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Students’ Solutions Manual for

Applied Linear Algebra by

Peter J. Olver and

Chehrzad Shakiban Second Edition

Undergraduate Texts in Mathematics Springer, New York, 2018. ISBN 978–3–319–91040–6

To the Student These solutions are a resource for students studying the second edition of our text Applied Linear Algebra , published by Springer in 2018. An expanded solutions manual is available for registered instructors of courses adopting it as the textbook.

Using the Manual The material taught in this book requires an active engagement with the exercises, and we urge you not to read the solutions in advance. Rather, you should use the ones in this manual as a means of verifying that your solution is correct. (It is our hope that all solutions appearing here are correct; errors should be reported to the authors.) If you get stuck on an exercise, try skimming the solution to get a hint for how to proceed, but then work out the exercise yourself. The more you can do on your own, the more you will learn. Please note: for students taking a course based on Applied Linear Algebra , copying solutions from this Manual can place you in violation of academic honesty. In particular, many solutions here just provide the final answer, and for full credit one must also supply an explanation of how this is found.

Acknowledgements We thank a number of people, who are named in the text, for corrections to the solutions manuals that accompanied the first edition. Of course, as authors, we take full responsibility for all errors that may yet appear. We encourage readers to inform us of any misprints, errors, and unclear explanations that they may find, and will accordingly update this manual on a timely basis. Corrections will be posted on the text’s dedicated web site: http://www.math.umn.edu/∼olver/ala2.html

Peter J. Olver University of Minnesota [email protected]

Cheri Shakiban University of St. Thomas [email protected] Minnesota, September 2018

Table of Contents Chapter 1.

Linear Algebraic Systems . . . . . . . . . . . . . . . . . 1

Chapter 2.

Vector Spaces and Bases . . . . . . . . . . . . . . . .

11

Chapter 3.

Inner Products and Norms . . . . . . . . . . . . . . .

19

Chapter 4.

Orthogonality . . . . . . . . . . . . . . . . . . . . .

28

Chapter 5.

Minimization and Least Squares . . . . . . . . . . . . .

36

Chapter 6.

Equilibrium

. . . . . . . . . . . . . . . . . . . . .

44

Chapter 7.

Linearity . . . . . . . . . . . . . . . . . . . . . . .

49

Chapter 8.

Eigenvalues and Singular Values . . . . . . . . . . . . .

58

Chapter 9.

Iteration . . . . . . . . . . . . . . . . . . . . . . .

70

Chapter 10. Dynamics

. . . . . . . . . . . . . . . . . . . . . .

81

Students’ Solutions Manual for Chapter 1: Linear Algebraic Systems

1.1.1. (b) Reduce the system to 6u + v = 5, − 52 v = 25 ; then use Back Substitution to solve for u = 1, v = −1. (d) Reduce the system to 2u − v + 2w = 2, − 32 v + 4w = 2, − w = 0; then solve for u = 31 , v = − 43 , w = 0.

(f ) Reduce the system to x + z − 2 w = − 3, − y + 3w = 1, − 4 z − 16 w = − 4, 6 w = 6; then solve for x = 2, y = 2, z = −3, w = 1. ♥ 1.1.3. (a) With Forward Substitution, we just start with the top equation and work down. Thus 2x = −6 so x = −3. Plugging this into the second equation gives 12 + 3y = 3, and so y = −3. Plugging the values of x and y in the third equation yields −3 + 4(−3) − z = 7, and so z = −22.

1.2.1. (a) 3 × 4,

(b) 7,

(c) 6,



1  1.2.2. Examples: (a)  4 7

2 5 8

!

(d) ( −2 0 1 2 ), 



3  6 , 9

!

1  (c)  4 7

2 5 8

(e) 

3 6 9

4  7 , 3

!



  



0  2 . −6

(e)



1



   2 .  

3

6 1 u 5 , x= , b= ; 3 −2 v 5       2 −1 2 u 2       , x =  v , b =  1 ; (d) A =  − 1 − 1 3       3 0 −2 w 1       1 0 1 −2 x −3        2 −1  y  −5  2 −1  . , x =  , b =  (f ) A =        2  0 −6 −4  z  2 1 3 2 −1 w 1

1.2.4. (b) A =

1.2.5. (b) u + w = −1, u + v = −1, v + w = 2. The solution is u = −2, v = 1, w = 1. (c) 3x1 − x3 = 1, −2 x1 − x2 = 0, x1 + x2 − 3 x3 = 1. The solution is x1 =



1.2.6. (a) I =

1

 0   0   0

0

0 1 0 0 0

0 0 1 0 0

0 0 0 1 0

1, 5

x2 = − 25 , x3 = −52 .



