b

12

210 1

[] ~

357 0

−

ªºª

=«»«

−

¬¼¬

bb x

suggested by Exercise 1 and solve

u

[x]

B

=

1

2

1.

2

c

ªºªº

=

«»«»

¬¼

4. As in Exercise 3,

12

1

[]

5

ª

=«

−

¬

bb x

and [x]

B

=

1

2

3.

2

c

−

ªºªº

=

«»«»

−

¬¼

5.

12

122 1

[]479~0

357 0

−

ªºª

«»«

=−

«»«

−−

¬¼¬

bb x

6.

12

375 1

[]230~0

452 0

−

ªºª

«»«

=−

«»«

−−

¬¼¬

bb x

7. Fig. 1 suggests that w = 2b

1

– b

2

an

[w]

B

= 2

1

ªº

«»

−

¬¼

and [x]

B

= 1.5

.5

ªº

«»

¬¼

. To

c

12

3

1

1.5 .5 1.5 .5

0

2

−

ª

ºª

+= +

«

»«

¬

¼¬

bb

Figure 1

Note

: Figures 1 and 2 display what Sec

t

8. Fig. 2 suggests that x = b

1

+ b

2

, y =

[x]

B

= 1

1

ªº

«»

¬¼

, [y]

B

= 1/3

1

ªº

«»

−

¬¼

, and [z]

B

b

2

b

1

x

w

0

2.9 • Sol

u

r

son Education, Inc. Publishing as Addison-Wesley.

b

1

+ c

2

b

2

, row reduce the augmented matrix:

1/2 0 1 0 1

~

7/2 7 0 1 2

−

ºª º

»« »

¼¬ ¼

. Or, one can write a matrix eq

u

sing the matrix inverse. In either case,

21 1 2 1 10 3

~~

39 0 714 0 1 2

−− −

ºª ºª º

»« »« »

−−

¼¬ ¼¬ ¼

,

22 104

11~011

11 000

−ºª º

»« »

−−

¼¬ ¼

. [x]

B

=

1

2

4.

1

c

ªºªº

=

«»«»

¬¼

15103

12~012

13 / 3 26 / 3 0 0 0

ºª º

»« »

¼¬ ¼

, [x]

B

=

1

2

c

ªº

ª

=

«»«

¬

¬¼

n

d x = 1.5b

1

+ .5b

2

, in which case,

c

onfirm [x]

B

, compute

1

4

2

1

ºªº

==

»«»

¼¬¼x

Figure 2

t

ion 4.4 calls B-graph paper.

1/3b

1

- b

2

, and z = –4/3 b

1

+2b

2

. If so, then

= 4/3

2

−

ªº

«»

¬¼

. To confirm [y]

B

and [z]

B

, compute

u

tions 155

uation as

3.

2

ª

º

»

¬

¼

156 CHAPTER 2 • Matrix Algebra

12

02 2

(1 / 3) (1 / 3) 32 1

−

ªº ªº ª º

−=−==

«» «» « »

−

¬¼ ¬¼ ¬ ¼

bb y

and

12

024

(4/3) 2 4/3 2

320

ªº ªº ª º

−+=−+==

«» «» « »

¬¼ ¬¼ ¬ ¼

bb z.

9. The information

132 6 133 2

39 1 5 005 7

~

26 19 000 5

515 0 14 00 0 0

A

−

ª

ºª º

«

»« »

−

«

»« »

=

«

»« »

−

«

»« »

¬

¼¬ ¼

is enough to see that columns 1, 3, and 4

of A form a basis for Col A:

12 6

315

,,.

219

5014

−

ªºª ºª º

«»« »« »

−

«»« »« »

¬¼¬ ¼¬ ¼

Columns 1, 2 and 4, of the echelon form certainly cannot span Col A since those vectors all have zero

in their fourth entries. For Nul A, use the reduced echelon form, augmented with a zero column to

insure that the equation Ax = 0 is kept in mind:

13 0 0 0

00 1 0 0

00 0 1 0

00 0 0 0

ªº

«»

¬¼

.

12

3

4

2

30

0

is the free variable

xx

x

+=

=

, x =

12

22

2

3

4

33

1.

00

xx

x

−−

ªº

ª

ºªº

«»

«

»«»

«»

«

»«»

==

«»

«

»«»

«»

«

»«»

¬

¼¬¼

¬¼

So

3

1

0

−

ªº

«»

¬¼

is

a basis for Nul A. From this information, dim Col A = 3 (because A has three pivot columns) and dim

Nul A = 1 (because the equation Ax = 0 has only one free variable).

10. The information

12 154 12120

21156 01103

~

20216 00010

31415 00001

A

−− −−

ª

ºª º

«

»« »

−

«

»« »

=

«

»« »

−−−

«

»« »

¬

¼¬ ¼

shows that columns 1, 2,4

and 5 of A form a basis for Col A:

1254

2156

,,,

2016

3115

−

ªºªºªºªº

«»«»«»«»

−

«»«»«»«»

−−

«»«»«»«»

¬¼¬¼¬¼¬¼

. For Nul A,

[]

10 1 000

011 000

~000 100

000 010

A

ªº

«»

¬¼

0

.

13

23

4

5

3

0

is a free variable

xx

x

+=

=

13

23

33

3

4

5

1

.

1

00

xx

x

−−

ªºªº ªº

«»«» «»

−−

«»«» «»

== =

«»«» «»

¬¼ ¬¼

¬¼

x Basis for Nul A:

1

0

−

ª

º

«

»

−

«

»

«

»

«

»

«

»

«

»

¬

¼

.

2.9 • Solutions 157

From this, dim Col A = 4 and dim Nul A = 1.

11. The information

2452 3 12514

3683 5 00505

~

0090 9 00000

367310 00000

A

−− −−

ª

ºª º

«

»« »

−−

«

»« »

=

«

»« »

«

»« »

−−− −−

¬

¼¬ ¼

shows that columns 1and

3 of A form a basis for Col A:

25

38

,.

09

37

−

ªºªº

«»«»

−

«»«»

−−

¬¼¬¼

For Nul A,

[]

120110

001010

~.

000000

A

ªº

«»

¬¼

0

12 45

35

24 5

20

0

, and are free variables

xx xx

xx

xx x

+++=

+=

1245

22

35245

44

55

2211

100

.

001

010

001

xxxx

xx

xxxxx

xx

−−− −−−

ªºª º ªº ªº ªº

«»« » «» «» «»

== −=++

−

«»« » «» «» «»

¬¼ ¬¼ ¬¼

¬¼¬ ¼

x Basis for Nul A:

211

100

,,.

001

010

00 1

−−−

ª

ºª ºª º

«

»« »« »

«

»« »« »

«

»« »« »

−

«

»« »« »

«

»« »« »

«

»« »« »

¬

¼¬ ¼¬ ¼

From this, dim Col A = 2 and dim Nul A = 3.

12. The information

12 4 4 6 1 284 6

51 9 210 0 234 1

~

46 91215 0 050 5

34 5 8 9 0 000 0

A

−−

ª

ºª º

«

»« »

−−

«

»« »

=

«

»« »

−−

«

»« »

−

¬

¼¬ ¼

shows that columns 1, 2,

and 3 of A form a basis for Col A:

12 4

51 9

,, .

46 9

34 5

−

ªºªºª º

«»«»« »

−

«»«»« »

−

«»«»« »

−

¬¼¬¼¬ ¼

For Nul A

[]

1000 00

0102 10

~.

0010 10

0000 00

A

ªº

«»

−

«»

¬¼

0

1

245

35

45

0

20

0

and are free variables

x

xxx

xx

=

++=

−=

1

245

3545

44

55

000

221

.

01

10

01

x

xxx

xxxx

xx

ªºª º ªº ª º

«»« » «» « »

−− −−

«»« » «» « »

== = +

«»« » «» « »

¬¼ ¬ ¼

¬¼¬ ¼

x Basis for Nul A:

00

21

,.

01

10

01

ª

ºª º

«

»« »

−−

«

»« »

«

»« »

«

»« »

«

»« »

«

»« »

¬

¼¬ ¼

158 CHAPTER 2 • Matrix Algebra

From this, dim Col A = 3 and dim Nul A = 2.