0  0   0 ,  0  1



O=

0

 0   0   0

0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

 c

2018



0  0   0 .  0  0

(b) I + O = I , I O = O I = O. No, it does not. 1

Peter J. Olver and Chehrzad Shakiban

2

Students’ Solutions Manual: Chapter 1

1.2.7. (b) undefined, 1.2.9.

!

3 6 0 , −1 4 2

(c)

(f )



1

 3 

7

11 −12 8



9  −12  , 8

1, 6, 11, 16. 

1

1.2.10. (a)



0 0 0

 0 

0

0  0 . −1

1.2.11. (a) True. 1.2.14. B =

x y 0 x

!

where x, y are arbitrary.

1.2.17. A On×p = Om×p ,

Ol×m A = Ol×n . 1 0 0 0

1.2.19. False: for example,

!

0 0 1 0

!

=

!

0 0 . 0 0

1.2.22. (a) A must be a square matrix. (b) By associativity, AA2 = AAA = A2 A = A3 . −1 3

1.2.25. (a) X =

!

1 . −2

♦ 1.2.29. (a) The ith entry of A z is 1 ai1 + 1 ai2 + · · · + 1 ain = ai1 + · · · + ain , which is 1 the ith row sum. (b) Each row of W has n − 1 entries equal to n and one entry equal 1−n 1 1 − n to n and so its row sums are (n − 1) n + n = 0. Therefore, by part (a), W z = 0. Consequently, the row sums of B = AW are the entries of B z = AW z = A0 = 0, and the result follows. ♥ 1.2.34. (a) This follows by direct computation. (b) (i ) −2 1 3 2

!

1 1

−2 0

!

=

!

−2 ( 1 −2 ) + 3

1.3.1. (b) u = 1, v = −1; (f ) a =

1, 3

(d) x1 =

b = 0, c = 

!

4, 3

d=

11 , 3 − 32 .

!

1 (1 0) = 2

−2 3

!

4 1 0 + −6 2 0

!

=

−1 5

!

4 . −6

2 x2 = − 10 3 , x3 = − 3 ; 

!

1 7  4 2R1 +R2 1 7  4   . Back Substitution yields x2 = 2, x1 = −10. −→ 0 5  10 −2 −9  2          1 −2 1  0 1 −2 1  0 3 R +R 1 −2 1  0 3    2 2   4R 1 +R 3    0  (c)  2 −8  8  2 −8  8  2 −8  8    −→  −→  0  0 .    0 0 1  3 0 −3 13  −9 −4 5 9  −9

1.3.2. (a)

Back Substitution yields z = 3, y = 16, x = 29.

c 

2018

Peter J. Olver and Chehrzad Shakiban

Students’ Solutions Manual: Chapter 1 

1   0  (e)   0 −4

0 1 −3 0

−2 0 2 0

0 −1 0 7

         

3



−1  2  0  −5

1  0   0

reduces to

0 −2 1 0 0 2 0 0

0

0 −1 −3 −5

         



−1  2 . 6  15

Back Substitution yields x4 = −3, x3 = − 32 , x2 = −1, x1 = −4.

1.3.3. (a) 3x + 2 y = 2, − 4 x − 3 y = −1; solution: x = 4, y = −5. !

!

2 1 2 1 −→ . 1 4 0 27    1 1 −2 3     (d) Not regular:  4 −1   −→  0  −2 0 3 −1 2

1.3.4. (a) Regular:

−i 1− i

1.3.5. (a)

1+ i 1

    

−1 −3 i

!

−i 0

−→



−2 0 5

3  5 . −7

1+ i 1 − 2i

    

−1 1 − 2i

!

;

use Back Substitution to obtain the solution y = 1, x = 1 − 2 i .

♦ 1.3.11. (a) Set lij =

(b) L =



  

  

aij , 0, 

0 0 0  1 0 0 , −2 0 0

i > j, i ≤ j,

,



uij = 3

 0 

D=

0

0 −4 0

  

aij ,

i < j,

0,

dij =

i ≥ j,





0  0 , 5

0  U = 0 0

1 0 0

  

aij ,

i = j,

0,

i 6= j.