13. The four vectors span the column space H of a matrix that can be reduced to echelon form:

1324 1324 1324 1324

3915 0057 0057 0057

~~~

2643 0005 0005 0005

412 2 7 0 010 9 0 0 0 5 0 0 0 0

−− −− −− −−

ªºªºªºªº

«»«»«»«»

−− − − −

«»«»«»«»

−−

«»«»«»«»

−−

«»«»«»«»

¬¼¬¼¬¼¬¼

Columns 1, 3, and 4 of the original matrix form a basis for H, so dim H = 3.

Note

: Either Exercise 13 or 14 should be assigned because there are always one or two students who

confuse Col A with Nul A. Or, they wrongly connect “set of linear combinations” with “parametric vector

form” (of the general solution of Ax = 0).

14. The five vectors span the column space H of a matrix that can be reduced to echelon form:

120 13 120 1 3 120 1 3

13 147 0 113 4 0113 4

~~

21376 033912 000410

3427 9 0221018 0000 0

−− −

ªºªºªº

«»«»«»

−− − − − − − −− −

«»«»«»

−− − − −

«»«»«»

−− −−−

¬¼¬¼¬¼

Columns 1,2 and 4 of the original matrix form a basis for H, so dim H = 3.

15. Col A = R

4

, because A has a pivot in each row and so the columns of A span R

4

. Nul A cannot equal

R

2

, because Nul A is a subspace of R

6

. It is true, however, that Nul A is two-dimensional. Reason: the

equation Ax = 0 has two free variables, because A has six columns and only four of them are pivot

columns.

16. Col A cannot be R

3

because the columns of A have four entries. (In fact, Col A is a 3-dimensional

subspace of R

4

, because the 3 pivot columns of A form a basis for Col A.) Since A has 7 columns and

3 pivot columns, the equation Ax = 0 has 4 free variables. So, dim Nul A = 4.

17. a. True. This is the definition of a B-coordinate vector.

b. False. Dimension is defined only for a subspace. A line must be through the origin in R

n

to be a

subspace of R

n

.

c. True. The sentence before Example 3 concludes that the number of pivot columns of A is the rank

of A, which is the dimension of Col A by definition.

d. True. This is equivalent to the Rank Theorem because rank A is the dimension of Col A.

e. True, by the Basis Theorem. In this case, the spanning set is automatically a linearly independent

set.

18. a. True. This fact is justified in the second paragraph of this section.

b. False. The dimension of Nul A is the number of free variables in the equation Ax = 0.

See Example 2.

c. True, by the definition of rank.

d. True. See the second paragraph after Fig. 1.

e. True, by the Basis Theorem. In this case, the linearly independent set is automatically a spanning

set.

2.9 • Solutions 159

19. The fact that the solution space of Ax = 0 has a basis of three vectors means that dim Nul A = 3.

Since a 5×7 matrix A has 7 columns, the Rank Theorem shows that rank A = 7 – dim Nul A = 4.

Note

: One can solve Exercises 19–22 without explicit reference to the Rank Theorem. For instance, in

Exercise 19, if the null space of a matrix A is three-dimensional, then the equation Ax = 0 has three free

variables, and three of the columns of A are nonpivot columns. Since a 5×7 matrix has seven columns, A

must have four pivot columns (which form a basis of Col A). So rank A = dim Col A = 4.

20. A 6×8 matrix A has 8 columns. By the Rank Theorem, rank A = 8 – dim Nul A. Since the null space

is three-dimensional, rank A = 5.

21. A 9×8 matrix has 8 columns. By the Rank Theorem, dim Nul A = 8 – rank A. Since the rank is seven,

dim Nul A = 1. That is, the dimension of the solution space of Ax = 0 is one.

22. Suppose that the subspace H = Span{v

1

, …, v

5

} is four-dimensional. If {v

1

, …, v

5

} were linearly

independent, it would be a basis for H. This is impossible, by the statement just before the definition

of dimension in Section 2.9, which essentially says that every basis of a p-dimensional subspace

consists of p vectors. Thus, {v

1

, …, v

5

} must be linearly dependent.

23. A 3×5 matrix A with a two-dimensional column space has two pivot columns. The remaining three

columns will correspond to free variables in the equation Ax = 0. So the desired construction is

possible. There are ten possible locations for the two pivot columns, one of which is

****

0***

00000

ªº

«»

¬¼



. A simple construction is to take two vectors in R

3

that are obviously not

linearly dependent, and put three copies of these two vectors in any order. The resulting matrix will

obviously have a two-dimensional column space. There is no need to worry about whether Nul A has

the correct dimension, since this is guaranteed by the Rank Theorem: dim Nul A = 5 – rank A.

24. A rank 1 matrix has a one-dimensional column space. Every column is a multiple of some fixed

vector. To construct a 3×4 matrix, choose any nonzero vector in R

3

, and use it for one column.

Choose any multiples of the vector for the other tthree columns.

25. The p columns of A span Col A by definition. If dim Col A = p, then the spanning set of p columns is

automatically a basis for Col A, by the Basis Theorem. In particular, the columns are linearly

independent.

26. If columns a

1

, a

3

, a

4,

a

5

, and a

7

of A are linearly independent and if dim Col A = 5, then {a

1

, a

3

, a

4

,a

5

,

a

7

} is a linearly independent set in a 5-dimensional column space. By the Basis Theorem, this set of

five vectors is a basis for the column space.

27. a. Start with B = [b

1

⋅ ⋅ ⋅ b

p

] and A = [a

1

⋅ ⋅ ⋅ a

q

], where q > p. For j = 1, …, q, the vector a

j

is

in W. Since the columns of B span W, the vector a

j

is in the column space of B. That is, a

j

= Bc

j

for some vector c

j

of weights. Note that c

j

is in R

p

because B has p columns.

b. Let C = [c

1

⋅ ⋅ ⋅ c

q

]. Then C is a p×q matrix because each of the q columns is in R

p

.

By hypothesis, q is larger than p, so C has more columns than rows. By a theorem, the columns of

C are linearly dependent and there exists a nonzero vector u in R

q

such that Cu = 0.

c. From part (a) and the definition of matrix multiplication

A = [a

1

⋅ ⋅ ⋅ a

q

] = [Bc

1

⋅ ⋅ ⋅ Bc

q

] = BC

160 CHAPTER 2 • Matrix Algebra

From part (b), Au = (BC)u = B(Cu) = B0 = 0. Since u is nonzero, the columns of A are linearly

dependent.

28. If A contained more vectors than B, then A would be linearly dependent, by Exercise 27, because B

spans W. Repeat the argument with B and A interchanged to conclude that B cannot contain more

vectors than A.

29. [M] Apply the matrix command ref or rref to the matrix [v

1

v

2

x]:

15 14 16 1 0 2

5100 011

~

12 13 11 0 0 0

7173 000

ªºªº

«»«»

−− −

«»«»

−

¬¼¬¼

The equation c

1

v

1

+ c

2

v

2

= x is consistent, so x is in the subspace H. Then c

1

= 2 and c

2

= -1. Thus,

the -coordinate of x is (2, -1).

30. [M] Apply the matrix command ref or rref to the matrix [v

1

v

2

v

3

x]:

68 911 100 2

304 2 010 1

~

97 8 17 001 1

433 8 0000

−− −

ªºªº

«»«»

−

«»«»

−−−

«»«»

−−

¬¼¬¼

The first three columns of [v

1

v

2

v

3

x] are pivot columns, so v

1

, v

2

and v

3

are linearly independent.

Thus v

1

, v

2

and v

3

form a basis B for the subspace H which they span. View [v

1

v

2

v

3

x] as an

augmented matrix for c

1

v

1

+ c

2

v

2

+ c

3

v

3

= x. The reduced echelon form shows that x is in H and

[x]

B

=

2

1.

1

−

ªº

«»

¬¼

Notes

: The Study Guide for Section 2.9 contains a complete list of the statements in the Invertible Matrix

Theorem that have been given so far. The format is the same as that used in Section 2.3, with three

columns statements that are logically equivalent for any m×n matrix and are related to existence concepts,

those that are equivalent only for any n×n matrix, and those that are equivalent for any n×p matrix and

are related to uniqueness concepts. Four statements are included that are not in the text’s official list of

statements, to give more symmetry to the three columns.

The Study Guide section also contains directions for making a review sheet for “dimension” and

“rank

Chapter 2 • Supplementary Exercises 161

Chapter 2 SUPPLEMENTARY EXERCISES

1. a. True. If A and B are m×n matrices, then B

T

has as many rows as A has columns, so AB

T

is

defined. Also, A

T

B is defined because A

T

has m columns and B has m rows.

b. False. B must have 2 columns. A has as many columns as B has rows.

c. True. The ith row of A has the form (0, …, d

i

, …, 0). So the ith row of AB is (0, …, d

i

, …, 0)B,

which is d

i

times the ith row of B.

d. False. Take the zero matrix for B. Or, construct a matrix B such that the equation Bx = 0 has

nontrivial solutions, and construct C and D so that C ≠ D and the columns of C – D satisfy the

equation Bx = 0. Then B(C – D) = 0 and BC = BD.

e. False. Counterexample: A = 10

00

ªº

«»

¬¼

and C = 00

01

ª

º

«

»

¬

¼.

f. False. (A + B)(A – B) = A

2

– AB + BA – B

2

. This equals A

2

– B

2

if and only if A commutes with B.

g. True. An n×n replacement matrix has n + 1 nonzero entries. The n×n scale and interchange

matrices have n nonzero entries.

h. True. The transpose of an elementary matrix is an elementary matrix of the same type.

i. True. An n×n elementary matrix is obtained by a row operation on I

n

.

j. False. Elementary matrices are invertible, so a product of such matrices is invertible. But not

every square matrix is invertible.

k. True. If A is 3×3 with three pivot positions, then A is row equivalent to I

3

.

l. False. A must be square in order to conclude from the equation AB = I that A is invertible.

m. False. AB is invertible, but (AB)

–1

= B

–1

A

–1

, and this product is not always equal to A

–1

B

–1

.

n. True. Given AB = BA, left-multiply by A

–1

to get B = A

–1

BA, and then right-multiply by A

–1

to

obtain BA

–1

= A

–1

B.

o. False. The correct equation is (rA)

–1

= r

–1

A

–1

, because

(rA)(r

–1

A

–1

) = (rr

–1

)(AA

–1

) = 1¸I = I.

p. True. If the equation Ax =

1

0

ªº

«»

¬¼

has a unique solution, then there are no free variables in this

equation, which means that A must have three pivot positions (since A is 3×3). By the Invertible

Matrix Theorem, A is invertible.

2. C = (C

–1

)

–1

= 75 7/25/2

1

64 3 2

2

−−

ªºª º

=

«»« »

−−

−¬¼¬ ¼

3.

2

000 000000 000

100, 100100 000

010 010010 100

AA

ªºªºªºªº

«»«»«»«»

== =

«»«»«»«»

¬¼¬¼¬¼¬¼

32

000000 000

100000 000

010100 000

AAA

ªºªºªº

«»«»«»

=⋅==

«»«»«»

¬¼¬¼¬¼

162 CHAPTER 2 • Matrix Algebra

Next,

22 22233

()( ) ( )IAIAA IAA AIAA IAA AA A IA−++ =++ −++ =++ −− − =−

.

Since A

3

= 0,

2

()( )IAIAA I−++ =

.

4. From Exercise 3, the inverse of I – A is probably I + A + A

2

+ ¸ ¸ ¸ + A

n–1

. To verify this, compute

11 11

()() ()

nn nnn

IAIA A IA A AIA A IAA IA

−− −−

−++ + =++ + −++ + =−=−"" "

If A

n

= 0, then the matrix B = I + A + A

2

+ ¸ ¸ ¸ + A

n–1

satisfies (I – A)B = I. Since I – A and B are

square, they are invertible by the Invertible Matrix Theorem, and B is the inverse of I – A.

5. A

2

= 2A – I. Multiply by A: A

3

= 2A

2

– A. Substitute A

2

= 2A – I: A

3

= 2(2A – I) – A = 3A – 2I.

Multiply by A again: A

4

= A(3A – 2I) = 3A

2

– 2A. Substitute the identity A

2

= 2A – I again:

Finally, A

4

= 3(2A – I) – 2A = 4A – 3I.

6. Let

10 01

and .

01 10

AB

ªº ªº

==

«» «»

−

¬¼ ¬¼

By direct computation, A

2

= I, B

2

= I, and AB =

01

10

ªº

«»

−

¬¼

= –

BA.

7. (Partial answer in Study Guide) Since A

–1

B is the solution of AX = B, row reduction of [A B] to

[I X] will produce X = A

–1

B. See Exercise 15 in Section 2.2.

[]

13 8 35 1 3 8 3 5 1 3 8 3 5

2411 15~0 2 5 7 5~0 1 3 6 1

12 5 34 0 1 3 6 1 0 2 5 7 5

AB

−− −

ªºªºªº

«»«»«»

=− − −−

«»«»«»

−− − − − −

¬¼¬¼¬¼

138 3 5 1303729 10010 1

~0 1 3 6 1~0 1 0 9 10~0 1 0 9 10

001 5 3 001 5 3 001 5 3

−−

ªºªºªº

«»«»«»

−

«»«»«»

−− −− −−

¬¼¬¼¬¼

Thus, A

–1

B =

10 1

910

53

−

ªº

«»

−−

¬¼

.

8. By definition of matrix multiplication, the matrix A satisfies

12 13

37 11

A

ªºªº

=

«»«»

¬¼¬¼

Right-multiply both sides by the inverse of

12

37

ª

º

«

»

¬

¼. The left side becomes A. Thus,

13 7 2 2 1

11 3 1 4 1

A

−−

ªºª ºª º

==

«»« »« »

−−

¬¼¬ ¼¬ ¼

9. Given

54 7 3

and

23 2 1

AB B

ªº ªº

==

«» «»

−

¬¼ ¬¼

, notice that ABB

–1

= A. Since det B = 7 – 6 =1,

11

13 54 1 3 313

and ( )

27 232 7 827

BAAB B

−−

−−−

ªº ªºªºªº

=== =

«» «»«»«»

−−− −

¬¼ ¬¼¬¼¬¼

2.9 • Solutions 163

Note:

Variants of this question make simple exam questions.

10. Since A is invertible, so is A

T

, by the Invertible Matrix Theorem. Then A

T

A is the product of

invertible matrices and so is invertible. Thus, the formula (A

T

A)

–1

A

T

makes sense. By Theorem 6 in

Section 2.2,

(A

T

A)

–1

¸A

T

= A

–1

(A

T

)

–1

A

T

= A

–1

I = A

–1

An alternative calculation: (A

T

A)

–1

A

T

¸A = (A

T

A)

–1

(A

T

A) = I. Since A is invertible, this equation shows

that its inverse is (A

T

A)

–1

A

T

.

11. a. For i = 1,…, n, p(x

i

) = c

0

+ c

1

x

i

+ ¸ ¸ ¸ +

1

1n

in

cx

−

=

0

1

row ( ) row ( )

n

ii

c

VV

c

−

ªº

«»

⋅=

«»

¬¼

c#

.

By a property of matrix multiplication, shown after Example 6 in Section 2.1, and the fact that c

was chosen to satisfy Vc= y,

row ( ) row ( ) row ( )

ii

VV y===ccy

Thus, p(x

i

) = y

i

. To summarize, the entries in Vc are the values of the polynomial p(x) at x

1

, …, x

n

.

b. Suppose x

1

, …, x

n

are distinct, and suppose Vc = 0 for some vector c. Then the entries in c are the

coefficients of a polynomial whose value is zero at the distinct points x

1

, ..., x

n

. However, a

nonzero polynomial of degree n – 1 cannot have n zeros, so the polynomial must be identically

zero. That is, the entries in c must all be zero. This shows that the columns of V are linearly

independent.

c. (Solution in Study Guide) When x

1

, …, x

n

are distinct, the columns of V are linearly independent,

by (b). By the Invertible Matrix Theorem, V is invertible and its columns span R

n

. So, for every

y = (y

1

, …, y

n

) in R

n

, there is a vector c such that Vc = y. Let p be the polynomial whose

coefficients are listed in c. Then, by (a), p is an interpolating polynomial for (x

1

, y

1

), …, (x

n

, y

n

).

12. If A = LU, then col

1

(A) = L¸col

1

(U). Since col

1

(U) has a zero in every entry except possibly the first,

L¸col

1

(U) is a linear combination of the columns of L in which all weights except possibly the first

are zero. So col

1

(A) is a multiple of col

1

(L).

Similarly, col

2

(A) = L¸col

2

(U), which is a linear combination of the columns of L using the first

two entries in col

2

(U) as weights, because the other entries in col

2

(U) are zero. Thus col

2

(A) is a

linear combination of the first two columns of L.

13. a. P

2

= (uu

T

)(uu

T

) = u(u

T

u)u

T

= u(1)u

T

= P, because u satisfies u

T

u = 1.

b. P

T

= (uu

T

)

T

= u

TT

u

T

= uu

T

= P

c. Q

2

= (I – 2P)(I – 2P) = I – I(2P) – 2PI + 2P(2P)

= I – 4P + 4P

2

= I, because of part (a).

14. Given

0

1

ªº

«»

=

«»

¬¼

u

, define P and Q as in Exercise 13 by

164 CHAPTER 2 • Matrix Algebra

[]

0000 100 000 10 0

00 0 1 0 0 0, 2 0 1 0 20 0 0 0 1 0

1001 001 001 00 1

T

PQI P

ªº ª º ª º ª º ª º

«» « » « » « » « »

== = =−=−=

«» « » « » « » « »

−

¬¼ ¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼

uu

If

1

5

3

ªº

«»

=

«»

¬¼

x

, then

0001 0 10 01 1

0005 0 and 0 1 05 5

0013 3 00 13 3

PQ

ª

ºª º ª º ª ºª º ª º

«

»« » « » « »« » « »

=== =

«

»« » « » « »« » « »

«

»« » « » « »« » « »

−−

¬

¼¬ ¼ ¬ ¼ ¬ ¼¬ ¼ ¬ ¼

xx

.

15. Left-multiplication by an elementary matrix produces an elementary row operation:

121321

~~ ~BEBEEBEEEBC=

so B is row equivalent to C. Since row operations are reversible, C is row equivalent to B.

(Alternatively, show C being changed into B by row operations using the inverse of the E

i

.)

16. Since A is not invertible, there is a nonzero vector v in R

n

such that Av = 0. Place n copies of v into

an n×n matrix B. Then AB = A[v ⋅ ⋅ ⋅ v] = [Av ⋅ ⋅ ⋅ Av] = 0.

17. Let A be a 6×4 matrix and B a 4×6 matrix. Since B has more columns than rows, its six columns are

linearly dependent and there is a nonzero x such that Bx = 0. Thus ABx = A0 = 0. This shows that the

matrix AB is not invertible, by the IMT. (Basically the same argument was used to solve Exercise 22

in Section 2.1.)

Note

: (In the Study Guide) It is possible that BA is invertible. For example, let C be an invertible 4×4

matrix and construct

1

and [ 0].

0

C

ABC

−

ªº

==

«»

¬¼ Then BA = I

4

, which is invertible.

18. By hypothesis, A is 5×3, C is 3×5, and CA = I

3

. Suppose x satisfies Ax = b. Then CAx = Cb. Since

CA = I, x must be Cb. This shows that Cb is the only solution of Ax = b.

19. [M] Let

.4 .2 .3

.3 .6 .3

.3 .2 .4

A

ªº

«»

=

«»

¬¼

. Then

2

.31 .26 .30

.39 .48 .39

.30 .26 .31

A

ª

º

«

»

=

«

»

«

»

¬

¼

. Instead of computing A

3

next, speed up

the calculations by computing

422 844

.2875 .2834 .2874 .2857 .2857 .2857

.4251 .4332 .4251 , .4285 .4286 .4285

.2874 .2834 .2875 .2857 .2857 .2857

AAA AAA

ªºªº

«»«»

== ==

«»«»

¬¼¬¼

To four decimal places, as k increases,

.2857 .2857 .2857

.4286 .4286 .4286

.2857 .2857 .2857

k

A

ªº

«»

→

«»

¬¼

, or, in rational format,

2/7 2/7 2/7

3/7 3/7 3/7

2/7 2/7 2/7

k

A

ª

º

«

»

→

«

»

«

»

¬

¼

.

If

2

0.2.3 .29 .18 .18

.1 .6 .3 , then .33 .44 .33

.9 .2 .4 .38 .38 .49

BB

ªº ª º

«» « »

==

«» « »

¬¼ ¬ ¼

,

2.9 • Solutions 165

48

.2119 .1998 .1998 .2024 .2022 .2022

.3663 .3784 .3663 , .3707 .3709 .3707

.4218 .4218 .4339 .4269 .4269 .4271

BB

ªºªº

«»«»

==

«»«»

¬¼¬¼

To four decimal places, as k increases,

.2022 .2022 .2022

.3708 .3708 .3708

.4270 .4270 .4270

k

B

ªº

«»

→

«»

¬¼

, or, in rational format,

18/89 18 /89 18 /89

33/89 33 /89 33/89

38/89 38 /89 38 /89

k

B

ª

º

«

»

→

«

»

«

»

¬

¼

.

20. [M] The 4×4 matrix A

4

is the 4×4 matrix of ones, minus the 4×4 identity matrix. The MATLAB

command is A4 = ones(4) – eye(4). For the inverse, use inv(A4).

1

44

0111 2/3 1/3 1/3 1/3

10 11 1/3 2/3 1/3 1/3

,

1101 1/3 1/3 2/3 1/3

1110 1/3 1/3 1/3 2/3

AA

−

ªºª º

«»« »

−

«»« »

==

«»« »

−

«»« »

−

¬¼¬ ¼

1

55

01111 3/4 1/4 1/4 1/4 1/4

10 111 1/4 3/4 1/4 1/4 1/4

,

11011 1/4 1/4 3/4 1/4 1/4

11101 1/4 1/4 1/4 3/4 1/4

11110 1/4 1/4 1/4 1/4 3/4

AA

−

ªºª º

«»« »

−

«»« »

==

−

«»« »

−

«»« »

−

¬¼¬ ¼

1

66

011111 4/5 1/5 1/5 1/5 1/5 1/5

10 1111 1/5 4/5 1/5 1/5 1/5 1/5

110111 1/5 1/5 4/5 1/5 1/5 1/5

,

111011 1/5 1/5 1/5 4/5 1/5 1/5

111101 1/5 1/5 1/5 1/5 4/5 1/5

111110 1/5 1/5 1/5 1/5 1/5 4/5

AA

−

ªºª º

«»« »

−

«»« »

−

==

«»« »

−

«»« »

−

«»« »

−

«»« »

¬¼¬ ¼

The construction of A

6

and the appearance of its inverse suggest that the inverse is related to I

6

. In

fact,

1

66

AI

−+

is 1/5 times the 6×6 matrix of ones. Let J denotes the n×n matrix of ones. The

conjecture is:

A

n

= J – I

n

and

11

1

nn

AJI

n

−=⋅−

−

Proof: (Not required) Observe that J

2

= nJ and A

n

J = (J – I )J = J

2

– J = (n – 1) J. Now compute

A

n

((n – 1)

–1

J – I) = (n – 1)

–1

A

n

J – A

n

= J – (J – I) = I

Since A

n

is square, A

n

is invertible and its inverse is (n – 1)

–1

J – I.

167

3.1 SOLUTIONS

Notes:

Some exercises in this section provide practice in computing determinants, while others allow the

student to discover the properties of determinants which will be studied in the next section. Determinants

are developed through the cofactor expansion, which is given in Theorem 1. Exercises 33–36 in this

section provide the first step in the inductive proof of Theorem 3 in the next section.

1. Expanding along the first row:

30 4 32 22 23

23 2 3 0 4 3(13)4(10)1

51 01 05

05 1

=−+=−+=

−−

−

Expanding along the second column:

12 22 32

30 4 22 34 34

23 2 (1) 0 (1) 3 (1) 5 3(3)5(2)1

01 01 22

05 1

+++

=−⋅ +−⋅ +−⋅ =−−− =

−−

−

2. Expanding along the first row:

051 30 40 4 3

4300 5 1 5(4)1(22)2

41 21 2 4

241

−−

−=−+=−+=

Expanding along the second column:

12 22 32

051 40 01 01

430(1)5 (1)(3) (1)4 5(4)3(2)4(4)2

21 21 40

241

++ +

−=−⋅ +−⋅− +−⋅ =−−−−−=

3. Expanding along the first row:

243 12 32 31

3122 (4) 3 2(9)4(5)(3)(11)5

41 11 14

141

−

=−− +=−+−+=−

−−

−

168 CHAPTER 3 • Determinants

Expanding along the second column:

12 22 32

243 32 23 23

312(1)(4) (1)1 (1)4 4(5)1(5)4(5)5

11 11 32

141

+++

−

=−⋅− +−⋅ +−⋅ =−+−− − =−

−−

−

4. Expanding along the first row:

135 11 21 21

2111 3 5 1(2)3(1)5(5)20

42 32 34

342

=−+=−− +=

Expanding along the second column:

12 22 32

135 21 15 15

211 (1) 3 (1) 1 (1) 4 3(1)1(13)4(9)20

32 32 21

342

+++

=−⋅ +−⋅ +−⋅ =−+−−−=

5. Expanding along the first row:

23 4 05 45 40

40 5 2 3 (4) 2(5)3(1)4(4) 23

16 56 51

51 6

−

=−+−=−−−− =−

6. Expanding along the first row:

524 35 05 03

0355 (2) 4 5(1)2(10)4(6)1

47 27 24

247

−

−−

−=−− +=++−=

−−

−

7. Expanding along the first row:

430 52 62 65

652 4 3 0 4(1)3(0)4

73 93 97

973

=−+=−=

8. Expanding along the first row:

816 03 43 4 0

4038 1 6 8(6)1(11)6(8)11

25 35 3 2

325

=−+=−+−=−

−−

−

9. First expand along the third row, then expand along the first row of the remaining matrix:

3.1 • Solutions 169

31 13

600 5 00 5

172 5 72

(1) 27 2 5 2(1) 5 10(1) 10

200 0 31

31 8

831 8

++

−

=−⋅ −=⋅− ⋅ ==

10. First expand along the second row, then expand along either the third row or the second column of

the remaining matrix.

23

1252 122

0030

(1) 32 6 5

2675 504

5044

+

−

=−⋅ −

−−

31 33

22 1 2

(3)(1) 5 (1) 4 (3)(5(2) 4(2)) 6

65 2 6

++

§·

−−

=−− ⋅ +−⋅ =−+−=−

¨¸

−−

©¹

or

23

1252 122

0030

(1) 32 6 5

2675 504

5044

+

−

=−⋅ −

−−

12 22

25 12

(3)(1) (2) (1) (6)

54 54

++

§·

=−− ⋅− +−⋅−

¨¸

©¹

()

(3)2(17) 6(6) 6=−−−−=−

11. There are many ways to do this determinant efficiently. One strategy is to always expand along the

first column of each matrix:

11 11

3584 23 7

0237 15

(1) 3 0 1 5 3(1) (2)

0015 02

00 2

0002

++

−

−−

=−⋅ =⋅− ⋅−

= 3(–2)(2) = –12

12. There are many ways to do this determinant efficiently. One strategy is to always expand along the

first row of each matrix:

11 11

4000 10 0

7100 30

(1) 4 6 3 0 4(1) (1)

2630 43

84 3

5843

++

−

=−⋅ =⋅− ⋅−

−

−−

= 4(–1)( –9) = 36

170 CHAPTER 3 • Determinants

13. First expand along either the second row or the second column. Using the second row,

23

40 7 3 5 40 3 5

00 2 0 0 73 4 8

(1) 2

73 6 4 8 50 2 3

50 5 2 3 00 1 2

00 9 1 2

+

−−

−

=−⋅

−−

−

Now expand along the second column to find:

23 22

40 3 5 435

73 4 8

(1) 2 2(1) 35 2 3

50 2 3 012

00 1 2

++

−

§·−

−¨¸

−⋅ =−− ⋅ −

¨¸

−¨¸

−

©¹

−

Now expand along either the first column or third row. The first column is used below.

22

435

2(1) 35 2 3

012

+

§·−

¨¸

−− ⋅ −

¨¸

−

©¹

11 21

23 35

6(1) 4 (1) 5 (6)(4(1) 5(1)) 6

12 12

++

§·−−

=−− ⋅ +−⋅ =−−=

¨¸

−−

©¹

14. First expand along either the fourth row or the fifth column. Using the fifth column,

35

63240 63 24

90410 90 41

(1) 1

85671 30 00

30000 42 32

42320

+

−

=−⋅

−

Now expand along the third row to find:

35 31

63 24 324

90 41

(1) 1 1(1) 30 4 1

30 00 232

42 32

++

§·

−¨¸

−⋅ =−⋅ −

¨¸

©¹

Now expand along either the first column or second row. The first column is used below.

31

324

1(1) 30 4 1

232

+

§·

¨¸

−⋅ −

¨¸

©¹

11 31

41 24

3(1) 3 (1) 2 (3)(3( 11) 2(18)) 9

32 41

++

§·−

=−⋅ +−⋅ =−+=

¨¸

−

©¹

15.

30 4

23 2

05 1

=

−

(3)(3)(–1) + (0)(2)(0) + (4)(2)(5) – (0)(3)(4) – (5)(2)(3) – (–1)(2)(0)

= –9 + 0 + 40 – 0 – 30 –0 = 1

3.1 • Solutions 171

16.

051

430

241

−=

(0)(–3)(1) + (5)(0)(2) + (1)(4)(4) – (2)(–3)(1) – (4)(0)(0) – (1)(4)(5)

= 0 + 0 + 16 – (–6) – 0 – 20 = 2

17.

243

312

141

−

=

−

(2)(1)(–1) + (–4)(2)(1) + (3)(3)(4) – (1)(1)(3) – (4)(2)(2) – (–1)(3)(–4)

= –2 + (–8) + 36 – 3 – 16 – 12 = –5

18.

135

211

342

=

(1)(1)(2) + (3)(1)(3) + (5)(2)(4) – (3)(1)(5) – (4)(1)(1) – (2)(2)(3)

= 2 + 9 + 40 – 15 – 4 – 12 = 20

19.

,

ab

ad bc

cd

=−

()

cd

cb da ad bc

ab

=−=−−

The row operation swaps rows 1 and 2 of the matrix, and the sign of the determinant is reversed.

20.

,

ab

ad bc

cd

=−

()() ()

ab

akd kcb kad kbc kad bc

kc kd

=−=−=−

The row operation scales row 2 by k, and the determinant is multiplied by k.

21.

3418 20 2,

56

=−=−

34

3(6 4 ) (5 3 )4 2

53 64

kk

=+ −+=−

++

The row operation replaces row 2 with k times row 1 plus row 2, and the determinant is unchanged.

22.

,

ab

ad bc

cd

=−

()()

akc bkd

akcdcbkd ad kcdbckcd adbc

cd

++

=+ −+=+−− =−

The row operation replaces row 1 with k times row 2 plus row 1, and the determinant is unchanged.

23.

111

3841(4)1(2)1(7)5,

232

−−=−+−=−

−

384 (4)(2)(7)5

232

kkk

kkk k

−−=−+−=−

−

The row operation scales row 1 by k, and the determinant is multiplied by k.

24.

322 (2) (6) (3)2 6 3,

656

abc

abc abc

=−+=−+

172 CHAPTER 3 • Determinants

322

3(6 5 ) 2(6 6 ) 2(5 6 ) 2 6 3

656

abc b c a c a b a b c

=−− −+−=−+−

The row operation swaps rows 1 and 2 of the matrix, and the sign of the determinant is reversed.

25. Since the matrix is triangular, by Theorem 2 the determinant is the product of the diagonal entries:

100

010 (1)(1)(1)1

01

k

==

26. Since the matrix is triangular, by Theorem 2 the determinant is the product of the diagonal entries:

100

010 (1)(1)(1)1

01

k

==

27. Since the matrix is triangular, by Theorem 2 the determinant is the product of the diagonal entries:

00

010 ()(1)(1)

001

k

kk

==

28. Since the matrix is triangular, by Theorem 2 the determinant is the product of the diagonal entries:

100

00(1)()(1)

001

kkk

==

29. A cofactor expansion along row 1 gives

010 10

100 1 1

01

001

=−=−

30. A cofactor expansion along row 1 gives

001 01

0101 1

10

100

==−

31. A 3 × 3 elementary row replacement matrix looks like one of the six matrices

100 100 100 100 10 1 0

10,010,010,01k,010,010

001 01 0 1 001 001 001

kk

k

kk

ªºªºªºªºªºªº

«»«»«»«»«»«»

¬¼¬¼¬¼¬¼¬¼¬¼

In each of these cases, the matrix is triangular and its determinant is the product of its diagonal

entries, which is 1. Thus the determinant of a 3 × 3 elementary row replacement matrix is 1.

3.1 • Solutions 173

32. A 3 × 3 elementary scaling matrix with k on the diagonal looks like one of the three matrices

00 100 100

010,0 0,010

001 001 00

k

ªºªºªº

«»«»«»

¬¼¬¼¬¼

In each of these cases, the matrix is triangular and its determinant is the product of its diagonal

entries, which is k. Thus the determinant of a 3 × 3 elementary scaling matrix with k on the diagonal

is k.

33.

01

,

10

Eªº

=«»

¬¼ ,

ab

A

cd

ªº

=«»

¬¼

cd

EA

ab

ªº

=«»

¬¼

det E = –1, det A = ad – bc,

det EA = cb – da = –1(ad – bc) = (det E)(det A)

34.

10

,

0

E

k

ªº

=«»

¬¼ ,

ab

A

cd

ªº

=«»

¬¼

ab

EA

kc kd

ªº

=«»

¬¼

det E = k, det A = ad – bc,

det EA = a(kd) – (kc)b = k(ad – bc) = (det E)(det A)

35.

1,

01

k

Eªº

=«»

¬¼ ,

ab

A

cd

ªº

=«»

¬¼

akc bkd

EA

cd

++

ªº

=«»

¬¼

det E = 1, det A = ad – bc,

det EA = (a + kc)d – c(b + kd) = ad + kcd – bc – kcd = 1(ad – bc) = (det E)(det A)

36.

10

,

1

E

k

ªº

=«»

¬¼ ,

ab

A

cd

ªº

=«»

¬¼

ab

EA

ka c kb d

ªº

=«»

++

¬¼

det E = 1, det A = ad – bc,

det EA = a(kb + d) – (ka + c)b = kab + ad – kab – bc = 1(ad – bc) = (det E)(det A)

37.

31

,

42

Aªº

=«»

¬¼

15 5

5,

20 10

Aªº

=«»

¬¼

det A = 2, det 5A = 50 ≠ 5det A

38.

,

ab

A

cd

ªº

=«»

¬¼

,

ka kb

kA

kc kd

ªº

=«»

¬¼

det A = ad – bc,

22

det ( )( ) ( )( ) ( ) det

kA ka kd kb kc k ad bc k A

=−=−=

39. a. True. See the paragraph preceding the definition of the determinant.

b . False. See the definition of cofactor, which precedes Theorem 1.

40. a. False. See Theorem 1.

b . False. See Theorem 2.

174 CHAPTER 3 • Determinants

41. The area of the parallelogram determined by

3,

0

ª

º

=

«

»

¬

¼

u

1,

2

ª

º

=

«

»

¬

¼

v

u + v, and 0 is 6, since the base of

the parallelogram has length 3 and the height of the parallelogram is 2. By the same reasoning, the

area of the parallelogram determined by

3,

0

ª

º

=

«

»

¬

¼

u

,

2

x

ª

º

=

«

»

¬

¼

x

u + x, and 0 is also 6.

Also note that

[]

31

det det 6,

02

ªº

==

«»

¬¼

uv

and

[]

3

det det 6.

02

x

ªº

==

«»

¬¼

ux

The determinant of the

matrix whose columns are those vectors which define the sides of the parallelogram adjacent to 0 is

equal to the area of the parallelogram

42. The area of the parallelogram determined by

a

b

ª

º

=

«

»

¬

¼

u

,

0

c

ª

º

=

«

»

¬

¼

v

, u + v, and 0 is cb, since the base of

the parallelogram has length c and the height of the parallelogram is b.

Also note that

[]

det det 0

ac

cb

b

ªº

==−

«»

¬¼

uv

, and

[]

det det .

0

ca

cb

b

ªº

==

«»

¬¼

vu

The determinant of

the matrix whose columns are those vectors which define the sides of the parallelogram adjacent to 0

either is equal to the area of the parallelogram or is equal to the negative of the area of the

parallelogram.

43. [M] Answers will vary. The conclusion should be that det (A + B) ≠ det A + det B.

44. [M] Answers will vary. The conclusion should be that det (AB) = (det A)(det B).

X

V

UU

X2

X1X1

2

12

1

11224

X

2

a

b

c

X

1

U

V

3.2 • Solutions 175

45. [M] Answers will vary. For 4 × 4 matrices, the conclusions should be that

det det ,

T

AA

=

det(–A) = det A, det(2A) = 16det A, and

4

det (10 ) 10 det

AA

=

. For 5 × 5 matrices, the conclusions

should be that

det det ,

T

AA

=

det(–A) = –det A, det(2A) = 32det A, and

5

det (10 ) 10 det .

AA

=

For 6

× 6 matrices, the conclusions should be that

det det

T

AA

=

, det(–A) = det A, det(2A) = 64det A, and

6

det (10 ) 10 det .

AA

=

46. [M] Answers will vary. The conclusion should be that

1

det 1/ det .

AA

−

=

3.2 SOLUTIONS

Notes:

This section presents the main properties of the determinant, including the effects of row

operations on the determinant of a matrix. These properties are first studied by examples in Exercises 1–

20. The properties are treated in a more theoretical manner in later exercises. An efficient method for

computing the determinant using row reduction and selective cofactor expansion is presented in this

section and used in Exercises 11–14. Theorems 4 and 6 are used extensively in Chapter 5. The linearity

property of the determinant studied in the text is optional, but is used in more advanced courses.

1. Rows 1 and 2 are interchanged, so the determinant changes sign (Theorem 3b.).

2. The constant 2 may be factored out of the Row 1 (Theorem 3c.).

3. The row replacement operation does not change the determinant (Theorem 3a.).

4. The row replacement operation does not change the determinant (Theorem 3a.).

5.

156156156

1440120123

279033003

−−−

−− =−=−=

−− −

6.

15 31 53 153 153

3330181260326032(6)(3)18

213 7 0 3 1 0 3 1 0 0 1

−− − −

−=−=−=−=−=−

−−−

7.

1302130213021302

2574017801780178

0

35210425003027003027

112304250030270000

−−

====

−−

−− − −

8.

1334133 41334

0125012 50125

0

2543012 50000

3752024100000

−−−

===

−−−

−−− −

176 CHAPTER 3 • Determinants

9.

1130 1130 1130 1130

015401540154 0154 (3) 3

12 850 155 00 0 1 00 3 5

312302730035 0001

−− −− −− −−

== =−=− − =

−−−

−− −−

10.

131021310213102

024160241602416

262390003500035

373870208100477

3552704821300001

−− −− −−

−−− −−− −−−

===

−−

−− − − − −

−

13 1 0 2

02 4 1 6

(24) 24

00 4 7 7

00 0 3 5

00 0 0 1

−−

−−−

−=−− =

−−

11. First use a row replacement to create zeros in the second column, and then expand down the second

column:

25 3 1 25 3 1 313

30 1 3 30 1 3 56 4 9

60 4 9 60 4 9 021

410 4 1 00 2 1

−− −−

−

−−

==−− −

−− −−

−−

Now use a row replacement to create zeros in the first column, and then expand down the first

column:

313 313 23

56 4 9 50 2 3 (5)(3) (5)(3)(8)120

21

021 021

−−

−

−− − =−− =−=−−=

12. First use a row replacement to create zeros in the fourth column, and then expand down the fourth

column:

1230 12 30 12 3

3430 34 30 33 4 3

5466 30 20 30 2

4243 42 43

−−

−

==

−−

Now use a row replacement to create zeros in the first column, and then expand down the first

column:

12 3 1 2 3 10 12

33 4 3 30 10 12 3(1) 3(1)(38)114

611

30 2 0 6 11

−−

==−=− − =

−−

−− −−

13. First use a row replacement to create zeros in the fourth column, and then expand down the fourth

column:

3.2 • Solutions 177

2541 2541 032

4762 0320 16 2 4

6240 6240 677

6770 6770

−−

==−−−

−− −−

−

−−

Now use a row replacement to create zeros in the first column, and then expand down the first

column:

032 032 32

16 2 4 16 2 4 (1)(6) (1)(6)(1)6

53

677 053

−− −−

−−

−−−=−−−=−− =−− =

−

14. First use a row replacement to create zeros in the third column, and then expand down the third

column:

3214 3214 133

1303 1303

19 0 0

3428 9000 344

3404 3404

−− − −− −

−

−−

==−

−− −

−

−−

Now expand along the second row:

133 33

19 0 0 1((9)) (1)(9)(0)0

44

344

−

−=− − ==

−

15.

55(7)35

555

abc abc

def def

ghi ghi

===

16.

333 3 3(7)21

abc abc

def def

gh i ghi

===

17.

7

abc abc

gh i de f

def ghi

=−=−

18.

(7) 7

ghi abc abc

abc gh i de f

def def ghi

§·

¨¸

=−=−− =−− =

¨¸

©¹

19.

222 2222 2(7)14

abcabcabc

da eb f c d e f d e f

ghighighi

++ += = ==

178 CHAPTER 3 • Determinants

20.

7

ad be c f a b c

defdef

ghighi

+++

==

21. Since

230

134 10

121

=−≠

, the matrix is invertible.

22. Since

50 1

1320

053

−

−−=

, the matrix is not invertible.

23. Since

2008

1750

0

3860

0754

−−

=

, the matrix is not invertible.

24. Since

473

605110

726

−−

−=≠

−

, the columns of the matrix form a linearly independent set.

25. Since

787

450 10

675

−

−=−≠

−−

, the columns of the matrix form a linearly independent set.

26. Since

3220

5610

0

6030

4703

−

−−

=

−

, the columns of the matrix form a linearly dependent set.

27. a. True. See Theorem 3.

b . True. See the paragraph following Example 2.

c . True. See the paragraph following Theorem 4.

d . False. See the warning following Example 5.

28. a. True. See Theorem 3.

b . False. See the paragraphs following Example 2.

c . False. See Example 3.

d . False. See Theorem 5.

29. By Theorem 6,

555

det (det ) ( 2) 32

BB

==−=−

.

3.2 • Solutions 179

30. Suppose the two rows of a square matrix A are equal. By swapping these two rows, the matrix A is

not changed so its determinant should not change. But since swapping rows changes the sign of the

determinant, det A = – det A. This is only possible if det A = 0. The same may be proven true for

columns by applying the above result to

T

A

and using Theorem 5.

31. By Theorem 6,

1

(det )(det ) det 1

AA I

−

==

, so

1

det 1/ det .

AA

−

=

32. By factoring an r out of each of the n rows,

det ( ) det .

n

rA r A

=

33. By Theorem 6, det AB = (det A)(det B) = (det B)(det A) = det BA.

34. By Theorem 6 and Exercise 31,

111

det ( ) (det )(det )(det ) (det )(det )(det )

PAP P A P P P A

−−−

==

1

(det ) (det ) 1det

det

PAA

P

§·

==

¨¸

©¹

det

A

=

35. By Theorem 6 and Theorem 5,

2

det (det )(det ) (det ) .

TT

UU U U U

==

Since ,

T

UU I=

det det 1

T

UU I

==

, so

2

(det ) 1.

U

=

Thus det U = ±1.

36. By Theorem 6

44

det (det )

AA

=

. Since

4

det 0

A

=

, then

4

(det ) 0

A

=

. Thus det A = 0, and A is not

invertible by Theorem 4.

37. One may compute using Theorem 2 that det A = 3 and det B = 8, while

60

17 4

AB ªº

=«»

¬¼

. Thus

det AB = 24 = 3 × 8 = (det A)(det B).

38. One may compute that det A = 0 and det B = –2, while

60

20

AB

ª

º

=

«

»

−

¬

¼

. Thus

det AB = 0 = 0 × –2 = (det A)(det B).

39. a. By Theorem 6, det AB = (det A)(det B) = 4 × –3 = –12.

b . By Exercise 32,

3

det 5 5 det 125 4 500

AA

==×=

.

c . By Theorem 5,

det det 3

T

BB

==−

.

d . By Exercise 31,

1

det 1/ det 1/ 4

AA

−

==

.

e . By Theorem 6,

333

det (det ) 4 64

AA

===

.

40. a. By Theorem 6, det AB = (det A)(det B) = –1 × 2 = –2.

b . By Theorem 6,

555

det (det ) 2 32

BB

===

.

c . By Exercise 32,

4

det 2 2 det 16 1 16

AA

==×− =−

.

d . By Theorems 5 and 6,

det (det )(det ) (det )(det ) 1 1 1

TT

AA A A A A

===− × − =

.

180 CHAPTER 3 • Determinants

e . By Theorem 6 and Exercise 31,

11

det (det )(det )(det ) (1/ det )(det )(det ) det 1

BAB B A B B A B A

−−

== ==−

.

41. det A = (a + e)d – c(b + f) = ad + ed – bc – cf = (ad – bc) + (ed – cf) = det B + det C.

42.

1

det ( ) (1 )(1 ) 1 det det

1

ab

AB adcb adadcb Aad B

cd

+

+= =+ +−=+ + + −=+++

+

, so

det (A + B) = det A + det B if and only if a + d = 0.

43. Compute det A by using a cofactor expansion down the third column:

11 13 2 2 23 3 3 33

det ( )det ( )det ( )detAuv A uv A uv A=+ −+++

1132 233 331132 233 33

det det det det det detuAu AuAvAv AvA=−++−+

det det

BC

=+

44. By Theorem 5,

det det ( ) .

T

AE AE

=

Since

()

TTT

AE E A

=

,

det det( ).

TT

AE E A

=

Now

T

E

is itself

an elementary matrix, so by the proof of Theorem 3,

det ( ) (det )(det ).

TT T T

EA E A

=

Thus it is true

that

det (det )(det ),

TT

AE E A

=

and by applying Theorem 5, det AE = (det E)(det A).

45. [M] Answers will vary, but will show that

det T

AA

always equals 0 while

det T

AA

should seldom

be zero. To see why

T

AA

- should not be invertible (and thus

det 0

T

AA

=

), let A be a matrix with

more columns than rows. Then the columns of A must be linearly dependent, so the equation Ax = 0

must have a non-trivial solution x. Thus

() () ,

TT T

AA A A A

===

xx00

and the equation

()

T

AA

=

x0

has a

non-trivial solution. Since

T

AA

is a square matrix, the Invertible Matrix Theorem now says that

T

AA

is not invertible. Notice that the same argument will not work in general for ,

T

AA since

T

A

has more rows than columns, so its columns are not automatically linearly dependent.

46. [M] One may compute for this matrix that det A = –4008 and cond A ≈ 16.3. Note that this is the

2

A

condition number, which is used in Section 2.3. Since det A ≠ 0, it is invertible and

1

837 181 207 297

750 574 30 654

1

171 195 87 1095

4008

21 187 81 639

A

−

−−−

ªº

«»

−−

«»

=−«»

−−

«»

−−

¬¼

The determinant is very sensitive to scaling, as

4

det10 10 det 40,080,000AA==−

and

det 0.1A=

4

(0.1) det 0.4008.A=−

The condition number is not changed at all by scaling:

cond(10A) = cond(0.1A) = cond A ≈ 16.3. When

4

AI=

, det A=1 and cond A = 1. As before the

determinant is sensitive to scaling:

4

det10 10 det 10,000AA==

and

4

det 0.1 (0.1) det 0.0001.AA==

Yet the condition number is not changed by scaling: cond(10A) = cond(0.1A) = cond A = 1.

3.3 • Solutions 181

3.3 SOLUTIONS

Notes:

This section features several independent topics from which to choose. The geometric

interpretation of the determinant (Theorem 10) provides the key to changes of variables in multiple

integrals. Students of economics and engineering are likely to need Cramer’s Rule in later courses.

Exercises 1–10 concern Cramer’s Rule, exercises 11–18 deal with the adjugate, and exercises 19–32

cover the geometric interpretation of the determinant. In particular, Exercise 25 examines students’

understanding of linear independence and requires a careful explanation, which is discussed in the Study

Guide. The Study Guide also contains a heuristic proof of Theorem 9 for 2 × 2 matrices.

1. The system is equivalent to Ax = b, where

57

24

A

ª

º

=

«

»

¬

¼

and

3

1

ª

º

=

«

»

¬

¼

b

. We compute

12 12

37 53

() , () ,det 6,det () 5,det () 1,

14 21

AA AAA

ªº ªº

=====−

«» «»

¬¼ ¬¼

bb bb

12

det ( ) det ( )51

,.

det 6 det 6

AA

xx

AA

=== =−

bb

2. The system is equivalent to Ax = b, where

41

52

A

ª

º

=

«

»

¬

¼

and

6

7

ª

º

=

«

»

¬

¼

b

. We compute

12 12

61 46

() , () ,det 3,det () 5,det () 2,

72 57

AA AAA

ªº ªº

=====−

«» «»

¬¼ ¬¼

bb bb

12

det ( ) det ( )52

,.

det 3 det 3

AA

xx

AA

=== =−

bb

3. The system is equivalent to Ax = b, where

32

56

A

−

ª

º

=

«

»

−

¬

¼

and

7

5

ª

º

=

«

»

−

¬

¼

b

. We compute

12 12

72 37

() , () ,det 8,det () 32,det () 20,

56 55

AA AAA

−

ªº ªº

=====

«» «»

−−−

¬¼ ¬¼

bb bb

12

det ( ) det ( )32 20 5

4, .

det 8 det 8 2

AA

xx

AA

==== ==

bb

4. The system is equivalent to Ax = b, where

53

31

A

−

ª

º

=

«

»

−

¬

¼

and

9

5

ª

º

=

«

»

−

¬

¼

b

. We compute

12 12

93 59

() , () ,det 4,det () 6,det () 2,

51 35

AA AAA

−

ªº ªº

===−==−

«» «»

−− −

¬¼ ¬¼

bb bb

12

det ( ) det ( )63 21

,.

det 4 2 det 4 2

AA

xx

AA

−

===−===

−−

bb

182 CHAPTER 3 • Determinants

5. The system is equivalent to Ax = b, where

210

30 1

012

A

ª

º

«

»

=−

«

»

«

»

¬

¼

and

7

8

3

ª

º

«

»

=−

«

»

«

»

−

¬

¼

b

. We compute

12 3

710 2 70 21 7

() 8 0 1, () 3 8 1, () 3 0 8,

312 0 32 01 3

AA A

ªºª ºª º

«»« »« »

=−=−− =−−

«»« »« »

−−−

¬¼¬ ¼¬ ¼

bb b

12 3

det 4,det ( ) 6,det ( ) 16,det ( ) 14,AA A A== = =−bb b

3

12

12 3

det ( )

det ( ) det ( )63 16 14 7

,4, .

det 4 2 det 4 det 4 2

A

AA

xx x

AA A

−

==== =====−

b

bb

6. The system is equivalent to Ax = b, where

211

102

313

A

ª

º

«

»

=−

«

»

«

»

¬

¼

and

4

2

ª

º

«

»

=

«

»

«

»

−

¬

¼

b

. We compute

12 3

411 2 41 21 4

() 2 0 2, () 1 2 2, () 1 0 2,

213 3 23 31 2

AA A

ªºª ºª º

«»« »« »

==−=−

«»« »« »

−−−

¬¼¬ ¼¬ ¼

bb b

123

det 4, det ( ) 16, det ( ) 52, det ( ) 4,AA A A==−==−bbb

3

12

123

det ( )

det ( ) det ( )16 52 4

4, 13, 1.

det 4 det 4 det 4

A

AA

xxx

AAA

−−

===−======−

b

bb

7. The system is equivalent to Ax = b, where

64

92

s

A

s

ª

º

=

«

»

¬

¼

and

5

2

ª

º

=

«

»

−

¬

¼

b

. We compute

12 12

54 6 5

() , () ,det () 10 8,det () 12 45.

22 9 2

s

AA AsAs

s

ªº ªº

== =+=− −

«» «»

−−

¬¼ ¬¼

bb bb

Since

22

det 12 36 12( 3) 0As s=−=−≠

for 3s≠±, the system will have a unique solution when

3s≠±. For such a system, the solution will be

12

22 22

det ( ) det ( )10 8 5 4 12 45 4 15

,.

det det

12( 3) 6( 3) 12( 3) 4( 3)

AAss ss

xx

AA

ss ss

++ −− −−

== = = = =

−− −−

bb

8. The system is equivalent to Ax = b, where

35

95

s

A

s

−

ª

º

=

«

»

¬

¼

and

3

2

ª

º

=

«

»

¬

¼

b

. We compute

12 1 2

35 33

() , () ,det () 15 10,det () 6 27.

25 92

s

AA AsAs

s

−

ªº ªº

== =+=−

«» «»

¬¼ ¬¼

bb b b

Since

22

det 15 45 15( 3) 0As s=+= +≠

for all values of s, the system will have a unique solution for

all values of s. For such a system, the solution will be

12

22 22

det ( ) det ( )15 10 3 2 627 29

,.

det det

15( 3) 3( 3) 15( 3) 5( 3)

AAss ss

xx

AA

ss ss

++ −−

== = = = =

++ ++

bb

3.3 • Solutions 183

9. The system is equivalent to Ax = b, where

2

36

ss

A

s

−

ª

º

=

«

»

¬

¼

and

1

4

−

ª

º

=

«

»

¬

¼

b

. We compute

12 12

12 1

() ,() ,det()2,det()4 3.

46 34

ss

AA AsAs

s

−− −

ªº ªº

== ==+

«» «»

¬¼ ¬¼

bb bb

Since

2

det 6 6 6 ( 1) 0As sss=+= +=

for s = 0, –1, the system will have a unique solution when s ≠ 0,

–1. For such a system, the solution will be

12

det ( ) det ( )21 43

,.

det 6 ( 1) 3( 1) det 6 ( 1)

AAss

xx

Ass s Ass

+

=== = =

++ +

bb

10. The system is equivalent to Ax = b, where

21

36

s

A

ss

ª

º

=

«

»

¬

¼

and

1

2

ª

º

=

«

»

¬

¼

b

. We compute

12 1 2

11 21

() , () ,det () 6 2,det () .

26 3 2

s

AA AsAs

ss

ªº ªº

== =−=

«» «»

¬¼ ¬¼

bb b b

Since

2

det 12 3 3 (4 1) 0Assss=−=−=

for s = 0,1/4, the system will have a unique solution when

s ≠ 0,1/4. For such a system, the solution will be

12

det ( ) det ( )62 1

,.

det 3 (4 1) det 3 (4 1) 3(4 1)

AAss

xx

Ass Ass s

−

== = = =

−−−

bb

11. Since det A = 3 and the cofactors of the given matrix are

11

00 0,

11

C==

12

30 3,

11

C=−=−

−

13

30 3,

11

C==

−

21

1,

11

C

−−

=−=

22

01 1,

11

C

−

==−

−

23

02

2,

11

C

−

=−=

−

31

21

0,

00

C

−−

==

32

01 3,

30

C

−

=−=−

33

02

6,

30

C

−

==

010

adj 3 1 3

326

A

ªº

«»

=−−−

«»

¬¼

and

1

01/30

1adj 1 1/ 3 1 .

det 12/32

AA

A

−

ª

º

«

»

==−− −

«

»

«

»

¬

¼

12. Since det A = 5 and the cofactors of the given matrix are

11

21 1,

10

C

−

==−

12

210,

00

C=−=

13

22

2,

01

C

−

==

21

13 3,

10

C=−=

22

13 0,

00

C==

23

11 1,

01

C=−=−

31

13 7,

21

C==

−

32

13 5,

21

C=−=

33

11 4,

22

C==−

−

137

adj 0 0 5

214

A

−

ªº

«»

=«»

«»

−−

¬¼

and

1

1/5 3/5 7/5

1adj 0 0 1 .

det 2/5 1/5 4/5

AA

A

−

ª

º

«

»

==

«

»

«

»

−−

¬

¼

184 CHAPTER 3 • Determinants

13. Since det A = 6 and the cofactors of the given matrix are

11

01 1,

11

C==−

12

11 1,

21

C=−=

13

10 1,

21

C==

21

54 1,

11

C=−=−

22

34 5,

21

C==−

23

35 7,

21

C=−=

31

54 5,

01

C==

32

341,

11

C=−=

33

35 5,

10

C==−

115

adj 1 5 1

17 5

A

−−

ªº

«»

=−

«»

−

¬¼

and

1

1/6 1/6 5/6

1adj 1/ 6 5/ 6 1/ 6 .

det 1/6 7/ 6 5/ 6

AA

A

−

−−

ª

º

«

»

==−

«

»

«

»

−

¬

¼

14. Since det A = –1 and the cofactors of the given matrix are

11

215,

34

C==

12

012,

24

C=−=

13

02 4,

23

C==−

21

67 3,

34

C=−=−

22

37 2,

24

C==−

23

36 3,

23

C=−=

31

67 8,

21

C==−

32

37 3,

01

C=−=−

33

36 6,

02

C==

538

adj 2 2 3

436

A

−−

ªº

«»

=−−

«»

−

¬¼

and

1

538

1adj 2 2 3 .

det 436

AA

A

−

ª

º

«

»

==−

«

»

«

»

−−

¬

¼

15. Since det A = 6 and the cofactors of the given matrix are

11

10 2,

32

C==

12

10 2,

22

C

−

=−=

−

13

11 1,

23

C

−

==−

−

21

00 0,

32

C=−=

22

30 6,

22

C==

−

23

30 9,

23

C=−=−

−

31

00 0,

10

C==

32

30 0,

10

C=−=

−

33

30 3,

11

C==

−

200

adj 2 6 0

193

A

ªº

«»

=«»

«»

−−

¬¼

and

1

1/3 0 0

1adj 1/ 3 1 0 .

det 1/6 3/2 1/2

AA

A

−

ª

º

«

»

==

«

»

«

»

−−

¬

¼

16. Since det A = –9 and the cofactors of the given matrix are

11

31 9,

03

C

−

==−

12

010,

03

C=−=

13

03

0,

00

C

−

==

21

24 6,

03

C=−=−

22

14 3,

03

C==

23

12 0,

00

C=−=

3.3 • Solutions 185

31

2414,

31

C==

−

32

14 1,

01

C=−=−

33

12 3,

03

C==−

−

9614

adj 0 3 1

003

A

−−

ªº

«»

=−

«»

−

¬¼

and

1

12/3 14/9

1adj 0 1/ 3 1/ 9 .

det 001/3

AA

A

−

ª

º

«

»

==−

«

»

«

»

¬

¼

17. Let

ab

A

cd

ªº

=

«»

¬¼

. Then the cofactors of A are

11

,Cdd==

12

,Ccc=−=−

21

Cbb=−=−

, and

22

Caa==

. Thus

adj db

A

ca

−

ªº

=

«»

−

¬¼

. Since det A = ad – bc, Theorem 8 gives that

1

11

adj

det

db

AA

ca

Aadbc

−

ªº

==

«»

−

¬¼

. This result is identical to that of Theorem 4 in Section 2.2.

18. Each cofactor of A is an integer since it is a sum of products of entries in A. Hence all entries in adj A

will be integers. Since det A = 1, the inverse formula in Theorem 8 shows that all the entries in

1

A

−

will be integers.

19. The parallelogram is determined by the columns of

56

24

A

ª

º

=

«

»

¬

¼

, so the area of the parallelogram is

|det A| = |8| = 8.

20. The parallelogram is determined by the columns of

14

35

A

−

ª

º

=

«

»

−

¬

¼

, so the area of the parallelogram

is |det A| = |–7| = 7.

21. First translate one vertex to the origin. For example, subtract (–1, 0) from each vertex to get a new

parallelogram with vertices (0, 0),(1, 5),(2, –4), and (3, 1). This parallelogram has the same area as

the original, and is determined by the columns of

12

54

A

ª

º

=

«

»

−

¬

¼

, so the area of the parallelogram is

|det A| = |–14| = 14.

22. First translate one vertex to the origin. For example, subtract (0, –2) from each vertex to get a new

parallelogram with vertices (0, 0),(6, 1),(–3, 3), and (3, 4). This parallelogram has the same area as

the original, and is determined by the columns of

63

13

A

−

ª

º

=

«

»

¬

¼

, so the area of the parallelogram is

|det A| = |21| = 21.

23. The parallelepiped is determined by the columns of

117

021

240

A

ª

º

«

»

=

«

»

«

»

−

¬

¼

, so the volume of the

parallelepiped is |det A| = |22| = 22.

186 CHAPTER 3 • Determinants

24. The parallelepiped is determined by the columns of

121

452

021

A

−−

ª

º

«

»

=−

«

»

«

»

−

¬

¼

, so the volume of the

parallelepiped is |det A| = |–15| = 15.

25. The Invertible Matrix Theorem says that a 3 × 3 matrix A is not invertible if and only if its columns

are linearly dependent. This will happen if and only if one of the columns is a linear combination of

the others; that is, if one of the vectors is in the plane spanned by the other two vectors. This is

equivalent to the condition that the parallelepiped determined by the three vectors has zero volume,

which is in turn equivalent to the condition that det A = 0.

26. By definition, p + S is the set of all vectors of the form p + v, where v is in S. Applying T to a typical

vector in p + S, we have T(p + v) = T(p) + T(v). This vector is in the set denoted by T(p) + T(S). This

proves that T maps the set p + S into the set T(p) + T(S). Conversely, any vector in T(p) + T(S) has

the form T(p) + T(v) for some v in S. This vector may be written as T(p + v). This shows that every

vector in T(p) + T(S) is the image under T of some point p + v in p + S.

27. Since the parallelogram S is determined by the columns of

22

35

−−

ª

º

«

»

¬

¼

, the area of S is

22

det | 4 | 4.

35

−−

ªº

=−=

«»

¬¼

The matrix A has

62

det 6

32

A

−

==

−

. By Theorem 10, the area of T(S)

is |det A|{area of S} = 6 ⋅ 4 = 24. Alternatively, one may compute the vectors that determine the

image, namely, the columns of

[]

12

6222 1822

3235 1216

A

−− − − −

ªºªºª º

==

«»«»« »

−

¬¼¬¼¬ ¼

bb

The determinant of this matrix is –24, so the area of the image is 24.

28. Since the parallelogram S is determined by the columns of

40

71

ª

º

«

»

−

¬

¼

, the area of S is

40

det | 4 | 4

71

ªº

==

«»

−

¬¼

. The matrix A has

72

det 5

11

A==

. By Theorem 10, the area of T(S) is

|det A|{area of S} =5 ⋅ 4 = 20. Alternatively, one may compute the vectors that determine the image,

namely, the columns of

[]

12

72 40 142

11 71 31

A

ªºª ºª º

==

«»« »« »

−−

¬¼¬ ¼¬ ¼

bb

The determinant of this matrix is 20, so the area of the image is 20.

29. The area of the triangle will be one half of the area of the parallelogram determined by

1

v

and

2

.v

By Theorem 9, the area of the triangle will be (1/2)|det A|, where

[]

12

.A=

vv

30. Translate R to a new triangle of equal area by subtracting

33

(, )xy

from each vertex. The new triangle

has vertices (0, 0),

131 3

(, )xxyy−−

, and

2323

(, ).xxyy−−

By Exercise 29, the area of the triangle

will be

3.3 • Solutions 187

13 23

1det .

2

xx x x

yy yy

−−

ªº

«»

−−

¬¼

Now consider using row operations and a cofactor expansion to compute the determinant in the

formula:

11 13 13

13 13

22 23 23

23 23

33 33

10

det 1 det 0det

11

xy xxyy xx yy

xy x y

−−

ªºª º

−−

ª

º

«»« »

=−−=

«

»

«»« »

−−

¬

¼

«»« »

¬¼¬ ¼

By Theorem 5,

13 13 13 23

23 23 13 23

det det

xx yy xx x x

xx yy yy yy

−− −−

ªºªº

=

«»«»

−− −−

¬¼¬¼

So the above observation allows us to state that the area of the triangle will be

11

13 23

22

13 23

33

1

11

det det 1

22

1

xy

xx x x xy

yy y y xy

ª

º

−−

ªº

«

»

=

«»

«

»

−−

¬¼

«

»

¬

¼

31. a. To show that T(S) is bounded by the ellipsoid with equation

2

22

312

222

1

xxx

abc

++=

, let

1

2

3

u

ªº

«»

=«»

«»

¬¼

u

and

let

1

2

3

x

xA

x

ªº

«»

==

«»

¬¼

xu

. Then

11

/

uxa=

,

22

/

uxb=

, and

33

/

uxc=

, and u lies inside S (or

222

123

1uuu++≤

) if and only if x lies inside T(S) (or

2

22

312

222

1

xxx

abc

++≤

).

b. By the generalization of Theorem 10,

{volume of ellipsoid} {volume of ( )}TS=

44

|det | {volumeof } 33

abc

ASabc

ππ

=⋅==

32. a. A linear transformation T that maps S onto S′will map

1

e

to

1

,v

2

e

to

2

,v

and

3

e

to

3

;v

that is,

11

()T=ev

,

22

()T=ev

, and

33

() .T=ev

The standard matrix for this transformation will be

[][]

123123

() ( ) () .AT T T==

ee evvv

188 CHAPTER 3 • Determinants

b. The area of the base of S is (1/2)(1)(1) = 1/2, so the volume of S is (1/3)(1/2)(1) = 1/6. By part a.

T(S) = S′, so the generalization of Theorem 10 gives that the volume of S′ is |det A|{volume of

S} = (1/6)|det A|.

33. [M] Answers will vary. In MATLAB, entries in B – inv(A) are approximately

15

10

−

or smaller.

34. [M] Answers will vary, as will the commands which produce the second entry of x. For example, the

MATLAB command is x2 = det([A(:,1) b A(:,3:4)])/det(A) while the Mathematica

command is x2 = Det[{Transpose[A][[1]],b,Transpose[A][[3]],

Transpose[A][[4]]}]/Det[A].

35. [M] MATLAB Student Version 4.0 uses 57,771 flops for inv A and 14,269,045 flops for the inverse

formula. The inv(A) command requires only about 0.4% of the operations for the inverse formula.

Chapter 3 SUPPLEMENTARY EXERCISES

1. a. True. The columns of A are linearly dependent.

b . True. See Exercise 30 in Section 3.2.

c. False. See Theorem 3(c); in this case

3

det 5 5 detAA=

.

d . False. Consider

20

01

A

ªº

=

«»

¬¼

,

10

03

B

ª

º

=

«

»

¬

¼

, and

30

04

AB

ª

º

+=

«

»

¬

¼

.

e . False. By Theorem 6,

33

det 2A=

.

f. False. See Theorem 3(b).

g . True. See Theorem 3(c).

h . True. See Theorem 3(a).

i. False. See Theorem 5.

j. False. See Theorem 3(c); this statement is false for n × n invertible matrices with n an even

integer.

k . True. See Theorems 6 and 5;

2

det (det )

T

AA A=

.

l. False. The coefficient matrix must be invertible.

m. False. The area of the triangle is 5.

n. True. See Theorem 6;

33

det (det )AA=

.

o. False. See Exercise 31 in Section 3.2.

p. True. See Theorem 6.

2.

12 13 14 12 13 14

15 16 17 3 3 3 0

18 19 20 6 6 6

==

3.

11 1

10 ()()0110

10 011

abc a bc abc

bac baab baca

cab caac

++ +

+= −−=−− −=

+−− −

Chapter 3 • Supplementary Exercises 189

4.

111 0

111

abcabc abc

ax bx cx x x x xy

aybycy y y y

+++= = =

+++

5.

91999 9992 405

90992 4050

(1) (1)(2)9 3 9

40050 9390 607

90390 6070

60070

=−=−−

45

(1)(2)(3) (1)(2)(3)(2) 12

67

=−− =−− −=−

6.

48885 4885 485

01000 6887 45

(1) (1)(2) 6 8 7 (1)( 2)( 3) (1)(2)( 3)( 2) 12

68887 0830 67

030

08830 0200

08200

===−=−−=

7. Expand along the first row to obtain

11 1 1

11

22 2 2

22

111

11 0.

11

1

xy xy y x

xy x y

xy y x

xy

=−+=

This is an equation of the form ax + by + c = 0, and since the points

11

(, )xy

and

22

(, )xy

are distinct,

at least one of a and b is not zero. Thus the equation is the equation of a line. The points

11

(, )xy

and

22

(, )xy

are on the line, because when the coordinates of one of the points are substituted for x and y,

two rows of the matrix are equal and so the determinant is zero.

8. Expand along the first row to obtain

11 1 1

11 11

111

11 1()()(1)0.

1001

01

xy xy y x

xy x y mxyxmy

mm

m

=−+=−− +=

This equation may

be rewritten as

11

0,mx y mx y−− +=

or

11

().yy mxx−=−

9.

22 2

222

11 1

det 1 0 0 ( )( )

0()()

10

aa a a aa

Tbb baba bababa

ca caca

cc cac a

==−−=−−+

−−+

−−

22

11

()()01 ()()01 ()()()

01 00

aa aa

baca ba baca ba bacacb

ca cb

=−− +=−− +=−−−

+−

190 CHAPTER 3 • Determinants

10. Expanding along the first row will show that

23

01 2 3

() det .ft V c ct ct ct==+++

By Exercise 9,

2

11

2

322213132

2

33

1

1()()()0

1

xx

cxxxxxxxx

xx

=−=−−−≠

since

1

x

,

2

x

, and

3

x

are distinct. Thus f (t) is a cubic polynomial. The points

1

(,0)x

,

2

(,0)x

, and

3

(,0)x

are on the graph of f, since when any of

1

x

,

2

x

or

3

x

are substituted for t, the matrix has two

equal rows and thus its determinant (which is f (t)) is zero. Thus

() 0

i

fx =

for i = 1, 2, 3.

11. To tell if a quadrilateral determined by four points is a parallelogram, first translate one of the

vertices to the origin. If we label the vertices of this new quadrilateral as 0,

1

v

,

2

v

, and

3

v

, then

they will be the vertices of a parallelogram if one of

1

v

,

2

v

, or

3

v

is the sum of the other two. In

this example, subtract (1, 4) from each vertex to get a new parallelogram with vertices 0 = (0, 0),

1

(2,1)=−v

,

2

(2,5)=v

, and

3

(4,4)=v

. Since

231

=+vvv

, the quadrilateral is a parallelogram as

stated. The translated parallelogram has the same area as the original, and is determined by the

columns of

[]

13

24

14

A

−

ªº

==

«»

¬¼

vv

, so the area of the parallelogram is |det A| = |–12| = 12.

12. A 2 × 2 matrix A is invertible if and only if the parallelogram determined by the columns of A has

nonzero area.

13. By Theorem 8,

1

(adj ) det

AAAAI

A

−

⋅==

. By the Invertible Matrix Theorem, adj A is invertible

and

1

(adj ) det

AA

A

−

=

.

14. a. Consider the matrix

k

AO

AOI

ªº

=

«»

¬¼

, where 1 ≤ k ≤ n and O is an appropriately sized zero matrix.

We will show that

det det

k

AA=

for all 1 ≤ k ≤ n by mathematical induction.

First let k = 1. Expand along the last row to obtain

(1)(1)

1

det det ( 1) 1 det det .

1

nn

AO

AAA

O

++ +

ªº

==−⋅⋅=

«»

¬¼

Now let 1 < k ≤ n and assume that

1

det det .

k

AA

−

=

Expand along the last row of

k

A

to obtain

()()

11

det det ( 1) 1 det det det .

nk nk

kkk

k

AO

AAAA

OI

+++

−−

ªº

==−⋅⋅==

«»

¬¼

Thus we have proven the

result, and the determinant of the matrix in question is det A.

b. Consider the matrix

k

IO

ACD

ªº

=

«»

¬¼

, where 1 ≤ k ≤ n,

k

C

is an n × k matrix and O is an

appropriately sized zero matrix. We will show that

det det

k

AD=

for all 1 ≤ k ≤ n by

mathematical induction.

Chapter 3 • Supplementary Exercises 191

First let k = 1. Expand along the first row to obtain

11

1

det det ( 1) 1 det det .

O

ADD

CD

+

ªº

==−⋅⋅=

«»

¬¼

Now let 1 < k ≤ n and assume that

1

det det .

k

AD

−

=

Expand along the first row of

k

A

to obtain

11

det det (1) 1det det det .

k

kkk

k

IO

AAAD

CD

+

−−

ªº

==−⋅⋅==

«»

¬¼

Thus we have proven the result, and

the determinant of the matrix in question is det D.

c. By combining parts a. and b., we have shown that

det det det (det )(det ).

AO AO I O

AD

CD O I CD

§·§·

ªº ªºªº

==

¨¸¨¸

«» «»«»

¬¼ ¬¼¬¼

©¹©¹

From this result and Theorem 5, we have

det det det (det )(det )

TT

AB AB A O

AD

OD OD BD

ªº

ªºªº

== =

«»

«»«»

¬¼¬¼ «»

¬¼

(det )(det ).AD=

15. a. Compute the right side of the equation:

IOAB A B

XIOY XAXBY

ªºªºª º

=

«»«»« »

+

¬¼¬¼¬ ¼

Set this equal to the left side of the equation:

so that

AB A B

XA C XB Y D

CD XAXBY

ªºª º

==+ =

«»« »

+

¬¼¬ ¼

Since XA = C and A is invertible,

1

.XCA

−

=

Since XB + Y = D,

1

YDXBDCAB

−

=−=−

. Thus

by Exercise 14(c),

11

det det det

IO A B

AB

CD CA I O D CA B

−−

ªºª º

ªº

=

«»« »

«»

−

¬¼

¬¼¬ ¼

1

(det )(det ( ))ADCAB

−

=−

b. From part a.,

11

det (det )(det ( )) det[ ( )]

AB

ADCAB ADCAB

CD

−−

ªº

=−=−

«»

¬¼

11

det[ ]det[ ]AD ACA B AD CAA B

−−

=−=−

det[ ]AD CB=−

16. a. Doing the given operations does not change the determinant of A since the given operations are

all row replacement operations. The resulting matrix is

00

00 0

ab ab

ab

bb b a

−−+…

ªº

«»

−−+…

«»

−…

«»

…

¬¼

## #%#

192 CHAPTER 3 • Determinants

b. Since column replacement operations are equivalent to row operations on

T

A

and

det det

T

AA=

,

the given operations do not change the determinant of the matrix. The resulting matrix is

00 0

23 (1)

ab

bbb anb

−…

ªº

«»

−…

«»

−…

«»

…+−

¬¼

###% #

c. Since the preceding matrix is a triangular matrix with the same determinant as A,

1

det ( ) ( ( 1) ).

n

Aab an b

−

=−+−

17. First consider the case n = 2. In this case

2

det (),det ,

0

ab b bb

BaabC ab b

aba

−

==−==−

so

222 21

det det det ( ) ()()()((21))ABCaababbababababa b

−

=+=−+−=−=−+=−+−

, and

the formula holds for n = 2.

Now assume that the formula holds for all (k – 1) × (k – 1) matrices, and let A, B, and C be k × k

matrices. By a cofactor expansion along the first column,

21

det ( ) ()()((2))()((2))

kk

ab b

ba b

Bab abab ak b ab ak b

bb a

−−

…

=−=−− +−=−+−

…

##%#

since the matrix in the above formula is a (k – 1) × (k – 1) matrix. We can perform a series of row

operations on C to “zero out” below the first pivot, and produce the following matrix whose

determinant is det C:

00

.

00

bb b

ab

…

ªº

«»

−…

«»

…−

«»

¬¼

##%#

Since this is a triangular matrix, we have found that

1

det ( )

k

Cbab

−

=−

. Thus

111

det det det ( ) ( ( 2) ) ( ) ( ) ( ( 1) ),

kkk

ABCabakbbab abakb

−−−

=+=−+−+−=−+−

which is what was to be shown. Thus the formula has been proven by mathematical induction.

18. [M] Since the first matrix has a = 3, b = 8, and n = 4, its determinant is

41 3

(3 8) (3 (4 1)8) ( 5) (3 24) ( 125)(27) 3375.

−

−+−=−+=−=−

Since the second matrix has a = 8, b =

3, and n = 5, its determinant is

51 4

(8 3) (8 (5 1)3) (5) (8 12) (625)(20) 12,500.

−

−+−=+= =

Chapter 3 • Supplementary Exercises 193

19. [M] We find that

11111

1111

111 12222

1222

122 1, 1, 1.

12333

1233

123 12344

1234 12345

== =

Our conjecture then is that

111 1

122 2

1.

123 3

123 n

…

=

…

## #%#

To show this, consider using row replacement operations to “zero out” below the first pivot. The

resulting matrix is

111 1

011 1

.

012 2

012 1n

…

ªº

«»

…

«»

…

«»

…−

¬¼

###% #

Now use row replacement operations to “zero out” below the second pivot, and so on. The final

matrix which results from this process is

111 1

011 1

,

001 1

000 1

…

ªº

«»

…

«»

…

«»

…

¬¼

###%#

which is an upper triangular matrix with determinant 1.

20. [M] We find that

1111 1

111 1

111 1333 3

1333

133 6, 18, 54.

1366 6

1366

136 136 9 9

1369 136 912

== =

Our conjecture then is that

194 CHAPTER 3 • Determinants

2

111 1

133 3

23 .

136 6

136 3( 1)

n

−

…

=⋅

…

…−

###% #

To show this, consider using row replacement operations to “zero out” below the first pivot. The

resulting matrix is

111 1

022 2

.

025 5

025 3( 1)1n

…

ªº

«»

…

«»

…

«»

…−−

¬¼

###% #

Now use row replacement operations to “zero out” below the second pivot. The matrix which results

from this process is

11111 1 1

02222 2 2

00333 3 3

.

00366 6 6

00369 9 9

0036912 3( 2)n

…

ªº

«»

…

«»

…

«»

…

«»

…

«»

…−

¬¼

##### #% #

This matrix has the same determinant as the original matrix, and is recognizable as a block matrix of

the form

,

AB

OD

ªº

«»

¬¼

where

333 3 3 1111 1

366 6 6 1222 2

11

and 3 .

369 9 9 1233 3

02

36912 3( 2) 1234 2

AD

nn

……

ª

ºª º

«

»« »

……

«

»« »

ªº

«

»« »

== =

……

«»

«

»« »

¬¼

«

»« »

«

»« »

…− −

¬

¼¬ ¼

### #% # ####% #

!

As in Exercise 14(c), the determinant of the matrix

AB

OD

ª

º

«

»

¬

¼

is (det A)(det D) = 2 det D.

Since D is an (n – 2) × (n – 2) matrix,

Chapter 3 • Supplementary Exercises 195

222

111 1 1

1222 2

det 3 3 (1) 3

1233 3

1234 2

nnn

D

n

−−−

…

===

…

…−

####% #

by Exercise 19. Thus the determinant of the matrix

AB

OD

ª

º

«

»

¬

¼

is

2

2det 2 3 .

n

D

−

=⋅

r

4.1 SOLUTIONS

Notes:

This section is designed to avoi

axioms on an array of sets. Theorem 1

p

set is a subspace. Students should be ta

u

(and the next few sections) emphasize

vectors do appear later in the chapter: t

h

polynomials are used in many sections

o

1. a. If u and v are in V, then their en

t

nonnegative, the vector u + v h

a

b. Example: If

2

ªº

=«»

¬¼

u

and c = –1

2. a. If

x

y

ªº

=«»

¬¼

u

is in W, then the vec

t

since xy ≥ 0.

b. Example: If

1

7

−

ªº

=«»

−

¬¼

u

and

ª

=«

¬

v

3. Example: If

.5

ªº

=«»

¬¼

u

and c = 4, the

n

multiplication, H is not a subspace

o

4. Note that u and v are on the line L,

u

v

L

u+v

r

son Education, Inc. Publishing as Addison-Wesley.

d the standard exercises in which a student is asked

p

rovides the main homework tool in this section for s

h

u

ght how to check the closure axioms. The exercises i

n

, to give students time to absorb the abstract co

n

h

e space of signals is used in Section 4.8, and the

o

f Chapters 4 and 6.

t

ries are nonnegative. Since a sum of nonnegative nu

m

a

s nonnegative entries. Thus u + v is in V.

, then u is in V but cu is not in V.

t

or

xcx

cc

ycy

ªº ª º

==

«» « »

¬¼ ¬ ¼

u

is in W because

2

()() (cx cy c x

y

=

2

3

ª

º

»

¬

¼

, then u and v are in W but u + v is not in W.

n

u is in H but cu is not in H. Since H is not closed u

n

o

f

2

.

but u + v is not.

197

to check ten

h

owing that a

n

this section

n

cepts. Other

spaces

n

of

m

bers is

)0

y

≥

n

der scalar

198 CHAPTER 4 • Vector Spaces

5. Yes. Since the set is

2

Span{ }t

, the set is a subspace by Theorem 1.

6. No. The zero vector is not in the set.

7. No. The set is not closed under multiplication by scalars which are not integers.

8. Yes. The zero vector is in the set H. If p and q are in H, then (p + q)(0) = p(0) + q(0) = 0 + 0 = 0,

so p + q is in H. For any scalar c, (cp)(0) = c ⋅ p(0) = c ⋅ 0 = 0, so cp is in H. Thus H is a subspace by

Theorem 1.

9. The set H = Span

{v}, where

2

5

3

−

ªº

«»

=«»

«»

¬¼

v

. Thus H is a subspace of

3

by Theorem 1.

10. The set H = Span

{v}, where

3

0

7

ªº

«»

=«»

«»

−

¬¼

v

. Thus H is a subspace of

3

by Theorem 1.

11. The set W = Span

{u, v}, where

2

1

0

ªº

«»

=−

«»

¬¼

u

and

3

0

2

ª

º

«

»

=

«

»

«

»

¬

¼

v

. Thus W is a subspace of

3

by Theorem 1.

12. The set W = Span

{u, v}, where

2

0

ªº

«»

=«»

«»

¬¼

u

and

4

0

3

5

ª

º

«

»

«

»

=

«

»

−

«

»

¬

¼

v

. Thus W is a subspace of

4

by Theorem 1.

13. a. The vector w is not in the set

123

{, , }vv v

. There are 3 vectors in the set

123

{, , }.vv v

b. The set

123

Span{ , , }vv v

contains infinitely many vectors.

c. The vector w is in the subspace spanned by

123

{, , }vv v

if and only if the equation

11 2 2 3 3

xx x++=vv vw

has a solution. Row reducing the augmented matrix for this system of

linear equations gives

1243 100 1

0121 0121,

1362 0000

ªºªº

«»«»

∼

«»«»

−

¬¼¬¼

so the equation has a solution and w is in the subspace spanned by

123

{, , }vv v

.

14. The augmented matrix is found as in Exercise 13c. Since

1241 100 5

0123 012 3,

13614 000 0

−

ªºªº

«»«»

∼

«»«»

−

¬¼¬¼

4.1 • Solutions 199

the equation

11 2 2 3 3

xx x++=vvvw

has a solution, the vector w is in the subspace spanned by

123

{, , }.vv v

15. Since the zero vector is not in W, W is not a vector space.

16. Since the zero vector is not in W, W is not a vector space.

17. Since a vector w in W may be written as

210

031

103

030

abc

−

ªº ªº ªº

«» «» «»

−

«» «» «»

=++

«» «» «»

−

«» «» «»

¬¼ ¬¼ ¬¼

w

210

031

,,

103

030

S

½−

ªºªºªº

°°

«»«»«»

−

°°

«»«»«»

=®¾

«»«»«»

−

°°

«»«»«»

°°

¬¼¬¼¬¼

¯¿

is a set that spans W.

18. Since a vector w in W may be written as

43 0

00 0

13 1

03 2

abc

ªº ªº ª º

«» «» « »

=++

«» «» « »

−

¬¼ ¬¼ ¬ ¼

w

43 0

00 0

,,

13 1

03 2

S

½

ªºªºª º

°°

«»«»« »

°°

«»«»« »

=®¾

«»«»« »

°°

«»«»« »

°°

−

¬¼¬¼¬ ¼

¯¿

is a set that spans W.

19. Let H be the set of all functions described by

12

() cos sin .yt c t c t

ωω

=+

Then H is a subset of the

vector space V of all real-valued functions, and may be written as H = Span

{cos

ω

t, sin

ω

t}. By

Theorem 1, H is a subspace of V and is hence a vector space.

20. a. The following facts about continuous functions must be shown.

1. The constant function f(t) = 0 is continuous.

2. The sum of two continuous functions is continuous.

3. A constant multiple of a continuous function is continuous.

b. Let H = {f in C[a, b]: f(a) = f(b)}.

1. Let g(t) = 0 for all t in [a, b]. Then g(a) = g(b) = 0, so g is in H.

2. Let g and h be in H. Then g(a) = g(b) and h(a) = h(b), and (g + h)(a) = g(a) + h(a) =

g(b) + h(b) = (g + h)(b), so g + h is in H.

3. Let g be in H. Then g(a) = g(b), and (cg)(a) = cg(a) = cg(b) = (cg)(b), so cg is in H.

200 CHAPTER 4 • Vector Spaces

Thus H is a subspace of C[a, b].

21. The set H is a subspace of

22

.M×

The zero matrix is in H, the sum of two upper triangular matrices is

upper triangular, and a scalar multiple of an upper triangular matrix is upper triangular.

22. The set H is a subspace of

24

.M×

The 2 × 4 zero matrix 0 is in H because F0 = 0. If A and B are

matrices in H, then F(A + B) = FA + FB = 0 + 0 = 0, so A + B is in H. If A is in H and c is a scalar,

then F(cA) = c(FA) = c0 = 0, so cA is in H.

23. a. False. The zero vector in V is the function f whose values f(t) are zero for all t in .

b . False. An arrow in three-dimensional space is an example of a vector, but not every arrow is a

vector.

c . False. See Exercises 1, 2, and 3 for examples of subsets which contain the zero vector but are not

subspaces.

d . True. See the paragraph before Example 6.

e . False. Digital signals are used. See Example 3.

24. a. True. See the definition of a vector space.

b . True. See statement (3) in the box before Example 1.

c . True. See the paragraph before Example 6.

d . False. See Example 8.

e . False. The second and third parts of the conditions are stated incorrectly. For example, part (ii)

does not state that u and v represent all possible elements of H.

25. 2, 4

26. a. 3

b . 5

c . 4

27. a. 8

b . 3

c . 5

d . 4

28. a. 4

b . 7

c . 3

d . 5

e . 4

29. Consider u + (–1)u. By Axiom 10, u + (–1)u = 1u + (–1)u. By Axiom 8, 1u + (–1)u = (1 + (–1))u =

0u. By Exercise 27, 0u = 0. Thus u + (–1)u = 0, and by Exercise 26 (–1)u = –u.

30. By Axiom 10 u = 1u. Since c is nonzero,

11cc

−

=

, and

1

()cc

−

=uu

. By Axiom 9,

11 1

() ()cc c c c

−− −

==uu0

since cu = 0. Thus

1

c−

==u00

by Property (2), proven in Exercise 28.

4.1 • Solutions 201

31. Any subspace H that contains u and v must also contain all scalar multiples of u and v, and hence

must also contain all sums of scalar multiples of u and v. Thus H must contain all linear

combinations of u and v, or Span

{u, v}.

Note:

Exercises 32–34 provide good practice for mathematics majors because these arguments involve

simple symbol manipulation typical of mathematical proofs. Most students outside mathematics might

profit more from other types of exercises.

32. Both H and K contain the zero vector of V because they are subspaces of V. Thus the zero vector of V

is in H ∩ K. Let u and v be in H ∩ K. Then u and v are in H. Since H is a subspace u + v is in H.

Likewise u and v are in K. Since K is a subspace u + v is in K. Thus u + v is in H ∩ K. Let u be in H

∩ K. Then u is in H. Since H is a subspace cu is in H. Likewise u is in K. Since K is a subspace cu is

in K. Thus cu is in H ∩ K for any scalar c, and H ∩ K is a subspace of V.

The union of two subspaces is not in general a subspace. For an example in

2

let H be the x-axis and

let K be the y-axis. Then both H and K are subspaces of

2

, but H ∪ K is not closed under vector

addition. The subset H ∪ K is thus not a subspace of

2

.

33. a. Given subspaces H and K of a vector space V, the zero vector of V belongs to H + K, because 0 is

in both H and K (since they are subspaces) and 0 = 0 + 0. Next, take two vectors in H + K, say

111

=+wuv

and

222

=+wuv

where

1

u

and

2

u

are in H, and

1

v

and

2

v

are in K. Then

121122 12 12

()()+=+++=+ ++ww uvu v uu vv

because vector addition in V is commutative and associative. Now

12

+uu

is in H and

12

+vv

is

in K because H and K are subspaces. This shows that

12

+ww

is in H + K. Thus H + K is closed

under addition of vectors. Finally, for any scalar c,

11111

()cc cc=+=+wuvuv

The vector

1

cu

belongs to H and

1

cv

belongs to K, because H and K are subspaces. Thus,

1

cw

belongs to H + K, so H + K is closed under multiplication by scalars. These arguments show that

H + K satisfies all three conditions necessary to be a subspace of V.

b. Certainly H is a subset of H + K because every vector u in H may be written as u + 0, where the

zero vector 0 is in K (and also in H, of course). Since H contains the zero vector of H + K, and H

is closed under vector addition and multiplication by scalars (because H is a subspace of V ), H is

a subspace of H + K. The same argument applies when H is replaced by K, so K is also a

subspace of H + K.

34. A proof that

11

Span{ , , , , , }

pq

HK+= … …uuvv

has two parts. First, one must show that H + K is a

subset of

11

Span{ , , , , , }.

pq

……uuvv

Second, one must show that

11

Span{ , , , , , }

pq

……uuvv

is a

subset of H + K.

(1) A typical vector in H has the form

11 pp

cc+…+uu

and a typical vector in K has the form

11

.

qq

dd+…+vv

The sum of these two vectors is a linear combination of

11

,, ,,,

pq

……uuvv

and so belongs to

11

Span{ , , , , , }.

pq

……uuvv

Thus H + K is a subset of

11

Span{ , , , , , }.

pq

……uuvv

(2) Each of the vectors

11

,, ,,,

pq

……uuvv

belongs to H + K, by Exercise 33(b), and so any linear

combination of these vectors belongs to H + K, since H + K is a subspace, by Exercise 33(a).

Thus,

11

Span{ , , , , , }

pq

……uuvv

is a subset of H + K.

202 CHAPTER 4 • Vector Spaces

35. [M] Since

84 79 1001

43 64 0102

,

32 54 0011

98187 0000

−−

ªºªº

«»«»

−− −

«»«»

∼

«»«»

−− − −

«»«»

−−

¬¼¬¼

w is in the subspace spanned by

123

{, , }.vv v

36. [M] Since

[]

3594 1001/5

8768 0102/5

,

5836 0013/5

2295 000 0

A

−−− −

ªºªº

«»«»

−− −

«»«»

=∼

«»«»

−−

«»«»

−−−

¬¼¬¼

y

y is in the subspace spanned by the columns of A.

37. [M] The graph of f(t) is given below. A conjecture is that f(t) = cos 4t.

The graph of g(t) is given below. A conjecture is that g(t) = cos 6t.

1 2 3 4 5 6

–1

–

0.5

1

1 2 3 4 5 6

–1

–

0.5

1

4.2 • Solutions 203

38. [M] The graph of f(t) is given below. A conjecture is that f(t) = sin 3t.

The graph of g(t) is given below. A conjecture is that g(t) = cos 4t.

The graph of h(t) is given below. A conjecture is that h(t) = sin 5t.

4.2 SOLUTIONS

Notes:

This section provides a review of Chapter 1 using the new terminology. Linear tranformations are

introduced quickly since students are already comfortable with the idea from

n

. The key exercises are

17–26, which are straightforward but help to solidify the notions of null spaces and column spaces.

Exercises 30–36 deal with the kernel and range of a linear transformation and are progressively more

advanced theoretically. The idea in Exercises 7–14 is for the student to use Theorems 1, 2, or 3 to

determine whether a given set is a subspace.

1. One calculates that

3531 0

6203 0,

8414 0

A

−−

ªºªºªº

«»«»«»

=−=

«»«»«»

−−

¬¼¬¼¬¼

w

so w is in Nul A.

2. One calculates that

2641 0

3251 0,

5411 0

A

ªºªºªº

«»«»«»

=−−=

«»«»«»

−−

¬¼¬¼¬¼

w

1 2 3 4 5 6

–1

–

0.5

1

1 2 3 4 56

–1

–

0.5

1

1 2 3 4 56

–1

–

0.5

1

204 CHAPTER 4 • Vector Spaces

so w is in Nul A.

3. First find the general solution of Ax = 0 in terms of the free variables. Since

[]

10 2 40

,

01 3 20

A

−

ªº

∼

«»

−

¬¼

0

the general solution is

134

24xxx=−

,

234

32xxx=−+

, with

3

x

and

4

x

free. So

1

2

34

3

4

24

32

,

10

01

x

xx

x

−

ªº ªº ªº

«» «» «»

−

«» «» «»

== +

«» «» «»

¬¼ ¬¼

¬¼

x

and a spanning set for Nul A is

24

32

,.

10

01

½−

ªºªº

°°

«»«»

−

°°

«»«»

®¾

«»«»

°°

«»«»

°°

¬¼¬¼

¯¿

4. First find the general solution of Ax = 0 in terms of the free variables. Since

[]

13000

,

00100

A

−

ªº

∼

«»

¬¼

0

the general solution is

12

3xx=

,

30x=

, with

2

x

and

4

x

free. So

1

2

24

3

4

30

10

,

00

01

x

xx

x

ªº ªº ªº

«» «» «»

== +

«» «» «»

¬¼ ¬¼

¬¼

x

and a spanning set for Nul A is

30

10

,.

00

01

½

ªºªº

°°

«»«»

°°

«»«»

®¾

«»«»

°°

«»«»

°°

¬¼¬¼

¯¿

5. First find the general solution of Ax = 0 in terms of the free variables. Since

[]

140200

001500,

000010

A

−

ªº

«»

∼−

«»

¬¼

0

the general solution is

124

42xxx=−

,

34

5xx=

,

50x=

, with

2

x

and

4

x

free. So

4.2 • Solutions 205

1

2

32 4

4

5

42

10

,

05

01

00

x

xx x

x

−

ªº ªº ª º

«» «» « »

== +

«» «» « »

¬¼ ¬ ¼

¬¼

x

and a spanning set for Nul A is

42

10

,.

05

01

00

½

−

ªºª º

°°

«»« »

°°

«»« »

°°

«»« »

®¾

«»« »

°°

«»« »

°°

«»« »

°°

¬¼¬ ¼

¯¿

6. First find the general solution of Ax = 0 in terms of the free variables. Since

[]

10 5 6 10

01 3 100,

00 0 000

A

−

ªº

«»

∼−

«»

¬¼

0

the general solution is

1345

56xxxx=−+−

,

234

3xxx=−

, with

3

x

,

4

x

, and

5

x

free. So

1

2

33 4 5

4

5

561

310

,

100

010

001

x

xx x x

x

−−

ªº ªº ªº ªº

«» «» «» «»

−

«» «» «» «»

== + +

«» «» «» «»

¬¼ ¬¼ ¬¼

¬¼

x

and a spanning set for Nul A is

561

310

,, .

100

010

001

½

−−

ªºªºªº

°°

«»«»«»

−

°°

«»«»«»

°°

«»«»«»

®¾

«»«»«»

°°

«»«»«»

°°

«»«»«»

°°

¬¼¬¼¬¼

¯¿

7. The set W is a subset of

3

. If W were a vector space (under the standard operations in

3

), then it

would be a subspace of

3

. But W is not a subspace of

3

since the zero vector is not in W. Thus W is

not a vector space.

8. The set W is a subset of

3

. If W were a vector space (under the standard operations in

3

), then it

would be a subspace of

3

. But W is not a subspace of

3

since the zero vector is not in W. Thus W is

not a vector space.

9. The set W is the set of all solutions to the homogeneous system of equations p – 3q – 4s = 0,

2p – s – 5r = 0. Thus W = Nul A, where

1340

2015

A

−−

ª

º

=

«

»

−−

¬

¼

. Thus W is a subspace of

4

by

Theorem 2, and is a vector space.

206 CHAPTER 4 • Vector Spaces

10. The set W is the set of all solutions to the homogeneous system of equations 3a + b – c = 0,

a + b + 2c – 2d = 0. Thus W = Nul A, where

31 1 0

11 2 2

A

−

ª

º

=

«

»

−

¬

¼

. Thus W is a subspace of

4

by

Theorem 2, and is a vector space.

11. The set W is a subset of

4

. If W were a vector space (under the standard operations in

4

), then it

would be a subspace of

4

. But W is not a subspace of

4

since the zero vector is not in W. Thus W is

not a vector space.

12. The set W is a subset of

4

. If W were a vector space (under the standard operations in

4

), then it

would be a subspace of

4

. But W is not a subspace of

4

since the zero vector is not in W. Thus W is

not a vector space.

13. An element w on W may be written as

1616

0101

1010

c

cd

d

−−

ªº ª º ª º

ªº

«» « » « »

=+ = «»

«» « » « »

¬¼

«» « » « »

¬¼ ¬ ¼ ¬ ¼

w

where c and d are any real numbers. So W = Col A where

16

01

10

A

−

ª

º

«

»

=

«

»

«

»

¬

¼

. Thus W is a subspace of

3

by Theorem 3, and is a vector space.

14. An element w on W may be written as

1313

1212

5151

s

st

t

−−

ªº ªºª º

ªº

«» «»« »

=+−=−«»

«» «»« »

¬¼

«» «»« »

−−

¬¼ ¬¼¬ ¼

w

where a and b are any real numbers. So W = Col A where

13

12

51

A

−

ª

º

«

»

=−

«

»

«

»

−

¬

¼

. Thus W is a subspace of

3

by Theorem 3, and is a vector space.

15. An element in this set may be written as

021021

112112

310310

211211

r

rs t s

t

ªº ª º ª º ª º

ª

º

«» « » « » « »

−−

«

»

«» « » « » « »

++=

«

»

«» « » « » « »

«

»

«» « » « » « »

¬

¼

−− −−

¬¼ ¬ ¼ ¬ ¼ ¬ ¼

where r, s and t are any real numbers. So the set is Col A where

021

112

310

211

A

ª

º

«

»

−

«

»

=

«

»

«

»

−−

¬

¼

.

16. An element in this set may be written as

4.2 • Solutions 207

110110

20 3203

13 3133

01 1011

b

bc d c

d

−−

ªº ª º ª º ª º

ª

º

«» « » « » « »

«

»

«» « » « » « »

++ =

«

»

«» « » « » « »

−−

«

»

«» « » « » « »

¬

¼

¬¼ ¬ ¼ ¬ ¼ ¬ ¼

where b, c and d are any real numbers. So the set is Col A where

110

203

133

011

A

−

ª

º

«

»

«

»

=

«

»

−

«

»

¬

¼

.

17. The matrix A is a 4 × 2 matrix. Thus

(a) Nul A is a subspace of

2

, and

(b) Col A is a subspace of

4

.

18. The matrix A is a 4 × 3 matrix. Thus

(a) Nul A is a subspace of

3

, and

(b) Col A is a subspace of

4

.

19. The matrix A is a 2 × 5 matrix. Thus

(a) Nul A is a subspace of

5

, and

(b) Col A is a subspace of

2

.

20. The matrix A is a 1 × 5 matrix. Thus

(a) Nul A is a subspace of

5

, and

(b) Col A is a subspace of

1

= .

21. Either column of A is a nonzero vector in Col A. To find a nonzero vector in Nul A, find the general

solution of Ax = 0 in terms of the free variables. Since

[]

12/30

000

,

000

A

−

ªº

«»

∼«»

«»

¬¼

0

the general solution is

12

(2 / 3)xx=

, with

2

x

free. Letting

2

x

be a

nonzero value (say

23x=

) gives the nonzero vector

1

2

3

x

ªºªº

==

«»«»

¬¼

x

which is in Nul A.

22. Any column of A is a nonzero vector in Col A. To find a nonzero vector in Nul A, find the general

solution of Ax = 0 in terms of the free variables. Since

[]

10 10

0110

,

0000

A

ªº

«»

∼«»

«»

¬¼

0

208 CHAPTER 4 • Vector Spaces

the general solution is

13

xx=−

,

23

xx=−

, with

3

x

free. Letting

3

x

be a nonzero value (say

3

1x=−

)

gives the nonzero vector

1

2

3

1

x

ªºªº

«»«»

==

«»«»

−

¬¼¬¼

x

which is in Nul A.

23. Consider the system with augmented matrix

[]

Aw

. Since

[]

121

,

000

A

−−

ªº

∼

«»

¬¼

w

the system is consistent and w is in Col A. Also, since

242 0

121 0

A

−

ªºªºªº

==

«»«»«»

−

¬¼¬¼¬¼

w

w is in Nul A.

24. Consider the system with augmented matrix

[]

Aw

. Since

[]

100 11

010 11

,

001 00

000 00

A

−

ªº

«»

−

«»

∼«»

«»

¬¼

w

the system is consistent and w is in Col A. Also, since

10 8 2 2 2 0

02222 0

11600 0

11022 0

A

−−−

ªºªºªº

«»«»«»

−

«»«»«»

==

«»«»«»

−

«»«»«»

−

¬¼¬¼¬¼

w

w is in Nul A.

25. a. True. See the definition before Example 1.

b . False. See Theorem 2.

c . True. See the remark just before Example 4.

d . False. The equation Ax = b must be consistent for every b. See #7 in the table on page 204.

e . True. See Figure 2.

f. True. See the remark after Theorem 3.

26. a. True. See Theorem 2.

b . True. See Theorem 3.

c . False. See the box after Theorem 3.

d . True. See the paragraph after the definition of a linear transformation.

e . True. See Figure 2.

f. True. See the paragraph before Example 8.

4.2 • Solutions 209

27. Let A be the coefficient matrix of the given homogeneous system of equations. Since Ax = 0 for

3

2

1

ªº

«»

=«»

«»

−

¬¼

x

, x is in Nul

A. Since Nul

A is a subspace of

3

, it is closed under scalar multiplication. Thus

30

10 20

10

ªº

«»

=«»

«»

−

¬¼

x

is also in Nul

A, and

1

30x=

,

2

20x=

,

3

10x=−

is also a solution to the system of

equations.

28. Let A be the coefficient matrix of the given systems of equations. Since the first system has a

solution, the constant vector

0

1

9

ªº

«»

=«»

«»

¬¼

b

is in Col

A. Since Col A is a subspace of

3

, it is closed under

scalar multiplication. Thus

0

55

45

ªº

«»

=«»

«»

¬¼

b

is also in Col A, and the second system of equations must thus

have a solution.

29. a. Since ,A=00

the zero vector is in Col A.

b . Since

(),

AA A AA

+= + +

xw xwxw

is in Col A.

c . Since

() (),

cA Ac cA

=

xxx

is in Col A.

30. Since

()

VW

T=00

, the zero vector

W

0

of W is in the range of T. Let T(x) and T(w) be typical

elements in the range of T. Then since

() ( ) ( ), () ( )

TT T TT

+=+ +

xwxwxw

is in the range of T and

the range of T is closed under vector addition. Let c be any scalar. Then since

() ( ), ()

cT T c cT

=

xxx

is in the range of T and the range of T is closed under scalar multiplication. Hence the range of T is a

subspace of W.

31. a. Let p and q be arbitary polynomials in

2

, and let c be any scalar. Then

()(0) (0)(0) (0) (0)

() ()()

()(1) (1)(1) (1) (1)

TTT

++

ªºª ºªºªº

+= = = + = +

«»« »«»«»

++

¬¼¬ ¼¬¼¬¼

pq p q p q

pq p q

pq p q p q

and

()(0) (0)

() ()

()(1) (1)

c

Tc c cT

c

ªºªº

===

«»«»

¬¼¬¼

pp

so T is a linear transformation.

b. Any quadratic polynomial q for which

(0) 0=

q

and

(1) 0=

q

will be in the kernel of T. The

polynomial q must then be a multiple of

() ( 1).

ttt

=−

p

Given any vector

1

2

x

ªº

«»

¬¼

in

2

, the

polynomial

121

()xxxt=+ −p

has

1

(0) x=p

and

2

(1) .x=p

Thus the range of T is all of

2

.

210 CHAPTER 4 • Vector Spaces

32. Any quadratic polynomial q for which

(0) 0=

q

will be in the kernel of T. The polynomial q must

then be

2

.

at bt

=+

q

Thus the polynomials

1

()tt=p

and

2

()

tt

=

p

span the kernel of T. If a vector

is in the range of T, it must be of the form

.

a

ª

º

«

»

¬

¼

If a vector is of this form, it is the image of the

polynomial

()

ta

=

p

in

2

. Thus the range of T is :real.

aa

a



½

ªº

°

®

¾

«»

°

¬¼

¯¿

33. a. For any A and B in

22

M×

and for any scalar c,

()()() ( )( )()()

TTTTT

TA B A B A B A B A B A A B B TA TB

+=+++ =++ + =+ ++ = +

and

()() ( ) ()

TT

TcA cA cA cT A

== =

so T is a linear transformation.

b. Let B be an element of

22

M×

with

,

T

BB=

and let

1

2

.

AB

=

Then

11 11 11

() ( )

22 22 22

TTT

TA A A B B B B B B B=+ = + = + = + =

c. Part b. showed that the range of T contains the set of all B in

22

M×

with

.

T

BB=

It must also be

shown that any B in the range of T has this property. Let B be in the range of T. Then B = T(A) for

some A in

22

.M×

Then

,

T

BAA=+

and

() ()

TTTTTTT T

BAA AA AAAAB

=+ = + = +=+ =

so B has the property that

.

T

BB=

d. Let

ab

Acd

ªº

=

«»

¬¼

be in the kernel of T. Then

() 0

T

TA A A

=+ =

, so

200

T

ab ac a cb

AA cd bd bc d

+

ªºªºª ºªº

+= + = =

«»«»« »«»

+

¬¼¬¼¬ ¼¬¼

Solving it is found that 0ad==

and cb=−. Thus the kernel of T is 0:real.

0

bb

b



½

ªº

°

®

¾

«»

−

°

¬¼

¯¿

34. Let f and g be any elements in C[0, 1] and let c be any scalar. Then T(f) is the antiderivative F of f

with F(0) = 0 and T(g) is the antiderivative G of g with G(0) = 0. By the rules for antidifferentiation

+FG

will be an antiderivative of

,+

fg

and

()(0)(0)(0)000.+=+=+=

FG F G

Thus

()()().

TTT

+= +

fg f g

Likewise cF will be an antiderivative of cf, and

()(0) (0) 00.

ccc

===

FF

Thus

() (),

Tc cT

=

ff

and T is a linear transformation. To find the kernel of T, we must find all

functions f in C[0,1] with antiderivative equal to the zero function. The only function with this

property is the zero function 0, so the kernel of T is {0}.

35. Since U is a subspace of V,

V

0

is in U. Since T is linear,

() .

VW

T=00

So

W

0

is in T(U). Let T(x) and

T(y) be typical elements in T(U). Then x and y are in U, and since U is a subspace of V, +xy

is also

in U. Since T is linear,

() () ( ).

TTT

+=+

xyxy

So

() ()

TT

+

xy

is in T(U), and T(U) is closed under

vector addition. Let c be any scalar. Then since x is in U and U is a subspace of V, cx is in U. Since T

4.2 • Solutions 211

is linear,

() ()

Tc cT

=

xx

and cT(x) is in T(U ). Thus T(U) is closed under scalar multiplication, and

T(U) is a subspace of W.

36. Since Z is a subspace of W,

W

0

is in Z. Since T is linear,

() .

VW

T=00

So

V

0

is in U. Let x and y be

typical elements in U. Then T(x) and T(y) are in Z, and since Z is a subspace of W,

() ()

TT

+

xy

is

also in Z. Since T is linear,

() () ( ).

TT T

+=+

xyxy

So

()

T

+

xy

is in Z, and +xy

is in U. Thus U is

closed under vector addition. Let c be any scalar. Then since x is in U, T(x) is in Z. Since Z is a

subspace of W, cT(x) is also in Z. Since T is linear,

() ( )

cT T c

=

xx

and T(cx) is in T(U). Thus cx is in

U and U is closed under scalar multiplication. Hence U is a subspace of V.

37. [M] Consider the system with augmented matrix

[]

.

Aw

Since

[]

100 1/95 1/95

010 39/19 20/19

,

001267/95 172/95

000 0 0

A

−

ªº

«»

−

«»

∼

«»

−

«»

¬¼

w

the system is consistent and w is in Col

A. Also, since

7641114

510210

911731 0

19 9 7 1 3 0

A

−

ªºªºªº

«»«»«»

−− −

«»«»«»

==

«»«»«»

−−−

«»«»«»

−−

«»«»«»

¬¼¬¼¬¼

w

w is not in Nul

A.

38. [M] Consider the system with augmented matrix

[]

Aw

. Since

[]

10 10 2

01 20 3

,

00 01 1

00 00 0

A

−−

ªº

«»

−−

«»

∼

«»

¬¼

w

the system is consistent and w is in Col

A. Also, since

85201 0

52122 0

10 8 6 3 1 0

32 100 0

A

−−

ªºªºªº

«»«»«»

−−

«»«»«»

==

«»«»«»

−−

«»«»«»

−

«»«»«»

¬¼¬¼¬¼

w

w is in Nul

A.

39. [M]

a. To show that

3

a

and

5

a

are in the column space of B, we can row reduce the matrices

[]

3

Ba

and

[]

3

Ba

:

212 CHAPTER 4 • Vector Spaces

[]

3

1001/3

0101/3

001 0

000 0

B

ªº

«»

∼

«»

¬¼

a

[]

5

100 10/3

010 26/3

001 4

000 0

B

ª

º

«

»

−

«

»

∼

«

»

−

«

»

«

»

¬

¼

a

Since both these systems are consistent,

3

a

and

5

a

are in the column space of B. Notice that the

same conclusions can be drawn by observing the reduced row echelon form for A:

101/30 10/3

011/30 26/3

00 01 4

00 00 0

A

ªº

«»

−

«»

∼

«»

−

«»

¬¼

b. We find the general solution of Ax = 0 in terms of the free variables by using the reduced row

echelon form of A given above:

135

(1/3) (10/3)xxx=−−

,

235

(1/3) (26/3)xxx=−+

,

45

4xx=

with

3

x

and

5

x

free. So

1

2

353

4

5

1/3 10/3

1/3 26/3

,

10

04

01

x

xxx

x

−−

ªº

ª

ºª º

«»

«

»« »

−

«»

«

»« »

«»

«

»« »

== +

«»

«

»« »

«»

«

»« »

«»

«

»« »

¬

¼¬ ¼

¬¼

x

and a spanning set for Nul A is

1/3 10/3

1/3 26/3

,.

10

04

01

½

−−

ªºª º

°°

«»« »

−

°°

«»« »

°°

«»« »

®¾

«»« »

°°

«»« »

°°

«»« »

°°

¬¼¬ ¼

¯¿

c. The reduced row echelon form of A shows that the columns of A are linearly dependent and do

not span

4

. Thus by Theorem 12 in Section 1.9, T is neither one-to-one nor onto.

4.3 • Solutions 213

40. [M] Since the line lies both in

12

Span{ , }H=vv

and in

34

Span{ , }K=vv

, w can be written both as

11 2 2

cc+vv

and

33 44

cc+vv

. To find w we must find the c

j

’s which solve

11 2 2 3 3 4 4

cc c c+−− =vvvv0

. Row reduction of

[]

12 3 4

−−

vv v v0

yields

51 2 00 100 10/30

33 1120 010 26/30,

84 5280 001 40

−−

ªºª º

«»« »

∼

«»« »

−−

¬¼¬ ¼

so the vector of c

j

’s must be a multiple of (10/3, –26/3, 4, 1). One simple choice is (10, –26, 12, 3),

which gives

1234

10 26 12 3 (24, 48, 24)=−=+=−−wv v vv

. Another choice for w is (1, –2, –1).

4.3 SOLUTIONS

Notes:

The definition for basis is given initially for subspaces because this emphasizes that the basis

elements must be in the subspace. Students often overlook this point when the definition is given for a

vector space (see Exercise 25). The subsection on bases for Nul A and Col A is essential for Sections 4.5

and 4.6. The subsection on “Two Views of a Basis” is also fundamental to understanding the interplay

between linearly independent sets, spanning sets, and bases. Key exercises in this section are Exercises

21–25, which help to deepen students’ understanding of these different subsets of a vector space.

1. Consider the matrix whose columns are the given set of vectors. This 3 × 3 matrix is in echelon form,

and has 3 pivot positions. Thus by the Invertible Matrix Theorem, its columns are linearly

independent and span

3

. So the given set of vectors is a basis for

3

.

2. Since the zero vector is a member of the given set of vectors, the set cannot be linearly independent

and thus cannot be a basis for

3

. Now consider the matrix whose columns are the given set of

vectors. This 3 × 3 matrix has only 2 pivot positions. Thus by the Invertible Matrix Theorem, its

columns do not span

3

.

3. Consider the matrix whose columns are the given set of vectors. The reduced echelon form of this

matrix is

132 101

011 011

341 000

−

ªºªº

«»«»

−∼ −

«»«»

−−

¬¼¬¼

so the matrix has only two pivot positions. Thus its columns do not form a basis for

3

; the set of

vectors is linearly independent and does not span

3

.

4. Consider the matrix whose columns are the given set of vectors. The reduced echelon form of this

matrix is

228 100

135 010

124 001

−

ªºªº

«»«»

−− ∼

«»«»

¬¼¬¼

so the matrix has three pivot positions. Thus its columns form a basis for

3

.

214 CHAPTER 4 • Vector Spaces

5. Since the zero vector is a member of the given set of vectors, the set cannot be linearly independent

and thus cannot be a basis for

3

. Now consider the matrix whose columns are the given set of

vectors. The reduced echelon form of this matrix is

3300 1000

3703 0100

0005 0001

−

ªºªº

«»«»

−−∼

«»«»

¬¼¬¼

so the matrix has a pivot in each row. Thus the given set of vectors spans

3

.

6. Consider the matrix whose columns are the given set of vectors. Since the matrix cannot have a pivot

in each row, its columns cannot span

3

; thus the given set of vectors is not a basis for

3

. The

reduced echelon form of the matrix is

14 10

2301

46 00

−

ªºªº

«»«»

∼

«»«»

−

¬¼¬¼

so the matrix has a pivot in each column. Thus the given set of vectors is linearly independent.

7. Consider the matrix whose columns are the given set of vectors. Since the matrix cannot have a pivot

in each row, its columns cannot span

3

; thus the given set of vectors is not a basis for

3

. The

reduced echelon form of the matrix is

26 10

3101

05 00

−

ªºªº

«»«»

−∼

«»«»

¬¼¬¼

so the matrix has a pivot in each column. Thus the given set of vectors is linearly independent.

8. Consider the matrix whose columns are the given set of vectors. Since the matrix cannot have a pivot

in each column, the set cannot be linearly independent and thus cannot be a basis for

3

. The reduced

echelon form of this matrix is

1020 1020

2310 0110

3151 0001

ªºªº

«»«»

−−∼

«»«»

−−

¬¼¬¼

so the matrix has a pivot in each row. Thus the given set of vectors spans

3

.

9. We find the general solution of Ax = 0 in terms of the free variables by using the reduced echelon

form of A:

10 2 2 10 20

0114 0110.

3173 0001

−− −

ªºªº

«»«»

∼

«»«»

−−

¬¼¬¼

So

13

2xx=

,

23

xx=−

,

4

0x=

, with

3

x

free. So

4.3 • Solutions 215

1

2

3

4

2

1,

1

0

x

xx

x

ªº ªº

«» «»

−

«» «»

==

«» «»

¬¼

x

and a basis for Nul A is

2

1.

1

0

½

ªº

°°

«»

−

°°

«»

®¾

«»

°°

«»

°°

¬¼

¯¿

10. We find the general solution of Ax = 0 in terms of the free variables by using the reduced echelon

form of A:

11 2 1 5 100 2 9

01 6 1 2 010 110.

00 8 016 001 0 2

−−

ªºªº

«»«»

−−∼ −

«»«»

−−

¬¼¬¼

So

145

29xxx=−+

,

24 5

10xx x=−

,

35

2xx=

, with

4

x

and

5

x

free. So

1

2

34 5

4

5

29

110

,

02

10

01

x

xx x

x

−

ªº ªº ª º

«» «» « »

−

«» «» « »

== +

«» «» « »

¬¼ ¬ ¼

¬¼

x

and a basis for Nul A is

29

110

,.

02

10

01

½

−

ªºª º

°°

«»« »

−

°°

«»« »

°°

«»« »

®¾

«»« »

°°

«»« »

°°

«»« »

°°

¬¼¬ ¼

¯¿

11. Let

[]

132

A

=−

. Then we wish to find a basis for Nul A. We find the general solution of Ax = 0

in terms of the free variables: x = 3y – 2z with y and z free. So

32

10,

01

x

yy z

z

−

ªº ªº ª º

«» «» « »

== +

«» «» « »

¬¼ ¬¼ ¬ ¼

x

and a basis for Nul A is

32

1, 0 .

01

½−

ªºª º

°°

«»« »

®¾

«»« »

°°

«»« »

¬¼¬ ¼

¯¿

216 CHAPTER 4 • Vector Spaces

12. We want to find a basis for the set of vectors in

2

in the line 3x + y = 0. Let

[]

31

A

=

. Then we

wish to find a basis for Nul A. We find the general solution of Ax = 0 in terms of the free variables: y

= – 3x with x free. So

1,

3

xx

y

ªº ª º

==

«» « »

−

¬¼ ¬ ¼

x

and a basis for Nul A is

1.

3

½

ªº

°°

®¾

«»

−

°°

¬¼

¯¿

13. Since B is a row echelon form of A, we see that the first and second columns of A are its pivot

columns. Thus a basis for Col A is

24

2, 6 .

38

½−

ªºªº

°°

«»«»

−

®¾

«»«»

°°

«»«»

−

¬¼¬¼

¯¿

To find a basis for Nul A, we find the general solution of Ax = 0 in terms of the free variables:

134

65,xxx=−−

234

(5/2) (3/2) ,xxx=−−

with

3

x

and

4

x

free. So

1

2

34

3

4

65

5/2 3/2 ,

10

01

x

xxx

x

−−

ªº ª º ª º

«» «»«»

−−

«» «»«»

== +

«» «»«»

«»«»

«» ¬¼¬¼

¬¼

x

and a basis for Nul A is

65

5/2 3/2

,.

10

01

½−−

ªºªº

°°

«»«»

−−

°°

«»«»

®¾

«»«»

°°

«»«»

°°

«»«»

¬¼¬¼

¯¿

14. Since B is a row echelon form of A, we see that the first, third, and fifth columns of A are its pivot

columns. Thus a basis for Col A is

138

108

,,.

239

309

½

ªºª ºªº

°°

«»« »«»

°°

«»« »«»

®¾

«»« »«»

−

°°

«»« »«»

°°

¬¼¬ ¼¬¼

¯¿

To find a basis for Nul A, we find the general solution of Ax = 0 in terms of the free variables,

mentally completing the row reduction of B to get:

124

22,xxx=−−

34

2,xx=

5

0,x=

with

2

x

and

4

x

free. So

4.3 • Solutions 217

1

2

32 4

4

5

22

10

,

02

01

00

x

xx x

x

−−

ªº ªº ªº

«» «» «»

== +

«» «» «»

¬¼ ¬¼

¬¼

x

and a basis for Nul A is

22

10

,.

02

01

00

½

−−

ªºªº

°°

«»«»

°°

«»«»

°°

«»«»

®¾

«»«»

°°

«»«»

°°

«»«»

°°

¬¼¬¼

¯¿

15. This problem is equivalent to finding a basis for Col A, where

[]

12345

A

=

vv vv v

. Since

the reduced echelon form of A is

10 2 2 3 10 200

01 2 1 1 01 200

,

22 810 6 00 010

33 0 3 9 00 001

ªºªº

«»«»

−−− −

«»«»

∼

«»«»

−− −

«»«»

¬¼¬¼

we see that the first, second, fourth and fifth columns of A are its pivot columns. Thus a basis for the

space spanned by the given vectors is

10 2 3

01 1 1

,, , .

2210 6

33 3 9

½

ªºªºªºªº

°°

«»«»«»«»

−−

°°

«»«»«»«»

®¾

«»«»«»«»

−−

°°

«»«»«»«»

°°

¬¼¬¼¬¼¬¼

¯¿

16. This problem is equivalent to finding a basis for Col A, where

[]

12345

A

=

vv vv v

. Since

the reduced echelon form of A is

12352 1005/2 0

001310103/41/2

,

00131001 3 1

12140 000 0 0

−−

ªºª º

«»« »

−−−

«»« »

∼

«»« »

−−

¬¼¬ ¼

we see that the first, second, and third columns of A are its pivot columns. Thus a basis for the space

spanned by the given vectors is

123

001

,, .

001

121

½−

ªº ª º ª º

°°

«» « » « »

−

°°

«» « » « »

®¾

«» « » « »

°°

«» « » « »

°°

−

¬¼ ¬ ¼ ¬ ¼

¯¿

17. [M] This problem is equivalent to finding a basis for Col A, where

[]

12345

A

=

vv vv v

.

Since the reduced echelon form of A is

218 CHAPTER 4 • Vector Spaces

24288 10010

00444 01020

,

42080 00110

64130 00001

04715100000

−− −

ªºªº

«»«»

−

«»«»

∼

−−

«»«»

−− −

«»«»

−

¬¼¬¼

we see that the first, second, third, and fifth columns of A are its pivot columns. Thus a basis for the

space spanned by the given vectors is

2428

0044

,,, .

4200

6410

0471

½

−−

ªºªºªºªº

°°

«»«»«»«»

−

°°

«»«»«»«»

°°

«»«»«»«»

−

®¾

«»«»«»«»

°°

−−

«»«»«»«»

°°

«»«»«»«»

−

°°

¬¼¬¼¬¼¬¼

¯¿

18. [M] This problem is equivalent to finding a basis for Col A, where

[]

12345

.

A

=

vv vv v

Since the reduced echelon form of A is

330 66 10000

202 23 01020

,

694140 00110

000 0100001

761130 00000

−−

ªºªº

«»«»

−

«»«»

∼

−−− −

«»«»

−

«»«»

−−

¬¼¬¼

we see that the first, second, third, and fifth columns of A are its pivot columns. Thus a basis for the

space spanned by the given vectors is

3306

2023

,,, .

6940

0001

7610

½

−−

ªºªºªºªº

°°

«»«»«»«»

°°

«»«»«»«»

°°

«»«»«»«»

−−

®¾

«»«»«»«»

°°

−

«»«»«»«»

°°

«»«»«»«»

−−

°°

¬¼¬¼¬¼¬¼

¯¿

19. Since

123

453 ,+−=vvv0

we see that each of the vectors is a linear combination of the others. Thus

the sets

12

{, },vv

13

{, },vv

and

23

{,}vv

all span H. Since we may confirm that none of the three

vectors is a multiple of any of the others, the sets

12

{, },vv

13

{, },vv

and

23

{,}vv

are linearly

independent and thus each forms a basis for H.

20. Since

123

2,−−=vv v 0

we see that each of the vectors is a linear combination of the others. Thus

the sets

12

{, },vv

13

{, },vv

and

23

{,}vv

all span H. Since we may confirm that none of the three

vectors is a multiple of any of the others, the sets

12

{, },vv

13

{, },vv

and

23

{,}vv

are linearly

independent and thus each forms a basis for H.

21. a. False. The zero vector by itself is linearly dependent. See the paragraph preceding Theorem 4.

b . False. The set

1

{, , }

p

…

bb

must also be linearly independent. See the definition of a basis.

c . True. See Example 3.

4.3 • Solutions 219

d . False. See the subsection “Two Views of a Basis.”

e . False. See the box before Example 9.

22. a. False. The subspace spanned by the set must also coincide with H. See the definition of a basis.

b . True. Apply the Spanning Set Theorem to V instead of H. The space V is nonzero because the

spanning set uses nonzero vectors.

c . True. See the subsection “Two Views of a Basis.”

d . False. See the two paragraphs before Example 8.

e . False. See the warning after Theorem 6.

23. Let

[]

1234

.

A

=

vv vv

Then A is square and its columns span

4

since

4

1234

Span{ , , , }.=vv v v

So its columns are linearly independent by the Invertible Matrix Theorem,

and

1234

{, , , }vv v v

is a basis for

4

.

24. Let

[

]

1

.

n

A

=…

vv

Then A is square and its columns are linearly independent, so its columns

span

n

by the Invertible Matrix Theorem. Thus

1

{, , }

n

…vv

is a basis for

n

.

25. In order for the set to be a basis for H,

123

{, , }vv v

must be a spanning set for H; that is,

123

Span{ , , }.H=vv v

The exercise shows that H is a subset of

123

Span{ , , }.vv v

but there are vectors

in

123

Span{ , , }vv v

which are not in H (

1

v

and

3

,v

for example). So

123

Span{ , , },H≠vv v

and

123

{, , }vv v

is not a basis for H.

26. Since sin t cos t = (1/2) sin 2t, the set {sin t, sin 2t} spans the subspace. By inspection we note that

this set is linearly independent, so {sin t, sin 2t} is a basis for the subspace.

27. The set {cos

ω

t, sin

ω

t} spans the subspace. By inspection we note that this set is linearly

independent, so {cos

ω

t, sin

ω

t} is a basis for the subspace.

28. The set

{, }

bt bt

ete

−−

spans the subspace. By inspection we note that this set is linearly independent,

so

{, }

bt bt

ete

−−

is a basis for the subspace.

29. Let A be the n × k matrix

[]

1k

…

vv

. Since A has fewer columns than rows, there cannot be a

pivot position in each row of A. By Theorem 4 in Section 1.4, the columns of A do not span

n

and

thus are not a basis for

n

.

30. Let A be the n × k matrix

[]

1k

…

vv

. Since A has fewer rows than columns, there cannot be a

pivot position in each column of A. By Theorem 8 in Section 1.7, the columns of A are not linearly

independent and thus are not a basis for

n

.

31. Suppose that

1

{, , }

p

…

vv

is linearly dependent. Then there exist scalars

1

,,

p

cc

…

not all zero with

11

.

pp

cc

+…+ =

vv0

Since T is linear,

11 1 1

()()()

pp p p

Tc c cT cT

+…+ = +…+

vvv v

and

220 CHAPTER 4 • Vector Spaces

11

()().

pp

Tc c T

+…+ = =

vv00

Thus

11

() ( )

pp

cT c T

+…+ =

vv0

and since not all of the

i

c

are zero,

1

{( ), , ( )}

p

TT

…

vv

is linearly dependent.

32. Suppose that

1

{( ), , ( )}

p

TT

…

vv

is linearly dependent. Then there exist scalars

1

,,

p

cc

…

not all zero

with

11

() ( ) .

pp

cT c T

+…+ =

vv0

Since T is linear,

11 1 1

()()()()

pp p p

Tc c cT cT T

+…+ = +…+ = =

vvv v00

Since T is one-to-one

11

()()

pp

Tc c T

+…+ =

vv0

implies that

11

.

pp

cc

+…+ =

vv0

Since not all of the

i

c

are zero,

1

{, , }

p

…

vv

is linearly dependent.

33. Neither polynomial is a multiple of the other polynomial. So

12

{, }pp

is a linearly independent set in

3

. Note:

12

{, }pp

is also a linearly independent set in

2

since

1

p

and

2

p

both happen to be in

2

.

34. By inspection,

312

=+ppp

, or

123

+−=ppp 0

. By the Spanning Set Theorem,

123 12

Span{ , , } Span{ , }=pp p pp

. Since neither

1

p

nor

2

p

is a multiple of the other, they are linearly

independent and hence

12

{, }pp

is a basis for

123

Span{ , , }.pp p

35. Let

13

{, }vv

be any linearly independent set in a vector space V, and let

2

v

and

4

v

each be linear

combinations of

1

v

and

3

.v

For instance, let

21

5=vv

and

413

.=+vvv

Then

13

{, }vv

is a basis for

1234

Span{ , , , }.vv vv

36. [M] Row reduce the following matrices to identify their pivot columns:

[]

123

10 3 10 3

224 011

,

011000

114 000

ªºªº

«»«»

−

«»«»

=∼

«»«»

−

«»«»

−−

¬¼¬¼

uu u

so

12

{, }uu

is a basis for H.

[]

123

221 100

234 010

,

126 001

362 000

−−

ªºªº

«»«»

−

«»«»

=∼

«»«»

−

«»«»

−−

¬¼¬¼

vv v

so

123

{, , }vv v

is a basis for K.

4.3 • Solutions 221

[]

123123

10 3 2 2 1

2242 34

011126

114 36 2

−−

ªº

«»

−

«»

=«»

−−

«»

−− −−

¬¼

uu u vv v

10 3 200

01 1 100

,

00 0 010

00 0 001

−

ªº

«»

−

«»

∼«»

«»

¬¼

so

1223

{, , , }uu v v

is a basis for H + K.

37. [M] For example, writing

12 3 4

sin cos 2 sin cos 0ct c t c t c t t⋅+⋅++ =

with t = 0, .1, .2, .3 gives the following coefficent matrix A for the homogeneous system Ac = 0 (to

four decimal places):

0sin0cos0 sin0cos0 0 0 1 0

.1 sin .1 cos .2 sin .1cos .1 .1 .0998 .9801 .0993 .

.2 sin .2 cos .4 sin .2 cos .2 .2 .1987 .9211 .1947

.3 sin .3 cos .6 sin .3 cos .3 .3 .2955 .8253 .2823

A

ªºªº

«»«»

==

«»«»

¬¼¬¼

This matrix is invertible, so the system Ac = 0 has only the trivial solution and {t, sin t, cos 2t,

sin t cos t} is a linearly independent set of functions.

38. [M] For example, writing

23456

12 3 4 5 6 7

1cos cos cos cos cos cos0

cc tc tc tc tc tc t

⋅+⋅+⋅+⋅+⋅+⋅+⋅=

with t = 0, .1, .2, .3, .4, .5, .6 gives the following coefficent matrix A for the homogeneous system Ac

= 0 (to four decimal places):

23456

234

1cos0 cos0 cos0 cos0 cos0 cos0

1cos.1cos.1cos.1cos.1cos.1cos.1

1cos.2cos.2cos.2cos.2cos.2cos.2

1cos.3cos.3cos.3cos.3cos.3cos.3

1cos.4cos.4cos.4cos.4cos.4cos.4

1cos.5cos.5cos.5cos.5

A=

56

23456

cos .5 cos .5

1cos.6cos.6cos.6cos.6cos.6cos.6

ªº

«»

¬¼

11 1 1 1 1 1

1 .9950 .9900 .9851 .9802 .9753 .9704

1.9801.9605.9414.9226.9042.8862

1.9553.9127.8719.8330.7958.7602

1 .9211 .8484 .7814 .7197 .6629 .6106

1 .8776 .7702 .6759 .5931 .5205 .4568

1 .8253 .6812 .5622 .4640 .3830 .3161

ª

«

=

¬

º

»

«»

¼

222 CHAPTER 4 • Vector Spaces

This matrix is invertible, so the system Ac = 0 has only the trivial solution and {1, cos t, cos

2

t, cos

3

t,

cos

4

t, cos

5

t, cos

6

t} is a linearly independent set of functions.

4.4 SOLUTIONS

Notes:

Section 4.7 depends heavily on this section, as does Section 5.4. It is possible to cover the

n

parts

of the two later sections, however, if the first half of Section 4.4 (and perhaps Example 7) is covered. The

linearity of the coordinate mapping is used in Section 5.4 to find the matrix of a transformation relative to

two bases. The change-of-coordinates matrix appears in Section 5.4, Theorem 8 and Exercise 27. The

concept of an isomorphism is needed in the proof of Theorem 17 in Section 4.8. Exercise 25 is used in

Section 4.7 to show that the change-of-coordinates matrix is invertible.

1. We calculate that

343

53 .

567

−

ªº ªºªº

=+=

«» «»«»

−−

¬¼ ¬¼¬¼

x

2. We calculate that

3426

(2) 5 .

211

−−

ªº ª º ª º

=−+=

«» « » « »

¬¼ ¬ ¼ ¬ ¼

x

3. We calculate that

15 47

12 00 (2)3 4.

32 03

−

ªº ªº ªºªº

«» «» «»«»

=−++−−=

«» «» «»«»

−

¬¼ ¬¼ ¬¼¬¼

x

4. We calculate that

23 48

(3) 2 20 (1) 1 5.

02 31

−

ªº ªº ªºªº

«» «» «»«»

=−++−−=−

«» «» «»«»

¬¼ ¬¼ ¬¼¬¼

x

5. The matrix

[]

12

bb x

row reduces to

10 2

,

01 1

ª

º

«

»

−

¬

¼

so

2

[] .

1

B

ª

º

=

«

»

−

¬

¼

x

6. The matrix

[]

12

bb x

row reduces to

10 3

,

01 2

ª

º

«

»

−

¬

¼

so

3

[] .

2

B

ª

º

=

«

»

−

¬

¼

x

7. The matrix

[]

123

bb b x

row reduces to

100 1

010 1,

001 3

−

ª

º

«

»

−

«

»

«

»

¬

¼

so

1

[] 1.

3

B

−

ª

º

«

»

=−

«

»

«

»

¬

¼

x

4.4 • Solutions 223

8. The matrix

[]

123

bb b x

row reduces to

100 1

010 1,

001 1

ª

º

«

»

−

«

»

«

»

¬

¼

so

1

[] 1.

1

B

ª

º

«

»

=−

«

»

«

»

¬

¼

x

9. The change-of-coordinates matrix from B to the standard basis in

2

is

[]

12

.

35

B

P

ªº

==

«»

−−

¬¼

bb

10. The change-of-coordinates matrix from B to the standard basis in

3

is

[]

123

321

022.

643

B

P

ª

º

«

»

==−

«

»

«

»

−

¬

¼

bbb

11. Since

1

B

P

−

converts x into its B-coordinate vector, we find that

1

13 2 532 5

[] .

25 5 215 1

BB

P

−

−−−

ªºªºªºªºªº

== = =

«»«»«»«»«»

−−−−−

¬¼¬¼¬¼¬¼¬¼

xx

12. Since

1

B

P

−

converts x into its B-coordinate vector, we find that

1

12 2 122 8

[] .

113 113 5

BB

P

−

−− −

ªºªºªºªºªº

== = =

«»«»«»«»«»

−−

¬¼¬¼¬¼¬¼¬¼

xx

13. We must find

1

c

,

2

c

, and

3

c

such that

22 2 2

123

(1 ) ( ) (1 2 ) ( ) 1 4 7 .ctcttc tt t tt++ ++ ++= =++

p

Equating the coefficients of the two polynomials produces the system of equations

13

23

12 3

1

24

7

cc

cc c

+=

++ =

We row reduce the augmented matrix for the system of equations to find

1011 100 2 2

0124 010 6,so[] 6.

1117 001 1 1

B

ªºª ºªº

«»« »«»

∼=

«»« »«»

−−

¬¼¬ ¼¬¼

p

One may also solve this problem using the coordinate vectors of the given polynomials relative to the

standard basis

2

{1, , } ;tt

the same system of linear equations results.

14. We must find

1

c

,

2

c

, and

3

c

such that

22 2 2

123

(1 ) ( ) (1 ) ( ) 2 3 6 .ctcttc tt t tt−+−+−+= =+−

p

Equating the coefficients of the two polynomials produces the system of equations

224 CHAPTER 4 • Vector Spaces

13

23

123

2

3

6

cc

cc c

+=

−=

−− +=−

We row reduce the augmented matrix for the system of equations to find

10 12 100 3 3

0113 0102,so[] 2.

1116 0011 1

B

ªºªºªº

«»«»«»

−∼ =

«»«»«»

−− − − −

¬¼¬¼¬¼

p

One may also solve this problem using the coordinate vectors of the given polynomials relative to the

standard basis

2

{1, , } ;tt

the same system of linear equations results.

15. a. True. See the definition of the B-coordinate vector.

b . False. See Equation (4).

c . False.

3

is isomorphic to

4

. See Example 5.

16. a. True. See Example 2.

b . False. By definition, the coordinate mapping goes in the opposite direction.

c . True. If the plane passes through the origin, as in Example 7, the plane is isomorphic to

2

.

17. We must solve the vector equation

123

1231

3871

xx x

−

ª

ºªºªºªº

++=

«

»«»«»«»

−−

¬

¼¬¼¬¼¬¼

. We row reduce the augmented

matrix for the system of equations to find

1231 1055

.

3871 0112

−−

ªºªº

∼

«»«»

−− −

¬¼¬¼

Thus we can let

13

55

xx

=+

and

23

2

xx

=−−

, where

3

x

can be any real number. Letting

30

x

=

and

31

x

=

produces two different ways to express

1

ª

º

«

»

¬

¼

as a linear combination of the other vectors:

12

52−vv

and

23

10 3−+

1

vvv

. There are infintely many correct answers to this problem.

18. For each k,

1

01 0

kkn

=⋅+⋅⋅⋅ +⋅+⋅⋅⋅+⋅bb b b

, so

[] (0,,1,,0) .

kB k

=…… =be

19. The set S spans V because every x in V has a representation as a (unique) linear combination of

elements in S. To show linear independence, suppose that

1

{, , }

n

S

=…vv

and that

11 nn

cc

+⋅⋅⋅ +=vv0

for some scalars

1

c

, …,

.

n

c

The case when

10

n

cc

=⋅⋅⋅ ==

is one possibility.

By hypothesis, this is the unique (and thus the only) possible representation of the zero vector as a

linear combination of the elements in S. So S is linearly independent and is thus a basis for V.

20. For w in V there exist scalars

1

k

,

2

k

,

3

k

, and

4

k

such that

11 2 2 3 3 4 4

kk k k

=+ + +wv v v v

(1)

because

1234

{, , , }vv vv

spans V. Because the set is linearly dependent, there exist scalars

1

c

,

2

c

,

3

c

,

and

4

c

not all zero, such that

11 2 2 3 3 4 4

cc c c

=+ + +0v v v v

(2)

Adding (1) and (2) gives

4.4 • Solutions 225

111 2 22 3 33 4 44

()( )( )( )

kc k c kc k c

=+= + + + + + + +ww0 v v v v

(3)

At least one of the weights in (3) differs from the corresponding weight in (1) because at least one of

the

i

c

is nonzero. So w is expressed in more than one way as a linear combination of

1

v

,

2

v

,

3

v

,

and

4.v

21. The matrix of the transformation will be

1

12 92

49 41

B

P

−

ª

ºª º

==

«

»« »

−

¬

¼¬ ¼

.

22. The matrix of the transformation will be

[]

1

.

Bn

P

−

=⋅⋅⋅bb

23. Suppose that

[] [ ]

BB

n

c

1

ªº

«»

==.

«»

¬¼

uw

#

By definition of coordinate vectors,

11

.

nn

cc

== +⋅⋅⋅ +uw b b

Since u and w were arbitrary elements of V, the coordinate mapping is one-to-one.

24. Given

1

(, , )

n

yy

=…y

in

n

, let

11 nn

yy

=+⋅⋅⋅ +ub b

. Then, by definition,

[]

B

=uy

. Since y was

arbitrary, the coordinate mapping is onto

n

.

25. Since the coordinate mapping is one-to-one, the following equations have the same solutions

1

,,

p

cc…

:

11

pp

cc+⋅⋅⋅ +=

uu0

(the zero vector in V ) (4)

[]

11 pp B

B

cc

ªº

+⋅⋅⋅ +=

¬¼

uu0

(the zero vector in

n

) (5)

Since the coordinate mapping is linear, (5) is equivalent to

11

0

[] [ ]

0

BppB

cc

ªº

«»

+⋅⋅⋅ +=

«»

¬¼

uu

#

(6)

Thus (4) has only the trivial solution if and only if (6) has only the trivial solution. It follows that

1

{, , }

p

…

uu

is linearly independent if and only if

1

{[ ] , ,[ ] }

BpB

…

uu

is linearly independent. This

result also follows directly from Exercises 31 and 32 in Section 4.3.

26. By definition, w is a linear combination of

1

,,

p

…

uu

if and only if there exist scalars

1

,,

p

cc…

such

that

11

pp

cc=+⋅⋅⋅ +

wu u

(7)

Since the coordinate mapping is linear,

11

[] [ ] [ ]

BBppB

cc=+⋅⋅ ⋅ +

wu u

(8)

Conversely, (8) implies (7) because the coordinate mapping is one-to-one. Thus w is a linear

combination of

1

,,

p

…

uu

if and only if

[]

B

w

is a linear combination of

1

[], ,[ ].

BpB

…

uu

226 CHAPTER 4 • Vector Spaces

Note:

Students need to be urged to write not just to compute in Exercises 27–34. The language in the

Study Guide solution of Exercise 31 provides a model for the students. In Exercise 32, students may have

difficulty distinguishing between the two isomorphic vector spaces, sometimes giving a vector in

3

as an

answer for part (b).

27. The coordinate mapping produces the coordinate vectors (1, 0, 0, 2), (2, 1, –3, 0), and (0, –1, 2, –1)

respectively. We test for linear independence of these vectors by writing them as columns of a matrix

and row reducing:

120 100

011010

.

032 001

201 000

ªºªº

«»«»

−

«»«»

∼

«»«»

−

«»«»

−

¬¼¬¼

Since the matrix has a pivot in each column, its columns (and thus the given polynomials) are

linearly independent.

28. The coordinate mapping produces the coordinate vectors (1, 0, –2, –1), (0, 1, 0, 2), and (1, 1, –2, 0)

respectively. We test for linear independence of these vectors by writing them as columns of a matrix

and row reducing:

10 1 100

01 1 010

.

20 2 001

12 0 000

ªºªº

«»«»

∼

«»«»

−−

«»«»

−

¬¼¬¼

Since the matrix has a pivot in each column, its columns (and thus the given polynomials) are

linearly independent.

29. The coordinate mapping produces the coordinate vectors (1, –2, 1, 0), (0, 1, –2, 1), and (1, –3, 3, –1)

respectively. We test for linear independence of these vectors by writing them as columns of a matrix

and row reducing:

101 101

213011

.

123 000

011000

ªºªº

«»«»

−− −

«»«»

∼

«»«»

−

«»«»

−

¬¼¬¼

Since the matrix does not have a pivot in each column, its columns (and thus the given polynomials)

are linearly dependent.

30. The coordinate mapping produces the coordinate vectors (8, –12, 6, –1), (9, –6, 1, 0), and (1, 6, –5,1)

respectively. We test for linear independence of these vectors by writing them as columns of a matrix

and row reducing:

891 101

12 6 6 0 1 1 .

615000

10 1 000

−

ªºªº

«»«»

−−

«»«»

∼

«»«»

−

«»«»

−

¬¼¬¼

Since the matrix does not have a pivot in each column, its columns (and thus the given polynomials)

are linearly dependent.

4.4 • Solutions 227

31. In each part, place the coordinate vectors of the polynomials into the columns of a matrix and reduce

the matrix to echelon form.

a.

1341 1341

3550 0473

5761 0000

−− − −

ªºªº

«»«»

−∼−−

«»«»

−− −

¬¼¬¼

Since there is not a pivot in each row, the original four column vectors do not span

3

. By the

isomorphism between

3

and

2

, the given set of polynomials does not span

2

.

b.

0132 122 0

5843 026 3

1220 0007/2

−−

ªºª º

«»« »

−−∼ −−

«»« »

−

¬¼¬ ¼

Since there is a pivot in each row, the original four column vectors span

3

. By the isomorphism

between

3

and

2

, the given set of polynomials spans

2

.

32. a. Place the coordinate vectors of the polynomials into the columns of a matrix and reduce the

matrix to echelon form:

10 1 101

011 011

133 001

ªºªº

«»«»

∼

«»«»

−− −

¬¼¬¼

The resulting matrix is invertible since it row equivalent to

3

.

I

The original three column vectors

form a basis for

3

by the Invertible Matrix Theorem. By the isomorphism between

3

and

2

, the

corresponding polynomials form a basis for

2

.

b. Since

[] (1,1,2),

B

=−q

12 3

2.=−++qpp p

One might do the algebra in

2

or choose to compute

10 11 1

0111 3.

1332 10

−

ªºªºªº

«»«»«»

=

«»«»«»

−− −

¬¼¬¼¬¼

This combination of the columns of the matrix corresponds to the

same combination of

1

,p

2

,p

and

3

.p

So

2

() 1 3 10 .ttt=+ −

q

33. The coordinate mapping produces the coordinate vectors (3, 7, 0, 0), (5, 1, 0, –2), (0, 1, –2, 0) and

(1, 16, –6, 2) respectively. To determine whether the set of polynomials is a basis for

3

, we

investigate whether the coordinate vectors form a basis for

4

. Writing the vectors as the columns of

a matrix and row reducing

3501 1002

71116 0101

,

0026 0013

0202 0000

ªºªº

«»«»

−

«»«»

∼

«»«»

−−

«»«»

−

«»«»

¬¼¬¼

we find that the matrix is not row equivalent to

4

.

I

Thus the coordinate vectors do not form a basis

for

4

. By the isomorphism between

4

and

3

, the given set of polynomials does not form a basis for

3

.

34. The coordinate mapping produces the coordinate vectors (5, –3, 4, 2), (9, 1, 8, –6), (6, –2, 5, 0), and

(0, 0, 0, 1) respectively. To determine whether the set of polynomials is a basis for

3

, we investigate

whether the coordinate vectors form a basis for

4

. Writing the vectors as the columns of a matrix,

and row reducing

228 CHAPTER 4 • Vector Spaces

5960 103/40

3120 011/40

4850 00 01

260100 00

ªºªº

«»«»

−−

«»«»

∼

«»«»

−

«»«»

¬¼¬¼

we find that the matrix is not row equivalent to I

4

. Thus the coordinate vectors do not form a basis for

4

. By the isomorphism between

4

and

3

, the given set of polynomials does not form a basis for

3

.

35. To show that x is in

12

Span{ , },

H

=vv

we must show that the vector equation

11 2 2

xx

+=vvx

has a

solution. The augmented matrix

[]

12

vv x

may be row reduced to show

11 14 19 1 0 5 / 3

5813 018/3

.

10 13 18 0 0 0

710 15 00 0

−

ªºªº

«»«»

−−−

«»«»

∼

«»«»

¬¼¬¼

Since this system has a solution, x is in H. The solution allows us to find the B-coordinate vector for

x: since

11 2 2 1 2

(5/3) (8/3)

xx

=+ =−+xv v v v

,

5/3

[] 8/3

B

−

ª

º

=

«

»

¬

¼

x

.

36. To show that x is in

123

Span{ , , }

H

=vv v

, we must show that the vector equation

11 2 2 3 3

xx x

++=vv vx

has a solution. The augmented matrix

[]

123

vv vx

may be row

reduced to show

6894 1003

4357 0105

.

9788 0012

4333 0000

−−

ªºªº

«»«»

−

«»«»

∼

«»«»

−−−

«»«»

−

«»«»

¬¼¬¼

The first three columns show that B is a basis for H. Moreover, since this system has a solution, x is

in H. The solution allows us to find the B-coordinate vector for x: since

11 2 2 3 3 1 2 3

35 2

xx x

=+ + =++xv v v vv v

,

3

[] 5.

2

B

ª

º

«

»

=

«

»

«

»

¬

¼

x

37. We are given that

1/2

[] 1/4,

1/6

B

ªº

«»

=«»

«»

¬¼

x

where

2.6 0 0

1.5 , 3 , 0 .

004.8

B



½

ª

ºªºª º

°

«

»«»« »

=−

®

¾

«

»«»« »

°

«

»«»« »

¬

¼¬¼¬ ¼

¯¿

To find the coordinates of x

relative to the standard basis in

3

, we must find x. We compute that

2.6 0 0 1/ 2 1.3

[] 1.5 3 0 1/4 0 .

004.81/6 0.8

BB

P

ªºªºªº

«»«»«»

==−=

«»«»«»

¬¼¬¼¬¼

xx

4.5 • Solutions 229

38. We are given that

1/2

[] 1/2,

1/3

B

ªº

«»

=«»

«»

¬¼

x

where

2.6 0 0

1.5 , 3 , 0 .

004.8

B



½

ª

ºªºª º

°

«

»«»« »

=−

®

¾

«

»«»« »

°

«

»«»« »

¬

¼¬¼¬ ¼

¯¿

To find the coordinates of x

relative to the standard basis in

3

, we must find x. We compute that

2.6 0 0 1/ 2 1.3

[] 1.5 3 0 1/2 0.75.

004.81/3 1.6

BB

P

ªºªºªº

«»«»«»

==−=

«»«»«»

¬¼¬¼¬¼

xx

4.5 SOLUTIONS

Notes:

Theorem 9 is true because a vector space isomorphic to

n

has the same algebraic properties as

n

; a proof of this result may not be needed to convince the class. The proof of Theorem 9 relies upon the

fact that the coordinate mapping is a linear transformation (which is Theorem 8 in Section 4.4). If you

have skipped this result, you can prove Theorem 9 as is done in Introduction to Linear Algebra by Serge

Lang (Springer-Verlag, New York, 1986). There are two separate groups of true-false questions in this

section; the second batch is more theoretical in nature. Example 4 is useful to get students to visualize

subspaces of different dimensions, and to see the relationships between subspaces of different

dimensions. Exercises 31 and 32 investigate the relationship between the dimensions of the domain and

the range of a linear transformation; Exercise 32 is mentioned in the proof of Theorem 17 in Section 4.8.

1. This subspace is

12

Span{ , },

H

=vv

where

1

0

ª

º

«

»

=

«

»

«

»

¬

¼

v

and

2

1.

3

−

ª

º

«

»

=

«

»

«

»

¬

¼

v

Since

1

v

and

2

v

are not

multiples of each other,

12

{, }vv

is linearly independent and is thus a basis for H. Hence the

dimension of H is 2.

2. This subspace is

12

Span{ , },

H

=vv

where

1

2

0

2

ª

º

«

»

=

«

»

«

»

−

¬

¼

v

and

2

0

4.

0

ª

º

«

»

=−

«

»

«

»

¬

¼

v

Since

1

v

and

2

v

are not

multiples of each other,

12

{, }vv

is linearly independent and is thus a basis for H. Hence the

dimension of H is 2.

3. This subspace is

123

Span{ , , },

H

=vv v

where

1

0

1,

0

1

ª

º

«

»

«

»

=

«

»

«

»

«

»

¬

¼

v

2

0

1,

1

2

ª

º

«

»

−

«

»

=

«

»

«

»

«

»

¬

¼

v

and

3

2

0.

3

0

ªº

«»

=«»

−

«»

¬¼

v

Theorem 4 in

Section 4.3 can be used to show that this set is linearly independent:

1

,≠v0

2

v

is not a multiple of

1

,v

and (since its first entry is not zero)

3

v

is not a linear combination of

1

v

and

2

.v

Thus

123

{, , }vv v

is linearly independent and is thus a basis for H. Alternatively, one can show that this set

is linearly independent by row reducing the matrix

[]

123

.

vv v0

Hence the dimension of the

subspace is 3.

230 CHAPTER 4 • Vector Spaces

4. This subspace is

12

Span{ , },

H

=vv

where

1

3

1

ª

º

«

»

−

«

»

=

«

»

«

»

¬

¼

v

and

2

0.

1

ª

º

«

»

«

»

=

«

»

−

«

»

¬

¼

v

Since

1

v

and

2

v

are not

multiples of each other,

12

{, }vv

is linearly independent and is thus a basis for H. Hence the

dimension of H is 2.

5. This subspace is

123

Span{ , , },

H

=vv v

where

1

2,

0

3

ª

º

«

»

«

»

=

«

»

«

»

−

¬

¼

v

2

0,

2

0

−

ª

º

«

»

«

»

=

«

»

−

«

»

¬

¼

v

and

3

0

5.

2

6

ªº

«»

=«»

«»

¬¼

v

The matrix A

with these vectors as its columns row reduces to

120 100

205 010

.

022 001

306 000

−

ª

ºª º

«

»« »

«

»« »

∼

«

»« »

−

«

»« »

−

¬

¼¬ ¼

There is a pivot in

each column, so

123

{, , }vv v

is linearly independent and is thus a basis for H. Hence the dimension

of H is 3.

6. This subspace is

123

Span{ , , },

H

=vv v

where

1

3

0,

7

3

ª

º

«

»

«

»

=

«

»

−

«

»

−

¬

¼

v

2

0

1,

6

0

ª

º

«

»

−

«

»

=

«

»

«

»

¬

¼

v

and

3

1

3.

5

1

−

ªº

«»

−

«»

=«»

«»

¬¼

v

The matrix A

with these vectors as its columns row reduces to

30 1 100

013 010

.

765 001

30 1 000

−

ª

ºª º

«

»« »

−−

«

»« »

∼

«

»« »

−

«

»« »

−

¬

¼¬ ¼

There is a pivot in

each column, so

123

{, , }vv v

is linearly independent and is thus a basis for H. Hence the dimension

of H is 3.

7. This subspace is H = Nul A, where

131

012.

021

A

−

ª

º

«

»

=−

«

»

«

»

−

¬

¼

Since

[]

1000

0100,

0010

A

ª

º

«

»

∼

«

»

«

»

¬

¼

0

the

homogeneous system has only the trivial solution. Thus H = Nul A = {0}, and the dimension of H is

0.

8. From the equation a – 3b + c = 0, it is seen that (a, b, c, d) = b(3, 1, 0, 0) + c(–1, 0, 1, 0) + d(0, 0, 0,

1). Thus the subspace is

123

Span{ , , },

H

=vv v

where

1

(3,1, 0,0),=v

2

(1,0,1,0),=−v

and

3

(0,0,0,1).=v

It is easily checked that this set of vectors is linearly independent, either by appealing

to Theorem 4 in Section 4.3, or by row reducing

[]

123

.

vv v0

Hence the dimension of the

subspace is 3.

4.5 • Solutions 231

9. This subspace is

:, in

a

Hbab

a

ªº

°«»

=®«»

°«»

¬¼

¯

12

Span{ , },

½

°

=

¾

°

¿

vv

where

1

0

1

ª

º

«

»

=

«

»

«

»

¬

¼

v

and

2

0

1.

0

ªº

«»

=«»

«»

¬¼

v

Since

1

v

and

2

v

are not multiples of each other,

12

{, }vv

is linearly independent and is thus a basis for H.

Hence the dimension of H is 2.

10. The matrix A with these vectors as its columns row reduces to

123 123

.

51015 0 0 0

−− −−

ªºªº

∼

«»«»

−

¬¼¬¼

There is one pivot column, so the dimension of Col A (which is the dimension of H) is 1.

11. The matrix A with these vectors as its columns row reduces to

13 2 5 10 10

01 12 01 10.

21 12 00 01

−

ªºªº

«»«»

−∼ −

«»«»

¬¼¬¼

There are three pivot columns, so the dimension of Col A (which is the dimension of the subspace

spanned by the vectors) is 3.

12. The matrix A with these vectors as its columns row reduces to

1323 1001

2635 0100.

0655 0011

−− − −

ªºªº

«»«»

−− ∼

«»«»

¬¼¬¼

There are three pivot columns, so the dimension of Col A (which is the dimension of the subspace

spanned by the vectors) is 3.

13. The matrix A is in echelon form. There are three pivot columns, so the dimension of Col A is 3.

There are two columns without pivots, so the equation Ax = 0 has two free variables. Thus the

dimension of Nul A is 2.

14. The matrix A is in echelon form. There are four pivot columns, so the dimension of Col A is 4. There

are three columns without pivots, so the equation Ax = 0 has three free variables. Thus the dimension

of Nul A is 3.

15. The matrix A is in echelon form. There are three pivot columns, so the dimension of Col A is 3.

There are two columns without pivots, so the equation Ax = 0 has two free variables. Thus the

dimension of Nul A is 2.

16. The matrix A row reduces to

32 10

.

65 01

ªºªº

∼

«»«»

−

¬¼¬¼

There are two pivot columns, so the dimension of Col A is 2. There are no columns without pivots,

so the equation Ax = 0 has only the trivial solution 0. Thus Nul A = {0}, and the dimension of Nul A

is 0.

232 CHAPTER 4 • Vector Spaces

17. The matrix A is in echelon form. There are three pivot columns, so the dimension of Col A is 3.

There are no columns without pivots, so the equation Ax = 0 has only the trivial solution 0. Thus Nul

A = {0}, and the dimension of Nul A is 0.

18. The matrix A is in echelon form. There are two pivot columns, so the dimension of Col A is 2. There

is one column without a pivot, so the equation Ax = 0 has one free variable. Thus the dimension of

Nul A is 1.

19. a. True. See the box before Example 5.

b . False. The plane must pass through the origin; see Example 4.

c . False. The dimension of

n

is n + 1; see Example 1.

d . False. The set S must also have n elements; see Theorem 12.

e . True. See Theorem 9.

20. a. False. The set

2

is not even a subset of

3

.

b . False. The number of free variables is equal to the dimension of Nul A; see the box before

Example 5.

c . False. A basis could still have only finitely many elements, which would make the vector space

finite-dimensional.

d . False. The set S must also have n elements; see Theorem 12.

e . True. See Example 4.

21. The matrix whose columns are the coordinate vectors of the Hermite polynomials relative to the

standard basis

23

{1, , , }tt t

of

3

is

10 2 0

02 0 12

.

00 4 0

00 0 8

A

−

ªº

«»

−

«»

=«»

«»

¬¼

This matrix has 4 pivots, so its columns are linearly independent. Since their coordinate vectors form

a linearly independent set, the Hermite polynomials themselves are linearly independent in

3

. Since

there are four Hermite polynomials and dim

3

= 4, the Basis Theorem states that the Hermite

polynomials form a basis for

3

.

22. The matrix whose columns are the coordinate vectors of the Laguerre polynomials relative to the

standard basis

23

{1, , , }tt t

of

3

is

112 6

01418

.

00 1 9

000 1

A

ªº

«»

−− −

«»

=«»

«»

−

«»

¬¼

This matrix has 4 pivots, so its columns are linearly independent. Since their coordinate vectors form

a linearly independent set, the Laguerre polynomials themselves are linearly independent in

3

. Since

there are four Laguerre polynomials and dim

3

= 4, the Basis Theorem states that the Laguerre

polynomials form a basis for

3

.

4.5 • Solutions 233

23. The coordinates of

23

() 1 8 8ttt=−++

p

with respect to B satisfy

2323

12 3 4

(1) (2 ) ( 2 4 ) ( 12 8 ) 1 8 8cctc tctt tt++−++−+=−++

Equating coefficients of like powers of t produces the system of equations

13

24

3

4

21

2120

48

88

cc

c

−=−

−=

=

Solving this system gives

1

3,

c

=

2

6,

c

=

3

2,

c

=

4

1,

c

=

and

3

6

[] .

2

1

B

ª

º

«

»

«

»

=

«

»

«

»

¬

¼

p

24. The coordinates of

2

() 5 5 2ttt=+ −

p

with respect to B satisfy

22

12 3

(1) (1 ) (2 4 ) 5 5 2cctc tt tt+−+−+=+−

Equating coefficients of like powers of t produces the system of equations

123

23

3

25

45

2

ccc

cc

c

++=

−− =

=−

Solving this system gives

1

6,

c

=

2

3,

c

=

3

2,

c

=−

and

6

[] 3.

2

B

ª

º

«

»

=

«

»

«

»

−

¬

¼

p

25. Note first that n ≥ 1 since S cannot have fewer than 1 vector. Since n ≥ 1, V ≠ 0. Suppose that S spans

V and that S contains fewer than n vectors. By the Spanning Set Theorem, some subset S′ of S is a

basis for V. Since S contains fewer than n vectors, and S′ is a subset of S, S′ also contains fewer

than n vectors. Thus there is a basis S′ for V with fewer than n vectors, but this is impossible by

Theorem 10 since dimV = n. Thus S cannot span V.

26. If dimV = dim H = 0, then V = {0} and H = {0}, so H = V. Suppose that dim V = dim H > 0. Then H

contains a basis S consisting of n vectors. But applying the Basis Theorem to V, S is also a basis for

V. Thus H = V = SpanS.

27. Suppose that dim = k < ∞. Now

n

is a subspace of for all n, and dim

k–1

= k, so dim

k–1

= dim

. This would imply that

k–1

= , which is clearly untrue: for example,

()

k

tt=

p

is in but not in

k–1

. Thus the dimension of cannot be finite.

28. The space C() contains as a subspace. If C( ) were finite-dimensional, then would also be

finite-dimensional by Theorem 11. But is infinite-dimensional by Exercise 27, so C( ) must also

be infinite-dimensional.

29. a. True. Apply the Spanning Set Theorem to the set

1

{, , }

p

…

vv

and produce a basis for V. This

basis will not have more than p elements in it, so dimV ≤ p.

234 CHAPTER 4 • Vector Spaces

b . True. By Theorem 11,

1

{, , }

p

…

vv

can be expanded to find a basis for V. This basis will have at

least p elements in it, so dimV ≥ p.

c . True. Take any basis (which will contain p vectors) for V and adjoin the zero vector to it.

30. a. False. For a counterexample, let v be a non-zero vector in

3

, and consider the set {v, 2v}. This is

a linearly dependent set in

3

, but dim

3

32=>

.

b . True. If dimV ≤ p, there is a basis for V with p or fewer vectors. This basis would be a spanning

set for V with p or fewer vectors. If necessary, vectors in V could be added to this spanning set to

give a spanning set for V with exactly p vectors, which contradicts the assumption.

c . False. For a counterexample, let v be a non-zero vector in

3

, and consider the set {v, 2v}. This is

a linearly dependent set in

3

with 3 – 1 = 2 vectors, and dim

3

3=

.

31. Since H is a nonzero subspace of a finite-dimensional vector space V, H is finite-dimensional and has

a basis. Let

1

{, , }

p

…

uu

be a basis for H. We show that the set

1

{( ), , ( )}

p

TT…

uu

spans T(H). Let y

be in T(H). Then there is a vector x in H with T(x) = y. Since x is in H and

1

{, , }

p

…

uu

is a basis for

H, x may be written as

11

pp

cc=+…+

xu u

for some scalars

1

,,.

p

cc…

Since the transformation T is

linear,

11 1 1

() ( ) ( ) ( )

pp p p

TTc c cT cT== +…+ = +…+

yx u u u u

Thus y is a linear combination of

1

(),,( )

p

TT…

uu

, and

1

{( ), , ( )}

p

TT…

uu

spans T(H). By the

Spanning Set Theorem, this set contains a basis for T(H). This basis then has not more than p vectors,

and dimT(H) ≤ p = dim H.

32. Since H is a nonzero subspace of a finite-dimensional vector space V, H is finite-dimensional and has

a basis. Let

1

{, }

p

…

uu

be a basis for H. In Exercise 31 above it was shown that

1

{( ), , ( )}

p

TT…

uu

spans T(H). In Exercise 32 in Section 4.3, it was shown that

1

{( ), , ( )}

p

TT…

uu

is linearly

independent. Thus

1

{( ), , ( )}

p

TT…

uu

is a basis for T(H), and dimT(H) = p = dim H.

33. [M]

a. To find a basis for

5

which contains the given vectors, we row reduce

99610000 1001/3001 3/7

74701000 010 0001 5/7

.

818001000011/3000 3/7

56500010 000 010322/7

77700001000 001953/7

−−

ªºª º

«»« »

−

«»« »

∼

−−−

«»« »

−

«»« »

−− −−

¬¼¬ ¼

The first, second, third, fifth, and sixth columns are pivot columns, so these columns of the

original matrix (

12323

{, , , ,}vv vee

) form a basis for

5

:

4.6 • Solutions 235

b. The original vectors are the first k columns of A. Since the set of original vectors is assumed

to be linearly independent, these columns of A will be pivot columns and the original set of

vectors will be included in the basis. Since the columns of A include all the columns of the

identity matrix, Col A =

n

.

34. [M]

a. The B-coordinate vectors of the vectors in C are the columns of the matrix

10 1 0 1 0 1

01 0 3 0 5 0

00 2 0 8 0 18

.

00 0 4 0 20 0

00 0 0 8 0 48

00 0 0 0 16 0

00 0 0 0 0 32

P

−−

ªº

«»

−

«»

−

«»

=−

«»

−

«»

¬¼

The matrix P is invertible because it is triangular with nonzero entries along its main

diagonal. Thus its columns are linearly independent. Since the coordinate mapping is an

isomorphism, this shows that the vectors in C are linearly independent.

b. We know that dim H = 7 because B is a basis for H. Now C is a linearly independent set, and

the vectors in C lie in H by the trigonometric identities. Thus by the Basis Theorem, C is a

basis for H.

4.6 SOLUTIONS

Notes:

This section puts together most of the ideas from Chapter 4. The Rank Theorem is the main result

in this section. Many students have difficulty with the difference in finding bases for the row space and

the column space of a matrix. The first process uses the nonzero rows of an echelon form of the matrix.

The second process uses the pivots columns of the original matrix, which are usually found through row

reduction. Students may also have problems with the varied effects of row operations on the linear

dependence relations among the rows and columns of a matrix. Problems of the type found in Exercises

19–26 make excellent test questions. Figure 1 and Example 4 prepare the way for Theorem 3 in Section

6.1; Exercises 27–29 anticipate Example 6 in Section 7.4.

1. The matrix B is in echelon form. There are two pivot columns, so the dimension of Col A is 2. There

are two pivot rows, so the dimension of Row A is 2. There are two columns without pivots, so the

equation Ax = 0 has two free variables. Thus the dimension of Nul A is 2. A basis for Col A is the

pivot columns of A:

14

1, 2 .

56

½−

ªºªº

°°

«»«»

−

®¾

«»«»

°°

«»«»

−

¬¼¬¼

¯¿

A basis for Row A is the pivot rows of B:

{

}

(1, 0, 1, 5), ( 0, 2, 5, 6) .−−−

To find a basis for Nul A row

reduce to reduced echelon form:

236 CHAPTER 4 • Vector Spaces

10 1 5

.

01 5/23

A

−

ªº

∼

«»

−

¬¼

The solution to

A=x0

in terms of free variables is

13 4

5

xx x

=−

,

234

(5/ 2) 3

xxx

=−

with

3

x

and

4

x

free. Thus a basis for Nul A is

15

5/2 3

,.

10

01

½

−

ªºªº

°°

«»«»

−

°°

«»«»

®¾

«»«»

°°

«»«»

°°

«»«»

¬¼¬¼

¯¿

2. The matrix B is in echelon form. There are three pivot columns, so the dimension of Col A is 3.

There are three pivot rows, so the dimension of Row A is 3. There are two columns without pivots,

so the equation

A=x0

has two free variables. Thus the dimension of Nul A is 2. A basis for Col A is

the pivot columns

of A:

14 2

26 3

,, .

33 3

30 0

½

ªºªºª º

°°

«»«»« »

−

°°

«»«»« »

®¾

«»«»« »

−

°°

«»«»« »

°°

¬¼¬¼¬ ¼

¯¿

A basis for Row A is the pivot rows of B:

{

}

(1, 3, 4, 1, 2), (0, 0,1, 1,1), (0, 0, 0, 0, 5) .−− −

To find a basis

for Nul A row reduce to reduced echelon form:

130 30

001 10

.

000 01

000 00

A

ªº

«»

−

«»

∼«»

«»

¬¼

The solution to

A=x0

in terms of free variables is

124

33

xxx

=−−

,

34

xx

=

,

5

0

x

=

, with

2

x

and

4

x

free. Thus a basis for Nul A is

33

10

,.

01

00

½

−−

ªºªº

°°

«»«»

°°

«»«»

°°

«»«»

®¾

«»«»

°°

«»«»

°°

«»«»

°°

¬¼¬¼

¯¿

3. The matrix B is in echelon form. There are three pivot columns, so the dimension of Col A is 3.

There are three pivot rows, so the dimension of Row A is 3. There are three columns without pivots,

so the equation

A=x0

has three free variables. Thus the dimension of Nul A is 3. A basis for Col A

is the pivot columns of A:

263

230

,, .

493

233

½

ªºªºªº

°°

«»«»«»

−−

°°

«»«»«»

®¾

«»«»«»

°°

«»«»«»

°°

−

¬¼¬¼¬¼

¯¿

4.6 • Solutions 237

A basis for Row A is the pivot rows of B:

{

}

(2,6, 6,6,3,6),(0,3,0,3,3,0),(0,0,0,0,3,0) .−

To find a

basis for Nul A row reduce to reduced echelon form:

10 3003

01 0100

.

00 0010

00 0000

A

−

ªº

«»

∼«»

«»

¬¼

The solution to

A=x0

in terms of free variables is

136

33,

xxx

=−

24

,

xx

=−

5

0,

x

=

with

3

x

,

4

x

,

and

6

x

free. Thus a basis for Nul A is

303

010

100

,, .

010

000

001

½

−

ªº ª ºª º

°°

«» « »« »

−

°°

«» « »« »

°°

«» « »« »

°°

«» « »« »

®¾

«» « »« »

°°

«» « »« »

°°

«» « »« »

°°

«» « »« »

°°

¬¼ ¬ ¼¬ ¼

¯¿

4. The matrix B is in echelon form. There are five pivot columns, so the dimension of Col A is 5. There

are five pivot rows, so the dimension of Row A is 5. There is one column without a pivot, so the

equation

A=x0

has one free variable. Thus the dimension of Nul A is 1. A basis for Col A is the

pivot columns of A:

11212

1232 3

,,,, .

11016

12230

12 12 1

½

−−

ªº ª º ª º ª º ª º

°°

«» « » « » « » « »

−− −

°°

«» « » « » « » « »

°°

«» « » « » « » « »

−

®¾

«» « » « » « » « »

°°

−−

«» « » « » « » « »

°°

«» « » « » « » « »

−−

°°

¬¼ ¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼

¯¿

A basis for Row A is the pivot rows of B:

{

}

(1, 1, 2, 0, 1, 2), (0, 1, 1, 0, 3, 1), (0, 0, 1, 1, 13, 1), (0, 0, 0, 0,1, 1), (0, 0, 0, 0, 0,1) .−− −−− −− −

To find a basis for Nul A row reduce to reduced echelon form:

100 100

010100

.

001100

000010

000001

A

ªº

«»

∼

«»

¬¼

The solution to

A=x0

in terms of free variables is

14

xx

=−

,

24

xx

=−

,

34

xx

=−

,

5

0

x

=

,

6

0

x

=

,

with

4

x

free. Thus a basis for Nul A is

238 CHAPTER 4 • Vector Spaces

1

1.

1

0

½

−

ªº

°°

«»

−

°°

«»

°°

«»

−

°°

«»

®¾

«»

°°

«»

°°

«»

°°

«»

°°

¬¼

¯¿

5. By the Rank Theorem,

dimNul 7 rank 7 3 4.AA=−=−=

Since

dimRow rank ,dimRow 3.AA A==

Since

rank dimCol dimRow ,

TT

AA A==

rank 3.

T

A

=

6. By the Rank Theorem,

dimNul 5 rank 5 2 3.AA=−=−=

Since

dimRow rank , dimRow 2.AA A==

Since

rank dimCol dimRow , rank 2.

TT T

AA AA== =

7. Yes, Col A =

4

. Since A has four pivot columns,

dimCol 4.A=

Thus Col A is a four-dimensional

subspace of

4

, and Col A =

4

.

No,

Nul A≠

3

. It is true that

dimNul 3A=

, but Nul A is a subspace of

7

.

8. Since A has four pivot columns,

rank 4,A=

and

dimNul 8 rank 8 4 4.AA=−=−=

No.

Col A≠

4

. It is true that

dimCol rank 4,AA==

but Col A is a subspace of

6

.

9. Since

dimNul 3, rank 6 dimNul 6 3 3.AA A==−=−=

So

dimCol rank 3.AA==

No.

Col A≠

3

. It is true that

dimCol rank 3,AA==

but Col A is a subspace of

4

.

10. Since

dimNul 5, rank 7 dimNul 7 5 2.AA A==−=−=

So

dimCol rank 2.AA==

11. Since

dimNul 3, rank 5 dimNul 5 3 2.AA A==−=−=

So

dimRow dimCol rank 2.AAA===

12. Since

dimNul 2, rank 4 dimNul 4 2 2.AA A==−=−=

So

dimRow dimCol rank 2.AAA===

13. The rank of a matrix A equals the number of pivot positions which the matrix has. If A is either a

75×

matrix or a

57×

matrix, the largest number of pivot positions that A could have is 5. Thus the

largest possible value for rank A is 5.

14. The dimension of the row space of a matrix A is equal to rank A, which equals the number of pivot

positions which the matrix has. If A is either a

54×

matrix or a

45×

matrix, the largest number of

pivot positions that A could have is 4. Thus the largest possible value for dimRow A is 4.

15. Since the rank of A equals the number of pivot positions which the matrix has, and A could have at

most 3 pivot positions,

rank 3.A≤

Thus

dimNul 7 rank 7 3 4.AA=−≥−=

16. Since the rank of A equals the number of pivot positions which the matrix has, and A could have at

most 5 pivot positions,

rank 5.A≤

Thus

dimNul 5 rank 5 5 0.AA=−≥−=

17. a. True. The rows of A are identified with the columns of

.

T

A

See the paragraph before Example 1.

b . False. See the warning after Example 2.

c . True. See the Rank Theorem.

d . False. See the Rank Theorem.

4.6 • Solutions 239

e . True. See the Numerical Note before the Practice Problem.

18. a. False. Review the warning after Theorem 6 in Section 4.3.

b . False. See the warning after Example 2.

c . True. See the remark in the proof of the Rank Theorem.

d . True. This fact was noted in the paragraph before Example 4. It also follows from the fact that the

rows of

T

A

are the columns of

() .

TT

AA=

e . True. See Theorem 13.

19. Yes. Consider the system as ,A=x0

where A is a

56×

matrix. The problem states that

dimNul 1A=

. By the Rank Theorem,

rank 6 dimNul 5.AA=−=

Thus

dim Col rank 5,AA==

and

since Col A is a subspace of

5

, Col A =

5

So every vector b in

5

is also in Col A, and ,A=xb

has

a solution for all b.

20. No. Consider the system as ,A=xb

where A is a

68×

matrix. The problem states that

dimNul 2.A=

By the Rank Theorem,

rank 8 dimNul 6.AA=−=

Thus

dimCol rank 6,AA==

and

since Col A is a subspace of

6

, Col A =

6

So every vector b in

6

is also in Col A, and

A=xb

has

a solution for all b. Thus it is impossible to change the entries in b to make

A=xb

into an

inconsistent system.

21. No. Consider the system as ,A=xb

where A is a

910×

matrix. Since the system has a solution for

all b in

9

, A must have a pivot in each row, and so

rank 9.A=

By the Rank Theorem,

dimNul 10 9 1.A=−=

Thus it is impossible to find two linearly independent vectors in Nul A.

22. No. Consider the system as ,A=x0

where A is a

10 12×

matrix. Since A has at most 10 pivot

positions,

rank 10.A≤

By the Rank Theorem,

dimNul 12 rank 2.AA=−≥

Thus it is impossible to

find a single vector in Nul A which spans Nul A.

23. Yes, six equations are sufficient. Consider the system as ,A=x0

where A is a

12 8×

matrix. The

problem states that

dimNul 2.A=

By the Rank Theorem,

rank 8 dimNul 6.AA=−=

Thus

dimCol rank 6.AA==

So the system

A=x0

is equivalent to the system ,B=x0

where B is an

echelon form of A with 6 nonzero rows. So the six equations in this system are sufficient to describe

the solution set of

.A=x0

24. Yes, No. Consider the system as ,A=xb

where A is a

76×

matrix. Since A has at most 6 pivot

positions,

rank 6.A≤

By the Rank Theorem,

dim Nul 6 rank 0.AA=−≥

If

dimNul 0,A=

then the

system

A=xb

will have no free variables. The solution to ,A=xb

if it exists, would thus have to be

unique. Since

rank 6,A≤

Col A will be a proper subspace of

7

. Thus there exists a b in

7

for

which the system

A=xb

is inconsistent, and the system

A=xb

cannot have a unique solution for

all b.

25. No. Consider the system as ,A=xb

where A is a

10 12×

matrix. The problem states that

dim Nul 3.A=

By the Rank Theorem,

dimCol rank 12 dimNul 9.AA A==−=

Thus Col A will be a

proper subspace of

10

Thus there exists a b in

10

for which the system

A=xb

is inconsistent, and

the system

A=xb

cannot have a solution for all b.

240 CHAPTER 4 • Vector Spaces

26. Consider the system ,A=x0

where A is a mn× matrix with .mn> Since the rank of A is the

number of pivot positions that A has and A is assumed to have full rank,

rank .An=

By the Rank

Theorem,

dimNul rank 0.An A=−=

So

Nul { },A=

0

and the system

A=x0

has only the trivial

solution. This happens if and only if the columns of A are linearly independent.

27. Since A is an m × n matrix, Row A is a subspace of

n

, Col A is a subspace of

m

, and Nul A is a

subspace of

n

. Likewise since

T

A

is an n × m matrix,

Row

T

A

is a subspace of

m

,

Col

T

A

is a

subspace of

n

, and

Nul

T

A

is a subspace of

m

. Since

Row Col

T

AA=

and

Col Row ,

T

AA=

there

are four dinstict subspaces in the list: Row A, Col A, Nul A, and

Nul .

T

A

28. a. Since A is an m × n matrix and dimRow A = rank A,

dimRow A + dimNul A = rank A + dimNul A = n.

b. Since

T

A

is an n × m matrix and

dimCol dimRow dimCol rank ,

TT

AAAA===

dimCol dimNul rank dimNul .

TT T

AAA Am+=+=

29. Let A be an m × n matrix. The system Ax = b will have a solution for all b in

m

if and only if A has a

pivot position in each row, which happens if and only if dimCol A = m. By Exercise 28 b., dimCol A

= m if and only if

dimNul 0

T

Amm=−=

, or

Nul { }.

T

A=

0

Finally,

Nul { }

T

A=

0

if and only if the

equation

T

A

=x0

has only the trivial solution.

30. The equation Ax = b is consistent if and only if

[]

rank rankAA=

b

because the two ranks will be

equal if and only if b is not a pivot column of

[]

.A

b

The result then follows from Theorem 2 in

Section 1.2.

31. Compute that

[]

2222

3333.

5555

T

abc

abc a b c

abc

ª

ºª º

«

»« »

=−=−−−

«

»« »

«

»« »

¬

¼¬ ¼

uv

Each column of

T

uv

is a multiple of u,

so

dimCol 1

T

=

uv

, unless a = b = c = 0, in which case

T

uv

is the 3 × 3 zero matrix and

dimCol 0.

T

=

uv

In any case,

rank dimCol 1

TT

=≤

uv uv

32. Note that the second row of the matrix is twice the first row. Thus if v = (1, –3, 4), which is the first

row of the matrix,

[]

1134

134 .

2268

T

−

ªº ª º

=−=

«» « »

−

¬¼ ¬ ¼

uv

33. Let

[]

123

,A=

uu u

and assume that rank A = 1. Suppose that

1

≠u0

. Then

1

{}u

is basis for Col

A, since Col A is assumed to be one-dimensional. Thus there are scalars x and y with

21

x

=uu

and

31

y

=uu

, and

1

,

T

A=

uv

where

1

.x

y

ªº

«»

=«»

«»

¬¼

v

4.6 • Solutions 241

If

1

=u0

but

2

≠u0

, then similarly

2

{}u

is basis for Col A, since Col A is assumed to be one-

dimensional. Thus there is a scalar x with

32

x

=uu

, and

2

,

T

A=

uv

where

0

1.

x

ªº

«»

=«»

«»

¬¼

v

If

12

==uu 0

but

3

,≠u0

then

3

,

T

A=

uv

where

0

0.

1

ª

º

«

»

=

«

»

«

»

¬

¼

v

34. Let A be an m × n matrix with of rank r > 0, and let U be an echelon form of A. Since A can be

reduced to U by row operations, there exist invertible elementary matrices

1

,,

p

EE…

with

1

().

p

EEAU⋅⋅⋅ =

Thus

1

(),

p

AE EU

−

=⋅⋅⋅

since the product of invertible matrices is invertible. Let

1

()

p

EE E

−

=⋅⋅⋅

; then A = EU. Let the columns of E be denoted by

1

,,.

m

…cc

Since the rank of A is

r, U has r nonzero rows, which can be denoted

1

,, .

TT

r

…

dd

By the column-row expansion of A

(Theorem 10 in Section 2.4):

[]

1

111

,

T

TTT

r

mrr

AEU

ªº

«»

== … = +…+

«»

¬¼

d

cc cdcd

0

#

which is the sum of r rank 1 matrices.

35. [M]

a. Begin by reducing A to reduced echelon form:

1013/20 50 3

0111/20 1/20 2

.

00 01 11/20 7

00 00 01 1

00 00 00 0

A

−

ªº

«»

∼−

«»

¬¼

A basis for Col A is the pivot columns of A, so matrix C contains these columns:

7953

4625

.

5752

3514

6849

C

−−

ªº

«»

−−−

«»

=−

«»

−−−

«»

−

¬¼

A basis for Row A is the pivot rows of the reduced echelon form of A, so matrix R contains

these rows:

242 CHAPTER 4 • Vector Spaces

1013/20 50 3

0111/20 1/20 2

.

00 01 11/20 7

00 00 01 1

R

−

ªº

«»

=«»

−

«»

¬¼

To find a basis for Nul A row reduce to reduced echelon form, note that the solution to Ax = 0

in terms of free variables is

1357

(13 / 2) 5 3 ,

xxxx

=−−+

2357

(11 / 2) (1 / 2) 2 ,

xxxx

=−−−

457

(11 / 2 ) 7 ,

xxx

=−

67

,

xx

=−

with

3

,

x

5

,

x

and

7

x

free. Thus matrix N is

13/ 2 5 3

11/ 2 1/ 2 2

100

.

011/2 7

010

001

N

−−

ªº

«»

−−−

«»

=−

«»

−

«»

¬¼

b. The reduced echelon form of

T

A

is

1000 2/11

0100 41/11

0010 0

,

0001 28/11

0000 0

T

A

−

ªº

«»

−

«»

∼«»

«»

¬¼

so the solution to

T

A

=x0

in terms of free variables is

15

(2/11) ,

xx

=

25

(41/11) ,

xx

=

3

0,

x

=

45

(28/11) ,

xx

=−

with

5

x

free. Thus matrix M is

2/11

41/11

.

0

28/11

1

M

ªº

«»

=«»

−

«»

¬¼

The matrix

T

SR N

ª

º

=

¬

¼ is 7 × 7 because the columns of

T

R

and N are in

7

and dimRow A

+ dimNul A = 7. The matrix

[]

TCM=

is 5 × 5 because the columns of C and M are in

5

and

dimCol dimNul 5.

T

AA+=

Both S and T are invertible because their columns are linearly

independent. This fact will be proven in general in Theorem 3 of Section 6.1.

36. [M] Answers will vary, but in most cases C will be 6 × 4, and will be constructed from the first 4

columns of A. In most cases R will be 4 × 7, N will be 7 × 3, and M will be 6 × 2.

37. [M] The C and R from Exercise 35 work here, and A = CR.

4.7 • Solutions 243

38. [M] If A is nonzero, then A = CR. Note that

[]

12 n

CR C C C=…rr r

, where

1

r

, …,

n

r

are the

columns of R. The columns of R are either pivot columns of R or are not pivot columns of R.

Consider first the pivot columns of R. The

th

i

pivot column of R is

i

e

, the

th

i

column in the

identity matrix, so

i

C

e

is the

th

i

pivot column of A. Since A and R have pivot columns in the same

locations, when C multiplies a pivot column of R, the result is the corresponding pivot column of A

in its proper location.

Now suppose

j

r

is a nonpivot column of R. Then

j

r

contains the weights needed to construct

the

th

j

column of A from the pivot columns of A, as is discussed in Example 9 of Section 4.3 and in

the paragraph preceding that example. Thus

j

r

contains the weights needed to construct the

th

j

column of A from the columns of C, and

.

jj

C=ra

4.7 SOLUTIONS

Notes:

This section depends heavily on the coordinate systems introduced in Section 4.4. The row

reduction algorithm that produces

cB

P

←

can also be deduced from Exercise 15 in Section 2.2, by row

reducing

.

CB

PPªº

¬¼

to

1

CB

IP P

−

ªº

¬¼

. The change-of-coordinates matrix here is interpreted in Section 5.4

as the matrix of the identity transformation relative to two bases.

1. a. Since

112

62=−bcc

and

212

94,=−bcc

1

6

[] ,

2

C

ª

º

=

«

»

−

¬

¼

b

2

9

[] ,

4

C

ª

º

=

«

»

−

¬

¼

b

and

69

.

24

CB

P

←

ªº

=

«»

−−

¬¼

b. Since

12

32,=−+xbb

3

[] 2

B

−

ªº

=

«»

¬¼

x

and

693 0

[] [] 242 2

CB

P

←

−

ª

ºª º ª º

== =

«

»« » « »

−− −

¬

¼¬ ¼ ¬ ¼

xx

2. a. Since

112

24=−+bcc

and

212

36,=−bcc

1

2

[] ,

4

C

−

ª

º

=

«

»

¬

¼

b

2

3

[] ,

6

C

ª

º

=

«

»

−

¬

¼

b

and

23

.

46

CB

P

←

−

ªº

=

«»

−

¬¼

b. Since

12

23,=+xb b

2

[] 3

B

ªº

=

«»

¬¼

x

and

232 5

[] [] 463 10

CB

P

←

−

ªºªºªº

== =

«»«»«»

−−

¬¼¬¼¬¼

xx

3. Equation (ii) is satisfied by P for all x in V.

4. Equation (i) is satisfied by P for all x in V.

244 CHAPTER 4 • Vector Spaces

5. a. Since

112

4,=−abb

2123

,=−++abbb

and

32 3

2,=−ab b

1

4

[] 1,

0

B

ª

º

«

»

=−

«

»

«

»

¬

¼

a

2

1

[] 1,

1

B

−

ªº

«»

=«»

«»

¬¼

a

3

0

[] 1,

2

B

ªº

«»

=«»

«»

−

¬¼

a

and

410

111.

012

BA

P

←

−

ªº

«»

=−

«»

−

¬¼

b. Since

123

34 ,=+ +xa aa

3

[] 4

1

A

ªº

«»

=«»

«»

¬¼

x

and

4103 8

[] 1 1 1 4 2

0121 2

BBA

P

←

−

ª

ºª º ª º

«

»« » « »

==−=

«

»« » « »

«

»« » « »

−

¬

¼¬ ¼ ¬ ¼

x

6. a. Since

1123

2,=−+fddd

223

3,=+fdd

and

313

32=−+fdd

,

1

2

[] 1,

1

D

ª

º

«

»

=−

«

»

«

»

¬

¼

f

2

0

[] 3,

1

D

ªº

«»

=«»

«»

¬¼

f

3

[] 0,

2

D

−

ªº

«»

=«»

«»

¬¼

f

and

20 3

13 0.

11 2

DF

P

←

−

ªº

«»

=−

«»

¬¼

b. Since

123

22,=−+xf f f

1

[] 2

2

F

ªº

«»

=−

«»

¬¼

x

and

20 3 1 4

[] [] 1 3 0 2 7

11 2 2 3

DF

P

←

−−

ªºªºªº

«»«»«»

==−−=−

«»«»«»

¬¼¬¼¬¼

xx

7. To find

,

CB

P

←

row reduce the matrix

[]

12 1 2

cc bb

:

[]

12 1 2

10 3 1

.

01 52

−

ªº

∼

«»

−

¬¼

cc bb

Thus

31

,

52

CB

P

←

−

ªº

=

«»

−

¬¼

and

1

21

.

53

BC C B

PP

−

←←

−

ª

º

==

«

»

−

¬

¼

8. To find

CB

P

←

, row reduce the matrix

[]

12 1 2

cc bb

:

[]

12 1 2

10 9 8

.

01 10 9

−

ªº

∼

«»

−

¬¼

cc bb

Thus

98

,

10 9

CB

P

←

−

ªº

=

«»

−

¬¼

and

1

98

.

10 9

BC C B

PP

−

←←

ª

º

==

«

»

¬

¼

4.7 • Solutions 245

9. To find

CB

P

←

, row reduce the matrix

[]

12 1 2

cc bb

:

[]

12 1 2

102 3

.

010 1

ªº

∼

«»

−

¬¼

cc bb

Thus

23

,

01

CB

P

←

ªº

=

«»

−

¬¼

and

1

1/ 2 3/ 2 .

01

BC C B

PP

−

←←

ª

º

==

«

»

−

¬

¼

10. To find

CB

P

←

, row reduce the matrix

[]

12 1 2

cc bb

:

[]

12 1 2

10 3 1

.

01 20

ªº

∼

«»

−

¬¼

cc bb

Thus

31

,

20

CB

P

←

ªº

=

«»

−

¬¼

and

1

01/2

.

13/2

BC C B

PP

−

←←

−

ª

º

==

«

»

¬

¼

11. a. False. See Theorem 15.

b. True. See the first paragraph in the subsection “Change of Basis in

n

.”

12. a. True. The columns of

CB

P

←

are coordinate vectors of the linearly independent set B. See the

second paragraph after Theorem 15.

b. False. The row reduction is discussed after Example 2. The matrix P obtained there satisfies

[] []

CB

P

=xx

13. Let

222

123

{, , }{12 ,35 4,2 3}Btttttt==−+−++bb b

and let

2

123

{, , }{1,, }.Ctt==cc c

The

C-coordinate vectors of

1

,b

2

,b

and

3

b

are

123

130

[] 2,[ ] 5,[] 2

143

CCC

ªº ªº ªº

«» «» «»

=−=−=

«» «» «»

¬¼ ¬¼ ¬¼

bb b

So

130

252

143

CB

P

←

ªº

«»

=−−

«»

¬¼

Let x = –1 + 2t. Then the coordinate vector

[]

B

x

satisfies

1

[] [] 2

0

BC

CB

P

←

−

ª

º

«

»

==

«

»

«

»

¬

¼

xx

This system may be solved by row reducing its augmented matrix:

1301 1005 5

2522 0102,so[] 2

1430 000 1 1

B

−

ªºªºªº

«»«»«»

−− ∼ − =−

«»«»«»

¬¼¬¼¬¼

x

246 CHAPTER 4 • Vector Spaces

14. Let

22

123

{, , }{13,2 5,12}Btttt==−+−+bb b

and let

2

123

{, , }{1,, }.Ctt==cc c

The C-coordinate

vectors of

1

b

,

2

b

, and

3

b

are

123

121

[] 0,[ ] 1,[] 2

350

CCC

ªº ªº ªº

«» «» «»

===

«» «» «»

−−

¬¼ ¬¼ ¬¼

bb b

So

121

012

350

CB

P

←

ªº

«»

=«»

«»

−−

¬¼

Let

2

.

t

=x

Then the coordinate vector

[]

B

x

satisfies

0

[] [] 0

1

BC

CB

P

←

ª

º

«

»

==

«

»

«

»

¬

¼

xx

This system may be solved by row reducing its augmented matrix:

1210 100 3 3

0120 0102,so[] 2

3501 000 1 1

B

ªºªºªº

«»«»«»

∼−=−

«»«»«»

−−

¬¼¬¼¬¼

x

and

22 2

3(1 3 ) 2(2 5 ) (1 2 ).ttttt=−−+−++

15. (a) B is a basis for V

(b) the coordinate mapping is a linear transformation

(c) of the product of a matrix and a vector

(d) the coordinate vector of v relative to B

16. (a)

11

[] []

CB

QQQ

1

ªº

«»

0

«»

===

«»

0

«»

¬¼

bb e

#

(b)

[]

kC

b

(c)

[] []

kC kB k

QQ

==bbe

17. [M]

a. Since we found P in Exercise 34 of Section 4.5, we can calculate that

4.7 • Solutions 247

1

32 0 16 0 12 0 10

032 024 020 0

0016016015

1.

00080100

32 0000406

0000020

0000001

P−

ª

º

«

»

«

»

«

»

«

»

=

«

»

«

»

«

»

«

»

«

»

¬

¼

b. Since P is the change-of-coordinates matrix from C to B,

1

P−

will be the change-of-

coordinates matrix from B to C. By Theorem 15, the columns of

1

P−

will be the C-coordinate

vectors of the basis vectors in B. Thus

2

1

cos (1 cos 2 )

2

tt=+

3

1

cos (3cos cos 3 )

4

ttt=+

4

1

cos (3 4 cos 2 cos 4 )

8

ttt=+ +

5

1

cos (10 cos 5cos 3 cos 5 )

16

tttt=++

6

1

cos (10 15cos 2 6cos 4 cos 6 )

32

tttt=+ + +

18. [M] The C-coordinate vector of the integrand is (0, 0, 0, 5, –6, 5, –12). Using

1

P−

from the previous

exercise, the B- coordinate vector of the integrand will be

1(0, 0, 0, 5, 6, 5, 12) ( 6, 55/ 8, 69 /8, 45/16, 3, 5/16, 3/8)P−

−−=−− − −

Thus the integral may be rewritten as

55 69 45 5 3

6cos cos2 cos33cos4 cos5cos6,

88 16 168

ttttttdt−+−+−+−

³

which equals

55 69 15 3 1 1

6sin sin2sin3sin4sin5sin6.

816 16 4 16 16

tt t tt t tC−+−+−+−+

19. [M]

a. If C is the basis

123

{, , },vv v

then the columns of P are

1

[],

C

u

2

[],

C

u

and

3

[].

C

u

So

[]

1231

[],

jC

=uvvvu

and

[][]

123 123

.P=uu u vv v

In the current exercise,

[]

123

287121 665

252350 590.

326461 21323

−−− − −−−

ªºªºªº

«»«»«»

=−− =−−

«»«»«»

¬¼¬¼¬¼

uu u

248 CHAPTER 4 • Vector Spaces

b. Analogously to part a.,

[][ ]

123 1 2 3

,P=vv v ww w

so

[]

123

=ww w

[]

1

123

.P

−

vv v

In the current exercise,

[]

1

123

287121

252350

326461

−

−−− −

ªºªº

«»«»

=−−

«»«»

¬¼¬¼

ww w

287585 283821

252353 9137.

326221 3 23

−−−

ªºªºª º

«»«»« »

=−−−=−− −

«»«»« »

−−− −

¬¼¬¼¬ ¼

20. a.

DB DCCB

PPP

←←←

=

Let x be any vector in the two-dimensional vector space. Since

CB

P

←

is the change-of-coordinates

matrix from B to C and

DC

P

←

is the change-of-coordinates matrix from C to D,

[] [] and[] [] []

CBDC B

CB DC DCCB

PPPP

←←←←

===xxx x x

But since

DB

P

←

is the change-of-coordinates matrix from B to D,

[] []

DB

P

←

=xx

Thus

[] []

BB

DB DCCB

PPP

←←←

=xx

for any vector

[]

B

x

in

2

, and

DB DCCB

PPP

←←←

=

b. [M] For example, let 73

,,

51

B½−

ªºª º

°°

=®¾

«»« »

−

°°

¬¼¬ ¼

¯¿

12

,,

52

C



½−

ª

ºª º

°

=

®

¾

«

»« »

−

°

¬

¼¬ ¼

¯¿

and 11

,.

85

D



½−

ªºªº

°

=

®

¾

«»«»

−

°

¬¼¬¼

¯¿

Then we

can calculate the change-of-coordinates matrices:

1273 1031 31

5251 0152 52

CB

P

←

−− − −

ªºªºªº

∼



=

«»«»«»

−−− −

¬¼¬¼¬¼

1112 100 8/3 0 8/3

8552 01114/3 114/3

DC

P

←

−− − −

ªºªºªº

∼



=

«»«»«»

−− − −

¬¼¬¼¬¼

1173 1040/316/3 40/316/3

8551 0161/325/3 61/325/3

DB

P

←

−− − −

ªºª ºªº

∼



=

«»« »«»

−− − −

¬¼¬ ¼¬¼

One confirms easily that

40 / 3 16 / 3 0 8/ 3 3 1

61/ 3 25 / 3 1 14 / 3 5 2

DB DCCB

PPP

←←←

−−−

ªºªºªº

== =

«»«»«»

−−−

¬¼¬¼¬¼

4.8 • Solutions 249

4.8 SOLUTIONS

Notes:

This is an important section for engineering students and worth extra class time. To spend only

one lecture on this section, you could cover through Example 5, but assign the somewhat lengthy

Example 3 for reading. Finding a spanning set for the solution space of a difference equation uses the

Basis Theorem (Section 4.5) and Theorem 17 in this section, and demonstrates the power of the theory of

Chapter 4 in helping to solve applied problems. This section anticipates Section 5.7 on differential

equations. The reduction of an

th

n

order difference equation to a linear system of first order difference

equations was introduced in Section 1.10, and is revisited in Sections 4.9 and 5.6. Example 3 is the

background for Exercise 26 in Section 6.5.

1. Let

2.

k

y=

Then

21

2822(2)8(2)

kk k

kkk

yyy

++

+−=+ −

22

2(2 2 8)

k

=+−

2(0) 0forall

k

k==

Since the difference equation holds for all k,

2

k

is a solution.

Let

(4)

k

y=−

. Then

21

28(4)2(4)8(4)

kkk

yyy ++

++

+−=−+−−−

2

(4)((4) 2(4) 8)

k

=−−+−−

(4)(0) 0forall

k

k=−=

Since the difference equation holds for all k,

(4)

k

−

is a solution.

2. Let

5.

k

y=

Then

2

225 5 25(5 )

kk

yy

+

−=−

2

5(5 25)

k

=−

5(0) 0forall

k

k==

Since the difference equation holds for all k,

5

k

is a solution.

Let

(5).

k

y=−

Then

2

225 ( 5) 25( 5)

kk

yy +

+

−=−−−

2

(5)((5) 25)

k

=−−−

(5)(0) 0forall

k

k=−=

Since the difference equation holds for all k,

(5)

k

−

is a solution.

250 CHAPTER 4 • Vector Spaces

3. The signals

2

k

and

(4)

k

−

are linearly independent because neither is a multiple of the other; that is,

there is no scalar c such that

2(4)

kk

c=−

for all k. By Theorem 17, the solution set H of the

difference equation

21

280

kkk

yyy

++

+−=

is two-dimensional. By the Basis Theorem, the two

linearly independent signals

2

k

and

(4)

k

−

form a basis for H.

4. The signals

5

k

and

(5)

k

−

are linearly independent because neither is a multiple of the other; that is,

there is no scalar c such that

5(5)

kk

c=−

for all k. By Theorem 17, the solution set H of the

difference equation

2

25 0

kk

yy

+

−=

is two-dimensional. By the Basis Theorem, the two linearly

independent signals

5

k

and

(5)

k

−

form a basis for H.

5. Let

(2).

k

y=−

Then

21

44(2)4(2)4(2)

kkk

yyy ++

++

++=−+−+−

2

(2)((2) 4(2) 4)

k

=−−+−+

(2)(0) 0forall

k

k=−=

Since the difference equation holds for all k,

(2)

k

−

is in the solution set H.

Let

(2).

k

yk=−

Then

21

44(2)(2)4(1)(2)4(2)

kkk

yyyk k k

++

++=+−++−+−

2

(2)(( 2)(2) 4( 1)(2) 4)

kkkk=−+−++−+

(2)(4 8 8 8 4)

k

kkk=−+−−+

(2)(0) 0forall

k

k=−=

Since the difference equation holds for all k,

(2)

k

k−

is in the solution set H.

The signals

(2)

k

−

and

(2)

k

k−

are linearly independent because neither is a multiple of the other;

that is, there is no scalar c such that

(2) (2)

kk

ck−=−

for all k and there is no scalar c such that

(2) (2)

kk

ck−=−

for all k . By Theorem 17, dim H = 2, so the two linearly independent signals

(2)

k

−

and

(2)

k

k−

form a basis for H by the Basis Theorem.

6. Let

2

4cos .

kk

k

y

π

=

Then

2

(2)

16 4 cos 16 4 cos

22

kk

yy

ππ

+

+§·

+= +

¨¸

©¹

2

(2)

44cos 16cos

22

k

kk

ππ

+

§·

=+

¨¸

©¹

16 4 cos cos

22

k

kk

ππ

π

§·

=⋅++

¨¸¨¸

©¹

16 4 (0) 0 for all

k

k=⋅=

since cos(t +

π

) = –cos t for all t. Since the difference equation holds for all k,

2

4cos

kk

π

is in the

solution set H.

4.8 • Solutions 251

Let

2

4sin .

kk

k

y

π

= Then

2

(2)

16 4 sin 16 4 sin

22

kk

yy

ππ

+

+§·

+= +

¨¸

©¹

2

(2)

44sin 16sin

22

k

kk

ππ

+

§·

=+

¨¸

©¹

16 4 sin sin

22

k

kk

ππ

π

§·

=⋅++

¨¸¨¸

©¹

16 4 (0) 0 for all

k

k=⋅=

since sin(t +

π

) = –sin t for all t. Since the difference equation holds for all k,

2

4sin

kk

π

is in the

solution set H.

The signals

2

4cos

kk

π

and

2

4sin

kk

π

are linearly independent because neither is a multiple of the

other. By Theorem 17, dim H = 2, so the two linearly independent signals

2

4cos

kk

π

and

2

4sin

kk

π

form a basis for H by the Basis Theorem.

7. Compute and row reduce the Casorati matrix for the signals

1,

k

2,

k

and

(2)

k

−

, setting k = 0 for

convenience:

00 0

11 1

22 2

12(2) 100

12(2) 010

001

12(2)

ªº

−ªº

«»

−∼

«»

−¬¼

«»

¬¼

This Casorati matrix is row equivalent to the identity matrix, thus is invertible by the IMT. Hence the

set of signals

{1 , 2 , ( 2 ) }

kk k

−

is linearly independent in . The exercise states that these signals are in

the solution set H of a third-order difference equation. By Theorem 17, dim H = 3, so the three

linearly independent signals

1,

k

2,

k

(2)

k

−

form a basis for H by the Basis Theorem.

8. Compute and row reduce the Casorati matrix for the signals

(1),

k

−

2,

k

and

3,

k

setting k = 0 for

convenience:

000

111

222

(1) 2 3 100

(1) 2 3 0 1 0

001

(1) 2 3

ªº

−ªº

«»

−∼

«»

−¬¼

«»

¬¼

This Casorati matrix is row equivalent to the identity matrix, thus is invertible by the IMT. Hence the

set of signals

{( 1) , 2 ,3 }

kkk

−

is linearly independent in . The exercise states that these signals are in

the solution set H of a third-order difference equation. By Theorem 17, dim H = 3, so the three

linearly independent signals

(1),

k

−

2,

k

and 3

k

form a basis for H by the Basis Theorem.

252 CHAPTER 4 • Vector Spaces

9. Compute and row reduce the Casorati matrix for the signals

2,

k

2

5cos ,

kk

and

2

5sin

kk

setting k = 0

for convenience:

00 0

11 1

22

22 2

25cos05sin0 100

25cos 5sin 010

001

25cos 5sin

ππ

ªº

«»

∼

«»

¬¼

«»

¬¼

This Casorati matrix is row equivalent to the identity matrix, thus is invertible by the IMT. Hence the

set of signals

22

{2 ,5 cos ,5 sin }

kk k

kk

ππ

is linearly independent in . The exercise states that these

signals are in the solution set H of a third-order difference equation. By Theorem 17, dim H = 3, so

the three linearly independent signals

2,

k

2

5cos ,

kk

and

2

5sin

kk

form a basis for H by the Basis

Theorem.

10. Compute and row reduce the Casorati matrix for the signals

(2),

k

−

(2),

k

k−

and

3k

setting k = 0 for

convenience:

000

111

222

(2) 0(2) 3 100

(2) 1(2) 3 0 1 0

001

(2) 2(2) 3

ªº

−− ªº

«»

−− ∼

«»

−− ¬¼

«»

¬¼

This Casorati matrix is row equivalent to the identity matrix, thus is invertible by the IMT. Hence the

set of signals

{( 2) , ( 2) , 3 }

kkk

k−−

is linearly independent in . The exercise states that these signals

are in the solution set H of a third-order difference equation. By Theorem 17, dim H = 3, so the three

linearly independent signals

(2),

k

−

(2),

k

k−

and

3k

form a basis for H by the Basis Theorem.

11. The solution set H of this third-order difference equation has dim H = 3 by Theorem 17. The two

signals

(1)

k

−

and

2

k

cannot possibly span a three-dimensional space, and so cannot be a basis for

H.

12. The solution set H of this fourth-order difference equation has dim H = 4 by Theorem 17. The two

signals

3k

and

(2)

k

−

cannot possibly span a four-dimensional space, and so cannot be a basis for H.

13. The auxiliary equation for this difference equation is

22/9 0.rr−+=

By the quadratic formula

(or factoring), r = 2/3 or r = 1/3, so two solutions of the difference equation are

(2/3)

k

and

(1 / 3)

k

.

The signals

(2/3)

k

and

(1 / 3)

k

are linearly independent because neither is a multiple of the other.

By Theorem 17, the solution space is two-dimensional, so the two linearly independent signals

(2/3)

k

and

(1 / 3)

k

form a basis for the solution space by the Basis Theorem.

14. The auxiliary equation for this difference equation is

2560.rr−+=

By the quadratic formula (or

factoring), r = 2 or r = 3, so two solutions of the difference equation are

2

k

and

3.

k

The signals

2

k

and

3k

are linearly independent because neither is a multiple of the other. By Theorem 17, the

solution space is two-dimensional, so the two linearly independent signals

2

k

and

3k

form a basis for

the solution space by the Basis Theorem.

4.8 • Solutions 253

15. The auxiliary equation for this difference equation is

2

620.rr+−=

By the quadratic formula (or

factoring), r = 1/2 or r = –2/3, so two solutions of the difference equation are

(1 / 2 )

k

and

(2/3).

k

−

The signals

(1 / 2 )

k

and

(2/3)

k

−

are linearly independent because neither is a multiple of the other.

By Theorem 17, the solution space is two-dimensional, so the two linearly independent signals

(1 / 2 )

k

and

(2/3)

k

−

form a basis for the solution space by the Basis Theorem.

16. The auxiliary equation for this difference equation is

225 0.r−=

By the quadratic formula (or

factoring), r = 5 or r = –5, so two solutions of the difference equation are

5k

and

(5).

k

−

The signals

5k

and

(5)

k

−

are linearly independent because neither is a multiple of the other. By Theorem 17, the

solution space is two-dimensional, so the two linearly independent signals

5k

and

(5)

k

−

form a basis

for the solution space by the Basis Theorem.

17. Letting a = .9 and b = 4/9 gives the difference equation

21

1.3 .4 1.

kkk

YYY

++

−+=

First we find a

particular solution

k

YT=

of this equation, where T is a constant. The solution of the equation T –

1.3T + .4T = 1 is T = 10, so 10 is a particular solution to

21

1.3 .4 1

kkk

YYY

++

−+=

. Next we solve the

homogeneous difference equation

21

1.3 .4 0.

kkk

YYY

++

−+=

The auxiliary equation for this difference

equation is

21.3 .4 0.rr−+=

By the quadratic formula (or factoring), r = .8 or r = .5, so two

solutions of the homogeneous difference equation are

.8k

and

.5 .

k

The signals

(.8)

k

and

(.5)

k

are

linearly independent because neither is a multiple of the other. By Theorem 17, the solution space is

two-dimensional, so the two linearly independent signals

(.8)

k

and

(.5)

k

form a basis for the

solution space of the homogeneous difference equation by the Basis Theorem. Translating the

solution space of the homogeneous difference equation by the particular solution 10 of the

nonhomogeneous difference equation gives us the general solution of

21

1.3 .4 1

kkk

YYY

++

−+=

:

12

(.8) (.5) 10.

kk

k

Yc c=+ +

As k increases the first two terms in the solution approach 0, so

k

Y

approaches 10.

18. Letting a = .9 and b = .5 gives the difference equation

21

1.35 .45 1.

kkk

YYY

++

−+=

First we find a

particular solution

k

YT=

of this equation, where T is a constant. The solution of the equation

T – 1.35T + .45T = 1 is T = 10, so 10 is a particular solution to

21

1.35 .45 1

kkk

YYY

++

−+=

. Next we

solve the homogeneous difference equation

21

1.35 .45 0.

kkk

YYY

++

−+=

The auxiliary equation for

this difference equation is

21.35 .45 0.rr−+=

By the quadratic formula (or factoring), r = .6 or

r = .75, so two solutions of the homogeneous difference equation are

.6k

and

.75 .

k

The signals

(.6)

k

and

(.75)

k

are linearly independent because neither is a multiple of the other. By Theorem 17, the

solution space is two-dimensional, so the two linearly independent signals

(.6)

k

and

(.75)

k

form a

basis for the solution space of the homogeneous difference equation by the Basis Theorem.

Translating the solution space of the homogeneous difference equation by the particular solution 10

of the nonhomogeneous difference equation gives us the general solution of

21

1.35 .45 1

kkk

YYY

++

−+=

:

12

(.6) (.75) 10.

kk

k

Yc c=+ +

19. The auxiliary equation for this difference equation is

2410.rr++=

By the quadratic formula,

23r=−+

or

23,r=−−

so two solutions of the difference equation are

(2 3)

k

−+

and

254 CHAPTER 4 • Vector Spaces

(2 3).

k

−−

The signals

(2 3)

k

−+

and

(2 3)

k

−−

are linearly independent because neither is a

multiple of the other. By Theorem 17, the solution space is two-dimensional, so the two linearly

independent signals

(2 3)

k

−+

and

(2 3)

k

−−

form a basis for the solution space by the Basis

Theorem. Thus a general solution to this difference equation is

12

(2 3) (2 3).

kk

k

yc c=−++−−

20. Let

23a=−+

and

23b=−−

. Using the solution from the previous exercise, we find that

11 2 5000ycacb=+ =

and

12

0.

NN

N

ycacb=+ =

This is a system of linear equations with variables

1

c

and

2

c

whose augmented matrix may be row reduced:

5000

10

5000

05000

01

N

NN

NN N

NN

b

ab ba ab

ab a

ba ab

ªº

«»

ªº−

«»

∼

«»

−

¬¼

«»

−

¬¼

so

12

5000 5000

,

NN

NN NN

ba

cc

ba ab ba ab

−

==

−−

(Alternatively, Cramer’s Rule may be applied to get the same solution). Thus

12

kk

k

ycacb=+

5000( )

kN Nk

NN

ab a b

ba ab

−

=

−

21. The smoothed signal

k

z has the following values:

1

(9 5 7) / 3 7,z=++ =

2

(5 7 3) / 3 5,z=++ =

3

(7 3 2) / 3 4,z=++ =

4

(3 2 4)/3 3,z=++ =

5

(2 4 6)/ 3 4,z=++ =

6

(4 6 5)/3 5,z=++ =

7

(6 5 7) / 3 6,z=++ =

8

(5 7 6) / 3 6,z=++ =

9

(7 6 8) / 3 7,z=++ =

10

(6 8 10)/ 3 8,z=++ =

11

(8 10 9) / 3 9,z=+ + =

12

(10 9 5) / 3 8,z=++ =

13

(9 5 7) / 3 7.z=++ =

22. a. The smoothed signal

k

z has the following values:

0210

.35 .5 .35 .35(0) .5(.7) .35(3) 1.4,zyyy=++= ++ =

1321

.35 .5 .35 .35( .7) .5(0) .35(.7) 0,zyyy=++=−++ =

2432

.35 .5 .35 .35( .3) .5( .7) .35(0) 1.4,zyyy=++=−+−+=−

3543

.35 .5 .35 .35( .7) .5( .3) .35( .7) 2,zyyy=++=−+−+−=−

4654

.35 .5 .35 .35(0) .5( .7) .35( .3) 1.4,zyyy=++= +−+−=−

5765

.35 .5 .35 .35(.7) .5(0) .35( .7) 0,zyyy=++= ++−=

2 4 6 8 10 12 14

2

4

6

8

10

original data

smoothed data

4.8 • Solutions 255

6876

.35 .5 .35 .35(3) .5(.7) .35(0) 1.4,zyyy=++= ++ =

7987

.35 .5 .35 .35(.7) .5(3) .35(.7) 2,zyyy=++= ++ =

81098

.35 .5 .35 .35(0) .5(.7) .35(3) 1.4,zyyy=++=++=…

b. This signal is two times the signal output by the filter when the input (in Example 3) was

y = cos(

π

t/4). This is expected because the filter is linear. The output from the input

2cos(

π

t/4) + cos(3

π

t/4) should be two times the output from cos(

π

t/4) plus the output from

cos(3

π

t/4) (which is zero).

23. a.

1

1.01 450,

kk

yy

+

−=−

0

10,000.y=

b. [M] MATLAB code to create the table:

p a y = 450, y = 10000, m = 0, table = [0;y]

while y>450

y = 1.01*y-pay

m = m+1

t a b l e = [table [m;y]]

e n d

m , y

Mathematica code to create the table:

pay = 450; y = 10000; m = 0; balancetable = {{0, y}};

W h i l e [ y > 4 5 0 , { y = 1.01*y - pay; m = m + 1,

A p p e n d T o [ b a l a n c e t a b l e , { m , y } ] } ] ;

m

y

c. [M] At the start of month 26, the balance due is $114.88. At the end of this month the unpaid

balance will be (1.01)($114.88)=$116.03. The final payment will thus be $116.03. The total paid

by the borrower is (25)($450.00)+$116.03=$11,366.03.

24. a.

1

1.005 200,

kk

yy

+

−=

0

1, 000.y=

b. [M] MATLAB code to create the table:

pay = 200, y = 1000, m = 0, table = [0;y]

f o r m = 1: 60

y = 1.005*y+pay

t a b l e = [table [m;y]]

e n d

i n t e r e s t = y-60*pay-1000

Mathematica code to create the table:

pay = 200; y = 1000; amounttable = {{0, y}};

D o [ { y = 1.005*y + pay;

A p p e n d T o [ a m o u n t t a b l e , { m , y } ] } , { m , 1 , 6 0 } ] ;

i n t e r e s t = y-60*pay-1000

256 CHAPTER 4 • Vector Spaces

c. [M] The total is $6213.55 at k = 24, $12,090.06 at k = 48, and $15,302.86 at k = 60. When k = 60,