−1  2 . 0

1.3.15. (a) Add −2 times the second row to the first row of a 2 × n matrix. (c) Add −5 times the third row to the second row of a 3 × n matrix.

1.3.16. (a)



1

 0   0

0

0 1 0 0

0 0 1 1



0  0 , 0  1

(c)



1

0 1 0 0

 0   0

0

0 0 1 0



3  0 . 0  1

1.3.20. (a) Upper triangular; (b) both upper and lower unitriangular; (d ) lower unitriangular. !

1.3.21. (a) L = 

1 0 , U = −1 1

!

1 3 , 0 3



1 0 0   (e) L =  −2 1 0 , U =  −1 −1 1 1.3.25. (a) aij = 0 for all i 6= j;



−1   0  0

(c) L =





1

  −1 

0 0  −3 0  . 0 2

1





0 0 −1 1    1 0 , U =  0 2 0 1 0 0



−1  0 , 3

(c) aij = 0 for all i > j and aii = 1 for all i. !

1 1 is regular. Only if the zero appear in the (1, 1) position 1 0 does it automatically preclude regularity of the matrix.

1.3.27. False. For instance

c 

2018

Peter J. Olver and Chehrzad Shakiban

4

Students’ Solutions Manual: Chapter 1 −1 ! 2 ,

1.3.32. (a) x =

(c) x =

3

1 0 −3 1

1.3.33. (a) L =

(c) L =



    

1 2 3 2 9

−



1   0 (e) L =   −1  0

0

!

0

, U=



  0 , 

1 5 3

 0    1 ,  



9

−2

1

−1

0 

0 0 1   0 0 0 , U =  0  1 0  0 −1 1

0 1 3 2 −21



x1 =  

0 2 0 0

−

 5 11  , x2 2 11  

=

x1 =

   2 ,  

x2 =

0     −1  ; x =   7 1 

−1 3 − 72 0

2

 

4



−2





 9  11 . 3 11

   −9 .  

−1

3



!

1 , x3 = 1

1

  1 ; 3   − 13

− 13

0

−1

   −1 .   5 2 

!

−1 3 ; 0 11

  = 0 

U

(e) x =

0



5 4  − 14   , 1  4 1 4



x2 =

1  14  − 5  14   1   14 1 2



    .   

1.4.1. (a) Nonsingular, (c ) nonsingular, (e ) singular. 1.4.2. (a) Regular and nonsingular, (c) nonsingular. 9 (d) x = − 13 2 , y = − 2 , z = −1, w = −3.

1.4.5. (b) x1 = 0, x2 = −1, x3 = 2;

1.4.6. True. All regular matrices are nonsingular.



0  1.4.11. (a)  0 1 

1 0 0

0

1 0 1.4.13. (b) 0 0 0  1 0  0 1   0 0 0 0  1   0



0  1 , (c) 0 0 0 0 1 0 0 0 1



0  0 , 1  0 0  0 ,  1 0





0 1 0 0

1 0 0 0

0 0 0 1

0  0 .  1 0

1 0 0 0  0  1   0 0

0 0 1 0 0 0 1 0

0 0 0 1 0 0 0 1

0  1 , 0  0 1  0 ,  0 0

     

    





0

 0   1

0 0  0   1 0



0 1 0 0 1 0 0 0

0 0 0 1 0 0 0 1



1  0 , 0  0 0  1 . 0  0

1.4.14. (a) True, since interchanging the same pair of rows twice brings you back to where you started. 1.4.16. (a)



0

 0 

1

0 1 0



1  0 . 0

(b) True.

c 

2018

Peter J. Olver and Chehrzad Shakiban

Students’ Solutions Manual: Chapter 1

5

1.4.19. Let ri , rj denote the rows of the matrix in question. After the first elementary row operation, the rows are ri and rj + ri . After the second, they are ri − (rj + ri ) = − rj and rj + ri . After the third operation, we are left with − rj and rj + ri + (− rj ) = ri .

0 1 1 0

1.4.21. (a)

(c)



0

 1 

0

(e)



0

 1   0

0

1.4.22. (b)



0

 0   1

0

!

0 2

!

1 −1



!

2 −1 , 0 1





1 0 0 1

=



!