the interest earned is $2302.86.

25. To show that

2

k

yk=

is a solution of

21

34107,

kkk

yyyk

++

+−=+

substitute

2

,

k

yk=

2

1

(1),

k

yk

+

=+

and

2

(2):

k

yk

+

=+

222

21

34(2)3(1)4

kk k

yykkk

++

+−=+ + + −

222

(44)3(21)4kk kk k=+++ ++−

222

443 634kk kk k=+++ ++−

10 7 for allkk=+

The auxiliary equation for the homogeneous difference equation

21

340

kkk

yyy

++

+−= is

2

340.rr+−= By the quadratic formula (or factoring), r = –4 or r = 1, so two solutions of the

difference equation are

(4)

k

−

and 1.

k

The signals

(4)

k

−

and

1k

are linearly independent because

neither is a multiple of the other. By Theorem 17, the solution space is two-dimensional, so the two

linearly independent signals

(4)

k

−

and

1k

form a basis for the solution space of the homogeneous

difference equation by the Basis Theorem. The general solution to the homogeneous difference

equation is thus

1212

(4) 1 (4) .

kk k

cccc−+⋅=−+

Adding the particular solution

2

k of the

nonhomogeneous difference equation, we find that the general solution of the difference equation

21

34107

kkk

yyyk

++

+−=+

is

2

12

(4) .

k

ykc c=+−+

26. To show that 1

k

yk=+ is a solution of

21

654,

kkk

yyy

++

−+=− substitute 1

k

yk=+ ,

1

1( 1) 2 ,

k

yk k

+

=+ + = + and

2

1( 2) 3

k

yk k

+

=+ + = + :

21

65(3)6(2)5(1)

kkk

yyy k k k

++

−+=+−++ +

312655kkk=+ −− ++

4forallk=−

The auxiliary equation for the homogeneous difference equation

21

650

kkk

yyy

++

−+=

is

2

650.rr−+= By the quadratic formula (or factoring), r = 1 or r = 5, so two solutions of the

difference equation are

1k

and 5.

k

The signals

1k

and 5

k

are linearly independent because neither

is a multiple of the other. By Theorem 17, the solution space is two-dimensional, so the two linearly

independent signals

1k

and 5

k

form a basis for the solution space of the homogeneous difference

equation by the Basis Theorem. The general solution to the homogeneous difference equation is thus

12

15.

kk

cc⋅+⋅

Adding the particular solution 1k+ of the nonhomogeneous difference equation, we

find that the general solution of the difference equation

21

654

kkk

yyy

++

−+=− is

12

115.

kk

k

ykcc=+ + ⋅+⋅

27. To show that 2

k

yk=− is a solution of

2

483

kk

yy k

+

−=−, substitute 2

k

yk=− and

2

(2)2

k

yk k

+

=+−=:

2

44(2)4883forall

kk

yykkkk k k

+

−=−−=−+=−

The auxiliary equation for the homogeneous difference equation

2

40

kk

yy

+

−= is

2

40.r−= By the

quadratic formula (or factoring), r = 2 or r =

−2, so two solutions of the difference equation are

2k

and

(2).

k

−

The signals

2k

and

(2)

k

−

are linearly independent because neither is a multiple of the

4.8 • Solutions 257

other. By Theorem 17, the solution space is two-dimensional, so the two linearly independent signals

2k

and

(2)

k

−

form a basis for the solution space of the homogeneous difference equation by the

Basis Theorem. The general solution to the homogeneous difference equation is thus

12

2(2).

kk

cc⋅+⋅−

Adding the particular solution 2k− of the nonhomogeneous difference equation,

we find that the general solution of the difference equation

2

483

kk

yy k

+

−=− is

12

22 (2).

kk

k

yk c c=−+⋅+⋅−

28. To show that 12

k

yk=+ is a solution of

2

25 30 52 ,

kk

yy k

+

+=+

substitute 12

k

yk=+ and

2

12( 2)52

k

yk k

+

=+ + = + :

2

25 5 2 25(1 2 ) 5 2 25 50 30 52 for all

kk

yyk kk k kk

+

+=+++=+++=+

The auxiliary equation for the homogeneous difference equation

2

25 0

kk

yy

+

+=

is

2

25 0.r+= By

the quadratic formula (or factoring), r = 5i or r = –5i, so two solutions of the difference equation are

22

5cos and 5sin

kk

. The signals

22

5cos and 5sin

kk

are linearly independent because neither is

a multiple of the other. By Theorem 17, the solution space is two-dimensional, so the two linearly

independent signals

22

5cos and 5sin

kk

form a basis for the solution space of the homogeneous

difference equation by the Basis Theorem. The general solution to the homogeneous difference

equation is thus

12

22

5cos 5sin .

kk

cc⋅+⋅

Adding the particular solution 12k+of the

nonhomogeneous difference equation, we find that the general solution of the difference equation

225 30 52

kk

yy k

++=+

is

12

22

12 5cos 5sin .

kk

k

ykc c=+ + ⋅+⋅

29. Let

1

2

3

k

y

+

ªº

«»

=«»

«»

¬¼

x

. Then

1

21

1

32

43

0100

0010 .

0001

2683

kk

yy

A

yy

+

++

+

++

ªº ªº

ªº

«» «»

«»

«» «»

«»

== =

«» «»

«»

«» «»

«»

−−

¬¼

¬¼ ¬¼

xx

30. Let

1

2

k

kk

k

y

+

ªº

«»

=«»

«»

¬¼

x

. Then

1

12 1

32

010

001 .

805

kk

kk k k

kk

yy

yyA

yy

+

++ +

++

ªº ªº

ªº

«» «»«»

== =

«» «»

«»

«» «»«»

−

¬¼¬¼ ¬¼

xx

31. The difference equation is of order 2. Since the equation

321

560

kkk

yyy

++ +

++=

holds for all k,

it holds if k is replaced by k − 1. Performing this replacement transforms the equation into

21

560,

kkk

yyy

++

++=

which is also true for all k. The transformed equation has order 2.

32. The order of the difference equation depends on the values of

1,a

2,a

and

3.a

If

30,a≠

then the

order is 3. If

30a=

and

20,a≠

then the order is 2. If

32

0aa==

and

10,a≠

then the order is 1.

If

321

0,aaa===

then the order is 0, and the equation has only the zero signal for a solution.

33. The Casorati matrix C(k) is

2

11

2| |

() (1) 2(1)| 1|

kk

yz kkk

Ck yz kkk

++

ªº

==

«»

+++

«»¬¼

¬¼

In particular,

258 CHAPTER 4 • Vector Spaces

00 1 2 4 8

(0) , ( 1) , and ( 2)

12 0 0 1 2

CC C

−−

ªº ª º ª º

=−=−=

«» « » « »

−

¬¼ ¬ ¼ ¬ ¼

none of which are invertible. In fact, C(k) is not invertible for all k, since

()

22

det ( ) 2 ( 1) | 1| 2( 1) | | 2 ( 1) | 1| ( 1) | |Ck k k k k k k kk k k k k=++−+=++−+

If k = 0 or k = –1, det C(k) = 0. If k > 0, then k + 1 > 0 and k| k + 1 | – (k + 1)| k | = k(k + 1) – (k + 1)k

= 0, so det C(k) = 0. If k < –1, then k + 1 < 0 and k| k + 1 | – (k + 1)| k | = –k(k + 1) + (k + 1)k = 0, so

det C(k) = 0. Thus detC(k)=0 for all k, and C(k) is not invertible for all k. Since C(k) is not invertible

for all k, it provides no information about whether the signals

{}

k

y

and

{}

k

z

are linearly dependent

or linearly independent. In fact, neither signal is a multiple of the other, so the signals

{}

k

y

and

{}

k

z

are linearly independent.

34. No, the signals could be linearly dependent, since the vector space V of functions considered on the

entire real line is not the vector space of signals. For example, consider the functions f (t) = sinπt,

g(t) = sin 2πt, and h(t) = sin 3πt. The functions f, g, and h are linearly independent in V since they

have different periods and thus no function could be a linear combination of the other two. However,

sampling the functions at any integer n gives f (n) = g(n) = h(n) = 0, so the signals are linearly

dependent in .

35. Let

{}

k

y

and

{}

k

z

be in , and let r be any scalar. The

th

k

term of

{}{}

kk

yz+

is

,

kk

yz+

while the

th

k

term of

{}

k

ry

is

.

k

ry

Thus

({ } { }) { }

kk kk

Ty z Ty z+=+

22 11

()()()

kk kk kk

yz ayz byz

++ ++

=++ +++

21 21

()()

kkkkkk

yaybyzazbz

++ ++

=+++++

{} {},and

kk

Ty Tz=+

({ }) { }

kk

Try Try=

21

()()

kk k

ry a ry b ry

++

=+ +

21

()

kkk

ry ay by

++

=++

{}

k

rT y=

so T has the two properties that define a linear transformation.

36. Let z be in V, and suppose that

p

x

in V satisfies

() .

p

T=xz

Let u be in the kernel of T; then T(u) =

0. Since T is a linear transformation,

()()() ,

pp

TTT+= + =+=ux u x 0zz

so the vector

p

=+xux

satisfies the nonhomogeneous equation T(x) = z.

37. We compute that

012 012 012 012

()(,,,) ((,,,)) (0,,,,)(,,,)TD y y y T D y y y T y y y y y y…= … = …= …

while

012 012 123 123

()(,,,) ((,,,)) (,,,)(0,,,,)DT y y y D T y y y D y y y y y y…= … = …= …

Thus TD = I (the identity transformation on

0

), while DT ≠ I.

4.9 • Solutions 259

4.9 SOLUTIONS

Notes:

This section builds on the population movement example in Section 1.10. The migration matrix is

examined again in Section 5.2, where an eigenvector decomposition shows explicitly why the sequence of

state vectors

k

x

tends to a steady state vector. The discussion in Section 5.2 does not depend on prior

knowledge of this section.

1. a. Let N stand for “News” and M stand for “Music.” Then the listeners’ behavior is given by the

table

From:

N M To:

.7 .6 N

.3 .4 M

so the stochastic matrix is

.7 .6 .

.3 .4

Pªº

=«»

¬¼

b. Since 100% of the listeners are listening to news at 8: 15, the initial state vector is

0

1

0

ªº

=«»

¬¼

x

.

c. There are two breaks between 8: 15 and 9: 25, so we calculate

2

x

:

10

.7 .6 1 .7

.3 .4 0 .3

P

ª

ºª º ª º

== =

«

»« » « »

¬

¼¬ ¼ ¬ ¼

xx

21

.7 .6 .7 .67

.3 .4 .3 .33

P

ª

ºª º ª º

== =

«

»« » « »

¬

¼¬ ¼ ¬ ¼

xx

Thus 33% of the listeners are listening to music at 9:25.

2. a. Let the foods be labelled “1,” “2,” and “3.” Then the animals’ behavior is given by the table

From:

1 2 3 To:

.6 .2 .2 1

.2 .6 .2 2

.2 .2 .6 3

so the stochastic matrix is

.6 .2 .2

.2 .6 .2

.2 .2 .6

P

ªº

«»

=«»

«»

¬¼

.

b. There are two trials after the initial trial, so we calculate

2

x

. The initial state vector is

1

0.

0

ªº

«»

¬¼

10

.6 .2 .2 1 .6

.2 .6 .2 0 .2

.2 .2 .6 0 .2

P

ª

ºª º ª º

«

»« » « »

== =

«

»« » « »

«

»« » « »

¬

¼¬ ¼ ¬ ¼

xx

260 CHAPTER 4 • Vector Spaces

21

.6 .2 .2 .6 .44

.2 .6 .2 .2 .28

.2 .2 .6 .2 .28

P

ª

ºª º ª º

«

»« » « »

== =

«

»« » « »

«

»« » « »

¬

¼¬ ¼ ¬ ¼

xx

Thus the probability that the animal will choose food #2 is .28.

3. a. Let H stand for “Healthy” and I stand for “Ill.” Then the students’ conditions are given by the

table

From:

H I To:

.95 .45 H

.05 .55 I

so the stochastic matrix is

.95 .45

.05 .55

Pªº

=«»

¬¼

.

b. Since 20% of the students are ill on Monday, the initial state vector is

0

.8

.2

ªº

=«»

¬¼

x

. For Tuesday’s

percentages, we calculate

1

x

; for Wednesday’s percentages, we calculate

2

x

:

10

.95 .45 .8 .85

.05 .55 .2 .15

P

ª

ºª º ª º

== =

«

»« » « »

¬

¼¬ ¼ ¬ ¼

xx

21

.95 .45 .85 .875

.05 .55 .15 .125

P

ª

ºª º ª º

== =

«

»« » « »

¬

¼¬ ¼ ¬ ¼

xx

Thus 15% of the students are ill on Tuesday, and 12.5% are ill on Wednesday.

c. Since the student is well today, the initial state vector is

0

1.

0

ª

º

=

«

»

¬

¼

x

We calculate

2

x

:

10

.95 .45 1 .95

.05 .55 0 .05

P

ª

ºª º ª º

== =

«

»« » « »

¬

¼¬ ¼ ¬ ¼

xx

21

.95 .45 .95 .925

.05 .55 .05 .075

P

ª

ºª º ª º

== =

«

»« » « »

¬

¼¬ ¼ ¬ ¼

xx

Thus the probability that the student is well two days from now is .925.

4. a. Let G stand for good weather, I for indifferent weather, and B for bad weather. Then the change

in the weather is given by the table

From:

G I B To:

.4 .5 .3 G

.3 .2 .4 I

.3 .3 .3 B

so the stochastic matrix is

.4 .5 .3

.3 .2 .4 .

.3 .3 .3

P

ªº

«»

=«»

«»

¬¼

4.9 • Solutions 261

b. The initial state vector is

.5

.5 .

0

ªº

«»

¬¼

We calculate

1

x

:

10

.4 .5 .3 .5 .45

.3 .2 .4 .5 .25

.3 .3 .3 0 .30

P

ª

ºª º ª º

«

»« » « »

== =

«

»« » « »

«

»« » « »

¬

¼¬ ¼ ¬ ¼

xx

Thus the chance of bad weather tomorrow is 30%.

c. The initial state vector is

0

.6 .

.4

ªº

«»

=«»

«»

¬¼

x

We calculate

2

x

:

10

.4 .5 .3 0 .42

.3 .2 .4 .6 .28

.3 .3 .3 .4 .30

P

ª

ºª º ª º

«

»« » « »

== =

«

»« » « »

«

»« » « »

¬

¼¬ ¼ ¬ ¼

xx

21

.4 .5 .3 .42 .398

.3 .2 .4 .28 .302

.3 .3 .3 .30 .300

P

ª

ºª º ª º

«

»« » « »

== =

«

»« » « »

«

»« » « »

¬

¼¬ ¼ ¬ ¼

xx

Thus the chance of good weather on Wednesday is 39.8%, or approximately 40%.

5. We solve Px = x by rewriting the equation as (P – I)x = 0, where

.9 .5 .

.9 .5

PI

−

ª

º

−=

«

»

−

¬

¼

Row reducing

the augmented matrix for the homogeneous system (P – I)x = 0 gives

.9 .5 0 1 5 / 9 0

.9 .5 0 0 0 0

−−

ªºªº

∼

«»«»

−

¬¼¬¼

Thus

1

2

5/9 ,

1

x

ªº ªº

==

«» «»

¬¼

x

and one solution is

5.

9

ª

º

«

»

¬

¼

Since the entries in

5

9

ª

º

«

»

¬

¼

sum to 14, multiply by

1/14 to obtain the steady-state vector

5/14 .

9/14

ª

º

=

«

»

¬

¼

q

6. We solve Px = x by rewriting the equation as (P – I)x = 0, where

.6 .8 .

.6 .8

PI

−

ª

º

−=

«

»

−

¬

¼

Row reducing

the augmented matrix for the homogeneous system (P – I)x = 0 gives

.6 .8 0 1 4 / 3 0

.6 .8 0 0 0 0

−−

ªºªº

∼

«»«»

−

¬¼¬¼

Thus

1

2

4/3 ,

1

x

ªº ªº

==

«» «»

¬¼

x

and one solution is

4.

3

ª

º

«

»

¬

¼

Since the entries in

4

3

ª

º

«

»

¬

¼

sum to 7, multiply by

1/7 to obtain the steady-state vector

4/7 .571 .

3/7 .429

ª

ºª º

=≈

«

»« »

¬

¼¬ ¼

q

262 CHAPTER 4 • Vector Spaces

7. We solve Px = x by rewriting the equation as (P – I)x = 0, where

.3 .1 .1

.2 .2 .2 .

.1 .1 .3

PI

−

ª

º

«

»

−=−

«

»

«

»

−

¬

¼

Row

reducing the augmented matrix for the homogeneous system (P – I)x = 0 gives

.3 .1 .1 0 1 0 1 0

.2 .2 .2 0 0 1 2 0

.1 .1 .3 0 0 0 0 0

−−

ªºªº

«»«»

−∼−

«»«»

−

¬¼¬¼

Thus

1

23

3

1

2,

1

x

xx

x

ªº ªº

«» «»

==

«» «»

¬¼ ¬¼

x

and one solution is

1

2.

1

ª

º

«

»

«

»

«

»

¬

¼

Since the entries in

1

2

1

ª

º

«

»

«

»

«

»

¬

¼

sum to 4, multiply by 1/4

to obtain the steady-state vector

1/ 4 .25

1/ 2 .5 .

1/ 4 .25

ªºªº

«»«»

==

«»«»

¬¼¬¼

q

8. We solve Px = x by rewriting the equation as (P – I)x = 0, where

.6 .5 .8

0.5.1.

.6 0 .9

PI

−

ª

º

«

»

−=−

«

»

«

»

−

¬

¼

Row

reducing the augmented matrix for the homogeneous system (P – I)x = 0 gives

.6 .5 .8 0 1 0 3 / 2 0

0.5.10 011/50

.6 0 .9 0 0 0 0 0

−−

ªºªº

«»«»

−∼−

«»«»

−

¬¼¬¼

Thus

1

23

3

3/2

1/5 ,

1

x

xx

x

ªº ªº

«» « »

==

«» «»

«» « »

¬¼¬¼

x

and one solution is

15

2.

10

ª

º

«

»

«

»

«

»

¬

¼

Since the entries in

15

2

10

ª

º

«

»

«

»

«

»

¬

¼

sum to 27, multiply

by 1/27 to obtain the steady-state vector

15 / 27 .556

2/27 .074 .

10 / 27 .370

ª

ºª º

«

»« »

=≈

«

»« »

«

»« »

¬

¼¬ ¼

q

9. Since

2.84 .2

.16 .8

Pªº

=«»

¬¼

has all positive entries, P is a regular stochastic matrix.

10. Since

11.7

0.7

k

P

ªº

−

=«»

«»

¬¼

will have a zero as its (2,1) entry for all k, P is not a regular stochastic

matrix.

11. a. From Exercise 1,

.7 .6 ,

.3 .4

Pªº

=«»

¬¼

so

.3 .6 .

.3 .6

PI

−

ª

º

−=

«

»

−

¬

¼

Solving (P – I)x = 0 by row reducing the

augmented matrix gives

4.9 • Solutions 263

.3 .6 0 1 2 0

.3 .6 0 0 0 0

−−

ªºªº

∼

«»«»

−

¬¼¬¼

Thus

1

2

2,

1

x

ªº ªº

==

«» «»

¬¼

x

and one solution is

2.

1

ª

º

«

»

¬

¼

Since the entries in

2

1

ª

º

«

»

¬

¼

sum to 3, multiply by 1/3

to obtain the steady-state vector

2/3 .667 .

1/3 .333

ªºª º

=≈

«»« »

¬¼¬ ¼

q

b. Since

2/3 ,

1/3

ªº

=«»

¬¼

q

2/3 of the listeners will be listening to the news at some time late in the day.

12. From Exercise 2,

.6 .2 .2

.2 .6 .2 ,

.2 .2 .6

P

ª

º

«

»

=

«

»

«

»

¬

¼

so

.4 .2 .2

.2 .4 .2 .

.2 .2 .4

PI

−

ª

º

«

»

−=−

«

»

«

»

−

¬

¼

Solving (P – I)x = 0 by row

reducing the augmented matrix gives

.4 .2 .2 0 1 0 1 0

.2 .4 .2 0 0 1 1 0

.2 .2 .4 0 0 0 0 0

−−

ªºªº

«»«»

−∼−

«»«»

−

¬¼¬¼

Thus

1

23

3

1

1,

1

x

xx

x

ªº ªº

«» «»

==

«» «»

¬¼ ¬¼

x

and one solution is

1

1.

1

ª

º

«

»

«

»

«

»

¬

¼

Since the entries in

1

ª

º

«

»

«

»

«

»

¬

¼

sum to 3, multiply by 1/3 to

obtain the steady-state vector

1/3 .333

1/3 .333 .

1/3 .333

ªºª º

«»« »

=≈

«»« »

¬¼¬ ¼

q

Thus in the long run each food will be preferred

equally.

13. a. From Exercise 3,

.95 .45 ,

.05 .55

Pªº

=«»

¬¼

so

.05 .45 .

.05 .45

PI

−

ª

º

−=

«

»

−

¬

¼

Solving (P – I)x = 0 by row

reducing the augmented matrix gives

.05 .45 0 1 9 0

.05 .45 0 0 0 0

−−

ªºªº

∼

«»«»

−

¬¼¬¼

Thus

1

2

9,

1

x

ªº

ª

º

==

«»

«

»

¬

¼

¬¼

x

and one solution is

9.

1

ª

º

«

»

¬

¼

Since the entries in

9

1

ª

º

«

»

¬

¼

sum to 10, multiply by

1/10 to obtain the steady-state vector

9/10 .9 .

1/10 .1

ª

ºªº

==

«

»«»

¬

¼¬¼

q

b. After many days, a specific student is ill with probability .1, and it does not matter whether that

student is ill today or not.

264 CHAPTER 4 • Vector Spaces

14. From Exercise 4,

.4 .5 .3

.3 .2 .4 ,

.3 .3 .3

P

ª

º

«

»

=

«

»

«

»

¬

¼

so

.6 .5 .3

.3 .8 .4 .

.3 .3 .7

PI

−

ª

º

«

»

−=−

«

»

«

»

−

¬

¼

Solving (P – I)x = 0 by row

reducing the augmented matrix gives

.6 .5 .3 0 1 0 4 / 3 0

.3 .8 .4 0 0 1 1 0

.3 .3 .7 0 0 0 0 0

−−

ªºªº

«»«»

−∼−

«»«»

−

¬¼¬¼

Thus

1

23

3

4/3

1,

1

x

xx

x

ªº ªº

«» « »

==

«» «»

«» « »

¬¼¬¼

x

and one solution is

4

3.

3

ª

º

«

»

«

»

«

»

¬

¼

Since the entries in

4

3

ª

º

«

»

«

»

«

»

¬

¼

sum to 10, multiply by

1/10 to obtain the steady-state vector

4/10 .4

3/10 .3 .

3/10 .3

ª

ºªº

«

»«»

==

«

»«»

«

»«»

¬

¼¬¼

q

Thus in the long run the chance that a day

has good weather is 40%.

15. [M] Let

.9821 .0029 ,

.0179 .9971

Pªº

=«»

¬¼

so

.0179 .0029 .

.0179 .0029

PI

−

ª

º

−=

«

»

−

¬

¼

Solving (P – I)x = 0 by row reducing

the augmented matrix gives

.0179 .0029 0 1 .162011 0

.0179 .0029 0 0 0 0

−−

ªºªº

∼

«»«»

−

¬¼¬¼

Thus

1

2

.162011 ,

1

x

ªº ªº

==

«» «»

¬¼

x

and one solution is

.162011 .

1

ª

º

«

»

¬

¼

Since the entries in

.162011

1

ªº

«»

¬¼

sum to

1.162011, multiply by 1/1.162011 to obtain the steady-state vector

.139423 .

.860577

ª

º

=

«

»

¬

¼

q

Thus about

13.9% of the total U.S. population would eventually live in California.

16. [M] Let

.90 .01 .09

.01 .90 .01 ,

.09 .09 .90

P

ªº

«»

=«»

«»

¬¼

so

.10 .01 .09

.01 .10 .01 .

.09 .09 .1

PI

−

ª

º

«

»

−=−

«

»

«

»

−

¬

¼

Solving (P – I)x = 0 by row

reducing the augmented matrix gives

.10 .01 .09 0 1 0 .919192 0

.01 .10 .01 0 0 1 .191919 0

.09 .09 .1 0 0 0 0 0

−−

ªºªº

«»«»

−∼−

«»«»

−

¬¼¬¼

4.9 • Solutions 265

Thus

1

23

3

.919192

.191919 ,

1

x

xx

x

ªº ª º

«» « »

==

«» « »

¬¼ ¬ ¼

x

and one solution is

.919192

.191919 .

1

ª

º

«

»

«

»

«

»

¬

¼

Since the entries in

.919192

.191919

1

ªº

«»

¬¼

sum

to 2.111111, multiply by 1/2.111111 to obtain the steady-state vector

.435407

.090909 .

.473684

ª

º

«

»

=

«

»

«

»

¬

¼

q

Thus on a

typical day, about (.090909)(2000) = 182 cars will be rented or available from the downtown

location.

17. a. The entries in each column of P sum to 1. Each column in the matrix P – I has the same entries as

in P except one of the entries is decreased by 1. Thus the entries in each column of P – I sum to 0,

and adding all of the other rows of P – I to its bottom row produces a row of zeros.

b . By part a., the bottom row of P – I is the negative of the sum of the other rows, so the rows of

P – I are linearly dependent.

c . By part b. and the Spanning Set Theorem, the bottom row of P – I can be removed and the

remaining (n – 1) rows will still span the row space of P – I. Thus the dimension of the row space

of P – I is less than n. Alternatively, let A be the matrix obtained from P – I by adding to the

bottom row all the other rows. These row operations did not change the row space, so the row

space of P – I is spanned by the nonzero rows of A. By part a., the bottom row of A is a zero row,

so the row space of P – I is spanned by the first (n – 1) rows of A.

d . By part c., the rank of P – I is less than n, so the Rank Theorem may be used to show that

dimNul(P – I) = n – rank(P – I) > 0. Alternatively the Invertible Martix Theorem may be used

since P – I is a square matrix.

18. If

α

=

β

= 0 then

10

.

01

P

ª

º

=

«

»

¬

¼

Notice that Px = x for any vector x in

2

, and that

1

0

ªº

«»

¬¼

and

0

1

ªº

«»

¬¼

are

two linearly independent steady-state vectors in this case.

If

α

≠ 0 or

β

≠ 0, we solve (P – I)x = 0 where

.PI

α

β

α

β

−

ª

º

−=

«

»

−

¬

¼

Row reducing the augmented

matrix gives

00

0000

αβ α β

αβ

−−

ªºªº

∼

«»«»

−

¬¼¬¼

So

12

,xx

α

β

=

and one possible solution is to let

1,x

β

=

2

x

α

=

. Thus

1

2

.

x

β

α

ªºªº

==

«»«»

¬¼

x

Since the

entries in

β

α

ªº

«»

¬¼

sum to

α

+

β

, multiply by 1/(

α

+

β

) to obtain the steady-state vector

1.

β

α

αβ

ª

º

=

«

»

+

¬

¼

q

19. a. The product Sx equals the sum of the entries in x. Thus x is a probability vector if and only if its

entries are nonnegative and Sx = 1.

b . Let

[]

12

,

n

P=…pp p

where

1

p

,

2

p

, …,

n

p

are probability vectors. By part a.,

[][]

12

11 1

n

SP S S S S=…=…=pp p

266 CHAPTER 4 • Vector Spaces

c . By part b., S(Px) = (SP)x = Sx = 1. The entries in Px are nonnegative since P and x have only

nonnegative entries. By part a., the condition S(Px) = 1 shows that Px is a probability vector.

20. Let

[]

12

,

n

P=…pp p

so

[]

2

12 .

n

PPPP P P== …

pp p

By Exercise 19c., the columns

of

2

P

are probability vectors, so

2

P

is a stochastic matrix.

Alternatively, SP = S by Exercise 19b., since P is a stochastic matrix. Right multiplication by P gives

2

,SP SP=

so SP = S implies that

2.SP S=

Since the entries in P are nonnegative, so are the entries

in

2

,P

and

2

P

is stochastic matrix.

21. [M]

a. To four decimal places,

23

.2779 .2780 .2803 .2941 .2817 .2817 .2817 .2814

.3368 .3355 .3357 .3335 .3356 .3356 .3355 .3352

,,

.1847 .1861 .1833 .1697 .1817 .1817 .1819 .1825

.2005 .2004 .2007 .2027 .2010 .2010 .2010 .2009

PP

ªºªº

«»«»

==

«»«»

¬¼¬¼

45

.2816 .2816 .2816 .2816

.3355 .3355 .3355 .3355

.1819 .1819 .1819 .1819

.2009 .2009 .2009 .2009

PP

ªº

«»

==

«»

¬¼

The columns of

k

P

are converging to a common vector as k increases. The steady state vector q

for P is

.2816

.3355 ,

.1819

.2009

ªº

«»

=«»

«»

¬¼

q

which is the vector to which the columns of

k

P

are converging.

b. To four decimal places,

10 20

.8222 .4044 .5385 .7674 .6000 .6690

.0324 .3966 .1666 , .0637 .2036 .1326 ,

.1453 .1990 .2949 .1688 .1964 .1984

QQ

ªºªº

«»«»

==

«»«»

¬¼¬¼

30 40

.7477 .6815 .7105 .7401 .7140 .7257

.0783 .1329 .1074 , .0843 .1057 .0960 ,

.1740 .1856 .1821 .1756 .1802 .1783

QQ

ªºªº

«»«»

==

«»«»

¬¼¬¼

50 60

.7372 .7269 .7315 .7360 .7320 .7338

.0867 .0951 .0913 , .0876 .0909 .0894 ,

.1761 .1780 .1772 .1763 .1771 .1767

QQ

ªºªº

«»«»

==

«»«»

¬¼¬¼

70 80

.7356 .7340 .7347 .7354 .7348 .7351

.0880 .0893 .0887 , .0881 .0887 .0884 ,

.1764 .1767 .1766 .1764 .1766 .1765

QQ

ªºªº

«»«»

==

«»«»

¬¼¬¼

Chapter 4 • Supplementary Exercises 267

116 117

.7353 .7353 .7353

.0882 .0882 .0882

.1765 .1765 .1765

QQ

ª

º

«

»

==

«

»

«

»

¬

¼

The steady state vector q for Q is

.7353

.0882

.1765

ª

º

«

»

=

«

»

«

»

¬

¼

q

Conjecture: the columns of

,

k

P

where P is a

regular stochastic matrix, converge to the steady state vector for P as k increases.

c. Let P be an n × n regular stochastic matrix, q the steady state vector of P, and

j

e

the

th

j

column

of the n × n identity matrix. Consider the Markov chain

{}

k

x

where

1

kk

P

+

=

xx

and

0

.

j

e=x

By

Theorem 18,

0

k

P=xx

converges to q as k → ∞. But

0

kk

j

PP=xe

, which is the

th

j

column of

.

k

P

Thus the

th

j

column of

k

P

converges to q as k → ∞; that is,

[]

k

P→…

qq q

.

22. [M] Answers will vary.

MATLAB Student Version 4.0 code for Method (1):

A=randstoc(32); flops(0);

tic, x=nulbasis(A-eye(32));

q=x/sum(x); toc, flops

MATLAB Student Version 4.0 code for Method (2):

A=randstoc(32); flops(0);

tic, B=A^100; q=B(: ,1); toc, flops

Chapter 4 SUPPLEMENTARY EXERCISES

1. a. True. This set is

1

Span{ , ... }

p

vv

, and every subspace is itself a vector space.

b. True. Any linear combination of

1

v

, …,

1p−

v

is also a linear combination of

1

v

, …,

1p−

v

,

p

v

using the zero weight on

.

p

v

c. False. Counterexample: Take

1

2

p

=vv

. Then

1

{,... }

p

vv

is linearly dependent.

d. False. Counterexample: Let

123

{, , }

ee e

be the standard basis for

3

. Then

12

{, }

ee

is a linearly

independent set but is not a basis for

3

.

e. True. See the Spanning Set Theorem (Section 4.3).

f. True. By the Basis Theorem, S is a basis for V because S spans V and has exactly p elements. So

S must be linearly independent.

g. False. The plane must pass through the origin to be a subspace.

h. False. Counterexample:

25 20

00 73

00 00

−

ªº

«»

¬¼

.

268 CHAPTER 4 • Vector Spaces

i. True. This statement appears before Theorem 13 in Section 4.6.

j. False. Row operations on A do not change the solutions of Ax = 0.

k. False. Counterexample:

12

36

Aªº

=«»

¬¼

; A has two nonzero rows but the rank of A is 1.

l. False. If U has k nonzero rows, then rank A = k and dimNul A = n – k by the Rank Theorem.

m. True. Row equivalent matrices have the same number of pivot columns.

n. False. The nonzero rows of A span Row A but they may not be linearly independent.

o. True. The nonzero rows of the reduced echelon form E form a basis for the row space of each

matrix that is row equivalent to E.

p. True. If H is the zero subspace, let A be the 3 × 3 zero matrix. If dim H = 1, let {v} be a basis

for H and set

[]

A=vvv

. If dim H = 2, let {u,v} be a basis for H and set

[]

A=uvv

,

for example. If dim H = 3, then H =

3

, so A can be any 3 × 3 invertible matrix. Or, let {u, v,

w} be a basis for H and set

[]

A=uvw

.

q. False. Counterexample:

100

010

Aªº

=«»

¬¼

. If rank A = n (the number of columns in A), then the

transformation x6Ax is one-to-one.

r. True. If x6Ax is onto, then Col A =

m

and rank A = m. See Theorem 12(a) in Section 1.9.

s. True. See the second paragraph after Theorem 15 in Section 4.7.

t. False. The

th

j

column of

CB

P

←

is

.

jC

ª

º

¬

¼

b

2. The set is SpanS, where

125

258

,, .

147

311

S

½−

ªºªºªº

°°

«»«»«»

−

°°

«»«»«»

=®¾

«»«»«»

−−

°°

«»«»«»

°°

«»«»«»

¬¼¬¼¬¼

¯¿

Note that S is a linearly dependent set, but each

pair of vectors in S forms a linearly independent set. Thus any two of the three vectors

1

2,

1

3

ªº

«»

−

«»

¬¼

2

5,

4

1

−

ª

º

«

»

«

»

«

»

−

«

»

«

»

¬

¼

5

8

7

1

ªº

«»

−

«»

¬¼

will be a basis for SpanS.

3. The vector b will be in

12

Span{ , }W=

uu

if and only if there exist constants

1

c

and

2

c

with

11 2 2

.cc+=

uub

Row reducing the augmented matrix gives

11

212

3123

21 21

42 04 2

65 00 2

bb

bbb

bbbb

−−

ªºª º

«»« »

∼+

«»« »

−− ++

¬¼¬ ¼

Chapter 4 • Supplementary Exercises 269

so

12

Span{ , }W=

uu

is the set of all

123

(, , )bb b

satisfying

123

20.bbb++=

4. The vector g is not a scalar multiple of the vector f, and f is not a scalar multiple of g, so the set

{f, g} is linearly independent. Even though the number g(t) is a scalar multiple of f(t) for each t, the

scalar depends on t.

5. The vector

1

p

is not zero, and

2

p

is not a multiple of

1

.

p

However,

3

p

is

12

22+

pp

, so

3

p

is

discarded. The vector

4

p

cannot be a linear combination of

1

p

and

2

p

since

4

p

involves

2

t

but

1

p

and

2

p

do not involve

2

.t

The vector

5

p

is

124

(3/ 2) (1/ 2)−+

ppp

(which may not be so easy to see

at first.) Thus

5

p

is a linear combination of

1

,

p

2

,

p

and

4

,

p

so

5

p

is discarded. So the resulting

basis is

124

{, , }.

pp p

6. Find two polynomials from the set

14

{,..., }

pp

that are not multiples of one another. This is easy,

because one compares only two polynomials at a time. Since these two polynomials form a linearly

independent set in a two-dimensional space, they form a basis for H by the Basis Theorem.

7. You would have to know that the solution set of the homogeneous system is spanned by two

solutions. In this case, the null space of the 18 × 20 coefficient matrix A is at most two-dimensional.

By the Rank Theorem, dimCol A = 20 – dimNul A ≥ 20 – 2 = 18. Since Col A is a subspace of

18

,

Col A =

18

. Thus Ax = b has a solution for every b in

18

.

8. If n = 0, then H and V are both the zero subspace, and H = V. If n > 0, then a basis for H consists of n

linearly independent vectors

1

,..., .

n

uu

These vectors are also linearly independent as elements of V.

But since dimV = n, any set of n linearly independent vectors in V must be a basis for V by the Basis

Theorem. So

1

,...,

n

uu

span V, and

1

Span{ , . .. , } .

n

HV==

uu

9. Let T:

n



→

m

be a linear transformation, and let A be the m × n standard matrix of T.

a. If T is one-to-one, then the columns of A are linearly independent by Theorem 12 in Section 1.9,

so dimNul A = 0. By the Rank Theorem, dimCol A = n – 0 = n, which is the number of columns

of A. As noted in Section 4.2, the range of T is Col A, so the dimension of the range of T is n.

b. If T maps

n

onto

m

, then the columns of A span

m

by Theoerm 12 in Section 1.9, so dimCol A

= m. By the Rank Theorem, dimNul A = n – m. As noted in Section 4.2, the kernel of T is Nul A,

so the dimension of the kernel of T is n – m. Note that n – m must be nonnegative in this case:

since A must have a pivot in each row, n ≥ m.

10. Let

1

{ , . .. , }.

p

S=vv

If S were linearly independent and not a basis for V, then S would not span V.

In this case, there would be a vector

1p+

v

in V that is not in

1

Span{ , .. . , }.

p

vv

Let

11

{ , .. . , , }.

pp

S

+

′=vvv

Then S′ is linearly independent since none of the vectors in S′ is a linear

combination of vectors that precede it. Since S′ has more elements than S, this would contradict the

maximality of S. Hence S must be a basis for V.

11. If S is a finite spanning set for V, then a subset of S is a basis for V. Denote this subset of S by .S′

Since S′ is a basis for V, S′ must span V. Since S is a minimal spanning set, S′ cannot be a proper

subset of S. Thus S′= S, and S is a basis for V.

270 CHAPTER 4 • Vector Spaces

12. a. Let y be in Col AB. Then y = ABx for some x. But ABx = A(Bx), so y = A(Bx), and y is in Col A.

Thus Col AB is a subspace of Col A, so rank AB = dimCol AB ≤ dimCol A = rank A by Theorem

11 in Section 4.5.

b. By the Rank Theorem and part a.:

rank rank( ) rank rank rank

TTTT

AB AB B A B B==≤=

13. By Exercise 12, rank PA ≤ rank A, and

11

rank rank( ) rank ( ) rankAPPAPPAPA

−−

== ≤

, so

rank PA = rank A.

14. Note that

() .

TTT

AQ Q A=

Since

T

Q

is invertible, we can use Exercise 13 to conclude that

rank( ) rank rank .

TTT T

AQ Q A A==

Since the ranks of a matrix and its transpose are equal (by the

Rank Theorem), rank AQ = rank A.

15. The equation AB = O shows that each column of B is in Nul A. Since Nul A is a subspace of

n

, all

linear combinations of the columns of B are in Nul A. That is, Col B is a subspace of Nul A. By

Theorem 11 in Section 4.5, rank B = dimCol B ≤ dimNul A. By this inequality and the Rank

Theorem applied to A,

n = rank A + dimNul A ≥ rank A + rank B

16. Suppose that

1

rank Ar=

and

2

rank Br=

. Then there are rank factorizations

11

ACR=

and

22

BCR=

of A and B, where

1

C

is

1

mr×

with rank

1

r

,

2

C

is

2

mr×

with rank

2

r

,

1

R

is

1

rn×

with rank

1

r

, and

2

R

is

2

rn×

with rank

2

.r

Create an

12

()mrr×+

matrix

[]

12

CCC=

and an

12

()rr n+×

matrix R

by stacking

1

R

over

2

.

R

Then

[]

1

11 2 2 1 2

2

R

ABCR CR C C CR

R

ªº

+= + = =

«»

¬¼

Since the matrix CR is a product, its rank cannot exceed the rank of either of its factors by Exercise

12. Since C has

12

rr+

columns, the rank of C cannot exceed

12

.rr+

Likewise R has

12

rr+

rows, so

the rank of R cannot exceed

12

.rr+

Thus the rank of A + B cannot exceed

12

rank rank ,rr A B+= +

or rank (A + B) ≤ rank A + rank B.

17. Let A be an m × n matrix with rank r.

(a) Let

1

A

consist of the r pivot columns of A. The columns of

1

A

are linearly independent, so

1

A

is an m × r matrix with rank r.

(b) By the Rank Theorem applied to

1

,A

the dimension of

1

RowA

is r, so

1

A

has r linearly

independent rows. Let

2

A

consist of the r linearly independent rows of

1

.A

Then

2

A

is an r × r

matrix with linearly independent rows. By the Invertible Matrix Theorem,

2

A

is invertible.

18. Let A be a 4 × 4 matrix and B be a 4 × 2 matrix, and let

03

,...,

uu

be a sequence of input vectors in

2

.

a. Use the equation

1kkk

AB

+

=+

xxu

for

0, . . . , 4,k=

with

0

.=

x0

1000

AB B=+=

xxuu

211 01

AB AB B=+= +

xxu uu

Chapter 4 • Supplementary Exercises 271

2

322 01 2 0 12

()AB AABB B AB ABB=+= + += + +xxu uu u u uu

2

433 0 12 3

()AB AAB ABB B=+= + + +xxu u uu u

32

0123

AB AB AB B=+++uuuu

3

2

23

1

0

BABABAB M

ªº

«»

ªº

==

¬¼

«»

¬¼

u

uu

u

Note that M has 4 rows because B does, and that M has 8 columns because B and each of the

matrices

k

AB

have 2 columns. The vector u in the final equation is in

8

, because each

k

u

is in

2

.

b. If (A, B) is controllable, then the controllability matrix has rank 4, with a pivot in each row, and

the columns of M span

4

. Therefore, for any vector v in

4

, there is a vector u in

8

such that

v = Mu. However, from part a. we know that

4

M

=

xu

when u is partitioned into a control

sequence

03

,,…

uu

. This particular control sequence makes

4

.=

xv

19. To determine if the matrix pair (A, B) is controllable, we compute the rank of the matrix

2

.BABAB

ªº

¬¼

To find the rank, we row reduce:

2

010100

1.9.81 010.

1.5.25 001

BABAB

ªºªº

«»«»

ªº

=−∼

¬¼

«»«»

¬¼¬¼

The rank of the matrix is 3, and the pair (A, B) is controllable.

20. To determine if the matrix pair (A, B) is controllable, we compute the rank of the matrix

2

.BABAB

ªº

¬¼

To find the rank, we note that :

2

1.5.19

1.7.45.

00 0

BABAB

ªº

«»

ªº

=

¬¼

«»

¬¼

The rank of the matrix must be less than 3, and the pair (A, B) is not controllable.

21. [M] To determine if the matrix pair (A, B) is controllable, we compute the rank of the matrix

23

.BABABAB

ªº

¬¼

To find the rank, we row reduce:

23

10 0 1 100 1

00 1 1.60101.6

.

011.6 .960011.6

11.6 .96 .024 0 0 0 0

BABABAB

−−

ªºªº

«»«»

−−

«»«»

ªº

=∼

¬¼

«»«»

−− −

«»«»

−−−

«»«»

¬¼¬¼

The rank of the matrix is 3, and the pair (A, B) is not controllable.

22. [M] To determine if the matrix pair (A, B) is controllable, we compute the rank of the matrix

23

.BABABAB

ªº

¬¼

To find the rank, we row reduce:

272 CHAPTER 4 • Vector Spaces

23

10 0 1 1000

00 1 .5 0100

.

01 .5 11.45 0010

1.511.45 10.275 0001

BABABAB

−

ªºªº

«»«»

−

«»«»

ªº

=∼

¬¼

«»«»

−

«»«»

−−

«»«»

¬¼¬¼

The rank of the matrix is 4, and the pair (A, B) is controllable.

r

5.1 SOLUTIONS

Notes

: Exercises 1–6 reinforce the d

e

eigenvectors and difference equations,

a

example and anticipates discussions of

d

1. The number 2 is an eigenvalue of

A

This equation is equivalent to

(−A

32 20

238 02

AI

ªºªº

ª

−=−=

«»«»«

¬¼¬¼

¬

The columns of A are obviously lin

e

so 2 is an eigenvalue of A.

2. The number −3 is an eigenvalue o

f

This equation is equivalent to

(A+

14 30

369 03

AI

−

ªºªº

+= + =

«»«»

¬¼¬¼

The columns of A are obviously lin

e

so

3−

is an eigenvalue of A.

3. Is

Ax

a multiple of x? Compute

1

6

ª

«

¬

eigenvalue

2−

.

4. Is

Ax

a multiple of x? Compute

5

3

ª

«

¬

eigenvalue

3

.

r

son Education, Inc. Publishing as Addison-Wesley.

e

finitions of eigenvalues and eigenvectors. The s

u

a

long with Exercises 33 and 34, refers to the chapter

d

ynamical systems in Sections 5.2 and 5.6.

A

if and only if the equation

2A=xx

has a nontrivial

s

2) .=

x

I0

Compute

12

36

ª

º

»

¬

¼

e

arly dependent, so

(2)AI−=

x

0

has a nontrivial sol

u

f

A if and only if the equation

3A=−xx

has a nontriv

i

3) .I=

x

0

Compute

24

612

ªº

«»

¬¼

e

arly dependent, so

(3)AI+=

x

0

has a nontrivial sol

u

1

11 2 1

2.

6

43 6 3

−−

ºª º ª º ª º

==−

»« » « » « »

−−

¼¬ ¼ ¬ ¼ ¬ ¼ So 1

3

ªº

«»

¬¼

is an eigenvector

5

21 3 1

3

61 3 1

ªº

«»

¬¼

−− −

ºª º ª º

==

»« » « »

¼¬ ¼ ¬ ¼ So 1

1

−

ªº

«»

¬¼

is an eigenvecto

r

273

u

bsection on

introductory

s

olution.

u

tion, and

i

al solution.

u

tion, and

of A with

r

of A with

274 CHAPTER 5 • Eigenvalues and Eigenvectors

5. Is

A

x

a multiple of x? Compute

4333 15 3

2322 10 52.

1021 5 1

−−

ªºªºªºªº

«»«»«»«»

−−−==−−

«»«»«»«»

−− −

¬¼¬¼¬¼¬¼

So

3

2

1

ªº

«»

−

«»

¬¼

is an eigenvector

of A for the eigenvalue

5−

.

6. Is

A

x

a multiple of x? Compute

367 1 5 1

327 2 13 2

564 2 1 2

λ

ªºªºªºªº

«»«»«»«»

−=≠−

«»«»«»«»

¬¼¬¼¬¼¬¼

So

1

2

ª

º

«

»

−

«

»

«

»

¬

¼

is not

an eigenvector of A.

7. To determine if 4 is an eigenvalue of A, decide if the matrix

4AI−

is invertible.

30 1 400 1 0 1

4231040211

34 5 004 3 4 1

AI

−−−

ªºªºª º

«»«»« »

−=−=−

«»«»« »

−−

¬¼¬¼¬ ¼

Invertibility can be checked in several ways, but since an eigenvector is needed in the event that one

exists, the best strategy is to row reduce the augmented matrix for

(4)AI−=

x0

:

10 10 10 10 10 10

2110 0110 0110

3410 0440 0000

−− −−

ªºªºªº

«»«»«»

−−−−−

«»«»«»

−

¬¼¬¼¬¼



The equation

(4)AI−=

x0

has a nontrivial solution, so 4 is an eigenvalue. Any nonzero solution of

(4)AI−=

x0

is a corresponding eigenvector. The entries in a solution satisfy

13

0xx+=

and

23

0,−− =xx with

3

x free. The general solution is not requested, so to save time, simply take any

nonzero value for

3

x to produce an eigenvector. If

3

1,=x then

(1 11).=−,−,

x

Note:

The answer in the text is

(1 1 1),,,−

written in this form to make the students wonder whether the

more common answer given above is also correct. This may initiate a class discussion of what answers

are “correct.”

8. To determine if 1 is an eigenvalue of A, decide if the matrix

AI−

is invertible.

423 100 323

013010 023

122 001 123

AI

−−

ªºªºªº

«»«»«»

−=−− =−

«»«»«»

−− −−

¬¼¬¼¬¼

Row reducing the augmented matrix

[(A ) ]I−

0

yields:

3230 1230 1000 10 00

0230 0230 0230 013/20

1230 0460 0000 00 00

−−

ªºªºªºª º

«»«»«»« »

−− −−

«»«»«»« »

−− −

¬¼¬¼¬¼¬ ¼



The equation

()AI−=

x0

has a nontrivial solution, so 1 is an eigenvalue. Any nonzero solution of

()AI−=

x0

is a corresponding eigenvector. The entries in a solution satisfy

1

0x= and

5.1 • Solutions 275

23

(3 / 2) 0,xx−= with

3

x free. The general solution is not requested, so to save time, simply take

any nonzero value for

3

x to produce an eigenvector. If

3

2,x= then

0

3.

2

ª

º

«

»

=

«

»

«

»

¬

¼

x

9. For 30 10 20

11

21 01 20

AI

λ

ª

ºª ºª º

=: −=−=

«

»« »« »

¬

¼¬ ¼¬ ¼

The augmented matrix for

()AI−=

x0

is 200

.

200

ª

º

«

»

¬

¼ Thus

1

0x= and

2

x is free. The general

solution of

()AI−=

x0

is

22

,ex where

2

0,

1

ª

º

=

«

»

¬

¼

e and so

2

e is a basis for the eigenspace

corresponding to the eigenvalue 1.

For 30 30 0 0

33

21 03 2 2

AI

λ

ª

ºª ºª º

=: −=−=

«

»« »« »

−

¬

¼¬ ¼¬ ¼

The equation

(3)AI−=

x0

leads to

12

22 0,xx−= so that

12

xx= and

2

x is free. The general

solution is

12

2

22

1.

1

xx

x

xx

ªºªº

«»«»

¬¼¬¼

ªº

==

«»

¬¼

So 1

1

ªº

«»

¬¼

is a basis for the eigenspace.

Note:

For simplicity, the answer omits the set brackets when listing a basis. I permit my students to

list a basis without the set brackets. Some instructors may prefer to include brackets.

10. For 42 10 12

55 5 .

31 01 36

AI

λ

−

ª

ºª ºª º

=−:+= + =

«

»« »« »

¬

¼¬ ¼¬ ¼

The augmented matrix for

(5)AI+=

x0

is 120 120

.

360 000

ª

ºª º

«

»« »

¬

¼¬ ¼

 Thus

12

2xx=− and

2

x is

free. The general solution is

12

2

22

22.

1

xx

x

xx

ªºª º

«»« »

¬¼¬ ¼

−−

ª

º

==

«

»

¬

¼

A basis for the eigenspace corresponding to

5−

is 2.

1

−

ªº

«»

¬¼

11. For

1

λ

=−

: 13 10 23

45 01 46

AI

−−

ª

ºª ºª º

+= + =

«

»« »« »

−−

¬

¼¬ ¼¬ ¼

The augmented matrix for

()AI+=

x0

is 230 13/20

.

460 0 0 0

−−

ª

ºª º

«

»« »

−

¬

¼¬ ¼

 Thus

12

(3 / 2)xx= and

2

x is free. The general solution is

12

2

22

(3 / 2) 3/2 .

1

xx

x

xx

ªº

«»

¬¼

ªº

ª

º

==

«»

«

»

¬

¼

¬¼

A basis for the eigenspace

corresponding to

1−

is 3/2 .

1

ªº

«»

¬¼

Another choice is 3.

2

ª

º

«

»

¬

¼

276 CHAPTER 5 • Eigenvalues and Eigenvectors

For

7

λ

=

: 13 70 63

745 07 42

AI

−−−

ª

ºª ºª º

−=−=

«

»« »« »

−−−

¬

¼¬ ¼¬ ¼

The augmented matrix for

(7)AI−=

x0

is 630 11/20

.

420 000

−−

ª

ºª º

«

»« »

−−

¬

¼¬ ¼

 Thus

12

(1/2)xx=− and

2

x

is free. The general solution is

12

2

22

(1/2) 1/ 2 .

1

xx

x

xx

ªº

«»

¬¼

−−

ªº

ª

º

==

«»

«

»

¬

¼

¬¼

A basis for the eigenspace

corresponding to 7 is 1/ 2 .

1

−

ª

º

«

»

¬

¼ Another choice is 1.

2

−

ª

º

«

»

¬

¼

12. For 41 30 11

33

36 03 33

AI

λ

ª

ºª ºª º

=: −=−=

«

»« »« »

¬

¼¬ ¼¬ ¼

The augmented matrix for

(3)AI−=

x0

is 110 110

.

330 000

ª

ºª º

«

»« »

¬

¼¬ ¼

 Thus

12

xx=− and

2

x is free.

A basis for the eigenspace corresponding to 1 is 1.

1

−

ª

º

«

»

¬

¼

For 41 70 3 1

77 .

36 07 3 1

AI

λ

−

ª

ºª ºª º

=: −=−=

«

»« »« »

−

¬

¼¬ ¼¬ ¼

The augmented matrix for

(7)AI−=

x0

is 310 11/30

.

310 0 00

−−

ª

ºª º

«

»« »

−

¬

¼¬ ¼

 Thus

12

(1 / 3)xx= and

2

x is free. The general solution is

12

2

22

(1 / 3) 1/3 .

1

xx

x

xx

ªºª º

«»« »

¬¼¬ ¼

ª

º

==

«

»

¬

¼

A basis for the eigenspace

corresponding to 7 is 1/3 .

1

ª

º

«

»

¬

¼ Another choice is 1.

3

ª

º

«

»

¬

¼

13. For λ = 1:

401 100 301

1210010200

201 001 200

AI

ªºªºªº

«»«»«»

−=−− =−

«»«»«»

−−

¬¼¬¼¬¼

The equations for

()AI−=

x0

are easy to solve:

13

1

30

2 0

xx

x

+=



½

®

¾

−=

¯¿

Row operations hardly seem necessary. Obviously

1

x is zero, and hence

3

x is also zero. There are

three-variables, so

2

x is free. The general solution of

()AI−=

x0

is

22

,ex where

2

0

1,

0

ªº

«»

=«»

«»

¬¼

e

and

so

2

e provides a basis for the eigenspace.

For λ = 2:

5.1 • Solutions 277

401 200 2 0 1

2210020210

201 002 2 0 1

AI

ªºªºª º

«»«»« »

−=−− =−−

«»«»« »

−−−

¬¼¬¼¬ ¼

2010 2010 0120

[( 2 ) ] 2 1 0 0 0 1 1 0 0 1 0

2010 0000 0000

AI

1/

ª

ºª ºª º

«

»« »« »

−=−− − 1−

«

»« »« »

«

»« »« »

−−

¬

¼¬ ¼¬ ¼

0

So

1323

(1 2) ,xxxx=−/,=

with

3

x free. The general solution of

(2)AI−=

x0

is

3

12

1.

1

−/

ªº

«»

¬¼

x

A nice

basis vector for the eigenspace is

1

2.

2

−

ªº

«»

¬¼

For λ = 3:

401 300 1 0 1

3210030220

201 003 2 0 2

ªºªºª º

«»«»« »

−=−− =−−

«»«»« »

−−−

¬¼¬¼¬ ¼

AI

10 10 1010 010

[( 3 ) ] 2 2 0 0 0 2 2 0 0 1 0

2020 0000 0000

1

ª

ºª ºª º

«

»« »« »

−=−− − 1−

«

»« »« »

«

»« »« »

−−

¬

¼¬ ¼¬ ¼

0

AI

So

1323

,=−,=xxxx

with

3

x free. A basis vector for the eigenspace is

1

1.

1

−

ª

º

«

»

«

»

«

»

¬

¼

14. For

401300 101

3(3)3303030333.

225003 222

AIAI

λ

−−

ªºªºªº

«»«»«»

=: −=−=−=−

«»«»«»

−−

¬¼¬¼¬¼

The augmented matrix for

(3)AI−=

x0

is

1010 1010 10 10

[( 3 ) ] 3 3 3 0 0 3 6 0 0 1 2 0

2220 0000 0000

AI

−−−

ª

ºª ºª º

«

»« »« »

−=−− −

«

»« »« »

«

»« »« »

−

¬

¼¬ ¼¬ ¼

0

Thus

132 3

2,xxx x=, = with

3

x free. The general solution of

(3)AI−=

x0

is

3

1

2.

1

x

ªº

«»

¬¼

A basis for the eigenspace corresponding to 2 is

1

2

1

ª

º

«

»

«

»

«

»

¬

¼

.

278 CHAPTER 5 • Eigenvalues and Eigenvectors

15. For

1110 1110

5[( 5) ] 2220 0000.

3330 0000

AI

λ

ªºªº

«»«»

=−:+ =

«»«»

¬¼¬¼

0

Thus

123

0,xx x++=

with

2

x and

3

x free. The general solution of

(5)AI+=

x0

is

23

2

3

11

10

01

xx

xx x

x

ªº

«»

¬¼

−− − −

ª

ºªº

«

»«»

==+.

«

»«»

«

»«»

¬

¼¬¼

xA basis for the

eigenspace corresponding to

5−

is

11

10

01

½

−−

ª

ºª º

°°

«

»« »

,

®¾

«

»« »

°°

«

»« »

¬

¼¬ ¼

¯¿

Note:

For simplicity, the text answer omits the set brackets. I permit my students to list a basis without

the set brackets. Some instructors may prefer to include brackets.

16. For

[]

10 10 10 10 10 10

1100 0110 0110

4(4) .

2110 0110 0000

4220 0220 0000

AI

λ

−−−

ªºªºªº

«»«»«»

−− −

«»«»«»

=: −==

«»«»«»

−− −

«»«»«»

−− −

¬¼¬¼¬¼

0

So

1323

,xxx x=, = with

3

x and

4

x

free variables. The general solution of

(4)AI−=

x0

is

13

23

34

33

44

10 10

Basis for the eigenspace

10 10

01 01

xx

ªºªº

«»«»

¬¼¬¼



½

ªº ªº ªºªº

°

«» «» «»«»

°

«» «» «»«»

=== + . : ,

®

¾

«» «» «»«»

°

«» «» «»«»

°

¬¼ ¬¼ ¬¼¬¼

¯¿

x

Note:

I urge my students always to include the extra column of zeros when solving a homogeneous

system. Exercise 16 provides a situation in which failing to add the column is likely to create problems

for a student, because the matrix

4AI−

itself has a column of zeros.

17. The eigenvalues of

00 0

03 4

00 2

ªº

«»

−

¬¼

are 0, 3, and

2−

, the numbers on the main diagonal, by Theorem

1.

18. The eigenvalues of

500

000

103

ªº

«»

−

¬¼

are 5, 0, and 3, the numbers on the main diagonal, by Theorem 1.

19. The matrix

123

ªº

«»

¬¼

is not invertible because its columns are linearly dependent. So the number 0

is an eigenvalue of the matrix. See the discussion following Example 5.

5.1 • Solutions 279

20. The matrix

222

A

ªº

«»

=«»

«»

¬¼

is not invertible because its columns are linearly dependent. So the

number 0 is an eigenvalue of A. Eigenvectors for the eigenvalue 0 are solutions of

A=x0

and

therefore have entries that produce a linear dependence relation among the columns of A. Any

nonzero vector (in

3

R

) whose entries sum to 0 will work. Find any two such vectors that are not

multiples; for instance,

1

2

ªº

«»

−

¬¼

and

1

1.

0

ªº

«»

−

«»

¬¼

21. a. False. The equation

A=λxx

must have a nontrivial solution.

b. True. See the paragraph after Example 5.

c. True. See the discussion of equation (3).

d. True. See Example 2 and the paragraph preceding it. Also, see the Numerical Note.

e. False. See the warning after Example 3.

22. a. False. The vector x in

A=λxx

must be nonzero.

b. False. See Example 4 for a two-dimensional eigenspace, which contains two linearly independent

eigenvectors corresponding to the same eigenvalue. The statement given is not at all the same as

Theorem 2. In fact, it is the converse of Theorem 2 (for the case

2r=

).

c. True. See the paragraph after Example 1.

d. False. Theorem 1 concerns a triangular matrix. See Examples 3 and 4 for counterexamples.

e. True. See the paragraph following Example 3. The eigenspace of A corresponding to

λ

is the null

space of the matrix

.−λAI

23. If a

22×

matrix A were to have three distinct eigenvalues, then by Theorem 2 there would

correspond three linearly independent eigenvectors (one for each eigenvalue). This is impossible

because the vectors all belong to a two-dimensional vector space, in which any set of three vectors is

linearly dependent. See Theorem 8 in Section 1.7. In general, if an nn× matrix has p distinct

eigenvalues, then by Theorem 2 there would be a linearly independent set of p eigenvectors (one for

each eigenvalue). Since these vectors belong to an n-dimensional vector space, p cannot exceed n.

24. A simple example of a

22×

matrix with only one distinct eigenvalue is a triangular matrix with the

same number on the diagonal. By experimentation, one finds that if such a matrix is actually a

diagonal matrix then the eigenspace is two dimensional, and otherwise the eigenspace is only one

dimensional.

Examples:

41

04

ªº

«»

¬¼

and 45

.

04

ªº

«»

¬¼

25. If is an eigenvalue of A, then there is a nonzero vector

x

such that

.=xxA

Since A is invertible,

11

(),

−−

=xxAA A

and so

1

().

−

=xxA

Since

≠x0

(and since A is invertible), cannot be zero.

Then

11

,A

−−

=xx

which shows that

1−

is an eigenvalue of

1

.

−

A

280 CHAPTER 5 • Eigenvalues and Eigenvectors

Note:

The Study Guide points out here that the relation between the eigenvalues of A and

1

A

−

is

important in the so-called inverse power method for estimating an eigenvalue of a matrix. See Section 5.8.

26. Suppose that

2

A

is the zero matrix. If

A=xx

for some

,≠x0

then

22

() () .AAAA A====xx xxx

Since x is nonzero, must be zero. Thus each eigenvalue of A is

zero.

27. Use the Hint in the text to write, for any

()() .

TT TT

AI A I A I,−=−=−

Since

( )

T

AI−

is

invertible if and only if

AI−

is invertible (by Theorem 6(c) in Section 2.2), it follows that

T

AI−

is not invertible if and only if

AI−

is not invertible. That is, is an eigenvalue of

T

A

if and only

if is an eigenvalue of A.

Note:

If you discuss Exercise 27, you might ask students on a test to show that A and

T

A

have the same

characteristic polynomial (discussed in Section 5.2). Since det

det ,=

T

AA

for any square matrix A,

det( )det( )det( ()) det( )

TTT

AI AI A I AI−=−=−=−.

28. If A is lower triangular, then

T

A

is upper triangular and has the same diagonal entries as A. Hence,

by the part of Theorem 1 already proved in the text, these diagonal entries are eigenvalues of .

T

A By

Exercise 27, they are also eigenvalues of A.

29. Let v be the vector in

n

R

whose entries are all ones. Then

.As=vv

30. Suppose the column sums of an nn× matrix A all equal the same number s. By Exercise 29 applied

to

T

A

in place of A, the number s is an eigenvalue of .

T

A By Exercise 27, s is an eigenvalue of A.

31. Suppose T reflects points across (or through) a line that passes through the origin. That line consists

of all multiples of some nonzero vector v. The points on this line do not move under the action of A.

So

() .=vvT

If A is the standard matrix of T, then

.=vvA

Thus v is an eigenvector of A

corresponding to the eigenvalue 1. The eigenspace is Span

{}.v

Another eigenspace is generated by

any nonzero vector u that is perpendicular to the given line. (Perpendicularity in

2

R

should be a

familiar concept even though orthogonality in

n

R

has not been discussed yet.) Each vector x on the

line through u is transformed into the vector

.−x

The eigenvalue is

1.−

32. Since T rotates points around a given line, the points on the line are not moved at all. Hence 1 is an

eigenvalue of the standard matrix A of T, and the corresponding eigenspace is the line the points are

being rotated around.

33. (The solution is given in the text.)

a. Replace k by

1k+

in the definition of

,

x

k

and obtain

11

11 2

.

kk

k

cc

λµ

++

+

=+xuv

b.

12

1

()

by linearity

since and are eigenvectors

+

=+

=

xuv

uv

uvuv

x

kk

k

kk

k

AAc c

cA c A

cc

λµ

λλ µµ

5.1 • Solutions 281

34. You could try to write

0

x as linear combination of eigenvectors,

1

.,,vv

p

…

If

1

λ,,λ

p

…

are

corresponding eigenvalues, and if

011

,

pp

cc=++"xv v

then you could define

11 1

kk

kppp

cc

λλ

=++xv v"

In this case, for

012 ,=,,,k…

11 1

11

11 1

1

()

Linearity

The are eigenvectors

kk

kppp

kk

pp p

kk

pp p i

k

AAc c

cA c A

cc

λλ

++

+

=++

=++ .

=

xv v

vv

vvv

x

"

35. Using the figure in the exercise, plot

()Tu

as

2,u

because u is an eigenvector for the eigenvalue 2 of

the standard matrix A. Likewise, plot

()

T

v

as

3,v

because v is an eigenvector for the eigenvalue 3.

Since T is linear, the image of w is

() ( ) () ().

=+= +

wuvuv

TT TT

36. As in Exercise 35,

()

T=−

uu

and

() 3

T=

vv

because u and v are eigenvectors for the eigenvalues

1−

and 3, respectively, of the standard matrix A. Since T is linear, the image of w is

() ( ) () ().

TT TT

=+= +wuvuv

Note

: The matrix programs supported by this text all have an eigenvalue command. In some cases, such

as MATLAB, the command can be structured so it provides eigenvectors as well as a list of the

eigenvalues. At this point in the course, students should not use the extra power that produces

eigenvectors. Students need to be reminded frequently that eigenvectors of A are null vectors of a

translate of A. That is why the instructions for Exercises 35–38 tell students to use the method of Example

4.

It is my experience that nearly all students need manual practice finding eigenvectors by the method

of Example 4, at least in this section if not also in Sections 5.2 and 5.3. However, [M] exercises do create

a burden if eigenvectors must be found manually. For this reason, the data files for the text include a

special command, nulbasis for each matrix program (MATLAB, Maple, etc.). The output of

nulbasis (A) is a matrix whose columns provide a basis for the null space of A, and these columns

are identical to the ones a student would find by row reducing the augmented matrix

[ ].0

A

With

nulbasis, student answers will be the same (up to multiples) as those in the text. I encourage my students

to use technology to speed up all numerical homework here, not just the

[]M

exercises,

37. [M] Let A be the given matrix. Use the MATLAB commands eig and nulbasis (or equivalent

commands). The command

ev = eig(A)

computes the three eigenvalues of A and stores them in a

vector ev. In this exercise,

(10 15 5).=,,

ev

The eigenspace for the eigenvalue 10 is the null space

of

10 .AI−

Use nulbasis to produce a basis for each null space:

3

2.

1

−

ª

º

«

»

«

»

«

»

¬

¼

nulbasis(A -ev(1)*eye(3))=

Basis for the eigenspace for

3

10 2

1

λ



½

−

ª

º

°

«

»

=:

®

¾

«

»

°

«

»

¬

¼

¯¿

282 CHAPTER 5 • Eigenvalues and Eigenvectors

For the next eigenvalue, 15, compute nulbasis

2

2.

1

ª

º

«

»

«

»

«

»

¬

¼

(A -ev(2)* eye(3))=

Basis for the eigenspace for

2

15 2

1

λ

½

ª

º

°°

«

»

=:

®¾

«

»

°°

«

»

¬

¼

¯¿

.

For the next eigenvalue, 5, compute nulbasis

1/2

1/2 .

1

−

ª

º

«

»

−

«

»

«

»

¬

¼

(A -ev(3)*eye(3))=

Scaling this vector by 2 to eliminate the fractions provides a basis for the eigenspace for

1

51

2

λ

½

−

ªº

°°

«»

=: −

®¾

«»

°°

«»

¬¼

¯¿

.

38. [M]

(2 2 1 1).=,−,−,

ev = eig(A)

For

2

λ

=

:

22

Basis:

11



½

ª

ºªº

°

«

»«»

°

«

»«»

.

®

¾

«

»«»

°

«

»«»

°

¬

¼¬¼

¯¿

nulbasis (A -ev(1)* eye(4))=

For

2

λ

=−

: nulbasis

0

1.

1

0

ª

º

«

»

«

»

«

»

«

»

¬

¼

(A -ev(2)* eye(4))=

Basis:

0

1

0



½

ª

º

°

«

»

°

«

»

®

¾

«

»

°

«

»

°

¬

¼

¯¿

For

1

λ

=−

: nulbasis

1

1.

0

1

ª

º

«

»

«

»

«

»

«

»

¬

¼

(A -ev(3)* eye(4))=

Basis:

1

0

1



½

ª

º

°

«

»

°

«

»

®

¾

«

»

°

«

»

°

¬

¼

¯¿

For

1

λ

=

: nulbasis

2

2.

0

1

ª

º

«

»

«

»

«

»

«

»

¬

¼

(A -ev(4)* eye(4))=

Basis:

2

0

1



½

ª

º

°

«

»

°

«

»

®

¾

«

»

°

«

»

°

¬

¼

¯¿

39. [M] For

4

λ

=−

, basis:

0

1

.

2

0

1

½

ªº

°°

«»

°°

«»

°°

«»

®¾

«»

°°

«»

°°

«»

°°

¬¼

¯¿

For

12,λ=

basis:

02

01

12

10

01



½

ª

ºªº

°

«

»«»

°

«

»«»

°

«

»«»

,

−

®

¾

«

»«»

°

«

»«»

°

«

»«»

°

¬

¼¬¼

¯¿

. For

8λ=−

, basis:

60

30

31

20

01

½

ªºª º

°°

«»« »

°°

«»« »

°°

«»« »

,−

®¾

«»« »

°°

«»« »

°°

«»« »

°°

¬¼¬ ¼

¯¿

.

5.2 • Solutions 283

40. [M] For

14

λ

=−

, basis:

110

06

,.

10

05

01

½

−

ªºªº

°°

«»«»

°°

«»«»

°°

«»«»

®¾

«»«»

°°

«»«»

°°

«»«»

°°

¬¼¬¼

¯¿

For

42,λ=

basis:

05

11

23

50

05



½

−

ª

ºª º

°

«

»« »

−

°

«

»« »

°

«

»« »

,

−−

®

¾

«

»« »

°

«

»« »

°

«

»« »

°

¬

¼¬ ¼

¯¿

.

Note:

Since so many eigenvalues in text problems are small integers, it is easy for students to form a

habit of entering a value for in nulbasis

λ(A - I)

based on a visual examination of the eigenvalues

produced by eig(A)when only a few decimal places for are displayed. Using

ev = eig(A)

and:

nulbasis

(A -ev(j)*eye(n))

help avoid problems caused by roundoff errors.

5.2 SOLUTIONS

Notes

: Exercises 9–14 can be omitted, unless you want your students to have some facility with

determinants of

33×

matrices. In later sections, the text will provide eigenvalues when they are needed

for matrices larger than

22.×

If you discussed partitioned matrices in Section 2.4, you might wish to

bring in Supplementary Exercises 12–14 in Chapter 5. (Also, see Exercise 14 of Section 2.4.)

Exercises 25 and 27 support the subsection on dynamical systems. The calculations in these exercises

and Example 5 prepare for the discussion in Section 5.6 about eigenvector decompositions.

1. 27 27 0 2 7 .

72 72 0 7 2

−

ªº ªºª ºª º

=,−=−=

«» «»« »« »

−

¬¼ ¬¼¬ ¼¬ ¼

AAI

λλ

λ

λλ

The characteristic polynomial is

22 2 2

det( ) (2 ) 7 4 4 49 4 45AI−λ =−λ − =−λ+λ− =λ−λ−

In factored form, the characteristic equation is

(9)(5)0,λ− λ+=

so the eigenvalues of A are 9 and

5.−

2. 41 4 1

.

61 6 1

AAI

λ

−− −− −

ªº ª º

=,−=

«» « »

−

¬¼ ¬ ¼

The characteristic polynomial is

2

det( ) ( 4 )(1 ) 6 ( 1) 3 2AI

λλλ λλ

−=−− − − ⋅− =++

Since

2

32(2)(1),

λλ λ λ

++=+ +

the eigenvalues of A are

2−

and

1−

.

3. 42 4 2 .

67 6 7

AAI

λ

−−−

ªº ª º

=,−=

«» « »

−

¬¼ ¬ ¼

The characteristic polynomial is

2

det( ) ( 4 )(7 ) (2)(6) 40AI−λ =−−λ −λ− =λ−3λ−

Use the quadratic formula to solve the characteristic equation and find the eigenvalues:

2

4 3 9 160 3 13 8, 5

222

b b ac

a

λ

−±−±+ ±

====−

284 CHAPTER 5 • Eigenvalues and Eigenvectors

4.82 8 2

.

33 3 3

AAI

λ

−

ªº ª º

=,−=

«» « »

−

¬¼ ¬ ¼

The characteristic polynomial of A is

2

det( ) (8 )(3 ) (3)(2) 11 18 ( 9)( 2).AI

λλλ λλλλ

−=−−− =−+=−−

The eigenvalues of

A

are

9 and 2.

5. 84 8 4 .

48 4 8

AAI

−λ

ªº ª º

=,−λ =

«» « »

−λ

¬¼ ¬ ¼

The characteristic polynomial of A is

2

det( ) (8 )(8 ) (4)(4) 16 48 ( 4)( 12)AI

λ

−λ =−λ −λ − =λ− λ+=λ− −

Thus, A has eigenvalues 4 and 12.

6. 92 9 2

.

25 2 5

AAI

λ

−−−

ªº ª º

=,−=

«» « »

−

¬¼ ¬ ¼

The characteristic polynomial is

2

det( ) (9 )(5 ) ( 2)(2) 14 49 ( 7)( 7)AI

λλλ λλλλ

−=−−−− =−+=−−

Thus A has only one eigenvalue, 7, with multiplicity 2.

7. 53 5 3 .

44 4 4

−

ªº ª º

=,−=

«» « »

−−−

¬¼ ¬ ¼

AAI

λ

The characteristic polynomial is

2

det( ) (5 )(4 ) (3)( 4) 9 32AI

λλλ λλ

−=−−−−=−+

Use the quadratic formula to solve det

()0AI

λ

−=

:

9814(32)

947

22

λ

±−±−

==

These values are complex numbers, not real numbers, so A has no real eigenvalues. There is no

nonzero vector x in

2

R

such that

,=xxA

λ

because a real vector

Ax

cannot equal a complex

multiple of x.

8. 43 4 3.

21 2 1

AAI

λ

−−−

ªº ª º

=,−=

«» « »

−

¬¼ ¬ ¼

The characteristic polynomial is

2

det( ) ( 4 )(1 ) (2)(3) 3 10 ( 5)( 2)AI

λλλ λλλλ

−=−− − − =+−=+ −

Thus, A has eigenvalues

5−

and 2.

9.

401

det( ) det 0 4 1 .

102

AI

λ

λλ

λ

−−

ªº

«»

−=−−

«»

−

¬¼

Using cofactor expansion down column 1, the

characteristic polynomial is

2

32

(4 ) 1 0 1

det( ) (4 ) det 0 1 det

0(2) (4)1

(4 )( 6 8) (4 )

(4 )( 6 9) (4 )(3 )(3 )

10 33 36

AI

λ

λλ

λλ λ λ

λλ λ λ λ λ

λλλ

−− −

ªºªº

−=−⋅ ++⋅

«»«»

−−−

¬¼¬¼

=−−++−

=−−+=−−−

=−+−+

5.2 • Solutions 285

This matrix has eigenvalue 4 with multiplicity 1, and eigenvalue 3 with multiplicity 2.

10.

311

det( ) det 0 5 0 .

207

AI

λ

λλ

λ

−

ªº

«»

−=−

«»

−−

¬¼

Using cofactor expansion down the first column, the

characteristic polynomial is

2

32

(5 ) 0 1 1

det( ) (3 ) det 0 ( 2)det

0(7) (5)0

(5 )( 10 21) 2(5 )

(5 )( 10 23)

15 73 115

AI

λ

λλ

λλ λ λ

λλ λ

−

ªºªº

−=−⋅ ++−

«»«»

−−

¬¼¬¼

=−−++ −

=−−+

=−+−+

11. The special arrangements of zeros in A makes a cofactor expansion along the first row highly

effective.

232

300 14

det( ) det 2 1 4 (3 ) det 04

104

(3 )(1 )(4 ) (3 )( 5 4) 8 19 12

AI

λ

λλλ

λ

λλλλλλλλ

−

ªº

−

ªº

«»

−=−=−«»

«» −

¬¼

«»

−

¬¼

=−−λ−=−−+=−+−+

If only the eigenvalues were required, there would be no need here to write the characteristic

polynomial in expanded form.

12. Make a cofactor expansion along the third row:

32

102 12 1 0

det( ) det 3 1 0 1 det (2 ) det

30 31

012

6(2 )(1 )(1 ) 2 4

AI

λ

λλ

λλ λ

λ

λλλλλλ

−−

ªº

−− −−

ª

ºªº

«»

−=−=−⋅ +−⋅

«

»«»

«» −

¬

¼¬¼

«»

−

¬¼

=+ −−− −=−+++

13. Make a cofactor expansion down the third column:

2

32

620 62

det( ) det 2 9 0 (3 ) det 29

583

(3 )[(6 )(9 ) ( 2)( 2)] (3 )( 15 50)

18 95 150 or (3 )( 5)( 10)

AI

λ

λλλ

λ

λλλ λλλ

λλ λ λλλ

−−

ªº

−−

ª

º

«»

−=−− =−⋅

«

»

«»

−−

¬

¼

«»

−

¬¼

=−−−−−−=−−+

=−+−+−−−

14. Make a cofactor expansion along the second column:

32

401 41 41

det( ) det 1 4 ( ) det 2 det

03 14

023

()[(4 )(3 )]2[(4 )41] 7 12 8 30

7430

AI

λ

λλ

λλλ

λ

λλλ λ λλλλ

λλλ

−−

ªº

−− −−

ª

ºª º

«»

−=−− =−⋅ −⋅

«

»« »

«» −−

¬

¼¬ ¼

«»

−

¬¼

=−⋅ − − − − −=−+−+−

=−+−−

15. Use the fact that the determinant of a triangular matrix is the product of the diagonal entries:

286 CHAPTER 5 • Eigenvalues and Eigenvectors

2

5502

02 3 6

det( ) det (5 ) (2 )(3 )

003 2

0005

AI

λ

λλλλ

λ

−

ªº

«»

−−

«»

−==−−−

«»

−−

«»

−

¬¼

The eigenvalues are 5, 5, 2, and 3.

16. The determinant of a triangular matrix is the product of its diagonal entries:

3000

62 0 0

det( ) det (3 )(2 )(6 )( 5 )

036 0

2335

AI

λ

λλλλλ

λ

−

ªº

«»

−

«»

−==−−−−−

«»

−

«»

−−

¬¼

The eigenvalues are 3, 2, 6, and

5.−

17. The determinant of a triangular matrix is the product of its diagonal entries:

22

30000

51 0 0 0

(3 ) (1 ) ( )

380 00

0721 0

41 9 23

λ

λλλ

λ

−

ªº

«»

−−

«»

=−−−

−

«»

−−

«»

−−−

¬¼

The eigenvalues are 3, 3, 1, 1, and 0.

18. Row reduce the augmented matrix for the equation (4)AI−=

x0

:

02330 02 3 30 02 3 00

02 30 00 360 00 300

000140 00 0 10 00 0 10

00020 00 0 00 00 0 00

hh h

ªºª ºª º

«»« »« »

−++

«»« »« »

−

¬¼¬ ¼¬ ¼



For a two-dimensional eigenspace, the system above needs two free variables. This happens if and

only if

3.h=−

19. Since the equation

12

det( ) ( )( ) ( )−λ =λ−λ λ −λ λ −λ"

n

AI holds for all

λ

, set

0λ=

and conclude

that

12

det .=λλ λ"

n

A

20. det( ) det( )

TTT

AI AI

−λ =−λ

5.2 • Solutions 287

det( ) Transpose property=−λ

T

AI

det( ) Theorem 3(c)=−AI

λ

21. a. False. See Example 1.

b. False. See Theorem 3.

c. True. See Theorem 3.

d. False. See the solution of Example 4.

22. a. False. See the paragraph before Theorem 3.

b. False. See Theorem 3.

c. True. See the paragraph before Example 4.

d. False. See the warning after Theorem 4.

23. If ,=AQR with Q invertible, and if

1

,=ARQ

then write

11

1

,

−−

==

AQQRQQAQ

which shows that

1

A is similar to A.

24. First, observe that if P is invertible, then Theorem 3(b) shows that

11

1det det( )(det)(det )

IPP PP

−−

== =

Use Theorem 3(b) again when

1

,

−

=

APBP

111

det det( ) (det )(det )(det ) (det )(det )(det ) det

APBP PBP BPP B

−−−

== = =

25. Example 5 of Section 4.9 showed that

11

,=vvA which means that

1

v is an eigenvector of A

corresponding to the eigenvalue 1.

a. Since A is a

22×

matrix, the eigenvalues are easy to find, and factoring the characteristic

polynomial is easy when one of the two factors is known.

2

63

det ( 6 )( 7 ) ( 3)( 4) 1 3 3 ( 1)( 3)

47

λλλ λ λ λλ

λ

.−.

ªº

=. −.−−..= −.+.=−−.

«»

..−

¬¼

The eigenvalues are 1 and .3. For the eigenvalue .3, solve

(3)

AI

−.=x0

:

63 3 0 3 30 110

4730440000

.−.. ..

ªºªºªº

=

«»«»«»

..−...

¬¼¬¼¬¼



Here

12

0,xx+=

with

2

x free. The general solution is not needed. Set

2

1x= to find an

eigenvector

2

1.

1

−

ª

º

=

«

»

¬

¼

v A suitable basis for

2

R

is

12

{}.,vv

b. Write

01 2

c=+xv v

: 12 37 1 .

12 47 1

//−

ªºª º ªº

=+

«»« » «»

//

¬¼¬ ¼ ¬¼

c By inspection, c is

114.−/

(The value of c depends

on how

2

v is scaled.)

c. For

12 ,=, ,

k… define

0

.=

xx

k

kA

Then

112121 2

() (3),Ac AcA c=+=+ =+.xvv vvv v

because

1

v and

2

v are eigenvectors. Again

21 1 2 1 21 2

((3)) (3) (3)(3)AA c AcA c== +. =+. =+...xx v v v vv v

288 CHAPTER 5 • Eigenvalues and Eigenvectors

Continuing, the general pattern is

12

(3) .

k

kc

=+.

xv v

As k increases, the second term tends to 0

and so

k

x tends to

1

.v

26. If

0,≠

a then

1

,

0

−

ªº

==

«»

−

¬¼



ab

AU

cd dcab

and

1

det ( )( ) .

−

=−=−

Aadcabadbc

If

0,=

a

then 0

0

bcd

AU

cd b

ªºªº

==

«»«»

¬¼¬¼

 (with one interchange), so

1

det ( 1) ( ) 0 .=−=−=−

Acbbcadbc

27. a.

11

,A=vv

22

5,A=.vv

33

2.A=.vv

b. The set

123

{},,vv v is linearly independent because the eigenvectors correspond to different

eigenvalues (Theorem 2). Since there are three vectors in the set, the set is a basis for

3

. So there

exist unique constants such that

0112233

,cc c=+ +xvvv

and

01 12 23 3

.

TTT T

cc c

=+ +

wx wv wv wv

Since

0

x and

1

v are probability vectors and since the entries in

2

v and

3

v sum to 0, the above

equation shows that

1

1.c=

c. By (b),

0112233

.=+ +xvvvcc c

Using (a),

01 12 23 3 12 23 3 1

(5) (2) as== + + =+.+.→→∞

xx v v vv v vv

kkk k k k

kAcAcAcA c c k

28. [M] Answers will vary, but should show that the eigenvectors of A are not the same as the

eigenvectors of ,

T

A

unless, of course, .=

T

AA

29. [M] Answers will vary. The product of the eigenvalues of A should equal det A.

30. [M] The characteristic polynomials and the eigenvalues for the various values of a are given in the

following table:

a Characteristic Polynomial Eigenvalues

31.8

23

426 4ttt−.−.+ −

312791 1279.,,−.

31.9

23

838 4ttt.−.+ − 2.7042, 1, .2958

32.0

23

25 4ttt−+− 2, 1, 1

32.1

23

32 62 4ttt.−.+ −

15 9747 1

i

.±. ,

32.2

23

44 74 4ttt.−.+ −

15 14663 1

i

.±. ,

The graphs of the characteristic polynomials are:

r

Notes

: An appendix in Section 5.3 of t

h

with integer coefficients, in case you

w

perhaps

44×

matrices.

The MATLAB box for Section 5.3

of the characteristic polynomial of the

m

the characteristic polynomial. (This is

n

have corresponding information. The a

p

list the coefficients of the characteristic

p

5.3 SOLUTIONS

1. 57 20

23 01

−

ªºªº

=,=,=

«»«»

¬¼¬¼

PDAPDP

14

37 160

,,

25 01

−

ªºªº

==

«»«»

−

¬¼¬¼

PD

a

2. 12 10

23 03

PDAPDP

−

ªºªº

=,=,=

«»«»

¬¼¬¼

14

32 10

,

21 081

PD

−

ªºªº

=,=

«»«»

−

¬¼¬¼

a

n

3.

1

10 0

210

k

kk

k

a

APDP

b

ªº

«»

−«»

«»

¬¼

ªº ª

==

«» «

−

¬¼ ¬

4.

1

32(3)

210(

k

kk

APDP

ª

«

−«

«

¬

−−

ªº

==

«»

−

¬¼

5. By the Diagonalization Theorem, e

i

correspond respectively to the eige

n

5.3 • Sol

u

r

son Education, Inc. Publishing as Addison-Wesley.

h

e Study Guide gives an example of factoring a cubi

c

w

ant your students to find integer eigenvalues of si

m

introduces the command poly (A), which lists th

e

m

atrix A, and it gives MATLAB code that will produ

c

n

eeded for Exercise 30.) The Maple and Mathematic

a

p

pendices for the TI calculators contain only the co

m

p

olynomial.

1

,

−

and

441

.

−

=APDP

We compute

nd

4

57160 3 7 226 525

230 1 2 5 90 209

A

−−

ª

ºª ºª º ª º

==

«

»« »« » « »

−−

¬

¼¬ ¼¬ ¼ ¬ ¼

1

,

−

and

441

.

−

=APDP

We compute

n

d

4

121 0 3 2 321 160

230812 1 480 239

A

−−

ª

ºª ºª º ª º

==

«

»« »« » « »

−−

¬

¼¬ ¼¬ ¼ ¬ ¼

10 0

.

21 22

k

kk

k

a

abb

ªº

«»

¬¼

º

=

»

−

¼

0123(3)4(2)6(3)6

23

2) 2(3) 2(2) 4(3) 3

kk

kkk

k

º

»

¼

ª

−−⋅−+⋅− ⋅− −

⋅

ªº

«

=

«»

−

«

−¬¼

−⋅− +⋅− ⋅− −

⋅

¬

i

genvectors form the columns of the left factor, and t

h

n

values on the diagonal of the middle factor.

u

tions 289

c

polynomial

m

ple

33×

or

e

coefficients

c

e a graph of

a

appendices

m

mands that

(2) .

(2)

k

º

⋅

−»

»

⋅

−¼

h

ey

290 CHAPTER 5 • Eigenvalues and Eigenvectors

110

2 1 31 1

101

−

ªº ªºªº

«» «»«»

=: ;=:−,−

«» «»«»

−

¬¼ ¬¼¬¼

6. As in Exercise 5, inspection of the factorization gives:

031

41 30 3

010

−

ªº ªºª º

«» «»« »

=: ;=: ,−

«» «»« »

¬¼ ¬¼¬ ¼

7. Since A is triangular, its eigenvalues are obviously

1.±

For λ = 1: 00

1.

62

ªº

−=«»

−

¬¼

AI The equation

(1)AI−=x0

amounts to

12

62 0,xx−= so

12

(1 3)xx=/

with

2

x free. The general solution is

2

13 ,

1

/

ª

º

«

»

¬

¼

x and a nice basis vector for the eigenspace is

1

1.

3

ª

º

=

«

»

¬

¼

v

For λ = −1: 20

1.

60

ª

º

+=

«

»

¬

¼

AI The equation

(1)AI+=x0

amounts to

1

20,=x so

1

0x= with

2

x

free. The general solution is

2

0,

1

ªº

«»

¬¼

x and a basis vector for the eigenspace is

2

0.

1

ªº

=«»

¬¼

v

From

1

v and

2

v construct

12

10

.

31

ªº

«»

¬¼

ª

º

==

«

»

¬

¼

vvP Then set 10

,

01

ª

º

=

«

»

−

¬

¼

D where the eigenvalues

in D correspond to

1

v and

2

v respectively.

8. Since A is triangular, its only eigenvalue is obviously 3.

For λ = 3: 02

3.

00

AI

ªº

−=«»

¬¼

The equation

(3)AI−=x0

amounts to

2

0,=x so

2

0x= with

1

x free.

The general solution is

1

1.

0

ªº

«»

¬¼

x Since we cannot generate an eigenvector basis for

2

, A is not

diagonalizable.

9. To find the eigenvalues of A, compute its characteristic polynomial:

22

2 1

det( )det (2 )(4 )(1)(1) 69( 3)

14

AI

−−

ªº

−==−−−−=−+= −

«»

−

¬¼

Thus the only eigenvalue of A is 3.

For λ = 3: 11

4.

11

−−

ªº

−=«»

¬¼

AI The equation

(3)AI−=x0

amounts to

12

0,+=xx so

12

xx=− with

2

x free. The general solution is

2

1.

1

−

ªº

«»

¬¼

x Since we cannot generate an eigenvector basis for

2

, A is

not diagonalizable.

5.3 • Solutions 291

10. To find the eigenvalues of A, compute its characteristic polynomial:

2

1 3

det( )det (1 )(2 )(3)(4) 310 ( 5)( 2)

42

AI

−

ªº

−==−−− =−−=−+

«»

−

¬¼

Thus the eigenvalues of A are 5 and

2−

.

For λ = −2: 33

2.

44

AI

ª

º

+=

«

»

¬

¼ The equation

(1)AI+=x0

amounts to

12

0,xx+=

so

12

xx=− with

2

x free. The general solution is

2

1,

1

x

−

ªº

«»

¬¼

and a nice basis vector for the eigenspace is

1

1.

1

−

ªº

=«»

¬¼

v

For λ = 5: 43

5.

43

AI

−

ªº

−=«»

−

¬¼

The equation

(3)AI−=x0

amounts to

12

43 0,xx−+=

so

12

(3 / 4)xx= with

2

x free. The general solution is

2

3/4 ,

1

x

ª

º

«

»

¬

¼ and a basis vector for the eigenspace is

2

3.

4

ªº

=«»

¬¼

v

From

1

v and

2

v construct

12

13

.

14

P

ªº

«»

¬¼

−

ª

º

==

«

»

¬

¼

vv Then set 20

,

05

D

−

ª

º

=

«

»

¬

¼ where the eigenvalues

in D correspond to

1

v and

2

v respectively.

11. The eigenvalues of A are given to be -1, and 5.

For λ = -1:

111

222,

333

AI

ªº

«»

+=

«»

¬¼

and row reducing

[]

AI+0

yields

1110

0000.

0000

ª

º

«

»

«

»

«

»

¬

¼

The general

solution is

23

11

10,

01

xx

−−

ªº ªº

«» «»

+

«» «»

¬¼ ¬¼

and a nice basis for the eigenspace is

{}

12

11

,1,0

01

½

−−

ªºªº

°°

«»«»

=®¾

«»«»

°°

«»«»

¬¼¬¼

¯¿

vv

.

For λ = 5:

511

5242,

333

AI

−

ªº

«»

−=−

«»

−

¬¼

and row reducing

[]

5AI−0

yields

10 130

01 2/30.

00 00

−/

ª

º

«

»

−

«

»

«

»

¬

¼

The

general solution is

3

13

2/3 ,

1

x

/

ª

º

«

»

«

»

«

»

¬

¼

and a nice basis vector for the eigenspace is

3

1

2.

3

ªº

«»

=«»

«»

¬¼

v

From

12

,vv

and

3

v construct

123

111

102.

013

P

ªº

«»

¬¼

−−

ª

º

«

»

==

«

»

«

»

¬

¼

vvv

Then set

D

=

100

010,

005

−

ªº

«»

−

«»

¬¼

where the eigenvalues in D correspond to

12

,vv

and

3

v respectively. Note that if changing the

order of the vectors in P also changes the order of the diagonal elements in D, and results in the

answer given in the text.

292 CHAPTER 5 • Eigenvalues and Eigenvectors

12. The eigenvalues of A are given to be 2 and 5.

For λ = 2:

111

2111,

111

AI

ªº

«»

−=«»

«»

¬¼

and row reducing

[]

2AI−0

yields

1110

0000.

0000

ª

º

«

»

«

»

«

»

¬

¼

The general

solution is

23

11

10,

01

xx

−−

ªº ªº

«» «»

+

«» «»

¬¼ ¬¼

and a nice basis for the eigenspace is

{}

12

11

,1,0.

01

½

−−

ªºªº

°°

«»«»

=®¾

«»«»

°°

«»«»

¬¼¬¼

¯¿

vv

.

For λ = 5:

21 1

5121,

11 2

AI

−

ªº

«»

−=−

«»

−

¬¼

and row reducing

[]

5AI−0

yields

10 10

01 10.

00 00

−

ªº

«»

−

«»

¬¼

The

general solution is

3

1

1,

1

x

ª

º

«

»

«

»

«

»

¬

¼

and a basis for the eigenspace is

3

1

1.

1

ª

º

«

»

=

«

»

«

»

¬

¼

v

From

12

,vv

and

3

v construct

123

111

101.

011

P

ªº

«»

¬¼

−−

ª

º

«

»

==

«

»

«

»

¬

¼

vvv

Then set

200

020,

005

D

ªº

«»

=«»

«»

¬¼

where

the eigenvalues in D correspond to

12

,vv

and

3

v respectively.

13. The eigenvalues of A are given to be 5 and 1.

For λ = 5:

321

5121,

123

AI

−−

ªº

«»

−=−−

«»

−−−

¬¼

and row reducing

[]

5AI−0

yields

1010

0110.

0000

ª

º

«

»

«

»

«

»

¬

¼

The

general solution is

3

1

1,

1

−

ª

º

«

»

−

«

»

«

»

¬

¼

x

and a basis for the eigenspace is

1

1.

1

−

ª

º

«

»

=−

«

»

«

»

¬

¼

v

For λ = 1:

121

1121,

121

−

ªº

«»

−=−

«»

−−

¬¼

AI

and row reducing

[]

AI−0

yields

12 10

00 00.

00 00

−

ª

º

«

»

«

»

«

»

¬

¼

The

general solution is

23

21

10,

01

−

ª

ºªº

«

»«»

+

«

»«»

«

»«»

¬

¼¬¼

xx

and a basis for the eigenspace is

23

21

{} 10.

01

½

−

ªºªº

°°

«»«»

,= ,

®¾

«»«»

°°

«»«»

¬¼¬¼

¯¿

vv

From

12

,vv

and

3

v construct

123

121

110.

101

ªº

«»

¬¼

−−

ª

º

«

»

==−

«

»

«

»

¬

¼

vvv

P

Then set

500

010,

001

ªº

«»

=«»

«»

¬¼

D

where the eigenvalues in D correspond to

12

,vv

and

3

v respectively.

5.3 • Solutions 293

14. The eigenvalues of A are given to be 2 and 3.

For λ = 2:

00 2

2112,

00 1

AI

−

ªº

«»

−=«»

«»

¬¼

and row reducing

[

]

2AI−0

yields

1100

0010.

0000

ª

º

«

»

«

»

«

»

¬

¼

The

general solution is

2

1

1,

0

x

−

ª

º

«

»

«

»

«

»

¬

¼

and a basis for the eigenspace is

1

0

−

ª

º

«

»

=

«

»

«

»

¬

¼

v

.

For λ = 3:

10 2

3102,

00 0

AI

−−

ªº

«»

−=«»

«»

¬¼

and row reducing

[]

3AI−0

yields

1020

0000.

0000

ª

º

«

»

«

»

«

»

¬

¼

The

general solution is

23

02

10,

01

xx

−

ª

ºªº

«

»«»

+

«

»«»

«

»«»

¬

¼¬¼

and a nice basis for the eigenspace is

23

02

{}10.

01

½

−

ªºª º

°°

«»« »

,= ,

®¾

«»« »

°°

«»« »

¬¼¬ ¼

¯¿

vv

From

12

,vv

and

3

v construct

123

10 2

11 0.

00 1

P

ªº

«»

¬¼

−−

ª

º

«

»

==

«

»

«

»

¬

¼

vvv

Then set

200

030,

003

D

ªº

«»

=«»

«»

¬¼

where the eigenvalues in D correspond to

12

,vv

and

3

v respectively.

15. The eigenvalues of A are given to be 0 and 1.

For λ = 0:

011

0121,

110

AI

−−

ªº

«»

−=«»

«»

−−

¬¼

and row reducing

[

]

0AI−0

yields

10 10

01 1 0.

00 0 0

−

ª

º

«

»

«

»

«

»

¬

¼

The

general solution is

3

1

1,

1

x

ª

º

«

»

−

«

»

«

»

¬

¼

and a basis for the eigenspace is

1

ª

º

«

»

=−

«

»

«

»

¬

¼

v

For λ = 1:

111

111,

111

AI

−−−

ªº

«»

−=«»

«»

−−−

¬¼

and row reducing

[]

AI−0

yields

1110

0000.

0000

ª

º

«

»

«

»

«

»

¬

¼

The

general solution is

23

11

10,

01

xx

−−

ª

ºªº

«

»«»

+

«

»«»

«

»«»

¬

¼¬¼

and a basis for the eigenspace is

23

11

{} 10.

01

½

−−

ªºªº

°°

«»«»

,= ,

®¾

«»«»

°°

«»«»

¬¼¬¼

¯¿

vv

From

12

,vv

and

3

v construct

123

111

110.

10 1

P

ªº

«»

¬¼

−−

ª

º

«

»

==−

«

»

«

»

¬

¼

vvv

Then set

000

010,

001

D

ªº

«»

=«»

«»

¬¼

where the eigenvalues in D correspond to

12

,vv

and

3

v respectively. Note that the answer for P

given in the text has the first column scaled by

1−

, which is also a correct answer, since any nonzero

multiple of an eiegnvector is an eigenvector.

16. The only eigenvalue of A given is 0.

294 CHAPTER 5 • Eigenvalues and Eigenvectors

For λ = 0:

12 3

0252,

13 1

AI

−

ªº

«»

−=−

«»

¬¼

and row reducing

[

]

0AI−0

yields

10 110

01 4 0.

00 0 0

−

ª

º

«

»

«

»

«

»

¬

¼

The

general solution is

2

11

4,

1

x

ª

º

«

»

−

«

»

«

»

¬

¼

and a basis for the eigenspace is

1

11

4.

1

ª

º

«

»

=−

«

»

«

»

¬

¼

v

Since

0=

has only a one-dimensional eigenspace, we can find at most one linearly independent

eigenvector for A, so A is not diagonalizable over the real numbers. The remaining eigenvalues are

complex, and this situation is dealt with in Section 5.

17. Since A is triangular, its eigenvalue is obviously 2.

For λ = 2:

000

2200,

220

AI

ªº

«»

−=«»

«»

¬¼

and row reducing

[]

2AI−0

yields

1000

0100.

0000

ª

º

«

»

«

»

«

»

¬

¼

The general

solution is

3

0

0,

1

x

ªº

«»

¬¼

and a basis for the eigenspace is

1

0

0.

1

ª

º

«

»

=

«

»

«

»

¬

¼

v

Since

2=

has only a one-dimensional eigenspace, we can find at most one linearly independent

eigenvector for A, so A is not diagonalizable.

18. The eigenvalues of A are given to be -2, -1 and 0.

For λ = -2:

422

2312,

220

AI

−−

ª

º

«

»

+= −−

«

»

«

»

−

¬

¼

and row reducing

[]

2AI+0

yields

10 10

01 10.

00 0 0

−

ª

º

«

»

−

«

»

«

»

¬

¼

The

general solution is

3

1

1,

1

x

ª

º

«

»

«

»

«

»

¬

¼

and a basis for the eigenspace is

1

ª

º

«

»

=

«

»

«

»

¬

¼

v

.

For λ = -1:

322

322,

221

AI

−−

ªº

«»

+= −−

«»

−−

¬¼

and row reducing

[]

AI+0

yields

10 1 0

01 1/20.

00 0 0

−

ª

º

«

»

−

«

»

«

»

¬

¼

The

general solution is

3

1

1/2 ,

1

x

ª

º

«

»

«

»

«

»

¬

¼

and a basis for the eigenspace is

2

1

2

ª

º

«

»

=

«

»

«

»

¬

¼

v

.

For λ = 0:

222

0332,

222

AI

−−

ªº

«»

−=−−

«»

−−

¬¼

and row reducing

[]

0AI−0

yields

1100

0010.

0000

−

ª

º

«

»

«

»

«

»

¬

¼

The

general solution is

2

1

1,

0

x

ª

º

«

»

«

»

«

»

¬

¼

and a nice basis vector for the eigenspace is

3

1

1.

0

ªº

«»

=«»

«»

¬¼

v

5.3 • Solutions 295

From

12

,vv

and

3

v construct

123

12 1

111.

120

P

ªº

«»

¬¼

ª

º

«

»

==

«

»

«

»

¬

¼

vvv

Then set

200

010,

000

D

−

ªº

«»

=−

«»

¬¼

where

the eigenvalues in D correspond to

12

,vv

and

3

v respectively.

19. Since A is triangular, its eigenvalues are 2, 3, and 5.

For

= 2:

3309

0112

2,

0000

−

ª

º

«

»

−

«

»

−=

«

»

«

»

«

»

¬

¼

AI

and row reducing

[]

2 AI−0

yields

10 1 10

011 20

.

000 00

ªº

«»

−

«»

¬¼

The general solution is

34

11

12

,

10

01

−−

ªº ªº

«» «»

−

«» «»

+

«» «»

¬¼ ¬¼

xx

and a nice basis for the eigenspace is

12

11

12

{} .

10

01

½−−

ªºªº

°°

«»«»

−

°°

«»«»

,= ,

®¾

«»«»

°°

«»«»

°°

«»«»

¬¼¬¼

¯¿

vv

For λ = 3:

2309

0012

3,

0010

000 1

−

ª

º

«

»

−

«

»

−=

«

»

−

«

»

−

«

»

¬

¼

AI

and row reducing

[]

3 AI−0

yields

132000

00100

.

00010

00000

−/

ªº

«»

¬¼

The general solution is

2

32

1,

0

/

ª

º

«

»

«

»

«

»

«

»

«

»

¬

¼

x

and a nice basis for the eigenspace is

3

2.

0

ªº

«»

=«»

«»

¬¼

v

For λ = 5:

0309

0212

5,

0030

0003

−

ª

º

«

»

−−

«

»

−=

«

»

−

«

»

−

«

»

¬

¼

AI

and row reducing

[]

5 AI−

0

yields

01000

00100

.

00010

00000

ªº

«»

¬¼

The general solution is

1

0,

0

ªº

«»

¬¼

x

and a basis for the eigenspace is

4

1

0.

0

ª

º

«

»

«

»

=

«

»

«

»

«

»

¬

¼

v

296 CHAPTER 5 • Eigenvalues and Eigenvectors

From

123

,,vv v

and

4

v construct

1234

1131

1220

.

1000

0100

ªº

¬¼

−−

ª

º

«

»

−

«

»

==

«

»

«

»

«

»

¬

¼

vv v v

P

Then set

2000

0200

,

0030

0005

ªº

«»

=«»

«»

¬¼

D

where the eigenvalues in D correspond to

12

,vv

,

3

v and

4

v respectively.

Note that this answer differs from the text. There,

[]

4312

P=

vvvv

and the entries in D are

rearranged to match the new order of the eigenvectors. According to the Diagonalization Theorem,

both answers are correct.

20. Since A is triangular, its eigenvalues are 2 and 3.

For λ = 2:

1000

0000

2,

0000

100 1

AI

ªº

«»

−=«»

«»

¬¼

and row reducing

[]

2 AI−

0

yields

10000

00010

.

00000

ª

º

«

»

«

»

«

»

«

»

¬

¼

The

general solution is

23

00

10

,

01

00

xx

ª

ºªº

«

»«»

«

»«»

+

«

»«»

«

»«»

¬

¼¬¼

and a basis for the eigenspace is

{}

12

00

10.

01

00

½

ªºªº

°°

«»«»

°°

«»«»

,= ,

®¾

«»«»

°°

«»«»

°°

¬¼¬¼

¯¿

vv

For λ = 3:

00 00

0100

3,

00 10

10 00

AI

ªº

«»

−

«»

−=«»

−

«»

¬¼

and row reducing

[]

3 AI−

0

yields

10000

01000

.

00100

00000

ªº

«»

¬¼

The general solution is

4

0

0,

0

1

x

ªº

«»

¬¼

and a basis for the eigenspace is

3

0

0.

0

1

ª

º

«

»

«

»

=

«

»

«

»

¬

¼

v

Since

3=

has only a one-dimensional eigenspace, we can find at most three linearly independent

eigenvectors for A, so A is not diagonalizable.

21. a. False. The symbol D does not automatically denote a diagonal matrix.

b. True. See the remark after the statement of the Diagonalization Theorem.

c. False. The

33×

matrix in Example 4 has 3 eigenvalues, counting multiplicities, but it is not

diagonalizable.

d. False. Invertibility depends on 0 not being an eigenvalue. (See the Invertible Matrix Theorem.)

A diagonalizable matrix may or may not have 0 as an eigenvalue. See Examples 3 and 5 for both

possibilities.

22. a. False. The n eigenvectors must be linearly independent. See the Diagonalization Theorem.

5.3 • Solutions 297

b. False. The matrix in Example 3 is diagonalizable, but it has only 2 distinct eigenvalues. (The

statement given is the converse of Theorem 6.)

c. True. This follows from

AP PD=

and formulas (1) and (2) in the proof of the Diagonalization

Theorem.

d. False. See Example 4. The matrix there is invertible because 0 is not an eigenvalue, but the matrix

is not diagonalizable.

23. A is diagonalizable because you know that five linearly independent eigenvectors exist: three in the

three-dimensional eigenspace and two in the two-dimensional eigenspace. Theorem 7 guarantees that

the set of all five eigenvectors is linearly independent.

24. No, by Theorem 7(b). Here is an explanation that does not appeal to Theorem 7: Let

1

v and

2

v be

eigenvectors that span the two one-dimensional eigenspaces. If v is any other eigenvector, then it

belongs to one of the eigenspaces and hence is a multiple of either

1

v or

2

.v So there cannot exist

three linearly independent eigenvectors. By the Diagonalization Theorem, A cannot be

diagonalizable.

25. Let

1

{}v be a basis for the one-dimensional eigenspace, let

2

v and

3

v form a basis for the two-

dimensional eigenspace, and let

4

v be any eigenvector in the remaining eigenspace. By Theorem 7,

1234

{ },,,vv v v is linearly independent. Since A is

44,×

the Diagonalization Theorem shows that

A is diagonalizable.

26. Yes, if the third eigenspace is only one-dimensional. In this case, the sum of the dimensions of the

eigenspaces will be six, whereas the matrix is

77.×

See Theorem 7(b). An argument similar to that

for Exercise 24 can also be given.

27. If A is diagonalizable, then

1

APDP

−

=

for some invertible P and diagonal D. Since A is invertible, 0

is not an eigenvalue of A. So the diagonal entries in D (which are eigenvalues of A) are not zero, and

D is invertible. By the theorem on the inverse of a product,

111111111

()()A PDP P D P PD P

−−−−−−−−−

== =

Since

1

D

−

is obviously diagonal,

1

A

−

is diagonalizable.

28. If A has n linearly independent eigenvectors, then by the Diagonalization Theorem,

1

APDP

−

=

for

some invertible P and diagonal D. Using properties of transposes,

11

()()

()

−−

==

TTTTT

TT

APDP PDP

PDPQDQ

where

1

().

−

=

T

QP

Thus

T

A

is diagonalizable. By the Diagonalization Theorem, the columns of Q

are n linearly independent eigenvectors of .

T

A

29. The diagonal entries in

1

D are reversed from those in D. So interchange the (eigenvector) columns of

P to make them correspond properly to the eigenvalues in

1

.D In this case,

11

11 30

and

21 05

PD

ªº ªº

==

«» «»

−−

¬¼ ¬¼

298 CHAPTER 5 • Eigenvalues and Eigenvectors

Although the first column of P must be an eigenvector corresponding to the eigenvalue 3, there is

nothing to prevent us from selecting some multiple of 1,

2

ª

º

«

»

−

¬

¼ say 3,

6

−

ª

º

«

»

¬

¼ and letting

2

31

.

61

−

ªº

=«»

−

¬¼

P

We now have three different factorizations or “diagonalizations” of A:

11 1

111 212

APDP PDP PDP

−− −

== =

30. A nonzero multiple of an eigenvector is another eigenvector. To produce

2

,P simply multiply one or

both columns of P by a nonzero scalar other than 1.

31. For a

22

×

matrix A to be invertible, its eigenvalues must be nonzero. A first attempt at a

construction might be something such as 23

,

04

ª

º

«

»

¬

¼ whose eigenvalues are 2 and 4. Unfortunately, a

22

×

matrix with two distinct eigenvalues is diagonalizable (Theorem 6). So, adjust the construction

to 23

,

02

ªº

«»

¬¼

which works. In fact, any matrix of the form 0

ab

a

ª

º

«

»

¬

¼ has the desired properties when a

and b are nonzero. The eigenspace for the eigenvalue a is one-dimensional, as a simple calculation

shows, and there is no other eigenvalue to produce a second eigenvector.

32. Any

22

×

matrix with two distinct eigenvalues is diagonalizable, by Theorem 6. If one of those

eigenvalues is zero, then the matrix will not be invertible. Any matrix of the form 00

ab

ªº

«»

¬¼

has the

desired properties when a and b are nonzero. The number a must be nonzero to make the matrix

diagonalizable; b must be nonzero to make the matrix not diagonal. Other solutions are 00

ab

ªº

«»

¬¼

and 0.

0

ªº

«»

¬¼

a

b

33.

9424

56 32 28 44 ,

14 14 6 14

42 33 21 45

A

−−−

ªº

«»

−−

«»

=«»

−− −

«»

−−

¬¼

ev = eig(A)=(13,-12,-12, 13)

,

nulbasis(A-ev(1)*eye(4))

0.5000 0.3333

0 1.3333

.

1.0000 0

0 1.0000

−

ª

º

«

»

−

«

»

=

«

»

«

»

¬

¼

A basis for the eigenspace of

11

04

13 is , .

20

03



½−

ª

ºª º

°

«

»« »

−

°

«

»« »

λ=

®

¾

«

»« »

°

«

»« »

°

¬

¼¬ ¼

¯¿

5.3 • Solutions 299

nulbasis(A-ev(2)*eye(4))

0.2857 0

10000 1.0000

10000 0

0 1.0000

ªº

«»

.−

«»

.

«»

¬¼

,

A basis for the eigenspace of

20

71

12 is , .

70

01



½

ª

ºª º

°

«

»« »

−

°

«

»« »

λ=−

®

¾

«

»« »

°

«

»« »

°

¬

¼¬ ¼

¯¿

Thus we construct

1120

0471

,

2070

0301

P

−

ªº

«»

−

«»

=«»

«»

¬¼

and

13 0 0 0

013 0 0

.

00 12 0

00 0 12

D

ªº

«»

=«»

−

«»

−

¬¼

Notice that the anwer in the text lists the eigenvector in an different

order in P, and hence the eigenvalues are listed in a different order in D. Both answers are correct.

34.

49782

790714

,

5105510

23704

313710 11

A

−−

ªº

«»

−−

«»

=−−

«»

−

«»

−− −

¬¼

ev = eig(A)

=(5,-2,-2,5,5)

,

nulbasis(A-ev(1)*eye(5))

2.0000 1.0000 2.0000

1.0000 1.0000 0

1.0000 0 0

01.0000 0

001.0000

−

ª

º

«

»

−

«

»

«

»

=,

«

»

«

»

«

»

¬

¼

A basis for the eigenspace of

212

110

5 is , .

100

010

001



½

−

ª

ºª ºª º

°

«

»« »« »

−

°

«

»« »« »

°

«

»« »« »

λ=,

®

¾

«

»« »« »

°

«

»« »« »

°

«

»« »« »

°

¬

¼¬ ¼¬ ¼

¯¿

nulbasis(A-ev(2)*eye(5))

0.4000 0.6000

1.4000 1.4000

1 0000 1.0000

10000 0

0 1.0000

−

ª

º

«

»

«

»

«

»

=−.−

«

»

.

«

»

«

»

¬

¼

,

300 CHAPTER 5 • Eigenvalues and Eigenvectors

A basis for the eigenspace of

23

77

2 is , .

55

50

05



½

−

ª

ºª º

°

«

»« »

°

«

»« »

°

«

»« »

λ=−−−

®

¾

«

»« »

°

«

»« »

°

«

»« »

°

¬

¼¬ ¼

¯¿

Thus we construct

21223

11077

100 5 5

01050

00105

P

−−

ª

º

«

»

−

«

»

«

»

=−−

«

»

«

»

«

»

¬

¼

and

500 0 0

050 0 0

.

005 0 0

000 2 0

000 0 2

D

ª

º

«

»

«

»

«

»

=

«

»

−

«

»

«

»

−

¬

¼

35.

13 12 9 15 9

659159

,

6125 69

6129 89

61212 62

A

−−

ªº

«»

−−

«»

=−−

«»

−−

«»

−−−

¬¼

ev = eig(A)

−

=(7, 14,-14,7,7)

,

nulbasis(A-ev(1)*eye(5))

2 0000 1 0000 1 5000

10000 0 0

010000 0

0010000

..−.

ª

º

«

»

.

«

»

«

»

=,

.

«

»

.

«

»

«

»

.

¬

¼

A basis for the eigenspace of

21 3

10 0

7 is , .

01 0

00 2



½

−

ª

ºªºª º

°

«

»«»« »

°

«

»«»« »

°

«

»«»« »

λ=,

®

¾

«

»«»« »

°

«

»«»« »

°

«

»«»« »

°

¬

¼¬¼¬ ¼

¯¿

nulbasis(A-ev(2)*eye(5))

10

01

10

01

ª

º

«

»

«

»

«

»

=,

−

«

»

«

»

«

»

¬

¼

A basis for the eigenspace of

10

14 is , .

01

10

01



½

ª

ºª º

°

«

»« »

°

«

»« »

°

«

»« »

λ=−−

®

¾

«

»« »

°

«

»« »

°

«

»« »

°

¬

¼¬ ¼

¯¿

5.3 • Solutions 301

Thus we construct

21 31 0

10 0 1 0

,

01 00 1

01 01 0

00 20 1

P

−

ª

º

«

»

«

»

«

»

=−

«

»

«

»

«

»

¬

¼

and

700 0 0

070 0 0

.

007 0 0

000 14 0

000 0 14

D

ª

º

«

»

«

»

«

»

=

«

»

−

«

»

«

»

−

¬

¼

36.

24 6 2 6 2

72 51 9 99 9

,

063156363

72 15 9 63 9

063216327

A

−

ªº

«»

−

«»

=−

«»

−

«»

−−

¬¼

ev = eig(A)=(24,-48,36,-48,36)

,

nulbasis(A-ev(1)*eye(5))

1

0

1

0

ª

º

«

»

«

»

«

»

=,

«

»

«

»

«

»

¬

¼

A basis for the eigenspace of

1

24 is .

0

1

0

ª

º

«

»

«

»

«

»

λ=

«

»

«

»

«

»

¬

¼

nulbasis(A-ev(2)*eye(5)),

A basis for the eigenspace of

00

10

48 is .

01

10

01



½

ª

ºª º

°

«

»« »

°

«

»« »

°

«

»« »

λ=−,−

®

¾

«

»« »

°

«

»« »

°

«

»« »

°

¬

¼¬ ¼

¯¿

nulbasis(A-ev(3)*eye(5))

1 0000 0 3333

0 0000 1 0000

30000 00000

10000 0

010000

.−.

ª

º

«

»

..

«

»

«

»

=..

«

»

.

«

»

«

»

.

¬

¼

,

A basis for the eigenspace of

11

03

36 is , .

30

10

03



½

−

ª

ºª º

°

«

»« »

°

«

»« »

°

«

»« »

λ=

®

¾

«

»« »

°

«

»« »

°

«

»« »

°

¬

¼¬ ¼

¯¿

302 CHAPTER 5 • Eigenvalues and Eigenvectors

Thus we construct

10 0 1 1

11 00 3

00 13 0

11 01 0

00 10 3

P

−

ª

º

«

»

«

»

«

»

=−

«

»

«

»

«

»

¬

¼

and

.

24 0 0 0 0

048000

00 4800

00 0360

00 0036

D

ª

º

«

»

−

«

»

«

»

=−

«

»

«

»

«

»

¬

¼

. Notice that

the anwer in the text lists the eigenvector in an different order in P, and hence the eigenvalues are listed

in a different order in D. Both answers are correct.

Notes

: For your use, here is another matrix with five distinct real eigenvalues. To four decimal places,

they are 11.0654, 9.8785, 3.8238,

37332,

−.

and

60345.−.

68530

73530

37535

04175

53208

−−

ªº

«»

−−

«»

−− −

«»

−−

«»

−−−

¬¼

The MATLAB box in the Study Guide encourages students to use eig (A) and nulbasis to

practice the diagonalization procedure in this section. It also remarks that in later work, a student may

automate the process, using the command []=PD eig (A). You may wish to permit students to use

the full power of eig in some problems in Sections 5.5 and 5.7.

5.4 SOLUTIONS

1. Since

1121

3

()3 5 [()] .

5

ªº

=−,=

«»

−

¬¼

bddb

D

TT

Likewise

212

() 6T=−+bdd

implies that

2

1

[( )] 6

D

T

−

ª

º

=

«

»

¬

¼

b

and

32

()4T=bd

implies that

3

0

[( )] .

4

ª

º

=

«

»

¬

¼

b

D

T Thus the matrix for T relative to

B

and

123

310

is [ ( )] [ ( )] [ ( )] .

564

ªº

¬¼

−

ªº

=«»

−

¬¼

bb b

DDD

DT T T

2.Since

1121

3

()3 3 [()] .

3

B

TT

ªº

=−,=

«»

−

¬¼

dbbd Likewise

212

() 2 5T=−+dbb

implies that

2

[( )] .

5

B

T

−

ª

º

=

«

»

¬

¼

d

Thus the matrix for T relative to

D

and

12

32

is [ ( )] [ ( )] .

35

BB

BT T

ªº

¬¼

−

ª

º

=

«

»

−

¬

¼

dd

5.4 • Solutions 303

3. a.

11232 1233123

() 0 0 ( ) 2 0 () 2 0 3TT T=−+, =−− +, = + +ebbbe bbbe bbb

b.

12 3

012

[( )] 0 [( )] 2 [( )] 0

103

BB B

TT T

−

ªº ª º ªº

«» « » «»

=, =−,=

«» « » «»

¬¼ ¬ ¼ ¬¼

ee e

.

c. The matrix for T relative to

!

and

123

012

is [ [ ( )] [ ( )] [ ( )] ] 0 2 0 .

103

BBB

BT T T

−

ªº

«»

=−

«»

¬¼

ee e

4. Let

12

{}=,ee! be the standard basis for . Since

11 2 2

23

[( )] ( ) [( )] ( ) ,

20

TT T T

−

ª

ºªº

==, = =

«

»«»

−

¬

¼¬¼

bb b b

!!

and

33

1

[( )] ( ) ,

5

TT

ªº

==

«»

¬¼

bb

!

the matrix for T relative to

B

and

123

is [[ ( )] [ ( )] [ ( )] ] =bb bTT T

!!!

! 231

.

205

−

ª

º

«

»

−

¬

¼

5. a.

223

() ( 3)(3 2 ) 9 3Tt tt ttt=+ −+=−++p

b. Let

p

and

q

be polynomials in

2

, and let

c

be any scalar. Then

(() ()) ( 3)[() ()] ( 3)() ( 3)()

(()) (())

Tt t t t t t t t t

Tt Tt

+=+ +=+ ++

=+

pq pq p q

pq

(())(3)[()] (3)()

[()]

Tc t t c t c t t

cT t

⋅=+ ⋅=⋅+

=⋅

ppp

p

and T is a linear transformation.

c. Let

2

{1 }Btt=,,

and

23

{1 } .=,,,Cttt

Since

11

3

1

() (1)( 3)(1) 3 [()] .

0

C

TTt tT

ªº

«»

==+ =+, =

«»

¬¼

bb

Likewise since

2

22

0

3

() ()( 3)() 3 [()] ,

1

0

C

TTtttttT

ª

º

«

»

«

»

==+ =+, =

«

»

«

»

¬

¼

bb

and since

304 CHAPTER 5 • Eigenvalues and Eigenvectors

2232

33

0

() ()( 3)() 3 [()] .

3

1

C

TTtttttT

ª

º

«

»

«

»

==+ =+, =

«

»

«

»

¬

¼

bb

Thus the matrix for T relative to B and

123

300

130

is [ [ ( )] [ ( )] [ ( )] ] .

013

001

CCC

CT T T

ª

º

«

»

«

»

=

«

»

«

»

¬

¼

bb b

6. a.

22 2 234

() (3 2 ) 2 (3 2 ) 3 2 7 4 2Tttttttttt=−++ −+=−+−+

p

b. Let

p

and

q

be polynomials in

2

, and let

c

be any scalar. Then

2

22

2

(() ()) [() ()] 2 [() ()]

[() 2 ()] [() 2 ()]

(()) (())

(())[()]2[()]

[() 2 ()]

[()]

Tt t t t t t t

ttt ttt

Tt Tt

Tc t c t t c t

ct tt

cT t

+=++ +

=+ ++

=+

⋅=⋅+⋅

=⋅+

=⋅

pq pq pq

ppqq

pq

pp p

pp

p

and T is a linear transformation.

c. Let

2

{1 }Btt=,,

and

234

{1 } .=,,,,Ctttt

Since

22

11

1

0

() (1)12(1)12 [()] .

2

0

C

TT t tT

ª

º

«

»

«

»

«

»

==+ =+, =

«

»

«

»

«

»

¬

¼

bb

Likewise since

23

22

0

1

() () (2)() 2[()] ,

0

2

0

C

TTttttttT

ª

º

«

»

«

»

«

»

==+ =+, =

«

»

«

»

«

»

¬

¼

bb and

since

22 222 4

33

0

() () (2)() 2 [()] .

1

0

2

C

TTttttttT

ª

º

«

»

«

»

«

»

==+ =+, =

«

»

«

»

«

»

¬

¼

bb Thus the matrix for T relative to

B

and

123

100

010

is [ [ ( )] [ ( )] [ ( )] ] .

201

020

002

CCC

CT T T

ª

º

«

»

«

»

«

»

=

«

»

«

»

«

»

¬

¼

bbb

5.4 • Solutions 305

7. Since

11

3

() (1)35[()] 5.

0

ª

º

«

»

==+, =

«

»

«

»

¬

¼

bb

B

TT tT

Likewise since

2

22

0

() () 2 4[()] 2,

4

ªº

«»

==−+, =−

«»

¬¼

bb

B

TTtttT

and since

22

33

0

() () [()] 0.

1

ªº

«»

==, =

«»

¬¼

bb

B

TTttT

Thus the

matrix representation of T relative to the basis

B

is

123

300

[( )] [( )] [( )] 5 2 0.

041

ªº

¬¼

ª

º

«

»

=−

«

»

«

»

¬

¼

bb b

BBB

TT T

Perhaps a faster way is to realize that the

information given provides the general form of

()Tp

as shown in the figure below:

22

01 2 0 0 1 12

coordinate coordinate

mapping mapping

00

multiplication

101

by[ ]

212

3(52)(4 )

3

52

4

ªº ª º

«» « »

¬¼ ¬ ¼

++ → +−++

→ −

+

B

T

aatat a a at aat

aa

aaa

The matrix that implements the multiplication along the bottom of the figure is easily filled in by

inspection:

00

101

212

3300

52 implies that [] 5 20

4041

B

aa

aaa T

aaa

ªºª º

«»« »

¬¼¬ ¼

???

ªº ª º

«» « »

??? = −=−

«» « »

??? +

¬¼ ¬ ¼

8. Since

12

4

[4 3 ] 3 ,

0

B

ª

º

«

»

−=−

«

»

«

»

¬

¼

bb

12 12

00 1 4 0

[(4 3 )] [][4 3 ] 2 1 2 3 5

13 1 0 5

BB B

TT

ª

ºª º ª º

«

»« » « »

−=−=−−=

«

»« » « »

«

»« » « »

−

¬

¼¬ ¼ ¬ ¼

bb bb

and

12 23

(4 3 ) 5 5 .

T

−=−bb b b

9. a.

53(1) 2

() 5 3(0) 5

53(1) 8

T

+−

ªºªº

«»«»

=+ =

«»«»

+

¬¼¬¼

p

b. Let p and q be polynomials in

2

, and let c be any scalar. Then

()(1) (1)(1) (1) (1)

()()(0) (0)(0) (0) (0) ()()

()(1) (1)(1) (1) (1)

+−−+−−−

ªºª ºªºªº

«»« »«»«»

+= + = + = + = +

«»« »«»«»

++

¬¼¬ ¼¬¼¬¼

pq p q p q

pq pq p q p q p q

pq p q p q

TTT

((1)) (1)

()(1)

() ((0)) (0) ()

( )(0)

((1)) (1)

()(1)

c

Tc c c cT

c

⋅− −

⋅−

ªºªº

ªº

«»«»

«»

⋅==⋅=⋅=⋅

⋅

«»«»

«»

«»«»

«»

⋅

¬¼

¬¼¬¼

pp

p

pppp

p

pp

p

306 CHAPTER 5 • Eigenvalues and Eigenvectors

and T is a linear transformation.

c. Let

2

{1 }=,,Btt

and

123

{}=,,ee e! be the standard basis for

3

. Since

11 2 2

11

[( )] ( ) (1) 1 [( )] ( ) () 0,

11

−

ªº ª º

«» « »

===, = ==

«» « »

¬¼ ¬ ¼

bb b b

TTT T TTt

!!

and

2

33

1

[( )] ( ) ( ) 0,

1

ªº

«»

===

«»

¬¼

bb

TTTt

!

the matrix for T relative to

B

and

!

is

123

111

[( )] [( )] [( )] 1 0 0.

111

ªº

¬¼

−

ªº

«»

=

«»

¬¼

bb b

TT T

!!!

10. a. Let p and q be polynomials in

3

, and let c be any scalar. Then

()(2) (2)(2)

()(3) (3)(3)

()

()(1) (1)(1)

()(0) (0)(0)

T

+−−+−

ª

ºªº

«

»«»

++

«

»«»

+= =

«

»«»

++

«

»

«»

++

¬¼

¬

¼

pq p q

pq pq p q

pq p q

(2) (2)

(3) (3) () ()

(1) (1)

(0) (0)

TT

−−

ªºªº

«»«»

=+=+

«»«»

¬¼¬¼

pq

pq pq

pq

()(2) ((2)) (2)

()(3) ((3)) (3)

() ()

()(1) ((1)) (1)

()(0) ((0)) (0)

cc

Tc c cT

cc

⋅− ⋅ − −

ªºªºªº

«»«»«»

⋅⋅

«»«»«»

⋅===⋅=⋅

«»«»«»

⋅⋅

«»«»«»

⋅⋅

¬¼¬¼¬¼

ppp

pp

ppp

and T is a linear transformation.

b. Let

23

{1 }=,, ,Bttt

and

1234

{}=,,,ee e e! be the standard basis for

4

. Since

2

11 2 2 3 3

124

139

[( )] ( ) (1) [( )] ( ) () [( )] ( ) ( ) ,

111

100

TTT T TTt TTTt

−

ªº ª º ª º

«» « » « »

===, = ==, = ==

«» « » « »

¬¼ ¬ ¼ ¬ ¼

bb b b bb

!! !

and

3

44

8

27

[( )] ( ) ( ) ,

1

0

TTTt

−

ªº

«»

===

«»

¬¼

bb

!

the matrix for T relative to

B

and

!

is

1234

1248

13927

[( )] [( )] [( )] [( )] .

1111

1000

TT TT

ªº

¬¼

−−

ª

º

«

»

«

»

=

«

»

«

»

¬

¼

bb bb

!!!!

11. Following Example 4, if

12

11

,

21

P

ªº

«»

¬¼

−−

ª

º

==

«

»

¬

¼

bb then the B-matrix is

5.4 • Solutions 307

1

114111 22

216121 0 1

PAP

−

−−−− −−

ªºªºªºªº

==

«»«»«»«»

−− −

¬¼¬¼¬¼¬¼

12. Following Example 4, if

12

01

,

12

P

ªº

«»

¬¼

−

ª

º

==

«

»

¬

¼

bb then the B-matrix is

1

21 6 20 1 4 0

10 4 01 2 2 2

PAP

−

−− − −

ªºª ºªºª º

==

«»« »«»« »

−−

¬¼¬ ¼¬¼¬ ¼

13. Start by diagonalizing A. The characteristic polynomial is

2

43( 1)( 3),−+= −−

so the

eigenvalues of A are 1 and 3.

For λ = 1: 11

.

33

−

ªº

−=«»

−

¬¼

AI The equation

()AI−=x0

amounts to

12

0,xx−+=

so

12

xx= with

2

x

free. A basis vector for the eigenspace is thus

1

1.

1

ª

º

=

«

»

¬

¼

v

For λ = 3: 31

3.

31

−

ªº

−=«»

−

¬¼

AI The equation

(3)AI−=x0

amounts to

12

30,xx−+=

so

12

(1 3)xx=/

with

2

x free. A nice basis vector for the eigenspace is thus

2

1.

3

ª

º

=

«

»

¬

¼

v

From

1

v and

2

v we may construct

12

11

13

P

ªº

«»

¬¼

ª

º

==

«

»

¬

¼

vv which diagonalizes A. By Theorem 8, the

basis

12

{}B=,vv has the property that the B-matrix of the transformation

Axx6

is a diagonal

matrix.

14. Start by diagonalizing A. The characteristic polynomial is

2

45( 5)( 1),−−=−+

so the

eigenvalues of A are 5 and

1.−

For λ = 5: 33

5.

33

AI

−

ªº

−=«»

−

¬¼

The equation

(5)AI−=x0

amounts to

12

0,xx−= so

12

xx= with

2

x free. A basis vector for the eigenspace is thus

1

1.

1

ª

º

=

«

»

¬

¼

v

For λ =

1−

:33

.

33

AI ªº

+=

«»

¬¼

The equation

()AI+=x0

amounts to

12

0,xx+=

so

12

xx=− with

2

x

free. A nice basis vector for the eigenspace is thus

2

1.

1

−

ª

º

=

«

»

¬

¼

v

From

1

v and

2

v we may construct

12

11

P

ªº

«»

¬¼

−

ª

º

==

«

»

¬

¼

vv which diagonalizes A. By Theorem 8,

the basis

12

{}B=,vv has the property that the B-matrix of the transformation

Axx6

is a diagonal

matrix.

308 CHAPTER 5 • Eigenvalues and Eigenvectors

15. Start by diagonalizing A. The characteristic polynomial is

2

310 ( +5)( 2),+−=−

so the

eigenvalues of A are −5 and 2.

For λ = 5: 62

5.

31

AI

ªº

+=

«»

¬¼

The equation

(5)AI+=x0

amounts to

12

30,xx+=

so

12

(1/3)xx=−

with

2

x free. A basis vector for the eigenspace is thus

1

1.

3

−

ª

º

=

«

»

¬

¼

v

For λ = 2: 12

2.

36

AI

−

ªº

−=«»

−

¬¼

The equation

(2)AI−=x0

amounts to

12

20,xx−= so

12

2xx=

with

2

x free. A basis vector for the eigenspace is thus

2

2.

1

ª

º

=

«

»

¬

¼

v

From

1

v and

2

v we may construct

12

31

P

ªº

«»

¬¼

−

ª

º

==

«

»

¬

¼

vv which diagonalizes A. By Theorem 8,

the basis

12

{}B=,vv has the property that the B-matrix of the transformation

Axx6

is a diagonal

matrix. Note that the solution in the text lists the vectors in the reverse order, which is also correct.

16. Start by diagonalizing A. The characteristic polynomial is

2

9+18 ( 3)( 6),−=−−

so the

eigenvalues of A are 3 and 6.

For λ = 3: 12

3.

12

AI

−

ªº

−=«»

−

¬¼

The equation

(3)AI−=x0

amounts to

12

20,xx−= so

12

2xx=

with

2

x free. A basis vector for the eigenspace is thus

1

2.

1

ª

º

=

«

»

¬

¼

v

For λ = 6: 22

6.

11

AI

−−

ªº

−=«»

−−

¬¼

The equation

(6)AI−=x0

amounts to

12

0,xx+=

so

12

xx=− with

2

x free. A basis vector for the eigenspace is thus

2

1.

1

−

ª

º

=

«

»

¬

¼

v

From

1

v and

2

v we may construct

12

21

11

P

ªº

«»

¬¼

−

ª

º

==

«

»

¬

¼

vv which diagonalizes A. By Theorem 8,

the basis

12

{}B=,vv has the property that the B-matrix of the transformation

Axx6

is a diagonal

matrix.

17. a. We compute that

11

411 3 3

12 1 3

Aªºªºªº

===

«»«»«»

−−−

¬¼¬¼¬¼

bb

so

1

b is an eigenvector of A corresponding to the eigenvalue 3. The characteristic polynomial of

A is

22

69( 3) ,−+= −

so 3 is the only eigenvalue for A. Now 11

3,

11

AI

ªº

−=«»

−−

¬¼

which

implies that the eigenspace corresponding to the eigenvalue 3 is one-dimensional. Thus the matrix

A is not diagonalizable.

5.4 • Solutions 309

b. Following Example 4, if

12

,

ªº

«»

¬¼

=bbP

then the B-matrix for T is

1

21 4 1 1 1 31

11 1 2 1 2 0 3

PAP

−

ªºª ºª ºªº

==

«»« »« »«»

−−

¬¼¬ ¼¬ ¼¬¼

18. If there is a basis

B

such that []

B

T is diagonal, then A is similar to a diagonal matrix, by the second

paragraph following Example 3. In this case, A would have three linearly independent eigenvectors.

However, this is not necessarily the case, because A has only two distinct eigenvalues.

19. If A is similar to B, then there exists an invertible matrix P such that

1

.

−

=PAP B

Thus B is invertible

because it is the product of invertible matrices. By a theorem about inverses of products,

1111111

() ,

−−−−−−−

==BPAP PAP

which shows that

1

A

−

is similar to

1

.

−

B

20. If

1

,

−

=APBP

then

211 11 121

()()() .

−− −− − −

===⋅⋅ =APBPPBP PBPPBPPBIBPPBP

So

2

A

is

similar to

2

.B

21. By hypothesis, there exist invertible P and Q such that

1

PBP A

−

=

and

1

.

−

=QCQ A

Then

11

.

−−

=PBP QCQ

Left-multiply by Q and right-multiply by

1

Q

−

to obtain

11 1 1

.

−− − −

=QP BPQ QQ CQQ

So

11 11 1

()(),

−− −− −

==CQPBPQ PQ BPQ

which shows that B is similar to C.

22. If A is diagonalizable, then

1

APDP

−

=

for some P. Also, if B is similar to A, then

1

BQAQ

−

=

for some Q. Then

11 11 1

()()()()()BQPDP Q QPDPQ QPDQP

−− −− −

== =

So B is diagonalizable.

23. If

0,=,≠xxxA

then

11

.PA P

−−

=xx

If

1

,

−

=BPAP

then

1111 1

() ()BP P APP P A P

−−−− −

===xxxx

(*)

by the first calculation. Note that

1

0,

−

≠xP

because

0≠x

and

1

P

−

is invertible. Hence (*) shows

that

1

P

−

x

is an eigenvector of B corresponding to . (Of course, is an eigenvalue of both A and B

because the matrices are similar, by Theorem 4 in Section 5.2.)

24. If

1

,

−

=APBP

then

11

rank rank ( ) rank ,

−−

==APBP BP

by Supplementary Exercise 13 in Chapter

4. Also,

1

rank rank ,

−

=BP B

by Supplementary Exercise 14 in Chapter 4, since

1

P

−

is invertible.

Thus

rank rank .=AB

25. If

1

,

−

=APBP

then

11

1

tr( ) tr(( ) ) tr( ( )) By the trace property

tr( ) tr( ) tr( )

APBP PPB

PPB IB B

−−

−

==

===

If B is diagonal, then the diagonal entries of B must be the eigenvalues of A, by the Diagonalization

Theorem (Theorem 5 in Section 5.3). So

tr tr {sum of the eigenvalues of }.==AB A

26. If

1

APDP

−

=

for some P, then the general trace property from Exercise 25 shows that

1

tr tr [( ) ]APDP

−

==

1

tr [ ] tr .

−

=PPD D

(Or, one can use the result of Exercise 25 that since A is

310 CHAPTER 5 • Eigenvalues and Eigenvectors

similar to D,

tr tr .=AD

) Since the eigenvalues of A are on the main diagonal of D, tr D is the sum

of the eigenvalues of A.

27. For each

() .,=bb

jj

jI

Since the standard coordinate vector of any vector in

n

is just the vector

itself,

[( )] .=bb

jj

I

ε

Thus the matrix for I relative to

B

and the standard basis

!

is simply

12

.

ªº

«»

¬¼

bb b

n

…

This matrix is precisely the change-of-coordinates matrix

B

P defined in Section

4.4.

28. For each

() ,,=bb

jj

jI

and

[( )] [ ].=bb

jC jC

I

By formula (4), the matrix for I relative to the bases

B

and C is

12

[] [] [ ]

CnC

C

M…

ªº

«»

¬¼

=bb b

. In Theorem 15 of Section 4.7, this matrix was

denoted by

CB

P

←

and was called the change-of-coordinates matrix from

B

to

.C

29. If

1

{},=,,bb

n

B…

then the B-coordinate vector of

j

b

is

,e

j

the standard basis vector for

n

. For

instance,

11 2

10 0=⋅+⋅++⋅bb b b

n

…

Thus

jjj

[( )] [ ] ,==bbe

BB

I

and

11

[] [( )] [( )] [ ]

BBnB n

II I I

ªº

¬¼

===bbee""

30. [M] If P is the matrix whose columns come from

,B

then the B-matrix of the transformation

Axx6

is

1

.

−

=DPAP

From the data in the text,

123

622 121

312 111

222 130

AP

ªº

«»

¬¼

−− −

ªº ªº

«» «»

=−,= = −,

«» «»

−

¬¼ ¬¼

bbb

331622121 210

1103 12111 030

211222130 014

D

−−−−−

ªºªºªºªº

«»«»«»«»

=−−−=

«»«»«»«»

−−

¬¼¬¼¬¼¬¼

31. [M] If P is the matrix whose columns come from

,B

then the B-matrix of the transformation

Axx6

is

1

.

−

=DPAP

From the data in the text,

123

74816 323

114 6 111

34519 330

131374816323 726

13 0114 6111 046

011334519330 001

ªº

«»

¬¼

−− − −−

ªº ªº

«» «»

=,==−,

«» «»

−− − −−

¬¼ ¬¼

−−−/−− − −− −−−

ªºª ºªºªº

«»« »«»«»

=−=−−

«»« »«»«»

−−/−− − −− −

¬¼¬ ¼¬¼¬¼

bbbAP

D

32. [M]

6409

3016

,

1210

4407

A

−

ªº

«»

−

«»

=«»

−−

«»

−

¬¼

5.4 • Solutions 311

ev

=

eig(A)

=

(5, 1, -2, -2)

nulbasis(A-ev(1)*eye(4))

10000

05000

10000

.

ª

º

«

»

.

«

»

=

«

»

−.

«

»

.

¬

¼

A basis for the eigenspace of

5=

is

1

2

1.

1

2

ª

º

«

»

«

»

=

«

»

−

«

»

¬

¼

b

nulbasis(A-ev(2)*eye(4))

1 0000

0 5000 .

3.5000

1.0000

.

ª

º

«

»

−.

«

»

=

«

»

−

«

»

¬

¼

A basis for the eigenspace of

1=

is

2

1.

7

2

ª

º

«

»

−

«

»

=

«

»

−

«

»

¬

¼

b

nulbasis(A-ev(3)*eye(4))

1 0000 1.5000

1 0000 0.7500

10000 0

0 1.0000

.

ª

º

«

»

.−

«

»

=

«

»

.

«

»

¬

¼

A basis for the eigenspace of

2=−

is

{}

34

16

13

,,.

10

04



½

ª

ºª º

°

«

»« »

−

°

«

»« »

=

®

¾

«

»« »

°

«

»« »

°

¬

¼¬ ¼

¯¿

bb

The basis

1234

{}B=,,,bb b b is a basis for

4

with the property that []

B

T is diagonal.

Note:

The Study Guide comments on Exercise 26 and tells students that the trace of any square matrix A

equals the sum of the eigenvalues of A, counted according to multiplicities. This provides a quick check

on the accuracy of an eigenvalue calculation. You could also refer students to the property of the

determinant described in Exercise 19 of Section 5.2.

312 CHAPTER 5 • Eigenvalues and Eigenvectors

5.5 SOLUTIONS _____________________________________________

1.12 1 2

13 13

AAI

λ

−−−

ªº ª º

=,−=

«» « »

−

¬¼ ¬ ¼

2

det( )(1 )(3 )(2) 4 5AI−=−−−−=−+

Use the quadratic formula to find the eigenvalues: 41620

2.

2

±−

==±i

λ

Example 2 gives a

shortcut for finding one eigenvector, and Example 5 shows how to write the other eigenvector with

no effort.

For λ = 2 + i: 12

(2 ) .

11

−− −

ªº

−+=

«»

−

¬¼

i

AiI

i The equation

( )AI−=x0

gives

12

(1 ) 2 0

(1 ) 0

ix x

xix

−− − =

+−=

As in Example 2, the two equations are equivalent—each determines the same relation between

1

x

and

2

.x So use the second equation to obtain

12

(1 ) ,=−−xix

with

2

x free. The general solution is

2

1,

1

−+

ªº

«»

¬¼

i

x and the vector

1

i−+

ªº

=«»

¬¼

v provides a basis for the eigenspace.

For λ = 2 – i: Let

1

2

1.

1

−−

ªº

==

«»

¬¼

vv

i

The remark prior to Example 5 shows that

2

v is

automatically an eigenvector for 2.+i In fact, calculations similar to those above would show that

2

{}v is a basis for the eigenspace. (In general, for a real matrix A, it can be shown that the set of

complex conjugates of the vectors in a basis of the eigenspace for

λ

is a basis of the eigenspace for

λ

.)

2. 33

.

33

A

−

ªº

=«»

¬¼

The characteristic polynomial is

2

618,−+

so the eigenvalues of A are

63672

33.

2i

±−

==±

For λ = 3 + 3i: 33

(3 3 ) .

33

i

AiI

i

−−

ªº

−+=

«»

−

¬¼

The equation

((33))AiI−+=x0

amounts to

12

0,xix−= so

12

xix= with

2

x free. A basis vector for the eigenspace is thus

1

.

1

i

ªº

=«»

¬¼

v

For λ = 3 – 3i: A basis vector for the eigenspace is

1

2

.

1

i−

ª

º

==

«

»

¬

¼

vv

5.5 • Solutions 313

3. 51

.

81

Aªº

=«»

−

¬¼

The characteristic polynomial is

2

613,−+

so the eigenvalues of A are

63652

32.

2i

±−

==±

For λ = 3 + 2i: 22 1

(3 2 ) .

822

i

AiI

i

−

ªº

−+=

«»

−−−

¬¼

The equation

((32))AiI−+=x0

amounts to

12

8(22) 0,xix−+−− = so

12

1

4

i

xx

−−

=

with

2

x free. A nice basis vector for the eigenspace is thus

1

1.

4

i−−

ªº

=«»

¬¼

v

For λ = 3 – 2i: A basis vector for the eigenspace is

1

2

1.

4

i−+

ª

º

==

«

»

¬

¼

vv

4. 12

.

13

A

−

ªº

=«»

¬¼

The characteristic polynomial is

2

45,−+

so the eigenvalues of A are

44

2.

2i

±−

==±

For λ = 2 + i: 12

(2 ) .

11

i

AiI

i

−− −

ªº

−+=

«»

−

¬¼

The equation

((2))AiI−+=x0

amounts to

12

(1 ) 0,xix+−= so

12

(1 )xix=−− with

2

x free. A basis vector for the eigenspace is thus

1

1.

1

i−+

ªº

=«»

¬¼

v

For λ = 2 – i: A basis vector for the eigenspace is

1

2

1.

1

i−−

ª

º

==

«

»

¬

¼

vv

5. 31

.

25

Aªº

=«»

−

¬¼

The characteristic polynomial is

2

817,−+

so the eigenvalues of A are

84

4.

2i

±−

==±

For λ = 4 + i: 11

(4 ) .

21

i

AiI

i

−−

ªº

−+=

«»

−−

¬¼

The equation

((4))AiI−+=x0

amounts to

12

2(1) 0,xix−+−= so

12

(1 )

2

i

xx

−

= with

1

x free. A basis vector for the eigenspace is thus

1

1.

2

i−

ªº

=«»

¬¼

v

For λ = 4 – i: A basis vector for the eigenspace is

1

2

1.

2

i+

ª

º

==

«

»

¬

¼

vv

314 CHAPTER 5 • Eigenvalues and Eigen

v

r

6. 75

.

13

A−

ªº

=«»

¬¼

The characteristic p

o

10 4 5.

2i

±−

==±

For λ = 5 + i: 2

(5 ) 1

i

AiI

−

ª

−+=

«

¬

12

(2 ) 0,xix−+=

so

12

(2 )xix=+

1

2.

1

i+

ªº

=«»

¬¼

v

For λ = 5 – i: A basis vector for t

h

7.

31

.

13

ªº

«»

¬¼

−

=A

From Example 6, t

h

Axx6

is

22

(3) 1

2

r=

||

=+=

xy-plane and use trigonometry:

ϕ

=

arctan

()ba/=

arctan

(1

/

Note:

Your students will want to kno

w

matrix of the form ab

ba

−

ªº

«»

¬¼

and simpl

y

the corresponding eigenvectors, 1

i

ªº

«»

−

¬¼

a

may have trouble keeping track of the c

o

8.

333

.

33 3

A

ªº

«»

¬¼

=−

The eigenvalu

e

is

22

(3) (3 3) 6.r=| |= + = Fr

o

(33/3) 3−=−π /

radians.

v

ectors

r

son Education, Inc. Publishing as Addison-Wesley.

o

lynomial is

2

10 26,−+

so the eigenvalues of A a

r

5.

2i

−º

»

−−

¼ The equation

((5))AiI−+=x0

amounts to

with

2

x free. A basis vector for the eigenspace is thu

s

h

e eigenspace is

1

2

2.

1

i−

ªº

==

«»

¬¼

vv

h

e eigenvalues are

3.±i

The scale factor for the tra

n

2

. For the angle of rotation, plot the point

()(3ab,= ,

3) 6

/

=π/

radians.

w

whether you permit them on an exam to omit calc

u

y

write the eigenvalues

.±abi

A similar question ma

y

a

nd 1,

ªº

«»

¬¼

i which are announced in the Practice Probl

e

o

rrespondence between eigenvalues and eigenvectors.

e

s are

3(33).i±

The scale factor for the transformati

o

m trigonometry, the angle of rotation

ϕ

is arctan

(ba/

r

e

s

n

sformation

1)

in the

u

lations for a

y

arise about

e

m. Students

o

n

Axx6

)=

arctan

5.5 • Solutions 315

9. 02

.

20

Aªº

=«»

−

¬¼

The eigenvalues are

2.i±

The scale factor for the transformation

Axx6

is

22

(0) (2) 2.r=

||

=+=

From trigonometry, the angle of rotation

ϕ

is arctan

()ba/=

arctan

() 2−∞ =−π /

radians.

10. 0.5

.

.5 0

Aªº

=«»

−

¬¼

The eigenvalues are

.5 .i±

The scale factor for the transformation

Axx6

is

22

(0) .5 .5.r=

||

=+=

From trigonometry, the angle of rotation

ϕ

is

arctan

()ba/=

arctan

() 2−∞ =−π/

radians.

11.

31

.

13

A

ªº

−

=«»

−−

«»

¬¼

The eigenvalues are

3.i−±

The scale factor for the transformation

Axx6

is

22

(3) (1) 2.r=

||

=+= From trigonometry, the angle of rotation

ϕ

is

arctan

()ba/=

arctan

(1 3 ) 5 6/=−π/

radians.

12.

33

.

33

A

ªº

−

=«»

«»

¬¼

The eigenvalues are

3(3).i±

The scale factor for the transformation

Axx6

is

22

3(3) 23.r=| |= + = From trigonometry, the angle of rotation

ϕ

is

arctan

()ba/=

arctan

(3/3) /6

π

=

radians.

13. From Exercise 1,

2,=±i

and the eigenvector 1

1

i−−

ª

º

=

«

»

¬

¼

v corresponds to

2.=−i

Since Re

1

−

ªº

=«»

¬¼

v and Im 1,

0

−

ª

º

=

«

»

¬

¼

v take 11

.

10

−−

ª

º

=

«

»

¬

¼

P Then compute

1

011211 0131 21

111310 1121 12

CPAP

−

−−− − − −

ªºªºªºªºªºªº

== = =

«»«»«»«»«»«»

−− −− −

¬¼¬¼¬¼¬¼¬¼¬¼

Actually, Theorem 9 gives the formula for C. Note that the eigenvector v corresponds to

abi−

instead of

.+abi

If, for instance, you use the eigenvector for

2,+i

your C will be 21

.

12

ªº

«»

−

¬¼

Notes:

The Study Guide points out that the matrix C is described in Theorem 9 and the first column of C

is the real part of the eigenvector corresponding to

,−abi

not

,+abi

as one might expect. Since students

may forget this, they are encouraged to compute C from the formula

1

,

−

=CPAP

as in the solution above.

The Study Guide also comments that because there are two possibilities for C in the factorization of a

22×

matrix as in Exercise 13, the measure of rotation of the angle associated with the transformation

Axx6

is determined only up to a change of sign. The “orientation” of the angle is determined by the

change of variable

.=xuP

See Figure 4 in the text.

316 CHAPTER 5 • Eigenvalues and Eigenvectors

14. 33

.

11

A

−

ªº

=«»

¬¼

The eigenvalues of A are

2(2),i=±

and the eigenvector 2i

i

ªº

+

=«»

¬¼

v corresponds

to

2(2).i=−

By Theorem 9, 21

[Re Im ] 01

P

ª

º

==

«

»

¬

¼

vv and

1

33 2 2

2/2 2/2 2 1

11

01 01

22

CPAP

−

ª

º

ªºªº

−−

ªº

−

== =

«

»

«»«»

«»

¬¼

«

»

¬¼¬¼

¬

¼

There are two choices for whice

eigenvalues to use and many choices for the muliple of the eigenvector to use,. The solutions can

look quite different. For example, for

22i

λ

=+

an eigenvector is

21(2)

1

ii

i

ªºª º

−++

−==⋅=⋅

«»« »

¬¼¬ ¼

vw Using w in Theorem 9, results in

[Re Im ]P==ww

12

10

ª

º

«

»

¬

¼

and

1

01

33 2 2

12 .

11

2/2 2/2 1 0 22

CPAP

−

ª

º

ªºªº

−

ªº

== =

«

»

«»«»

«»

−−

¬¼

«

»

¬¼¬¼

¬

¼

15. 05

.

22

Aªº

=«»

−

¬¼

The eigenvalues of A are

13,i=±

and the eigenvector 3

2

i

+

ªº

=«»

¬¼

v corresponds to

13.i=+

By Theorem 9,

[Re Im ]P==vv

31

02

ª

º

«

»

¬

¼and

1

210531 13

1.

032202 31

6

CPAP

−

ªºªºªºªº

== =

«»«»«»«»

−−

¬¼¬¼¬¼¬¼

There are many choices for the muliple of the

eigenvector to use, and the solutions can look quite different. For example,

313

22

ii

i

+−

ªºª º

==⋅=⋅

«»« »

¬¼¬ ¼

vw

. Using w in Theorem 9, results in

[Re Im ]P==ww

13

20

−

ªº

«»

¬¼

and

1

03 051 3 13

1.

21 222 0 31

6

CPAP

−

ªºªºªºªº

== =

«»«»«»«»

−− −

¬¼¬¼¬¼¬¼

16. 42

.

16

A

−

ªº

=«»

¬¼

The eigenvalues of A are

5,i=±

and the eigenvector 1i

i

+

ª

º

=

«

»

−

¬

¼

v corresponds to

5.i=+

By Theorem 9,

[]

11

Re Im 01

P

ª

º

==

«

»

−

¬

¼

vv and

1

114211 51

011601 15

CPAP

−

ªºªºªºªº

== =

«»«»«»«»

−−−

¬¼¬¼¬¼¬¼

. There are many choices for the muliple of the

eigenvector to use, and the solutions can look quite different. For example,

5.5 • Solutions 317

11

1

ii

i

+−+

ªº ª º

==−⋅ =−⋅

«» « »

−

¬¼ ¬ ¼

vw

. Using w in Theorem 9, results in

[Re Im ]P==ww

11

10

−

ªº

«»

¬¼

and

1

014 2 1 1 5 1

.

111 6 10 15

CPAP

−

−−

ªºª ºª ºª º

== =

«»« »« »« »

−

¬¼¬ ¼¬ ¼¬ ¼

17. 11 4 .

20 5

A

−−

ªº

=«»

¬¼

The eigenvalues of A are

34,i=−±

and the eigenvector

12

5

i

+

ªº

=«»

−

¬¼

v corresponds to

34.i=−+

By Theorem 9,

[]

12

Re Im 05

Pªº

==

«»

−

¬¼

vv and

1

5211412 34

1

0120505 43

5

CPAP

−

−− −

ªºª ºªºª º

== =

«»« »«»« »

−−−−

¬¼¬ ¼¬¼¬ ¼

. There are many choices for the muliple

of the eigenvector to use, and the solutions can look quite different. For example,

12 2

55

ii

i

+−+

ªº ª º

==−⋅ =−⋅

«» « »

−

¬¼ ¬ ¼

vw

. Using w in Theorem 9, results in

[Re Im ]P==ww

21

50

−

ªº

«»

¬¼

and

1

01 11 4 21 3 4

1.

52 20 5 50 4 3

5

CPAP

−

−−− −

ªºª ºª ºª º

== =

«»« »« »« »

−−

¬¼¬ ¼¬ ¼¬ ¼

18. 35

.

25

A

−

ªº

=«»

¬¼

The eigenvalues of A are

43,i=±

and the eigenvector 3

2

i

+

ªº

=«»

−

¬¼

v corresponds to

43.i=+

By Theorem 9,

[]

31

Re Im 02

P

ª

º

==

«

»

−

¬

¼

vv and

1

213531 43

1

032502 34

6

CPAP

−

ªºªºªºªº

== =

«»«»«»«»

−−−

¬¼¬¼¬¼¬¼

. There are many choices for the muliple of

the eigenvector to use, and the solutions can look quite different. For example,

313

22

ii

i

+−+

ªº ª º

== = −⋅ =−⋅

«» « »

−

¬¼ ¬ ¼

vw

. Using w in Theorem 9, results in

[Re Im ]P==ww

13

20

−

ª

º

«

»

¬

¼ and

1

033 5 13 43

1.

212 5 20 34

6

CPAP

−

−−

ª

ºª ºª º ª º

== =

«

»« »« » « »

−

¬

¼¬ ¼¬ ¼ ¬ ¼

19. 152 7 .

56 4

.−.

ªº

=«»

..

¬¼

A The characteristic polynomial is

2

192 1,−.+

so the eigenvalues of A are

96 28 .=. ±. i

To find an eigenvector corresponding to

96 28 ,.−.i

we compute

56 28 7

(96 28) 56 56 28

i

AiI

i

.+. −.

ªº

−.−.=

«»

.−.+.

¬¼

318 CHAPTER 5 • Eigenvalues and Eigenvectors

The equation

((9628))AiI−.−.=x0

amounts to

12

56 ( 56 28 ) 0,.+−.+. =xix

so

12

((2 ) 2)xix=−/

with

2

x free. A nice eigenvector corresponding to

96 28i.−.

is thus 2.

2

−

ª

º

=

«

»

¬

¼

vi By Theorem 9,

[]

21

Re Im 20

P

−

ª

º

==

«

»

¬

¼

vv and

1

01152 72 1 96 28

1

22 56 42 0 28 96

2

CPAP

−

.−.−.−.

ª

ºª ºª º ª º

== =

«

»« »« » « »

−.. ..

¬

¼¬ ¼¬ ¼ ¬ ¼

20. 38

.

45

A

−−

ªº

=«»

¬¼

The eigenvalues of A are

14,i=±

and the eigenvector 1i

i

+

ªº

=«»

−

¬¼

v corresponds to

14.i=+

By Theorem 9,By Theorem 9,

[]

11

Re Im 01

P

ª

º

==

«

»

−

¬

¼

vv and

1

113811 14

014501 41

CPAP

−

−−

ªºª ºªºªº

== =

«»« »«»«»

−−−

¬¼¬ ¼¬¼¬¼

. There are many choices for the muliple

of the eigenvector to use, and the solutions can look quite different. For example,

11

1

ii

i

+−+

ªº ª º

==−⋅ =−⋅

«» « »

−

¬¼ ¬ ¼

vw

. Using w in Theorem 9, results in

[Re Im ]P==ww

11

10

−

ª

º

«

»

¬

¼ and

1

01 3 8 11 14

.

11 4 5 10 4 1

CPAP

−

−−−

ª

ºª ºª º ª º

== =

«

»« »« » « »

−

¬

¼¬ ¼¬ ¼ ¬ ¼

21. The first equation in (2) is

12

(3 6) 6 0.−.+. −.=ix x We solve this for

2

x to find that

211

(( 3 6 ) 6) (( 1 2 ) 2) .=−.+. /. = −+/xixix

Letting

1

2,=x we find that 2

12i

ª

º

=

«

»

−+

¬

¼

y is an eigenvector

for the matrix A. Since

1

224

12 12

12 5

55

i

ii

i

−−

ªº ªº

−+−+

== =

«» «»

−+

¬¼ ¬¼

yv

the vector y is a complex

multiple of the vector

1

v used in Example 2.

22. Since

() () ()()===,xxxxxAA

µµ µ µµ

is an eigenvector of A.

23. (a) properties of conjugates and the fact that

TT

=xx

(b)

=xxAA

and A is real

(c)

T

Axx

is a scalar and hence may be viewed as a

11×

matrix

(d) properties of transposes

(e)

T

AA

=

and the definition of q

24.

()

TT T

A=λ=λ⋅xxx x xx

because x is an eigenvector. It is easy to see that

T

xx

is real (and positive)

because zz is nonnegative for every complex number z. Since

T

Axx

is real, by Exercise 23, so is

.

Next, write

,=+xu vi

where u and v are real vectors. Then

() andAA i AiA i=+=+ =+xuvuv xuv

The real part of Ax is Au because the entries in A, u, and v are all real. The real part of

x

is

u

because and the entries in u and v are real. Since Ax and

x

are equal, their real parts are equal,

5.5 • Solutions 319

too. (Apply the corresponding statement about complex numbers to each entry of Ax.) Thus

,=uuA

which shows that the real part of x is an eigenvector of A.

25. Write

Re (Im ),=+xx xi

so that

(Re ) (Im ).=+xx xAA iA

Since A is real, so are

(Re )Ax

and

(Im ).xA

Thus

(Re )Ax

is the real part of Ax and

(Im )Ax

is the imaginary part of Ax.

26. a. If

,=−abi

then

Re Im

()(Re Im )

( Re Im ) ( Im Re )

Av Av

Aabii

ab iab

==−+

=++ −

vv v v

vv vv

    

By Exercise 25,

(Re ) Re Re Im

(Im ) Im Re Im

AAab

AAba

== +

==−+

vvvv

vv vv

b. Let

[]

Re Im .=vvP

By (a),

(Re ) (Im )

ab

APAP

ba

−

ªº ª º

=, =

«» « »

¬¼ ¬ ¼

vv

So

[]

(Re ) (Im )=

ªº

−−

ªº ª º ª º

===

«»

«» « » « »

¬¼ ¬ ¼ ¬ ¼

¬¼

vvAP A A

ab ab

PP P PC

ba ba

27.

26 33 23 20

68113

[] 14 19 16 3

20 20 20 14

A

ªº

«»

−−−−

«»

=«»

−−−

«»

−−−−

¬¼

M

The MATLAB command [V D] = eig(A) returns

V = -0.5709 - 0.4172i -0.5709 + 0.4172i -0.3708 + 0.0732i -0.3708 - 0.0732i

0.4941 - 0.0769i 0.4941 + 0.0769i 0.4440 + 0.2976i 0.4440 - 0.2976i

0.0769 + 0.4941i 0.0769 - 0.4941i -0.4440 - 0.2976i -0.4440 + 0.2976i

-0.0000 - 0.0000i -0.0000 + 0.0000i 0.2976 - 0.4440i 0.2976 + 0.4440i

D = -2.0000 + 5.0000i 0 0 0

0 -2.0000 - 5.0000i 0 0

0 0 -4.0000 +10.0000i 0

0 0 0 -4.0000 -10.0000i

The eigenvalues of A are the elements listed on the diagonal of D. The corresponding

eigenvectors are listed in the corresponding columns of V. To get a nice eigenvector for

25i

λ

=−+

take v

1

=V(1:4,1)/V(3,1), resulting in

1

0

i

−+

ª

º

«

»

−

«

»

=

«

»

«

»

¬

¼

v

For

410,iλ=−+

take v

2

=V(1:4,3)/V(4,3), and then multiply the vector by 2 resulting in

320 CHAPTER 5 • Eigenvalues and Eigenvectors

v

2

=

1

2

i

−−

ªº

«»

−

«»

¬¼

.

Hence by Theorem 9,

1122

1111

0102

Re Im Re Im 100 2

0020

P

ªº

«»

¬¼

−−−

ª

º

«

»

−

«

»

==

«

»

−

«

»

¬

¼

vvvv

and

25 00

52 00

.

00 410

00104

C

−

ªº

«»

−−

«»

=«»

−

«»

−−

¬¼

Other choices are possible, but C must equal

1

.

−

PAP

28.

7112017

20 40 86 74

[] 051010

10 28 60 53

A

ªº

«»

−−−−

«»

=«»

−− −

«»

¬¼

M

The matlab command [V D] = eig(A) returns

V= -0.2132 + 0.2132i -0.2132 - 0.2132i 0.1085 - 0.3254i 0.1085 + 0.3254i

0 - 0.8528i 0 + 0.8528i 0.3254 + 0.1085i 0.3254 - 0.1085i

0 0 0 - 0.5423i 0 + 0.5423i

0 + 0.4264i 0 - 0.4264i -0.2169 + 0.6508i -0.2169 - 0.6508i

D = 2.0000 + 5.0000i 0 0 0

0 2.0000 - 5.0000i 0 0

0 0 3.0000 + 1.0000i 0

0 0 0 3.0000 - 1.0000

The eigenvalues of A are the elements listed on the diagonal of D. The corresponding

eigenvectors are listed in the corresponding columns of V. To get a nice eigenvector for

25i

λ

=+

take v

1

=V(1:4,1)/V(4,1), and then multiply the vector by 2 resulting in

1

4

0

2

i+

ªº

«»

−

«»

=«»

«»

¬¼

v

For

3,i=+

take v

2

=V(1:4,3)/V(4,3), and then multiply the vector by 4 resulting in

2

3

4

i

−

ªº

«»

−

«»

=«»

−+

«»

¬¼

v

5.6 • Solutions 321

Hence by Theorem 9,

1122

11 2 0

40 0 2

Re Im Re Im 00 3 1

20 4 0

P

ªº

«»

¬¼

−

ª

º

«

»

−−

«

»

==

«

»

−

«

»

¬

¼

vvvv

and

25 00

52 00

.

00 31

00 13

C

ªº

«»

−

«»

=«»

«»

−

¬¼

Other choices are possible, but C must equal

1

.

−

PAP

5.6 SOLUTIONS

1. The exercise does not specify the matrix A, but only lists the eigenvalues 3 and 1/3, and the

corresponding eigenvectors

1

ªº

=«»

¬¼

v and

2

1.

1

−

ª

º

=

«

»

¬

¼

v Also,

0

9.

1

ª

º

=

«

»

¬

¼

x

a. To find the action of A on

0

,x express

0

x in terms of

1

v and

2

.v That is, find

1

c and

2

c such

that

01122

.=+xvvcc

This is certainly possible because the eigenvectors

1

v and

2

v are linearly

independent (by inspection and also because they correspond to distinct eigenvalues) and hence

form a basis for

2

.R (Two linearly independent vectors in

2

R

automatically span

2

.R) The row

reduction

120

119 105

111 014

ªº

«»

¬¼

−

ªºª º

=«»« »

−

¬¼¬ ¼

vvx  shows that

012

54.=−xvv

Since

1

v and

2

v are eigenvectors (for the eigenvalues 3 and 1/3):

10 1 2 1 2

15 4 3 49 3

54 534(13)15 4 3 41 3

AAA

−//

ª

ºª ºª º

== −=⋅−⋅/=−=

«

»« »« »

//

¬

¼¬ ¼¬ ¼

xx v v v v

b. Each time A acts on a linear combination of

1

v and

2

,v the

1

v term is multiplied by the

eigenvalue 3 and the

2

v term is multiplied by the eigenvalue 1/3:

22

21 1 2 1 2

[5 3 4(1 3) ] 5(3) 4(1 3)AA==⋅−/= −/xx v v v v

In general,

12

5(3) 4(1 3) ,=−/xv v

kk

k

for

0.≥k

2. The vectors

123

013

357

−

ªº ªº ªº

«» «» «»

=,=,=−

«» «» «»

−−

¬¼ ¬¼ ¬¼

vv v

are eigenvectors of a

33×

matrix A, corresponding to

eigenvalues 3, 4/5, and 3/5, respectively. Also,

0

2

5.

3

−

ª

º

«

»

=−

«

»

«

»

¬

¼

x

To describe the solution of the equation

1

(12),

+

==,,xx

kk

Ak …

first write

0

x in terms of the eigenvectors.

10123

230

1232 1002

0135 0101 2 2

3573 0002

ªº

«»

¬¼

−−

ªºªº

«»«»

=−− =++

«»«»

−−

¬¼¬¼

vvvx x vv v

322 CHAPTER 5 • Eigenvalues and Eigenvectors

Then,

11231231 2 3

(2 2 ) 2 2 2 3 (4 5) 2 (3 5) .AAAA=++=++=⋅+/ +⋅/xvvvvvvv v v

In general,

12 3

23 (45) 2(35) .

kk k

k

=⋅+/ +⋅/xv v v

For all k sufficiently large,

1

23 23 0

3

kk

k

ªº

«»

≈⋅ =⋅«»

«»

−

¬¼

xv

3.

2

54

det( ) ( 5 )(1 1 ) 08 1 6 63.

211

..

ªº

=,−=. −.−+. = −.+.

«»

−..

¬¼

AAI

λλλλλ

This characteristic polynomial

factors as

(9)(7),−.−.

λλ

so the eigenvalues are .9 and .7. If

1

v and

2

v denote corresponding

eigenvectors, and if

01122

,=+xvvcc

then

1112211221122

() (9)(7)Ac c cA c A c c=+=+=.+.xvv vv v v

and for

1,≥

k

112 2

(9) (7)

kk

k

cc=. + .xvv

For any choices of

1

c and

2

,c both the owl and wood rat populations decline over time.

4.

2

54

det( ) ( 5 )(1 1 ) ( 4)( 125) 1 6 6.

125 1 1

..

ªº

=,−=. −.−−.. = −.+.

«»

−..

¬¼

AAI

λλλ λλ

This characteristic

polynomial factors as

(1)( 6),−−.

λλ

so the eigenvalues are 1 and .6. For the eigenvalue 1, solve

540 540

()0 .

125 1 0 0 0 0

−.. −

ªºªº

−=:

«»«»

−..

¬¼¬¼

xAI A basis for the eigenspace is

1

4.

5

ªº

=«»

¬¼

v Let

2

v be an

eigenvector for the eigenvalue .6. (The entries in

2

v are not important for the long-term behavior of

the system.) If

01122

,=+xvvcc

then

11122 112 2

(6) ,=+ =+.xv vv vcA c A c c and for k sufficiently

large,

1221

44

(6)

55

k

cc c

ªº ªº

=+.≈

«» «»

¬¼ ¬¼

xv

Provided that

1

0,≠c the owl and wood rat populations each stabilize in size, and eventually the

populations are in the ratio of 4 owls for each 5 thousand rats. If some aspect of the model were to

change slightly, the characteristic equation would change slightly and the perturbed matrix A might

not have 1 as an eigenvalue. If the eigenvalue becomes slightly larger than 1, the two populations

will grow; if the eigenvalue becomes slightly less than 1, both populations will decline.

5.

2

43

det( ) 1 6 5775.

325 1 2

..

ªº

=,−=−.+.

«»

−..

¬¼

AAI

λλ λ

The quadratic formula provides the roots of the

characteristic equation:

2

16 16 4(5775) 16 25 105 and 55

22

λ

.± . −..± .

===..

Because one eigenvalue is larger than one, both populations grow in size. Their relative sizes are

determined eventually by the entries in the eigenvector corresponding to 1.05. Solve

(105) :−.=x0AI

1

65 3 0 13 6 0 6

An eigenvector is

325 15 0 0 0 0 13

−.. −

ªºªº ªº

.=.

«»«» «»

−..

¬¼¬¼ ¬¼

v

5.6 • Solutions 323

Eventually, there will be about 6 spotted owls for every 13 (thousand) flying squirrels.

6. When 43

5,

512

..

ªº

=. , =«»

−..

¬¼

pA and

2

det( ) 1 6 63 ( 9)( 7).−=−.+.=−.−.AI

λλ λ λ λ

The eigenvalues of A are .9 and .7, both less than 1 in magnitude. The origin is an attractor for the

dynamical system and each trajectory tends toward 0. So both populations of owls and squirrels

eventually perish.

The calculations in Exercise 4 (as well as those in Exercises 27 and Exercise 33 in Section 5.2) show

that if the largest eigenvalue of A is 1, then in most cases the population vector

k

x will tend toward a

multiple of the eigenvector corresponding to the eigenvalue 1. [If

1

v and

2

v are eigenvectors, with

1

v corresponding to

1,=

λ

and if

01122

,=+xvvcc

then

k

x tends toward

11

,vc provided

1

c is not

zero.] So the problem here is to determine the value of the predation parameter p such that the largest

eigenvalue of A is 1. Compute the characteristic polynomial:

2

43

det ( 4 )(1 2 ) 3 1 6 ( 48 3 )

12 pp

p

λλλ λλ

λ

.−.

ªº

=. −.−+. = −.+.+.

«»

−.−

¬¼

By the quadratic formula,

2

16 16 4(48 3 )

2

p

λ

.± . −.+.

=

The larger eigenvalue is 1 when

2

16 16 4(48 3 ) 2 and 256 192 12 4pp.+ . −.+. = .−.−.=.

In this case,

64 1 2 16,.−.=.p

and

4.p=.

7. a. The matrix A in Exercise 1 has eigenvalues 3 and 1/3. Since

31

||

>

and

13 1,

|

/

|

<

the origin is a

saddle point.

b. The direction of greatest attraction is determined by

2

1,

1

−

ª

º

=

«

»

¬

¼

v the eigenvector corresponding to

the eigenvalue with absolute value less than 1. The direction of greatest repulsion is determined

by

1

1,

1

ªº

=«»

¬¼

v the eigenvector corresponding to the eigenvalue greater than 1.

c. The drawing below shows: (1) lines through the eigenvectors and the origin, (2) arrows toward

the origin (showing attraction) on the line through

2

v and arrows away from the origin (showing

repulsion) on the line through

1

,v (3) several typical trajectories (with arrows) that show the

general flow of points. No specific points other than

1

v and

2

v were computed. This type of

324 CHAPTER 5 • Eigenvalues and Eigen

v

r

drawing is about all that one ca

n

Note:

If you wish your class to sketch

t

beyond the discussion in the text. The f

o

Sketching trajectories for a dynami

c

difficult than the sketch in Exercise

7

trajectories “bend” as they move towar

d

Section 5.6 through a quarter-turn and

r

figure corresponds to the matrix A with

diagonal matrix, with positive diagonal

or on curves whose equations have th

e

initial point

0

.x (See Encounters with C

h

150.)

8. The matrix from Exercise 2 has eig

e

and the others are less than one in

m

repulsion is the line through the ori

g

direction of greatest attraction is th

e

smallest eigenvalue 3/5.

9.

2

17 3 det( )

12 8

AAI

.−.

ªº

=,−λ =

−

«»

−..

¬¼

2

25 25 4(1) 25

2

.± . −.±

==

The origin is a saddle point becaus

e

than 1 in magnitude. The direction

o

found below. Solve (2)AI

ª

−=:

«

¬

x0

1

1.

1

−

ªº

=«»

¬¼

v The direction of greates

v

ectors

r

son Education, Inc. Publishing as Addison-Wesley.

n

make without using a computer to plot points.

t

rajectories for anything except saddle points, you wi

l

o

llowing remarks from the Study Guide are relevant.

c

al system in which the origin is an attractor or a rep

e

7

. There has been no discussion of the direction i

n

d

or away from the origin. For instance, if you rotat

e

r

elabel the axes so that

1

x is on the horizontal axis,

t

the diagonal entries .8 and .64 interchanged. In gen

e

entries a and d, unequal to 1, then the trajectories li

e

form

21

(),=

s

xrx

where

(ln ) (ln )sda=/

and r de

p

Ch

aos, by Denny Gulick, New York: McGraw-Hill, 19

e

nvalues 3, 4/5, and 3/5. Since one eigenvalue is grea

t

m

agnitude, the origin is a saddle point. The direction

o

g

in and the eigenvector

(1 0 3),,−

for the eigenvalue 3.

e

line through the origin and the eigenvector

(3 37

)

−,−,

25 10

−

.+=

225 25 15 2 and 5

22

..±.

==.

e

one eigenvalue is greater than 1 and the other eigenv

o

f greatest repulsion is through the origin and the eig

e

330110

,

12 12 0 0 0 0

−.−.

ª

ºª º

»« »

−.−.

¬

¼¬ ¼

 so x

1

= –x

2

, and x

2

i

s

t attraction is through the origin and the eigenvector

v

l

l need to go

e

llor is more

n

which the

e

Figure 1 of

t

hen the new

e

ral, if A is a

e

on the axes

p

ends on the

92, pp. 147–

t

er than 1

o

f greatest

The

)

for the

alue is less

e

nvector

1

v

s

free. Take

2

v

found

5.6 • Solutions 325

below. Solve 12 3 0 1 25 0

(5) ,

12 3 0 0 0 0

.−.−.

ªºªº

−.=:

«»«»

−..

¬¼¬¼

x0 AI so

12

25 ,xx=. and

2

x is free. Take

2

1.

4

ªº

=«»

¬¼

v

10.

2

34

det( )14 45 0

311

AAI

..

ªº

=,−=−.+.=

«»

−..

¬¼

2

14 14 4(45) 14 16 14 4 5 and 9

222

.± . −..± . .±.

====..

The origin is an attractor because both eigenvalues are less than 1 in magnitude. The direction of

greatest attraction is through the origin and the eigenvector

1

v found below. Solve

240 1 20

(5) ,

360 0 00

−.. −

ªºªº

−.=:

«»«»

−..

¬¼¬¼

x0 AI so

12

2,=xx

and

2

x is free. Take

1

2.

1

ªº

=«»

¬¼

v

11.

2

45

det( )17 72 0

413

AAI

..

ªº

=,−=−.+.=

«»

−..

¬¼

2

17 17 4(72) 17 01 17 1 8 and 9

222

.± . −..± . .±.

====..

The origin is an attractor because both eigenvalues are less than 1 in magnitude. The direction of

greatest attraction is through the origin and the eigenvector

1

v found below. Solve

450 1 1250

(8) ,

450 0 00

−.. −.

ªºª º

−.=:

«»« »

−..

¬¼¬ ¼

x0 AI so

12

125 ,=.xx

and

2

x is free. Take

1

5.

4

ªº

=«»

¬¼

v

12.

2

56

det( )19 88 0

314

AAI

..

ªº

=,−=−.+.=

«»

−..

¬¼

2

19 19 4(88) 19 09 19 3 8 and 1 1

222

.± . −..± . .±.

====..

The origin is a saddle point because one eigenvalue is greater than 1 and the other eigenvalue is less

than 1 in magnitude. The direction of greatest repulsion is through the origin and the eigenvector

1

v

found below. Solve 660 1 10

(11) ,

330 0 00

−.. −

ªºªº

−.=:

«»«»

−..

¬¼¬¼

x0 AI so

12

,=xx

and

2

x is free. Take

1

1.

1

ªº

=«»

¬¼

v The direction of greatest attraction is through the origin and the eigenvector

2

v found

below. Solve 360 1 20

(8) ,

360 0 00

−.. −

ªºªº

−.=:

«»«»

−..

¬¼¬¼

x0 AI so

12

2,=xx

and

2

x is free. Take

2

2.

1

ªº

=«»

¬¼

v

326 CHAPTER 5 • Eigenvalues and Eigenvectors

13.

2

83

det( )23 132 0

415

AAI

..

ªº

=,−=−.+.=

«»

−..

¬¼

2

23 23 4(132) 23 01 23 1 11 and 12

222

.± . −..± . .±.

====..

The origin is a repellor because both eigenvalues are greater than 1 in magnitude. The direction of

greatest repulsion is through the origin and the eigenvector

1

v found below. Solve

430 1 750

(12) ,

430 0 00

−.. −.

ªºª º

−.=:

«»« »

−..

¬¼¬ ¼

x0 AI so

12

75 ,=.xx

and

2

x is free. Take

1

3.

4

ªº

=«»

¬¼

v

14.

2

17 6 det( )24 143 0

47

AAI

..

ªº

=,−=−.+.=

«»

−..

¬¼

2

24 24 4(143) 24 04 24 2 11 and 13

222

.± . −..± . .±.

====..

The origin is a repellor because both eigenvalues are greater than 1 in magnitude. The direction of

greatest repulsion is through the origin and the eigenvector

1

v found below. Solve

4601150

(13) ,

460000

.. .

ªºªº

−.=:

«»«»

−.−.

¬¼¬¼

x0 AI so

12

15 ,=−.xx

and

2

x is free. Take

1

3.

2

−

ªº

=«»

¬¼

v

15.

40 2

383.

325

..

ªº

«»

=. . .

«»

...

¬¼

A

Given eigenvector

1

6

3

.

ª

º

«

»

=.

«

»

«

»

.

¬

¼

v

and eigenvalues .5 and .2. To find the eigenvalue for

1

,v compute

111

40 21 1

3836 61 Thusisaneigenvectorfor 1

3253 3

A

λ

....

ªºªºªº

«»«»«»

=. . . . =. =⋅=.

«»«»«»

.... .

¬¼¬¼¬¼

vvv

13

23 2

3

1020 10 20 2 2

For 5 3 3 3 0 0 1 3 0 3 . Set 3

3200 00 00 isfree 1

−.. −=

ªºªº ªº

«»«» «»

=. : . . . , =−=−.

«»«» «»

..

¬¼¬¼ ¬¼

v

xx

x

λ

13

23

3

2020 1010 1

For 2 3 6 3 0 0 1 0 0 , 0 . Set 0

3230 0000 is free 1

.. =−−

ªºªº ªº

«»«» «»

=. : . . . = =

«»«» «»

...

¬¼¬¼ ¬¼

v

xx

x

λ

Given

0

(0 3 7),=,.,.x find weights such that

011 233

.=++xvvvccc

1230

12 10 1001

6303 0101.

3117 0003

ªº

«»

¬¼

.−

ªºªº

«»«»

=. −..

«»«»

.. .

¬¼¬¼

vvvx 

5.6 • Solutions 327

01 2 3

11 2 31 2 3

12 3 1

13

1 3 1( 5) 3( 2) and

1( 5) 3( 2) As increases approaches

kk

AA A

k

=+. +.

=+. +. =+..+..,

=+.. +.. . , .

xv v v

xv v vv v v

xv v v x v

16. [M]

90 01 09 1 0000

01 90 01 0 8900 To four decimal places

09 09 90 8100

09192 9199

01919 Exact 1999

10000 1

... .

ªº ªº

«» «»

=. . . ⋅=. . ,

«» «»

... .

¬¼ ¬¼

./

ªº ªº

«» «»

.. :/

«» «»

.

¬¼ ¬¼

1

2

ev = eig(A)

v=nulbasis(A-eye(3))=

v=nulbasis(A-ev(2)

A

1

0

1

0

1

−

ªº

«»

¬¼

−

ªº

«»

¬¼

3

*eye(3))=

v=nulbasis(A-ev(3)*eye(3))=

The general solution of the dynamical system is

11 2 2 3 3

(89) (81) .=+. +.xv v v

kk

k

cc c

Note:

When working with stochastic matrices and starting with a probability vector (having nonnegative

entries whose sum is 1), it helps to scale

1

v to make its entries sum to 1. If

1

(91 209 19 209 99 209),=/,/,/v or

(435 091 474).,.,.

to three decimal places, then the weight

1

c above

turns out to be 1. See the text’s discussion of Exercise 27 in Section 5.2.

17. a. 016

38

A.

ªº

=«»

..

¬¼

b.

2

16

det 8 48 0.

38

−.

ªº

=−.−.=

«»

..−

¬¼

λλλ

λ

The eigenvalues of A are given by

2

8(8)4(48)

8256816

12 and 4

222

λ

.± −.−−..± . .±.

====.−.

The numbers of juveniles and adults are increasing because the largest eigenvalue is greater than

1. The eventual growth rate of each age class is 1.2, which is 20% per year.

To find the eventual relative population sizes, solve

(12) :−.=x0AI

12

1

2

(4 3)

12 16 0 1 43 0 4

Set

is free

340000 3

xx

x

=/

−.. −/

ªºªº ªº

..=.

«»«» «»

.−.

¬¼¬¼ ¬¼

v

Eventually, there will be about 4 juveniles for every 3 adults.

328 CHAPTER 5 • Eigenvalues and Eigenvectors

c. [M] Suppose that the initial populations are given by

0

(15 10).=,x The Study Guide describes

how to generate the trajectory for as many years as desired and then to plot the values for each

population. Let (j a ).=,x

kkk

Then we need to plot the sequences {j } {a } {j a },,,+

kkkk

and

{j a }./

kk

Adjacent points in a sequence can be connected with a line segment. When a sequence is

plotted, the resulting graph can be captured on the screen and printed (if done on a computer) or

copied by hand onto paper (if working with a graphics calculator).

18. a.

0042

60 0

07595

A

.

ªº

«»

=.

«»

..

¬¼

b.

00774 04063

11048

i

.+.

ªº

«»

.−.

«»

.

¬¼

ev = eig(A)=

The long-term growth rate is 1.105, about 10.5 % per year.

03801

0 2064

10000

.

ª

º

«

»

=.

«

»

«

»

.

¬

¼

v=nulbasis(A-ev(3)*eye(3))

For each 100 adults, there will be approximately 38 calves and 21 yearlings.

Note:

The MATLAB box in the Study Guide and the various technology appendices all give directions

for generating the sequence of points in a trajectory of a dynamical system. Details for producing a

graphical representation of a trajectory are also given, with several options available in MATLAB, Maple,

and Mathematica.

5.7 SOLUTIONS

1. From the “eigendata” (eigenvalues and corresponding eigenvectors) given, the eigenfunctions for the

differential equation

A′=xx

are

4

1

t

ev

and

2

.v

t

e

The general solution of

A′=xx

has the form

42

12

31

11

tt

cece

−−

ªº ªº

+

«» «»

¬¼ ¬¼

The initial condition 6

(0) 1

−

ªº

=«»

¬¼

x determines

1

c and

2

:c

4(0) 2(0)

12

316

111

316 1052

111 0132

−−−

ªº ªº ªº

+=

«» «» «»

¬¼ ¬¼ ¬¼

−−− /

ªºªº

«»«»

−/

¬¼¬¼



cece

Thus

12

52 32,=/, =−/cc and

42

31

53

() .

11

22

−−

ª

ºªº

=−

«

»«»

¬

¼¬¼

x

tt

te e

5.7 • Solutions 329

2. From the eigendata given, the eigenfunctions for the differential equation

A′=xx

are

3

1

t

e

−

v

and

1

2.

−

v

t

e

The general solution of

A′=xx

has the form

31

12

11

tt

cece

−−

−

ªº ªº

+

«» «»

¬¼ ¬¼

The initial condition 2

(0) 3

ªº

=«»

¬¼

x determines

1

c and

2

c:

3(0) 1(0)

12

112

113

112 1 0 12

11 3 0 1 52

ce ce

−−

−

ªº ªº ªº

+=

«» «» «»

¬¼ ¬¼ ¬¼

−/

ªºª º

«»« »

/

¬¼¬ ¼



Thus

12

12 52,=/, = /cc and

3

11

15

() .

11

22

−−

−

ªº ªº

=+

«» «»

¬¼ ¬¼

x

tt

tee

3.

2

23

det( )1( 1)( 1) 0.

12

ªº

=,−=−=−+=

«»

−−

¬¼

AAI Eigenvalues: 1 and

1.−

For λ = 1: 130 130

,

130 000

ªºªº

«»«»

−−

¬¼¬¼

 so

12

3xx=− with

2

x free. Take

2

1x= and

1

3.

1

−

ªº

=«»

¬¼

v

For λ = –1: 330 110

,

110 000

ªºªº

«»«»

−−

¬¼¬¼

 so

12

xx=− with

2

x free. Take

2

1x= and

2

1.

1

−

ªº

=«»

¬¼

v

For the initial condition 3

(0) ,

2

ªº

=«»

¬¼

x find

1

c and

2

c such that

11 2 2

(0) :+=vvxcc

12

313 1052

(0) 112 0192

−− −/

ªºª º

=

ªº

«»« »

¬¼ /

¬¼¬ ¼

vvx 

Thus

12

52 92,=−/, = /cc

and 31

59

() .

22

11

−

−−

ªº ªº

=−+

«» «»

¬¼ ¬¼

x

tt

tee

Since one eigenvalue is positive and the other is negative, the origin is a saddle point of the

dynamical system described by

.′=xxA

The direction of greatest attraction is the line through

2

v

and the origin. The direction of greatest repulsion is the line through

1

v and the origin.

4.

2

25

det( ) 2 3( 1)( 3) 0.

14

AAI

−−

ªº

=,−=−−=+ −=

«»

¬¼ Eigenvalues:

1−

and 3.

For λ = 3: 550 110

,

110 000

−−

ªºªº

«»«»

¬¼¬¼

 so

12

xx=− with

2

x free. Take

2

1x= and

1

1.

1

−

ªº

=«»

¬¼

v

For λ = –1: 150 150

,

150 000

−−

ªºªº

«»«»

¬¼¬¼

 so

12

5xx=− with

2

x free. Take

2

1x= and

2

5.

1

−

ªº

=«»

¬¼

v

330 CHAPTER 5 • Eigenvalues and Eigenvectors

For the initial condition 3

(0) ,

2

ªº

=«»

¬¼

x find

1

c and

2

c such that

11 2 2

(0)cc+=vvx

:

12

153 10134

(0) 112 0154

−− /

ªºª º

=

ªº

«»« »

¬¼ −/

¬¼¬ ¼

vvx 

Thus

12

13 4 5 4,=/,=−/cc and

3

15

13 5

() .

11

44

−

−−

ªº ªº

=−

«» «»

¬¼ ¬¼

x

tt

tee

Since one eigenvalue is positive and the other is negative, the origin is a saddle point of the

dynamical system described by

.′=xxA

The direction of greatest attraction is the line through

2

v

and the origin. The direction of greatest repulsion is the line through

1

v and the origin.

5. 71

,

33

−

ªº

=«»

¬¼

A det

2

( ) 10 24 ( 4)( 6) 0.−=−+=−−=AI

Eigenvalues: 4 and 6.

For λ = 4: 310 1130

,

310 0 00

−−/

ªºª º

«»« »

−

¬¼¬ ¼

 so

12

(1 3)xx=/ with

2

x free. Take

2

3x= and

1

1.

3

ª

º

=

«

»

¬

¼

v

For λ = 6: 110 110

,

330 000

−−

ªºªº

«»«»

−

¬¼¬¼

 so

12

xx= with

2

x free. Take

2

1x= and

2

1.

1

ªº

=«»

¬¼

v

For the initial condition 3

(0) ,

2

ªº

=«»

¬¼

x find

1

c and

2

c such that

11 2 2

(0) :+=vvxcc

12

113 10 12

(0) 312 0 1 72

−/

ªºª º

=

ªº

«»« »

¬¼ /

¬¼¬ ¼

vvx 

Thus

12

12 72,=−/, = /cc

and

46

11

17

() .

31

22

ª

ºªº

=−+

«

»«»

¬

¼¬¼

x

tt

tee

Since both eigenvalues are positive, the origin is a repellor of the dynamical system described by

.′=xxA

The direction of greatest repulsion is the line through

2

v and the origin.

6. 12

,

34

−

ªº

=«»

−

¬¼

A det

2

( ) 3 2( 1)( 2) 0.−=++=+ +=AI

Eigenvalues:

1−

and

2.−

For λ = –2: 320 1230

,

320 0 00

−−/

ªºª º

«»« »

−

¬¼¬ ¼

 so

12

(2 3)xx=/ with

2

x free. Take

2

3x= and

1

2.

3

ª

º

=

«

»

¬

¼

v

For λ = –1: 220 110

,

330 000

−−

ªºªº

«»«»

−

¬¼¬¼

 so

12

xx= with

2

x free. Take

2

1x= and

2

1.

1

ªº

=«»

¬¼

v

For the initial condition 3

(0) ,

2

ªº

=«»

¬¼

x find

1

c and

2

c such that

11 2 2

(0)cc+=vvx

:

12

213 10 1

[(0)]

312 0 1 5

−

ªºª º

=«»« »

¬¼¬ ¼

vvx 

5.7 • Solutions 331

Thus

12

15,=−,=cc

and

2

21

() 5 .

31

−−

ªº ªº

=−+

«» «»

¬¼ ¬¼

x

tt

te e

Since both eigenvalues are negative, the origin is an attractor of the dynamical system described by

.′=xxA

The direction of greatest attraction is the line through

1

v and the origin.

7. From Exercise 5, 71

,

33

−

ª

º

=

«

»

¬

¼

A with eigenvectors

1

3

ª

º

=

«

»

¬

¼

v and

2

1

ª

º

=

«

»

¬

¼

v corresponding to

eigenvalues 4 and 6 respectively. To decouple the equation

,′=xxA

set

12

11

[]

31

ªº

==

«»

¬¼

vvP and let

40

,

06

ªº

=«»

¬¼

D so that

1

APDP

−

=

and

1

.

−

=DPAP

Substituting

() ()tPt=xy

into

A

′=

xx

we have

1

() () ()

−

== =yy yy

dPAPPDPPPD

dt

Since P has constant entries,

() (()),=yy

dd

dt dt

PP

so that left-multiplying the equality

(())=yy

d

dt

PPD

by

1

P

−

yields ,′=yyD or

11

22

() ()

40

() ()

06

′

ªº ªºªº

=

«» «»

«»

′¬¼¬¼ ¬¼

yt yt

8. From Exercise 6, 12

,

34

−

ª

º

=

«

»

−

¬

¼

A with eigenvectors

1

2

3

ª

º

=

«

»

¬

¼

v and

2

1

ª

º

=

«

»

¬

¼

v corresponding to

eigenvalues 2− and 1− respectively. To decouple the equation ,′=xxA set

12

21

31

P

ªº

«»

¬¼

ªº

==

«»

¬¼

vv

and let 20

,

01

−

ªº

=«»

−

¬¼

D so that

1

APDP

−

= and

1

.

−

=DPAP

Substituting () ()tPt=xy into A′=xx

we have

1

() () ()

dPAPPDPPPD

dt

−

== =yy yy

Since P has constant entries,

()

() (),=yy

dd

dt dt

PP

so that left-multiplying the equality

()

d

dt

PPD=yy

by

1

P

−

yields ,′=yyD or

11

22

() ()

20

() ()

01

′−

ªº ªºªº

=

«» «»

«»

′−

¬¼¬¼ ¬¼

yt yt

9. 32

.

11

−

ªº

=«»

−−

¬¼

A An eigenvalue of A is

2i−+

with corresponding eigenvector 1.

1

−

ªº

=«»

¬¼

vi The

complex eigenfunctions

t

e

λ

v and

λ

v

t

e form a basis for the set of all complex solutions to

.′=xxA

The general complex solution is

(2 ) (2 )

12

11

it it

ii

ce c e

−+−−

−+

ªº ªº

+

«» «»

¬¼ ¬¼

332 CHAPTER 5 • Eigenvalues and Eigenvectors

where

1

c and

2

c are arbitrary complex numbers. To build the general real solution, rewrite

(2 )it

e

−+

v

as:

(2 ) 2 2

2

22

11

(cos sin )

11

cos cos sin sin

cos sin

cos sin sin cos

cos sin

−+−−

−

−−

ªº ªº

== +

«» «»

¬¼ ¬¼

ªº

−+−

=«»

+

¬¼

+−

ªºªº

=+

«»«»

¬¼¬¼

v

it t it t

t

tt

ii

eeeetit

ti ti ti t

e

ti t

tt t t

ei e

tt

The general real solution has the form

22

12

cos sin sin cos

cos sin

tt

tt t t

cec e

tt

−−

+−

ªºªº

+

«»«»

¬¼¬¼

where

1

c and

2

c now are real numbers. The trajectories are spirals because the eigenvalues are

complex. The spirals tend toward the origin because the real parts of the eigenvalues are negative.

10. 31

.

21

ªº

=«»

−

¬¼

A An eigenvalue of A is

2i+

with corresponding eigenvector 1.

2

+

ªº

=«»

−

¬¼

v

i

The complex

eigenfunctions

t

ev and

t

ev

form a basis for the set of all complex solutions to

.A′=xx

The

general complex solution is

(2 ) (2 )

12

11

22

it it

ii

cec e

+−

ªº ªº

+

«» «»

−−

¬¼ ¬¼

where

1

c and

2

c are arbitrary complex numbers. To build the general real solution, rewrite

(2 )it

e

+

v

as:

(2 ) 2 2

2

22

11

(cos sin )

22

cos cos sin sin

2cos 2 sin

cos sin sin cos

2cos 2sin

it t it t

t

tt

ii

eeeetit

ti ti ti t

e

tit

tt t t

ei e

tt

+

++

ªº ªº

== +

«» «»

−−

¬¼ ¬¼

ªº

+++

=«»

−−

¬¼

−+

ªºªº

=+

«»«»

−−

¬¼¬¼

v

The general real solution has the form

22

12

cos sin sin cos

2cos 2sin

tt

tt t t

cec e

tt

−+

ªºªº

+

«»«»

−−

¬¼¬¼

where

1

c and

2

c now are real numbers. The trajectories are spirals because the eigenvalues are

complex. The spirals tend away from the origin because the real parts of the eigenvalues are positive.

11. 39

.

23

−−

ªº

=«»

¬¼

A An eigenvalue of A is 3i with corresponding eigenvector 33.

2

−+

ªº

=«»

¬¼

v

i

The

complex eigenfunctions

t

ev and

t

ev

form a basis for the set of all complex solutions to

.A′=xx

The general complex solution is

5.7 • Solutions 333

(3 ) ( 3 )

12

33 33

22

it it

ii

cec e

−

−+−−

ªº ªº

+

«» «»

¬¼ ¬¼

where

1

c and

2

c are arbitrary complex numbers. To build the general real solution, rewrite

(3 )it

ev

as:

(3 ) 33(cos3 sin 3 )

2

3cos3 3sin3 3sin3 3cos3

2cos3 2sin3

it i

etit

tt t t

i

tt

−+

ªº

=+

«»

¬¼

−− −+

ªºªº

=+

«»«»

¬¼¬¼

v

The general real solution has the form

12

3cos3 3sin3 3sin3 3cos3

2cos3 2sin3

tt t t

cc

tt

−− −+

ªºªº

+

«»«»

¬¼¬¼

where

1

c and

2

c now are real numbers. The trajectories are ellipses about the origin because the real

parts of the eigenvalues are zero.

12. 710

.

45

−

ªº

=«»

−

¬¼

A An eigenvalue of A is

12i−+

with corresponding eigenvector 3.

2

−

ªº

=«»

¬¼

v

i

The

complex eigenfunctions

t

ev and

t

ev

form a basis for the set of all complex solutions to

.A′=xx

The general complex solution is

(12) (12)

12

33

22

it it

ii

ce c e

−+−−

−+

ªº ªº

+

«» «»

¬¼ ¬¼

where

1

c and

2

c are arbitrary complex numbers. To build the general real solution, rewrite

(12)it

e

−+

v

as:

(12) 3(cos 2 sin 2 )

2

3cos2 sin 2 3sin2 cos2

2cos2 2sin2

it t

tt

i

eetit

tt t t

ei e

tt

−+−

−−

−

ªº

=+

«»

¬¼

+−

ªºªº

=+

«»«»

¬¼¬¼

v

The general real solution has the form

12

3cos2 sin2 3sin 2 cos2

2cos2 2sin2

tt

tt t t

cec e

tt

−−

+−

ªºªº

+

«»«»

¬¼¬¼

where

1

c and

2

c now are real numbers. The trajectories are spirals because the eigenvalues are

complex. The spirals tend toward the origin because the real parts of the eigenvalues are negative.

13. 43

.

62

−

ªº

=«»

−

¬¼

A An eigenvalue of A is

13i+

with corresponding eigenvector 1.

2

+

ªº

=«»

¬¼

v

i

The

complex eigenfunctions

t

ev and

t

ev

form a basis for the set of all complex solutions to

.A′=xx

The general complex solution is

(1 3 ) (1 3 )

12

11

22

it it

ii

cec e

+−

ªº ªº

+

«» «»

¬¼ ¬¼

334 CHAPTER 5 • Eigenvalues and Eigenvectors

where

1

c and

2

c are arbitrary complex numbers. To build the general real solution, rewrite

(1 3 )it

e

+

v

as:

(1 3 ) 1(cos 3 sin 3 )

2

cos 3 sin 3 sin 3 cos3

2cos3 2sin3

it t

tt

i

eetit

tt t t

ei e

tt

++

ªº

=+

«»

¬¼

−+

ªºªº

=+

«»«»

¬¼¬¼

v

The general real solution has the form

12

cos 3 sin 3 sin 3 cos3

2cos3 2sin3

tt

tt t t

cec e

tt

−+

ªºªº

+

«»«»

¬¼¬¼

where

1

c and

2

c now are real numbers. The trajectories are spirals because the eigenvalues are

complex. The spirals tend away from the origin because the real parts of the eigenvalues are positive.

14. 21

.

82

−

ªº

=«»

−

¬¼

A An eigenvalue of A is 2i with corresponding eigenvector 1.

4

−

ªº

=«»

¬¼

v

i

The complex

eigenfunctions

t

ev and

t

ev

form a basis for the set of all complex solutions to

.A′=xx

The

general complex solution is

(2 ) ( 2 )

12

11

44

it it

ii

cec e

−

−+

ªº ªº

+

«» «»

¬¼ ¬¼

where

1

c and

2

c are arbitrary complex numbers. To build the general real solution, rewrite

(2 )it

ev

as:

(2 ) 1(cos 2 sin 2 )

4

cos 2 sin 2 sin 2 cos 2

4cos2 4sin2

it i

etit

tt t t

i

tt

−

ªº

=+

«»

¬¼

+−

ªºªº

=+

«»«»

¬¼¬¼

v

The general real solution has the form

12

cos 2 sin 2 sin 2 cos 2

4cos2 4sin2

tt t t

cc

tt

+−

ªºªº

+

«»«»

¬¼¬¼

where

1

c and

2

c now are real numbers. The trajectories are ellipses about the origin because the real

parts of the eigenvalues are zero.

15. [M]

8126

212.

7125

−− −

ªº

«»

=«»

«»

¬¼

A

The eigenvalues of A are:

ev = eig(A)=

1.0000

-1.0000

-2.0000

nulbasis(A-ev(1)*eye(3))

=

5.7 • Solutions 335

-1.0000

0.2500

1.0000

so that

1

4

1

4

−

ªº

«»

=«»

«»

¬¼

v

nulbasis(A-ev(2)*eye(3))

=

-1.2000

0.2000

1.0000

so that

2

6

1

5

−

ªº

«»

=«»

«»

¬¼

v

nulbasis (A-ev(3)*eye(3))

=

-1.0000

0.0000

1.0000

so that

3

1

0

1

−

ªº

«»

=«»

«»

¬¼

v

Hence the general solution is

2

12 3

46 1

() 1 1 0 .

45 1

tt t

tc ec e c e

−−

−− −

ªº ªº ªº

«» «» «»

=+ +

«» «» «»

¬¼ ¬¼ ¬¼

x

The origin is a saddle point.

A solution with

1

0c= is attracted to the origin while a solution with

23

0cc==

is repelled.

336 CHAPTER 5 • Eigenvalues and Eigenvectors

16. [M]

61116

254.

4510

−−

ªº

«»

=−

«»

−−

¬¼

A

The eigenvalues of A are:

ev = eig(A)=

4.0000

3.0000

2.0000

nulbasis(A-ev(1)*eye(3))

=

2.3333

-0.6667

1.0000

so that

1

7

2

3

ªº

«»

=−

«»

¬¼

v

nulbasis(A-ev(2)*eye(3))

=

3.0000

-1.0000

1.0000

so that

2

3

1

ªº

«»

=−

«»

¬¼

v

nulbasis(A-ev(3)*eye(3))

=

2.0000

0.0000

1.0000

so that

3

2

0

1

ªº

«»

=«»

«»

¬¼

v

Hence the general solution is

432

123

732

() 2 1 0 .

311

ttt

tc e c e c e

ª

ºªºªº

«

»«»«»

=−+−+

«

»«»«»

«

»«»«»

¬

¼¬¼¬¼

x

The origin is a repellor,

because all eigenvalues are positive. All trajectories tend away from the origin.

5.7 • Solutions 337

17. [M]

30 64 23

11 23 9 .

6154

ªº

«»

=−− −

«»

¬¼

A

The eigenvalues of A are:

ev = eig(A)=

5.0000 + 2.0000i

5.0000 - 2.0000i

1.0000

nulbasis(A-ev(1)*eye(3))

=

7.6667 - 11.3333i

-3.0000 + 4.6667i

1.0000

so that

1

23 34

914

3

i

−

ªº

«»

=−+

«»

¬¼

v

nulbasis (A-ev(2)*eye(3))

=

7.6667 + 11.3333i

-3.0000 - 4.6667i

1.0000

so that

2

23 34

914

3

i

+

ªº

«»

=−−

«»

¬¼

v

nulbasis (A-ev(3)*eye(3))

=

-3.0000

1.0000

so that

3

1

−

ªº

«»

=«»

«»

¬¼

v

Hence the general complex solution is

(5 2 ) (5 2 )

123

23 34 23 34 3

() 9 14 9 14 1

331

it it t

ii

tc ie c ie c e

+−

−+−

ªº ªº ªº

«» «» «»

=−++−− +

«» «» «»

¬¼ ¬¼ ¬¼

x

Rewriting the first eigenfunction yields

555

23 34 23cos 2 34sin 2 23sin 2 34cos 2

914 (cos2 sin2) 9cos2 14sin2 9sin2 14cos2

33cos23sin2

ttt

itttt

ie t i t t te i t te

tt

−+−

ªº ª ºª º

«» « »« »

−++=−− +−+

«» « »« »

¬¼ ¬ ¼¬ ¼

Hence the general real solution is

338 CHAPTER 5 • Eigenvalues and Eigenvectors

55

123

23cos 2 34sin 2 23sin 2 34cos 2 3

() 9cos2 14sin2 9sin2 14cos2 1

3cos2 3sin2 1

ttt

tt t t

tc t te c t te c e

tt

+−−

ªºªºªº

«»«»«»

=−− +−++

«»«»«»

¬¼¬¼¬¼

x

where

12

,,cc and

3

c are real. The origin is a repellor, because the real parts of all eigenvalues are

positive. All trajectories spiral away from the origin.

18. [M]

53 30 2

90 52 3 .

20 10 2

A

−−

ªº

«»

=−−

«»

−

¬¼

The eigenvalues of A are:

ev = eig(A)=

-7.0000

5.0000 + 1.0000i

5.0000 - 1.0000i

nulbasis(A-ev(1)*eye(3))

=

0.5000

1.0000

0.0000

so that

1

2

0

ªº

«»

=«»

«»

¬¼

v

nulbasis(A-ev(2)*eye(3))

=

0.6000 + 0.2000i

0.9000 + 0.3000i

1.0000

so that

2

62

93

10

i

+

ªº

«»

=+

«»

¬¼

v

nulbasis(A-ev(3)*eye(3))

=

0.6000 - 0.2000i

0.9000 - 0.3000i

1.0000

so that

3

62

93

10

i

−

ªº

«»

=−

«»

¬¼

v

Hence the general complex solution is

7(5)(5)

12 3

162 62

() 2 9 3 9 3

010 10

titit

ii

tc e c ie c ie

−+−

+−

ªº ª º ª º

«» « » « »

=+++−

«» « » « »

¬¼ ¬ ¼ ¬ ¼

x

5.7 • Solutions 339

Rewriting the second eigenfunction yields

555

62 6cos 2sin 6sin 2cos

93 (cos sin) 9cos 3sin 9sin 3cos

10 10cos 10sin

+−+

ªº ª º ª º

«» « » « »

++=−++

«» « » « »

¬¼ ¬ ¼ ¬ ¼

ttt

itttt

ie t i t t te i t te

tt

Hence the general real solution is

755

12 3

16cos2sin6sin2cos

() 2 9cos 3sin 9sin 3cos

010cos 10sin

ttt

tt t t

tc e c t te c t te

tt

−

−+

ªº ª º ª º

«» « » « »

=+−++

«» « » « »

¬¼ ¬ ¼ ¬ ¼

x

where

12

,,cc and

3

c are real. When

23

0cc==

the trajectories tend toward the origin, and in other

cases the trajectories spiral away from the origin.

19. [M] Substitute

121

15 13 4,=/, =/, =RRC

and

2

3C= into the formula for A given in Example 1, and

use a matrix program to find the eigenvalues and eigenvectors:

11 2 1

234 1 3

525

11 2 2

A

−/−

ªº ªº ªº

=,=−.: = , =−.: =

«» «» «»

−

¬¼ ¬¼ ¬¼

vv

The general solution is thus

525

12

13

() .

22

−.−.

−

ªº ª º

=+

«» « »

¬¼ ¬ ¼

x

tt

tc e c e The condition 4

(0) 4

ªº

=«»

¬¼

x implies that

1

2

13 4

.

22 4

ªº

«»

¬¼

−

ªºªº

=

«»«»

¬¼¬¼

c

By a matrix program,

1

52c=/ and

2

12,=−/c so that

1525

2

() 13

51

()

() 22

22

tt

vt

te e

vt

−.−.

−

ªº ªº ªº

== −

«» «» « »

¬¼ ¬ ¼¬¼

x

20. [M] Substitute

121

115 13 9,RRC=/ , =/, = and

2

2C= into the formula for A given in Example 1,

and use a matrix program to find the eigenvalues and eigenvectors:

11 2 2

213 1 2

125

32 32 3 3

A

−/−

ªº ªº ªº

=,=−:= , =−.: =

«» «» «»

/−/

¬¼ ¬¼ ¬¼

vv

The general solution is thus

25

12

() .

33

−−.

−

ªº ª º

=+

«» « »

¬¼ ¬ ¼

x

tt

tc e c e The condition 3

(0) 3

ªº

=«»

¬¼

x implies

that

1

2

12 3

.

33 3

ªº

«»

¬¼

−

ªºªº

=

«»«»

¬¼¬¼

c

By a matrix program,

1

53c=/ and

2

23,=−/c so that

125

2

() 12

52

()

() 33

33

tt

vt

te e

vt

−−.

−

ªº ªº ªº

== −

«» «» « »

¬¼ ¬ ¼¬¼

x

21. [M] 18

.

55

−−

ªº

=«»

−

¬¼

A Using a matrix program we find that an eigenvalue of A is

36i−+

with

corresponding eigenvector 26.

5

+

ªº

=«»

¬¼

v

i

The conjugates of these form the second

eigenvalue-eigenvector pair. The general complex solution is

340 CHAPTER 5 • Eigenvalues and Eigenvectors

(36) (36)

12

26 26

() 55

it it

ii

tc e c e

−+−−

+−

ªº ªº

=+

«» «»

¬¼ ¬¼

x

where

1

c and

2

c are arbitrary complex numbers. Rewriting the first eigenfunction and taking its real

and imaginary parts, we have

(36) 3

33

26 (cos 6 sin 6 )

5

2cos6 6sin6 2sin6 6cos6

5cos6 5sin6

−+−

−−

+

ªº

=+

«»

¬¼

−+

ªºªº

=+

«»«»

¬¼¬¼

vit t

tt

i

eetit

tt t t

ei e

tt

The general real solution has the form

33

12

2cos6 6sin6 2sin6 6cos6

() 5cos6 5sin6

tt

tt t t

tc e c e

tt

−−

−+

ªºªº

=+

«»«»

¬¼¬¼

x

where

1

c and

2

c now are real numbers. To satisfy the initial condition 0

(0) ,

15

ªº

=«»

¬¼

x we solve

12

260

5015

cc

ªº ªº ª º

+=

«» «» « »

¬¼ ¬¼ ¬ ¼

to get

12

31.=, =−cc We now have

33 3

() 2cos6 6sin6 2sin6 6cos6 20sin6

() 3

() 5cos6 5sin6 15cos6 5sin6

−− −

−+−

ªº ªºªºª º

== −=

«» «»«»« »

−

¬¼¬¼¬ ¼

¬¼

x

Ltt t

C

it tt t t t

tee e

vt tttt

22. [M] 02

.

48

ªº

=«»

−.−.

¬¼

A Using a matrix program we find that an eigenvalue of A is

48i−.+.

with

corresponding eigenvector 12.

1

−−

ªº

=«»

¬¼

v

i

The conjugates of these form the second eigenvalue-

eigenvector pair. The general complex solution is

(48) (48)

12

12 12

() 11

it it

ii

tc e c e

−.+. −.−.

−− −+

ªº ªº

=+

«» «»

¬¼ ¬¼

x

where

1

c and

2

c are arbitrary complex numbers. Rewriting the first eigenfunction and taking its real

and imaginary parts, we have

(48) 4

44

12 (cos 8 sin 8 )

1

cos 8 2sin 8 sin 8 2 cos 8

cos 8 sin 8

it t

tt

i

eetit

tt t t

ei e

tt

−.+. −.

−.−.

−−

ªº

=.+.

«»

¬¼

−.+ . −.−.

ªºªº

=+

«»«»

..

¬¼¬¼

v

The general real solution has the form

44

12

cos 8 2sin 8 sin 8 2cos 8

() cos 8 sin 8

tt

tt t t

tc e c e

tt

−.−.

−.+ . −.−.

ªºªº

=+

«»«»

..

¬¼¬¼

x

5.8 • Solutions 341

where

1

c and

2

c now are real numbers. To satisfy the initial condition 0

(0) ,

12

ªº

=«»

¬¼

x we solve

12

12 0

1012

cc

−−

ªºªº ªº

+=

«»«» «»

¬¼ ¬¼¬¼

to get

12

12 6.=, =−cc We now have

444

() cos 8 2sin 8 sin 8 2cos 8 30sin 8

() 12 6

() cos 8 sin 8 12cos 8 6sin 8

Lttt

C

it tt t t t

te e e

vt tttt

−.−.−.

−.+ . −.−..

ªº ªºªºªº

== −=

«» «»«»«»

...−.

¬¼¬¼¬ ¼

¬¼

x

5.8 SOLUTIONS

1. The vectors in the given sequence approach an eigenvector

1

.v The last vector in the sequence,

4

1,

3326

ªº

=«»

.

¬¼

x is probably the best estimate for

1

.v To compute an estimate for

1

,λ examine

4

49978 .

16652

.

ªº

=«»

.

¬¼

xA This vector is approximately

11

.λv From the first entry in this vector, an estimate

of

1

λ is 4.9978.

2. The vectors in the given sequence approach an eigenvector

1

.v The last vector in the sequence,

4

2520 ,

1

−.

ªº

=«»

¬¼

x is probably the best estimate for

1

.v To compute an estimate for

1

,λ examine

4

12536 .

50064

−.

ªº

=«»

.

¬¼

xA This vector is approximately

11

.λv From the second entry in this vector, an

estimate of

1

λ is 5.0064.

3. The vectors in the given sequence approach an eigenvector

1

.v The last vector in the sequence,

4

5188 ,

1

.

ªº

=«»

¬¼

x is probably the best estimate for

1

.v To compute an estimate for

1

,λ examine

4

4594 .

9075

.

ªº

=«»

.

¬¼

xA This vector is approximately

11

.λv From the second entry in this vector, an estimate

of

1

λ is .9075.

4. The vectors in the given sequence approach an eigenvector

1

.v The last vector in the sequence,

4

1,

7502

ªº

=«»

.

¬¼

x is probably the best estimate for

1

.v To compute an estimate for

1

,λ examine

4

4012 .

3009

−.

ªº

=«»

−.

¬¼

xA This vector is approximately

11

.λv From the first entry in this vector, an estimate

of

1

λ is

4012.−.

342 CHAPTER 5 • Eigenvalues and Eigenvectors

5. Since

5

24991

31241

Aªº

=«»

−

¬¼

x is an estimate for an eigenvector, the vector

24991 7999

1

31241 1

31241

−.

ªºªº

=−=

«»«»

−

¬¼¬¼

v is a vector with a 1 in its second entry that is close to an

eigenvector of A. To estimate the dominant eigenvalue

1

λ of A, compute 40015 .

50020

.

ªº

=«»

−.

¬¼

vA From the

second entry in this vector, an estimate of

1

λ is

50020.−.

6. Since

5

2045

4093

A

−

ªº

=«»

¬¼

x is an estimate for an eigenvector, the vector 2045 4996

1

4093 1

4093

−−.

ªºª º

==

«»« »

¬¼¬ ¼

v is

a vector with a 1 in its second entry that is close to an eigenvector of A. To estimate the dominant

eigenvalue

1

λ of A, compute 20008 .

40024

−.

ªº

=«»

.

¬¼

vA From the second entry in this vector, an estimate of

1

λ is 4.0024.

7. [M]

0

67 1

.

85 0

ªº ªº

=,=

«» «»

¬¼ ¬¼

xA The data in the table below was calculated using Mathematica, which

carried more digits than shown here.

k 0 1 2 3 4 5

k

x 1

0

ªº

«»

¬¼

75

1

.

ªº

«»

¬¼

1

9565

ªº

«»

.

¬¼

9932

1

.

ª

º

«

»

¬

¼ 1

9990

ª

º

«

»

.

¬

¼ .9998

1

ªº

«»

¬¼

k

Ax 6

8

ªº

«»

¬¼

11 5

11 0

.

ªº

«»

.

¬¼

12 6957

12 7826

.

ªº

«»

.

¬¼

12 9592

12 9456

.

ª

º

«

»

.

¬

¼ 12 9927

12 9948

.

ª

º

«

»

.

¬

¼ 12 9990

12 9987

.

ªº

«»

.

¬¼

k

µ

8 11.5 12.7826 12.9592 12.9948 12.9990

The actual eigenvalue is 13.

8. [M]

0

21 1

.

45 0

ªº ªº

=,=

«» «»

¬¼ ¬¼

xA The data in the table below was calculated using Mathematica, which

carried more digits than shown here.

k 0 1 2 3 4 5

k

x 1

0

ªº

«»

¬¼

5

1

.

ªº

«»

¬¼

2857

1

.

ªº

«»

¬¼

2558

1

.

ª

º

«

»

¬

¼ 2510

1

.

ª

º

«

»

¬

¼ .2502

1

ª

º

«

»

¬

¼

k

Ax 2

4

ªº

«»

¬¼

2

7

ªº

«»

¬¼

15714

61429

.

ªº

«»

.

¬¼

15116

60233

.

ª

º

«

»

.

¬

¼ 15019

60039

.

ª

º

«

»

.

¬

¼ 15003

60006

.

ª

º

«

»

.

¬

¼

k

µ

4 7 6.1429 6.0233 6.0039 6.0006

The actual eigenvalue is 6.

5.8 • Solutions 343

9. [M]

0

8012 1

121 0.

030 0

ªºªº

«»«»

=−,=

«»«»

¬¼¬¼

xA

The data in the table below was calculated using Mathematica,

which carried more digits than shown here.

k 0 1 2 3 4 5 6

k

x

1

0

ªº

«»

¬¼

1

125

0

ªº

«»

.

«»

¬¼

1

0938

0469

ªº

«»

.

«»

.

¬¼

1

1004

0328

ª

º

«

»

.

«

»

«

»

.

¬

¼

1

0991

0359

ª

º

«

»

.

«

»

«

»

.

¬

¼

1

0994

0353

ªº

«»

.

«»

.

¬¼

1

0993

0354

ªº

«»

.

«»

.

¬¼

k

Ax

8

1

0

ªº

«»

¬¼

8

75

375

ªº

«»

.

«»

.

¬¼

85625

8594

2812

.

ªº

«»

.

«»

.

¬¼

83942

8321

3011

.

ª

º

«

»

.

«

»

«

»

.

¬

¼

84304

8376

2974

.

ª

º

«

»

.

«

»

«

»

.

¬

¼

84233

8366

2981

.

ª

º

«

»

.

«

»

«

»

.

¬

¼

84246

8368

2979

.

ªº

«»

.

«»

.

¬¼

k

µ

8 8 8.5625 8.3942 8.4304 8.4233 8.4246

Thus

5

8 4233

µ

=. and

6

84246.=.

µ

The actual eigenvalue is

(7 97) 2,+/

or 8.42443 to five

decimal places.

10. [M]

0

12 2 1

11 9 0.

01 9 0

−

ªºªº

«»«»

=,=

«»«»

¬¼¬¼

xA

The data in the table below was calculated using Mathematica,

which carried more digits than shown here.

k 0 1 2 3 4 5 6

k

x

1

0

ªº

«»

¬¼

1

0

ª

º

«

»

«

»

«

»

¬

¼

1

6667

3333

ªº

«»

.

«»

.

¬¼

3571

1

7857

.

ª

º

«

»

«

»

«

»

.

¬

¼

0932

1

9576

.

ª

º

«

»

«

»

«

»

.

¬

¼

0183

1

9904

.

ª

º

«

»

«

»

«

»

.

¬

¼

0038

1

9982

.

ªº

«»

.

¬¼

k

Ax

1

0

ªº

«»

¬¼

3

2

1

ª

º

«

»

«

»

«

»

¬

¼

1 6667

4 6667

36667

.

ªº

«»

.

«»

.

¬¼

7857

84286

80714

.

ª

º

«

»

.

«

»

«

»

.

¬

¼

1780

9 7119

9 6186

.

ª

º

«

»

.

«

»

«

»

.

¬

¼

0375

9 9319

9 9136

.

ª

º

«

»

.

«

»

«

»

.

¬

¼

0075

9 9872

9 9834

.

ªº

«»

.

«»

.

¬¼

k

µ

1 3 4.6667 8.4286 9.7119 9.9319 9.9872

Thus

5

99319=.

µ

and

6

99872.=.

µ

The actual eigenvalue is 10.

11. [M]

0

52 1

.

22 0

ªº ªº

=,=

«» «»

¬¼ ¬¼

xA The data in the table below was calculated using Mathematica, which

carried more digits than shown here.

k 0 1 2 3 4

k

x 1

0

ªº

«»

¬¼

1

4

ªº

«»

.

¬¼

1

4828

ª

º

«

»

.

¬

¼ 1

4971

ª

º

«

»

.

¬

¼ 1

4995

ª

º

«

»

.

¬

¼

344 CHAPTER 5 • Eigenvalues and Eigenvectors

k

Ax 5

2

ªº

«»

¬¼

58

28

.

ªº

«»

.

¬¼

59655

29655

.

ªº

«»

.

¬¼

59942

29942

.

ª

º

«

»

.

¬

¼ 59990

29990

.

ª

º

«

»

.

¬

¼

k

µ

5 5.8 5.9655 5.9942 5.9990

()

k

Rx 5 5.9655 5.9990 5.99997 5.9999993

The actual eigenvalue is 6. The bottom two columns of the table show that ()

k

Rx estimates the

eigenvalue more accurately than .

k

µ

12. [M]

0

32 1

.

22 0

−

ªºªº

=,=

«»«»

¬¼¬¼

xA The data in the table below was calculated using Mathematica,

which carried more digits than shown here.

k 0 1 2 3 4

k

x 1

0

ªº

«»

¬¼

1

6667

ªº

«»

−.

¬¼

1

4615

ª

º

«

»

−.

¬

¼ 1

5098

ª

º

«

»

−.

¬

¼ 1

4976

ªº

«»

−.

¬¼

k

Ax 3

2

−

ªº

«»

¬¼

43333

20000

−.

ª

º

«

»

.

¬

¼ 39231

20000

−.

ª

º

«

»

.

¬

¼ 40196

20000

−.

ª

º

«

»

.

¬

¼ 39951

20000

−.

ªº

«»

.

¬¼

k

µ

3−

43333−.

39231−.

40196−.

39951−.

()

k

Rx

3−

39231−.

39951−.

39997−.

3 99998−.

The actual eigenvalue is

4.−

The bottom two columns of the table show that ()

k

Rx estimates the

eigenvalue more accurately than .

k

µ

13. If the eigenvalues close to 4 and

4−

have different absolute values, then one of these is a strictly

dominant eigenvalue, so the power method will work. But the power method depends on powers of

the quotients

21

λ/λ and

31

λ/λ going to zero. If

21

|

λ/λ

|

is close to 1, its powers will go to zero

slowly, and the power method will converge slowly.

14. If the eigenvalues close to 4 and

4−

have the same absolute value, then neither of these is a strictly

dominant eigenvalue, so the power method will not work. However, the inverse power method may

still be used. If the initial estimate is chosen near the eigenvalue close to 4, then the inverse power

method should produce a sequence that estimates the eigenvalue close to 4.

15. Suppose

,=λxxA

with

0.≠x

For any

().,−=λ−xx xAI

αα α

If

α

is not an eigenvalue of A, then

AI

α

−

is invertible and

α

λ−

is not 0; hence

111

()() and ()()AI AI

αα α α

−−−

=−λ− λ−=−xxxx

This last equation shows that x is an eigenvector of

1

()AI

α

−

corresponding to the eigenvalue

1

().

−

λ−

α

16. Suppose that

µ

is an eigenvalue of

1

()AI

α

−

with corresponding eigenvector x. Since

1

() ,

−

−=xxAI

α

µ

()()()()()()AI A I A

α

µµ

α

µµ

α

µ

=−=−=−xxxxxx

5.8 • Solutions 345

Solving this equation for Ax, we find that

11

()

§· § ·

=+=+

¨¸ ¨ ¸

©¹ © ¹

xxx xA

αµα

µµ

Thus

(1 )

α

µ

λ=+/

is an eigenvalue of A with corresponding eigenvector x.

17. [M]

0

10 8 4 1

813 4 0 33.

454 0

−−

ªºªº

«»«»

=−,= ,=.

«»«»

−

¬¼¬¼

xA

α

The data in the table below was calculated using

Mathematica, which carried more digits than shown here.

k 0 1 2

k

x

1

0

ªº

«»

¬¼

1

7873

0908

ªº

«»

.

«»

.

¬¼

1

7870

0957

ª

º

«

»

.

«

»

«

»

.

¬

¼

k

y

26 0552

20 5128

23669

.

ªº

«»

.

«»

.

¬¼

47 1975

37 1436

45187

.

ªº

«»

.

«»

.

¬¼

47 1233

37 0866

45083

.

ª

º

«

»

.

«

»

«

»

.

¬

¼

k

µ

26.0552 47.1975 47.1233

k

ν

3.3384 3.32119 3.3212209

Thus an estimate for the eigenvalue to four decimal places is 3.3212. The actual eigenvalue is

(25 337) 2,−/

or 3.3212201 to seven decimal places.

18. [M]

0

8012 1

121 0 14.

030 0

ªºªº

«»«»

=−,= ,=−.

«»«»

¬¼¬¼

xA

α

The data in the table below was calculated using

Mathematica, which carried more digits than shown here.

k 0 1 2 3 4

k

x

1

0

ªº

«»

¬¼

1

3646

7813

ªº

«»

.

«»

−.

¬¼

1

3734

7855

ª

º

«

»

.

«

»

«

»

−.

¬

¼

1

3729

7854

ª

º

«

»

.

«

»

«

»

−.

¬

¼

1

3729

7854

ª

º

«

»

.

«

»

«

»

−.

¬

¼

k

y

40

14 5833

31 25

ªº

«»

.

«»

−.

¬¼

38 125

14 2361

29 9479

−.

ªº

«»

−.

«»

.

¬¼

41 1134

15 3300

32 2888

−.

ª

º

«

»

−.

«

»

«

»

.

¬

¼

40 9243

15 2608

32 1407

−.

ª

º

«

»

−.

«

»

«

»

.

¬

¼

40 9358

15 2650

32 1497

−.

ª

º

«

»

−.

«

»

«

»

.

¬

¼

k

µ

40

38 125−.

41 1134−.

40 9243−.

40 9358−.

k

ν

1375−.

1 42623−.

142432−.

142444−.

1 42443−.

346 CHAPTER 5 • Eigenvalues and Eigenvectors

Thus an estimate for the eigenvalue to four decimal places is

14244.−.

The actual eigenvalue is

(7 97) 2,−/

or

1 424429−.

to six decimal places.

19. [M]

0

10 7 8 7 1

756 5 0

.

86109 0

75910 0

ªºªº

«»«»

=,=

«»«»

¬¼¬¼

xA

(a) The data in the table below was calculated using Mathematica, which carried more digits than

shown here.

k 0 1 2 3

k

x

1

0

ª

º

«

»

«

»

«

»

«

»

«

»

¬

¼

1

7

8

7

ªº

«»

.

«»

.

«»

.

«»

¬¼

988679

709434

1

932075

.

ª

º

«

»

.

«

»

«

»

«

»

.

«

»

¬

¼

961467

691491

1

942201

.

ª

º

«

»

.

«

»

«

»

«

»

.

«

»

¬

¼

k

Ax

10

7

8

7

ªº

«»

¬¼

26 2

18 8

26 5

24 7

.

ªº

«»

.

«»

.

«»

.

«»

¬¼

29 3774

21 1283

30 5547

28 7887

.

ª

º

«

»

.

«

»

«

»

.

«

»

.

«

»

¬

¼

29 0505

20 8987

30 3205

28 6097

.

ª

º

«

»

.

«

»

«

»

.

«

»

.

«

»

¬

¼

k

µ

10 26.5 30.5547 30.3205

k 4 5 6 7

k

x

958115

689261

1

943578

.

ªº

«»

.

«»

.

«»

¬¼

957691

688978

1

943755

.

ªº

«»

.

«»

.

«»

¬¼

957637

688942

1

943778

.

ª

º

«

»

.

«

»

«

»

«

»

.

«

»

¬

¼

957630

688938

1

943781

.

ª

º

«

»

.

«

»

«

»

«

»

.

«

»

¬

¼

k

Ax

29 0110

20 8710

30 2927

28 5889

.

ªº

«»

.

«»

.

«»

.

«»

¬¼

29 0060

20 8675

30 2892

28 5863

.

ªº

«»

.

«»

.

«»

.

«»

¬¼

29 0054

20 8671

30 2887

28 5859

.

ª

º

«

»

.

«

»

«

»

.

«

»

.

«

»

¬

¼

29 0053

20 8670

30 2887

28 5859

.

ª

º

«

»

.

«

»

«

»

.

«

»

.

«

»

¬

¼

k

µ

30.2927 30.2892 30.2887 30.2887

Thus an estimate for the eigenvalue to four decimal places is 30.2887. The actual eigenvalue is

30.2886853 to seven decimal places. An estimate for the corresponding eigenvector is

957630

688938 .

1

943781

.

ªº

«»

.

«»

.

«»

¬¼

5.8 • Solutions 347

(b) The data in the table below was calculated using Mathematica, which carried more digits than

shown here.

k 0 1 2 3 4

k

x

1

0

ªº

«»

¬¼

609756

1

243902

146341

−.

ªº

«»

−.

«»

.

«»

¬¼

604007

1

251051

148899

−.

ª

º

«

»

«

»

«

»

−.

«

»

.

«

»

¬

¼

603973

1

251134

148953

−.

ª

º

«

»

«

»

«

»

−.

«

»

.

«

»

¬

¼

603972

1

251135

148953

−.

ª

º

«

»

«

»

«

»

−.

«

»

.

«

»

¬

¼

k

y

25

41

10

6

ªº

«»

−

«»

−

«»

¬¼

59 5610

98 6098

24 7561

14 6829

−.

ªº

«»

.

«»

−.

«»

.

«»

¬¼

59 5041

98 5211

24 7420

14 6750

−.

ª

º

«

»

.

«

»

«

»

−.

«

»

.

«

»

¬

¼

59 5044

98 5217

24 7423

14 6751

−.

ª

º

«

»

.

«

»

«

»

−.

«

»

.

«

»

¬

¼

59 5044

98 5217

24 7423

14 6751

−.

ª

º

«

»

.

«

»

«

»

−.

«

»

.

«

»

¬

¼

k

µ

41−

98.6098 98.5211 98.5217 98.5217

k

ν

0243902−.

.0101410 .0101501 .0101500 .0101500

Thus an estimate for the eigenvalue to five decimal places is .01015. The actual eigenvalue is

.01015005 to eight decimal places. An estimate for the corresponding eigenvector is

603972

1.

251135

148953

−.

ªº

«»

−.

«»

.

«»

¬¼

20. [M]

0

1232 1

2121311 0

.

2302 0

4572 0

ªºªº

«»«»

=,=

«»«»

−

«»«»

¬¼¬¼

xA

(a) The data in the table below was calculated using Mathematica, which carried more digits than

shown here.

k 0 1 2 3 4

k

x

1

0

ª

º

«

»

«

»

«

»

«

»

«

»

¬

¼

25

5

1

.

ªº

«»

.

«»

−.

«»

¬¼

159091

1

272727

181818

.

ª

º

«

»

«

»

«

»

.

«

»

.

«

»

¬

¼

187023

1

170483

442748

.

ª

º

«

»

«

»

«

»

.

«

»

.

«

»

¬

¼

184166

1

180439

402197

.

ª

º

«

»

«

»

«

»

.

«

»

.

«

»

¬

¼

k

Ax

1

2

4

ª

º

«

»

«

»

«

»

−

«

»

«

»

¬

¼

175

11

3

2

.

ªº

«»

¬¼

334091

17 8636

304545

790909

.

ª

º

«

»

.

«

»

«

»

.

«

»

.

«

»

¬

¼

358397

19 4606

351145

782697

.

ª

º

«

»

.

«

»

«

»

.

«

»

.

«

»

¬

¼

352988

19 1382

343606

780413

.

ª

º

«

»

.

«

»

«

»

.

«

»

.

«

»

¬

¼

k

µ

4 11 17.8636 19.4606 19.1382

348 CHAPTER 5 • Eigenvalues and Eigenvectors

k 5 6 7 8 9

k

x

184441

1

179539

407778

.

ªº

«»

.

«»

.

«»

¬¼

184414

1

179622

407021

.

ªº

«»

.

«»

.

«»

¬¼

184417

1

179615

407121

.

ª

º

«

»

«

»

«

»

.

«

»

.

«

»

¬

¼

184416

1

179615

407108

.

ª

º

«

»

«

»

«

»

.

«

»

.

«

»

¬

¼

184416

1

179615

407110

.

ªº

«»

.

«»

.

«»

¬¼

k

Ax

353861

19 1884

344667

781010

.

ªº

«»

.

«»

.

«»

.

«»

¬¼

353732

19 1811

344521

780905

.

ªº

«»

.

«»

.

«»

.

«»

¬¼

353750

19 1822

344541

780921

.

ª

º

«

»

.

«

»

«

»

.

«

»

.

«

»

¬

¼

353748

19 1820

344538

780919

.

ª

º

«

»

.

«

»

«

»

.

«

»

.

«

»

¬

¼

353748

19 1820

344539

780919

.

ªº

«»

.

«»

.

«»

.

¬¼

k

µ

19.1884 19.1811 19.1822 19.1820 19.1820

Thus an estimate for the eigenvalue to four decimal places is 19.1820. The actual eigenvalue is

19.1820368 to seven decimal places. An estimate for the corresponding eigenvector is

184416

1.

179615

407110

.

ªº

«»

.

«»

.

«»

¬¼

(b) The data in the table below was calculated using Mathematica, which carried more digits than

shown here.

k 0 1 2

k

x

1

0

ª

º

«

»

«

»

«

»

«

»

«

»

¬

¼

1

226087

921739

660870

ªº

«»

.

«»

−.

«»

.

«»

¬¼

1

222577

917970

660496

ª

º

«

»

.

«

»

«

»

−.

«

»

.

«

»

¬

¼

k

y

115

26

106

76

ª

º

«

»

«

»

«

»

−

«

»

«

»

¬

¼

81 7304

18 1913

75 0261

53 9826

.

ªº

«»

.

«»

−.

«»

.

«»

¬¼

81 9314

18 2387

75 2125

54 1143

.

ª

º

«

»

.

«

»

«

»

−.

«

»

.

«

»

¬

¼

k

µ

115 81.7304 81.9314

k

ν

.00869565 .0122353 .0122053

Thus an estimate for the eigenvalue to four decimal places is .0122. The actual eigenvalue is

.01220556 to eight decimal places. An estimate for the corresponding eigenvector is

1

222577 .

917970

660496

ªº

«»

.

«»

−.

«»

.

«»

¬¼

Chapter 5 • Supplementary Exercises 349

21. (a) 80 5

.

02 5

..

ªºªº

=,=

«»«»

..

¬¼¬¼

xA Here is the sequence

k

Ax

for

15:=,k…

432256204816384

102004000800016

.. . . .

ªºª ºª ºª ºª º

,, , ,

«»« »« »« »« »

.. . . .

¬¼¬ ¼¬ ¼¬ ¼¬ ¼

Notice that

5

Ax

is approximately

4

8( )..xA

Conclusion: If the eigenvalues of A are all less than 1 in magnitude, and if

0,≠x

then

k

Ax

is