1 0 1 −3 1      3 0 =0  0 2 0 1 0 2 0

0 0 1 0 0 1 0 0 1 0 0



1 0 0 0

0 0 1 0  2 0 3 1   0 4 −1  1 1 7 −1 2

1 0 0 0

0 0   0 0   0  1 1 1



0 1 2





 5   2 ;

x=

3

1 0 2 0   0  0 1 −3 0 0 9 1

0 1 0   0 0 1 =  2   2 −5 3 7 −29 

1 −1 1  1 1 0 = −1 1 −3   2 −1 1



solution: x = 4, y = 0, z = 1, w = 1.

1

1

 , 

x



0 0 1 3

0 1 1 3

 0   0



0 1  0 0   0  0 1 0

0 0 1 5 2



 = 



−1  1 ; 0







4 −1 2 −1     −1  1 0 0 , x =  .    0 3 −4   1 0 0 1 3



1 0   0 0   0  0 0 1



−1 1 0 0

1 −3  1 0 ; −2 1  3 0 2

1.4.26. False. Changing the permuation matrix typically changes the pivots.

1.5.1. (b)



2

 3 

2



!

0 1 , 1 0

1.5.3. (a)



3 −1 −1 1    −4 2 1 1 =  −1 0 1 2

1 2 1

(c)

!

1 2 , 0 1

(e)



1 0



1

0 1 −6 0

 0   0

0

−1

1.5.6. (a) A

=

!

1.5.9. (a)

0

 0   0

1

!

1 , B −1 =  −1 −

2 3 1 3

0 0 1 0

0 1 0 0



1  0 , 0  0

1.5.13. Since c is a scalar,

(c)



1

 0   0

0

1 −1 A c

!

0 0 0 1

0 1 0 0

(c A) =

0 0 1 0



−1 2 0



2 −1  3 1  2 1

1 2 1



1  1 . 2

0  0 .  0 1

 1 3  . 1 3

 1 3  . − 32

0 2 1 , C −1 = B −1 A−1 =  3 0 1

(b) C = 

−1 2







3 0    −4 0 =   −1 1

0 1 0

 0 



0  0 . 1  0 1 c A−1 A = I . c

c 

2018

Peter J. Olver and Chehrzad Shakiban

6

Students’ Solutions Manual: Chapter 1 1.5.16. If all the diagonal entries are nonzero, then D−1 D = I . On the other hand, if one of diagonal entries is zero, then all the entries in that row are zero, and so D is singular. ♦ 1.5.19. (a) A = I −1 A I . (b) If B = S −1 AS, then A = S B S −1 = T −1 B T , where T = S −1 . 

 ! 1  0 

! −1  1 0 1 . = 1.5.21. (a) B A =  0 1 1 1 (b) AX = I does not have a solution. Indeed, the first column of this matrix equation is     ! 1 −1 1  x    1 0 =  the linear system   , which has no solutions since x − y = 1, y = 0, 0  y 1 1 0 and x + y = 0 are incompatible.

1 −1

1.5.25. (b)





−

1.5.31. (b)

!









−

−

1  0 (f )   2 2 

i 2 1 2

5 17 1 17

1 0

(d) no inverse;

!

0 −8  0 1 0  0 1 0  1 2 0   1 1 0 3   0 1 0  0   0 0 0 1



1.5.32. (b)

 3 8 ; −81

1 8 3 8

1 0 3 1  1 0  (f )  3 1 0 0

1.5.26. (b)

1.5.28. (a)

1 0 −1 1

 1 2  , − 2i

(f )

!

!

    

− 85

− 21 7 8

1 8 1 2 −83

5 8 − 12 1 8



  .  

1 3 1 3 = ; (d) not possible; 0 1 3 1     0 1 0 0 1 0 0 1 0 0         0   0 1 0   0 −1 0   0 1 0 1 0 3 1 0 0 1 0 0 8     1 2 3 1 2 0 0 0         1 4   0 1 0  =  3 5 5 . 2 1 2 0 0 1 0 1 

i  (c)  1 − i  −1



0 −i −1

−1  1  . −i

    2 2 17  2   , = (d) 3 12 2 17   

4 0 1 1    11  0 −1 −1  =   −1 −1 0  −7   6 −1 −1 −1

 1  4 ; 1 4



    



  

3  1 . 4  −2

(d) singular matrix; (f )...