approximately an eigenvector for large k.

(b) 10 5

.

08 5

.

ªºªº

=,=

«»«»

..

¬¼¬¼

xA Here is the sequence

k

Ax

for

15:=,k…

55 5 5 5

432256204816384

.. . . .

ªºª ºª ºª ºª º

,, , ,

«»« »« »« »« »

.. . . .

¬¼¬ ¼¬ ¼¬ ¼¬ ¼

Notice that

k

Ax

seems to be converging to 5.

0

.

ª

º

«

»

¬

¼

Conclusion: If the strictly dominant eigenvalue of A is 1, and if x has a component in the

direction of the corresponding eigenvector, then

{}

k

Ax

will converge to a multiple of that

eigenvector.

c. 80 5

.

02 5

.

ªºªº

=,=

«»«»

.

¬¼¬¼

xA Here is the sequence

k

Ax

for

15:=,k…

432256204816384

12 4 8 16

ªºª ºª ºª ºª º

,, , ,

«»« »« »« »« »

¬¼¬ ¼¬ ¼¬ ¼¬ ¼

Notice that the distance of

k

Ax

from either eigenvector of A is increasing rapidly as k increases.

Conclusion: If the eigenvalues of A are all greater than 1 in magnitude, and if x is not an

eigenvector, then the distance from

k

Ax

to the nearest eigenvector will increase as

.→∞k

Chapter 5 SUPPLEMENTARY EXERCISES

1. a. True. If A is invertible and if

1

A

=⋅

xx

for some nonzero x, then left-multiply by

1

A

−

to obtain

1

,

−

=xxA

which may be rewritten as

1

1.

−

=⋅xxA Since x is nonzero, this shows 1 is an

eigenvalue of

1

.

−

A

b. False. If A is row equivalent to the identity matrix, then A is invertible. The matrix in Example 4

of Section 5.3 shows that an invertible matrix need not be diagonalizable. Also, see Exercise 31

in Section 5.3.

c. True. If A contains a row or column of zeros, then A is not row equivalent to the identity matrix

and thus is not invertible. By the Invertible Matrix Theorem (as stated in Section 5.2), 0 is an

eigenvalue of A.

350 CHAPTER 5 • Eigenvalues and Eigenvectors

d. False. Consider a diagonal matrix D whose eigenvalues are 1 and 3, that is, its diagonal entries

are 1 and 3. Then

2

D

is a diagonal matrix whose eigenvalues (diagonal entries) are 1 and 9. In

general, the eigenvalues of

2

A

are the squares of the eigenvalues of A.

e. True. Suppose a nonzero vector x satisfies

,=xxA

λ

then

22

() ()AAAA A

λλ λ

====xx xxx

This shows that x is also an eigenvector for

2

A

f. True. Suppose a nonzero vector x satisfies ,=xxA

λ

then left-multiply by

1

A

−

to obtain

11

() .

−−

==xxxAA

λλ

Since A is invertible, the eigenvalue λ is not zero. So

11

,

−−

λ=xxA

which

shows that x is also an eigenvector of

1

.

−

A

g. False. Zero is an eigenvalue of each singular square matrix.

h. True. By definition, an eigenvector must be nonzero.

i. False. Let

20

02

Aªº

=«»

¬¼

then and are eigenvectors of A for the eigenvalue 2,

and they are linearly independent.

j. True. This follows from Theorem 4 in Section 5.2

k. False. Let A be the 33× matrix in Example 3 of Section 5.3. Then A is similar to a diagonal

matrix D. The eigenvectors of D are the columns of

3

,I but the eigenvectors of A are entirely

different.

l. False. Let 20

.

03

ªº

=«»

¬¼

A Then

1

0

ª

º

=

«

»

¬

¼

e and

2

0

1

ª

º

=

«

»

¬

¼

e are eigenvectors of A, but

12

+ee

is not.

(Actually, it can be shown that if two eigenvectors of A correspond to distinct eigenvalues, then

their sum cannot be an eigenvector.)

m. False. All the diagonal entries of an upper triangular matrix are the eigenvalues of the matrix

(Theorem 1 in Section 5.1). A diagonal entry may be zero.

n. True. Matrices A and

T

A

have the same characteristic polynomial, because

det( ) det( ) det( ),−λ =−λ =−λ

TT

AI AI AI

by the determinant transpose property.

o. False. Counterexample: Let A be the 55× identity matrix.

p. True. For example, let A be the matrix that rotates vectors through 2π/ radians about the origin.

Then Ax is not a multiple of x when x is nonzero.

q. False. If A is a diagonal matrix with 0 on the diagonal, then the columns of A are not linearly

independent.

r. True. If

1

A

λ

=xx

and

2

,=xxA

λ

then

12

λλ

=xx

and

12

().−=x0

λλ

If ,≠x0 then

1

λ

must equal

2

.

λ

s. False. Let A be a singular matrix that is diagonalizable. (For instance, let A be a diagonal matrix

with 0 on the diagonal.) Then, by Theorem 8 in Section 5.4, the transformation Axx6 is

represented by a diagonal matrix relative to a coordinate system determined by eigenvectors of

A.

1

0

ª

º

=

«

»

¬

¼

e2

0

1

ª

º

=

«

»

¬

¼

e

Chapter 5 • Supplementary Exercises 351

t. True. By definition of matrix multiplication,

11

22

[][ ]

nn

AAI A A A A== =ee e e e e""

If =ee

jjj

Ad for

1,=, ,

j…n

then A is a diagonal matrix with diagonal entries

1

.,,

n

d…d

u. True. If

1

,

−

=BPDP

where D is a diagonal matrix, and if

1

,

−

=AQBQ then

11 1

()()(),AQPDPQ QPDQP

−− −

==

which shows that A is diagonalizable.

v. True. Since B is invertible, AB is similar to

1

(),

−

BABB which equals BA.

w. False. Having n linearly independent eigenvectors makes an nn× matrix diagonalizable (by the

Diagonalization Theorem 5 in Section 5.3), but not necessarily invertible. One of the eigenvalues

of the matrix could be zero.

x. True. If A is diagonalizable, then by the Diagonalization Theorem, A has n linearly independent

eigenvectors

1

,,vv

n

… in .R

n

By the Basis Theorem,

1

{},,vv

n

… spans .R

n

This means that

each vector in

n

R

can be written as a linear combination of

1

.,,vv

n

…

2. Suppose B≠x0

and =λxxAB for some λ. Then

() .=λ

xxAB

Left-multiply each side by B and

obtain

() () ().=λ=λ

xxxBA B B B

This equation says that Bx is an eigenvector of BA, because

.≠x0B

3. a. Suppose ,=λxxA with .≠x0

Then

(5 ) 5 5 (5 ) .−=−=−λ =−λ

xx xxx xIA A

The eigenvalue

is

5.−λ

b.

222

(5 3 ) 5 3 ( ) 5 3( ) (5 3 ) .−+=−+=−λ +λ=−λ+λxx x x x x x xIAA AAA The eigenvalue is

2

53 .−λ+λ

4. Assume that A

λ

=xx

for some nonzero vector x. The desired statement is true for 1,=m by the

assumption about

λ

. Suppose that for some 1,≥k the statement holds when .=mk

That is, suppose

that .=xx

kk

A

λ

Then

1

()()

kkk

AAAA

λ

+

==xx x

by the induction hypothesis. Continuing,

11

,

++

==xxx

kkk

AA

λλ

because x is an eigenvector of A corresponding to A. Since x is nonzero, this

equation shows that

1

k

λ

+

is an eigenvalue of

1

,

+k

A

with corresponding eigenvector x. Thus the

desired statement is true when 1.=+mk By the principle of induction, the statement is true for each

positive integer m.

5. Suppose ,=λxxA with .≠x0

Then

2

01 2

2

01 2

2

01 2

() ( )

()

=++ ++

=+ + ++

=+λ+λ++λ=λ

xx

xx x x

xx x x x

n

pA cI cA cA … cA

ccAcA…cA

cc c …c p

So

()λ

p

is an eigenvalue of

().

pA

6. a. If

1

,

−

=APDP

then

1

,

−

=

kk

APDP

and

21 121

21

53 5 3

(5 3 )

−−−

−

=−+= −+

=−+

BIAA PIP PDP PDP

PI D DP

Since D is diagonal, so is

2

53 .−+IDD

Thus B is similar to a diagonal matrix.

352 CHAPTER 5 • Eigenvalues and Eigenvectors

b.

121 1

01 2

21

01 2

1

()

−− −

−

=+ + ++

=++++

=

"

n

pA cI cPDP cPDP cPDP

PcI cD cD cD P

Pp D P

This shows that

()

pA

is diagonalizable, because

()

pD

is a linear combination of diagonal

matrices and hence is diagonal. In fact, because D is diagonal, it is easy to see that

(2) 0

() 0(7)

p

pD p

ªº

=«»

¬¼

7. If

1

,

−

=APDP

then

1

() () ,

−

=pA PpDP as shown in Exercise 6. If the

(),

jj

entry in D is λ, then the

(),

jj

entry in

k

D

is

,λ

k

and so the

(),

jj

entry in

()

pD

is

().λ

p

If p is the characteristic

polynomial of A, then

() 0λ=

p

for each diagonal entry of D, because these entries in D are the

eigenvalues of A. Thus

()

pD

is the zero matrix. Thus

1

() 0 0.

−

=⋅⋅ =pA P P

8. a. If

λ

is an eigenvalue of an nn× diagonalizable matrix A, then

1

APDP

−

=

for an invertible

matrix P and an nn× diagonal matrix D whose diagonal entries are the eigenvalues of A. If the

multiplicity of

λ

is n, then

λ

must appear in every diagonal entry of D. That is, .=DI

λ

In this

case,

111

() .

−−−

====APIP PIP PP I

λλ λλ

b. Since the matrix 31

03

Aªº

=«»

¬¼

is triangular, its eigenvalues are on the diagonal. Thus 3 is an

eigenvalue with multiplicity 2. If the

22×

matrix A were diagonalizable, then A would be 3I, by

part (a). This is not the case, so A is not diagonalizable.

9. If

IA

−

were not invertible, then the equation

().−=

x0IA

would have a nontrivial solution x. Then

A−=xx0

and 1,=⋅xxA which shows that A would have 1 as an eigenvalue. This cannot happen if

all the eigenvalues are less than 1 in magnitude. So

IA

−

must be invertible.

10. To show that

k

A

tends to the zero matrix, it suffices to show that each column of

k

A

can be made as

close to the zero vector as desired by taking k sufficiently large. The jth column of A is ,e

j

A where

j

e is the jth column of the identity matrix. Since A is diagonalizable, there is a basis for

n

consisting of eigenvectors

1

,,,vv

n

… corresponding to eigenvalues

1

.λ,,λ

n

… So there exist scalars

1

,,,

n

c…c such that

11

(an eigenvector decomposition of )=++ev v e

jnn j

…

cc

Then, for 12 ,=,,k…

11 1

() ( ) ()=λ++ λ∗ev v"

kk k

jnnn

Ac c

If the eigenvalues are all less than 1 in absolute value, then their kth powers all tend to zero. So

()∗

shows that

k

j

Ae

tends to the zero vector, as desired.

11. a. Take x in H. Then

c=xu

for some scalar c. So

() ( ) ( )(),===λ=λ

xu u u uAAc cA c c

which

shows that

Ax

is in H.

Chapter 5 • Supplementary Exercises 353

b. Let x be a nonzero vector in K. Since K is one-dimensional, K must be the set of all scalar

multiples of x. If K is invariant under A, then

Ax

is in K and hence

Ax

is a multiple of x. Thus x

is an eigenvector of A.

12. Let U and V be echelon forms of A and B, obtained with r and s row interchanges, respectively, and

no scaling. Then det ( 1) det

r

AU=− and det ( 1) det

s

BV=−

Using first the row operations that reduce A to U, we can reduce G to a matrix of the form

.

0

ªº

′=«»

¬¼

UY

G

B Then, using the row operations that reduce B to V, we can further reduce

G′

to

.

0

ªº

′′ =«»

¬¼

UY

G

V There will be rs+ row interchanges, and so

det det ( 1) det

00

+

ªº ªº

==−

«» «»

¬¼ ¬¼

rs

AX UY

G

BV

Since 0

ª

º

«

»

¬

¼

UY

V is upper triangular, its determinant

equals the product of the diagonal entries,

and since U and V are upper triangular, this product also equals (det U ) (det V ). Thus

det ( 1) (det )(det ) (det )(det )

+

=−=

rs

GUVAB

For any scalar

λ

, the matrix

−λGI

has the same partitioned form as G, with

−λAI

and

−λBI

as

its diagonal blocks. (Here

I

represents various identity matrices of appropriate sizes.) Hence the

result about det G shows that

det( ) det( ) det( )−λ =−λ ⋅ −λ

GI AI BI

13. By Exercise 12, the eigenvalues of A are the eigenvalues of the matrix

[

]

3 together with the

eigenvalues of 52

.

43

−

ªº

«»

−

¬¼

The only eigenvalue of

[

]

3 is 3, while the eigenvalues of 52

43

−

ªº

«»

−

¬¼

are

1 and 7. Thus the eigenvalues of A are 1, 3, and 7.

14. By Exercise 12, the eigenvalues of A are the eigenvalues of the matrix 15

24

ª

º

«

»

¬

¼ together with the

eigenvalues of 74

.

31

−−

ªº

«»

¬¼

The eigenvalues of 15

24

ª

º

«

»

¬

¼ are

1−

and 6, while the eigenvalues of

74

31

−−

ªº

«»

¬¼

are

5−

and

1.−

Thus the eigenvalues of A are

15,−,−

and 6, and the eigenvalue

1−

has

multiplicity 2.

15. Replace a by

a

λ

−

in the determinant formula from Exercise 16 in Chapter 3 Supplementary

Exercises.

1

det( ) ( ) [ ( 1) ]

−

−λ =−−λ −λ+−

n

AI ab a n b

This determinant is zero only if

0−−λ=ab

or

(1) 0.−λ+−=

anb

Thus

λ

is an eigenvalue of A if

and only if

λ=−ab

or

(1).λ=+ −

an

From the formula for

det( )−λ

AI

above, the algebraic

multiplicity is

1n−

for

ab−

and 1 for

(1).+−

an b

16. The

33×

matrix has eigenvalues

12−

and

1(2)(2),+

that is,

1−

and 5. The eigenvalues of the

55×

matrix are

73−

and

7(4)(3),+

that is 4 and 19.

354 CHAPTER 5 • Eigenvalues and Eigenvectors

17. Note that

2

11 22 12 21 11 22 11 22 12 21

det( ) ( )( ) ( ) ( )−λ =−λ −λ − =λ− +λ+−AI a a aa a a aa aa

2

(tr ) det ,=λ− λ+AA

and use the quadratic formula to solve the characteristic equation:

2

tr (tr ) 4det

2

±−

=AA A

λ

The eigenvalues are both real if and only if the discriminant is nonnegative, that is,

2

(tr ) 4det 0.−≥AA

This inequality simplifies to

2

(tr ) 4detAA≥ and

2

det .

2

§·

≥

¨¸

©¹

trA A

18. The eigenvalues of A are 1 and .6. Use this to factor A and

.

k

A

1310 23

1

2206 21

4

131 0 23

1

22 21

4

06

23

13

1

22

42(6) (6)

26(6) 33(6)

1

444(6) 62(6)

23

1 as

46

4

ªº

«»

¬¼

ªº

«»

¬¼

ªº

«»

¬¼

−−

ªºªºªº

=⋅

«»«»«»

.−−

¬¼¬¼¬¼

−−

ªº ªº

=⋅

«» «»

−−

.

¬¼ ¬¼

−−

ªº

=«»

−⋅.−.

¬¼

−+. −+.

=−.−.

−−

ªº

→→

«»

¬¼

k

kk

A

k∞

19.

2

01

det( ) 6 5 ( )

65

ªº

=;−λ =−λ+λ=λ

«»

−

¬¼

pp

CCIp

20.

010

001;

24 26 9

ªº

«»

=«»

«»

−

¬¼

p

C

23

det( ) 24 26 9 ( )−λ =−λ+λ−λ=λ

p

CI p

21. If p is a polynomial of order 2, then a calculation such as in Exercise 19 shows that the characteristic

polynomial of

p

C

is

2

() (1) (),λ=−λpp

so the result is true for

2.=n

Suppose the result is true for

nk=

for some

2,≥k

and consider a polynomial p of degree

1.+k

Then expanding

det( )−λ

p

CI

by cofactors down the first column, the determinant of

−λ

p

CI

equals

1

0

12

10

()det (1)

01

+

−λ

ªº

«»

−λ +−

«»

−− −−λ

«»

¬¼

"

##

"

k

a

aa a

Chapter 5 • Supplementary Exercises 355

The

kk×

matrix shown is

,−λ

q

CI

where

1

12

() .

−

=+ ++ +

"

kk

k

qt a at at t

By the induction

assumption, the determinant of

−λ

q

CI

is

(1) ().−λ

kq

Thus

1

0

11

01

1

det( ) ( 1) ( )( 1) ( )

(1) [ ( )]

(1) ()

+

+−

+

−λ =−+−λ − λ

=−+λ++λ+λ

=−λ

"

kk

p

kkk

k

CI a q

aa a

p

So the formula holds for

1nk=+

when it holds for

.=nk

By the principle of induction, the formula

for

det( )−λ

p

CI

is true for all

2.≥n

22. a.

012

010

001

p

C

aaa

ªº

«»

¬¼

=

−−−

b. Since

λ

is a zero of p,

23

01 2

0+λ+λ+λ=aa a

and

23

01 2

.−−λ−λ=λaa a

Thus

22

2

201 2

1

p

C

aa a

ªºªº

ªº

«»«»

«»

«»«»

«»

«»«»

«»

«»«»

«»

«»«»

«» 3

«»«»

«»

¬¼

¬¼¬¼

λλ

λ=λ=λ

−−λ−λ

λλ

That is,

22

(1 ) (1 ),,λ,λ=,λ,λ

p

C

λ

which shows that

2

(1 ),λ,λ

is an eigenvector of

p

C

corresponding to the eigenvalue

λ

.

23. From Exercise 22, the columns of the Vandermonde matrix V are eigenvectors of

,

p

C

corresponding

to the eigenvalues

123

λ,λ,λ

(the roots of the polynomial p). Since these eigenvalues are distinct, the

eigenvectors from a linearly independent set, by Theorem 2 in Section 5.1. Thus V has linearly

independent columns and hence is invertible, by the Invertible Matrix Theorem. Finally, since the

columns of V are eigenvectors of

,

p

C

the Diagonalization Theorem (Theorem 5 in Section 5.3)

shows that

1

p

VCV

−

is diagonal.

24. [M] The MATLAB command roots(p) requires as input a row vector p whose entries are the

coefficients of a polynomial, with the highest order coefficient listed first. MATLAB constructs a

companion matrix

p

C

whose characteristic polynomial is p, so the roots of p are the eigenvalues of

.

p

C

The numerical values of the eigenvalues (roots) are found by the same QR algorithm used by

the command eig(A).

25. [M] The MATLAB command [P D] = e i g ( A ) produces a matrix P, whose condition number is

8

16 10 ,.×

and a diagonal matrix D, whose entries are almost 2, 2, 1. However, the exact eigenvalues

of A are 2, 2, 1, and A is not diagonalizable.

26. [M] This matrix may cause the same sort of trouble as the matrix in Exercise 25. A matrix program

that computes eigenvalues by an interative process may indicate that A has four distinct eigenvalues,

all close to zero. However, the only eigenvalue is 0, with multiplicity 4, because

4

0.=A

r

6.1 SOLUTIONS

Notes:

The first half of this section is

c

concepts of orthogonality and orthogon

a

an important general fact, but is needed

in Section 7.4. The optional material on

1. Since 1

2

−

ªº

=«»

¬¼

u and 4,

6

ªº

=«»

¬¼

v

⋅

=

uu

2. Since

3

1

5

ªº

«»

=−

«»

−

¬¼

w

and

6

2,

3

ªº

«»

=−

«»

¬¼

x

⋅w

and

51

.

35 7

⋅

==

⋅

xw

ww

3. Since

3

1,

5

ªº

«»

=−

«»

−

¬¼

w

22

3(1)⋅=+−ww

4. Since 1,

2

−

ªº

=«»

¬¼

u

22

(1) 2⋅=−+

=

uu

5. Since 1

2

−

ªº

=«»

¬¼

u and 4,

6

ªº

=«»

¬¼

v u ⋅ v

48/13

2.

612/13

13

ªº ª º

⋅

§·

==

¨¸ «» « »

⋅

©¹ ¬¼ ¬ ¼

uv v

vv

r

son Education, Inc. Publishing as Addison-Wesley.

c

omputational and is easily learned. The second half

c

a

l complements, which are essential for later work.

T

only for Supplementary Exercise 13 at the end of th

e

angles is not used later. Exercises 27–31 concern fact

s

22

(1) 2 5=−+=

, v ⋅ u = 4(–1) + 6(2) = 8, and

8

5

⋅

=

⋅

vu

uu

222

3(1)(5)35=+−+−=w

, x ⋅ w = 6(3) + (–2)(–1) +

2

(5) 35+−=

, and

3/35

11/35 .

1/7

ªº

«»

=−

«»

⋅«»

−

¬¼

w

ww

5

=

and 1/5

1.

2/5

−

ªº

=«»

⋅¬¼

u

uu

= (–1)(4) + 2(6) = 8,

22

4652,⋅=+=vv and

357

c

oncerns the

T

heorem 3 is

e

chapter and

s

used later.

8

.

5

3(–5) = 5,

358 CHAPTER 6 • Orthogonality and Least Squares

6. Since

6

2

3

ªº

«»

=−

«»

¬¼

x

and

3

1,

5

ªº

«»

=−

«»

−

¬¼

w

x ⋅ w = 6(3) + (–2)(–1) + 3(–5) = 5,

222

6(2)349,⋅=+−+=xx

and

630/49

5210/49.

49 315/49

ªºª º

⋅

§· «»« »

=−=−

¨¸ «»« »

⋅

©¹ «»« »

¬¼¬ ¼

xw x

xx

7. Since

3

1,

5

ªº

«»

=−

«»

−

¬¼

w

222

|| || 3 ( 1) ( 5) 35.=⋅=+−+−=www

8. Since

6

2,

3

ªº

«»

=−

«»

¬¼

x

222

|| || 6 ( 2) 3 49 7.=⋅=+−+= =xxx

9. A unit vector in the direction of the given vector is

22

30 30 3/ 5

11

40 40 4 / 5

50

(30) 40

−−−

ª

ºªºªº

==

«

»«»«»

¬

¼¬¼¬¼

−+

10. A unit vector in the direction of the given vector is

22 2

6/ 61

66

11

444/61

61

(6) 4 (3) 33

361

ª

º

−

−−

ªº ªº

«

»

«» «»

==

«

»

«» «»

«

»

−++−«» «»

−−

−

¬¼ ¬¼

«

»

¬

¼

11. A unit vector in the direction of the given vector is

222

7/ 69

7/4 7/4

11

1/2 1/2 2/ 69

69/16

(7 / 4) (1/ 2) 1 11

4/ 69

ª

º

ªº ªº

«

»

«» «»

==

«

»

«» «»

«

»

++

«» «»

¬¼ ¬¼

«

»

¬

¼

12. A unit vector in the direction of the given vector is

22

8/3 8/3 4/5

11

223/5

100 / 9

(8 / 3) 2

ªº ªºªº

==

«» «»«»

¬¼ ¬¼¬¼

+

13. Since

10

3

ªº

=«»

−

¬¼

x and

1,

5

−

ªº

=«»

−

¬¼

y

22 2

|| || [10 ( 1)] [ 3 ( 5)] 125−=−− +−−− =xy

and

dist ( , ) 125 5 5.==xy

6.1 • Solutions 359

14. Since

0

5

2

ªº

«»

=−

«»

¬¼

u

and

4

1,

8

−

ªº

«»

=−

«»

¬¼

z

22 22

|| || [0 ( 4)] [ 5 ( 1)] [2 8] 68−=−− +−−− +−=uz

and

dist ( , ) 68 2 17.==uz

15. Since a ⋅ b = 8(–2) + (–5)( –3) = –1 ≠ 0, a and b are not orthogonal.

16. Since u ⋅ v = 12(2) + (3)( –3) + (–5)(3) = 0, u and v are orthogonal.

17. Since u ⋅ v = 3(–4) + 2(1) + (–5)( –2) + 0(6) = 0, u and v are orthogonal.

18. Since y ⋅ z = (–3)(1) + 7(–8) + 4(15) + 0(–7) = 1 ≠ 0, y and z are not orthogonal.

19. a. True. See the definition of || v ||.

b . True. See Theorem 1(c).

c . True. See the discussion of Figure 5.

d . False. Counterexample:

11

.

00

ªº

«»

¬¼

e . True. See the box following Example 6.

20. a. True. See Example 1 and Theorem 1(a).

b . False. The absolute value sign is missing. See the box before Example 2.

c . True. See the defintion of orthogonal complement.

d . True. See the Pythagorean Theorem.

e . True. See Theorem 3.

21. Theorem 1(b):

() () ( )

TTTTT

+⋅=+ = + = + =⋅+⋅uvw uvw u vwuwvwuwvw

The second and third equalities used Theorems 3(b) and 2(c), respectively, from Section 2.1.

Theorem 1(c):

() () ( ) ( )

TT

cccc⋅===⋅uv uv uv uv

The second equality used Theorems 3(c) and 2(d), respectively, from Section 2.1.

22. Since u ⋅ u is the sum of the squares of the entries in u, u ⋅ u ≥ 0. The sum of squares of numbers is

zero if and only if all the numbers are themselves zero.

23. One computes that u ⋅ v = 2(–7) + (–5)( –4) + (–1)6 = 0,

2222

|| || 2 ( 5) ( 1) 30,=⋅=+−+−=uuu

2222

|| || ( 7) ( 4) 6 101,=⋅=−+−+=vvv

and

2

|| || ( ) ( )+=+⋅+=uv uv uv

222

(2 ( 7)) ( 5 ( 4)) ( 1 6) 131.+−+−+−+−+=

24. One computes that

222

|| || ( ) ( ) 2 || || 2 || ||+=+⋅+=⋅+⋅+⋅=+⋅+uv uv uv uu uvvv u uv v

and

360 CHAPTER 6 • Orthogonality and Least Squares

222

|| || ( ) ( ) 2 || || 2 || ||−=−⋅− =⋅− ⋅+⋅=−⋅+uv uv uv uu uvvv u uv v

so

222 22 222

|| || || || || || 2 || || || || 2 || || 2|| || 2 || ||++−=+⋅++−⋅+= +uv uv u uv v u uv v u v

25. When

,

a

b

ªº

=«»

¬¼

v the set H of all vectors x

y

ª

º

«

»

¬

¼

that are orthogonal to v is the subspace of vectors whose

entries satisfy ax + by = 0. If a ≠ 0, then x = – (b/a)y with y a free variable, and H is a line through

the origin. A natural choice for a basis for H in this case is

.

b

a



½−

ª

º

°

®

¾

«

»

°

¬

¼

¯¿

If a = 0 and b ≠ 0, then by = 0.

Since b ≠ 0, y = 0 and x is a free variable. The subspace H is again a line through the origin. A

natural choice for a basis for H in this case is

1,

0



½

ª

º

°

®

¾

«

»

°

¬

¼

¯¿

but

b

a



½−

ª

º

°

®

¾

«

»

°

¬

¼

¯¿

is still a basis for H since a = 0

and b ≠ 0. If a = 0 and b = 0, then H =

2

since the equation 0x + 0y = 0 places no restrictions on x or

y.

26. Theorem 2 in Chapter 4 may be used to show that W is a subspace of

3

, because W is the null space

of the 1 × 3 matrix

.

T

u

Geometrically, W is a plane through the origin.

27. If y is orthogonal to u and v, then y ⋅ u = y ⋅ v = 0, and hence by a property of the inner product,

y ⋅ (u + v) = y ⋅ u + y ⋅ v = 0 + 0 = 0. Thus y is orthogonal to u + v.

28. An arbitrary w in Span{u, v} has the form

12

cc=+wuv

. If y is orthogonal to u and v, then

u ⋅ y = v ⋅ y = 0. By Theorem 1(b) and 1(c),

12 1 2

()()()000cc c c⋅=+ ⋅=⋅+⋅=+=wy u v y uy vy

29. A typical vector in W has the form

11

.

pp

cc=+…+wv v

If x is orthogonal to each

,

j

v

then by

Theorems 1(b) and 1(c),

11 1 1

()()()0

pp p p

ccc c⋅=+…+ ⋅=⋅+…+ ⋅=wx v v x v x v x

So x is orthogonal to each w in W.

30. a. If z is in

,

W

⊥

u is in W, and c is any scalar, then (cz) ⋅ u = c(z ⋅ u) = c

0 = 0. Since u is any

element of W, c

z is in

.W⊥

b. Let

1

z

and

2

z

be in

.W⊥

Then for any u in W,

12 1 2

() 000.+⋅=⋅+⋅=+=zzuzuzu

Thus

12

+zz

is in

.W⊥

c. Since 0 is orthogonal to every vector, 0 is in

.W⊥

Thus

W⊥

is a subspace.

31. Suppose that x is in W and

.W⊥

Since x is in

,

W

⊥

x is orthogonal to every vector in W, including x

itself. So x ⋅ x = 0, which happens only when x = 0.

6.2 • Solutions 361

32. [M]

a. One computes that

1234

|| || || || || || || || 1====aa aa

and that

0

ij

⋅=aa

for i ≠ j.

b. Answers will vary, but it should be that || Au || = || u || and || Av || = || v ||.

c. Answers will again vary, but the cosines should be equal.

d. A conjecture is that multiplying by A does not change the lengths of vectors or the angles

between vectors.

33. [M] Answers to the calculations will vary, but will demonstrate that the mapping

()T⋅

§·

=¨¸

⋅

©¹

xv

xx v

vv

6

(for v ≠ 0) is a linear transformation. To confirm this, let x and y be in

n

, and

let c be any scalar. Then

() ()()

()T+⋅⋅+⋅

§·§ ·

+= =

¨¸¨ ¸

⋅⋅

©¹© ¹

xyv xv yv

xy v v

vv vv () ()TT

⋅⋅

§·§·

=+=+

¨¸¨¸

⋅⋅

©¹©¹

xv yv

vvxy

vv vv

and

() ( )

() ()

cc

Tc c cT

⋅⋅ ⋅

§·§·§·

====

¨¸¨¸¨¸

⋅⋅⋅

©¹©¹©¹

xv xv xv

xvvvx

vv vv vv

34. [M] One finds that

51

1050 1/3

14

,01104/3

10

0001 1/3

01

03

NR

−

ªº

«» −

−ªº

«»

== −

«»

−¬¼

«»

¬¼

The row-column rule for computing RN produces the 3 × 2 zero matrix, which shows that the rows of

R are orthogonal to the columns of N. This is expected by Theorem 3 since each row of R is in Row

A and each column of N is in Nul A.

6.2 SOLUTIONS

Notes:

The nonsquare matrices in Theorems 6 and 7 are needed for the QR factorization in Section 6.4. It

is important to emphasize that the term orthogonal matrix applies only to certain square matrices. The

subsection on orthogonal projections not only sets the stage for the general case in Section 6.3, it also

provides what is needed for the orthogonal diagonalization exercises in Section 7.1, because none of the

eigenspaces there have dimension greater than 2. For this reason, the Gram-Schmidt process (Section 6.4)

is not really needed in Chapter 7. Exercises 13 and 14 are good preparation for Section 6.3.

1. Since

13

4420,

37

−

ªºªº

«»«»

⋅− =≠

«»«»

−−

¬¼¬¼

the set is not orthogonal.

362 CHAPTER 6 • Orthogonality and Least Squares

2. Since

10 1 5 0 5

21 2 2 1 20,

12 1 1 2 1

−−

ªºªºªºªºªºªº

«»«»«»«»«»«»

−⋅ =−⋅− =⋅− =

«»«»«»«»«»«»

¬¼¬¼¬¼¬¼¬¼¬¼

the set is orthogonal.

3. Since

63

31300,

91

−

ªºªº

«»«»

−⋅ =−≠

«»«»

−

¬¼¬¼

the set is not orthogonal.

4. Since

20 2 4 0 4

50 5 2 0 20,

30 3 6 0 6

ªºªºªºªºªºªº

«»«»«»«»«»«»

−⋅ =−⋅− =⋅− =

«»«»«»«»«»«»

−−

¬¼¬¼¬¼¬¼¬¼¬¼

the set is orthogonal.

5. Since

31 33 13

23 28 38

0,

13 17 37

34 30 40

−−

ªºªºªºªºªºªº

«»«»«»«»«»«»

−−

«»«»«»«»«»«»

⋅=⋅=⋅=

«»«»«»«»«»«»

−−

«»«»«»«»«»«»

¬¼¬¼¬¼¬¼¬¼¬¼

the set is orthogonal.

6. Since

43

1332 0,

35

81

−

ªºªº

«»«»

⋅=−≠

«»«»

−

«»«»

−

«»«»

¬¼¬¼

the set is not orthogonal.

7. Since

12

12 12 0,⋅=−=uu

12

{, }uu

is an orthogonal set. Since the vectors are non-zero,

1

u

and

2

u

are linearly independent by Theorem 4. Two such vectors in

2

automatically form a basis for

2

. So

12

{, }uu

is an orthogonal basis for

2

. By Theorem 5,

12

1212

11 2 2

1

32

⋅⋅

=+ =+

⋅⋅

xu xu

xu uuu

uu u u

8. Since

12 660,⋅=−+=uu

12

{, }uu

is an orthogonal set. Since the vectors are non-zero,

1

u

and

2

u

are linearly independent by Theorem 4. Two such vectors in

2

automatically form a basis for

2

. So

12

{, }uu

is an orthogonal basis for

2

. By Theorem 5,

12

1212

11 2 2

33

24

⋅⋅

=+ =−+

⋅⋅

xu xu

xu uuu

uu u u

9. Since

12 13 23

0,⋅=⋅=⋅=uu uu u u

,

123

{, }uu u

is an orthogonal set. Since the vectors are non-zero,

1,u

2,u

and

3

u

are linearly independent by Theorem 4. Three such vectors in

3

automatically form

a basis for

3

. So

123

{, , }uu u

is an orthogonal basis for

3

. By Theorem 5,

3

12

123123

11 2 2 3 3

532

22

⋅⋅⋅

=+ + =−+

⋅⋅ ⋅

xuxu xu

xu u uuuu

uu u u u u

6.2 • Solutions 363

10. Since

12 13 23

0,⋅=⋅=⋅=uu uu u u

,

123

{, }uuu

is an orthogonal set. Since the vectors are non-zero,

1,u

2,u

and

3

u

are linearly independent by Theorem 4. Three such vectors in

3

automatically form

a basis for

3

. So

123

{, , }uu u

is an orthogonal basis for

3

. By Theorem 5,

3

12

123123

11 2 2 3 3

411

333

⋅⋅⋅

=+ + =++

⋅⋅ ⋅

xuxu xu

xu u uuuu

uu u u uu

11. Let

1

7

ªº

=«»

¬¼

y and

4.

2

−

ª

º

=

«

»

¬

¼

u The orthogonal projection of y onto the line through u and the origin is

the orthogonal projection of y onto u, and this vector is

2

1

ˆ1

2

−

ªº

⋅

===

«»

⋅¬¼

yu

yuu

uu

12. Let

1

ªº

=«»

−

¬¼

y and

1.

3

−

ª

º

=

«

»

¬

¼

u The orthogonal projection of y onto the line through u and the origin is

the orthogonal projection of y onto u, and this vector is

2/5

2

ˆ6/5

5

ªº

⋅

==−=«»

−

⋅¬¼

yu

yuu

uu

13. The orthogonal projection of y onto u is

4/5

13

ˆ7/5

65

−

ªº

⋅

==−=«»

⋅¬¼

yu

yuu

uu

The component of y orthogonal to u is

ˆ14/5

ªº

−=«»

8/5

¬¼

yy

Thus

ˆˆ

−4/5 14/5

ª

ºª º

=+(−)= +

«

»« »

7/5 8/5

¬

¼¬ ¼

yy yy .

14. The orthogonal projection of y onto u is

14 / 5

2

ˆ2/5

5

ªº

⋅

===

«»

⋅¬¼

yu

yuu

uu

The component of y orthogonal to u is

ˆ

−4/5

ªº

−=«»

28/5

¬¼

yy

Thus

ˆˆ .

14/5 −4/5

ª

ºª º

=+(−)= +

«

»« »

2/5 28/5

¬

¼¬ ¼

yy yy

15. The distance from y to the line through u and the origin is ||y –

ˆ

y

||. One computes that

383/5

3

ˆ164/5

10

ªº ªº ª º

⋅

−=−=−=

«» «» « »

−

⋅¬¼ ¬¼ ¬ ¼

yu

yyy u

uu

364 CHAPTER 6 • Orthogonality and Least Squares

so

ˆ

|| −

||

=9/25+16/25=1yy

is the desired distance.

16. The distance from y to the line through u and the origin is ||y –

ˆ

y

||. One computes that

316

ˆ3

923

−−

ªº ªºªº

⋅

−=−=−=

«» «»«»

⋅¬¼ ¬¼¬¼

yu

yyy u

uu

so

ˆ

|| −

||

=36+9=35yy

is the desired distance.

17. Let

1/3

1/3 ,

1/3

ªº

«»

=«»

«»

¬¼

u

1/2

0.

1/2

−

ªº

«»

=«»

«»

¬¼

v

Since u ⋅ v = 0, {u, v} is an orthogonal set. However,

2

|| || 1/ 3=⋅=uuu

and

2

|| || 1/ 2,=⋅=vvv

so {u, v} is not an orthonormal set. The vectors u and v may be normalized to

form the orthonormal set

3/3 2/2

,3/3,0

|| || || || 3/3 2/2



½

ªº

−

°

«»

½

°

=«»

®¾® ¾

«»

¯¿

«»

°

«»

¬¼

°

¬¼

¯¿

uv

18. Let

0

1,

0

ªº

«»

=«»

«»

¬¼

u

0

1.

0

ªº

«»

=−

«»

¬¼

v

Since u ⋅ v = –1 ≠ 0, {u, v} is not an orthogonal set.

19. Let

.6 ,

.8

−

ªº

=«»

¬¼

u

.8 .

.6

ªº

=«»

¬¼

v

Since u ⋅ v = 0, {u, v} is an orthogonal set. Also,

2

|| || 1=⋅=uuu

and

2

|| || 1,=⋅=vvv

so {u, v} is an orthonormal set.

20. Let

2/3

1/3 ,

2/3

−

ªº

«»

=«»

«»

¬¼

u

1/3

2/3 .

0

ª

º

«

»

=

«

»

«

»

¬

¼

v

Since u ⋅ v = 0, {u, v} is an orthogonal set. However,

2

|| || 1=⋅=uuu

and

2

|| || 5/9,=⋅=vvv

so {u, v} is not an orthonormal set. The vectors u and v may be normalized

to form the orthonormal set

1/ 5

2/3

,1/3,2/5

|| || || || 2/3 0



½

ªº

−

ªº

°

«»

½

°

«»

=«»

®¾® ¾

«»

¯¿ «»

°

«»

¬¼

«»

°

¬¼

¯¿

uv

21. Let

1/ 10

3/ 20 ,

3/ 20

ªº

«»

=

«»

¬¼

u

3/ 10

1/ 20 ,

1/ 20

ª

º

«

»

=−

«

»

«

»

−

«

»

¬

¼

v and

0

1/ 2 .

1/ 2

ª

º

«

»

=−

«

»

«

»

¬

¼

w Since u ⋅ v = u ⋅ w = v ⋅ w = 0, {u, v, w} is an

orthogonal set. Also,

2

|| || 1,=⋅=uuu

2

|| || 1,=⋅=vvv

and

2

|| || 1,=⋅=www

so {u, v, w} is an

orthonormal set.

6.2 • Solutions 365

22. Let

1/ 18

4/ 18 ,

1/ 18

ªº

«»

=

«»

¬¼

u

1/ 2

0,

1/ 2

ª

º

«

»

=

«

»

«

»

−

¬

¼

v and

2/3

1/3 .

2/3

−

ª

º

«

»

=

«

»

«

»

−

¬

¼

w Since u ⋅ v = u ⋅ w = v ⋅ w = 0, {u, v, w} is an

orthogonal set. Also,

2

|| || 1,=⋅=uuu

2

|| || 1,=⋅=vvv

and

2

|| || 1,=⋅=www

so {u, v, w} is an

orthonormal set.

23. a. True. For example, the vectors u and y in Example 3 are linearly independent but not orthogonal.

b . True. The formulas for the weights are given in Theorem 5.

c . False. See the paragraph following Example 5.

d . False. The matrix must also be square. See the paragraph before Example 7.

e . False. See Example 4. The distance is ||y –

ˆ

y||.

24. a. True. But every orthogonal set of nonzero vectors is linearly independent. See Theorem 4.

b . False. To be orthonormal, the vectors is S must be unit vectors as well as being orthogonal to each

other.

c . True. See Theorem 7(a).

d . True. See the paragraph before Example 3.

e . True. See the paragraph before Example 7.

25. To prove part (b), note that

()( )()( )

TTTT

UU UU UU⋅====⋅xy xyx yxyxy

because

T

UU I=

. If y = x in part (b), (Ux) ⋅ (Ux) = x ⋅ x, which implies part (a). Part (c) of the

Theorem follows immediately fom part (b).

26. A set of n nonzero orthogonal vectors must be linearly independent by Theorem 4, so if such a set

spans W it is a basis for W. Thus W is an n-dimensional subspace of

n

, and

W=

n

.

27. If U has orthonormal columns, then

T

UU I=

by Theorem 6. If U is also a square matrix, then the

equation

T

UU I=

implies that U is invertible by the Invertible Matrix Theorem.

28. If U is an n × n orthogonal matrix, then

1T

IUU UU

−

==

. Since U is the transpose of

,

T

U Theorem

6 applied to

T

U

says that

T

U

has orthogonal columns. In particular, the columns of

T

U

are linearly

independent and hence form a basis for

n

by the Invertible Matrix Theorem. That is, the rows of U

form a basis (an orthonormal basis) for

n

.

29. Since U and V are orthogonal, each is invertible. By Theorem 6 in Section 2.2, UV is invertible and

111

() (),

TT T

UV V U V U UV

−−−

===

where the final equality holds by Theorem 3 in Section 2.1. Thus

UV is an orthogonal matrix.

30. If U is an orthogonal matrix, its columns are orthonormal. Interchanging the columns does not

change their orthonormality, so the new matrix – say, V – still has orthonormal columns. By

Theorem 6,

.

T

VV I=

Since V is square,

1T

VV

−

=

by the Invertible Matrix Theorem.

366 CHAPTER 6 • Orthogonality and Least Squares

31. Suppose that

ˆ.

⋅

=

⋅

yu

uu

Replacing u by cu with c ≠ 0 gives

2

22

() ( ) ( ) ˆ

() ()

()() () ()

cc c

cc

cc cc

⋅⋅⋅⋅

====

⋅⋅

yu yu yu yu

uuuuy

uu uu

So

ˆ

y does not depend on the choice of a nonzero u in the line L used in the formula.

32. If

12

0⋅=

vv , then by Theorem 1(c) in Section 6.1,

11 2 2 1 1 2 2 12 1 2 12

()( )[( )] ( ) 00cc cc cc cc⋅=⋅=⋅==

vv vv vv

33. Let L = Span{u}, where u is nonzero, and let

()T

⋅

=

⋅

xu

uu

. For any vectors x and y in

n

and any

scalars c and d, the properties of the inner product (Theorem 1) show that

()

cd

Tc d

+⋅

+=

⋅

xyu

xy u

uu

cd

⋅+⋅

=

⋅

xu yu

u

uu

cd

⋅⋅

=+

⋅⋅

xu yu

uu

uu uu

() ()cT dT=+xy

Thus T is a linear transformation. Another approach is to view T as the composition of the following

three linear mappings: x6a = x ⋅ v, a 6b = a / v ⋅ v, and b 6bv.

34. Let L = Span{u}, where u is nonzero, and let

() refl 2proj

LL

T== −

xy yy

. By Exercise 33, the

mapping

projL

yy

6

is linear. Thus for any vectors y and z in

n

and any scalars c and d,

()2proj()()

L

Tc d c d c d+= +−+

yz yz yz

2( proj proj )

LL

cd cd=+−−

yzyz

2proj 2 proj

LL

ccdd=−+−

yy z z

(2 proj ) (2 proj )

LL

cd=−+−

yy zz

() ()cT d T=+yz

Thus T is a linear transformation.

35. [M] One can compute that

4

100 .

T

AA I=

Since the off-diagonal entries in

T

AA are zero, the columns

of A are orthogonal.

6.3 • Solutions 367

36. [M]

a. One computes that

4

,

T

UU I=

while

82 0 20 8 6 20 24 0

04224 020 62032

20 24 58 20 0 32 0 6

802082242060

1

620 02418 0 820

100

20 6 32 20 0 58 0 24

24 20 0 6 8 0 18 20

032 6 020242042

T

UU

−

ªº

«»

−−

«»

−

«»

−

§·

«»

=¨¸

«»

−−

©¹

«»

−

«»

−−

«»

−−

«»

¬¼

The matrices

T

UU

and

T

UU

are of different sizes and look nothing like each other.

b. Answers will vary. The vector

T

UU=py

is in Col

U because

()

T

UU=py

. Since the columns of

U are simply scaled versions of the columns of A, Col

U = Col

A. Thus each p is in Col A.

c. One computes that

T

U=

z0.

d. From (c), z is orthogonal to each column of A. By Exercise 29 in Section 6.1, z must be

orthogonal to every vector in Col A; that is, z is in

(Col ) .A

⊥

6.3 SOLUTIONS

Notes:

Example 1 seems to help students understand Theorem 8. Theorem 8 is needed for the Gram-

Schmidt process (but only for a subspace that itself has an orthogonal basis). Theorems 8 and 9 are

needed for the discussions of least squares in Sections 6.5 and 6.6. Theorem 10 is used with the QR

factorization to provide a good numerical method for solving least squares problems, in Section 6.5.

Exercises 19 and 20 lead naturally into consideration of the Gram-Schmidt process.

1. The vector in

4

Span{ }

u is

4

444

44

10

6

72 22

36

2

ªº

«»

−

⋅«»

===

«»

−

⋅«»

«»

¬¼

xu uuu

uu

Since

4

11 2 2 3 3 4

44

,cc c ⋅

=+ + +

⋅

xu

xu u u u

uu

the vector

4

44

10 10 0

862

224

022

ª

ºª ºª º

«

»« »« »

−−−

⋅

«

»« »« »

−=−=

«

»« »« »

−

⋅

«

»« »« »

−

«

»« »« »

¬

¼¬ ¼¬ ¼

xu

uu

is in

123

Span{ , , }.

uu u

368 CHAPTER 6 • Orthogonality and Least Squares

2. The vector in

1

Span{ }

u is

1

111

11

2

4

14 22

7

2

ªº

«»

⋅«»

===

«»

⋅

«»

¬¼

vu uuu

uu

Since

1

1223344

11

,ccc

⋅

=+++

⋅

vu

xuuuu

uu

the vector

1

11

42 2

54 1

32 5

32 1

ª

ºªºª º

«

»«»« »

⋅

«

»«»« »

−=−=

«

»«»« »

−−

⋅

«

»«»« »

«

»«»« »

¬

¼¬¼¬ ¼

vu

uu

is in

234

Span{ , , }.

uuu

3. Since

12 110 0,⋅=−++ =

uu

12

{, }

uu is an orthogonal set. The orthogonal projection of y onto

12

Span{ , }

uu is

12

1212

11 2 2

111

35 3 5

ˆ114

22 2 2

000

−−

ª

ºªºªº

⋅⋅

«

»«»«»

=+ =+=+=

«

»«»«»

⋅⋅

«

»«»«»

¬

¼¬¼¬¼

yu yu

yu uuu

uu u u

4. Since

12 12 12 0 0,⋅=−++=

uu

12

{, }

uu is an orthogonal set. The orthogonal projection of y onto

12

Span{ , }

uu is

12

1212

11 2 2

346

30 15 6 3

ˆ433

25 25 5 5

000

−

ª

ºªºªº

⋅⋅

«

»«»«»

=+ =−=−=

«

»«»«»

⋅⋅

«

»«»«»

¬

¼¬¼¬¼

yu yu

yu uuu

uu u u

5. Since

12

314 0,⋅=+−=

uu

12

{, }

uu is an orthogonal set. The orthogonal projection of y onto

12

Span{ , }

uu is

12

1212

11 2 2

311

7151 5

ˆ112

14 6 2 2

226

−

ª

ºªºªº

⋅⋅

«

»«»«»

=+ =−=−− − =

«

»«»«»

⋅⋅

«

»«»«»

−

¬

¼¬¼¬¼

yu yu

yu uuu

uu u u

6. Since

12

0110,⋅=−+=

uu

12

{, }

uu is an orthogonal set. The orthogonal projection of y onto

12

Span{ , }

uu is

12

1212

11 2 2

406

27 5 3 5

ˆ114

18 2 2 2

111

−

ª

ºªºªº

⋅⋅

«

»«»«»

=+ =−+=−−+=

«

»«»«»

⋅⋅

«

»«»«»

¬

¼¬¼¬¼

yu yu

yu uuu

uu u u

7. Since

12

5380,⋅=+−=

uu

12

{, }

uu is an orthogonal set. By the Orthogonal Decomposition

Theorem,

6.3 • Solutions 369

12

1212

11 2 2

10 / 3 7 / 3

2

ˆˆ

02/3, 7/3

38/3 7/3

−

ª

ºªº

⋅⋅

«

»«»

=+ =+= =−=

«

»«»

⋅⋅

«

»«»

¬

¼¬¼

yu yu

yu uuu zyy

uu u u

and y =

ˆ

y+ z, where

ˆ

y is in W and z is in

.W⊥

8. Since

12 132 0,⋅=−+−=

uu

12

{, }

uu is an orthogonal set. By the Orthogonal Decomposition

Theorem,

12

1212

11 2 2

3/2 5/2

1

ˆˆ

27/2, 1/2

212

−

ª

ºªº

⋅⋅

«

»«»

=+ =+= =−=

«

»«»

⋅⋅

«

»«»

¬

¼¬¼

yu yu

yu uuu zyy

uu u u

and y =

ˆ

y+ z, where

ˆ

y is in W and z is in

.W⊥

9. Since

12 13 23

0,⋅=⋅=⋅=

uu uu u u

123

{, , }

uu u is an orthogonal set. By the Orthogonal

Decomposition Theorem,

312

123123

11 2 2 3 3

22

41

22

ˆˆ

2,

03

33

01

ª

ºªº

«

»«»

−

⋅

⋅⋅

«

»«»

=+ + =+−==−=

«

»«»

⋅⋅ ⋅

«

»«»

−

«

»«»

¬

¼¬¼

yu

yu yu

yu u uuuuzyy

uu uu uu

and y =

ˆ

y+ z, where

ˆ

y is in W and z is in

.W⊥

10. Since

12 13 23

0,⋅=⋅=⋅=

uu uu u u

123

{, , }

uu u is an orthogonal set. By the Orthogonal

Decomposition Theorem,

312

123123

11 2 2 3 3

52

22

114 5

ˆˆ

,

32

333

60

−

ª

ºªº

«

»«»

⋅

⋅⋅

«

»«»

=+ + =+−==−=

«

»«»

⋅⋅ ⋅

«

»«»

«

»«»

¬

¼¬¼

yu

yu yu

yu u uuuuzyy

uu u u uu

and y =

ˆ

y+ z, where

ˆ

y is in W and z is in

.W⊥

11. Note that

1

v and

2

v are orthogonal. The Best Approximation Theorem says that

ˆ

y, which is the

orthogonal projection of y onto

12

Span{ , },W=

vv is the closest point to y in W. This vector is

12

1212

11 2 2

3

1

13

ˆ1

22

1

ª

º

«

»

−

⋅⋅

«

»

=+ =+=

«

»

⋅⋅

«

»

−

«

»

¬

¼

yv yv

yv vvv

vv v v

12. Note that

1

v and

2

v are orthogonal. The Best Approximation Theorem says that

ˆ

y, which is the

orthogonal projection of y onto

12

Span{ , },W=

vv is the closest point to y in W. This vector is

370 CHAPTER 6 • Orthogonality and Least Squares

12

1212

11 2 2

1

5

ˆ31 3

9

−

ª

º

«

»

−

⋅⋅

«

»

=+ =+=

«

»

−

⋅⋅

«

»

«

»

¬

¼

yv yv

yv vvv

vv v v

13. Note that

1

v and

2

v are orthogonal. By the Best Approximation Theorem, the closest point in

12

Span{ , }

vv to z is

12

1212

11 2 2

1

3

27

ˆ2

33

3

−

ª

º

«

»

−

⋅⋅

«

»

=+ =−=

«

»

−

⋅⋅

«

»

«

»

¬

¼

zv zv

zv vvv

vv v v

14. Note that

1

v and

2

v are orthogonal. By the Best Approximation Theorem, the closest point in

12

Span{ , }

vv to z is

12

1212

11 2 2

1

0

1

ˆ01/2

2

3/2

ª

º

«

»

⋅⋅

«

»

=+ =+=

«

»

−

⋅⋅

«

»

−

«

»

¬

¼

zv zv

zv vvv

vv v v

15. The distance from the point y in

3

to a subspace W is defined as the distance from y to the closest

point in W. Since the closest point in W to y is

ˆproj ,

W

=

yy

the desired distance is || y –

ˆ

y||. One

computes that

32

ˆˆ

90,

16

ªº ªº

«» «»

=−,−=

«» «»

−

¬¼ ¬¼

yyy

and

ˆ

|| 40 10.−|| = = 2yy

16. The distance from the point y in

4

to a subspace W is defined as the distance from y to the closest

point in W. Since the closest point in W to y is

ˆproj ,

W

=

yy

the desired distance is || y –

ˆ

y||. One

computes that ˆˆ

,

−14

ªº ªº

«» «»

−54

«» «»

=,−=

«» «»

−34

«» «»

94

«» «»

¬¼ ¬¼

yyy

and || y –

ˆ

y|| = 8.

17. a.

8/9 2/9 2/9

10

,2/95/94/9

01 2/9 4/9 5/9

TT

UU UU

−

ªº

ªº «»

==−

«» «»

¬¼ «»

¬¼

b . Since

2

,

T

UU I=

the columns of U form an orthonormal basis for W, and by Theorem 10

8/9 2/9 2/9 4 2

proj 2 / 9 5 / 9 4 / 9 8 4 .

2/9 4/9 5/9 1 5

T

WUU

−

ª

ºª º ª º

«

»« » « »

==−=

«

»« » « »

«

»« » « »

¬

¼¬ ¼ ¬ ¼

yy

6.3 • Solutions 371

18. a.

[]

1/10 3/10

11, 3/10 9/10

TT

UU UU

−

ªº

== =

«»

−

¬¼

b . Since

1,

T

UU

=

1

{}

u forms an orthonormal basis for W, and by Theorem 10

1/10 3/10 7 2

proj .

3/10 9/10 9 6

T

W

UU

−−

ª

ºª º ª º

== =

«

»« » « »

−

¬

¼¬ ¼ ¬ ¼

yy

19. By the Orthogonal Decomposition Theorem,

3

u is the sum of a vector in

12

Span{ , }W=

uu and a

vector v orthogonal to W. This exercise asks for the vector v:

33312

000

11

proj 0 2 / 5 2 / 5

315 14/51/5

W

ª

ºª ºª º

§·

«

»« »« »

=−=−− +=−− =

¨¸

«

»« »« »

©¹

«

»« »« »

¬

¼¬ ¼¬ ¼

vu u u u u

Any multiple of the vector v will also be in

.W⊥

20. By the Orthogonal Decomposition Theorem,

4

u is the sum of a vector in

12

Span{ , }W=

uu and a

vector v orthogonal to W. This exercise asks for the vector v:

44412

000

11

proj 1 1/ 5 4 / 5

630 02/52/5

W

ª

ºª ºª º

§·

«

»« »« »

=−=−− =−=

¨¸

«

»« »« »

©¹

«

»« »« »

−

¬

¼¬ ¼¬ ¼

vu u u u u

Any multiple of the vector v will also be in

.W⊥

21. a. True. See the calculations for

2

z in Example 1 or the box after Example 6 in Section 6.1.

b . True. See the Orthogonal Decomposition Theorem.

c . False. See the last paragraph in the proof of Theorem 8, or see the second paragraph after the

statement of Theorem 9.

d . True. See the box before the Best Approximation Theorem.

e . True. Theorem 10 applies to the column space W of U because the columns of U are linearly

independent and hence form a basis for W.

22. a. True. See the proof of the Orthogonal Decomposition Theorem.

b . True. See the subsection “A Geometric Interpretation of the Orthogonal Projection.”

c . True. The orthgonal decomposition in Theorem 8 is unique.

d . False. The Best Approximation Theorem says that the best approximation to y is

proj .

W

y

e . False. This statement is only true if x is in the column space of U. If n > p, then the column space

of U will not be all of

n

, so the statement cannot be true for all x in

n

.

23. By the Orthogonal Decomposition Theorem, each x in

n

can be written uniquely as x = p + u, with

p in Row A and u in

(Row ) .A

⊥

By Theorem 3 in Section 6.1,

(Row ) Nul ,AA

⊥

=

so u is in Nul

A.

Next, suppose Ax = b is consistent. Let x be a solution and write x = p + u as above. Then

Ap = A(x – u) = Ax – Au = b – 0 = b, so the equation Ax = b has at least one solution p in Row A.

Finally, suppose that p and

1

p are both in Row

A and both satisfy Ax = b. Then

1

−

pp

is in

Nul (Row ) ,AA

⊥

=

since

11

()AAA−=−=−=

pp p p bb0

. The equations

1()=+−1

pp pp and

372 CHAPTER 6 • Orthogonality and Least Squares

p = p + 0 both then decompose p as the sum of a vector in Row

A and a vector in

(Row )A

⊥

. By

the uniqueness of the orthogonal decomposition (Theorem 8),

1,=

pp

and p is unique.

24. a. By hypothesis, the vectors

1

w, …,

p

w

are pairwise orthogonal, and the vectors

1

v, …,

q

v

are

pairwise orthogonal. Since

i

w is in W for any i and

j

v

is in

W⊥

for any j,

0

ij

⋅=wv

for any i

and j. Thus

11

{,, ,,,}

pq

……wwvv

forms an orthogonal set.

b. For any y in

n

, write y =

ˆ

y+ z as in the Orthogonal Decomposition Theorem, with

ˆ

y in

W and z in

W⊥

. Then there exist scalars

1

,,

p

cc…

and

1

,,

q

dd…

such that

ˆ

=+=

yyz

11 11pp qq

ccdd+…+ + +…+wwvv

. Thus the set

11

{,, ,,,}

pq

……wwvv

spans

n

.

c . The set

11

{,, ,,,}

pq

……wwvv

is linearly independent by (a) and spans

n

by (b), and is thus a

basis for

n

. Hence

dim dim dimWWpq

⊥

+=+=

n

.

25. [M] Since

4

T

UU I=

, U has orthonormal columns by Theorem 6 in Section 6.2. The closest point to

y in Col U is the orthogonal projection

ˆ

y of y onto Col U. From Theorem 10,

ˆUU Τ

1.2

ªº

«»

.4

«»

1.2

«»

1.2

«»

==

«»

.4

«»

1.2

«»

.4

«»

.4

«»

¬¼

yy

26. [M] The distance from b to Col

U is || b –

ˆ

b||, where

ˆ.UU Τ

=

bb One computes that

ˆˆˆ

UU Τ

.2 .8

ªº ªº

«» «»

.92 .08

«» «»

.44 .56

«» «»

10

112

«» «»

==,−=,||−||=

«» «»

−.2 −.8 5

«» «»

−.44 −.56

«» «»

.6 −1.6

«» «»

−.92 −.08

«» «»

¬¼ ¬¼

bb bb bb

which is 2.1166 to four decimal places.

6.4 • Solutions 373

6.4 SOLUTIONS

Notes:

The QR factorization encapsulates the essential outcome of the Gram-Schmidt process, just as the

LU factorization describes the result of a row reduction process. For practical use of linear algebra, the

factorizations are more important than the algorithms that produce them. In fact, the Gram-Schmidt

process is not the appropriate way to compute the QR factorization. For that reason, one should consider

deemphasizing the hand calculation of the Gram-Schmidt process, even though it provides easy exam

questions.

The Gram-Schmidt process is used in Sections 6.7 and 6.8, in connection with various sets of

orthogonal polynomials. The process is mentioned in Sections 7.1 and 7.4, but the one-dimensional

projection constructed in Section 6.2 will suffice. The QR factorization is used in an optional subsection

of Section 6.5, and it is needed in Supplementary Exercise 7 of Chapter 7 to produce the Cholesky

factorization of a positive definite matrix.

1. Set

11

=

vx

and compute that

21

22 12 1

11

1

35.

3

−

ª

º

⋅

«

»

=−=−=

«

»

⋅

«

»

−

¬

¼

xv

vx vx v

vv Thus an orthogonal basis for W

is

31

0, 5 .

13

½−

ªºªº

°°

«»«»

®¾

«»«»

°°

«»«»

−−

¬¼¬¼

¯¿

2. Set

11

=

vx

and compute that

21

22 12 1

11

5

14.

28

ª

º

⋅

«

»

=−=−=

«

»

⋅

«

»

−

¬

¼

xv

vx vx v

vv Thus an orthogonal basis for W

is

05

4, 4 .

28

½

ªºª º

°°

«»« »

®¾

«»« »

°°

«»« »

−

¬¼¬ ¼

¯¿

3. Set

11

=

vx

and compute that

21

22 12 1

11

3

13/2 .

23/2

ª

º

⋅

«

»

=−=−=

«

»

⋅

«

»

¬

¼

xv

vx vx v

vv Thus an orthogonal basis for

W is

23

5,3/2 .

13/2

½

ªºª º

°°

«»« »

−

®¾

«»« »

°°

«»« »

¬¼¬ ¼

¯¿

4. Set

11

=

vx

and compute that

21

22 12 1

11

3

(2) 6.

3

ª

º

⋅

«

»

=−=−− =

«

»

⋅

«

»

¬

¼

xv

vx vx v

vv Thus an orthogonal basis for

W is

33

4,6 .

53

½

ªºªº

°°

«»«»

−

®¾

«»«»

°°

«»«»

¬¼¬¼

¯¿

374 CHAPTER 6 • Orthogonality and Least Squares

5. Set

11

=

vx

and compute that

21

22 12 1

11

5

1

2.

4

1

ª

º

«

»

⋅

«

»

=−=−=

«

»

−

⋅

«

»

−

«

»

¬

¼

xv

vx vx v

vv Thus an orthogonal basis for W

is

15

41

,.

04

11

½

ªºªº

°°

«»«»

−

°°

«»«»

®¾

«»«»

−

°°

«»«»

°°

−

«»«»

¬¼¬¼

¯¿

6. Set

11

=

vx

and compute that

21

22 12 1

11

4

6

(3) .

3

0

ª

º

«

»

⋅

«

»

=−=−− =

«

»

−

⋅

«

»

«

»

¬

¼

xv

vx vx v

vv Thus an orthogonal basis for

W is

34

16

,.

23

10

½

ªºªº

°°

«»«»

−

°°

«»«»

®¾

«»«»

−

°°

«»«»

°°

−

«»«»

¬¼¬¼

¯¿

7. Since

1

|| || 30=v

and

2

|| || 27/ 2 3 6 / 2,==v

an orthonormal basis for W is

12

2/ 30 2/ 6

,5/30,1/6.

|| || || || 1/ 30 1/ 6

½

ªºªº

°°

«»«»

½

°°

=−

«»«»

®¾® ¾

«»«»

¯¿

°°

«»«»

°°

¬¼¬¼

¯¿

vv

8. Since

1

|| || 50=v

and

2

|| || 54 3 6,==v

an orthonormal basis for W is

12

3/ 50 1/ 6

,4/50,2/6.

|| || || || 5/ 50 1/ 6

½

ªºªº

°°

«»«»

½

°°

=−

«»«»

®¾® ¾

«»«»

¯¿

°°

«»«»

°°

¬¼¬¼

¯¿

vv

9. Call the columns of the matrix

1

x,

2

x, and

3

x and perform the Gram-Schmidt process on these

vectors:

11

=

vx

21

22 12 1

11

1

3

(2) 3

1

ª

º

«

»

⋅

«

»

=−=−− =

«

»

⋅

«

»

−

«

»

¬

¼

xv

vx vx v

vv

6.4 • Solutions 375

31 32

33 1 23 1 2

11 2 2

3

1

31

1

22

3

−

ª

º

«

»

⋅⋅ §·

«

»

=−− =−−− =

¨¸

«

»

⋅⋅ ©¹

«

»

«

»

¬

¼

xv xv

vx v vx v v

vv v v

Thus an orthogonal basis for W is

313

13 1

,, .

13 1

313

½−

ªºªºªº

°°

«»«»«»

°°

«»«»«»

®¾

«»«»«»

−

°°

«»«»«»

°°

−

«»«»«»

¬¼¬¼¬¼

¯¿

10. Call the columns of the matrix

1

x,

2

x, and

3

x and perform the Gram-Schmidt process on these

vectors:

11

=

vx

21

22 12 1

11

3

1

(3) 1

1

ª

º

«

»

⋅

«

»

=−=−− =

«

»

⋅

«

»

−

«

»

¬

¼

xv

vx vx v

vv

31 32

33 1 23 1 2

11 2 2

1

15

3

22

1

−

ª

º

«

»

−

⋅⋅

«

»

=−− =−− =

«

»

⋅⋅

«

»

−

«

»

¬

¼

xv xv

vx v vx v v

vv v v

Thus an orthogonal basis for W is

131

311

,, .

113

111

½−−

ªºªºªº

°°

«»«»«»

−

°°

«»«»«»

®¾

«»«»«»

°°

«»«»«»

°°

−−

«»«»«»

¬¼¬¼¬¼

¯¿

11. Call the columns of the matrix

1

x,

2

x, and

3

x and perform the Gram-Schmidt process on these

vectors:

11

=

vx

21

22 12 1

11

3

0

(1) 3

3

ª

º

«

»

«

»

⋅

«

»

=−=−− =

⋅

«

»

−

«

»

«

»

¬

¼

xv

vx vx v

vv

31 3 2

33 1 23 1 2

11 2 2

2

0

1

42

32

2

ª

º

«

»

«

»

⋅⋅ §·

«

»

=−− =−−− =

¨¸

⋅⋅ ©¹

«

»

«

»

«

»

−

¬

¼

xv xv

vx v vx v v

vv v v

376 CHAPTER 6 • Orthogonality and Least Squares

Thus an orthogonal basis for W is

132

100

,, .

132

½

ªºªºªº

°°

«»«»«»

−

°°

«»«»«»

°°

«»«»«»

−

®¾

«»«»«»

°°

−

«»«»«»

°°

«»«»«»

−

°°

¬¼¬¼¬¼

¯¿

12. Call the columns of the matrix

1

x,

2

x, and

3

x and perform the Gram-Schmidt process on these

vectors:

11

=

vx

21

22 12 1

11

1

42

1

−

ªº

«»

⋅«»

=−=−=

⋅«»

«»

¬¼

xv

vx vx v

vv

31 32

33 1 23 1 2

11 2 2

3

73 0

22 3

3

ª

º

«

»

«

»

⋅⋅

«

»

=−− =−− =

⋅⋅

«

»

−

«

»

«

»

¬

¼

xv xv

vx v vx v v

vv v v

Thus an orthogonal basis for W is

113

,, .

020

113

½

−

ªºªºªº

°°

«»«»«»

−

°°

«»«»«»

°°

«»«»«»

®¾

«»«»«»

°°

−

«»«»«»

°°

«»«»«»

°°

¬¼¬¼¬¼

¯¿

13. Since A and Q are given,

59

5/6 1/6 3/6 1/6 1 7 6 12

1/6 5/ 6 1/ 6 3/ 6 3 5 0 6

15

T

RQA

ªº

«»

−

ªºªº

«»

== =

«»«»

«»

−−−

¬¼¬¼

«»

¬¼

14. Since A and Q are given,

23

2/7 5/7 2/7 4/7 5 7 7 7

5/7 2/7 4/7 2/7 2 2 0 7

46

T

RQA

−

ªº

«»

−

ªºªº

«»

== =

«»«»

«»

−−

¬¼¬¼

«»

¬¼

15. The columns of Q will be normalized versions of the vectors

1

v,

2

v, and

3

v found in Exercise 11.

Thus

6.4 • Solutions 377

1/ 5 1/2 1/ 2

1/ 5 0 0 5545

,062

1/ 5 1/2 1/ 2

004

1/ 5 1/2 1/ 2

T

QRQA

ªº

«»

ª

º

−

«»

−

«

»

«»

===−

−

«

»

«»

«

»

«»

−

¬

¼

«»

−

¬¼

16. The columns of Q will be normalized versions of the vectors

1

v,

2

v, and

3

v found in Exercise 12.

Thus

1/2 1/(2 2) 1/ 2

287

1/2 1/(2 2) 1/ 2

,02232

01/20

006

1/2 1/(2 2) 1/ 2

T

QRQA

ªº

−

«»

ª

º

−

«»

«

»

«»

===

«

»

«»

«

»

«»

−

¬

¼

«»

¬¼

17. a. False. Scaling was used in Example 2, but the scale factor was nonzero.

b . True. See (1) in the statement of Theorem 11.

c . True. See the solution of Example 4.

18. a. False. The three orthogonal vectors must be nonzero to be a basis for a three-dimensional

subspace. (This was the case in Step 3 of the solution of Example 2.)

b . True. If x is not in a subspace w, then x cannot equal

projW

x, because

projW

x is in W. This idea

was used for

1k+

v in the proof of Theorem 11.

c . True. See Theorem 12.

19. Suppose that x satisfies Rx = 0; then Q Rx = Q0 = 0, and Ax = 0. Since the columns of A are linearly

independent, x must be 0. This fact, in turn, shows that the columns of R are linearly indepedent.

Since R is square, it is invertible by the Invertible Matrix Theorem.

20. If y is in Col

A, then y = Ax for some x. Then y = QRx = Q(Rx), which shows that y is a linear

combination of the columns of Q using the entries in Rx as weights. Conversly, suppose that y = Qx

for some x. Since R is invertible, the equation A = QR implies that

1

QAR

−

=

. So

11

(),AR A R

−−

==yx x

which shows that y is in Col A.

21. Denote the columns of Q by

1

{, , }

n

…

qq

. Note that n ≤ m, because A is m × n and has linearly

independent columns. The columns of Q can be extended to an orthonormal basis for

m

as follows.

Let

1

f be the first vector in the standard basis for

m

that is not in

1

Span{ , , },

nn

W=…

qq

let

11 1

proj

n

W

=−uf f

, and let

11 1

/

|| ||.

n+=

quu

Then

11

{, , , }

nn+

…

qqq

is an orthonormal basis for

11 1

Span{ , , , }.

nnn

W++

=…

qqq Next let

2

f be the first vector in the standard basis for

m

that is

not in

1n

W+

, let

1

22 2

proj ,

n

W

+

=−uf f

and let

22 2

/

|| || .

n+=

quu

Then

112

{, , , , }

nn n++

…

qqqq

is an

orthogonal basis for

21 12

Span{ , , , , }.

nnnn

W+++

=…

qqqq This process will continue until m – n vectors

have been added to the original n vectors, and

11

{, , , , , }

nn m+

……

qqq q

is an orthonormal basis for

m

.

378 CHAPTER 6 • Orthogonality and Least Squares

Let

[]

01nm

Q

+

=…qq

and

[]

10

QQQ=

. Then, using partitioned matrix multiplication,

1.

R

QQRA

O

ªº

==

«»

¬¼

22. We may assume that

1

{, , }

p

…uu

is an orthonormal basis for W, by normalizing the vectors in the

original basis given for W, if necessary. Let U be the matrix whose columns are

1

,, .

p

…uu

Then, by

Theorem 10 in Section 6.3,

() proj ( )

T

W

TUU==xxx

for x in

n

. Thus T is a matrix transformation

and hence is a linear transformation, as was shown in Section 1.8.

23. Given A = QR, partition

[]

12

AAA=

, where

1

A

has p columns. Partition Q as

[]

12

QQQ=

where

1

Q

has p columns, and partition R as

11 12

22

,

RR

ROR

ª

º

=

«

»

¬

¼

where

11

R

is a p × p matrix. Then

[][ ] [ ]

11 12

12 12 111112222

22

RR

AAA QRQQ QR QR QR

OR

ªº

=== = +

«»

¬¼

Thus

1111

.AQR=

The matrix

1

Q

has orthonormal columns because its columns come from Q. The

matrix

11

R

is square and upper triangular due to its position within the upper triangular matrix R. The

diagonal entries of

11

R

are positive because they are diagonal entries of R. Thus

111

QR

is a QR

factorization of

1

A

.

24. [M]

Call the columns of the matrix

1

x,

2

x,

3

x, and

4

x and perform the Gram-Schmidt process on

these vectors:

11

=

vx

21

22 12 1

11

3

(1) 3

0

3

ª

º

«

»

«

»

⋅

«

»

=−=−− =−

⋅

«

»

«

»

«

»

¬

¼

xv

vx vx v

vv

31 32

33 1 23 1 2

11 2 2

6

0

14

6

23

6

0

ª

º

«

»

«

»

⋅⋅ §· §·

«

»

=−− =−− −− =

¨¸ ¨¸

⋅⋅ ©¹ ©¹

«

»

«

»

«

»

¬

¼

xv xv

vx v vx v v

vv v v

43

41 4 2

44 1 2 34 1 2 3

11 2 2 3 3

11

(1)

22

⋅

⋅⋅ §·

=−− − =−−−−−

¨¸

⋅⋅ ⋅ ©¹

xv

xv xv

vx v v vx v v v

vv v v v v

0

5

0

5

ªº

«»

=«»

«»

−

¬¼

6.5 • Solutions 379

Thus an orthogonal basis for W is

10 3 6 0

2305

,,, .

6360

16 0 6 0

2305

½

−

ªºªºªºªº

°°

«»«»«»«»

°°

«»«»«»«»

°°

«»«»«»«»

−−

®¾

«»«»«»«»

°°

«»«»«»«»

°°

«»«»«»«»

−

°°

¬¼¬¼¬¼¬¼

¯¿

25. [M] The columns of Q will be normalized versions of the vectors

1

v,

2

v, and

3

v found in Exercise

24. Thus

1/2 1/ 2 1/ 3 0 20 20 10 10

1/10 1/2 0 1/ 2 06 8 6

,

3/10 1/2 1/ 3 0 006333

4/5 0 1/ 3 0 00 052

1/10 1/2 0 1/ 2

T

QRQA

ªº

−

«»

−−

ª

º

«»

«

»

−−

«»

«

»

===

−−

«»

«

»

−

«»

«

»

«»

«

»

¬

¼

«»

−

¬¼

26. [M] In MATLAB, when A has n columns, suitable commands are

Q = A(:,1)/norm(A(:,1))

% The first column of Q

for j=2: n

v=A(:,j) – Q*(Q’*A(:,j))

Q(:,j)=v/norm(v)

% Add a new column to Q

end

6.5 SOLUTIONS

Notes:

This is a core section – the basic geometric principles in this section provide the foundation

for all the applications in Sections 6.6–6.8. Yet this section need not take a full day. Each example

provides a stopping place. Theorem 13 and Example 1 are all that is needed for Section 6.6. Theorem 15,

however, gives an illustration of why the QR factorization is important. Example 4 is related to Exercise

17 in Section 6.6.

1. To find the normal equations and to find ˆ

x, compute

12

121 6 11

23

233 1122

13

T

AA

−

ªº

−− −

ªº ªº

«»

=−=

«» «»

«»

−−

¬¼ ¬¼

«»

−

¬¼

4

121 4

1

233 11

2

T

A

ªº

−− −

ªºªº

«»

==

«»«»

«»

−

¬¼¬¼

«»

¬¼

b

380 CHAPTER 6 • Orthogonality and Least Squares

a. The normal equations are

()

TT

AA A=xb

:

1

2

611 4

.

11 22 11

x

−−

ªº

ª

ºªº

=

«»

«

»«»

−

¬

¼¬¼¬¼

b. Compute

1

611 4 22114

1

ˆ

x( ) 11 22 11 11 6 11

11

TT

AA A

−

−− −

ªºªºªºªº

== =

«»«»«»«»

−

¬¼¬¼¬¼¬¼

b

33 3

1

22 2

11

ªºªº

==

«»«»

¬¼¬¼

2. To find the normal equations and to find

ˆ,

x compute

21

222 128

20

103 810

23

T

AA

ªº

−

ªº ªº

«»

=−=

«» «»

«»

¬¼ ¬¼

«»

¬¼

5

222 24

8

103 2

1

T

A

−

ªº

−−

ªºªº

«»

==

«»«»

«» −

¬¼¬¼

«»

¬¼

b

a. The normal equations are

()

TT

AA A=xb

:

1

2

12 8 24 .

810 2

x

−

ªº

ª

ºªº

=

«»

«

»«»

−

¬

¼¬¼¬¼

b. Compute

1

12 8 24 10 8 24

1

ˆ

x( ) 810 2 8 12 2

56

TT

AA A

−

−−−

ªºªºª ºªº

== =

«»«»« »«»

−− −

¬¼¬¼¬ ¼¬¼

b

224 4

1

168 3

56

−−

ªºªº

==

«»«»

¬¼¬¼

3. To find the normal equations and to find ˆ

x, compute

12

110212 66

22350 3 642

25

T

AA

−

ªº

«»

−−

ªºªº

«»

==

«»«»

«»

−

¬¼¬¼

«»

¬¼

3

11021 6

22354 6

2

T

A

ªº

«»

−

ªºªº

«»

==

«»«»

«»

−−−

¬¼¬¼

«»

¬¼

b

a. The normal equations are

()

TT

AA A=xb

:

1

2

66 6

642 6

x

ªº

ª

ºªº

=

«»

«

»«»

−

¬

¼¬¼¬¼

b. Compute

6.5 • Solutions 381

66 6 4266

1

ˆ642 6 6 6 6

216

TT

−1

−

ªºªºª ºªº

=(ΑΑ)Α==

«»«»« »«»

−−−

¬¼¬¼¬ ¼¬¼

xb

288 4 / 3

1

72 1/ 3

216

ªºª º

==

«»« »

−−

¬¼¬ ¼

4. To find the normal equations and to find ˆ

x, compute

13

111 33

11

311 311

11

T

AA

ªº

ªº ªº

«»

=−=

«» «»

«»

−

¬¼ ¬¼

«»

¬¼

5

111 6

1

311 14

0

T

A

ªº

ªºªº

«»

==

«»«»

«»

−

¬¼¬¼

«»

¬¼

b

a. The normal equations are

()

TT

AA A=xb

:

1

2

33 6

311 14

x

ªº

ª

ºªº

=

«»

«

»«»

¬

¼¬¼¬¼

b. Compute

6

ˆ11 14 14

TT

−1

33 11−36

ªºªºª ºªº

1

=(ΑΑ)Α==

«»«»« »«»

3−33

24

¬¼¬¼¬ ¼¬¼

xb

24 1

1

24 1

24

ªºªº

==

«»«»

¬¼¬¼

5. To find the least squares solutions to Ax = b, compute and row reduce the augmented matrix for the

system

TT

AA A=

xb:

42214 10 1 5

220 4 01 1 3

20210 00 0 0

TT

AA A

ª

ºª º

«

»« »

ªº

=∼−−

¬¼

«

»« »

«

»« »

¬

¼¬ ¼

b

so all vectors of the form

51

ˆ31

01

x

3

−

ªº ªº

«» «»

=−+

«» «»

¬¼ ¬¼

x

are the least-squares solutions of Ax = b.

6. To find the least squares solutions to Ax = b, compute and row reduce the augmented matrix for the

system

TT

AA A=

xb:

63327 10 1 5

33012 01 1 1

30315 00 0 0

TT

AA A

ª

ºª º

«

»« »

ªº

=∼−−

¬¼

«

»« »

«

»« »

¬

¼¬ ¼

b

so all vectors of the form

51

ˆ11

01

x

3

−

ªº ªº

«» «»

=−+

«» «»

¬¼ ¬¼

x

are the least-squares solutions of Ax = b.

382 CHAPTER 6 • Orthogonality and Least Squares

7. From Exercise 3,

12

,

03

25

A

−

ª

º

«

»

−

«

»

=

«

»

«

»

«

»

¬

¼

3

1,

4

2

ª

º

«

»

«

»

=

«

»

−

«

»

«

»

¬

¼

b

and

ˆ.

4/3

ª

º

=

«

»

−1/3

¬

¼

x

Since

ˆ0

2

A

1−2323−1

ªº ªºªºªºªº

«» «»«»«»«»

−124/3 1 −21−3

ªº

«» «»«»«»«»

−=−=−=

«»

«» «»«»«»«»

3−1/3 −4−1−43

¬¼

«» «»«»«»«»

25 2 1 −1

«» «»«»«»«»

¬¼ ¬¼¬¼¬¼¬¼

xb

the least squares error is

ˆ

|| .A−

||

=20=25xb

8. From Exercise 4,

13

11,

11

A

ª

º

«

»

=−

«

»

«

»

¬

¼

5

1,

0

ªº

«»

=«»

«»

¬¼

b

and

ˆ.

1

ª

º

=

«

»

1

¬

¼

x

Since

13 5 4 5 1

1

ˆ11 1 0 1 1

1

11 0 2 0 2

A

−

ªºªºªºªºªº

ªº

«»«»«»«»«»

−=−−=−=−

«»

«»«»«»«»«»

¬¼

«»«»«»«»«»

¬¼¬¼¬¼¬¼¬¼

xb

the least squares error is

ˆ

|| .A−

||

=6xb

9. (a) Because the columns

1

a and

2

a of A are orthogonal, the method of Example 4 may be used to

find

ˆ

b, the orthogonal projection of b onto Col A:

12

1212

11 2 2

151

21 2 1

ˆ311

77 7 7

240

ª

ºªºªº

⋅⋅

«

»«»«»

=+ =+=+=

«

»«»«»

⋅⋅

«

»«»«»

−

¬

¼¬¼¬¼

ba ba

ba aaa

aa a a

(b) The vector ˆ

x contains the weights which must be placed on

1

a and

2

a to produce

ˆ

b. These

weights are easily read from the above equation, so

ˆ.

2/7

ª

º

=

«

»

1/7

¬

¼

x

10. (a) Because the columns

1

a and

2

a of A are orthogonal, the method of Example 4 may be used to

find

ˆ

b, the orthogonal projection of b onto Col A:

12

1212

11 2 2

124

11

ˆ33141

22

124

ª

ºªºªº

⋅⋅

«

»«»«»

=+ =+=−+=−

«

»«»«»

⋅⋅

«

»«»«»

¬

¼¬¼¬¼

ba ba

ba aaa

aa a a

(b) The vector ˆ

x contains the weights which must be placed on

1

a and

2

a to produce

ˆ

b. These

weights are easily read from the above equation, so

ˆ.

3

ª

º

=

«

»

1/ 2

¬

¼

x

11. (a) Because the columns

1

a,

2

a and

3

a of A are orthogonal, the method of Example 4 may be used

to find

ˆ

b, the orthogonal projection of b onto Col A:

6.5 • Solutions 383

3

12

123123

11 2 2 3 3

21

ˆ0

33

⋅

⋅⋅

=+ + =++

⋅⋅ ⋅

ba

ba ba

ba a aaaa

aa a a a a

40 13

15 11

21

0

6104

33

1151

ªº ª º ª º ª º

«» « » « » « »

−

«» « » « » « »

=+ + =

«» « » « » « »

−−−

«» « » « » « »

¬¼ ¬ ¼ ¬ ¼ ¬ ¼

(b) The vector ˆ

x contains the weights which must be placed on

1

a,

2

a, and

3

a to produce

ˆ

b. These

weights are easily read from the above equation, so

ˆ.

2/3

ª

º

«

»

=0

«

»

«

»

1/ 3

¬

¼

x

12. (a) Because the columns

1

a,

2

a and

3

a of A are orthogonal, the method of Example 4 may be used

to find

ˆ

b, the orthogonal projection of b onto Col A:

3

12

123123

11 2 2 3 3

114 5

ˆ

33 3

⋅

⋅⋅ §·

=+ + =++−

¨¸

⋅⋅ ⋅ ©¹

ba

ba ba

ba a aaa a

aa a a a a

1105

1012

1145

0113

333

11 16

ªº ªº ªºªº

«» «» «»«»

−

«» «» «»«»

=+−=

«» «» «»«»

−−

«» «» «»«»

¬¼ ¬¼ ¬¼¬¼

(b) The vector ˆ

x contains the weights which must be placed on

1

a,

2

a, and

3

a to produce

ˆ

b. These

weights are easily read from the above equation, so

ˆ.

1/3

ª

º

«

»

= 14/3

«

»

«

»

−5/3

¬

¼

x

13. One computes that

11 0

11 , 2 , || || 40

11 6

AA A

ªº ªº

«» «»

=−−=−=

«» «»

−

¬¼ ¬¼

ubu bu

74

12 , 3 , || || 29

72

AAA

ªº ªº

«» «»

=−−=−=

«» «»

−

¬¼ ¬¼

vbvbv

Since Av is closer to b than Au is, Au is not the closest point in Col A to b. Thus u cannot be a least-

squares solution of Ax = b.

14. One computes that

32

8, 4,|| || 24

22

AA A

ªº ª º

«» « »

=−=−−=

«» « »

¬¼ ¬ ¼

ubu bu

384 CHAPTER 6 • Orthogonality and Least Squares

72

2, 2,|| || 24

84

AA A

−

ªº ª º

«» « »

=−=−=

«» « »

−

¬¼ ¬ ¼

vbv bv

Since Au and Au are equally close to b, and the orthogonal projection is the unique closest point in

Col A to b, neither Au nor Av can be the closest point in Col A to b. Thus neither u nor v can be a

least-squares solution of Ax = b.

15. The least squares solution satisfies

ˆ.

T

RQ=xb

Since

35

01

R

ª

º

=

«

»

¬

¼

and

7

1

T

Q

ªº

=«»

−

¬¼

b

, the augmented

matrix for the system may be row reduced to find

35 7 10 4

01 1 01 1

T

RQ

ªºªº

ªº

=∼

«»«»

¬¼ −−

¬¼¬¼

b

and so

ˆ4

ªº

=«»

−1

¬¼

x

is the least squares solution of Ax = b.

16. The least squares solution satisfies

ˆ.

T

RQ=xb

Since

23

05

R

ª

º

=

«

»

¬

¼

and

17 / 2

9/2

T

Q

ªº

=«»

¬¼

b

, the augmented

matrix for the system may be row reduced to find

2317/2 102.9

05 9/2 01 .9

T

RQ

ªºªº

ªº

=∼

«»«»

¬¼

¬¼¬¼

b

and so

ˆ2.9

ªº

=«»

.9

¬¼

x

is the least squares solution of Ax = b.

17. a. True. See the beginning of the section. The distance from Ax to b is || Ax – b ||.

b . True. See the comments about equation (1).

c . False. The inequality points in the wrong direction. See the definition of a least-squares solution.

d . True. See Theorem 13.

e . True. See Theorem 14.

18. a. True. See the paragraph following the definition of a least-squares solution.

b . False. If ˆ

x is the least-squares solution, then Aˆ

x is the point in the column space of A closest to

b. See Figure 1 and the paragraph preceding it.

c . True. See the discussion following equation (1).

d . False. The formula applies only when the columns of A are linearly independent. See Theorem

14.

e . False. See the comments after Example 4.

f. False. See the Numerical Note.

19. a. If Ax = 0, then

.

TT

AA A==

x00 This shows that Nul A is contained in

Nul .

T

AA

b . If

,

T

AA

=x0

then

0.

TT T

AA ==

xxx0 So

()()0,

T

AA=xx

which means that

2

|| || 0,A=x

and

hence Ax = 0. This shows that

Nul

T

AA

is contained in Nul A.

6.6 • Solutions 385

20. Suppose that Ax = 0. Then

.

TT

AA A==

x00 Since

T

AA is invertible, x must be 0. Hence the

columns of A are linearly independent.

21. a. If A has linearly independent columns, then the equation Ax = 0 has only the trivial solution. By

Exercise 19, the equation

T

AA =

x0 also has only the trivial solution. Since

T

AA is a square

matrix, it must be invertible by the Invertible Matrix Theorem.

b . Since the n linearly independent columns of A belong to

m

, m could not be less than n.

c . The n linearly independent columns of A form a basis for Col A, so the rank of A is n.

22. Note that

T

AA has n columns because A does. Then by the Rank Theorem and Exercise 19,

rank dim Nul dim Nul rank

TT

AA n AA n A A=−=−=

23. By Theorem 14,

ˆˆ.

TT

AAAAA

−1

==( )bx b

The matrix

1

()

TT

AAA A

−

is sometimes called the hat-

matrix in statistics.

24. Since in this case

,

T

AA I

=

the normal equations give

ˆ.

T

A=

xb

25. The normal equations are

22 6

,

22 6

x

y

ªºªºªº

=

«»«»«»

¬¼¬¼¬¼

whose solution is the set of all (x, y) such that x + y =

3. The solutions correspond to the points on the line midway between the lines x + y = 2 and x + y =

4.

26. [M] Using .7 as an approximation for

2/2,

02

.353535aa=≈

and

1.5.a=

Using .707 as an

approximation for

2/2

,

02

.35355339aa=≈

,

1.5.a=

6.6 SOLUTIONS

Notes:

This section is a valuable reference for any person who works with data that requires statistical

analysis. Many graduate fields require such work. Science students in particular will benefit from

Example 1. The general linear model and the subsequent examples are aimed at students who may take a

multivariate statistics course. That may include more students than one might expect.

1. The design matrix X and the observation vector y are

10 1

11 1

,,

12 2

13 2

X

ªºªº

«»«»

==

«»«»

¬¼¬¼

y

and one can compute

1

46 6 .9

ˆ

,,()

614 11 .4

TTTT

XX X XX X

−

ªº ªº ªº

====

«» «» «»

¬¼ ¬¼ ¬¼

yy

β

The least-squares line

01

yx

ββ

=+

is thus y = .9 + .4x.

2. The design matrix X and the observation vector y are

386 CHAPTER 6 • Orthogonality and Least Squares

11 0

12 1

,,

14 2

15 3

X

ªºªº

«»«»

==

«»«»

¬¼¬¼

y

and one can compute

1

412 6 .6

ˆ

,,()

12 46 25 .7

TTTT

XX X XX X

−

ªº ªº ªº

====

«» «» «»

¬¼ ¬¼ ¬¼

yy

β

The least-squares line

01

yx

ββ

=+

is thus y = –.6 + .7x.

3. The design matrix X and the observation vector y are

11 0

10 1

,,

11 2

12 4

X

−

ªºªº

«»«»

==

«»«»

¬¼¬¼

y

and one can compute

1

42 7 1.1

ˆ

,,()

26 10 1.3

TT TT

XX X XX X

−

ªº ªº ªº

=== =

«» «» «»

¬¼ ¬¼ ¬¼

yy

β

The least-squares line

01

yx

ββ

=+

is thus y = 1.1 + 1.3x.

4. The design matrix X and the observation vector y are

12 3

13 2

,,

15 1

16 0

X

ªºªº

«»«»

==

«»«»

¬¼¬¼

y

and one can compute

1

416 6 4.3

ˆ

,,()

16 74 17 .7

TTTT

XX X XX X

−

ªº ªº ªº

====

«» «» «»

−

¬¼ ¬¼ ¬¼

yy

β

The least-squares line

01

yx

ββ

=+

is thus y = 4.3 – .7x.

5. If two data points have different x-coordinates, then the two columns of the design matrix X cannot

be multiples of each other and hence are linearly independent. By Theorem 14 in Section 6.5, the

normal equations have a unique solution.

6. If the columns of X were linearly dependent, then the same dependence relation would hold for the

vectors in

3

formed from the top three entries in each column. That is, the columns of the matrix

2

11

2

22

2

33

1

xx

ªº

«»

¬¼

would also be linearly dependent, and so this matrix (called a Vandermonde matrix)

would be noninvertible. Note that the determinant of this matrix is

213132

()()()0xxxxxx−−−≠

since

1

x

,

2

x

, and

3

x

are distinct. Thus this matrix is invertible, which means that the columns of X

6.6 • Solutions 387

are in fact linearly independent. By Theorem 14 in Section 6.5, the normal equations have a unique

solution.

7. a. The model that produces the correct least-squares fit is y = X

β

+



where

1

2

1

3

2

4

5

11 1.8

24 2.7

,,,and

39 3.4

416 3.8

525 3.9

X

β

ª

º

ªºªº

«

»

«»«»

«

»

«»«»

ªº

«

»

«»«»

====

«»

«

»

«»«»

¬¼

«

»

«»«»

«

»

«»«»

¬¼¬¼

¬

¼

y



β

b. [M] One computes that (to two decimal places)

1.76

ˆ,

.20

ª

º

=

«

»

−

¬

¼

β

so the desired least-squares equation

is

2

1.76 .20yxx=−

.

8. a. The model that produces the correct least-squares fit is y = X

β

+ where

23

11 1 1 1 1

2

23 3

,, ,and

nn

nnn

xx x y

X

y

xxx

β

ªº

ªº ªº ªº

«»

«» «» «»

====

«»

«» «» «»

«»

«» «» «»

¬¼ ¬¼ ¬¼

¬¼

y### # #



β

b . [M] For the given data,

416 64 1.58

636216 2.08

8 64 512 2.5

10 100 1000 2.8

and

12 144 1728 3.1

14 196 2744 3.4

16 256 4096 3.8

18 324 5832 4.32

X

ªºªº

«»«»

==

«»«»

¬¼¬¼

y

so

1

.5132

ˆ() .03348,

.001016

TT

XX X

−

ªº

«»

==−

«»

¬¼

y

β

and the least-squares curve is

23

.5132 .03348 .001016 .yx x x=−+

9. The model that produces the correct least-squares fit is y = X

β

+ where

1

2

3

cos 1 sin 1 7.9

cos 2 sin 2 , 5.4 , , and

cos 3 sin 3 .9

A

XB

ª

º

ªºªº

ªº

«

»«»«»

====

«»

«

»

«»«»

¬¼

«

»«»«»

−

¬¼¬¼

¬

¼

y







β

388 CHAPTER 6 • Orthogonality and Least Squares

10. a. The model that produces the correct least-squares fit is y = X

β

+ where

.02(10) .07(10)

1

.02(11) .07(11)

2

.02(12) .07(12)

3

.02(14) .07(14) 4

.02(15) .07(15) 5

21.34

20.68

,,,and,

20.05

18.87

18.30

A

B

ee

ee M

Xee M

ee

−−

ªº

ª

º

ªº

«»

«

»

«»

«

»

«»ªº

«»

«

»

«»

====

«»

«

»

«»¬¼

«»

«

»

«»

«

»

«»

¬¼

¬

¼

«»

¬¼

y







β

b. [M] One computes that (to two decimal places) 19.94

ˆ,

10.10

ª

º

=

«

»

¬

¼

β

so the desired least-squares

equation is

.02 .07

19.94 10.10 .

tt

ye e

−−

=+

11. [M] The model that produces the correct least-squares fit is y = X

β

+ where

1

2

3

4

5

13cos.88 3

12.3cos1.1 2.3

,,,and

11.65cos1.42 1.65

11.25cos1.77 1.25

11.01cos2.14 1.01

Xe

β

ª

º

ªºªº

«

»

«»«»

«

»

«»«»

ªº

«

»

«»«»

====

«»

«

»

«»«»

¬¼

«

»

«»«»

«

»

«»«»

¬¼¬¼

¬

¼

y







β

One computes that (to two decimal places) 1.45

ˆ

.811

ª

º

=

«

»

¬

¼

β

. Since e = .811 < 1 the orbit is an ellipse. The

equation r =

β

/ (1 – e cos

ϑ

) produces r = 1.33 when

ϑ

= 4.6.

12. [M] The model that produces the correct least-squares fit is y = X

β

+ , where

1

2

0

3

1

4

5

13.78 91

14.11 98

,,,and

14.39 103

14.73 110

14.88 112

X

β

ª

º

ªºªº

«

»

«»«»

«

»

«»«»

ªº

«

»

«»«»

====

«»

«

»

«»«»

¬¼

«

»

«»«»

«

»

«»«»

¬¼¬¼

¬

¼

y

β







One computes that (to two decimal places) 18.56

ˆ

19.24

ª

º

=

«

»

¬

¼

β

, so the desired least-squares equation is

p = 18.56 + 19.24 ln w. When w = 100, p ≈ 107 millimeters of mercury.

6.6 • Solutions 389

13. [M]

a. The model that produces the correct least-squares fit is y = X

β

+ where

23

10 0 0

0

11 1 1

8.8

122 2 29.9

133 3 62.0

144 4 104.7

155 5 159.1

166 6

,222.0

294.5

177 7

380.4

188 8

471.1

199 9

571.7

11010 10 686.8

11111 11 809.2

11212 12

X

ªº

«»

ª

«»

«

«»

«

«»

«

«»

«

«»

«

«»

«

«»

«

«»

==

«»

¬

«»

¬¼

y

0

1

2

3

4

0

5

1

6

2

7

3

8

9

10

11

12

,,and

β

ª

º

«

»

«

»

«

»

«

»

«

»

«

»

«

»

»ªº

«

»

«»«»

«

»

«»«»

==

«

»

«»«»

«

»

«»«»

¬¼

«

»

«»

«

»

«»

«

»

«»

«

»

«»

«

»

«»

«

»

«»

«

»

«»

«

»

¼

¬

¼

β







One computes that (to four decimal places)

.8558

4.7025

ˆ,

5.5554

.0274

−

ª

º

«

»

«

»

=

«

»

«

»

−

«

»

¬

¼

β

so the desired least-squares

polynomial is

23

() .8558 4.7025 5.5554 .0274 .yt t t t=−++ −

b. The velocity v(t) is the derivative of the position function y(t), so

2

() 4.7025 11.1108 .0822 ,vt t t=+ −

and v(4.5) = 53.0 ft/sec.

14. Write the design matrix as

[]

.1x

Since the residual vector = y – X

ˆ

β

is orthogonal to Col X,

ˆˆ

0()()

TT

XX=⋅=⋅− =−11y 1y1

ββ

0

10101

1

ˆˆˆ ˆ ˆ

() ˆ

n

yynx yn xnynnx

β

ββ β β

β

ªº

=+…+ −=−− =−−

«»

¬¼

«»

¬¼

¦¦ ¦

This equation may be solved for y to find

01

ˆˆ

.yx

ββ

=+

15. From equation (1) on page 369,

1

2

1

11

1

T

n

xnx

XX xx xx

x

ªº

ª

º

…

ªº

«»

==

«

»

«»

…

«

»

¬¼

¬

¼

«»

¬¼

¦

##

1

11

T

n

yy

Xxy

xx

y

ªºªº

…

ªº

«»

==

«»

…

¬¼

«»

¬¼

¦

y#

390 CHAPTER 6 • Orthogonality and Least Squares

The equations (7) in the text follow immediately from the normal equations

.

TT

XX X=y

β

16. The determinant of the coefficient matrix of the equations in (7) is

22

().nx x−

¦

Using the 2 × 2

formula for the inverse of the coefficient matrix,

2

0

22

1

ˆ1

ˆ()

y

xx

xy

xn

nx x

β

ªº ªº

ª

º

−

=

«» «»

«

»

−

−«»

«»

¬

¼

¬¼

¦

¦¦

¦

¦¦

Hence

2

01

22 22

()()()() ()()

ˆˆ

,

() ()

xy xxy nxyxy

nx x nx x

ββ

−−

==

−−

¦¦ ¦¦ ¦ ¦¦

¦¦ ¦¦

Note:

A simple algebraic calculation shows that

10

ˆˆ

() ,yxn

ββ

−=

¦

which provides a simple

formula for

0

ˆ

β

once

1

ˆ

β

is known

17. a. The mean of the data in Example 1 is 5.5,x= so the data in mean-deviation form are (–3.5, 1),

(–.5, 2), (1.5, 3), (2.5, 3), and the associated design matrix is

13.5

1.5

.

11.5

12.5

X

−

ª

º

«

»

−

«

»

=

«

»

«

»

«

»

¬

¼

The columns of X are

orthogonal because the entries in the second column sum to 0.

b. The normal equations are

,

TT

XX X=y

β

or

0

1

40 9

.

021 7.5

β

ªº

ª

ºªº

=

«»

«

»«»

¬

¼¬¼¬¼ One computes that

9/4

ˆ,

5/14

ªº

=«»

¬¼

β

so the desired least-squares line is

*

(9/ 4) (5/14) (9/ 4) (5/14)( 5.5).yx x=+ =+ −

18. Since

1

2

1

11

1

T

n

xnx

XX xx xx

x

ªº

ª

º

…

ªº

«»

==

«

»

«»

…

«

»

¬¼

¬

¼

«»

¬¼

¦

##

T

XX is a diagonal matrix when

0.x=

¦

19. The residual vector = y –

ˆ

X

β

is orthogonal to Col X, while

ˆ

y

=X

ˆ

β

is in Col X. Since and

ˆ

y

are

thus orthogonal, apply the Pythagorean Theorem to these vectors to obtain

22222 2

ˆˆ

SS(T) || || || || || || || || || || || || SS(R) SS(E)XX

ββ

==+= += +−=+yy y y

20. Since

ˆ

β

satisfies the normal equations,

ˆ,

TT

XX X=y

β

and

2

ˆˆˆˆˆˆ

|| || ( ) ( )

TTTTT

XXX XXX===y

ββββββ

Since

2

ˆ

|| || SS(R)X=

β

and

2

|| || SS(T)

T

==yy y

, Exercise 19 shows that

ˆ

SS(E) SS(T) SS(R)

TTT

X=−=−yy y

β

6.7 • Solutions 391

6.7 SOLUTIONS

Notes

: The three types of inner products described here (in Examples 1, 2, and 7) are matched by

examples in Section 6.8. It is possible to spend just one day on selected portions of both sections.

Example 1 matches the weighted least squares in Section 6.8. Examples 2–6 are applied to trend analysis

in Seciton 6.8. This material is aimed at students who have not had much calculus or who intend to take

more than one course in statistics.

For students who have seen some calculus, Example 7 is needed to develop the Fourier series in

Section 6.8. Example 8 is used to motivate the inner product on C[a, b]. The Cauchy-Schwarz and

triangle inequalities are not used here, but they should be part of the training of every mathematics

student.

1. The inner product is

11 2 2

,4 5xy xy xy¢²=+

. Let x = (1, 1), y = (5, –1).

a. Since

2

|| || , 9,xx=¢²=x

|| x || = 3. Since

2

|| || , 105,yy=¢²=y

|| || 105.=y

Finally,

22

|, |15 225.xy¢²==

b. A vector z is orthogonal to y if and only if ¢x, y² = 0, that is,

12

20 5 0,zz−=

or

12

4.zz=

Thus

all multiples of 1

4

ª

º

«

»

¬

¼ are orthogonal to y.

2. The inner product is

11 2 2

,4 5.xy xy xy¢²=+

Let x = (3, –2), y = (–2, 1). Compute that

2

|| || , 56,xx=¢²=x

2

|| || , 21,yy=¢²=y

22

|| || || || 56 21 1176=⋅=xy

, ¢x, y² = –34, and

2

|, |1156xy¢²=

.

Thus

222

|, | ||||||||,xy¢²≤xy

as the Cauchy-Schwarz inequality predicts.

3. The inner product is ¢ p, q² = p(–1)q(–1) + p(0)q(0) + p(1)q(1), so

2

4,54 3(1)4(5)5(1)28tt¢+−²=+ +=

.

4. The inner product is ¢ p, q² = p(–1)q(–1) + p(0)q(0) + p(1)q(1), so

22

3,32tt t

¢

−+²=

(4)(5) 0(3) 2(5) 10.−++=−

5. The inner product is ¢ p, q² = p(–1)q(–1) + p(0)q(0) + p(1)q(1), so

222

,4,434550pp t t¢²=¢++²=+ +=

and

|| || , 50 5 2ppp=¢²==

. Likewise

22222

,54,5415127qq t t¢²=¢−−²=+ +=

and

|| || , 27 3 3qqq=¢²==

.

6. The inner product is ¢ p, q² = p(–1)q(–1) + p(0)q(0) + p(1)q(1), so

22

,3,3pp t t t t

¢

²=¢−−²=

222

(4) 0 2 20−++=

and

|| || , 20 2 5.ppp=¢²==

Likewise

22

,32,32qq t t

¢

²=¢++²=

222

53559++=

and

|| || , 59.qqq=¢²=

7. The orthogonal projection

ˆ

q

of q onto the subspace spanned by p is

,28 5614

ˆ(4 )

,50 2525

qp

qp t t

pp

¢²

==+=+

¢²

8. The orthogonal projection

ˆ

q

of q onto the subspace spanned by p is

392 CHAPTER 6 • Orthogonality and Least Squares

22

,10 31

ˆ(3 )

,20 22

qp

qptttt

pp

¢²

==−−=−+

¢²

9. The inner product is ¢p, q² = p(–3)q(–3) + p(–1)q(–1) + p(1)q(1) + p(3)q(3).

a. The orthogonal projection

ˆ

p

2

of

2

p

onto the subspace spanned by

0

p

and

1

p

is

20 21

201

00 11

,,200

ˆ(1) 5

,,420

pp pp

ppp t

pp pp

¢²

=+=+=

¢²¢²

b. The vector

2

ˆ

qp p t

2

=−=−5

will be orthogonal to both

0

p

and

1

p

and

01

{,,}ppq

will be an

orthogonal basis for

012

Span{ , , }.ppp

The vector of values for q at (–3, –1, 1, 3) is (4, –4, –4, 4),

so scaling by 1/4 yields the new vector

2

(1/ 4)( 5).qt=−

10. The best approximation to

3

pt=

by vectors in

01

Span{ , , }Wppq=

will be

2

01

00 11

,,,01640541

ˆproj (1) ( )

,,,420445

W

pp pp pq t

pp p p q t t

pp pp qq

§·

¢² ¢² ¢² −

== + + =++ =

¨¸

¢²¢²¢² ©¹

11. The orthogonal projection of

3

pt=

onto

012

Span{ , , }Wppp=

will be

2

012

01 2

00 11 22

,,,034017

ˆproj (1) ( ) ( 2)

,,,510145

W

pp pp pp

pp p p p tt t

pp pp pp

¢² ¢² ¢²

== + + =++−=

¢²¢²¢²

12. Let

012

Span{ , , }.Wppp=

The vector

3

proj (17 / 5)

W

pp pt t=−=−

will make

0123

{,, ,}pppp

an orthogonal basis for the subspace

3

of

4

. The vector of values for

3

p

at (–2, –1, 0, 1, 2) is

(–6/5, 12/5, 0, –12/5, 6/5), so scaling by 5/6 yields the new vector

3

(5/ 6)( (17 / 5) )ptt=−=

3

(5/ 6) (17 / 6) .tt−

13. Suppose that A is invertible and that ¢u, v² = (Au) ⋅ (Av) for u and v in

n

. Check each axiom in the

definition on page 376, using the properties of the dot product.

i. ¢u, v² = (Au) ⋅ (Av) = (Av) ⋅ (Au) = ¢v, u²

ii. ¢u + v, w² = (A(u + v)) ⋅ (Aw) = (Au + Av) ⋅ (Aw) = (Au) ⋅ (Aw) + (Av) ⋅ (Aw) = ¢u, w² + ¢v, w²

iii. ¢c

u, v² = (A(

cu)) ⋅ (Av) = (c(Au)) ⋅ (Av) = c((Au) ⋅ (Av)) = c¢u, v²

iv.

2

,()()||||0,AA A¢²=⋅=≥uu u u u

and this quantity is zero if and only if the vector Au is 0. But

Au = 0 if and only u = 0 because A is invertible.

14. Suppose that T is a one-to-one linear transformation from a vector space V into

n

and that ¢u, v² =

T(u) ⋅ T(v) for u and v in

n

. Check each axiom in the definition on page 376, using the properties of

the dot product and T. The linearity of T is used often in the following.

i. ¢u, v² = T(u) ⋅ T(v) = T(v) ⋅ T(u) = ¢v, u²

ii. ¢u + v, w² = T(u + v) ⋅ T(w) = (T(u) + T(v)) ⋅ T(w) = T(u) ⋅ T(w) + T(v) ⋅ T(w) = ¢u, w² + ¢v, w²

iii. ¢cu, v² = T(cu) ⋅ T(v) = (cT(u)) ⋅ T(v) = c(T(u) ⋅ T(v)) = c¢u, v²

iv.

2

,()()||()||0,TT T¢²=⋅=≥uu u u u

and this quantity is zero if and only if u = 0 since T is a one-

to-one transformation.

6.7 • Solutions 393

15. Using Axioms 1 and 3, ¢u, c

v² = ¢c

v, u² = c¢v, u² = c¢u, v².

16. Using Axioms 1, 2 and 3,

2

|| || , , ,−=¢−−²=¢−²−¢−²uv uvuv uuv vuv

,,,,,2,,=¢²−¢²−¢²+¢²=¢²−¢²+¢²uu uv vu vv uu uv vv

22

|| || 2 , || ||=−¢²+uuvv

Since {u, v} is orthonormal,

22

|| || || || 1==uv

and ¢u, v² = 0. So

2

|| || 2.−=uv

17. Following the method in Exercise 16,

2

|| || , , ,+=¢++²=¢+²+¢+²uv uvuv uuv vuv

,,,,,2,,=¢²+¢²+¢²+¢²=¢²+¢²+¢²uu uv vu vv uu uv vv

22

|| || 2 , || ||=+¢²+uuvv

Subtracting these results, one finds that

22

|| || || || 4 , ,+−− =¢²uv uv uv

and dividing by 4 gives the

desired identity.

18. In Exercises 16 and 17, it has been shown that

22 2

|| || || || 2 , || ||−=−¢²+uv u uv v

and

2

|| ||+=uv

22

|| || 2 , || || .+¢²+uuvv

Adding these two results gives

2222

|| || || || 2 || || 2 || || .++−=+uv uv u v

19. let

a

b

ªº

=«»

«»

¬¼

u

and

.

b

a

ª

º

=

«

»

«

»

¬

¼

v

Then

2

|| || ,ab=+u

2

|| || ,ab=+v

and

,2.ab¢²=uv

Since a and b are

nonnegative,

|| || ,ab=+u

|| || .ab=+v

Plugging these values into the Cauchy-Schwarz

inequality gives

2|,|||||||||ab a b a b a b=¢²≤=+ +=+uv u v

Dividing both sides of this equation by 2 gives the desired inequality.

20. The Cauchy-Schwarz inequality may be altered by dividing both sides of the inequality by 2 and then

squaring both sides of the inequality. The result is

222

,||||||||

24

¢²

§·

≤

¨¸

©¹

uv u v

Now let a

b

ªº

=«»

¬¼

u and 1

1

ªº

=«»

¬¼

v. Then

222

|| || ,ab=+u

2

|| || 2=v

, and ¢u, v² = a + b. Plugging these

values into the inequality above yields the desired inequality.

21. The inner product is

1

0

,()().fg ftgtdt¢²=

³

Let

2

() 1 3 ,ft t=−

3

() .gt t t=−

Then

11

23 53

00

,(13)()34 0fg t t t dt t t tdt¢²=−−=−+=

³³

22. The inner product is

1

0

,()().fg ftgtdt¢²=

³

Let f (t) = 5t – 3,

32

() .gt t t=−

Then

394 CHAPTER 6 • Orthogonality and Least Squares

11

32 4 3 2

00

,(53)()5830fg t t t dt t t tdt¢²=−− =−+=

³³

23. The inner product is

1

0

,()(),fg ftgtdt¢²=

³

so

11

22 4 2

00

,(13) 9614/5,ff t dt t t dt¢²=−=−+=

³³

and

|| || , 2 / 5.fff=¢²=

24. The inner product is

1

0

,()(),fg ftgtdt¢²=

³

so

11

322 6 54

00

,() 2 1/105,gg t t dt t t tdt¢²=−=−+=

³³

and

|| || , 1/ 105.ggg=¢²=

25. The inner product is

1

,()().fg ftgtdt

−

¢²=

³

Then 1 and t are orthogonal because

1

1, 0.ttdt

−

¢²==

³

So 1 and t can be in an orthogonal basis for

2

Span{1, , }.tt

By the Gram-Schmidt process, the third

basis element in the orthogonal basis can be

22

2

,1 ,

1

1,1 ,

ttt

tt

¢²¢²

−−

¢² ¢²

Since

1

22

1

,1 2 / 3,ttdt

−

¢²==

³

1

1,1 1 2,dt

−

¢²==

³

and

1

23

1

,0,tt tdt

−

¢

²==

³

the third basis element can

be written as

2

(1 / 3).t−

This element can be scaled by 3, which gives the orthogonal basis as

2

{1, , 3 1} .tt−

26. The inner product is

2

,()().fg ftgtdt

−

¢²=

³

Then 1 and t are orthogonal because

2

1, 0.ttdt

−

¢²==

³

So 1 and t can be in an orthogonal basis for

2

Span{1, , }.tt

By the Gram-Schmidt process, the third

basis element in the orthogonal basis can be

22

2

,1 ,

1

1,1 ,

ttt

tt

¢²¢²

−−

¢² ¢²

Since

2

22

2

,1 16/3,ttdt

−

¢²==

³

2

1, 1 1 4,dt

−

¢²==

³

and

2

23

2

,0,tt tdt

−

¢

²==

³

the third basis element can

be written as

2

(4/3).t−

This element can be scaled by 3, which gives the orthogonal basis as

2

{1, , 3 4} .tt−

27. [M] The new orthogonal polynomials are multiples of

3

17 5tt−+

and

24

72 155 35 .tt−+

These

polynomials may be scaled so that their values at –2, –1, 0, 1, and 2 are small integers.

28. [M] The orthogonal basis is

0

() 1,ft=

1

() cos ,ft t=

2

() cos (1/2) (1/2)cos2,ft t t=−=

and

3

() cos (3/4)cos (1/4)cos3.ft t t t=−=

6.8 • Solutions 395

6.8 SOLUTIONS

Notes

: The connections between this section and Section 6.7 are described in the notes for that section.

For my junior-senior class, I spend three days on the following topics: Theorems 13 and 15 in Section 6.5,

plus Examples 1, 3, and 5; Example 1 in Section 6.6; Examples 2 and 3 in Section 6.7, with the

motivation for the definite integral; and Fourier series in Section 6.8.

1. The weighting matrix W, design matrix X, parameter vector

β

, and observation vector y are:

0

1

10000 1 2 0

02000 1 1 0

,,,

00200 1 0 2

00020 1 1 4

00001 1 2 4

WX

β

−

ªºªºªº

«»«»«»

−

«»«»«»

ªº

«»«»«»

====

«»

«»«»«»

¬¼

«»«»«»

¬¼¬¼¬¼

y

β

The design matrix X and the observation vector y are scaled by W:

12 0

22 0

,

20 4

22 8

12 4

WX W

−

ªºªº

«»«»

−

«»«»

==

«»«»

¬¼¬¼

y

Further compute

14 0 28

() ,()

016 24

TT

WX WX WX W

ªº ªº

==

«» «»

¬¼ ¬¼

y

and find that

1

1/14 0 28 2

ˆ(( ) ) ( ) 01/1624 3/2

TT

WX WX WX W

−

ªºªºªº

===

«»«»«»

¬¼¬¼¬¼

y

β

Thus the weighted least-squares line is y = 2 + (3/2)x.

2. Let X be the original design matrix, and let y be the original observation vector. Let W be the

weighting matrix for the first method. Then 2W is the weighting matrix for the second method. The

weighted least-squares by the first method is equivalent to the ordinary least-squares for an equation

whose normal equation is

ˆ

() ()

TT

WX WX WX W=y

β

(1)

while the second method is equivalent to the ordinary least-squares for an equation whose normal

equation is

ˆ

(2 ) (2 ) (2 ) (2 )

TT

WX W X WX W=y

β

(2)

Since equation (2) can be written as

ˆ

4( ) 4( ) ,

TT

WX WX WX W

β

=y

it has the same solutions as

equation (1).

396 CHAPTER 6 • Orthogonality and Least Squares

3. From Example 2 and the statement of the problem,

0

() 1,pt=

1

() ,pt t=

2

() 2,pt t=−

3

() (5/6) (17/6),pt t t=−

and g = (3, 5, 5, 4, 3). The cubic trend function for g is the orthogonal

projection

ˆ

p

of g onto the subspace spanned by

0

,p

1

,p

2

,p

and

3

:p

03

12

01 2 3

00 11 22 33

,,

ˆ,,, ,

gp gp

pp p p p

pp pp pp pp

¢² ¢²

=+++

¢²¢²¢²¢²

()

23

20 1 7 2 5 17

(1) 2

51014 1066

tt tt

−− §·

=++ −+−

¨¸

©¹

()

23 23

11 15 17 21 1

42 5

10 2 5 6 6 3 2 6

tt t t ttt

§·

=−− −+−=−− +

¨¸

©¹

This polynomial happens to fit the data exactly.

4. The inner product is ¢ p, q² = p(–5)q(–5) + p(–3)q(–3) + p(–1)q(–1) + p(1)q(1) + p(3)q(3) + p(5)q(5).

a. Begin with the basis

2

{1, , }tt

for

2

. Since 1 and t are orthogonal, let

0

() 1pt=

and

1

() .pt t=

Then the Gram-Schmidt process gives

22

222

2

,1 , 70 35

() 1

1,1 , 6 3

ttt

pt t t t t

tt

¢²¢²

=−−=−=−

¢² ¢²

The vector of values for

2

p

is (40/3, –8/3, –32/3, –32/3, –8/3, 40/3), so scaling by 3/8 yields the

new function

22

2

(3/ 8)( (35/3)) (3/8) (35/8).pt t=−=−

b. The data vector is g = (1, 1, 4, 4, 6, 8). The quadratic trend function for g is the orthogonal

projection

ˆ

p

of g onto the subspace spanned by

0

p

,

1

p

and

2

p

:

2

012

01 2

00 11 22

,,,24506335

ˆ(1)

,,,6708488

gp gp gp

pp p p tt

pp pp pp

¢² ¢² ¢² §·

=++=++−

¨¸

¢²¢²¢² ©¹

22

513 35595 3

47148 8 167112

tt tt

§·

=+ + −=++

¨¸

©¹

5. The inner product is

2

0

,()().fg ftgtdt

π

¢²=

³

Let m ≠ n. Then

22

00

1

sin , sin sin sin cos(( ) ) cos(( ) ) 0

2

mt nt mt nt dt m n t m n t dt

ππ

¢²==−− +=

³³

Thus sin mt and sin nt are orthogonal.

6. The inner product is

2

0

,()().fg ftgtdt

π

¢²=

³

Let m and n be positive integers. Then

22

00

1

sin ,cos sin cos sin(( ) ) sin(( ) ) 0

2

mt nt mt nt dt m n t m n t dt

ππ

¢²==++−=

³³

Thus sinmt and cosnt are orthogonal.

7. The inner product is

2

0

,()().fg ftgtdt

π

¢²=

³

Let k be a positive integer. Then

22

00

1

|| cos || cos ,cos cos 1 cos 2

2

kt kt kt kt dt kt dt

ππ

π

=¢²==+=

³³

6.8 • Solutions 397

and

22

00

1

|| sin || sin ,sin sin 1 cos 2

2

kt kt kt kt dt kt dt

ππ

π

=¢²==−=

³³

8. Let f(t) = t – 1. The Fourier coefficients for f are:

22

0

00

11 1

() 1 1

22 2

aftdt t dt

ππ

π

ππ

==−=−+

³³

and for k > 0,

22

00

11

()cos ( 1)cos 0

k

aftktdttktdt

ππ

==−=

³³

22

00

11 2

()sin ( 1)sin

k

bftktdttktdt

k

ππ

==−=−

³³

The third-order Fourier approximation to f is thus

012 3

2

sin sin 2 sin 3 1 2 sin sin 2 sin 3

23

abtb tb t t t t

π

++ + =−+−−−

9. Let f(t) = 2

π

– t. The Fourier coefficients for f are:

22

0

00

11 1

() 2

22 2

aftdt tdt

ππ

==−=

³³

and for k > 0,

22

00

11

()cos (2 )cos 0

k

aftktdt tktdt

ππ

π

ππ

==−=

³³

22

00

11 2

()sin (2 )sin

k

bftktdt tktdt

k

ππ

π

ππ

==−=

³³

The third-order Fourier approximation to f is thus

012 3

2

sin sin 2 sin 3 2 sin sin 2 sin 3

23

abtb tb t t t t

π

++ + =+++

10. Let 1for0

() .

1for 2

t

ft t

π

ππ

≤<



=®−≤<

¯ The Fourier coefficients for f are:

22

0

00

11 1 1

() 0

22 2 2

aftdt dt dt

πππ

π

πππ

==−=

³³³

and for k > 0,

22

00

111

()cos cos cos 0

k

aftktdt ktdt ktdt

πππ

π

πππ

==−=

³³³

22

00

4/( ) for odd

111

()sin sin sin 0foreven

k

kk

bftktdt ktdt ktdt k

πππ

π

πππ



==−=®

¯

³³³

The third-order Fourier approximation to f is thus

13

44

sin sin 3 sin sin 3

3

btb t t t

ππ

+=+

398 CHAPTER 6 • Orthogonality and Least Squares

11. The trigonometric identity

2

cos 2 1 2 sintt=−

shows that

2

11

sin cos 2

22

tt=−

The expression on the right is in the subspace spanned by the trigonometric polynomials of order 3 or

less, so this expression is the third-order Fourier approximation to

2

sin t

.

12. The trigonometric identity

3

cos 3 4 cos 3 costtt=−

shows that

3

31

cos cos cos 3

44

tt t=+

The expression on the right is in the subspace spanned by the trigonometric polynomials of order 3 or

less, so this expression is the third-order Fourier approximation to

3

cos .t

13. Let f and g be in C [0, 2π] and let m be a nonnegative integer. Then the linearity of the inner product

shows that

¢( f + g), cos mt² = ¢ f, cos mt² + ¢g, cos mt², ¢( f + g), sin mt² = ¢ f, sin mt² + ¢ g, sin mt²

Dividing these identities respectively by ¢cos mt, cos mt² and ¢sin mt, sin mt² shows that the Fourier

coefficients

m

a

and

m

b

for f + g are the sums of the corresponding Fourier coefficients of f and of g.

14. Note that g and h are both in the subspace H spanned by the trigonometric polynomials of order 2 or

less. Since h is the second-order Fourier approximation to f, it is closer to f than any other function in

the subspace H.

15. [M] The weighting matrix W is the 13 × 13 diagonal matrix with diagonal entries 1, 1, 1, .9, .9, .8, .7,

.6, .5, .4, .3, .2, .1. The design matrix X, parameter vector

β

, and observation vector y are:

23

23 0

23 1

2

23

3

23

10 0 0

0.0

11 1 1

8.8

122 2 29.9

133 3 62.0

144 4 104.7

155 5 159.1

166 6

,,

222.0

294.5

177 7

380.4

188 8

199 9

11010 10

11111 11

11212 12

X

β

ªº

«»

ªº

«»

===

«»

¬¼

«»

¬¼

y

β

471.1

571.7

686.8

809.2

ª

º

«

»

«

»

«

»

«

»

«

»

«

»

«

»

«

»

«

»

«

»

«

»

«

»

«

»

«

»

«

»

«

»

«

»

«

»

«

»

¬

¼

The design matrix X and the observation vector y are scaled by W:

6.8 • Solutions 399

1.0 0.0 0.0 0.0

1.0 1.0 1.0 1.0

1.0 2.0 4.0 8.0

.9 2.7 8.1 24.3

.9 3.6 14.4 57.6

.8 4.0 20.0 100.0

.7 4.2 25.2 151.2

.6 4.2 29.4 205.8

.5 4.0 32.0 256.0

.4 3.6 32.4 291.6

.3 3.0 30.0 300.0

.2 2.2 24.2 266.2

.1 1.2 14.4 172.8

WX

ªº

«»

«

=«

«

¬¼

0.00

8.80

29.90

55.80

94.23

127.28

,155.40

176.70

190.20

188.44

171.51

137.36

80.92

W

ª

º

«

»

«»

»

«»

»

«»

»

«»

»

«»

»

«»

»

«»

»

«»

=

»

«»

»

«»

»

«»

»

«»

»

«»

»

«»

»

«»

»

«»

»

«»

»

«»

¬

¼

y

Further compute

6.66 22.23 120.77 797.19 747.844

22.23 120.77 797.19 5956.13 4815.438

() ,()

120.77 797.19 5956.13 48490.23 35420.468

797.19 5956.13 48490.23 420477.17 285262.440

T T

WX WX WX W

ªºªº

«»«»

==

«»«»

¬¼¬¼

y

and find that

1

0.2685

3.6095

ˆ(( ) ) ( ) 5.8576

0.0477

TT

WX WX WX W

−

ªº

«»

==

«»

−

«»

¬¼

y

β

Thus the weighted least-squares cubic is

23

() .2685 3.6095 5.8576 .0477 .ygt t t t==−++ −

The

velocity at t = 4.5 seconds is g'(4.5) = 53.4 ft./sec. This is about 0.7% faster than the estimate

obtained in Exercise 13 of Section 6.6.

16. [M] Let 1for0

() .

1for 2

t

ft t

π

ππ

≤<



=®−≤<

¯ The Fourier coefficients for f have already been found to be

0

k

a=

for all k ≥ 0 and 4/( ) for odd .

0foreven

k

kk

bk

π



=®

¯ Thus

45

44 44 4

() sin sin3 and () sin sin3 sin5

335

ft t t ft t t t

ππ ππ π

=+ =+ +

A graph of

4

f

over the interval [0, 2

π

] is

400 CHAPTER 6 • Orthogonality and Least Squares

A graph of

5

f

over the interval [0, 2

π

] is

A graph of

5

f

over the interval [–2

π

, 2

π

] is

Chapter 6 SUPPLEMENTARY EXERCISES

1. a. False. The length of the zero vector is zero.

b. True. By the displayed equation before Example 2 in Section 6.1, with c = –1,

|| –x || = || (–1)x || =| –1 ||| x || = || x ||.

c. True. This is the definition of distance.

d. False. This equation would be true if r|| v || were replaced by | r ||| v ||.

e. False. Orthogonal nonzero vectors are linearly independent.

f. True. If x ⋅ u = 0 and x ⋅ v = 0, then x ⋅ (u – v) = x ⋅ u – x ⋅ v = 0.

g. True. This is the “only if” part of the Pythagorean Theorem in Section 6.1.

h. True. This is the “only if” part of the Pythagorean Theorem in Section 6.1 where v is replaced

by –v, because

2

|| ||−v

is the same as

2

|| ||v

.

i. False. The orthogonal projection of y onto u is a scalar multiple of u, not y (except when y

itself is already a multiple of u).

1

0.5

–0.5

–1

234 56

1

0.5

–0.5

–1

1234 56

1

0.5

–0.5

–1

–6 –4 –2 246

Chapter 6 • Supplementary Exercises 401

j. True. The orthogonal projection of any vector y onto W is always a vector in W.

k. True. This is a special case of the statement in the box following Example 6 in Section 6.1 (and

proved in Exercise 30 of Section 6.1).

l. False. The zero vector is in both W and

.W

⊥

m. True. See Exercise 32 in Section 6.2. If

0,

ij

⋅=vv

then

()( ) ( ) 00.

ii j j ij i j ij

cc cc cc⋅=⋅==vv vv

n. False. This statement is true only for a square matrix. See Theorem 10 in Section 6.3.

o. False. An orthogonal matrix is square and has orthonormal columns.

p. True. See Exercises 27 and 28 in Section 6.2. If U has orthonormal columns, then

.

T

UU I=

If

U is also square, then the Invertible Matrix Theorem shows that U is invertible and

1

.

T

UU

−

=

In this case, ,

T

UU I= which shows that the columns of

T

U

are orthonormal; that is, the rows

of U are orthonormal.

q. True. By the Orthogonal Decomposition Theorem, the vectors

proj

Wv

and

proj

W

−

vv

are

orthogonal, so the stated equality follows from the Pythagorean Theorem.

r. False. A least-squares solution is a vector

ˆ

x

(not A

ˆ

x

) such that A

ˆ

x

is the closest point to b

in Col A.

s. False. The equation

ˆAA A

Τ−1Τ

=( )xb

describes the solution of the normal equations, not the

matrix form of the normal equations. Furthermore, this equation makes sense only when

T

AA

is invertible.

2. If

12

{, }

vv

is an orthonormal set and

11 2 2

,cc=+

xv v

then the vectors

11

c

v

and

22

c

v

are orthogonal

(Exercise 32 in Section 6.2). By the Pythagorean Theorem and properties of the norm

22222222

11 2 2 11 2 2 1 1 2 2 1 2

|| || || || || || || || ( || ||) ( || ||) | | | |cc c c c c c c=+ = + = + =+xvv v v v v

So the stated equality holds for p = 2. Now suppose the equality holds for p = k, with k ≥ 2. Let

11

{, , }

k

+

…

vv

be an orthonormal set, and consider

11 1 1 1 1

,

kk k k k k k

ccc c

++ ++

=+…+ + =+

xv v v u v

where

11

.

kkk

cc=+…+

uv v

Observe that

k

u

and

11

kk

c

++

v

are orthogonal because

1

0

jk+

⋅=vv

for j

= 1,…,k. By the Pythagorean Theorem and the assumption that the stated equality holds for k, and

because

222 2

11 1 1 1

|| || | | || || | | ,

kk k k k

cc c

++ + + +

==vv

222222

11 11 1 1

|| || || || || || || || | | | |

kkk k kk k

cccc

++ ++ +

=+ = + = +…+xu v u v

Thus the truth of the equality for p = k implies its truth for p = k + 1. By the principle of induction,

the equality is true for all integers p ≥ 2.

3. Given x and an orthonormal set

1

{, , }

p

…vv

in

n

, let

ˆ

x

be the orthogonal projection of x onto the

subspace spanned by

1

,,

p

…vv

. By Theorem 10 in Section 6.3,

11

ˆ() ( ).

pp

=⋅+…+ ⋅xxvv xvv

By

Exercise 2,

22 2

1

ˆ

|| || | | | | .

p

=⋅+…+ ⋅xxv xv

Bessel’s inequality follows from the fact that

22

ˆ

|| || || || ,≤xx

which is noted before the proof of the Cauchy-Schwarz inequality in Section 6.7.

4. By parts (a) and (c) of Theorem 7 in Section 6.2,

1

{,, }

k

UU…

vv

is an orthonormal set in

n

. Since

there are n vectors in this linearly independent set, the set is a basis for

n

.

402 CHAPTER 6 • Orthogonality and Least Squares

5. Suppose that (U x)⋅(U y) = x⋅y for all x, y in

n

, and let

1

,,

n

…

ee

be the standard basis for

n

. For

j = 1, …, n,

j

Ue

is the jth column of U. Since

2

|| || ( ) ( ) 1,

jjjjj

UUU=⋅=⋅=eeeee

the columns of U

are unit vectors; since

()() 0

jkjk

UU⋅=⋅=eeee

for j ≠ k, the columns are pairwise orthogonal.

6. If Ux = λx for some x ≠ 0, then by Theorem 7(a) in Section 6.2 and by a property of the norm,

|| x || = || Ux || = || λx || = | λ ||| x ||, which shows that | λ | = 1, because x ≠ 0.

7. Let u be a unit vector, and let

2.

T

QI=−uu

Since

() ,

TT TT T T

==uu u u uu

(2 ) 2( ) 2

TTTTTT

QI I I Q=−=−=−=uu uu uu

Then

22

(2 ) 2 2 4( )( )

TTTTTT

QQ Q I I==−=−−+uu uu uu uu uu

Since u is a unit vector, 1,

T

=⋅=uu uu so

()()()() ,

TT T T T

==uu uu u u u u uu

and

224

TTTT

QQ I I=−−+=uu uu uu

Thus Q is an orthogonal matrix.

8. a. Suppose that x ⋅ y = 0. By the Pythagorean Theorem,

22 2

|| || || || || || .+=+xyxy

Since T preserves

lengths and is linear,

22 2 2

|| ( ) || || ( ) || || ( ) || || ( ) ( ) ||TTT TT+=+=+xyxyxy

This equation shows that T(x) and T(y) are orthogonal, because of the Pythagorean Theorem.

Thus T preserves orthogonality.

b . The standard matrix of T is

[]

1

() ( )

n

TT…ee

, where

1

,,

n

…

ee

are the columns of the identity

matrix. Then

1

{( ), , ( )}

n

TT…

ee

is an orthonormal set because T preserves both orthogonality and

lengths (and because the columns of the identity matrix form an orthonormal set). Finally, a

square matrix with orthonormal columns is an orthogonal matrix, as was observed in Section 6.2.

9. Let W = Span{u, v}. Given z in

n

, let

ˆproj .

W

=

zz

Then

ˆ

z

is in Col A, where

[]

.A=uv

Thus

there is a vector, say,

ˆ

x

in

2

, with A

ˆ

x

=

ˆ

z

. So,

ˆ

x

is a least-squares solution of Ax = z. The normal

equations may be solved to find

ˆ

x

, and then

ˆ

z

may be found by computing Aˆ.x

10. Use Theorem 14 in Section 6.5. If c ≠ 0, the least-squares solution of Ax = c

b is given by

1

() (),

TT

AA A c

−

b

which equals

1

() ,

TT

cAA A

−

b

by linearity of matrix multiplication. This solution is

c times the least-squares solution of Ax = b.

11. Let

,

x

y

z

ªº

«»

=«»

«»

¬¼

x

,

a

b

c

ªº

«»

=«»

«»

¬¼

b

1

2,

5

ªº

«»

=−

«»

¬¼

v

and

125

125.

125

T

A

ªº −

ª

º

«»

«

»

==−

«»

«

»

«»

«

»

−

¬

¼

«»

¬¼

v

Then the given set of equations is

Ax = b, and the set of all least-squares solutions coincides with the set of solutions of the normal

equations

TT

AA A=

xb

. The column-row expansions of

T

AA and

T

A

b

give

3, ( )

TTTTTT

AA A a b c a b c=++= =++=++vv vv vv vv b v v v v

Chapter 6 • Supplementary Exercises 403

Thus

3( ) 3 ( ) 3( )

TTTT

AA ===xvvxvvx vxv

since

T

vx

is a scalar, and the normal equations have

become

3( ) ( ) ,

T

abc=++vxv v

so

3( ) ,

T

abc=++vx

or

()/3.

T

abc=++vx

Computing

T

vx

gives

the equation x – 2y + 5z = (a + b + c)/3 which must be satisfied by all least-squares solutions to Ax =

b.

12. The equation (1) in the exercise has been written as Vλ = b, where V is a single nonzero column

vector v, and b = Av. The least-squares solution

ˆ

λ

of Vλ = b is the exact solution of the normal

equations

.

TT

VV Vλ=

b

In the original notation, this equation is

.

TT

Aλ=

vv v v

Since

T

vv

is

nonzero, the least squares solution

ˆ

λ

is

/

().

TT

Avvvv

This expression is the Rayleigh quotient

discussed in the Exercises for Section 5.8.

13. a. The row-column calculation of Au shows that each row of A is orthogonal to every u in Nul A. So

each row of A is in

(Nul ) .A

⊥

Since

(Nul )A

⊥

is a subspace, it must contain all linear

combinations of the rows of A; hence

(Nul )A

⊥

contains Row A.

b . If rank A = r, then dim

Nul A = n – r by the Rank Theorem. By Exercsie 24(c) in Section 6.3,

dimNul dim(Nul ) ,AAn

⊥

+=

so

dim(Nul )A

⊥

must be r. But Row A is an r-dimensional

subspace of

(Nul )A

⊥

by the Rank Theorem and part (a). Therefore,

Row (Nul ) .AA

⊥

=

c . Replace A by

T

A in part (b) and conclude that

Row (Nul ) .

TT

AA

⊥

=

Since

Row Col ,

T

AA=

Col (Nul ) .

T

AA

⊥

=

14. The equation Ax = b has a solution if and only if b is in Col A. By Exercise 13(c), Ax = b has a

solution if and only if b is orthogonal to

Nul .

T

A

This happens if and only if b is orthogonal to all

solutions of

.

T

A=

x0

15. If

T

AURU=

with U orthogonal, then A is similar to R (because U is invertible and

1

T

UU

−

=

), so A

has the same eigenvalues as R by Theorem 4 in Section 5.2. Since the eigenvalues of R are its n real

diagonal entries, A has n real eigenvalues.

16. a. If

[]

12

,

n

U=…uu u

then

[]

11 2

.

n

AU A A=λ…uu u

Since

1

u

is a unit vector and

2

,,

n

…

uu

are orthogonal to

1

,

u

the first column of

T

UAU

is

11 1 1 11

() .

TT

UUλ=λ=λuue

b . From (a),

1

****

0

T

UAU A

λ

ªº

«»

=«»

«»

¬¼

#

View

T

UAU

as a 2 × 2 block upper triangular matrix, with

1

A

as the (2, 2)-block. Then from

Supplementary Exercise 12 in Chapter 5,

11 1 11 1 1

det( ) det(( ) ) det( ) ( ) det( )

T

nnn

UAU I I A I A I

−−

−λ =λ−λ ⋅ −λ =λ−λ⋅ −λ

This shows that the eigenvalues of ,

T

UAU

namely,

1

,, ,

n

λ…λ

consist of

1

λ

and the eigenvalues

of

1

A

. So the eigenvalues of

1

A

are

2

,,.

n

λ…λ

404 CHAPTER 6 • Orthogonality and Least Squares

17. [M] Compute that || ∆x ||/|| x || = .4618 and

4

cond( ) (|| || / || ||) 3363 (1.548 10 ) .5206A

−

×=××=bb∆

. In

this case, || ∆x ||/|| x || is almost the same as cond(A) × || ∆b ||/|| b ||.

18. [M] Compute that || ∆x ||/|| x || = .00212 and cond(A) × (|| ∆b ||/|| b ||) = 3363 × (.00212) ≈ 7.130. In

this case, || ∆x ||/|| x || is almost the same as || ∆b ||/|| b ||, even though the large condition number

suggests that || ∆x ||/|| x || could be much larger.

19. [M] Compute that

8

|| || / || || 7.178 10

−

=×xx∆

and

4

cond( ) (|| || / || ||) 23683 (2.832 10 )A

−

×=××=bb∆

6.707.

Observe that the relative change in x is much smaller than the relative change in b. In fact the

theoretical bound on the relative change in x is 6.707 (to four significant figures). This exercise

shows that even when a condition number is large, the relative error in the solution need not be as

large as you suspect.

20. [M] Compute that || ∆x ||/|| x || = .2597 and

5

cond( ) (|| || / || ||) 23683 (1.097 10 ) .2598A

−

×=××=bb∆

.

This calculation shows that the relative change in x, for this particular b and ∆b, should not exceed

.2598. In this case, the theoretical maximum change is almost acheived.

r

7.1 SOLUTIONS

Notes:

Students can profit by reviewin

working on this section. Theorems 1 an

d

sections that follow. Note that symmetri

c

text have real entries, as mentioned at

been constructed so that mastery of the

G

Theorem 2 is easily proved for the

2

If

,

ab

Acd

ªº

=«»

¬¼

then

(

1

2

a

λ=

If b = 0 there is nothing to prove. Other

w

diagonalizable. In each case, an eigenve

c

1. Since

35 ,

57

T

AA

ªº

==

«»

−

¬¼

the matri

2. Since

35 ,

53

T

AA

−

ªº

=≠

«»

−

¬¼

the matri

3. Since

22 ,

44

T

AA

ªº

=≠

«»

¬¼

the matri

x

4. Since

083

802 ,

320

T

AA

ªº

«»

=−=

«»

−

¬¼

the

5. Since

620

062 ,

006

T

AA

−

ªº

«»

=−≠

«»

−

¬¼

th

r

son Education, Inc. Publishing as Addison-Wesley.

n

g Section 5.3 (focusing on the Diagonalization The

o

d

2 and the calculations in Examples 2 and 3 are imp

o

c matri

x

means real symmetric matrix, because all m

a

the beginning of this chapter. The exercises in this

s

G

ra

m

-Schmidt process is not needed.

2

× 2 case:

)

22

()4.

a

dad b+± −+

w

ise, there are two distinct eigenvalues, so A must be

c

tor for λ is .

d

b

−λ

ªº

«»

−

¬¼

i

x is symmetric.

i

x is not symmetric.

x

is not symmetric.

matrix is symmetric.

h

e matrix is not symmetric.

405

o

rem) before

o

rtant for the

a

trices in the

s

ection have

406 CHAPTER 7 • Symmetric Matrices and Quadratic Forms

6. Since A is not a square matrix

T

AA≠

and the matrix is not symmetric.

7. Let .6 .8 ,

.8 .6

Pªº

=«»

−

¬¼

and compute that

2

.6 .8 .6 .8 1 0

.8 .6 .8 .6 0 1

T

PP I

ªºªºªº

===

«»«»«»

−−

¬¼¬¼¬¼

Since P is a square matrix, P is orthogonal and

1

.6 .8 .

.8 .6

T

PP

−

ª

º

==

«

»

−

¬

¼

8. Let

1/ 2 1/ 2 ,

1/ 2 1/ 2

P

ªº

−

=«»

«»

¬¼

and compute that

2

1/ 2 1/ 2 1/ 2 1/ 2 1 0

01

1/ 2 1/ 2 1/ 2 1/ 2

T

PP I

ªºªº

−ªº

===

«»«»

«»

−¬¼

«»«»

¬¼¬¼

Since P is a square matrix, P is orthogonal and

1

1/ 2 1/ 2 .

1/ 2 1/ 2

T

PP

−

ª

º

==

«

»

−

«

»

¬

¼

9. Let 52

,

25

P

−

ªº

=«»

¬¼

and compute that

2

52 52 29 0

25 25 029

T

PP I

−−

ªºªºª º

==≠

«»«»« »

¬¼¬¼¬ ¼

Thus P is not orthogonal.

10. Let

122

212,

221

P

−

ªº

«»

=−

«»

−

¬¼

and compute that

3

122122 900

212212 090

221221 009

T

PP I

−−

ªºªºªº

«»«»«»

=−−=≠

«»«»«»

−−

¬¼¬¼¬¼

Thus P is not orthogonal.

11. Let

2/3 2/3 1/3

01/52/5,

5/3 4/ 45 2/ 45

P

ªº

«»

=−

«»

−−

¬¼

and compute that

7.1 • Solutions 407

3

2/3 0 5/3 2/3 2/3 1/3 100

2/3 1/ 5 4/ 45 0 1/ 5 2/ 5 0 1 0

001

1/3 2/ 5 2/ 45 5 /3 4/ 45 2/ 45

T

PP I

ªº

«»

=−−==

«»

−− − − ¬¼

«»

¬¼

Since P is a square matrix, P is orthogonal and

1

2/3 0 5/3

2/3 1/ 5 4/ 45 .

1/3 2/ 5 2/ 45

T

PP

−

ª

º

«

»

== −

«

»

«

»

−−

«

»

¬

¼

12. Let

.5 .5 .5 .5

.5 .5 .5 .5 ,

.5 .5 .5 .5

P

−−

ªº

«»

−−

«»

=«»

«»

−−

«»

¬¼

and compute that

4

.5 .5 .5 .5 .5 .5 .5 .5 1 0 0 0

.5 .5 .5 .5 .5 .5 .5 .5 0 1 0 0

.5 .5 .5 .5 .5 .5 .5 .5 0 0 1 0

.5 .5 .5 .5 .5 .5 .5 .5 0 0 0 1

T

PP I

−− −−

ªºªºªº

«»«»«»

−−

«»«»«»

===

«»«»«»

−−

«»«»«»

−−−−

«»«»«»

¬¼¬¼¬¼

Since P is a square matrix, P is orthogonal and

1

.5 .5 .5 .5

.5 .5 .5 .5 .

.5 .5 .5 .5

T

PP

−

−−

ª

º

«

»

«

»

==

«

»

−−

«

»

−−

«

»

¬

¼

13. Let 31

.

13

Aªº

=«»

¬¼

Then the characteristic polynomial of A is

22

(3 ) 1 6 8 ( 4)( 2),−λ − =λ−λ+=λ− λ−

so the eigenvalues of A are 4 and 2. For λ = 4, one computes that a basis for the eigenspace is 1,

1

ª

º

«

»

¬

¼

which can be normalized to get

1

1/ 2 .

1/ 2

ªº

=«»

«»

¬¼

u

For λ = 2, one computes that a basis for the eigenspace

is 1,

1

−

ªº

«»

¬¼

which can be normalized to get

2

1/ 2 .

1/ 2

ª

º

−

=

«

»

«

»

¬

¼

u

Let

[]

12

1/ 2 1/ 2 4 0

and 02

1/ 2 1/ 2

PD

ªº

−

ª

º

== =

«»

«

»

¬

¼

«»

¬¼

uu

Then P orthogonally diagonalizes A, and

1

APDP

−

=

.

408 CHAPTER 7 • Symmetric Matrices and Quadratic Forms

14. Let 15

.

51

Aªº

=«»

¬¼

Then the characteristic polynomial of A is

22

(1 ) 25 2 24 ( 6)( 4),−λ − =λ−λ− =λ− λ+

so the eigenvalues of A are 6 and –4. For λ = 6, one

computes that a basis for the eigenspace is 1,

1

ª

º

«

»

¬

¼ which can be normalized to get

1

1/ 2 .

1/ 2

ªº

=«»

«»

¬¼

u

For

λ = –4, one computes that a basis for the eigenspace is 1,

1

−

ª

º

«

»

¬

¼ which can be normalized to get

2

1/ 2 .

1/ 2

ªº

−

=«»

«»

¬¼

u

Let

[]

12

1/ 2 1/ 2 6 0

and 04

1/ 2 1/ 2

PD

ªº

−

ª

º

== =

«»

«

»

−

¬

¼

«»

¬¼

uu

Then P orthogonally diagonalizes A, and

1.APDP

−

=

15. Let 16 4 .

41

A

−

ªº

=«»

−

¬¼

Then the characteristic polynomial of A is

2

(16 )(1 ) 16 17 ( 17)−λ −λ − =λ− λ=λ− λ

, so the eigenvalues of A are 17 and 0. For λ = 17, one

computes that a basis for the eigenspace is 4,

1

−

ª

º

«

»

¬

¼ which can be normalized to get

1

4/ 17 .

1/ 17

ªº

−

=«»

«»

¬¼

u

For λ = 0, one computes that a basis for the eigenspace is 1

4

ª

º

«

»

¬

¼, which can be normalized to get

2

1/ 17 .

4/ 17

ªº

=«»

«»

¬¼

u

Let

[]

12

4/ 17 1/ 17 17 0

and 00

1/ 17 4/ 17

PD

ªº

−

ª

º

== =

«»

«

»

¬

¼

«»

¬¼

uu

Then P orthogonally diagonalizes A, and

1.APDP

−

=

16. Let 724

.

24 7

A

−

ªº

=«»

¬¼

Then the characteristic polynomial of A is

2

(7 )(7 ) 576 625−−λ −λ− =λ− =

(25)(25)λ− λ+

, so the eigenvalues of A are 25 and –25. For λ = 25, one computes that a basis for

the eigenspace is 3,

4

ªº

«»

¬¼ which can be normalized to get

1

3/5 .

4/5

ª

º

=

«

»

¬

¼

u

For λ = –25, one computes that a

basis for the eigenspace is 4,

3

−

ªº

«»

¬¼

which can be normalized to get

2

4/5 .

3/5

−

ª

º

=

«

»

¬

¼

u

Let

7.1 • Solutions 409

[]

12

3/5 4/5 25 0

and

4/5 3/5 0 25

PD

−

ª

ºªº

== =

«

»«»

−

¬

¼¬¼

uu

Then P orthogonally diagonalizes A, and

1

APDP

−

=

.

17. Let

113

131.

311

A

ªº

«»

=«»

«»

¬¼

The eigenvalues of A are 5, 2, and –2. For λ = 5, one computes that a basis for

the eigenspace is

1

1,

1

ªº

«»

¬¼

which can be normalized to get

1

1/ 3

1/ 3 .

1/ 3

ª

º

«

»

=

«

»

«

»

«

»

¬

¼

u

For λ = 2, one computes that a

basis for the eigenspace is

1

2,

1

ªº

«»

−

«»

¬¼

which can be normalized to get

2

1/ 6

2/ 6 .

1/ 6

ª

º

«

»

=−

«

»

«

»

«

»

¬

¼

u

For λ = –2, one

computes that a basis for the eigenspace is

1

0,

1

−

ª

º

«

»

«

»

«

»

¬

¼

which can be normalized to get

3

1/ 2

0.

1/ 2

ªº

−

«»

=«»

«»

¬¼

u

Let

[]

123

1/ 3 1/ 6 1/ 2 50 0

1/ 3 2/ 6 0 and 0 2 0

00 2

1/ 3 1/ 6 1 2

PD

ªº

−

ª

º

«»

«

»

==−=

«»

«

»

«»

«

»

−

¬

¼

«»

¬¼

uu u

Then P orthogonally diagonalizes A, and

1

APDP

−

=

.

18. Let

2360

36 23 0 .

003

A

−−

ªº

«»

=−−

«»

¬¼

The eigenvalues of A are 25, 3, and –50. For λ = 25, one computes that a

basis for the eigenspace is

4

3,

0

−

ªº

«»

¬¼

which can be normalized to get

1

4/5

3/5 .

0

−

ª

º

«

»

=

«

»

«

»

¬

¼

u For λ = 3, one

computes that a basis for the eigenspace is

0

0,

1

ª

º

«

»

«

»

«

»

¬

¼

which is of length 1, so

2

0

0.

1

ªº

«»

=«»

«»

¬¼

u For λ = –50, one

computes that a basis for the eigenspace is

3

4,

0

ª

º

«

»

«

»

«

»

¬

¼

which can be normalized to get

3

3/5

4/5 .

0

ªº

«»

=«»

«»

¬¼

u Let

410 CHAPTER 7 • Symmetric Matrices and Quadratic Forms

[]

123

4/5 0 3/5 25 0 0

3/5 0 4/5 and 0 3 0

01 0 00 50

PD

−

ªºªº

«»«»

== =

«»«»

−

¬¼¬¼

uu u

Then P orthogonally diagonalizes A, and

1

APDP

−

=

.

19. Let

324

262.

423

A

−

ªº

«»

=−

«»

¬¼

The eigenvalues of A are 7 and –2. For λ = 7, one computes that a basis for

the eigenspace is

11

2,0 .

01

½−

ªºªº

°°

«»«»

®¾

«»«»

°°

«»«»

¬¼¬¼

¯¿

This basis may be converted via orthogonal projection to an

orthogonal basis for the eigenspace:

14

2,2 .

05

½−

ª

ºªº

°°

«

»«»

®¾

«

»«»

°°

«

»«»

¬

¼¬¼

¯¿

These vectors can be normalized to get

1

1/ 5

2/ 5 ,

0

ªº

−

«»

=«»

«»

¬¼

u

2

4/ 45

2/ 45 .

5/ 45

ª

º

«

»

=

«

»

«

»

«

»

¬

¼

u

For λ = –2, one computes that a basis for the eigenspace is

2

1,

2

−

ªº

«»

−

«»

¬¼

which can be normalized to get

3

2/3

1/3 .

2/3

−

ªº

«»

=−

«»

¬¼

u Let

[]

123

1/ 5 4/ 45 2/3 70 0

2/ 5 2/ 45 1/3 and 0 7 0

00 2

05/45 2/3

PD

ªº

−−

ª

º

«»

«

»

== −=

«»

«

»

«»

«

»

−

¬

¼

«»

¬¼

uu u

Then P orthogonally diagonalizes A, and

1

APDP

−

=

.

20. Let

744

450.

409

A

−

ªº

«»

=−

«»

¬¼

The eigenvalues of A are 13, 7, and 1. For λ = 13, one computes that a basis

for the eigenspace is

2

1,

2

ª

º

«

»

−

«

»

«

»

¬

¼

which can be normalized to get

1

2/3

1/3 .

2/3

ª

º

«

»

=−

«

»

«

»

¬

¼

u For λ = 7, one computes

7.1 • Solutions 411

that a basis for the eigenspace is

1

2,

2

−

ªº

«»

¬¼

which can be normalized to get

2

1/3

2/3 .

2/3

−

ªº

«»

=«»

«»

¬¼

u For λ = 1, one

computes that a basis for the eigenspace is

2

2,

1

ª

º

«

»

«

»

«

»

−

¬

¼

which can be normalized to get

3

2/3

2/3 .

1/3

ªº

«»

=«»

«»

−

¬¼

u Let

[]

123

2/3 1/3 2/3 13 0 0

1/3 2/3 2/3 and 0 7 0

2/3 2/3 1/3 0 0 1

PD

−

ªºªº

«»«»

==−=

«»«»

−

¬¼¬¼

uu u

Then P orthogonally diagonalizes A, and

1

APDP

−

=

.

21. Let

4131

1413

.

3141

1314

A

ªº

«»

=«»

«»

¬¼

The eigenvalues of A are 9, 5, and 1. For λ = 9, one computes that a basis

for the eigenspace is

1

1,

1

ª

º

«

»

«

»

«

»

«

»

«

»

¬

¼

which can be normalized to get

1

1/ 2

1/ 2 .

1/ 2

ª

º

«

»

«

»

=

«

»

«

»

«

»

¬

¼

u

For λ = 5, one computes that a

basis for the eigenspace is

1

1,

1

−

ªº

«»

−

«»

¬¼

which can be normalized to get

2

1/2

1/2 .

1/2

−

ª

º

«

»

«

»

=

«

»

−

«

»

«

»

¬

¼

u

For λ = 1, one

computes that a basis for the eigenspace is

10

01

,.

10

01



½−

ª

ºª º

°

«

»« »

−

°

«

»« »

®

¾

«

»« »

°

«

»« »

°

«

»« »

¬

¼¬ ¼

¯¿

This basis is an orthogonal basis for the

eigenspace, and these vectors can be normalized to get

3

1/ 2

0,

1/ 2

0

ª

º

−

«

»

«

»

=

«

»

«

»

«

»

¬

¼

u

4

0

1/ 2 .

0

1/ 2

ªº

«»

−

«»

=«»

«»

¬¼

u

Let

[]

1234

1/2 1/2 1/ 2 0 9000

1/2 1/2 0 1/ 2 0 5 0 0

and 0010

1/2 1/2 1/ 2 0

0001

1/2 1/2 0 1/ 2

PD

ªº

−−

ª

º

«»

«

»

−

«»

«

»

== =

«»

«

»

−

«»

«

»

«»

«

»

¬

¼

¬¼

uu uu

412 CHAPTER 7 • Symmetric Matrices and Quadratic Forms

Then P orthogonally diagonalizes A, and

1

APDP

−

=

.

22. Let

2000

0101

.

0020

0101

A

ªº

«»

=«»

«»

¬¼

The eigenvalues of A are 2 and 0. For λ = 2, one computes that a basis for

the eigenspace is

100

010

,, .

001

010

½

ªºªºªº

°°

«»«»«»

°°

«»«»«»

®¾

«»«»«»

°°

«»«»«»

°°

«»«»«»

¬¼¬¼¬¼

¯¿

This basis is an orthogonal basis for the eigenspace, and these

vectors can be normalized to get

1

0,

0

ª

º

«

»

«

»

=

«

»

«

»

«

»

¬

¼

u

2

0

1/ 2 ,

0

1/ 2

ª

º

«

»

«

»

=

«

»

«

»

«

»

¬

¼

u

and

3

0

0.

1

0

ª

º

«

»

«

»

=

«

»

«

»

«

»

¬

¼

u

For λ = 0, one computes

that a basis for the eigenspace is

0

1,

0

1

ªº

«»

−

«»

¬¼

which can be normalized to get

4

0

1/ 2 .

0

1/ 2

ªº

«»

−

«»

=«»

«»

¬¼

u

Let

[]

1234

100 0 2000

01/2 0 1/2 0 200

and

001 0 0020

0000

01/2 0 1/2

PD

ªº

ª

º

«»

«

»

−

«»

«

»

== =

«»

«

»

«»

«

»

«

»

«»

¬

¼

¬¼

uu u u

Then P orthogonally diagonalizes A, and

1

APDP

−

=

.

23. Let

311

131

113

A

ªº

«»

=«»

«»

¬¼

. Since each row of A sums to 5,

131115 1

11311551

111315 1

A

ª

ºª ºªºªº ªº

«

»« »«»«» «»

===

«

»« »«»«» «»

«

»« »«»«» «»

¬

¼¬ ¼¬¼¬¼ ¬¼

and

1

ªº

«»

¬¼

is an eigenvector of A with corresponding eigenvalue λ = 5. The eigenvector may be

normalized to get

1

1/ 3

ª

º

«

»

=

«

»

«

»

«

»

¬

¼

u

. For λ = 2, one computes that a basis for the eigenspace is

11

1, 0 ,

01

½−−

ªºªº

°°

«»«»

®¾

«»«»

°°

«»«»

¬¼¬¼

¯¿

so λ = 2 is an eigenvalue of A. This basis may be converted via orthogonal

7.1 • Solutions 413

projection to an orthogonal basis

11

1, 1

02

½−−

ªºªº

°°

«»«»

−

®¾

«»«»

°°

«»«»

¬¼¬¼

¯¿

for the eigenspace, and these vectors can be

normalized to get

2

1/ 2

0

ª

º

−

«

»

=

«

»

«

»

«

»

¬

¼

u

and

3

1/ 6

1/ 6 .

2/ 6

ª

º

−

«

»

=−

«

»

«

»

«

»

¬

¼

u

Let

[]

123

1/ 3 1/ 2 1/ 6 500

1/ 3 1/ 2 1/ 6 and 0 2 0

002

1/ 3 0 2/ 6

PD

ªº

−−

ª

º

«»

«

»

== −=

«»

«

»

«»

«

»

¬

¼

«»

¬¼

uu u

Then P orthogonally diagonalizes A, and

1

APDP

−

=

.

24. Let

542

452.

222

A

−−

ªº

«»

=−

«»

−

¬¼

One may compute that

220 2

220102

110 1

A

−− −

ª

ºª º ªº

«

»« » «»

==

«

»« » «»

«

»« » «»

¬

¼¬ ¼ ¬¼

, so

1

2

1

−

ªº

«»

=«»

«»

¬¼

v is an

eigenvector of A with associated eigenvalue

110λ=

. Likewise one may compute that

11 1

1111

00 0

A

ªº ªº ªº

«» «» «»

==

«» «» «»

¬¼ ¬¼ ¬¼

, so

1

0

ª

º

«

»

«

»

«

»

¬

¼

is an eigenvector of A with associated eigenvalue

21λ=

. For

21λ=

, one

computes that a basis for the eigenspace is

11

1,0 .

02



½

ª

ºªº

°

«

»«»

®

¾

«

»«»

°

«

»«»

¬

¼¬¼

¯¿

This basis may be converted via orthogonal

projection to an orthogonal basis for the eigenspace:

{}

23

11

,1,1.

04



½

ª

ºª º

°

«

»« »

=−

®

¾

«

»« »

°

«

»« »

¬

¼¬ ¼

¯¿

vv

The eigenvectors

1

v

,

2

v

, and

3

v

may be normalized to get the vectors

1

2/3

2/3 ,

1/3

−

ª

º

«

»

=

«

»

«

»

¬

¼

u

2

1/ 2

1/ 2 ,

0

ª

º

«

»

=

«

»

«

»

«

»

¬

¼

u

and

3

1/ 18

1/ 18 .

4/ 18

ªº

«»

=−

«»

¬¼

u

Let

[]

123

2/3 1/ 2 1/ 18 10 0 0

2/3 1/ 2 1/ 18 and 0 1 0

001

1/3 0 4/ 18

PD

ªº

−

ª

º

«»

«

»

== −=

«»

«

»

«»

«

»

¬

¼

«»

¬¼

uu u

Then P orthogonally diagonalizes A, and

1

APDP

−

=

.

25. a. True. See Theorem 2 and the paragraph preceding the theorem.

414 CHAPTER 7 • Symmetric Matrices and Quadratic Forms

b . True. This is a particular case of the statement in Theorem 1, where u and v are nonzero.

c . False. There are n real eigenvalues (Theorem 3), but they need not be distinct (Example 3).

d . False. See the paragraph following formula (2), in which each u is a unit vector.

26. a. True. See Theorem 2.

b . True. See the displayed equation in the paragraph before Theorem 2.

c . False. An orthogonal matrix can be symmetric (and hence orthogonally diagonalizable), but not

every orthogonal matrix is symmetric. See the matrix P in Example 2.

d . True. See Theorem 3(b).

27. Since A is symmetric,

()

TTTTTTT

BAB BAB BAB==

, and

T

BAB

is symmetric. Applying this result

with A = I gives

T

BB

is symmetric. Finally,

()

TT TT T T

BB B B BB==

, so

T

BB

is symmetric.

28. Let A be an n × n symmetric matrix. Then

() () ()

TTTT

AA AAA⋅====⋅xy xy x y x y x y

since

T

AA=

.

29. Since A is orthogonally diagonalizable,

1

APDP

−

=

, where P is orthogonal and D is diagonal. Since

A is invertible,

11111

()APDP PDP

−−−−−

==

. Notice that

1

D

−

is a diagonal matrix, so

1

A

−

is

orthogonally diagonalizable.

30. If A and B are orthogonally diagonalizable, then A and B are symmetric by Theorem 2. If AB = BA,

then

() ()

TTTT

AB BA A B AB===

. So AB is symmetric and hence is orthogonally diagonalizable by

Theorem 2.

31. The Diagonalization Theorem of Section 5.3 says that the columns of P are linearly independent

eigenvectors corresponding to the eigenvalues of A listed on the diagonal of D. So P has exactly k

columns of eigenvectors corresponding to λ. These k columns form a basis for the eigenspace.

32. If

1

,APRP

−

=

then

1

.PAP R

−

=

Since P is orthogonal,

T

R

PAP=

. Hence

()

TTTTTTT

RPAPPAP== =

,

T

PAP R=

which shows that R is symmetric. Since R is also upper triangular, its entries above the

diagonal must be zeros to match the zeros below the diagonal. Thus R is a diagonal matrix.

33. It is previously been found that A is orthogonally diagonalized by P, where

[]

123

1/ 2 1/ 6 1/ 3 800

1/ 2 1/ 6 1/ 3 and 0 6 0

003

02/61/3

PD

ªº

−−

ª

º

«»

«

»

==−=

«»

«

»

«»

«

»

¬

¼

«»

¬¼

uu u

Thus the spectral decomposition of A is

111 2 2 2 3 3 3 11 2 2 3 3

86 3

TTTTTT

A=+ + =++uu u u uu uu u u uu

7.1 • Solutions 415

1/ 2 1/ 2 0 1/6 1/6 2/ 6 1/3 1/3 1/3

81/2 1/20 61/6 1/6 2/6 31/31/31/3

000 2/62/64/61/31/31/3

−−

ªºª ºªº

«»« »«»

=−+−+

«»« »«»

−−

¬¼¬ ¼¬¼

34. It is previously been found that A is orthogonally diagonalized by P, where

[]

123

1/ 2 1/ 18 2/3 70 0

04/18 1/3and 070

00 2

1/ 2 1/ 18 2/3

PD

ªº

−−

ª

º

«»

«

»

== −=

«»

«

»

«»

«

»

−

¬

¼

«»

¬¼

uu u

Thus the spectral decomposition of A is

111 2 2 2 33 3 11 2 2 3 3

77 2

TTTTTT

A=+ + =+−uu u u u u uu u u u u

1/ 2 0 1/ 2 1/18 4/18 1/18 4/9 2/9 4/9

700 074/1816/18 4/1822/9 1/9 2/9

1/ 2 0 1/ 2 1/18 4/18 1/18 4/9 2/9 4/9

−− −

ªºª ºª º

«»« »« »

=+−−−

«»« »« »

−−−

¬¼¬ ¼¬ ¼

35. a. Given x in

n

,

() ()(),

TTT

b===xuuxuux uxu

because

T

ux

is a scalar. So Bx = (x ⋅ u)u. Since

u is a unit vector, Bx is the orthogonal projection of x onto u.

b . Since

() ,

TTTTTTT

BB====uu u u uu

B is a symmetric matrix. Also,

2

()()()

TT TT T

BB====uu uu u u u u uu

because

1.

T=uu

c . Since

1

T=uu

,

() ()(1)

TT

B====uuuuuuuu u

, so u is an eigenvector of B with corresponding

eigenvalue 1.

36. Given any y in

n

, let

ˆ

y

= By and z = y –

ˆ

y

. Suppose that

T

BB=

and

2

BB=

. Then

.

T

BB BB B==

a. Since

ˆˆ ˆ

()()()() () 0

TTTTT

BBBBBBBBB⋅=−⋅ =⋅−⋅=−=−=zy y y y y y y y y y y y y y y y

, z is

orthogonal to

ˆ.y

b. Any vector in W = Col B has the form Bu for some u. Noting that B is symmetric, Exercise 28

gives

( y –

ˆ

y

) ⋅ (Bu) = [B(y –

ˆ

y

)] ⋅ u = [By – BBy] ⋅ u = 0

since

2.BB=

So y –

ˆ

y

is in

,W

⊥

and the decomposition y =

ˆ

y

+ (y –

ˆ

y

) expresses y as the sum

of a vector in W and a vector in

.W⊥

By the Orthogonal Decomposition Theorem in Section 6.3,

this decomposition is unique, and so

ˆ

y

must be

proj .

Wy

416 CHAPTER 7 • Symmetric Matrices and Quadratic Forms

37. [M] Let

5296

2569

.

9652

6925

A

−

ªº

«»

−

«»

=«»

−

«»

−

«»

¬¼

The eigenvalues of A are 18, 10, 4, and –12. For λ = 18, one

computes that a basis for the eigenspace is

1

1,

1

−

ª

º

«

»

«

»

«

»

−

«

»

«

»

¬

¼

which can be normalized to get

1

1/ 2

1/ 2 .

1/ 2

−

ªº

«»

=«»

−

«»

¬¼

u

For

λ = 10, one computes that a basis for the eigenspace is

1

ª

º

«

»

«

»

«

»

«

»

«

»

¬

¼

, which can be normalized to get

2

1/ 2

1/ 2 .

1/ 2

ªº

«»

=«»

«»

¬¼

u

For λ = 4, one computes that a basis for the eigenspace is

1

ª

º

«

»

«

»

«

»

−

«

»

−

«

»

¬

¼

, which can be

normalized to get

3

1/ 2

1/ 2 .

1/ 2

ª

º

«

»

«

»

=

«

»

−

«

»

−

«

»

¬

¼

u

For λ = –12, one computes that a basis for the eigenspace is

1

1,

1

ªº

«»

−

«»

−

«»

¬¼

which can be normalized to get

4

1/2

1/2 .

1/2

ªº

«»

−

«»

=«»

−

«»

¬¼

u

Let

[]

1234

1/ 2 1/ 2 1/ 2 1/ 2

P

−

ªº

«»

−

«»

==

«»

−−−

«»

−

«»

¬¼

uu uu

and

18 0 0 0

010 0 0

.

004 0

000 12

D

ª

º

«

»

«

»

=

«

»

«

»

−

«

»

¬

¼

Then P

orthogonally diagonalizes A, and

1

APDP

−

=

.

38. [M] Let

.38 .18 .06 .04

.18 .59 .04 .12 .

.06 .04 .47 .12

.04 .12 .12 .41

A

−−−

ªº

«»

−−

«»

=«»

−− −

«»

−−

«»

¬¼

The eigenvalues of A are .25, .30, .55, and .75. For λ =

.25, one computes that a basis for the eigenspace is

4

2,

2

1

ª

º

«

»

«

»

«

»

«

»

«

»

¬

¼

which can be normalized to get

1

.8

.4 .

.4

.2

ª

º

«

»

«

»

=

«

»

«

»

«

»

¬

¼

u

For

7.1 • Solutions 417

λ = .30, one computes that a basis for the eigenspace is

1

2,

2

4

−

ª

º

«

»

−

«

»

«

»

«

»

«

»

¬

¼

which can be normalized to get

2

.2

.4 .

.4

.8

−

ªº

«»

−

«»

=«»

«»

¬¼

u

For λ = .55, one computes that a basis for the eigenspace is

2

1,

4

2

ª

º

«

»

−

«

»

«

»

−

«

»

«

»

¬

¼

which can be

normalized to get

3

.4

.2 .

.8

.4

ª

º

«

»

−

«

»

=

«

»

−

«

»

«

»

¬

¼

u

For λ = .75, one computes that a basis for the eigenspace is

2

4,

1

2

−

ªº

«»

−

«»

¬¼

which can be normalized to get

4

.4

.8 .

.2

.4

−

ªº

«»

=«»

−

«»

¬¼

u

Let

[]

1234

.8 .2 .4 .4

.4 .4 .2 .8

.4 .4 .8 .2

.2 .8 .4 .4

P

−−

ª

º

«

»

−−

«

»

==

«

»

−−

«

»

«

»

¬

¼

uu uu

and

.25 0 0 0

0.30 0 0

.

00.550

000.75

D

ªº

«»

=«»

«»

¬¼

Then P orthogonally diagonalizes A, and

1

APDP

−

=

.

39. [M] Let

.31 .58 .08 .44

.58 .56 .44 .58 .

.08 .44 .19 .08

.44 .58 .08 .31

A

ªº

«»

−−

«»

=«»

−

«»

−−

«»

¬¼

The eigenvalues of A are .75, 0, and –1.25. For λ = .75, one

computes that a basis for the eigenspace is

13

02

,.

02

10



½

ª

ºªº

°

«

»«»

°

«

»«»

®

¾

«

»«»

°

«

»«»

°

«

»«»

¬

¼¬¼

¯¿

This basis may be converted via orthogonal

projection to the orthogonal basis

13

04

,.

04

13

½

ªºª º

°°

«»« »

°°

«»« »

®¾

«»« »

°°

«»« »

°°

−

«»« »

¬¼¬ ¼

¯¿

These vectors can be normalized to get

1

1/ 2

0,

0

1/ 2

ªº

«»

=«»

«»

¬¼

u

2

3/ 50

4/ 50 .

4/ 50

3/ 50

ªº

«»

=«»

«»

−

¬¼

u

For λ = 0, one computes that a basis for the eigenspace is

2

1,

4

2

−

ªº

«»

−

«»

¬¼

418 CHAPTER 7 • Symmetric Matrices and Quadratic Forms

which can be normalized to get

3

.4

.2 .

.8

.4

−

ªº

«»

−

«»

=«»

«»

¬¼

u

For λ = –1.25, one computes that a basis for the

eigenspace is

2

4,

1

2

−

ªº

«»

−

«»

¬¼

which can be normalized to get

4

.4

.8 .

.2

.4

−

ª

º

«

»

«

»

=

«

»

−

«

»

«

»

¬

¼

u

Let

[]

1234

1/ 2 3/ 50 .4 .4

04/50 .2.8

04/50 .8.2

1/ 2 3/ 50 .4 .4

P

ªº

−−

«»

−

«»

==

«»

−

«»

−

¬¼

uu uu

and

.75 0 0 0

0.750 0

000 0

0001.25

D

ª

º

«

»

«

»

=

«

»

«

»

−

«

»

¬

¼

.

Then P orthogonally diagonalizes A, and

1

APDP

−

=

.

40. [M] Let

10 2 2 6 9

210 2 6 9

.

22106 9

66626 9

999919

A

−

ªº

«»

−

«»

=−

«»

−−−

«»

−

¬¼

The eigenvalues of A are 8, 32, –28, and 17. For λ = 8,

one computes that a basis for the eigenspace is

11

10

,.

01

00



½

−

ª

ºª º

°

«

»« »

−

°

«

»« »

°

«

»« »

®

¾

«

»« »

°

«

»« »

°

«

»« »

°

¬

¼¬ ¼

¯¿

This basis may be converted via

orthogonal projection to the orthogonal basis

11

,.

02

00



½

ª

ºª º

°

«

»« »

−

°

«

»« »

°

«

»« »

−

®

¾

«

»« »

°

«

»« »

°

«

»« »

°

¬

¼¬ ¼

¯¿

These vectors can be normalized to get

1

1/ 2

,

0

ªº

«»

−

«»

=«»

«»

¬¼

u

2

1/ 6

.

2/ 6

0

ª

º

«

»

«

»

«

»

=−

«

»

«

»

«

»

«

»

¬

¼

u

For λ = 32, one computes that a basis for the eigenspace is

1

,

1

3

0

ªº

«»

−

«»

¬¼

7.1 • Solutions 419

which can be normalized to get

3

1/ 12

.

1/ 12

3/ 12

0

ªº

«»

=«»

«»

−

«»

¬¼

u

For λ = –28, one computes that a basis for the

eigenspace is

1

,

1

4

ªº

«»

−

¬¼

which can be normalized to get

4

1/ 20

.

1/ 20

4/ 20

ª

º

«

»

«

»

«

»

=

«

»

«

»

«

»

«

»

−

¬

¼

u

For λ = 17, one computes that

a basis for the eigenspace is

1

,

1

ªº

«»

¬¼

which can be normalized to get

5

1/ 5

.

1/ 5

ª

º

«

»

«

»

«

»

=

«

»

«

»

«

»

«

»

¬

¼

u

Let

[]

12345

1/ 2 1/ 6 1/ 12 1/ 20 1/ 5

02/6 1/12 1/201/5

003/121/201/5

00 04/201/5

P

ªº

«»

−

«»

==

−

«»

−

«»

−

¬¼

uu u u u

and

80 0 0 0

08 0 0 0

.

0032 0 0

00 0 28 0

00 0 017

D

ªº

«»

=«»

−

«»

¬¼

Then P orthogonally diagonalizes A, and

1

APDP

−

=

.

420 CHAPTER 7 • Symmetric Matrices and Quadratic Forms

7.2 SOLUTIONS

Notes:

This section can provide a good conclusion to the course, because the mathematics here is widely

used in applications. For instance, Exercises 23 and 24 can be used to develop the second derivative test

for functions of two variables. However, if time permits, some interesting applications still lie ahead.

Theorem 4 is used to prove Theorem 6 in Section 7.3, which in turn is used to develop the singular value

decomposition.

1. a.

[]

2

12

12 1 122

2

51/3 5(2/3)

1/3 1

Tx

Axx x xxx

x

ªº

==++

«»

¬¼

xx

b . When

6,

1

ªº

=«»

¬¼

x

22

5(6) (2 / 3)(6)(1) (1) 185.

T

A=+ +=xx

c . When

1,

3

ªº

=«»

¬¼

x

22

5(1) (2 / 3)(1)(3) (3) 16.

T

A=+ +=xx

2. a.

[]

1

222

123 2 1 23 12 23

3

430

321 4 2 6 2

011

T

x

Axxx x x xx xx xx

x

ªºªº

«»«»

==++++

«»«»

¬¼¬¼

xx

b . When

2

1,

5

ªº

«»

=−

«»

¬¼

x

222

4(2) 2( 1) (5) 6(2)( 1) 2( 1)(5) 21.

T

A=+−++ −+−=xx

c . When

1/ 3

1/ 3 ,

1/ 3

ªº

«»

=«»

«»

¬¼

x

222

4(1/ 3) 2(1/ 3) (1/ 3) 6(1/ 3)(1/ 3) 2(1/ 3)(1/ 3) 5.

T

A=+++ + =xx

3. a. The matrix of the quadratic form is

10 3 .

33

−

ª

º

«

»

−−

¬

¼

b . The matrix of the quadratic form is

53/2

.

3/2 0

ª

º

«

»

¬

¼

4. a. The matrix of the quadratic form is

20 15 / 2 .

15/ 2 10

ª

º

«

»

−

¬

¼

b . The matrix of the quadratic form is

01/2

.

1/2 0

ª

º

«

»

¬

¼

5. a. The matrix of the quadratic form is

832

37 1.

213

−

ª

º

«

»

−−

«

»

«

»

−−

¬

¼

7.2 • Solutions 421

b . The matrix of the quadratic form is

023

204.

340

ª

º

«

»

−

«

»

«

»

−

¬

¼

6. a. The matrix of the quadratic form is

55/2 3/2

5/2 1 0 .

3/2 0 7

−

ª

º

«

»

−

«

»

«

»

−

¬

¼

b . The matrix of the quadratic form is

020

202.

021

−

ª

º

«

»

−

«

»

«

»

¬

¼

7. The matrix of the quadratic form is

15

.

51

A

ª

º

=

«

»

¬

¼

The eigenvalues of A are 6 and –4. An eigenvector

for λ = 6 is

1,

1

ªº

«»

¬¼

which may be normalized to

1

1/ 2 .

1/ 2

ª

º

=

«

»

«

»

¬

¼

u

An eigenvector for λ = –4 is

1,

1

−

ªº

«»

¬¼

which may be normalized to

2

1/ 2 .

1/ 2

ªº

−

=«»

«»

¬¼

u

Then

1

APDP

−

=

, where

[]

12

1/ 2 1/ 2

P

ªº

−

==

«»

¬¼

uu

and

60

.

04

D

ª

º

=

«

»

−

¬

¼

The desired change of variable is x = Py, and

the new quadratic form is

22

12

()() 6 4

TT TTT

APAP PAP D y y====−xx y y y yyy

8. The matrix of the quadratic form is

944

470.

4011

A

−

ª

º

«

»

=−

«

»

«

»

¬

¼

The eigenvalues of A are 3, 9, and 15. An

eigenvector for λ = 3 is

2

2,

1

−

ªº

«»

−

«»

¬¼

which may be normalized to

1

2/3

2/3 .

1/3

−

ª

º

«

»

=−

«

»

«

»

¬

¼

u An eigenvector for λ = 9

is

1

2,

2

−

ªº

«»

¬¼

which may be normalized to

2

1/3

2/3 .

2/3

−

ª

º

«

»

=

«

»

«

»

¬

¼

u An eigenvector for λ = 15 is

2

1,

2

ªº

«»

−

«»

¬¼

which may

be normalized to

3

2/3

1/3 .

2/3

ª

º

«

»

=−

«

»

«

»

¬

¼

u Then

1

APDP

−

=

, where

[]

123

2/3 1/3 2/3

2/3 2/3 1/3

1/3 2/3 2/3

P

−−

ª

º

«

»

==−−

«

»

«

»

¬

¼

uu u and

30 0

09 0.

0015

D

ª

º

«

»

=

«

»

«

»

¬

¼

The desired change of variable

is x = Py, and the new quadratic form is

422 CHAPTER 7 • Symmetric Matrices and Quadratic Forms

22 2

12 3

()() 3 9 15

TT TTT

APAP PAP D y y y====++xx y y y yyy

9. The matrix of the quadratic form is

32

.

26

A

−

ª

º

=

«

»

−

¬

¼

The eigenvalues of A are 7 and 2, so the

quadratic form is positive definite. An eigenvector for λ = 7 is

1,

2

−

ª

º

«

»

¬

¼

which may be normalized to

1

1/ 5 .

2/ 5

ªº

−

=«»

«»

¬¼

u

An eigenvector for λ = 2 is

2,

1

ª

º

«

»

¬

¼

which may be normalized to

2

2/ 5 .

1/ 5

ªº

=«»

«»

¬¼

u

Then

1

APDP

−

=

, where

[]

12

1/ 5 2/ 5

2/ 5 1/ 5

P

ªº

−

==

«»

¬¼

uu

and

70

.

02

D

ª

º

=

«

»

¬

¼

The desired change of

variable is x = Py, and the new quadratic form is

22

12

()() 7 2

TT TTT

APAP PAP D y y====+xx y y y yyy

10. The matrix of the quadratic form is

94

.

43

A

−

ª

º

=

«

»

−

¬

¼

The eigenvalues of A are 11 and 1, so the

quadratic form is positive definite. An eigenvector for λ = 11 is

2,

1

ª

º

«

»

−

¬

¼

which may be normalized to

1

2/ 5 .

1/ 5

ªº

=«»

−

«»

¬¼

u

An eigenvector for λ = 1 is

1

2

ª

º

«

»

¬

¼

, which may be normalized to

2

1/ 5 .

2/ 5

ªº

=«»

«»

¬¼

u

Then

1

APDP

−

=

, where

[]

12

2/ 5 1/ 5

1/ 5 2/ 5

P

ªº

==

«»

−

«»

¬¼

uu

and

11 0 .

01

D

ª

º

=

«

»

¬

¼

The desired change of

variable is x = Py, and the new quadratic form is

22

12

()() 11

TT TTT

APAP PAP D yy====+xx y y y yyy

11. The matrix of the quadratic form is

25

.

52

A

ª

º

=

«

»

¬

¼

The eigenvalues of A are 7 and –3, so the quadratic

form is indefinite. An eigenvector for λ = 7 is

1,

1

ª

º

«

»

¬

¼

which may be normalized to

1

1/ 2 .

1/ 2

ªº

=«»

«»

¬¼

u

An

eigenvector for λ = –3 is

1,

1

−

ªº

«»

¬¼

which may be normalized to

2

1/ 2 .

1/ 2

ª

º

−

=

«

»

«

»

¬

¼

u

Then

1

APDP

−

=

,

where

[]

12

1/ 2 1/ 2

P

ªº

−

==

«»

¬¼

uu

and

70

.

03

D

ª

º

=

«

»

−

¬

¼

The desired change of variable is x

= Py, and the new quadratic form is

22

12

()() 7 3

TT TTT

APAP PAP D yy====−xx y y y yyy

7.2 • Solutions 423

12. The matrix of the quadratic form is

52

.

22

A

−

ª

º

=

«

»

−

¬

¼

The eigenvalues of A are –1 and –6, so the

quadratic form is negative definite. An eigenvector for λ = –1 is

1,

2

ª

º

«

»

¬

¼

which may be normalized to

1

1/ 5 .

2/ 5

ªº

=«»

«»

¬¼

u

An eigenvector for λ = –6 is

2,

1

−

ª

º

«

»

¬

¼

which may be normalized to

2

2/ 5 .

1/ 5

ªº

−

=«»

«»

¬¼

u

Then

1

APDP

−

=

, where

[]

12

1/ 5 2/ 5

2/ 5 1/ 5

P

ªº

−

==

«»

¬¼

uu

and

10

06

D

−

ª

º

=

«

»

−

¬

¼

. The desired change of

variable is x = Py, and the new quadratic form is

22

12

()() 6

TT TTT

APAP PAP D y y====−−xx y y y yyy

13. The matrix of the quadratic form is

13

.

39

A

−

ª

º

=

«

»

−

¬

¼

The eigenvalues of A are 10 and 0, so the

quadratic form is positive semidefinite. An eigenvector for λ = 10 is

1,

3

ª

º

«

»

−

¬

¼

which may be

normalized to

1

1/ 10 .

3/ 10

ªº

=«»

−

«»

¬¼

u

An eigenvector for λ = 0 is

3,

1

ª

º

«

»

¬

¼

which may be normalized to

2

3/ 10 .

1/ 10

ªº

=«»

«»

¬¼

u

Then

1

APDP

−

=

, where

[]

12

1/ 10 3/ 10

3/ 10 1/ 10

P

ª

º

==

«

»

−

«

»

¬

¼

uu

and

10 0 .

00

D

ªº

=«»

¬¼

The

desired change of variable is x = Py, and the new quadratic form is

2

1

()() 10

TT TTT

APAP PAP D y====xx y y y yy y

14. The matrix of the quadratic form is

83

.

30

A

ª

º

=

«

»

¬

¼

The eigenvalues of A are 9 and –1, so the quadratic

form is indefinite. An eigenvector for λ = 9 is

3,

1

ª

º

«

»

¬

¼

which may be normalized to

1

3/ 10 .

1/ 10

ªº

=«»

«»

¬¼

u

An

eigenvector for λ = –1 is

1,

3

−

ªº

«»

¬¼

which may be normalized to

2

1/ 10 .

3/ 10

ª

º

−

=

«

»

«

»

¬

¼

u

Then

1

APDP

−

=

,

where

[]

12

3/ 10 1/ 10

1/ 10 3/ 10

P

ª

º

−

==

«

»

«

»

¬

¼

uu

and

90

.

01

D

ª

º

=

«

»

−

¬

¼

The desired change of variable is x =

Py, and the new quadratic form is

22

12

()() 9

TT TTT

APAP PAP D yy====−xx y y y yyy

424 CHAPTER 7 • Symmetric Matrices and Quadratic Forms

15. [M] The matrix of the quadratic form is

2222

2600

.

2093

2039

A

−

ª

º

«

»

−

«

»

=

«

»

−

«

»

−

«

»

¬

¼

The eigenvalues of A are 0, –6, –

8, and –12, so the quadratic form is negative semidefinite. The corresponding eigenvectors may be

computed:

301 0

121 0

0: , 6: , 8: , 12 :

111 1

−

ªº ª º ª º ª º

«» « » « » « »

−

«» « » « » « »

==−=−=−

«» « » « » « »

−

«» « » « » « »

¬¼ ¬ ¼ ¬ ¼ ¬ ¼

These eigenvectors may be normalized to form the columns of P, and

1

APDP

−

=

, where

3/ 12 0 1/2 0 000 0

1/ 12 2/ 6 1/ 2 0 0 6 0 0

and 008 0

1/ 12 1/ 6 1/ 2 1/ 2

00012

1/ 12 1/ 6 1/ 2 1/ 2

PD

ªº

−

ª

º

«»

«

»

−−

«»

«

»

==

«»

«

»

−

«»

«

»

«»

−

«

»

¬

¼

¬¼

The desired change of variable is x = Py, and the new quadratic form is

22 2

23 4

()() 6 8 12

TT TTT

APAP PAP D yy y====−− −xx y y y yyy

16. [M] The matrix of the quadratic form is

43/2 0 2

3/2 4 2 0 .

0243/2

203/24

A

−

ª

º

«

»

«

»

=

«

»

«

»

−

«

»

¬

¼

The eigenvalues of A are

13/2 and 3/2, so the quadratic form is positive definite. The corresponding eigenvectors may be

computed:

43 4 3

05 0 5

13/ 2 : , , 3/2: ,

34 3 4

50 5 0

½ ½−

ªºªº ªºªº

°°° °

«»«» «»«»

−

°°° °

«»«» «»«»

==

®¾® ¾

«»«» «»«»

−

°°° °

«»«» «»«»

°°° °

«»«» «»«»

¬¼¬¼ ¬¼¬¼

¯¿¯ ¿

Each set of eigenvectors above is already an orthogonal set, so they may be normalized to form the

columns of P, and

1

APDP

−

=

, where

3/ 50 4/ 50 3/ 50 4/ 50 13 / 2 0 0 0

5/ 50 0 5/ 50 0 0 13/2 0 0

and 003/20

4/ 50 3/ 50 4/ 50 3/ 50

0003/2

05/50 05/50

PD

ªº

−

ª

º

«»

«

»

−

«»

«

»

==

«»

«

»

−

«»

«

»

«»

«

»

¬

¼

¬¼

The desired change of variable is x = Py, and the new quadratic form is

2222

1234

13 13 3 3

()() 2222

TT TTT

APAP PAP D y y y y====+++xx y y y yyy

7.2 • Solutions 425

17. [M] The matrix of the quadratic form is

19/2 0 6

9/2 1 6 0 .

0619/2

609/21

A

−

ª

º

«

»

«

»

=

«

»

«

»

−

«

»

¬

¼

The eigenvalues of A are

17/2 and –13/2, so the quadratic form is indefinite. The corresponding eigenvectors may be

computed:

43 4 3

05 0 5

17 / 2 : , , 13/ 2 : ,

34 3 4

50 5 0

½  ½−

ªºªº ªºªº

°° ° °

«»«» «»«»

−

°° ° °

«»«» «»«»

==−

®¾ ® ¾

«»«» «»«»

−

°° ° °

«»«» «»«»

°° ° °

«»«» «»«»

¬¼¬¼ ¬¼¬¼

¯¿ ¯ ¿

Each set of eigenvectors above is already an orthogonal set, so they may be normalized to form the

columns of P, and

1

APDP

−

=

, where

3/ 50 4/ 50 3/ 50 4/ 50 17 / 2 0 0 0

5/ 50 0 5/ 50 0 0 17/2 0 0

and 0013/2 0

4/ 50 3/ 50 4/ 50 3/ 50

00 013/2

05/50 05/50

PD

ªº

−

ª

º

«»

«

»

−

«»

«

»

==

«»

«

»

−

«»

«

»

«»

−

«

»

¬

¼

¬¼

The desired change of variable is x = Py, and the new quadratic form is

2222

1234

17 17 13 13

()() 2222

TT TTT

APAP PAP D y y y y====+−−xx y y y yyy

18. [M] The matrix of the quadratic form is

11 6 6 6

6100

.

6001

6010

A

−−−

ª

º

«

»

−−

«

»

=

«

»

−−

«

»

−−

«

»

¬

¼

The eigenvalues of A are 17, 1, –

1, and –7, so the quadratic form is indefinite. The corresponding eigenvectors may be computed:

30 0 1

10 2 1

17 : , 1: , 1: , 7:

11 11

−

ªº ªº ªº ªº

«» «» «» «»

−

«» «» «» «»

===−=−

«» «» «» «»

−

«» «» «» «»

¬¼ ¬¼ ¬¼ ¬¼

These eigenvectors may be normalized to form the columns of P, and

1

APDP

−

=

, where

3/ 12 0 0 1/2 17 0 0 0

1/ 12 0 2/ 6 1/ 2 0 1 0 0

and 00 1 0

1/ 12 1/ 2 1/ 6 1/2

00 0 7

1/ 12 1/ 2 1/ 6 1/2

PD

ªº

−

ª

º

«»

«

»

−

«»

«

»

==

«»

«

»

−

«»

«

»

«»

−

¬

¼

¬¼

The desired change of variable is x = Py, and the new quadratic form is

222 2

123 4

()() 17 7

TT TTT

APAP PAP D yyy y====+−−xx y y y yyy

426 CHAPTER 7 • Symmetric Matrices and Quadratic Forms

19. Since 8 is larger than 5, the

2

x

term should be as large as possible. Since

22

12

1xx+=

, the largest

value that

2

x

can take is 1, and

10x=

when

21x=

. Thus the largest value the quadratic form can

take when

1

T

=

xx

is 5(0) + 8(1) = 8.

20. Since 5 is larger in absolute value than –3, the

2

1

x

term should be as large as possible. Since

22

12

1xx+=

, the largest value that

1

x

can take is 1, and

20x=

when

11x=

. Thus the largest value

the quadratic form can take when

1

T

=

xx

is 5(1) – 3(0) = 5.

21. a. True. See the definition before Example 1, even though a nonsymmetric matrix could be used to

compute values of a quadratic form.

b . True. See the paragraph following Example 3.

c . True. The columns of P in Theorem 4 are eigenvectors of A. See the Diagonalization Theorem in

Section 5.3.

d . False. Q(x) = 0 when x = 0.

e . True. See Theorem 5(a).

f. True. See the Numerical Note after Example 6.

22. a. True. See the paragraph before Example 1.

b . False. The matrix P must be orthogonal and make

T

PAP

diagonal. See the paragraph before

Example 4.

c . False. There are also “degenerate” cases: a single point, two intersecting lines, or no points at all.

See the subsection “A Geometric View of Principal Axes.”

d . False. See the definition before Theorem 5.

e . True. See Theorem 5(b). If

T

A

xx

has only negative values for x ≠ 0, then

T

A

xx

is negative

definite.

23. The characteristic polynomial of A may be written in two ways:

22

det( )det ()

ab

AI ad adb

bd

−

ªº

−==−++−

«»

−

¬¼

and

2

12 1212

()( ) ( )−−=−++

The coefficients in these polynomials may be equated to obtain

12

ad+=+

and

12

=

2

detad b A−=

.

24. If det A > 0, then by Exercise 23,

12 0>

, so that

1

and

2

have the same sign; also,

2

det 0ad A b=+>

.

a . If det A > 0 and a > 0, then d > 0 also, since ad > 0. By Exercise 23,

12 0ad+=+>

. Since

1

and

2

have the same sign, they are both positive. So Q is positive definite by Theorem 5.

b . If det A > 0 and a < 0, then d < 0 also, since ad > 0. By Exercise 23,

12 0ad+=+<

. Since

1

and

2

have the same sign, they are both negative. So Q is negative definite by Theorem 5.

c . If det A < 0, then by Exercise 23,

12 0<

. Thus

1

and

2

have opposite signs. So Q is

indefinite by Theorem 5.

7.3 • Solutions 427

25. Exercise 27 in Section 7.1 showed that

T

BB

is symmetric. Also

() || ||0

TT T

BB B B B==≥xxxxx

, so

the quadratic form is positive semidefinite, and the matrix

T

BB

is positive semidefinite. Suppose

that B is square and invertible. Then if

0,

TT

BB =xx

|| Bx || = 0 and Bx = 0. Since B is invertible, x =

0. Thus if x ≠ 0,

0

TT

BB >xx

and

T

BB

is positive definite.

26. Let

,

T

APDP=

where

1.

T

PP

−

=

The eigenvalues of A are all positive: denote them

1,,.

n

…

Let C

be the diagonal matrix with

1

,,

n

…

on its diagonal. Then

2T

DC CC==

. If

T

BPCP=

, then B

is positive definite because its eigenvalues are the positive numbers on the diagonal of C. Also

()()( )()

TTTTTTTTTTTT

BB PCP PCP P CP PCP PCCP PDP A== ===

since

.

T

PP I=

27. Since the eigenvalues of A and B are all positive, the quadratic forms

T

Axx

and

T

Bxx

are positive

definite by Theorem 5. Let x ≠ 0. Then

0

TA>xx

and

0

TB>xx

, so

() 0

TTT

AB A B+= + >xxxxxx

,

and the quadratic form

()

T

AB+xx

is positive definite. Note that A + B is also a symmetric matrix.

Thus by Theorem 5 all the eigenvalues of A + B must be positive.

28. The eigenvalues of A are all positive by Theorem 5. Since the eigenvalues of

1

A

−

are the reciprocals

of the eigenvalues of A (see Exercise 25 in Section 5.1), the eigenvalues of

1

A

−

are all positive. Note

that

1

A

−

is also a symmetric matrix. By Theorem 5, the quadratic form

1T

A

−

xx

is positive definite.

7.3 SOLUTIONS

Notes:

Theorem 6 is the main result needed in the next two sections. Theorem 7 is mentioned in Example

2 of Section 7.4. Theorem 8 is needed at the very end of Section 7.5. The economic principles in Example

6 may be familiar to students who have had a course in macroeconomics.

1. The matrix of the quadratic form on the left is

520

262.

027

A

ª

º

«

»

=−

«

»

«

»

−

¬

¼

The equality of the quadratic

forms implies that the eigenvalues of A are 9, 6, and 3. An eigenvector may be calculated for each

eigenvalue and normalized:

1/3 2/3 2/3

9: 2/3 , 6: 1/3 , 3: 2/3

2/3 1/3 1/3

−

ªº ªº ªº

«» «» «»

===

«» «» «»

−

¬¼ ¬¼ ¬¼

The desired change of variable is x = Py, where

1/3 2/3 2/3

2/3 1/3 2/3 .

2/3 2/3 1/3

P

−

ª

º

«

»

=

«

»

«

»

−

¬

¼

428 CHAPTER 7 • Symmetric Matrices and Quadratic Forms

2. The matrix of the quadratic form on the left is

311

122.

122

A

ª

º

«

»

=

«

»

«

»

¬

¼

The equality of the quadratic forms

implies that the eigenvalues of A are 5, 2, and 0. An eigenvector may be calculated for each

eigenvalue and normalized:

1/ 3 2/ 6 0

5: 1/ 3 , 2: 1/ 6 , 0: 1/ 2

1/ 3 1/ 6 1/ 2

ªº ª º

−

ª

º

«» « »

«

»

== =−

«» « »

«

»

«» « »

«

»

«» « »

¬

¼

¬¼ ¬ ¼

The desired change of variable is x = Py, where

1/ 3 2/ 6 0

1/ 3 1/ 6 1/ 2 .

1/ 3 1/ 6 1/ 2

P

ª

º

−

«

»

=−

«

»

«

»

«

»

¬

¼

3. (a) By Theorem 6, the maximum value of

T

Axx

subject to the constraint

1

T

=xx

is the greatest

eigenvalue

1

of A. By Exercise 1,

19.=

(b) By Theorem 6, the maximum value of

T

Axx

subject to the constraint

1

T

=xx

occurs at a unit

eigenvector u corresponding to the greatest eigenvalue

1

of A. By Exercise 1,

1/3

2/3 .

2/3

ªº

«»

=±«»

«»

−

¬¼

u

(c) By Theorem 7, the maximum value of

T

Axx

subject to the constraints

1

T

=xx

and

0

T=xu

is

the second greatest eigenvalue

2

of A. By Exercise 1,

26.=

4. (a) By Theorem 6, the maximum value of

T

Axx

subject to the constraint

1

T

=xx

is the greatest

eigenvalue

1

of A. By Exercise 2,

15.=

(b) By Theorem 6, the maximum value of

T

Axx

subject to the constraint

1

T

=xx

occurs at a unit

eigenvector u corresponding to the greatest eigenvalue

1

of A. By Exercise 2,

1/ 3

1/ 3 .

1/ 3

ªº

«»

=±«»

«»

¬¼

u

(c) By Theorem 7, the maximum value of

T

Axx

subject to the constraints

1

T

=xx

and

0

T=xu

is

the second greatest eigenvalue

2

of A. By Exercise 2,

22.=

5. The matrix of the quadratic form is

52

.

25

A

−

ª

º

=

«

»

−

¬

¼

The eigenvalues of A are

17=

and

23.=

(a) By Theorem 6, the maximum value of

T

Axx

subject to the constraint

1

T

=xx

is the greatest

eigenvalue

1

of A, which is 7.

7.3 • Solutions 429

(b) By Theorem 6, the maximum value of

T

Axx

subject to the constraint

1

T

=xx

occurs at a unit

eigenvector u corresponding to the greatest eigenvalue

1

of A. One may compute that

1

−

ªº

«»

¬¼

is

an eigenvector corresponding to

17,=

so

1/ 2 .

1/ 2

ª

º

−

=±

«

»

«

»

¬

¼

u

(c) By Theorem 7, the maximum value of

T

Axx

subject to the constraints

1

T

=xx

and

0

T=xu

is

the second greatest eigenvalue

2

of A, which is 3.

6. The matrix of the quadratic form is

73/2

.

3/2 3

A

ª

º

=

«

»

¬

¼

The eigenvalues of A are

115/ 2=

and

25/2.=

(a) By Theorem 6, the maximum value of

T

Axx

subject to the constraint

1

T

=xx

is the greatest

eigenvalue

1

of A, which is 15/2.

(b) By Theorem 6, the maximum value of

T

Axx

subject to the constraint

1

T

=xx

occurs at a unit

eigenvector u corresponding to the greatest eigenvalue

1

of A. One may compute that

3

1

ªº

«»

¬¼

is an

eigenvector corresponding to

115 / 2,=

so

3/ 10 .

1/ 10

ª

º

=±

«

»

«

»

¬

¼

u

(c) By Theorem 7, the maximum value of

T

Axx

subject to the constraints

1

T

=xx

and

0

T=xu

is

the second greatest eigenvalue

2

of A, which is 5/2.

7. The eigenvalues of the matrix of the quadratic form are

12,=

21,=−

and

34.=−

By Theorem

6, the maximum value of

T

Axx

subject to the constraint

1

T

=xx

occurs at a unit eigenvector u

corresponding to the greatest eigenvalue

1

of A. One may compute that

1/2

1

ª

º

«

»

«

»

«

»

¬

¼

is an eigenvector

corresponding to

12,=

so

1/3

2/3 .

2/3

ªº

«»

=±«»

«»

¬¼

u

8. The eigenvalues of the matrix of the quadratic form are

19,=

and

23.=−

By Theorem 6, the

maximum value of

T

Axx

subject to the constraint

1

T

=xx

occurs at a unit eigenvector u

corresponding to the greatest eigenvalue

1

of A. One may compute that

1

0

1

−

ª

º

«

»

«

»

«

»

¬

¼

and

2

1

0

−

ªº

«»

¬¼

are linearly

independent eigenvectors corresponding to

19,=

so u can be any unit vector which is a linear

combination of

1

0

1

−

ªº

«»

¬¼

and

2

1.

0

−

ªº

«»

¬¼

Alternatively, u can be any unit vector which is orthogonal to the

430 CHAPTER 7 • Symmetric Matrices and Quadratic Forms

eigenspace corresponding to the eigenvalue

23.=−

Since multiples of

1

2

1

ª

º

«

»

«

»

«

»

¬

¼

are eigenvectors

corresponding to

23,=−

u can be any unit vector orthogonal to

1

2.

1

ª

º

«

»

«

»

«

»

¬

¼

9. This is equivalent to finding the maximum value of

T

Axx

subject to the constraint

1.

T=xx

By

Theorem 6, this value is the greatest eigenvalue

1

of the matrix of the quadratic form. The matrix of

the quadratic form is

71

,

13

A

−

ªº

=«»

−

¬¼

and the eigenvalues of A are

1

55,=+

2

55.=−

Thus

the desired constrained maximum value is

1

55.=+

10. This is equivalent to finding the maximum value of

T

Axx

subject to the constraint

1

T

=xx

. By

Theorem 6, this value is the greatest eigenvalue

1

of the matrix of the quadratic form. The matrix of

the quadratic form is

31

,

15

A

−−

ªº

=«»

−

¬¼

and the eigenvalues of A are

1

117,=+

2

117.=−

Thus

the desired constrained maximum value is

1

117.=+

11. Since x is an eigenvector of A corresponding to the eigenvalue 3, Ax = 3x, and

(3 )

TT

A==xxx x

2

3( ) 3 || || 3

T

==xx x

since x is a unit vector.

12. Let x be a unit eigenvector for the eigenvalue λ. Then

( ) ()

TT T

A===xxx x xx

since

1

T

=xx

.

So λ must satisfy m ≤ λ ≤ M.

13. If m = M, then let t = (1 – 0)m + 0M = m and

.

n

=xu

Theorem 6 shows that

.

T

nn

Am=uu

Now

suppose that m < M, and let t be between m and M. Then 0 ≤ t – m ≤ M – m and 0 ≤ (t – m)/(M – m)

≤ 1. Let

α = (t – m)/(M – m), and let

1

1.

n

αα

=−+xuu

The vectors

1

n

α

−u

and

1

α

u

are orthogonal

because they are eigenvectors for different eigenvectors (or one of them is 0). By the Pythagorean

Theorem

22222

11

|| || || 1 || || || |1 ||| || | ||| || (1 ) 1

T

nn

αααα αα

==−+=−+=−+=xx x u u u u

since

n

u

and

1

u

are unit vectors and 0 ≤ α ≤ 1. Also, since

n

u

and

1

u

are orthogonal,

11

(1 ) (1 )

TT

nn

AA

αα αα

=−+−+xx u u u u

11

(1 )( 1 )

T

nn

mM

αα α α

=−+−+uu u u

11

|1 | | | (1 )

TT

nn

mM mMt

αα αα

=−+=−+=uu uu

Thus the quadratic form

T

Axx

assumes every value between m and M for a suitable unit vector x.

7.3 • Solutions 431

14. [M] The matrix of the quadratic form is

01/23/2 15

1/ 2 0 15 3/ 2 .

3/2 15 0 1/2

15 3/ 2 1/ 2 0

A

ª

º

«

»

«

»

=

«

»

«

»

«

»

¬

¼

The eigenvalues of A are

117,=

213,=

314,=−

and

416.=−

(a) By Theorem 6, the maximum value of

T

Axx

subject to the constraint

1

T

=xx

is the greatest

eigenvalue

1

of A, which is 17.

(b) By Theorem 6, the maximum value of

T

Axx

subject to the constraint

1

T

=xx

occurs at a unit

eigenvector u corresponding to the greatest eigenvalue

1

of A. One may compute that

1

ªº

«»

¬¼

is an

eigenvector corresponding to

117,=

so

1/2

1/2 .

1/2

ª

º

«

»

«

»

=±

«

»

«

»

«

»

¬

¼

u

(c) By Theorem 7, the maximum value of

T

Axx

subject to the constraints

1

T

=xx

and

0

T=xu

is

the second greatest eigenvalue

2

of A, which is 13.

15. [M] The matrix of the quadratic form is

03/25/27/2

3/2 0 7/2 5/2 .

5/2 7/2 0 3/2

7/2 5/2 3/2 0

A

ª

º

«

»

«

»

=

«

»

«

»

«

»

¬

¼

The eigenvalues of A are

115 / 2,=

21/ 2,=−

35/2,=−

and

49/2.=−

(a) By Theorem 6, the maximum value of

T

Axx

subject to the constraint

1

T

=xx

is the greatest

eigenvalue

1

of A, which is 15/2.

(b) By Theorem 6, the maximum value of

T

Axx

subject to the constraint

1

T

=xx

occurs at a unit

eigenvector u corresponding to the greatest eigenvalue

1

of A. One may compute that

1

ªº

«»

¬¼

is an

eigenvector corresponding to

115 / 2,=

so

1/2

1/2 .

1/2

ª

º

«

»

«

»

=±

«

»

«

»

«

»

¬

¼

u

(c) By Theorem 7, the maximum value of

T

Axx

subject to the constraints

1

T

=xx

and

0

T=xu

is

the second greatest eigenvalue

2

of A, which is –1/2.

432 CHAPTER 7 • Symmetric Matrices and Quadratic Forms

16. [M] The matrix of the quadratic form is

4355

3033

.

5301

5310

A

−−−

ª

º

«

»

−−−

«

»

=

«

»

−− −

«

»

−− −

«

»

¬

¼

The eigenvalues of A are

19,=

23,=

31,=

and

49.=−

(a) By Theorem 6, the maximum value of

T

Axx

subject to the constraint

1

T

=xx

is the greatest

eigenvalue

1

of A, which is 9.

(b) By Theorem 6, the maximum value of

T

Axx

subject to the constraint

1

T

=xx

occurs at a unit

eigenvector u corresponding to the greatest eigenvalue

1

of A. One may compute that

2

0

1

−

ªº

«»

¬¼

is

an eigenvector corresponding to

19,=

so

2/ 6

0.

1/ 6

ª

º

−

«

»

«

»

=±

«

»

«

»

«

»

¬

¼

u

(c) By Theorem 7, the maximum value of

T

Axx

subject to the constraints

1

T

=xx

and

0

T=xu

is the

second greatest eigenvalue

2

of A, which is 3.

17. [M] The matrix of the quadratic form is

6222

210 0 0

.

20133

20313

A

−−−−

ª

º

«

»

−−

«

»

=

«

»

−−

«

»

−−

«

»

¬

¼

The eigenvalues of A are

14,=−

210,=−

312,=−

and

416.=−

(a) By Theorem 6, the maximum value of

T

Axx

subject to the constraint

1

T

=xx

is the greatest

eigenvalue

1

of A, which is –4.

(b) By Theorem 6, the maximum value of

T

Axx

subject to the constraint

1

T

=xx

occurs at a unit

eigenvector u corresponding to the greatest eigenvalue

1

of A. One may compute that

3

1

−

ªº

«»

¬¼

is

an eigenvector corresponding to

14,=−

so

3/ 12

1/ 12 .

1/ 12

ª

º

−

«

»

«

»

=±

«

»

«

»

«

»

¬

¼

u

(c) By Theorem 7, the maximum value of

T

Axx

subject to the constraints

1

T

=xx

and

0

T=xu

is

the second greatest eigenvalue

2

of A, which is –10.

7.4 • Solutions 433

7.4 SOLUTIONS

Notes:

The section presents a modern topic of great importance in applications, particularly in computer

calculations. An understanding of the singular value decomposition is essential for advanced work in

science and engineering that requires matrix computations. Moreover, the singular value decomposition

explains much about the structure of matrix transformations. The SVD does for an arbitrary matrix almost

what an orthogonal decomposition does for a symmetric matrix.

1. Let

10

.

03

Aªº

=«»

−

¬¼

Then

10

,

09

T

AA ªº

=«»

¬¼

and the eigenvalues of

T

AA

are seen to be (in decreasing

order)

19=

and

21.=

Thus the singular values of A are

1

93

σ

==

and

2

11.

σ

==

2. Let

50

.

00

A

−

ªº

=«»

¬¼

Then

25 0 ,

00

T

AA ªº

=«»

¬¼

and the eigenvalues of

T

AA

are seen to be (in decreasing

order)

125=

and

20.=

Thus the singular values of A are

1

25 5

σ

==

and

2

00.

σ

==

3. Let

61

.

06

A

ªº

=«»

«»

¬¼

Then

66

,

67

T

AA

ªº

=«»

«»

¬¼

and the characteristic polynomial of

T

AA

is

2

13 36 ( 9)( 4),−+=−−

and the eigenvalues of

T

AA

are (in decreasing order)

19=

and

24.=

Thus the singular values of A are

1

93

σ

==

and

2

42.

σ

==

4. Let

32

.

03

A

ªº

=«»

«»

¬¼

Then

323

,

23 7

T

AA

ªº

=«»

«»

¬¼

and the characteristic polynomial of

T

AA

is

2

10 9( 9)( 1),−+= −−

and the eigenvalues of

T

AA

are (in decreasing order)

19=

and

21.=

Thus the singular values of A are

1

93

σ

==

and

2

11.

σ

==

5. Let

30

.

00

A

−

ªº

=«»

¬¼

Then

90

,

00

T

AA ªº

=«»

¬¼

and the eigenvalues of

T

AA

are seen to be (in decreasing

order)

19=

and

20.=

Associated unit eigenvectors may be computed:

10

9: , 0:

01

ªº ªº

==

«» «»

¬¼ ¬¼

Thus one choice for V is

10

.

01

Vªº

=«»

¬¼

The singular values of A are

1

93

σ

==

and

2

00.

σ

==

Thus the matrix Σ is

30

.

00

ªº

Σ=«»

¬¼

Next compute

11

1

0

A

σ

−

ªº

==

«»

¬¼

uv

Because Av

2

= 0, the only column found for U so far is u

1

. Find the other column of U is found by

extending {u

1

} to an orthonormal basis for

2

. An easy choice is u

2

=

0.

1

ª

º

«

»

¬

¼

434 CHAPTER 7 • Symmetric Matrices and Quadratic Forms

Let

10

.

01

U

−

ªº

=«»

¬¼

Thus

103010

010001

T

AU V

−

ªºªºªº

=Σ=«»«»«»

¬¼¬¼¬¼

6. Let

20

.

01

A

−

ªº

=«»

−

¬¼

Then

40

,

01

T

AA ªº

=«»

¬¼

and the eigenvalues of

T

AA

are seen to be (in decreasing

order)

14=

and

21.=

Associated unit eigenvectors may be computed:

10

4: , 1:

01

ªº ªº

==

«» «»

¬¼ ¬¼

Thus one choice for V is

10

.

01

Vªº

=«»

¬¼

The singular values of A are

1

42

σ

==

and

2

11.

σ

==

Thus the matrix Σ is

20

.

01

ªº

Σ=«»

¬¼

Next compute

11 2 2

12

10

11

,

01

AA

σσ

−

ªº ªº

== = =

«» «»

−

¬¼ ¬¼

uv u v

Since

12

{, }uu

is a basis for

2

, let

10

.

01

U

−

ª

º

=

«

»

−

¬

¼

Thus

102010

010101

T

AU V

−

ªºªºªº

=Σ=«»«»«»

−

¬¼¬¼¬¼

7. Let

21

.

22

A

−

ªº

=«»

¬¼

Then

82

,

25

T

AA ªº

=«»

¬¼

and the characteristic polynomial of

T

AA

is

2

13 36 ( 9)( 4),−+=−−

and the eigenvalues of

T

AA

are (in decreasing order)

19=

and

24.=

Associated unit eigenvectors may be computed:

2/ 5 1/ 5

9: , 4:

1/ 5 2/ 5

ªº ª º

−

==

«» « »

¬¼ ¬ ¼

Thus one choice for V is

2/ 5 1/ 5 .

1/ 5 2/ 5

V

ªº

−

=«»

«»

¬¼

The singular values of A are

1

93

σ

==

and

2

42.

σ

==

Thus the matrix Σ is

30

.

02

ª

º

Σ=

«

»

¬

¼

Next compute

11 2 2

12

1/ 5 2/ 5

11

,

2/ 5 1/ 5

AA

σσ

ªº ª º

−

== = =

«» « »

¬¼ ¬ ¼

uv u v

Since

12

{, }uu

is a basis for

2

, let

1/ 5 2/ 5 .

2/ 5 1/ 5

U

ª

º

−

=

«

»

«

»

¬

¼

Thus

7.4 • Solutions 435

1/ 5 2/ 5 3 0 2/ 5 1/ 5

02

2/ 5 1/ 5 1/ 5 2/ 5

T

AU V

ªºªº

−ªº

=Σ=«»«»

«»

−

¬¼

«»«»

¬¼¬¼

8. Let

23

.

02

Aªº

=«»

¬¼

Then

46

,

613

T

AA ªº

=«»

¬¼

and the characteristic polynomial of

T

AA

is

2

17 16 ( 16)( 1),−+=−−

and the eigenvalues of

T

AA

are (in decreasing order)

116=

and

21.=

Associated unit eigenvectors may be computed:

1/ 5 2/ 5

16 : , 1:

2/ 5 1/ 5

ªº ª º

−

==

«» « »

¬¼ ¬ ¼

Thus one choice for V is

1/ 5 2/ 5 .

2/ 5 1/ 5

V

ªº

−

=«»

«»

¬¼

The singular values of A are

1

16 4

σ

==

and

2

11.

σ

==

Thus the matrix Σ is

40

.

01

ª

º

Σ=

«

»

¬

¼

Next compute

11 2 2

12

2/ 5 1/ 5

11

,

1/ 5 2/ 5

AA

σσ

ªº ª º

−

== = =

«» « »

¬¼ ¬ ¼

uv u v

Since

12

{, }uu

is a basis for

2

, let

2/ 5 1/ 5 .

1/ 5 2/ 5

U

ª

º

−

=

«

»

«

»

¬

¼

Thus

2/ 5 1/ 5 4 0 1/ 5 2/ 5

01

1/ 5 2/ 5 2/ 5 1/ 5

T

AU V

ªºªº

−ªº

=Σ=«»«»

«»

−

¬¼

«»«»

¬¼¬¼

9. Let

71

00.

55

A

ªº

«»

=«»

«»

¬¼

Then 74 32 ,

32 26

T

AA

ªº

=«»

¬¼

and the characteristic polynomial of

T

AA

is

2

100 900 ( 90)( 10),−+=−−

and the eigenvalues of

T

AA

are (in decreasing order)

190=

and

210.=

Associated unit eigenvectors may be computed:

2/ 5 1/ 5

90 : , 10 :

1/ 5 2/ 5

ªº ª º

−

==

«» « »

¬¼ ¬ ¼

Thus one choice for V is

2/ 5 1/ 5 .

1/ 5 2/ 5

V

ªº

−

=«»

«»

¬¼

The singular values of A are

1

90 3 10

σ

==

and

2

10.

σ

=

Thus the matrix Σ is

310 0

010.

00

ªº

«»

Σ=«»

«»

¬¼

Next compute

436 CHAPTER 7 • Symmetric Matrices and Quadratic Forms

11 2 2

12

1/ 2 1/ 2

11

0, 0

1/ 2 1/ 2

AA

σσ

ªº ª º

−

«» « »

== = =

«» « »

¬¼ ¬ ¼

uv u v

Since

12

{, }uu

is not a basis for

3

, we need a unit vector

3

u

that is orthogonal to both

1

u

and

2.u

The vector

3

u

must satisfy the set of equations

1

0

T

=ux

and

2

0.

T

=ux

These are equivalent to the

linear equations

123

3

123

00

,so 1 ,and 1

00

xxx

ªº ªº

++ = «» «»

==

«» «»

−++ = «» «»

¬¼ ¬¼

xu

Therefore let

1/ 2 1/ 2 0

001

1/ 2 1/ 2 0

U

ªº

−

«»

=«»

«»

¬¼

. Thus

310 0

1/ 2 1/ 2 0 2/ 5 1/ 5

001010

1/ 5 2/ 5

00

1/ 2 1/ 2 0

T

AU V

ªº

−«»

ª

º

«»

=Σ=«»

«

»

«»

−

«»

«

»

¬

¼

«»

¬¼

10. Let

42

21.

00

A

−

ªº

«»

=−

«»

¬¼

Then

20 10 ,

10 5

T

AA

−

ªº

=«»

−

¬¼

and the characteristic polynomial of

T

AA

is

2

25 (25)−=−

, and the eigenvalues of

T

AA

are (in decreasing order)

125=

and

20.=

Associated unit eigenvectors may be computed:

2/ 5 1/ 5

25 : , 0:

1/ 5 2/ 5

ªº ªº

==

«» «»

−

«» «»

¬¼ ¬¼

Thus one choice for V is

2/ 5 1/ 5 .

1/ 5 2/ 5

V

ªº

=«»

−

«»

¬¼

The singular values of A are

1

25 5

σ

==

and

2

00.

σ

==

Thus the matrix Σ is

50

00.

00

ª

º

«

»

Σ=

«

»

«

»

¬

¼

Next compute

11

1

2/ 5

11/ 5

0

A

σ

ªº

«»

==

«»

¬¼

uv

Because Av

2

= 0, the only column found for U so far is u

1

. Find the other columns of U found by

extending {u

1

} to an orthonormal basis for

3

. In this case, we need two orthogonal unit vectors u

2

and u

3

that are orthogonal to u

1

. Each vector must satisfy the equation

1

0,

T

=ux

which is equivalent

to the equation 2x

1

+ x

2

= 0. An orthonormal basis for the solution set of this equation is

7.4 • Solutions 437

23

1/ 5 0

2/ 5 , 0 .

01

ªº

«»

=−=

«»

¬¼

«»

¬¼

uu

Therefore, let

2/ 5 1/ 5 0

1/ 5 2/ 5 0 .

001

U

ªº

«»

=−

«»

¬¼

Thus

2/ 5 1/ 5 0 502/ 5 1/ 5

1/ 5 2/ 5 0 0 0

1/ 5 2/ 5

00100

T

AU V

ªº

«»

ª

º

−

«»

=Σ=−

«»

«

»

«»

«

»

¬

¼

«»

¬¼

«»

¬¼

11. Let

31

62.

62

A

−

ªº

«»

=−

«»

−

¬¼

Then

81 27 ,

27 9

T

AA

−

ªº

=«»

−

¬¼

and the characteristic polynomial of

T

AA

is

2

90 (90),−=−

and the eigenvalues of

T

AA

are (in decreasing order)

190=

and

20.=

Associated unit eigenvectors may be computed:

3/ 10 1/ 10

90 : , 0: .

1/ 10 3/ 10

ªºªº

==

«»«»

−

«»«»

¬¼¬¼

Thus one choice for V is

3/ 10 1/ 10 .

1/ 10 3/ 10

V

ªº

=«»

−

«»

¬¼

The singular values of A are

1

90 3 10

σ

==

and

2

00.

σ

==

Thus the matrix Σ is

310 0

00.

00

ª

º

«

»

Σ=

«

»

«

»

¬

¼

Next compute

11

1

1/3

12/3

2/3

A

σ

−

ªº

«»

==

«»

¬¼

uv

Because Av

2

= 0, the only column found for U so far is u

1

. The other columns of U can be found by

extending {u

1

} to an orthonormal basis for

3

. In this case, we need two orthogonal unit vectors u

2

and u

3

that are orthogonal to u

1

. Each vector must satisfy the equation

1

0,

T

=ux

which is equivalent

to the equation

123

220.xx x−++=

An orthonormal basis for the solution set of this equation is

23

2/3 2/3

1/3 , 2/3 .

2/3 1/3

ªº ªº

«» «»

=−=

«» «»

−

¬¼ ¬¼

uu

Therefore, let

1/3 2/3 2/3

2/3 1/3 2/3 .

2/3 2/3 1/3

U

−

ªº

«»

=−

«»

−

¬¼

Thus

438 CHAPTER 7 • Symmetric Matrices and Quadratic Forms

1/3 2/3 2/3 3 10 0 3/ 10 1/ 10

2/3 1/3 2/3 0 0

1/ 10 3/ 10

2/3 2/3 1/3 0 0

T

AU V

ªº

−

ªº

ª

º

−

«»

=Σ=−

«

»

«»

«

»

¬

¼

«»

−

¬¼

12. Let

11

01.

11

A

ªº

«»

=«»

«»

−

¬¼

Then

20

,

03

T

AA ªº

=«»

¬¼

and the eigenvalues of

T

AA

are seen to be (in decreasing

order)

13=

and

22.=

Associated unit eigenvectors may be computed:

01

3: , 2:

10

ªº ªº

==

«» «»

¬¼ ¬¼

Thus one choice for V is

01

.

10

Vªº

=«»

¬¼

The singular values of A are

1

3

σ

=

and

2

2.

σ

=

Thus the

matrix Σ is

30

02.

00

ªº

«»

Σ=«»

«»

¬¼

Next compute

11 2 2

12

1/ 3 1/ 2

11

1/ 3 , 0

1/ 3 1/ 2

AA

σσ

ªº

ª

º

«»

«

»

== = =

«»

«

»

«»

«

»

−

«»

¬

¼

¬¼

uv u v

Since

12

{, }uu

is not a basis for

3

, we need a unit vector

3

u

that is orthogonal to both

1

u

and

2.u

The vector

3

u

must satisfy the set of equations

1

0

T

=ux

and

2

0.

T

=ux

These are equivalent to the

linear equations

123

3

123

1/ 6

1

0,so 2 ,and 2/ 6

00

11/ 6

xx x

xxx

ª

º

ªº

«

»

++ = «»

=−=−

«

»

«»

+−=

«

»

«»

¬¼

«

»

¬

¼

xu

. Therefore let

1/ 3 1/ 2 1/ 6

1/ 3 0 2/ 6 .

1/ 3 1/ 2 1/ 6

U

ª

º

«

»

=−

«

»

«

»

−

«

»

¬

¼

Thus

1/ 3 1/ 2 1/ 6 3 0

01

1/ 3 0 2/ 6 0 2 10

00

1/ 3 1/ 2 1/ 6

T

AU V

ªºªº

«»«»

ª

º

=Σ=−

«»«»

«

»

¬

¼«»«»

−

«»«»

¬¼¬¼

13. Let

32 2

.

23 2

Aªº

=«»

−

¬¼

Then

32

23,

22

T

A

ªº

«»

=«»

«»

−

¬¼

17 8 ,

817

TT T T

AA AA

ª

º

==

«

»

¬

¼ and the eigenvalues of

TT T

AA

are seen to be (in decreasing order)

125=

and

29.=

Associated unit eigenvectors may

be computed:

7.4 • Solutions 439

1/ 2 1/ 2

25: , 9:

1/ 2 1/ 2

ªº ª º

−

==

«» « »

¬¼ ¬ ¼

Thus one choice for V is

1/ 2 1/ 2 .

1/ 2 1/ 2

V

ªº

−

=«»

«»

¬¼

The singular values of

T

A

are

1

25 5

σ

==

and

2

93.

σ

==

Thus the matrix Σ is

50

03.

00

ª

º

«

»

Σ=

«

»

«

»

¬

¼

Next compute

11 2 2

12

1/ 2 1/ 18

11

1/ 2 , 1/ 18

04/ 18

TT

AA

σσ

ª

ºªº

−

«

»«»

== = =

«

»«»

«

»«»

−

«

»«»

¬

¼¬¼

uv u v

Since

12

{, }uu

is not a basis for

3

, we need a unit vector

3

u

that is orthogonal to both

1

u

and

2.u

The vector

3

u

must satisfy the set of equations

1

0

T

=ux

and

2

0.

T

=ux

These are equivalent to the

linear equations

12 3

3

12 3

22/3

00,so 2 ,and 2/3

4011/3

xx x

−−

ªº ª º

++ =«» « »

==

«» « »

−+−=«» « »

¬¼ ¬ ¼

xu

Therefore let

1/ 2 1/ 18 2/3

1/ 2 1/ 18 2/3 .

04/18 1/3

U

ªº

−−

«»

=«»

«»

−

«»

¬¼

Thus

1/ 2 1/ 18 2/3 501/ 2 1/ 2

1/ 2 1/ 18 2/3 0 3 1/ 2 1/ 2

00

04/181/3

TT

AUV

ªº

−−

ªº

«»

ª

º

«»

=Σ=«»

«

»

«»

−

«»

«

»

¬

¼

«»

−¬¼

«»

¬¼

An SVD for A is computed by taking transposes:

1/ 2 1/ 2 0

1/ 2 1/ 2 5 0 0 1/ 18 1/ 18 4/ 18

030

1/ 2 1/ 2 2/3 2/3 1/3

A

ª

º

«

»

ªº

−ªº

=−−

«

»

«»

¬¼

«

»

«»

¬¼ −

«

»

¬

¼

14. From Exercise 7,

T

AUV=Σ

with

2/ 5 1/ 5 .

1/ 5 2/ 5

V

ª

º

−

=

«

»

«

»

¬

¼

Since the first column of V is unit

eigenvector associated with the greatest eigenvalue

1

of

,

T

AA

so the first column of V is a unit

vector at which || Ax || is maximized.

15. a. Since A has 2 nonzero singular values, rank A = 2.

440 CHAPTER 7 • Symmetric Matrices and Quadratic Forms

b . By Example 6,

12

.40 .78

{, } .37, .33

.84 .52

½−

ªºªº

°°

«»«»

=−

®¾

«»«»

°°

«»«»

−−

¬¼¬¼

¯¿

uu

is a basis for Col A and

3

.58

{} .58

.58

½

ªº

°°

«»

=−

®¾

«»

°°

«»

¬¼

¯¿

v

is a basis

for Nul A.

16. a. Since A has 2 nonzero singular values, rank A = 2.

b . By Example 6,

12

.86 .11

{, } .31, .68

.41 .73

½−−

ªºªº

°°

«»«»

=®¾

«»«»

°°

«»«»

−

¬¼¬¼

¯¿

uu

is a basis for Col A and

34

.65 .34

.08 .42

{, } ,

.16 .84

.73 .08

½−

ªºªº

°°

«»«»

°°

«»«»

=®¾

«»«»

−−

°°

«»«»

°°

−−

«»«»

¬¼¬¼

¯¿

vv

is a basis for Nul A.

17. Let

1

.

T

AUV UV

−

=Σ=Σ

Since A is square and invertible, rank A = n, and all of the entries on the

diagonal of Σ must be nonzero. So

111111

() .

T

AUV VUVU

−−−−−−

=Σ=Σ=Σ

18. First note that the determinant of an orthogonal matrix is ±1, because

1det det

T

IUU== =

2

(det )(det ) (det ) .

T

UU U=

Suppose that A is square and

.

T

AUV=Σ

Then Σ is square, and

1

det (det )(det )(det ) det

T

n

AU V

σσ

=Σ=± Σ=± …

.

19. Since U and V are orthogonal matrices,

1

() () ()

TTTTTTTTTT

AA U V U V V UU V V V V V

−

=ΣΣ=ΣΣ=ΣΣ =ΣΣ

If

1,,

r

σσ

…

are the diagonal entries in Σ, then

T

ΣΣ

is a diagonal matrix with diagonal entries

22

1

,,

r

σσ

…

and possibly some zeros. Thus V diagonalizes

T

AA

and the columns of V are

eigenvectors of

T

AA

by the Diagonalization Theorem in Section 5.3. Likewise

1

() () ()

TTTTTTT TT T

AA U V U V U V V U U U U U

−

=ΣΣ=ΣΣ =ΣΣ =ΣΣ

so U diagonalizes

T

AA

and the columns of U must be eigenvectors of

T

AA

. Moreover, the

Diagonalization Theorem states that

22

1

,,

r

σσ

…

are the nonzero eigenvalues of

T

AA

. Hence

1,,

r

σσ

…

are the nonzero singular values of A.

20. If A is positive definite, then

T

APDP=

, where P is an orthogonal matrix and D is a diagonal matrix.

The diagonal entries of D are positive because they are the eigenvalues of a positive definite matrix.

Since P is an orthogonal matrix,

T

PP I=

and the square matrix

T

P

is invertible. Moreover,

111

() ( ) (),

TTT

PPPP

−−−

===

so

T

P

is an orthogonal matrix. Thus the factorization

T

APDP=

has

the properties that make it a singular value decomposition.

21. Let

.

T

AUV=Σ

The matrix PU is orthogonal, because P and U are both orthogonal. (See Exercise

29 in Section 6.2). So the equation

()

T

PA PU V=Σ

has the form required for a singular value

decomposition. By Exercise 19, the diagonal entries in Σ are the singular values of PA.

7.4 • Solutions 441

22. The right singular vector

1

v

is an eigenvector for the largest eigenvector

1

of

.

T

AA

By Theorem 7

in Section 7.3, the second largest eigenvalue

2

is the maximum of

()

TT

AAxx

over all unit vectors

orthogonal to

1

v

. Since

2

()||||,

TT

AA A=xxx

the square root of

2,

which is the second largest

singular value of A, is the maximum of || Ax || over all unit vectors orthogonal to

1.v

23. From the proof of Theorem 10,

[]

11

.

rr

U

σσ

Σ=… …uu00

The column-row expansion

of the product

()

T

UVΣ

shows that

1

111

() ()

T

TT

T

rrr

T

n

AUV U

σσ

ªº

«»

=Σ=Σ=+…+

«»

¬¼

v

uv uv

v

#

where r is the rank of A.

24. From Exercise 23,

111

.

TT T

rrr

A

σσ

=+…+vu vu

Then since

0for ,

1for

T

ij

≠



=®=

¯

uu

111

()()()

TT T T T

jrrrjjjjjjjjjjj

A

σσ σσ σ

=+…+ = = =uvu vuu vuuvuu v

25. Consider the SVD for the standard matrix A of T, say

T

AUV=Σ

. Let

1

{, , }

n

B=…vv

and

1

{, , }

m

C=…uu

be bases for

n

and

m

constructed respectively from the columns of V and U. Since

the columns of V are orthogonal,

T

jj

V=ve

, where

j

e

is the jth column of the n × n identity matrix.

To find the matrix of T relative to B and C, compute

()

T

jj j jjjjjjj

TAUVUU U

σσ σ

==Σ=Σ===vv v e e eu

so

[( )]

jC jj

T

σ

=ve

. Formula (4) in the discussion at the beginning of Section 5.4 shows that the

“diagonal” matrix Σ is the matrix of T relative to B and C.

26. [M] Let

18 13 4 4

219 412

.

14 11 12 8

221 4 8

A

−−

ªº

«»

−

«»

=«»

−−

«»

−

«»

¬¼

Then

528 392 224 176

392 1092 176 536 ,

224 176 192 128

176 536 128 288

T

AA

−−

ª

º

«

»

−−

«

»

=

«

»

−−

«

»

−−

«

»

¬

¼

and the

eigenvalues of

T

AA

are found to be (in decreasing order)

11600,=

2400,=

3100,=

and

40.=

Associated unit eigenvectors may be computed:

1234

.4 .8 .4 .2

.8 .4 .2 .4

:,:,:,:

.2 .4 .8 .4

.4 .2 .4 .8

−−

ªº ªº ªº ªº

«» «» «» «»

−−

«» «» «» «»

−−

«» «» «» «»

¬¼ ¬¼ ¬¼ ¬¼

442 CHAPTER 7 • Symmetric Matrices and Quadratic Forms

Thus one choice for V is

.4 .8 .4 .2

.8 .4 .2 .4 .

.2 .4 .8 .4

.4 .2 .4 .8

V

−−

ªº

«»

−−

«»

=«»

−−

«»

¬¼

The singular values of A are

140,

σ

=

120,

σ

=

310,

σ

=

and

40.

σ

=

Thus the matrix Σ is

40 0 0 0

020 00

.

00100

0000

ª

º

«

»

«

»

Σ=

«

»

«

»

«

»

¬

¼

Next compute

11 2 2

12

.5 .5

11

,,

.5 .5

AA

σσ

−

ªº ª º

«» « »

== = =

«» « »

−

«» « »

¬¼ ¬ ¼

uv u v

33

3

.5

1

.5

A

σ

−

ª

º

«

»

«

»

==

«

»

«

»

−

«

»

¬

¼

uv

Because Av

4

= 0, only three columns of U have been found so far. The last column of U can be found

by extending {u

1

, u

2

, u

3

} to an orthonormal basis for

4

. The vector u

4

must satisfy the set of

equations

1

0,

T

=ux

2

0,

T

=ux

and

3

0.

T

=ux

These are equivalent to the linear equations

1234

1234 4

1234

1.5

01.5

0,so ,and .

1.5

01.5

xx xx

xx x x

xx xx

−−

ªº ª º

+++= «» « »

−−

«» « »

−+−+= = =

«» « »

−++−=«» « »

«» « »

¬¼ ¬ ¼

xu

Therefore, let

.5 .5 .5 .5

.5 .5 .5 .5 .

.5 .5 .5 .5

U

−−−

ªº

«»

−

«»

=«»

−

«»

−

«»

¬¼

Thus

.5 .5 .5 .5 40 0 0 0 .4 .8 .2 .4

.5 .5 .5 .5 0 20 0 0 .8 .4 .4 .2

.5 .5 .5 .5 0 0 10 0 .4 .2 .8 .4

.5 .5 .5 .5 0 0 0 0 .2 .4 .4 .8

T

AUV

−−− − −

ªºªºªº

«»«»«»

−

«»«»«»

=Σ=«»«»«»

−−−

«»«»«»

−−−

«»«»«»

¬¼¬¼¬¼

7.4 • Solutions 443

27. [M] Let

68454

27564

.

01822

12448

A

−− −

ªº

«»

−−

«»

=«»

−−

«»

−− −

«»

¬¼

Then

41 32 38 14 8

32 118 3 92 74

,

38 3 121 10 52

14 92 10 81 72

8745272100

T

AA

−− −

ª

º

«

»

−−−

«

»

«

»

=−− −

«

»

−−

«

»

«

»

−−−

¬

¼

and the

eigenvalues of

T

AA

are found to be (in decreasing order)

1270.87,=

2147.85,=

323.73,=

418.55,=

and

50.=

Associated unit eigenvectors may be computed:

12345

.10 .39 .74 .41 .36

.61 .29 .27 .50 .48

:,:,:,:,:

.21 .84 .07 .45 .19

.52 .14 .38 .23 .72

.55 .19 .49 .58 .29

−−− −

ªº ªºªº ªº ªº

«» «»«» «» «»

−−−

«» «»«» «» «»

−− −

«» «»«» «» «»

−− −−

«» «»«» «» «»

−−

¬¼ ¬¼¬¼ ¬¼ ¬¼

Thus one choice for V is

.10 .39 .74 .41 .36

.61 .29 .27 .50 .48

.

.21 .84 .07 .45 .19

.52 .14 .38 .23 .72

.55 .19 .49 .58 .29

V

−−− −

ªº

«»

−−−

«»

=−− −

«»

−− −−

«»

−−

¬¼

The nonzero singular values of A

are

116.46,

σ

=

112.16,

σ

=

34.87,

σ

=

and

44.31.

σ

=

Thus the matrix Σ is

16.46 0 0 0 0

012.16 0 00

.

004.8700

0004.310

ªº

«»

Σ=«»

«»

¬¼

Next compute

11 2 2

12

.57 .65

.63 .24

11

,,

.07 .63

.51 .34

AA

σσ

−−

ªº ªº

«» «»

−

«» «»

== = =

«» «»

−

«» «»

−

«» «»

¬¼ ¬¼

uv u v

33 4 4

34

.42 .27

.68 .29

11

,

.53 .56

.29 .73

AA

σσ

−

ª

ºªº

«

»«»

−−

«

»«»

== = =

«

»«»

−

«

»«»

−−

«

»«»

¬

¼¬¼

uv uv

Since

1234

{, , , }uu uu

is a basis for

4

, let

.57 .65 .42 .27

.63 .24 .68 .29 .

.07 .63 .53 .56

.51 .34 .29 .73

U

−−−

ª

º

«

»

−−−

«

»

=

«

»

−−

«

»

−−−

«

»

¬

¼

Thus

T

AUV=Σ

444 CHAPTER 7 • Symmetric Matrices and Quadratic Forms

.10 .61 .21 .52 .55

.57 .65 .42 .27 16.46 0 0 0 0 .39 .29 .84 .14 .19

.63 .24 .68 .29 0 12.16 0 0 0

=.74 .27 .07 .38 .49

.07 .63 .53 .56 0 0 4.87 0 0 .41 .50 .45 .23 .58

.51 .34 .29 .73 0 0 0 4.31 0 .36 .4

−−−

ªºª º

−−−

«»« »

−−−

«»« »

−−−

«»« »

−−

«»« »

−−−

«»« »

¬¼¬ ¼

−−8.19.72.29

ª

º

«

»

«

»

«

»

«

»

«

»

«

»

−−−

¬

¼

28. [M] Let

4037

6999

.

751019

124 1

A

−−

ªº

«»

−

«»

=«»

−

«»

−−

¬¼

Then

102 91 0 108

91 110 39 16 ,

039206246

108 16 246 492

T

AA

−

ª

º

«

»

−−

«

»

=

«

»

«

»

−

¬

¼

and the eigenvalues of

T

AA

are found to be (in decreasing order)

1649.9059,=

2218.0033,=

339.6345,=

and

42.4564.=

The singular values of A are thus

125.4933,

σ

=

214.7649,

σ

=

36.2956,

σ

=

and

41.5673.

σ

=

The condition number

14

/

6.266.

σσ

≈

29. [M] Let

531 7 9

642 8 8

.

75310 9

964 9 5

852 11 4

A

ªº

«»

−

«»

=

«»

−−

«»

¬¼

Then

255 168 90 160 47

168 111 60 104 30

,

90 60 34 39 8

160 104 39 415 178

47 30 8 178 267

T

AA

ª

º

«

»

«

»

«

»

=

«

»

«

»

«

»

¬

¼

and the

eigenvalues of

T

AA

are found to be (in decreasing order)

1672.589,=

2280.745,=

3127.503,=

41.163,=

and

7

5

1.428 10 .

−

=×

The singular values of A are thus

125.9343,

σ

=

216.7554,

σ

=

311.2917,

σ

=

41.07853,

σ

=

and

5.000377928.

σ

=

The condition number

15

/

68,622.

σσ

=

7.5 SOLUTIONS

Notes:

The application presented here has turned out to be of interest to a wide variety of students,

including engineers. I cover this in Course Syllabus 3 described in the front mater of the text, but I only

have time to mention the idea briefly to my other classes.

1. The matrix of observations is

19 22 6 3 2 20

12 6 9 15 13 5

Xªº

=«»

¬¼

and the sample mean is

72 12

1.

60 10

6

Mªºªº

==

«»«»

¬¼¬¼

The mean-deviation form B is obtained by subtracting M from each column of X,

so

710 6 9 10 8

.

2415 35

B

−−−

ªº

=«»

−− −

¬¼

The sample covariance matrix is

430 135 86 27

11

135 80 27 16

61 5

T

SBB

−−

ªºªº

== =

«»«»

−−

−¬¼¬¼

7.5 • Solutions 445

2. The matrix of observations is

1526 73

3116 81511

Xªº

=«»

¬¼

and the sample mean is

24 4

1.

54 9

6

Mªºªº

==

«»«»

¬¼¬¼

The mean-deviation form B is obtained by subtracting M from each column of X,

so

31 2 23 1

.

62 3 16 2

B

−− −

ªº

=«»

−−−

¬¼

The sample covariance matrix is

28 40 5.6 8

11

40 90 8 18

61 5

T

SBB

ª

ºª º

== =

«

»« »

−

¬

¼¬ ¼

3. The principal components of the data are the unit eigenvectors of the sample covariance matrix S.

One computes that (in descending order) the eigenvalues of

86 27

27 16

S

−

ª

º

=

«

»

−

¬

¼

are

195.2041=

and

26.79593.=

One further computes that corresponding eigenvectors are

1

2.93348

1

−

ªº

=«»

¬¼

v

and

2

.340892 .

1

ªº

=«»

¬¼

v

These vectors may be normalized to find the principal components, which are

1

.946515

.322659

ªº

=«»

−

¬¼

u

for

195.2041=

and

2

.322659

.946515

ª

º

=

«

»

¬

¼

u

for

26.79593.=

4. The principal components of the data are the unit eigenvectors of the sample covariance matrix S.

One computes that (in descending order) the eigenvalues of

5.6 8

818

S

ª

º

=

«

»

¬

¼

are

121.9213=

and

21.67874.=

One further computes that corresponding eigenvectors are

1

.490158

1

ªº

=«»

¬¼

v

and

2

2.04016 .

1

−

ªº

=«»

¬¼

v

These vectors may be normalized to find the principal components, which are

1

.44013

.897934

ªº

=«»

¬¼

u

for

121.9213=

and

2

.897934

.44013

−

ª

º

=

«

»

¬

¼

u

for

21.67874.=

5. [M] The largest eigenvalue of

164.12 32.73 81.04

32.73 539.44 249.13

81.04 249.13 189.11

S

ªº

«»

=«»

«»

¬¼

is

1677.497,=

and the first

principal component of the data is the unit eigenvector corresponding to

1

, which is

1

.129554

.874423

.467547

ªº

«»

=«»

«»

¬¼

u. The fraction of the total variance that is contained in this component is

1/ tr( ) 677.497 / (164.12 539.44 189.11) .758956S=++=

so 75.8956% of the variance of the data is

contained in the first principal component.

446 CHAPTER 7 • Symmetric Matrices and Quadratic Forms

6. [M] The largest eigenvalue of

29.64 18.38 5.00

18.38 20.82 14.06

5.00 14.06 29.21

S

ªº

«»

=«»

«»

¬¼

is

151.6957,=

and the first principal

component of the data is the unit eigenvector corresponding to

1,

which is

1

.615525

.599424 .

.511683

ªº

«»

=«»

«»

¬¼

u Thus

one choice for the new variable is

1123

.615525 .599424 .511683 .yxxx=+ +

The fraction of the total

variance that is contained in this component is

1/tr( ) 51.6957/(29.64 20.82 29.21) .648872,S=++=

so 64.8872% of the variance of the data is

explained by

1.y

7. Since the unit eigenvector corresponding to

195.2041=

is

1

.946515 ,

.322659

ª

º

=

«

»

−

¬

¼

u

one choice for the

new variable is

112

.946515 .322659 .yxx=−

The fraction of the total variance that is contained in this

component is

1/tr( ) 95.2041/(86 16) .933374,S=+=

so 93.3374% of the variance of the data is

explained by

1.y

8. Since the unit eigenvector corresponding to

121.9213=

is

1

.44013 ,

.897934

ª

º

=

«

»

¬

¼

u

one choice for the new

variable is

11 2

.44013 .897934 .yx x=+

The fraction of the total variance that is contained in this

component is

1/tr( ) 21.9213/(5.6 18) .928869,S=+=

so 92.8869% of the variance of the data is

explained by

1.y

9. The largest eigenvalue of

520

262

027

S

ªº

«»

=«»

«»

¬¼

is

19,=

and the first principal component of the data is

the unit eigenvector corresponding to

1,

which is

1

1/3

2/3 .

2/3

ª

º

«

»

=

«

»

«

»

¬

¼

u Thus one choice for y is

123

(1 / 3) ( 2 / 3) ( 2 / 3) ,yx x x=+ +

and the variance of y is

19.=

10. [M] The largest eigenvalue of

542

4114

245

S

ªº

«»

=«»

«»

¬¼

is

115,=

and the first principal component of the

data is the unit eigenvector corresponding to

1,

which is

1

1/ 6

2/ 6 .

1/ 6

ª

º

«

»

=

«

»

«

»

«

»

¬

¼

u

Thus one choice for y is

123

(1 / 6 ) ( 2 / 6 ) (1 / 6 ) ,yx x x=+ +

and the variance of y is

115.=

11. a. If w is the vector in

N

with a 1 in each position, then

[]

11NN

…=+…+=XXwXX0

since

the

k

X

are in mean-deviation form. Then

Chapter 7 • Supplementary Exercises 447

[] [ ]

11 1

TTT T

NNN

PPP P

ªº

…= … = … ==

¬¼

YYwX XwX Xw00

Thus

1,

N

+…+ =YY0

and the

k

Y

are in mean-deviation form.

b. By part a., the covariance matrix

SY

of

1,,

N

…YY

is

[][]

11

1

T

NN

SN

=… …

−

Y

YYYY

[][]

11

1()

1

TTT

NN

PP

N

=… …

−

XXXX

[][]

11

1

T

TT

NN

PPPSP

N

§·

=……=

¨¸

−

©¹

XXXX

since the

k

X

are in mean-deviation form.

12. By Exercise 11, the change of variables X = PY changes the covariance matrix S of X into the

covariance matrix

T

PSP

of Y. The total variance of the data as described by Y is

tr( ).

T

PSP

However, since

T

PSP

is similar to S, they have the same trace (by Exercise 25 in Section 5.4). Thus

the total variance of the data is unchanged by the change of variables X = PY.

13. Let M be the sample mean for the data, and let

ˆ.

kk

=−XXM

Let

1

ˆˆ

N

B

ª

º

=…

¬

¼

XX

be the

matrix of observations in mean-deviation form. By the row-column expansion of

,

T

BB

the sample

covariance matrix is

1

T

SBB

N

=

−

1

ˆ

1ˆˆ

1ˆ

T

N

T

N

ªº

«»

ªº

=…

«»

¬¼

−«»

¬¼

X

XX

X

#

1

1ˆˆ

1

NN

TT

kk k k

kk

NN

==1

1

==(−)( −)

−−1

¦¦

XX X M X M

Chapter 7 SUPPLEMENTARY EXERCISES

1. a. True. This is just part of Theorem 2 in Section 7.1. The proof appears just before the statement

of the theorem.

b. False. A counterexample is

01

.

10

A

−

ªº

=«»

¬¼

c. True. This is proved in the first part of the proof of Theorem 6 in Section 7.3. It is also a

consequence of Theorem 7 in Section 6.2.

d. False. The principal axes of

T

Axx

are the columns of any orthogonal matrix P that

diagonalizes A. Note: When A has an eigenvalue whose eigenspace has dimension greater than

1, the principal axes are not uniquely determined.

448 CHAPTER 7 • Symmetric Matrices and Quadratic Forms

e. False. A counterexample is

11

.

11

P

−

ªº

=«»

¬¼

The columns here are orthogonal but not

orthonormal.

f. False. See Example 6 in Section 7.2.

g. False. A counterexample is

20

03

Aªº

=«»

−

¬¼

and

1.

0

ª

º

=

«

»

¬

¼

x

Then

20

TA=>xx

, but

T

Axx

is an

indefinite quadratic form.

h. True. This is basically the Principal Axes Theorem from Section 7.2. Any quadratic form can be

written as

T

Axx

for some symmetric matrix A.

i. False. See Example 3 in Section 7.3.

j. False. The maximum value must be computed over the set of unit vectors. Without a restriction

on the norm of x, the values of

T

Axx

can be made as large as desired.

k. False. Any orthogonal change of variable x = Py changes a positive definite quadratic form into

another positive definite quadratic form. Proof: By Theorem 5 of Section 7.2., the classification

of a quadratic form is determined by the eigenvalues of the matrix of the form. Given a form

,

T

Axx

the matrix of the new quadratic form is

1

,PAP

−

which is similar to A and thus has the

same eigenvalues as A.

l. False. The term “definite eigenvalue” is undefined and therefore meaningless.

m. True. If x = Py, then

1

()()

TT TTT

APAP PAP PAP

−

===xx y y y yy y

.

n. False. A counterexample is

11

.

11

U

−

ª

º

=

«

»

−

¬

¼

The columns of U must be orthonormal to make

T

UU x

the orthogonal projection of x onto Col U.

o. True. This follows from the discussion in Example 2 of Section 7.4., which refers to a proof

given in Example 1.

p. True. Theorem 10 in Section 7.4 writes the decomposition in the form

,

T

UVΣ

where U and V

are orthogonal matrices. In this case,

T

V

is also an orthogonal matrix. Proof: Since V is

orthogonal, V is invertible and

1.

T

VV

−=

Then

11

() ( ) (),

TTTT

VVV

−−

==

and since V is square

and invertible,

T

V

is an orthogonal matrix.

q. False. A counterexample is

20

.

01

A

ª

º

=

«

»

¬

¼

The singular values of A are 2 and 1, but the singular

values of

T

AA

are 4 and 1.

2. a. Each term in the expansion of A is symmetric by Exercise 35 in Section 7.1. The fact that

()

TT T

BC B C+=+

implies that any sum of symmetric matrices is symmetric, so A is

symmetric.

b . Since

11

1

T

=uu

and

1

0

T

j

=uu

for j ≠ 1,

11111 11111 111

()()() ()

TTTT

nnn nn n

A=+…+ =+…+ =uuuu uuuuuu uuuu

Since

1≠u0

,

1

is an eigenvalue of A. A similar argument shows that

j

is an eigenvalue of A

for j = 2, …, n.

Chapter 7 • Supplementary Exercises 449

3. If rank A = r, then dim

Nul A = n – r by the Rank Theorem. So 0 is an eigenvalue of A with

multiplicity n – r, and of the n terms in the spectral decomposition of A exactly n – r are zero. The

remaining r terms (which correspond to nonzero eigenvalues) are all rank 1 matrices, as mentioned

in the discussion of the spectral decomposition.

4. a. By Theorem 3 in Section 6.1,

(Col ) Nul Nul

T

AAA

⊥

==

since

.

T

AA=

b . Let y be in

n

. By the Orthogonal Decomposition Theorem in Section 6.3, y =

ˆ

y

+ z, where

ˆ

y

is

in Col A and z is in

(Col ) .A

⊥

By part a., z is in Nul A.

5. If Av = λv for some nonzero λ, then

11

(),AA

−−

==vv v

which shows that v is a linear

combination of the columns of A.

6. Because A is symmetric, there is an orthonormal eigenvector basis

1

{, , }

n

…uu

for

n

. Let r = rank A.

If r = 0, then A = O and the decomposition of Exercise 4(b) is y = 0 + y for each y in

n

; if r = n then

the decomposition is y = y + 0 for each y in

n

.

Assume that 0 < r < n. Then dim

Nul A = n – r by the Rank Theorem, and so 0 is an eigenvalue of A

with multiplicity n – r. Hence there are r nonzero eigenvalues, counted according to their

multiplicities. Renumber the eigenvector basis if necessary so that

1,,

r

…uu

are the eigenvectors

corresponding to the nonzero eigenvalues. By Exercise 5,

1,,

r

…uu

are in Col A. Also,

1,,

rn+…uu

are in Nul A because these vectors are eigenvectors corresponding to the eigenvalue 0. For y in

n

,

there are scalars

1,,

n

cc…

such that

11 1 1

ˆ

rr r r nn

ccc c

++

=+…+ + +…+

Z

yu u u u

y

   

This provides the decomposition in Exercise 4(b).

7. If

T

ARR=

and R is invertible, then A is positive definite by Exercise 25 in Section 7.2.

Conversely, suppose that A is positive definite. Then by Exercise 26 in Section 7.2,

T

ABB=

for

some positive definite matrix B. Since the eigenvalues of B are positive, 0 is not an eigenvalue of B

and B is invertible. Thus the columns of B are linearly independent. By Theorem 12 in Section 6.4, B

= QR for some n × n matrix Q with orthonormal columns and some upper triangular matrix R with

positive entries on its diagonal. Since Q is a square matrix,

,

T

QQ I=

and

()()

TT TTT

ABB QR QR RQQRRR== = =

and R has the required properties.

8. Suppose that A is positive definite, and consider a Cholesky factorization of

T

ARR=

with R upper

triangular and having positive entries on its diagonal. Let D be the diagonal matrix whose diagonal

entries are the entries on the diagonal of R. Since right-multiplication by a diagonal matrix scales the

columns of the matrix on its left, the matrix

1T

LRD

−

=

is lower triangular with 1’s on its diagonal.

If U = DR, then

1.

T

ARDDRLU

−

==

9. If A is an m × n matrix and x is in

n

, then

2

()()|| || 0.

TT T

AA A A A==≥xxxx x

Thus

T

AA

is positive

semidefinite. By Exercise 22 in Section 6.5,

rank rank .

T

AA A=

450 CHAPTER 7 • Symmetric Matrices and Quadratic Forms

10. If rank G = r, then dim

Nul G = n – r by the Rank Theorem. Hence 0 is an eigenvalue of G with

multiplicity n – r, and the spectral decomposition of G is

111

TT

rrr

G=+…+uu uu

Also

1,,

r

…

are positive because G is positive semidefinite. Thus

()

11 11

TT

rr rr

G=+…+uu u u

By the column-row expansion of a matrix product,

T

GBB=

where B is the n × r matrix

11

.

rr

Bªº

=…

¬¼

uu

Finally,

T

GAA=

for

.

T

AB=

11. Let

T

AUV=Σ

be a singular value decomposition of A. Since U is orthogonal,

T

UU I=

and

TT

AUUUV PQ=Σ=

where

1T

PUU UU

−

=Σ=Σ

and

.

T

QUV=

Since Σ is symmetric, P is

symmetric, and P has nonnegative eigenvalues because it is similar to Σ, which is diagonal with

nonnegative diagonal entries. Thus P is positive semidefinite. The matrix Q is orthogonal since it is

the product of orthogonal matrices.

12. a. Because the columns of

r

V

are orthonormal,

11

()( )( )

TT TT

rr r r r r rr

AA U DV V D U U DD U U U

+−−

===yyyy

Since

T

rr

UU y

is the orthogonal projection of y onto

Col r

U

by Theorem 10 in Section 6.3, and

since

Col Col

r

UA=

by (5) in Example 6 of Section 7.4,

AA

+

y

is the orthogonal projection of

y onto Col A.

b . Because the columns of

r

U

are orthonormal,

11

()()( )

TT T T

rrrr r r rr

AA VD U UDV VD DV VV

+−−

===xxxx

Since

T

rr

VV x

is the orthogonal projection of x onto

Col r

V

by Theorem 10 in Section 6.3, and

since

Col Row

r

VA=

by (8) in Example 6 of Section 7.4,

AA

+

x

is the orthogonal projection of

x onto Row A.

c . Using the reduced singular value decomposition, the definition of

A

+

, and the associativity of

matrix multiplication gives:

11

()( )()( )()

TTT TT

rr r r rr r r rr

AA A U DV V D U U DV U DD U U DV

+−−

==

1TT

rrrr

UDD DV UDV A

−

===

1111

()()()( )()

TT T T T

rrrrrr r rrr

AAA VD U UDV VD U VD DV VD U

++ −−−−

==

11 1TT

rrrr

VD DD U VD U A

−− − +

===

13. a. If b = Ax, then

.AAA

++ +

==xb x

By Exercise 12(a),

+

x

is the orthogonal projection of x onto

Row A.

b. From part (a) and Exercise 12(c),

()() .AAAA AAAA

++ +

====xxxxb

c. Let Au = b. Since

+

x

is the orthogonal projection of x onto Row A, the Pythagorean Theorem

shows that

22 2 2

|| || || || || || || || ,

+++

=+−≥ux ux x

with equality only if

.

+

=ux

Chapter 7 • Supplementary Exercises 451

14. The least-squares solutions of Ax = b are precisely the solutions of Ax =

ˆ,b

where

ˆ

b

is the

orthogonal projection of b onto Col A. From Exercise 13, the minimum length solution of Ax =

ˆ

b

is

ˆ,A

+

b

so

ˆ

A+b

is the minimum length least-squares solution of Ax = b. However,

ˆAA+

=bb

by

Exercise 12(a) and hence

ˆ

AAAA

+++ +

==Αbbb

by Exercise 12(c). Thus

A+b

is the minimum

length least-squares solution of Ax = b.

15. [M] The reduced SVD of A is

,

T

rr

AUDV=

where

.966641 .253758 .034804 9.84443 0 0

.185205 .786338 .589382 ,02.624660,

.125107 .398296 .570709 001.09467

.125107 .398296 .570709

r

UD

−

ªº

ª

º

«»

−−

«

»

«»

==

«

»

«»

−

«

»

«»

¬

¼

−

«»

¬¼

.313388 .009549 .633795

and .633380 .023005 .313529

.633380 .023005 .313529

.035148 .999379 .002322

r

V

−

ªº

«»

−

«»

=−−

«»

−

«»

¬¼

So the pseudoinverse

1T

rr

AVDU

+−

=

may be calculated, as well as the solution

ˆA+

=xb

for the

system Ax = b:

.05 .35 .325 .325

ˆ

,

.05 .15 .175 .175

.10 .30 .150 .150

A

+

−− .7

ªºªº

«»«»

−− .7

«»«»

==

−−−−.8

«»«»

−.8

«»«»

−− − .6

¬¼¬¼

x

Row reducing the augmented matrix for the system

ˆ

T

A=zx

shows that this system has a solution, so

ˆ

x is in

Col Row .

T

AA=

A basis for Nul A is

12

01

{, } , ,

10

00



½

−

ª

ºª º

°

«

»« »

°

«

»« »

°

«

»« »

=

®

¾

«

»« »

°

«

»« »

°

«

»« »

°

¬

¼¬ ¼

¯¿

aa

and an arbitrary element of

Nul A is

12

.cd=+ua a

One computes that

ˆ

|| ,

||

=131/50x

while

ˆ

|| .cd

22

+

||

=(131/50)+2 +2xu

Thus if u ≠ 0, || ˆ

x|| < || ˆ

x + u ||, which confirms that ˆ

x is the minimum length solution to Ax = b.

16. [M] The reduced SVD of A is

,

T

rr

AUDV=

where

.337977 .936307 .095396 12.9536 0 0

.591763 .290230 .752053 ,01.445530,

.231428 .062526 .206232 00.337763

.694283 .187578 .618696

r

UD

−

ªº

ª

º

«»

−

«

»

«»

==

«

»

«»

−−−

«

»

«»

¬

¼

−−−

«»

¬¼

452 CHAPTER 7 • Symmetric Matrices and Quadratic Forms

.690099 .721920 .050939

00 0

and .341800 .387156 .856320

.637916 .573534 .513928

00 0

r

V

−

ªº

«»

=−

«»

¬¼

So the pseudoinverse

1T

rr

AVDU

+−

=

may be calculated, as well as the solution

ˆA+

=xb

for the

system Ax = b:

.5 0 .05 .15

00 0 0

ˆ

,

02 .5 1.5

.5 1 .35 1.05

00 0 0

A

+

−− 2.3

ªºªº

«»«»

0

«»«»

==

5.0

«»«»

−− − −.9

«»«»

0

¬¼¬¼

x

Row reducing the augmented matrix for the system

ˆ

T

A=zx

shows that this system has a solution, so

ˆ

x is in

Col Row

T

AA=

. A basis for Nul A is

12

00

10

{, } , ,

00

01



½

ª

ºªº

°

«

»«»

°

«

»«»

°

«

»«»

=

®

¾

«

»«»

°

«

»«»

°

«

»«»

°

¬

¼¬¼

¯¿

aa

and an arbitrary element of

Nul A is

12

.cd=+ua a

One computes that

ˆ

|| ,

||

=311/10x

while

ˆ

|| .cd

22

+||=(311/10)+ +xu

Thus if u ≠ 0, || ˆ

x|| < || ˆ

x+ u ||, which confirms that ˆ

x is the minimum length solution to Ax = b.

!

453

!

! "#$!%$&'$()*!&+!

,$-(&)!./0-$1!

!

8.1 SOLUTIONS

Notes

. This section introduces a special kinds of linear combination used to describe the sets created

when a subspace is shifted away from the origin. An affine combination is a linear combination in which

the coefficients sum to one. Theorems 1, 3, and 4 connect affine combinations directly to linear

combinations. There are several approaches to solving many of the exercises in this section, and some of

the alternatives are presented here.

1.

12 3 4

12035

,,4,,

22473

−

ªº ª º ªº ªº ªº

== = ==

«» « » «» «» «»

¬¼ ¬ ¼ ¬¼ ¬¼ ¬¼

vv v v y

21 31 41 1

3124

,,,

0251

−−

ªº ªº ªº ªº

−=−=−=−=

«» «» «» «»

¬¼ ¬¼ ¬¼ ¬¼

vv vv vv yv

Solve c

2

(v

2

− v

1

) + c

3

(v

3

− v

1

) + c

4

(v

4

− v

1

) = y − v

1

by row reducing the augmented matrix.

3124 312 4 304.54.5 101.51.5

0251 012.5.5 012.5.5 012.5 .5

−− −− − − −

ªºª ºª ºª º

«»« »« »« »

¬¼¬ ¼¬ ¼¬ ¼



The general solution is c

2

= 1.5c

4

−1.5, c

3

= −2.5c

4

+ .5, with c

4

free. When c

4

= 0,

y − v

1

= −1.5(v

2

− v

1

) + .5(v

3

− v

1

) and y = 2v

1

− 1.5v

2

+ .5v

3

If c

4

= 1, then c

2

= 0 and

y − v

1

= −2(v

3

− v

1

) + 1(v

4

− v

1

) and y = 2v

1

− 2v

3

+ v

4

If c

4

= 3, then

y − v

1

= 3(v

2

− v

1

) − 7(v

3

− v

1

) + 3(v

4

− v

1

) and y = 2v

1

+ 3v

2

− 7v

3

+ 3v

4

Of course, many other answers are possible. Note that in all cases, the weights in the linear

combination sum to one.

2.

12 3

1135

,,,

1227

−

ªº ª º ª º ª º

== ==

«» « » « » « »

¬¼ ¬ ¼ ¬ ¼ ¬ ¼

vv v y

, so

21 31 1

22 4

,,and

11 6

−

ª

ºªº ªº

−=−=−=

«

»«» «»

¬

¼¬¼ ¬¼

vv vv yv

Solve c

2

(v

2

– v

1

) + c

3

(v

3

– v

1

) = y – v

1

by row reducing the augmented matrix:

454 CHAPTER 8 • The Geometry of Vector Spaces

!

224 112 102

~~

116 028 014

−−−

ªºªºªº

«»«»«»

¬¼¬¼¬¼

The general solution is c

2

= 2 and c

3

= 4, so y – v

1

= 2(v

2

– v

1

) + 4(v

3

– v

1

) and

y = –5v

1

+ 2v

2

+ 4v

3

. The weights sum to one, so this is an affine sum.

3. Row reduce the augmented matrix [v

2

-v

1

v

3

-v

1

y-v

1

] to find a solution for writing y-v

1

in terms of

v

2

-v

1

and v

3

-v

2

. Then solve for y to get y = −3v

1

+ 2v

2

+ 2v

3

. The weights sum to one, so this is an

affine sum.

4. Row reduce the augmented matrix [v

2

-v

1

v

3

-v

1

y-v

1

] to find a solution for writing y-v

1

in terms of

v

2

-v

1

and v

3

-v

2

. Then solve for y to get y = 2.6v

1

− .4v

2

− 1.2v

3

. The weights sum to one, so this is

an affine sum.

5. Since {b

1

, b

2

, b

3

} is an orthogonal basis, use Theorem 5 from Section 6.2 to write

12 3

11 2 2 3 3

= +

jj j

j

+

pb pb pb

pbbb

bb bb bb

<< <

a. p

1

= 3b

1

– b

2

– b

3

∈ aff S since the coefficients sum to one.

b. p

2

= 2b

1

+ 0b

2

+ b

3

∉ aff S since the coefficients do not sum to one.

c. p

3

= – b

1

+ 2b

2

+ 0b

3

∈ aff S since the coefficients sum to one.

6. Since {b

1

, b

2

, b

3

} is an orthogonal basis, use Theorem 5 from Section 6.2 to write

12 3

11 2 2 3 3

= +

jj j

j

+

pb pb pb

pbbb

bb bb bb

<< <

a. p

1

= – 4b

1

+2b

2

+3b

3

∈ aff S since the coefficients sum to one.

b. p

2

= .2b

1

+ .5b

2

+ .3b

3

∈ aff S since the coefficients sum to one.

c. p

3

= b

1

+ b

2

+ b

3

∉ aff S since the coefficients do not sum to one.

7. The matrix [v

1

v

2

v

3

p

1

p

2

p

3

] row reduces to

100 2 2 2

010 1 4 2

001 1 3 2

000 0 0 5

ª

º

«

»

−

«

»

«

»

−

«

»

−

«

»

¬

¼

.

Parts a., b., and c. use columns 4, 5, and 6, respectively, as the “augmented” column.

a. p

1

= 2v

1

+ v

2

− v

3

, so p

1

is in Span S. The weights do not sum to one, so p

1

∉ aff S.

b. p

2

= 2v

1

− 4v

2

+ 3v

3

, so p

2

is in Span S. The weights sum to one, so p

2

∈ aff S.

c. p

3

∉ Span S because 0 v −5, so p

3

cannot possibly be in aff S.

8.1 • Solutions 455

!

8. The matrix [v

1

v

2

v

3

p

1

p

2

p

3

] row reduces to

100 3 0 2

010 10 6

001 1 0 3

000 0 1 0

−

ª

º

«

»

−

«

»

«

»

−

«

»

«

»

¬

¼

.

Parts a., b., and c. use columns 4, 5, and 6, respectively, as the “augmented” column.

a. p

1

= 3v

1

− v

2

+ v

3

, so p

1

is in Span S. The weights do not sum to one, so p

1

∉ aff S.

b. p

2

∉ Span S because 0 v 1 (column 5 is the augmented column), so p

2

cannot possibly be

in aff S.

c. p

3

= −2v

1

+ 6v

2

− 3v

3

, so p

3

is in Span S. The weights sum to one, so p

3

∈ aff S.

9. Choose v

1

and v

2

to be any two point on the line x=x

3

u+p. For example, take x

3

=0 and x

3

=1 to get

1

3

0

−

ªº

=«»

¬¼

v

and

2

1

2

ª

º

=

«

»

−

¬

¼

v

respectively. Other answers are possible.

10. Choose !

"

!and !

#

!to be any two point on the line x=x

3

u+p. For example, take x

3

=0 and x

3

=1 to get

1

3

4

ªº

«»

=−

«»

¬¼

v

and

2

6

2

ª

º

«

»

=−

«

»

«

»

¬

¼

v

respectively. Other answers are possible.

11. a. True. See the definition at the beginning of this section.

b. False. The weights in the linear combination must sum to one. See the definition.

c. True. See equation (1).

d. False. A flat is a translate of a subspace. See the definition prior to Theorem 3.

e. True. A hyperplane in 

3

has dimension 2, so it is a plane. See the definition prior to Theorem 3.

12. a. False. If S = {x}, then aff S = {x}. See the definition at the beginning of this section.

b. True. Theorem 2.

c. True. See the definition prior to Theorem 3.

d. False. A flat of dimension 2 is called a hyperplane only if the flat is considered a subset of 

3

. In

general, a hyperplane is a flat of dimension n − 1. See the definition prior to Theorem 3.

e. True. A flat through the origin is a subspace translated by the 0 vector.

13. Span {v

2

− v

1

, v

3

− v

1

} is a plane if and only if {v

2

− v

1

, v

3

− v

1

} is linearly independent. Suppose c

2

and c

3

satisfy c

2

(v

2

− v

1

) + c

3

(v

3

− v

1

) = 0. Then c

2

v

2

+ c

3

v

3

− (c

2

+ c

3

)v

1

= 0. Then c

2

= c

3

= 0,

because {v

1

, v

2

, v

3

} is a linearly independent set. This shows that {v

2

− v

1

, v

3

− v

1

} is a linearly

independent set. Thus, Span {v

2

− v

1

, v

3

− v

1

} is a plane in 

3

.

14. Since {v

1

, v

2

, v

3

} is a basis for 

3

, the set W = Span {v

2

− v

1

, v

3

− v

1

} is a plane in 

3

, by

Exercise 13. Thus, W + v

1

is a plane parallel to W that contains v

1

. Since v

2

= (v

2

− v

1

) + v

1

, W +

v

1

contains v

2

. Similarly, W + v

1

contains v

3

. Finally, Theorem 1 shows that aff {v

1

, v

2

, v

3

} is the

plane W + v

1

that contains v

1

, v

2

, and v

3

.

15. Let S = {x : Ax = b}. To show that S is affine, it suffices to show that S is a flat, by Theorem 3.

Let W = {x : Ax = 0}. Then W is a subspace of 

n

, by Theorem 2 in Section 4.2 (or Theorem 12

in Section 2.8). Since S = W + p, where p satisfies Ap = b, by Theorem 6 in Section 1.5, S is a

translate of W, and hence S is a flat.

456 CHAPTER 8 • The Geometry of Vector Spaces

!

16. Suppose p, q ∈ S and t ∈ . Then, by properties of the dot product (Theorem 1 in Section 6.1),

[(1 − t)p + t q] ⋅ v = (1 − t)(p ⋅ v) + t (q ⋅ v) = (1 − t)k + t k = k

Thus, [(1 − t)p + t q] ∈ S, by definition of S. This shows that S is an affine set.

17. A suitable set consists of any three vectors that are not collinear and have 5 as their third entry. If

5 is their third entry, they lie in the plane x

3

= 5. If the vectors are not collinear, their affine hull

cannot be a line, so it must be the plane. For example use

101

0,1,1 .

555

S



½

ª

ºªºªº

°

«

»«»«»

=

®

¾

«

»«»«»

°

«

»«»«»

¬

¼¬¼¬¼

¯¿

18. A suitable set consists of any four vectors that lie in the plane 2x

1

+ x

2

− 3x

3

= 12 and are not col-

linear. If the vectors are not collinear, their affine hull cannot be a line, so it must be the plane.

For example use

60 0 3

0,12, 0 , 3 .

00 4 1

S



½

ª

ºª ºª ºª º

°

«

»« »« »« »

=

®

¾

«

»« »« »« »

°

«

»« »« »« »

−−

¬

¼¬ ¼¬ ¼¬ ¼

¯¿

19. If p, q ∈ f (S), then there exist r, s ∈ S such that f (r) = p and f (s) = q. Given any t ∈ , we must

show that z = (1 − t)p + t q is in f (S). Since f is linear,

z = (1 − t)p + t q = (1 − t) f (r) + t f (s) = f ((1 − t)r + t s)

Since S is affine, (1 − t)r + t s ∈ S. Thus, z is in S and f (S) is affine.

20. Given an affine set T, let S = {x ∈ 

n

: f (x) ∈ T}. Consider x, y ∈ S and t ∈ . Then

f ((1 − t)x + t y) = (1 − t) f (x) + t f (y)

But f (x) ∈ T and f (y) ∈ T, so (1 − t) f (x) + t f (y) ∈ T because T is an affine set. It follows that

[(1 − t)x + t y] ∈ S. This is true for all x, y ∈ S and t ∈ , so S is an affine set.

21. Since B is affine, Theorem 1 implies that B contains all affine combinations of points of B. Hence

B contains all affine combinations of points of A. That is, aff A  B.

22. Since B ⊂ aff B, we have A ⊂ B ⊂ aff B. But aff B is an affine set, so Exercise 21 implies

aff A ⊂ aff B.

23. Since A  (A  B), it follows from Exercise 22 that aff A  aff (A  B).

Similarly, aff B  aff (A  B), so [aff A  aff B]  aff (A  B).

24. One possibility is to let A ={(0, 1)} and B = {(1, 0)}. Then (aff A)  (aff B) consists of the two

coordinate axes, but aff (A  B) = 

2

.

25. Since (A  B) ⊆ A, it follows from Exercise 22 that aff (A  B)  aff A.

Similarly, aff (A  B)  aff B, so aff (A  B)  (aff A  aff B).

26. One possibility is to let A = {(0, 1)} and B = {(0, 2)}. Then both aff A and aff B are equal to the

x-axis. But A  B = ∅, so aff (A  B) = ∅.

8.2 • Solutions 457

!

8.2 SOLUTIONS

Notes:

Affine dependence and independence are developed in this section. Theorem 5 links affine

independence to linear independence. This material has important applications to computer graphics.

1. Let

123

302

,,.

360

ªº ªº ªº

===

«» «» «»

−

¬¼ ¬¼ ¬¼

vvv

Then

21 31

31

,.

93

−−

ª

ºªº

−=−=

«

»«»

¬

¼¬¼

vv vv

Since v

3

− v

1

is a multiple

of v

2

− v

1

, these two points are linearly dependent. By Theorem 5, {v

1

, v

2

, v

3

} is affinely dependent.

Note that (v

2

− v

1

) − 3(v

3

− v

1

) = 0. A rearrangement produces the affine dependence relation 2v

1

+ v

2

− 3v

3

= 0. (Note that the weights sum to zero.) Geometrically, v

1

, v

2

, and v

3

are collinear.

2.

123 2131

25 3 3 5

,, . ,

14 2 3 3

−−

ªº ªº ª º ªº ª º

=== −=−=

«» «» « » «» « »

−−

¬¼ ¬¼ ¬ ¼ ¬¼ ¬ ¼

vv v vvvv

. Since v

3

− v

1

and v

2

− v

1

are not

multiples, they are linearly independent. By Theorem 5, {v

1

, v

2

, v

3

} is affinely independent.

3. The set is affinely independent. If the points are called v

1

, v

2

, v

3

, and v

4

, then row reduction of

[v

1

v

2

v

3

v

4

] shows that {v

1

, v

2

, v

3

} is a basis for 

3

and v

4

= 16v

1

+ 5v

2

− 3v

3

. Since there is

unique way to write v

4

in terms of the basis vectors, and the weights in the linear combination do not

sum to one, v

4

is not an affine combination of the first three vectors.

4. Name the points v

1

, v

2

, v

3

, and v

4

. Then

21 31 41

230

8, 7, 2

496

ª

ºªºªº

«

»«»«»

−=−−=−−=

«

»«»«»

«

»«»«»

−−

¬

¼¬¼¬¼

vv vv vv

. To study

the linear independence of these points, row reduce the augmented matrix for Ax = 0:

2300 2 300 2300 10.60

8720~0 520~0520~01.40

4960 01560 0000 00 00

−

ªºªºªºªº

«»«»«»«»

−−

«»«»«»«»

−− − −

¬¼¬¼¬¼¬¼

. The first three columns

are linearly dependent, so {v

1

, v

2

, v

3

, v

4

} is affinely dependent, by Theorem 5. To find the affine

dependence relation, write the general solution of this system: x

1

= .6x

3

, x

2

= −.4x

3

, with x

3

free. Set

x

3

= 5, for instance. Then x

1

= 3, x

2

= −2, and x

3

= 5. Thus, 3(v

2

− v

1

) − 2(v

3

− v

1

) + 5(v

4

− v

1

) = 0.

Rearrange to obtain −6v

1

+ 3v

2

− 2v

3

+ 5v

4

= 0.

Alternative solution: Name the points v

1

, v

2

, v

3

, and v

4

. Use Theorem 5(d) and study the

homogeneous forms of the points. The first step is to move the bottom row of ones (in the

augmented matrix) to the top to simplify the arithmetic:

[]

1234

1111 1001.2

2012 010.6

~~

5327 001.4

376 3 000 0

ª

ºª º

«

»« »

−− −

«

»« »

«

»« »

−−

«

»« »

−−

«

»« »

¬

¼¬ ¼

vv v v

  

Thus, x

1

+ 1.2x

4

= 0, x

2

− .6x

4

= 0, and x

3

+ .4x

4

= 0, with x

4

free. Take x

4

= 5, for example, and

get x

1

= −6, x

2

= 3, and x

3

= 2−. An affine dependence relation is −6v

1

+ 3v

2

− 2v

3

+ 5v

4

= 0.

5. −

4v

1

+ 5v

2

– 4v

3

+ 3v

4

= 0 is an affine dependence relation. It can be found by row reducing the

matrix

[]

1234

,vv v v

  

and proceeding as in the solution to Exercise 4.

!

458 CHAPTER 8 • The Geometry of Vector Spaces

!

6. The set is affinely independent, as the following calculation with homogeneous forms shows:

[]

1234

1111 1000

1023 0100

~~

3155 0010

1220 0001

ª

ºª º

«

»« »

«

»« »

«

»« »

−

«

»« »

−

«

»« »

¬

¼¬ ¼

vv vv

  

Alternative solution: Row reduction of [v

1

v

2

v

3

v

4

] shows that {v

1

, v

2

, v

3

} is a basis for 

3

and

v

4

= −2v

1

+ 1.5v

2

+ 2.5v

3

, but the weights in the linear combination do not sum to one, so this v

4

is

not an affine combination of the basis vectors and hence the set is affinely independent.

Note: A potential exam question might be to change the last entry of v

4

from 0 to 1 and again ask if

the set is affinely independent. Notice that row reduction of this new set of vectors [v

1

v

2

v

3

v

4

]

shows that {v

1

, v

2

, v

3

} is a basis for 

3

and v

4

= −3v

1

+ v

2

+ 3v

3

is an affine combination of the basis.

7. Denote the given points as v

1

, v

2

, v

3

, and p. Row reduce the augmented matrix for the equation

x

1

vɫ

1

+ x

2

vɫ

2

+ x

3

vɫ

3

= p.ɫ Remember to move the bottom row of ones to the top as the first step to

simplify the arithmetic by hand.

[]

123

1111 1002

12 15 0 104

~~

1124 0011

2022 0000

1102 0000

−

ª

ºª º

«

»« »

«

»« »

«

»« »

−−

«

»« »

−−

«

»« »

«

»« »

¬

¼¬ ¼

vv vp



 

Thus, x

1

= −2, x

2

= 4, x

3

= −1, and pɫ = −2vɫ

1

+ 4vɫ

2

− vɫ

3

, so p = −2v

1

+ 4v

2

− v

3

, and the barycentric

coordinates are (−2, 4, −1).

Alternative solution: Another way that this problem can be solved is by “translating” it to the

origin. That is, compute v

2

– v

1

, v

3

– v

1

, and p – v

1

, find weights c

2

and c

3

such that

c

2

(v

2

– v

1

) + c

3

(v

3

– v

1

) = p – v

1

and then write p = (1 – c

2

– c

3

)v

1

+ c

2

v

2

+ c

3

v

3

. Here are the calculations for Exercise 7:

21

21 1

11 2

02 2

11 0

ªº ª º ª º

«» « » « »

−

«» « » « »

−=−=

«» « » « »

−

«» « » « »

¬¼ ¬ ¼ ¬ ¼

vv

,

31

11 0

21 3

22 4

01 1

ª

ºª º ª º

«

»« » « »

−

«

»« » « »

−=−=

«

»« » « »

−−

«

»« » « »

−

¬

¼¬ ¼ ¬ ¼

vv

,

1

51 4

41 5

22 4

21 1

ªºªº ª º

«»«» « »

−

«»«» « »

−=−=

«»«» « »

−−

«»«» « »

¬¼¬¼ ¬ ¼

pv

[]

2131 1

104 104

235 011

~~

244 000

011 000

ª

ºª º

«

»« »

−

«

»« »

−−−

«

»« »

−− −

«

»« »

−

¬

¼¬ ¼

vvvvpv

Thus p – v

1

= 4(v

2

– v

1

) – 1(v

3

– v

1

), and p = –2 v

1

+ 4v

2

– v

3

.

!

8.2 • Solutions 459

!

8. Denote the given points as v

1

, v

2

, v

3

, and p. Row reduce the augmented matrix for the equation

x

1

vɫ

1

+ x

2

vɫ

2

+ x

3

vɫ

3

= pɫ.

[]

123

1111 1002

0111 0101

~~

1141 0010

2064 0000

1250 0000

ªºªº

«»«»

−−

«»«»

−−−

«»«»

¬¼¬¼

vv vp



 

Thus. pɫ = 2vɫ

1

− vɫ

2

+ 0vɫ

3

, so p = 2v

1

− v

2

. The barycentric coordinates are (2 , −1, 0).

Notice v

3

= 3v

1

+ v

2

9. a. True. Theorem 5 uses the point v

1

for the translation, but the paragraph after the theorem

points out that any one of the points in the set can be used for the translation.

b. False, by (d) of Theorem 5.

c. False. The weights in the linear combination must sum to zero, not one. See the definition at the

beginning of this section.

d. False. The only points that have barycentric coordinates determined by S belong to aff S. See the

definition after Theorem 6.

e. True. The barycentric coordinates have some zeros on the edges of the triangle and are only

positive for interior points. See Example 6.

10. a. False. By Theorem 5, the set of homogeneous forms must be linearly dependent, too.

b. True. If one statement in Theorem 5 is false, the other statements are false, too.

c. False. Theorem 6 applies only when S is affinely independent.

d. False. The color interpolation applies only to points whose barycentric coordinates are

nonnegative, since the colors are formed by nonnegative combinations of red, green, and blue.

See Example 5.

e. True. See the discussion of Fig. 5.

11. When a set of five points is translated by subtracting, say, the first point, the new set of four

points must be linearly dependent, by Theorem 8 in Section 1.7, because the four points are in 

3

.

By Theorem 5, the original set of five points is affinely dependent.

12. Suppose v

1

, …, v

p

are in 

n

and p ≥ n + 2. Since p −

1 ≥ n + 1, the points v

2

− v

1

, v

3

− v

1

, … , v

p

− v

1

are linearly dependent, by Theorem 8 in Section 1.7. By Theorem 5, {v

1

, v

2

, …, v

p

} is affinely

dependent.

13. If {v

1

, v

2

} is affinely dependent, then there exist c

1

and c

2

, not both zero, such that c

1

+ c

2

= 0, and

c

1

v

1

+ c

2

v

2

= 0. Then c

1

= – c

2

0 and c

1

v

1

= – c

2

v

2

= c

1

v

2

, which implies that v

1

= v

2

.

Conversely, if v

1

= v

2

, let c

1

= 1 and c

2

= −1. Then c

1

v

1

+ c

2

v

2

= v

1

+ (−1)v

1

= 0 and c

1

+ c

2

= 0,

which shows that {v

1

, v

2

} is affinely dependent.

14. Let S

1

consist of three (distinct) points on a line through the origin. The set is affinely dependent

because the third point is on the line determined by the first two points. Let S

2

consist of two

(distinct) points on a line through the origin. By Exercise 13, the set is affinely independent

460 CHAPTER 8 • The Geometry of Vector Spaces

!

because the two points are distinct. (A correct solution should include a justification for the sets

presented.)

15. a. The vectors v

2

− v

1

=

1

2

ªº

«»

¬¼

and v

3

− v

1

=

3

2

ª

º

«

»

−

¬

¼

are not multiples and hence are linearly

independent. By Theorem 5, S is affinely independent.

b.

()

12

695 11

888 22

,, , 0,, ,↔− ↔pp

()()

34

5657

14 1

888 888

,, , ,,↔−− ↔−pp

,

()

5

11

5488

,,↔p

c. p

6

is (−, −, +), p

7

is (0, +, −), and p

8

is (+, +, −).

16. a. The vectors v

2

− v

1

=

1

4

ªº

«»

¬¼

and v

3

− v

1

=

4

2

ª

º

«

»

¬

¼

are not

multiples and hence are linearly independent. By

Theorem 5, S is affinely independent.

b.

55103

24 2 22

12 3

777 7 7 7 777

(,,), (, ,), (,,)↔− ↔ − ↔pp p

c.

45

(,,), (,,),↔+−− ↔ ++−pp

67

(,,), (,0,).↔+++ ↔− +pp

See the figure to the right. Actually,

19 3 5 3

212

45

14 14 14 14 14 14

93 3

21

67

14 14 14 2 2

(, , ), (, , ),

(, ,), ( ,0,).

↔−− ↔ −

↔↔−

pp

17. Suppose S = {b

1

, …, b

k

} is an affinely independent set. Then (7) has a solution, because p is in

aff S. Hence (8) has a solution. By Theorem 5, the homogeneous forms of the points in S are

linearly independent. Thus (8) has a unique solution. Then (7) also has a unique solution,

because (8) encodes both equations that appear in (7).

The following argument mimics the proof of Theorem 7 in Section 4.4. If S = {b

1

, …, b

k

} is

an affinely independent set, then scalars c

1

, …, c

k

exist that satisfy (7), by definition of aff S.

Suppose x also has the representation

x = d

1

b

1

+ … + d

k

b

k

and d

1

+ … + d

k

= 1 (7a)

for scalars d

1

, …, d

k

. Then subtraction produces the equation

0 = x − x = (c

1

− d

1

)b

1

+ … + (c

k

−

d

k

)b

k

( 7 b )

The weights in (7b) sum to zero because the c’s and the d’s separately sum to one. This is

impossible, unless each weight in (8) is zero, because S is an affinely independent set. This

proves that c

i

= d

i

for i = 1, …, k.

18. Let

.

x

y

z

ªº

«»

=«»

«»

¬¼

p

Then

00 0

0010

00 0

xa

xyz xyz

yb

abc abc

zc

ªº ªº ªº ªº ªº

§·

«» «» «» «» «»

=+++−− −

¨¸

«» «» «» «» «»

©¹

«» «» «» «» «»

¬¼ ¬¼ ¬¼ ¬¼ ¬¼

So the barycentric coordi-

nates are x/a, y/b, z/c, and 1 − x/a − y/b − z/c. This holds for any nonzero choices of a, b, and c.

19. If {p

1

, p

2

, p

3

} is an affinely dependent set, then there exist scalars c

1,

c

2

, and c

3

, not all zero, such

that c

1

p

1

+ c

2

p

2

+ c

3

p

3

= 0 and c

1

+ c

2

+ c

3

= 0. But then, applying the transformation f,

p

2

v

3

v

1

v

2

p

1

p

3

p

4

p

5

!

p

6

p

7

8.2 • Solutions 461

!

11 2 2 3 3 112233

() ( ) () ( ) ()cf c f cf f c c c f++ =++==ppp ppp00

,

since f is linear. This shows that

123

{( ), ( ), ( )}ff fpp p

is also affinely dependent.

20. If the translated set {p

1

+ q, p

2

+ q, p

3

+ q} were affinely dependent, then there would exist real

numbers c

1,

c

2

, and c

3

, not all zero and with c

1

+ c

2

+ c

3

= 0, such that

c

1

(p

1

+ q) + c

2

(p

2

+ q) + c

3

(p

3

+ q) = 0.

But then,

c

1

p

1

+ c

2

p

2

+ c

3

p

3

+ (c

1

+ c

2

+ c

3

)q = 0.

Since c

1

+ c

2

+ c

3

= 0, this implies c

1

p

1

+ c

2

p

2

+ c

3

p

3

= 0, which would make {p

1

, p

2

, p

3

} affinely

dependent. But {p

1

, p

2

, p

3

} is affinely independent, so the translated set must in fact be affinely

independent, too.

21. Let

11 1

22 2

,, and .

ab c

ªº ªº ªº

== =

«» «» «»

¬¼ ¬¼ ¬¼

ab c

Then det

[aɫ bɫ cɫ] =

111 12

222 12

12

1

det det 1

111 1

abc aa

abc bb

cc

ª

ºª º

«

»« »

=

«

»« »

«

»« »

¬

¼¬ ¼

,

by using the transpose property of the determinant (Theorem 5 in Section 3.2). By Exercise 30 in

Section 3.3, this determinant equals 2 times the area of the triangle with vertices at a, b, and c.

22. If p is on the line through a and b, then p is an affine combination of a and b, so pɫ is a linear

combination of aɫ and bɫ. Thus the columns of [aɫ bɫ pɫ] are linearly dependent. So the determinant

of this matrix is zero.

23. If [aɫ bɫ cɫ]

r

s

t

ªº

«»

¬¼

= pɫ, then Cramer’s rule gives r = det

[pɫ bɫ cɫ] /

det

[aɫ bɫ cɫ]. By Exercise 21, the

numerator of this quotient is twice the area of pbc, and the denominator is twice the area of abc.

This proves the formula for r. The other formulas are proved using Cramer’s rule for s and t.

24. Let p = (1 − x)q + x

a, where q is on the line segment from b to c. Then, because the determinant

is a linear function of the first column when the other columns are fixed (Section 3.2),

det

[pɫ bɫ cɫ] = det

[(1 − x)qɫ + x

aɫ bɫ cɫ] = (1 − x)

·

det

[qɫ bɫ cɫ] + x

·

det

[aɫ bɫ cɫ]

Now, [qɫ bɫ cɫ] is a singular matrix because qɫ is a linear combination of bɫ and cɫ. So det

[qɫ bɫ cɫ] =

0 and det

[pɫ bɫ cɫ] = x

·

det

[aɫ bɫ cɫ] .

462 CHAPTER 8 • The Geometry of Vector Spaces

!

8.3 SOLUTIONS

Notes:

The notion of convexity is introduced in this section and has important applications in computer

graphics. Bézier curves are introduced in Exercises 21-24 explored in greater detail in Section 8.6.

1. The set 0:0 1Vy

y

½

ªº

°°

=≤<

®¾

«»

°°

¬¼

¯¿

is the vertical line segment from (0,0) to

(0,1) that includes (0,0) but not (0,1). The convex hull of S includes

each line segment from a point in V to the point (2,0), as shown in the

figure. The dashed line segment along the top of the shaded region

indicates that this segment is not in conv S, because (0,1) is not in S.

2. a. Conv S includes all points p of the form

1/ 2 1/ 2 ( 1/ 2)

(1 ) 21/ 2(21/)

xtx

tt

xtx

+−

ªºªºª º

=−+=

«»«»« »

−−

¬¼¬¼¬ ¼

p

, where

1/ 2 and 0 1.xt≥≤≤

Notice that if t=a/x, then

2

1/ 2 /(2 )

() 22/ /

aa x

xax ax

+−

ªº

=«»

−−

¬¼

p

and

1/ 2

lim ( ) 2

x

a

x

→∞

+

ª

º

=

«

»

¬

¼

p

, establishing that there are points

arbitrarily close to the line y=2 in conv S. Since the curve y=1/x is in S, the line segments

between y=2 and y=1/x are also included in conv S, whenever

1/ 2.x≥

b. Recall that for any integer n,

()

sin( 2 ) sinxn x

π

+=

. Then

22

(1 ) conv S.

sin( ) sin( 2 ) sin( )

xxnxnt

tt

xxn x

ππ

π

++

ªºª ºª º

=−+=∈

«»« »« »

+

¬¼¬ ¼¬ ¼

p

Notice that sin(x) is always a number between -1 and 1. For a

fixed x and any real number r, an integer n and a number t (with

01t≤≤

) can be chosen so that

2.rx nt

π

=+

c. Conv S includes all points p of the form

0

(1 ) 0

xx

tt t

xx

ª

ºªº

ªº

=−+=

«

»«»

«»

¬¼

¬

¼¬¼

p

, where

0 and 0 t 1.x≥≤≤

Letting t=a/x,

lim 0

x

a

→∞

ªº

=«»

¬¼

p

establishing that there are point arbitrarily close to y=0 in the set.

3. From Exercise 5, Section 8.1,

a. p

1

= 3b

1

– b

2

– b

3

∉

conv S since some of the coefficients are negative.

b. p

2

= 2b

1

+ 0b

2

+ b

3

∉ conv S since the coefficients do not sum to one.

c. p

3

= – b

1

+ 2b

2

+ 0b

3

∉

conv S since some of the coefficients are negative.

"! #

"

"! #!

"

$!

8.3 • Solutions 463

!

4. From Exercise 5, Section 8.1,

a. p

1

= – 4b

1

+2b

2

+3b

3

∉

conv S since some of the coefficients are negative.

b. p

2

= .2b

1

+ .5b

2

+ .3b

3

∈ conv S since the coefficients are nonnegative and sum to one.

c. p

3

= b

1

+ b

2

+ b

3

∉ conv S since the coefficients do not sum to one.

5. Row reduce the matrix

[]

123412

vv vvpp



  

to obtain the barycentric coordinates

p

1

=

11 2 1

123 4

63 3 6

−+++vv v v

, so p

1

∉ conv S, and p

2

=

11 1 1

12 3 4

33 6 6

+++vv v v

, so p

2

∈ conv S.

6. Let W be the subspace spanned by the orthogonal set S = {v

1

, v

2

, v

3

}. As in Example 1, the

barycentric coordinates of the points p

1

, …, p

4

with respect to S are easy to compute, and they

determine whether or not a point is in Span S, aff S, or conv S.

a.

13

11 1 2

1123

11 2 2 3 3

proj

W

⋅

⋅⋅

=+ +

⋅⋅ ⋅

pv

pv pv

pvvv

vv v v vv

31

2

5

2

1

202

2

021

11

120

22

212

−

ªº

−

ªº ªº ªº «»

«» «» «»

−«»

«» «» «»

=−+= =

«»

−

«» «» «»

−«»

«» «» «» «»

¬¼ ¬¼ ¬¼ ¬¼

p

This shows that p

1

is in W = Span S. Also, since the coefficients sum to 1, p

1

is an aff S.

However, p

1

is not in conv S, because the coefficients are not all nonnegative.

b. Similarly,

99 9

44 2

21231232

11 1

proj 99 9 442

W

=++ =++=pvvvvvvp

. This shows that p

2

lies in Span S. Also, since the coefficients sum to 1, p

2

is in aff S. In fact, p

2

is in conv S,

because the coefficients are also nonnegative.

c.

31231233

9918

proj 2

999

W=+−=+−=pvvvvvvp

. Thus p

3

is in Span S. However, since

the coefficients do not sum to one, p

3

is not in aff S and certainly not in conv S.

d.

41234

68 8

proj 999

W=++ ≠pvvvp

. Since

4

proj

W

p

is the closest point in Span S to p

4

, the

point p

4

is not in Span S. In particular, p

4

cannot be in aff S or conv S.

7.

1231234

1242320

,,,,,,

0311202

−

ªº ªº ªº ªº ªº ªº ªº

=======

«» «» «» «» «» «» «»

¬¼ ¬¼ ¬¼ ¬¼ ¬¼ ¬¼ ¬¼

vvvpppp

, T = {v

1

, v

2

, v

3

}

a. Use an augmented matrix (with four augmented columns) to write the homogeneous forms of

p

1

, …, p

4

in terms of the homogeneous forms of v

1

, v

2

, and v

3

, with the first step interchanging

rows 1 and 3:

464 CHAPTER 8 • The Geometry of Vector Spaces

!

1231234

111

322

3

11 1

62 4 4

3

11 1

22 4 4

100 0

1111111

~1242320~010

0311202 001

ª

º

ªº

«

»

«»

ªº

«

»

−−

¬¼

«»

«

»

«»

¬¼

«

»

−

¬

¼

vv vpppp

   

The first four columns reveal that

11 1

1

123

36 2

++=vv vp

 

and

11 1

1231

36 2

++=vv vp

. Thus column 4

contains the barycentric coordinates of p

1

relative to the triangle determined by T. Similarly,

column 5 (as an augmented column) contains the barycentric coordinates of p

2

, column 6

contains the barycentric coordinates of p

3

, and column 7 contains the barycentric coordinates of

p

4

.

b. p

3

and p

4

are outside conv T, because in each case at least one of the barycentric coordinates is

negative. p

1

is inside conv T, because all of its barycentric coordinates are positive. p

2

is on the

edge

12

vv

of conv T, because its its barycentric coordinates are nonnegative and its first

coordinate is 0.

8. a. The barycentric coordinates of p

1

, p

2

, p

3

, and p

4

are, respectively,

()

3

12 2

13 13 13

,, ,−

()

83

2

13 13 13

,, ,

()

21

33

,0, , and

()

95

1

13 13 13

,,.−

b. The point p

1

and p

4

are outside conv T since they each have a negative coordinate. The point p

2

is

inside conv T since the coordinates are positive, and p

3

is on the edge

13

vv

of conv T.

9. The points p

1

and p

3

are outside the tetrahedron conv S since their barycentric coordinates contain

negative numbers. The point p

2

is on the face containing the vertices v

2

, v

3

, and v

4

since its first

barycentric coordinate is zero and the rest are positive. The point p

4

is inside conv S since all its

barycentric coordinates are positive. The point p

5

is on the edge between v

1

and v

3

since the first and

third barycentric coordinates are positive and the rest are zero.

10. The point q

1

is inside conv S because the barycentric coordinates are all positive. The point q

2

is

outside conv S because it has one negative barycentric coordinate. The point q

4

is outside conv S for

the same reason. The point q

3

is on the edge between v

2

and v

3

because

()

31

44

0, , , 0

shows that q

3

is

a convex combination of v

2

and v

3

. The point q

5

is on the face containing the vertices v

1

, v

2

, and v

3

because

()

111

333

,,,0

shows that q

5

is a convex combination of those vertices.

11. a. False. In order for y to be a convex combination, the c’s must also all be nonnegative. See the

definition at the beginning of this section.

b. False. If S is convex, then conv S is equal to S. See Theorem 7.

c. False. For example, the union of two distinct points is not convex, but the individual points are.

12. a. True. See the definition prior to Theorem 7.

b. True. Theorem 9.

c. False. The points do not have to be distinct. For example, S might consist of two points in 

5

. A

point in conv S would be a convex combination of these two points. Caratheodory’s Theorem

requires n + 1 or fewer points.

8.3 • Solutions 465

!

13. If p, q ∈ f

(S), then there exist r, s ∈ S such that f

(r) = p and f

(s) = q. The goal is to show that the

line segment y = (1 − t)p + t

q, for 0 ≤ t ≤ 1, is in f

(S). Since

f

is linear,

y = (1 − t)p + t

q = (1 − t)

f

(r) + t

f

(s) = f

((1 − t)r + t

s)

Since S is convex, (1 − t)r + t

s ∈ S for 0 ≤ t ≤ 1. Thus y ∈ f (S) and f

(S) is convex.

14. Suppose r, s ∈ S and 0 ≤ t ≤ 1. Then, since f is a linear transformation,

f

[(1 − t)r + t

s ] = (1 − t)

f

(r) + t

f

(s)

But

f

(r) ∈ T and f

(s) ∈ T, so (1 − t)

f

(r) + t

f

(s) ∈ T since T is a convex set. It follows that

(1 − t)r + t

s ∈ S, because S consists of all points that f maps into T. This shows that S is convex.

15. It is straightforward to confirm the equations in the problem: (1)

11 1 1

1234

33 6 6

+++ =vv v vp

and

(2) v

1

– v

2

+ v

3

– v

4

= 0. Notice that the coefficients of v

1

and v

3

in equation (2) are positive. With

the notation of the proof of Caratheodory’s Theorem, d

1

= 1 and d

3

= 1. The corresponding

coefficients in equation (1) are

1

13

c=

and

1

36

c=

. The ratios of these coefficients are

1

11 3

/cd=

and

1

33 6

/cd=

. Use the smaller ratio to eliminate v

3

from equation (1). That is, add

1

6

−

times

equation (2) to equation (1):

11 11 11 11 1 1 1

1234124

36 36 66 66 6 2 3

()()()()=−++ +−++ = + +pvv v vvvv

To obtain the second combination, multiply equation (2) by –1 to reverse the signs so that d

2

and d

4

become positive. Repeating the analysis with these terms eliminates the v

4

term resulting in

11 1

123

26 3

=+ +pv v v

.

16.

1234

103 11

,,, ,

031 12

−

ªº ªº ªº ªº ªº

=====

«» «» «» «» «»

−

¬¼ ¬¼ ¬¼ ¬¼ ¬¼

vvvv p

It is straightforward to confirm the

equations in the problem: (1)

72 37

11

1234

121 121 121 11

+++=vv vvp

and (2) 10v

1

– 6v

2

+ 7v

3

– 11v

4

=

0.

Notice that the coefficients of v

1

and v

3

in equation (2) are positive. With the notation of the proof of

Caratheodory’s Theorem, d

1

= 10 and d

3

= 7. The corresponding coefficients in equation (1) are

1

1121

c=

and

37

3121

c=

. The ratios of these coefficients are

11

11 121 1210

/10cd=÷=

and

37 37

33 121 847

/7cd=÷=

. Use the smaller ratio to eliminate v

1

from equation (1). That is, add

1

1210

−

times equation (2) to equation (1):

10 72 6 37 7 3 3

11111

1234234

121 1210 121 1210 121 1210 11 1210 5 10 10

()()()()=−++ +−++ = + +pvv vvvvv

To obtain the second combination, multiply equation (2) by –1 to reverse the signs so that d

2

and d

4

become positive. Repeating the analysis with these terms eliminates the v

4

term resulting in

10 72 6 37 7 6

111114

1234123

121 121 121 121 121 121 11 121 11 11 11

()()()()=+ +−++ +−=++pvv vvvvv

466 CHAPTER 8 • The Geometry of Vector Spaces

!

17. Suppose A  B, where B is convex. Then, since B is convex, Theorem 7 implies that B contains

all convex combinations of points of B. Hence B contains all convex combinations of points of A.

That is, conv A  B.

18. Suppose A  B. Then A  B  conv B. Since conv B is convex, Exercise 17 shows that

conv A  conv B.

19 a. Since A  (A  B), Exercise 18 shows that conv A  conv (A  B). Similarly,

conv B  conv (A  B). Thus, [(conv A)  (conv B)]  conv (A  B).

b. One possibility is to let A be two adjacent corners of a square and B be the other two corners.

Then (conv A)  (conv B) consists of two opposite sides of the square, but conv (A  B) is the

whole square.

20. a. Since (A  B)  A, Exercise 18 shows that conv (A  B)  conv A. Similarly,

conv (A  B)  conv B. Thus, conv (A  B)  [(conv A)  (conv B)].

b. One possibility is to let A be a pair of opposite vertices of a square and let B be the other pair

of opposite vertices. Then conv A and conv B are intersecting diagonals of the square.

A  B is the empty set, so conv (A  B) must be empty, too. But conv A  conv B contains

the single point where the diagonals intersect. So conv (A  B) is a proper subset of conv A

 conv B.

21. 22.

!

23. g(t) = (1 − t)f

0

(t) + t

f

1

(t)

= (1 − t)[(1 − t)p

0

+ t

p

1

] + t[(1 − t)p

1

+ t

p

2

] = (1 − t)

2

p

0

+ 2t(1 − t)p

1

+ t

2

p

2

.

The sum of the weights in the linear combination for g is (1 − t)

2

+ 2t(1 − t) + t

2

, which equals

(1 − 2t + t

2

) + (2t − 2t

2

) + t

2

= 1. The weights are each between 0 and 1 when 0 ≤ t ≤ 1, so g(t) is in

conv{p

0

, p

1

, p

2

}.

24. h(t) = (1 − t)g

1

(t) + t

g

2

(t). Use the representation for g

1

(t) from Exercise 23, and the analogous

representation for g

2

(t), based on the control points p

1

, p

2

, and p

3

, and obtain

h(t) = (1 − t)[(1 − t)

2

p

0

+ 2t(1 − t)p

1

+ t

2

p

2

] + t

[(1 − t)

2

p

1

+ 2t(1 − t)p

2

+ t

2

p

3

]

= (1 − t)

3

p

0

+ 2t(1 − 2t + t

2

)p

1

+ (t

2

− t

3

)p

2

+ t

(1 − 2t + t

2

)

p

1

+ 2t

2

(1 − t)p

2

+ t

3

p

3

= (1 − 3t + 3t

2

− t

3

)p

0

+ (2t − 4t

2

+ 2t

3

)p

1

+ (t

2

− t

3

)p

2

+ (t − 2t

2

+ t

3

)p

1

+ (2t

2

− 2t

3

)p

2

+ t

3

p

3

= (1 − 3t + 3t

2

− t

3

)p

0

+ (3t − 6t

2

+ 3t

3

)p

1

+ (3t

2

− 3t

3

)p

2

+ t

3

p

3

By inspection, the sum of the weights in this linear combination is 1, for all t. To show that the

weights are nonnegative for 0 ≤ t ≤ 1, factor the coefficients and write

!

"

!

#

!

$

!

()

1

02

f!

()

1

12

f

()

1

2

g!

!

"

!

#

!

$

!

()

3

04

f!

()

3

14

f

()

3

4

g

8.4 • Solutions 467

!

h(t) = (1 − t)

3

p

0

+ 3t(1 − t)

2

p

1

+ 3t

2

(1 − t)p

2

+ t

3

p

3

for 0 ≤ t ≤ 1

Thus, h(t) is in the convex hull of the control points p

0

, p

1

, p

2

, and p

3

.

8.4 SOLUTIONS

Notes:

In this section lines and planes are generalized to higher dimensions using the notion of

hyperplanes. Important topological ideas such as open, closed, and compact sets are introduced.

1. Let

12

13

and .

41

−

ªº ªº

==

«» «»

¬¼ ¬¼

vv

Then

21

314

.

14 3

−

ª

ºªºª º

−=−=

«

»«»« »

−

¬

¼¬¼¬ ¼

vv

Choose n to be a vector orthogonal to

21

−vv

, for example let

3

4

ªº

=«»

¬¼

n

. Then f

(x

1

, x

2

) = 3x

1

+ 4x

2

and d = f

(v

1

) = 3(–1) + 4(4) = 13.

This is easy to check by verifying that f

(v

2

) is also 13.

2. Let

12

and .

41

−

ªº ª º

==

«» « »

−

¬¼ ¬ ¼

vv

Then

21

21 3

.

14 5

−−

ª

ºªºª º

−=−=

«

»«»« »

−−

¬

¼¬¼¬ ¼

vv

Choose n to be a vector orthogonal

to

21

−vv

, for example let

5

3

ªº

=«»

−

¬¼

n

. Then f

(x

1

, x

2

) = 5x

1

− 3x

2

and d = f

(v

1

) = 5(1) − 3(4) =

7−

.

3. a. The set is open since it does not contain any of its boundary points.

b. The set is closed since it contains all of its boundary points.

c. The set is neither open nor closed since it contains some, but not all, of its boundary points.

d. The set is closed since it contains all of its boundary points.

e. The set is closed since it contains all of its boundary points.

4. a. The set is closed since it contains all of its boundary points.

b. The set is open since it does not contain any of its boundary points.

c. The set is neither open nor closed since it contains some, but not all, of its boundary points.

d. The set is closed since it contains all of its boundary points.

e. The set is open since it does not contain any of its boundary points.

5. a. The set is not compact since it is not closed, however it is convex.

b. The set is compact since it is closed and bounded. It is also convex.

c. The set is not compact since it is not closed, however it is convex.

d. The set is not compact since it is not bounded. It is not convex.

e. The set is not compact since it is not bounded, however it is convex.

6. a. The set is compact since it is closed and bounded. It is not convex.

b. The set is not compact since it is not closed. It is not convex.

468 CHAPTER 8 • The Geometry of Vector Spaces

!

c. The set is not compact since it is not closed, however it is convex.

d. The set is not compact since it is not bounded. It is convex.

!!"# The set is not compact since it is not closed. It is not convex.

7. a. Let

12 3

12 1

1, 4, 2,

31 5

a

b

c

−

ªº ªº ª º ªº

«» «» « » «»

== =−=

«» «» « » «»

¬¼ ¬¼ ¬ ¼ ¬¼

vv v n

and compute the translated points

21 31

12

3, 3

22

−

ª

ºªº

«

»«»

−=−=−

«

»«»

«

»«»

−

¬

¼¬¼

vv vv

.

To solve the system of equations (v

2

– v

1

) · n = 0 and (v

3

– v

1

) · n = 0, reduce the augmented

matrix for a system of two equations with three variables.

[1 3 2] 0,

a

b

c

ªº

«»

−=

«»

¬¼

[2 3 2] 0

a

b

c

ªº

«»

−− =

«»

¬¼

.

Row operations show that

1320 1000

~

2320 0320

−

ªºªº

«»«»

−− −

¬¼¬¼

. A suitable normal vector

is

0

2

3

ªº

«»

=«»

«»

¬¼

n

.

b. The linear functional is

123 2 3

(, , ) 2 3fxx x x x=+

, so d = f

(1, 1, 3) = 2 + 9 = 11. As a check,

evaluate f at the other two points on the hyperplane: f

(2, 4, 1) = 8 + 3 = 11 and

f

(–1 , –2 , 5) = – 4 + 15 = 11.

8. a. Find a vector in the null space of the transpose of [v

2

-v

1

v

3

-v

1

]. For example, take

4

3.

6

ªº

«»

=«»

«»

−

¬¼

n

b. f

(x) = 4x

1

+ 3x

2

− 6x

3

, d = f

(v

1

) =−8

9. a. Find a vector in the null space of the transpose of [v

2

-v

1

v

3

-v

1

v

4

-v

1

]. For example, take

n =

3

1.

2

1

ªº

«»

−

«»

¬¼

b. f

(x) = 3x

1

− x

2

+ 2x

3

+ x

4

, d = f

(v

1

) =5

#

8.4 • Solutions 469

!

10. a. Find a vector in the null space of the transpose of [v

2

-v

1

v

3

-v

1

v

4

-v

1

]. For example, take

n =

2

3.

5

1

−

ªº

«»

−

«»

¬¼

b. f

(x) = −2x

1

+ 3x

2

− 5x

3

+ x

4

, d = f

(v

1

) = 4

11.

12 3

2; 0 2; 5 2; 2 2; 2.==< => =−<=np n0 nv nv nv

<< < < <

Hence v

2

is on the same side of H

as 0, v

1

is on the other side, and v

3

is in H.

12. Let H = [ f : d ], where f (x

1

, x

2

, x

3

) = 3x

1

+ x

2

− 2x

3

. f (a

1

) = −5, f (a

2

) = 4. f (a

3

) =3, f (b

1

) =7,

f (b

2

) = 4, and f (b

3

) = 6. Choose d = 4 so that all the points in A are in or on one side of H and all

the points in B are in or on the other side of H. There is no hyperplane parallel to H that strictly

separates A and B because both sets have a point at which f takes on the value of 4. There may be

(and in fact is) a hyperplane that is not parallel to H that strictly separates A and B.

13. H

1

= {x : n

1

⋅ x = d

1

} and H

2

= {x : n

2

⋅ x = d

2

}. Since p

1

∈ H

1

, d

1

= n

1

⋅ p

1

= 4. Similarly,

d

2

= n

2

⋅ p

2

= 22. Solve the simultaneous system [1 2 4 2]x = 4 and [2 3 1 5]x = 22:

12424 1 0 10 4 32

~

231522 0 1 7 114

−

ªºª º

«»« »

−−

¬¼¬ ¼

The general solution provides one set of vectors, p, v

1

, and v

2

. Other choices are possible.

34 3142

32 10 4

14 7 1 ,

010

00 1

xx xx

−

ªº ªº ªº

«» «» «»

−−

«» «» «»

=+ + =++

«» «» «»

¬¼ ¬¼ ¬¼

xpvv

where

12

32 10 4

14 7 1

,,

010

00 1

−

ª

ºªº ªº

«

»«» «»

−−

«

»«» «»

===

«

»«» «»

«

»«» «»

¬

¼¬¼ ¬¼

pvv

Then H

1

∩ H

2

= {x : x = p + x

3

v

1

+ x

4

v

2

}.

14. Since each of F

1

and F

2

can be described as the solution sets of A

1

x=b

1

and A

2

x=b

2

respectively,

where A

1

and A

2

have rank 2, their intersection is described as the solution set to

11

22

A

ªº ªº

=

«» «»

¬¼ ¬¼

b

xb

.

Since

1

2

2rank 4

A

§·

ªº

≤≤

¨¸

«»

¬¼

©¹

, the solution set will have dimensions 6 −2=4,6 −3=3, or 6 −4=2.

15. f (x

1

, x

2

, x

3

) =Ax= x

1

– 3x

2

+ 4x

3

– 2x

4

and d =b= 5

16. f (x

1

, x

2

, x

3

) = Ax= 2x

1

+ 5x

2

– 3x

3

+ 6x

5

and d = b= 0

17. Since by Theorem 3 in Section 6.1, Row B=(Nul B)

⊥

, choose a nonzero vector n∈ Nul B . For

example take n=

1

2

1

ªº

«»

−

«»

¬¼

. Then f (x

1

, x

2

, x

3

) = x

1

– 2x

2

+ x

3

and d = 0

470 CHAPTER 8 • The Geometry of Vector Spaces

!

18. Since by Theorem 3, Section 6.1, Row B=(Nul B)

⊥

, choose a nonzero vector n∈ Nul B . For

example take n=

11

4

1

−

ªº

«»

¬¼

. Then f (x

1

, x

2

, x

3

) = –11x

1

+ 4x

2

+ x

3

and d = 0

19. Theorem 3 in Section 6.1 says that (Col B)

⊥

= Nul B

T

. Since the two columns of B are clearly linear

independent, the rank of B is 2, as is the rank of B

T

. So dim Nul B

T

= 1, by the Rank Theorem, since

there are three columns in B

T

. This means that Nul B

T

is one-dimensional and any nonzero vector n

in Nul B

T

will be orthogonal to H and can be used as its normal vector. Solve the linear system

B

T

x = 0 by row reduction to find a basis for Nul B

T

:

1470 1050

~

0260 0130

−

ªºªº

«»«»

−−

¬¼¬¼



5

3

1

−

ª

º

«

»

=

«

»

«

»

¬

¼

n

!Now, let f

(x

1

, x

2

, x

3

) = –5x

1

+ 3x

2

+ x

3

. Since the hyperplane H is a subspace, it goes through the

origin and d must be 0.

The solution is easy to check by evaluating f at each of the columns of B.

20. Since by Theorem 3, Section 6.1, Col B=(Nul B

T

)

⊥

, choose a nonzero vector n∈ Nul B

T

. For

example take n=

6

2

1

−

ªº

«»

¬¼

. Then f (x

1

, x

2

, x

3

) = – 6x

1

+ 2x

2

+ x

3

and d = 0

21. a. False. A linear functional goes from 

n

to . See the definition at the beginning of this section.

b. False. See the discussion of (1) and (4). There is a 1×n matrix A such that f (x) = Ax for all x in



n

. Equivalently, there is a point n in 

n

such that f (x) = n ⋅ x for all x in 

n

.

c. True. See the comments after the definition of strictly separate.

d. False. See the sets in Figure 4.

22. a. True. See the statement after (3).

b. False. The vector n must be nonzero. If n = 0, then the given set is empty if d ≠ 0 and the set

is all of 

n

if d = 0.

c. False. Theorem 12 requires that the sets A and B be convex. For example, A could be the

boundary of a circle and B could be the center of the circle.

d. False. Some other hyperplane might strictly separate them. See the caution at the end of

Example 8.

23. Notice that the side of the triangle closest to p is

23

vv

. A vector orthogonal to

23

vv

.is n=

3

2

ªº

«»

−

¬¼

.

Take f (x

1

, x

2

) = 3x

1

– 2x

2

. Then f (v

2

) = f (v

3

)=9 and f (p)=10 so any d satisfying 9 < d < 10 will

work. There are other possible answers.

!

8.4 • Solutions 471

!

24. Notice that the side of the triangle closest to p is

13

vv

A vector orthogonal to

13

vv

is n=

2

3

−

ªº

«»

¬¼

.

Take f (x

1

, x

2

) = −2x

1

+3x

2

. Then f (v

1

) = f (v

3

)=4 and f (p)=5 so any d satisfying 4 < d < 5 will

work. There are other possible answers.

25. Let L be the line segment from the center of B(0, 3) to the center of B(p, 1). This is on the line

through the origin in the direction of p. The length of L is (4

2

+ 1

2

)

1/2

4.1231. This exceeds the

sum of the radii of the two disks, so the disks do not touch. If the disks did touch, there would be no

hyperplane (line) strictly separating them, but the line orthogonal to L through the point of tangency

would (weakly) separate them. Since the disks are separated slightly, the hyperplane need not be

exactly perpendicular to L, but the easiest one to find is a hyperplane H whose normal vector is p.

So define f by f

(x) = p

⋅

x.

To find d, evaluate

f

at any point on L that is between the two disks. If the disks were tangent, that

point would be three-fourths of the distance between their centers, since the radii are 3 and 1. Since

the disks are slightly separated, the distance is about 4.1231. Three-fourths of this distance is greater

than 3, and one-fourth of this distance is greater than 1. A suitable value of d is f

(q), where q =

(.25)0 + (.75)p = (3, .75). So d = p

⋅

q = 4(3) + 1(.75) = 12.75.

26. The normal to the separating hyperplane has the direction of the line segment between p and q. So,

let n = p − q =

4

2

ªº

«»

−

¬¼

. The distance between p and q is

20

, which is more than the sum of the radii

of the two balls. The large ball has center q. A point three-fourths of the distance from q to p will be

greater than 3 units from q and greater than 1 unit from p. This point is

x = .75p + .25q =

625.0

.75 .25

131.5

ªº ªº ª º

+=

«» «» « »

¬¼ ¬¼ ¬ ¼

Compute n

⋅

x = 17. The desired hyperplane is :4 2 17

xxy

y



½

ªº

°

−=

®

¾

«»

°

¬¼

¯¿

.

27. Exercise 2(a) in Section 8.3 gives one possibility. Or let S = {(x, y) : x

2

y

2

= 1 and y > 0}. Then

conv S is the upper (open) half-plane.

28. One possibility is B = {(x, y) : x

2

y

2

= 1 and y > 0} and A = {(x, y) : |

x

| ≤ 1 and y = 0}.

29. Let x, y ∈ B(

p,

δ

) and suppose z = (1 − t)

x + t

y, where 0 ≤ t ≤ 1. Then

||z − p|| = ||

[(1 − t)

x + t

y] − p|| = ||(1 − t)(x − p) + t

(y − p)||

≤ (1 − t)

||x − p|| + t

||y − p|| < (1 − t)

δ

+ t

δ

=

δ

where the first inequality comes from the Triangle Inequality (Theorem 17 in Section 6.7) and the

second inequality follows from x, y ∈ B(

p,

δ

). It follows that z ∈ B(

p,

δ

) and B(

p,

δ

) is convex.

30. Let S be a bounded set. Then there exists a

δ

> 0 such that S



B(0,

δ

). But B(0,

δ

) is

convex by Exercise 29, so Theorem 9 in Section 8.3 (or Exercise 17 in Section 8.3) implies that

conv S



B(

p,

δ

) and conv S is bounded.

472 CHAPTER 8 • The Geometry of Vector Spaces

!

8.5 SOLUTIONS

Notes:

A polytope is the convex hull of a finite number of points. Polytopes and simplices are important

in linear programming, which has numerous applications in engineering design and business

management. The behavior of functions on polytopes is studied in this section.

1. Evaluate each linear functional at each of the three extreme points of S. Then select the extreme

point(s) that give the maximum value of the functional.

a. f

(p

1

) = 1, f

(p

2

) = –1, and f

(p

3

) = –3, so m = 1 at p

1

.

b. f

(p

1

) = 1, f

(p

2

) = 5, and f

(p

3

) = 1, so m = 5 at p

2

.

c. f

(p

1

) = –3, f

(p

2

) = –3, and f

(p

3

) = 5, so m = 5 at p

3

"!

!"# Evaluate each linear functional at each of the three extreme points of S. Then select the point(s) that

give the maximum value of the functional.

a. f

(p

1

) = −1, f

(p

2

) = 3, and f

(p

3

) = 3, so m = 3 on the set conv

{p

2

, p

3

}.

b. f

(p

1

) = 1, f

(p

2

) = 1, and f

(p

3

) = −1, so m = 1 on the set conv

{p

1

, p

2

}.

c. f

(p

1

) = –1, f

(p

2

) = –3, and f

(p

3

) = 0, so m = 0 at p

3

"!

3. Evaluate each linear functional at each of the three extreme points of S. Then select the point(s) that

give the minimum value of the functional.

a. f

(p

1

) = 1, f

(p

2

) = –1, and f

(p

3

) = –3, so m = –3 at the point p

3

b. f

(p

1

) = 1, f

(p

2

) = 5, and f

(p

3

) = 1, so m = 1 on the set conv

{p

1

, p

3

}.

c. f

(p

1

) = –3, f

(p

2

) = –3, and f

(p

3

) = 5, so m = –3 on the set conv

{p

1

, p

2

}.

4. Evaluate each linear functional at each of the three extreme points of S. Then select the point(s) that

give the maximum value of the functional.

a. f

(p

1

) = −1, f

(p

2

) = 3, and f

(p

3

) = 3, so m = –1 at the point p

1

.

b. .f

(p

1

) = 1, f

(p

2

) = 1, and f

(p

3

) = −1, so m = –1 at the point p

3

.

c. f

(p

1

) = –1, f

(p

2

) = –3, and f

(p

3

) = 0, so m = –3 at the point p

2

.

5. The two inequalities are (a) x

1

+ 2x

2

≤ 10 and (b) 3x

1

+ x

2

≤ 15. Line (a) goes from (0,5) to (10,0).

Line (b) goes from (0,15) to (5,0). One vertex is (0,0). The x

1

-intercepts (when x

2

= 0) are 10 and 5,

so (5,0) is a vertex. The x

2

-intercepts (when x

1

= 0) are 5 and 15, so (0,5) is a vertex. The two lines

intersect at (4,3) so (4,3) is a vertex. The minimal representation is 0540

,,,

0035



½

ª

ºªºªºªº

°

®

¾

«

»«»«»«»

°

¬

¼¬¼¬¼¬¼

¯¿

8.5 • Solutions 473

!

6. The two inequalities are (a) 2x

1

+ 3x

2

≤ 18 and (b) 4x

1

+ x

2

≤ 16. Line (a) goes from (0,6) to (9,0).

Line (b) goes from (0,16) to (4,0). One vertex is (0,0). The x

1

-intercepts (when x

2

= 0) are 9 and 4, so

(4,0) is a vertex. The x

2

-intercepts (when x

1

= 0) are 6 and 16, so (0,6) is a vertex. The two lines

intersect at (3,4) so (3,4) is a vertex. The minimal representation is 0430

,,,

0046



½

ª

ºªºªºªº

°

®

¾

«

»«»«»«»

°

¬

¼¬¼¬¼¬¼

¯¿

7. The three inequalities are (a) x

1

+ 3x

2

≤ 18, (b) x

1

+ x

2

≤ 10, and (c) 4x

1

+ x

2

≤ 28. Line (a) goes from

(0,6) to (18,0). Line (b) goes from (0,10) to (10,0). And line (c) goes from (0,28) to (7,0). One

vertex is (0,0). The x

1

-intercepts (when x

2

= 0) are 18, 10, and 7, so (7,0) is a vertex. The x

2

-

intercepts (when x

1

= 0) are 6, 10, and 28, so (0,6) is a vertex. All three lines go through (6,4), so

(6,4) is a vertex. The minimal representation is 0760

,,,

0046



½

ª

ºªºªºªº

°

®

¾

«

»«»«»«»

°

¬

¼¬¼¬¼¬¼

¯¿

.

8. The three inequalities are (a) 2x

1

+ x

2

≤ 8, (b) x

1

+ x

2

≤ 6, and (c) x

1

+ 2x

2

≤ 7. Line (a) goes from

(0,8) to (4,0). Line (b) goes from (0,6) to (6,0). And line (c) goes from (0,3.5) to (7,0). One vertex is

(0,0). The x

1

-intercepts (when x

2

= 0) are 4, 6, and 7, so (4,0) is a vertex. The x

2

-intercepts (when x

1

= 0) are 8, 6, and 3.5, so (0,3.5) is a vertex. All three lines go through (3,2), so (3,2) is a vertex. The

minimal representation is 043 0

,,,

0023.5

½

ªºªºªºª º

°°

®¾

«»«»«»« »

°°

¬¼¬¼¬¼¬ ¼

¯¿

9. The origin is an extreme point, but it is not a vertex. It is an

extreme point since it is not in the interior of any line segment

that lies in S. It is not a vertex since the only supporting

hyperplane (line) containing the origin also contains the line

segment from (0,0) to (3,0).

10. One possibility is a ray. It has an extreme point at one end.

11. One possibility is to let S be a square that includes part of the boundary but not all of it. For example,

include just two adjacent edges. The convex hull of the profile P is a triangular region.

12. a. f

0

(S

5

) = 6, f

1

(S

5

) = 15, f

2

(S

5

) = 20, f

3

(S

5

) = 15, f

4

(S

5

) = 6, and 6 − 15 + 20 − 15 + 6 = 2.

b.

f

0

f

1

f

2

f

3

f

4

S

1

2

S

2

3 3

S

3

4 6 4

S

4

5 10 10 5

S

5

6 15 20 15 6

!

""#$%!#!!&!

474 CHAPTER 8 • The Geometry of Vector Spaces

!

1

() ,

1

n

k

n

fS k

+

§·

=¨¸

+

©¹

where

!

!( )!

aa

bba b

§·

=

¨¸ −

©¹

is the binomial coefficient.

13. a. To determine the number of k-faces of the 5-dimensional hypercube C

5

, look at the pattern that is

followed in building C

4

from C

3

. For example, the 2-faces in C

4

include the 2-faces of C

3

and

the 2-faces in the translated image of C

3

. In addition, there are the 1-faces of C

3

that are

“stretched” into 2-faces. In general, the number of k-faces in C

n

equals twice the number of k-

faces in C

n – 1

plus the number of (k – 1)-faces in C

n – 1

. Here is the pattern: f

k

(C

n

) = 2 f

k

(C

n – 1

)

+ f

k – 1

(C

n – 1

). For k = 0, 1, …, 4, and n = 5, this gives f

0

(C

5

) = 32, f

1

(C

5

) = 80, f

2

(C

5

) = 80,

f

3

(C

5

) = 40, and f

4

(C

5

) = 10. These numbers satisfy Euler’s formula since, 32 − 80 + 80 − 40 +

10 = 2.

b. The general formula is

!

()2 ,where !( )!

nnk

k

na

a

fC kb

ba b

−

§· §·

==

¨¸ ¨¸ −

©¹ ©¹

is the binomial coefficient.!

14. a.!X

1

is a line segment X

2

is a parallelogram

b. f

0

(X

3

) = 6, f

1

(X

3

) = 12, f

2

(X

3

) = 8. X

3

is an octahedron.

c. f

0

(X

4

) = 8, f

1

(X

4

) = 24, f

2

(X

4

) = 32, f

3

(X

4

) = 16, 8 − 24 + 32 − 16 = 0

d.

1

()2 ,

1

nk

k

n

fX k

+§·

=¨¸

+

©¹

0 ≤ k ≤ n

−1, where

!

!( )!

aa

bba b

§·

=

¨¸ −

©¹

is the binomial coefficient.

15. a. f

0

(P

n

) = f

0

(Q) +

1

b. f

k

(P

n

) = f

k

(Q) + f

k

−

1

(Q)

c. f

n

−

1

(P

n

) = f

n

−

2

(Q) +

1

16. a. True. See the definition at the beginning of this section.

b. True. See the definition after Example 1.

c. False. S must be compact. See Theorem 15.

d. True. See the comment after Fig. 7.

17. a. False. It has six facets (faces).

b. True. See Theorem 14.

c. False. The maximum is always attained at some extreme point, but there may be other points that

are not extreme points at which the maximum is attained. See Theorem 16.

d. True. Follows from Euler’s formula with n = 2.

18. Let v be an extreme point of the convex set S and let T = {y ∈ S : y ≠ v}. If y and z are in T, then

S

⊆

yz

since S is convex. But since v is an extreme point of S, v ∉

,yz

so

.Tyz 

Thus T is

convex.

!" #

"!

#

"!

#

#!

8.6 • Solutions 475

!

Conversely, suppose v ∈ S, but v is not an extreme point of S. Then there exist y and z in S such

that v ∈

yz

, with v ≠ y and v ≠ z. It follows that y and z are in T, but

.Tyz 

Hence T is not

convex.

19. Let S be convex and let x ∈ cS + dS, where c > 0 and d > 0. Then there exist s

1

and s

2

in S such that

x = cs

1

+ ds

2

. But then

12 1 2

()

cd

cd cd

§·

=+ =+ +

¨¸

++

©¹

xss s s

.

Now show that the expression on the right side is a member of (c + d)S.

For the converse, pick an typical point in (c + d)S and show it is in cS + dS.

20. For example, let S = {1, 2} in

1



. Then 2S = {2, 4}, 3S = {3, 6} and (2 + 3)S = {5, 10}.

However, 2S + 3S = {2, 4} + {3, 6} = {2 + 3, 4 + 3, 2 + 6, 4 + 6} = {5, 7, 8, 10} ≠ (2 + 3)S.

21. Suppose A and B are convex. Let x, y ∈ A

+

B. Then there exist a, c ∈ A and b, d ∈ B such that

x = a + b and y = c + d. For any t such that 0 ≤ t ≤ 1, we have

[][ ]

(1 ) (1 )( ) ( )

(1 ) (1 )

tt t t

tt tt

=−+=−++ +

=−++−+

wxy abcd

ac bd

But (1 − t)a + tc ∈ A since A is convex, and (1 − t)b + td ∈ B since B is convex. Thus w is in A + B,

which shows that A + B is convex.

22. a. Since each edge belongs to two facets, kr is twice the number of edges: k

r = 2e. Since each edge

has two vertices, s

v = 2e.

b. v − e + r = 2, so

22 11 11

2

ee

sse

kk

e−+=+=+

c. A polygon must have at least three sides, so k ≥ 3. At least three edges meet at each vertex,

so s ≥ 3. But both k and s cannot both be greater than 3, for then the left side of the equation

in (b) could not exceed 1!2.

When k = 3, we get

11 1

6

,

se

−=

so s = 3, 4, or 5. For these values, we get e = 6, 12, or 30,

corresponding to the tetrahedron, the octahedron, and the icosahedron, respectively.

When s = 3, we get

11 1

6

,

e

k

−=

so k = 3, 4, or 5 and e = 6, 12, or 30, respectively.

These values correspond to the tetrahedron, the cube, and the dodecahedron.

8.6 SOLUTIONS

Notes:

This section moves beyond lines and planes to the study of some of the curves that are used to

model surfaces in engineering and computer aided design. Notice that these curves have a matrix

representation.

1. The original curve is x(t) = (1 – t)

3

p

0

+ 3t(1 – t)

2

p

1

+ 3t

2

(1 – t)p

2

+ t

3

p

3

(0 < t < 1). Since the

curve is determined by its control points, it seems reasonable that to translate the curve, one

should translate the control points. In this case, the new Bézier curve y(t) would have the

equation

476 CHAPTER 8 • The Geometry of Vector Spaces

!

y(t) = (1 – t)

3

(p

0

+ b) + 3t(1 – t)

2

(p

1

+ b) + 3t

2

(1 – t)(p

2

+ b) + t

3

(p

3

+ b)

= (1 – t)

3

p

0

+ 3t(1 – t)

2

p

1

+ 3t

2

(1 – t)p

2

+ t

3

p

3

+ (1 – t)

3

b + 3t(1 – t)

2

b + 3t

2

(1 – t)b + t

3

b

A routine algebraic calculation verifies that (1 – t)

3

+ 3t(1 – t)

2

+ 3t

2

(1 – t) + t

3

= 1 for all t.

Thus y(t) = x(t) + b for all t, and translation by b maps a Bézier curve into a Bézier curve.

2. a. Equation (15) reveals that each polynomial weight is nonnegative for 0 < t < 1, since 4 − 3t >

0. For the sum of the coefficients, use (15) with the first term expanded: 1 – 3t + 6t

2

− t

3

.

The 1 here plus the 4 and 1 in the coefficients of p

1

and p

2

, respectively, sum to 6, while the

other terms sum to 0. This explains the 1/6 in the formula for x(t), which makes the

coefficients sum to 1. Thus, x(t) is a convex combination of the control points for 0 < t < 1.

b. Since the coefficients inside the brackets in equation (14) sum to 6, it follows that

[]

332 32 3

11

66

6(1)(364)(3331)ttt tttt

ª

º

==−+−++−++++

¬

¼

bb b b bb

and hence x(t) + b

may be written in a similar form, with p

i

replaced by p

i

+ b for each i. This shows that

x(t) + b is a cubic B-spline with control points p

i

+ b for i = 0, …, 3.

3. a. x'

(t) = (–3 + 6t – 3t

2

)p

0

+ (3 –12t + 9t

2

)p

1

+ (6t – 9t

2

)p

2

+ 3t

2

p

3

, so x'

(0) = –3p

0

+ 3p

1

=3(p

1

– p

0

),

and x'

(1) = –3p

2

+ 3p

3

= 3(p

3

– p

2

). This shows that the tangent vector x'

(0) points in the

direction from p

0

to p

1

and is three times the length of p

1

– p

0

. Likewise, x'

(1) points in the

direction from p

2

to p

3

and is three times the length of p

3

– p

2

. In particular, x'

(1) = 0 if and only

if p

3

= p

2

.

b. x''

(t) = (6 – 6t)p

0

+ (–12 + 18t)p

1

+ (6 – 18t)p

2

+ 6tp

3

, so that

x''

(0) = 6p

0

– 12p

1

+ 6p

2

= 6(p

0

– p

1

) + 6(p

2

– p

1

)

and x''

(1) = 6p

1

– 12p

2

+ 6p

3

= 6(p

1

– p

2

) + 6(p

3

– p

2

)

For a picture of x''

(0), construct a coordinate system with the origin at p

1

, temporarily, label p

0

as p

0

− p

1

, and label p

2

as p

2

− p

1

. Finally, construct a line from this new origin through the sum

of p

0

− p

1

and p

2

− p

1

, extended out a bit. That line points in the direction of x''

(0).

4. a. x'

(t) =

()()()

2222

01 23

1

6

363 912 963 3tt t t tt t

ªº

−+−+−+−++ +

¬¼

pp pp

x'

(0) =

()

20

1

2

−pp

and x'

(1) =

()

31

1

2

−pp

(Verify that, in the first part of Fig. 10, a line drawn through p

0

and p

2

is parallel to the

tangent line at the beginning of the B-spline.)

When x'

(0) and x'

(1) are both zero, the figure collapses and the convex hull of the set of

control points is the line segment between p

0

and p

3

, in which case x(t) is a straight line.

Where does x(t) start? In this case,

x(t) =

32 32

03

1

6

(4 6 2) (4 6 4)tt tt

ªº

−++ + −+

¬¼

pp

x(0) =

12

03

33

+pp

and x(1) =

21

03

33

+pp

0 = p

1

p

0

– p

1

p

2

– p

1

w

1

01 21 6

()()(0)

′′

=−+−=wpp pp x

8.6 • Solutions 477

!

The curve begins closer to p

3

and finishes closer to p

0

. Could it turn around during its travel?

Since x'

(t) = 2t(1 − t)(p

0

− p

3

), the curve travels in the direction p

0

− p

3

, so when x'

(0) = x'

(1)

= 0, the curve always moves away from p

3

toward p

0

for 0 < t < 1.

b. x''

(t) = (1 – t)p

0

+ (–2 + 3t)p

1

+ (1 – 3t)p

2

+ tp

3

x''

(0) = p

0

– 2p

1

+ p

2

= (p

0

− p

1

) + (p

2

− p

1

)

and x''

(1) = p

1

– 2p

2

+ p

3

= (p

1

− p

2

) + (p

3

− p

2

)

For a picture of x''

(0), construct a coordinate system with the origin at p

1

, temporarily, label p

0

as

p

0

− p

1

, and label p

2

as p

2

− p

1

. Finally, construct a line from this new origin to the sum of p

0

− p

1

and p

2

− p

1

. That segment represents x''

(0).

For a picture of x''

(1), construct a coordinate system with the origin at p

2

, temporarily, label p

1

as

p

1

− p

2

, and label p

3

as p

3

− p

2

. Finally, construct a line from this new origin to the sum of

p

1

− p

2

and p

3

− p

2

. That segment represents x''

(1).

5. a. From Exercise 3(a) or equation (9) in the text,

x'

(1) = 3(p

3

– p

2

)

Use the formula for x'(0), with the control points from y(t), and obtain

y'

(0) = –3p

3

+ 3p

4

= 3(p

4

– p

3

)

For C

1

continuity, 3(p

3

– p

2

) = 3(p

4

– p

3

), so p

3

= (p

4

+ p

2

)/2, and p

3

is the midpoint of the

line segment from p

2

to p

4

.

b. If x'

(1) = y'

(0) = 0, then p

2

= p

3

and p

3

= p

4

. Thus, the “line segment” from p

2

to p

4

is just

the point p

3

. [Note: In this case, the combined curve is still C

1

continuous, by definition.

However, some choices of the other control points, p

0

, p

1

, p

5

, and p

6

can produce a curve

with a visible “corner” at p

3

, in which case the curve is not G

1

continuous at p

3

.]

6. a. With x(t) as in Exercise 2,

x(0) = (p

0

+ 4p

1

+ p

2

)/6 and x(1) = (p

1

+ 4p

2

+ p

3

)/6

Use the formula for x(0), but with the shifted control points for y(t), and obtain

y(0) = (p

1

+ 4p

2

+ p

3

)/6

This equals x(1), so the B-spline is G

0

continuous at the join point.

b. From Exercise 4(a),

x'

(1) = (p

3

– p

1

)/2 and x'

(0) = (p

2

– p

0

)/2

Use the formula for x'

(0) with the control points for y(t), and obtain

y'

(0) = (p

3

– p

1

)/2 = x'

(1)

Thus the B-spline is C

1

continuous at the join point.

7. From Exercise 3(b),

x!!

(0) = 6(p

0

– p

1

) + 6(p

2

– p

1

) and x!!

(1) = 6(p

1

– p

2

) + 6(p

3

– p

2

)

Use x!!

(0) with the control points for y(t), to get

y!!

(0) = 6(p

3

– p

4

) + 6(p

5

– p

4

)

p

2

= 0

p

3

– p

2

p

1

– p

2

w

12 32

()()(1)

′′

=−+−=wpp pp x

478 CHAPTER 8 • The Geometry of Vector Spaces

!

Set x''

(1) = y''

(0) and divide by 6, to get

(p

1

– p

2

) + (p

3

– p

2

) = (p

3

– p

4

) + (p

5

– p

4

) (*)

Since the curve is C

1

continuous at p

3

, the point p

3

is the midpoint of the segment from p

2

to p

4

, by

Exercise 5(a). Thus

1

324

2

()=+ppp

, which leads to p

4

– p

3

= p

3

– p

2

. Substituting into (*) gives

(p

1

– p

2

) + (p

3

– p

2

) = –(p

3

– p

2

) + p

5

– p

4

(p

1

– p

2

) + 2(p

3

– p

2

) + p

4

= p

5

Finally, again from C

1

continuity, p

4

= p

3

+ p

3

– p

2

. Thus,

p

5

= p

3

+ (p

1

– p

2

) + 3(p

3

– p

2

)

So p

4

and p

5

are uniquely!"#$#%&'(#"!)*!!

+

,!!

-

,!.("!!

/

0!!!1(2*!!

3

!4.(!)#!4567#(!.%)'$%.%'2*0!

8. From Exercise 4(b), x!!

(0) = p

0

– 2p

1

+ p

2

and x!!

(1) = p

1

– 2p

2

+ p

3

. Use the formula for x!!

(0), with

the shifted control points for y(t), to get

y!!

(0) = p

1

– 2p

2

+ 2p

3

= x!!

(1)

Thus the curve has C

2

continuity at x(1).

9. Write a vector of the polynomial weights for x(t), expand the polynomial weights and factor the

vector as M

B

u(t):

234

2

234

3

34

4

14 6 4 1

412 12 4 0412124

00 6126

6126

00 0 44

44

00 0 01

tt tt

t

tt tt

t

ttt

t

tt

t

ªº

−+−+

ª

º

−−

ªº

«»

«

»

«»

−+−−−

«

»

«»

«

»

«»

=−

−+

«»

«

»

«»

«

»

−

«»

−

«»

«

»

«»

¬¼

«

»

¬

¼

«»

¬¼

, M

B

=

14 6 4 1

0412124

00 6126

00 0 44

00 0 01

−−

ª

º

«

»

−−

«

»

«

»

−

«

»

−

«

»

«

»

¬

¼

10. Write a vector of the polynomial weights for x(t), expand the polynomial weights, taking care to

write the terms in ascending powers of t, and factor the vector as M

S

u(t):

23

2

23

3

13 3 1

1331

46 3 4 0 6 3

11

1333

66

13 3 3

0001

ttt

t

tt

t

tt t

t

ªº

−+−

ª

º

−−

ªº

«»

«

»

«»

−+−

«

»

«»

=

«»

«

»

«»

−

++ −

«»

«

»

«»

«

»

¬¼

¬

¼

«»

¬¼

= M

S

u(t), M

S

=

1331

4063

1

1333

6

0001

−−

ª

º

«

»

−

«

»

«

»

−

«

»

«

»

¬

¼

11. a. True. See equation (2).

b. False. Example 1 shows that the tangent vector x′(t) at p

0

is two times the directed line segment

from p

0

to p

1

.

c. True. See Example 2.

12. a. False. The essential properties are preserved under translations as well as linear transformations.

See the comment after Figure 1.

b. True. This is the definition of G

0

continuity at a point.

c. False. The Bézier basis matrix is a matrix of the polynomial coefficients of the control points.

See the definition before equation (4).

13. a. From (12),

111

10 10 1 0

222

()−=−=−qq pp p p

. Since

1

001 10

2

, ( )==+qpq pp

.

b. From (13), 8(q

3

– q

2

) = –p

0

– p

1

+ p

2

+ p

3

. So 8q

3

+ p

0

+ p

1

– p

2

– p

3

= 8q

2

.

c. Use (8) to substitute for 8q

3

, and obtain

8.6 • Solutions 479

!

8q

2

= (p

0

+ 3p

1

+ 3p

2

+ p

3

) + p

0

+ p

1

– p

2

– p

3

= 2p

0

+ 4p

1

+ 2p

2

Then divide by 8, regroup the terms, and use part (a) to obtain

111 11 11 11

2012 01 12 1 12

424 44 44 24

11

112

22

()()()

(( ))

=++= + + + =+ +

=++

qppp pp pp q pp

qpp

14. a. 3(r

3

− r

2

) = z!(1), by (9) with z!(1) and r

j

in place of x!(1) and p

j

.

z!(1) = .5x!(1), by (11) with t = 1.

.5x!(1) = (.5)3(p

3

− p

2

), by (9).

b. From part (a), 6(r

3

− r

2

) = 3(p

3

− p

2

),

11 11

32 3 2 3 3 2 2

22 22

, and .−=−−+=rr p p r p p r

Since r

3

= p

3

, this equation becomes

1

232

2()=+rpp

.

c. 3(r

1

− r

0

) = z!

(0), by (9) with z!(0) and r

j

in place of x!(0) and p

j

.

z!

(0) = .5x!(.5), by (11) with t = 0.

d. Part (c) and (10) show that 3(r

1

− r

0

) =

3

8

(−p

0

− p

1

+ p

2

+ p

3

). Multiply by

8

3

and rearrange to

obtain 8r

1

= −p

0

− p

1

+ p

2

+ p

3

+ 8r

0

.

e. From (8), 8r

0

= p

0

+ 3p

1

+ 3p

2

+ p

3

. Substitute into the equation from part (d), and obtain

8r

1

= 2p

1

+ 4p

2

+ 2p

3

. Divide by 8 and use part (b) to obtain

()

11 1 11 1 11 1

112 3 12 23 12 2

42 4 44 4 22 2

() ()=+ + = + + += ++rpp p p p pp pp r<

Interchange the terms on the right, and obtain r

1

=

11

212

22

[( )]++rpp

.

15. a. From (11), y!(1) = .5x!(.5) = z!(0).

b. Observe that y!(1) = 3(q

3

– q

2

). This follows from (9), with y(t) and its control points in

place of x(t) and its control points. Similarly, for z(t) and its control points, z!(0) = 3(r

1

– r

0

).

By part (a) 3(q

3

– q

2

) = 3(r

1

– r

0

). Replace r

0

by q

3

, and obtain q

3

– q

2

= r

1

– q

3

, and hence

q

3

= (q

2

+ r

1

)/2.

c. Set q

0

= p

0

and r

3

= p

3

. Compute q

1

= (p

0

+ p

1

)/2 and r

2

= (p

2

+ p

3

)/2. Compute m = (p

1

+ p

2

)/2.

Compute q

2

= (q

1

+ m)/2 and r

1

= (m + r

2

)/2. Compute q

3

= (q

2

+ r

1

)/2 and set r

0

= q

3

.

16. A Bézier curve is completely determined by its four control points. Two are given directly: p

0

=

x(0) and p

3

= x(1). From equation (9), x!

!

(0) = 3(p

1

– p

0

) and x!

!

(1) = 3(p

3

– p

2

). Solving gives

p

1

= p

0

+

1

3

x!

!

(0) and p

2

= p

3

−

1

3

x!

!

(1)

17. a. The quadratic curve is w(t) = (1 – t)

2

p

0

+ 2t(1 − t)p

1

+ t

2

p

2

. From Example 1, the tangent

vectors at the endpoints are w!

(0) = 2p

1

− 2p

0

and w!

(1) = 2p

2

− 2p

1

. Denote the control

points of x(t) by r

0

, r

1

, r

2

, and r

3

. Then

r

0

= x(0) = w(0) = p

0

and r

3

= x(1) = w(1) = p

2

From equation (9) or Exercise 3(a) (using r

i

in place of p

i

) and Example 1,

–3r

0

+ 3r

1

= x!

(0) = w!

!

(0) = 2p

1

− 2p

0

so −p

0

+ r

1

=

10

22

3

−pp

and

480 CHAPTER 8 • The Geometry of Vector Spaces

!

r

1

=

01

2

3

+pp

(i)

Similarly, using the tangent data at t = 1, along with equation (9) and Example 1, yields

–3r

2

+ 3r

3

= x!

(1) = w!

!

(1) = 2p

2

− 2p

1

,

−r

2

+ p

2

=

21

22

3

−pp

, r

2

=

21

2

22

3

−

pp

p

, and

r

2

=

12

2

3

+pp

(ii)

b. Write the standard formula (7) in this section, with r

i

in place of p

i

for i = 1, …, 4, and then

replace r

0

and r

3

by p

0

and p

2

, respectively:

x(t) = (1 – 3t + 3t

2

– t

3

)p

0

+ (3t – 6t

2

+ 3t

3

)r

1

+ (3t

2

– 3t

3

)r

2

+ t

3

p

2

(iii)

Use the formulas (i) and (ii) for r

1

and r

2

to examine the second and third terms in (iii):

23 23 23

12

101

33

23 2 3

01

(3 6 3 ) (3 6 3 ) (3 6 3 )

(2 ) (24 2)

tt t tt t tt t

ttt tt t

−+=−++−+

=−++−+

rp p

pp

23 23 23

21

212

33

23 23

12

(3 3 ) (3 3 ) (3 3 )

(2 2 ) ( )

tt tt tt

tt tt

−=−+−

=−+−

rpp

pp

When these two results are substituted in (iii), the coefficient of p

0

is

(1 − 3t + 3t

2

− t

3

) + (t − 2t

2

+ t

3

) = 1 − 2t + t

2

= (1 − t)

2

The coefficient of p

1

is

(2t − 4t

2

+ 2t

3

) + (2t

2

− 2t

3

) = 2t − 2t

2

= 2t(1 − t)

The coefficient of p

2

is (t

2

− t

3

) + t

3

= t

2

. So x(t) = (1 − t)

2

p

0

+ 2t(1 − t)p

1

+ t

2

p

2

, which

shows that x(t) is the quadratic Bézier curve w(t).

18.

0

01

012

0123

33

363

33

ªº

«»

−+

«»

−+

«»

−+−+

«»

¬¼

p

pp

ppp

pppp

Solutions Manual For Linear Algebra And Its Applications (4th Edition)

Solutions%20Manual%20for%20Linear%20Algebra%20and%20Its%20Applications%20(4th%20Edition)

Solutions%20Manual%20for%20Linear%20Algebra%20and%20Its%20Applications%20(4th%20Edition)

Navigation menu

Versions of this User Manual:

Views

Navigation