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
−
⎡⎤
⎢⎥
−=
⎢⎥
⎢⎥
⎣
⎡
⎢
⎢
⎢
⎣
⎡
⎢
⎢
⎢
⎣
and row reducing
[
]
2AI−0
yields
1100
0010.
0000
⎡
⎤
⎢
⎥
⎢
⎥
⎢
⎥
The
For λ = 3:
10 2
3102,
00 0
AI
−−
⎡⎤
⎢⎥
−=
⎢⎥
⎢⎥
⎣⎦
and row reducing
[]
3AI−0
yields
1020
0000.
0000
⎡
⎤
⎢
⎥
⎢
⎥
⎢
⎥
⎣
⎦
The
⎡
⎢
⎢
⎢
⎣
⎡
⎢
⎢
⎢
⎣
15. The eigenvalues of A are given to be 0 and 1.
For λ = 0:
011
0121,
110
AI
−−
⎡⎤
⎢⎥
−=
⎢⎥
⎢⎥
−−
⎣⎦
and row reducing
[
]
0AI−0
yields
10 10
01 1 0.
00 0 0
−
⎡
⎤
⎢
⎥
⎢
⎥
⎢
⎥
⎣
⎦
The
⎡
⎢
⎢
⎢
⎣
⎡
⎢
⎢
⎢
⎣
For λ = 1:
111
111,
111
AI
−−−
⎡⎤
⎢⎥
−=
⎢⎥
⎢⎥
−−−
⎣⎦
and row reducing
[]
AI−0
yields
1110
0000.
0000
⎡
⎤
⎢
⎥
⎢
⎥
⎢
⎥
⎣
⎦
The
⎡
⎢
⎢
⎢
⎣
From
12
,vv
and
3
v construct
123
111
110.
10 1
P
⎡⎤
⎢⎥
⎣⎦
−−
⎡
⎤
⎢
⎥
==−
⎢
⎥
⎢
⎥
⎣
⎦
vvv
Then set
000
010,
001
D
⎡⎤
⎢⎥
=⎢⎥
⎢⎥
⎣⎦
16. The only eigenvalue of A given is 0.
294 CHAPTER 5 • Eigenvalues and Eigenvectors
For λ = 0:
12 3
0252,
13 1
AI
−
⎡⎤
⎢⎥
−= −
⎢⎥
⎢⎥
⎣
⎡
⎢
⎢
⎢
⎣
⎡
⎢
⎢
⎢
⎣
and row reducing
[
]
0AI−0
yields
10 110
01 4 0.
00 0 0
−
⎡
⎤
⎢
⎥
⎢
⎥
⎢
⎥
The
17. Since A is triangular, its eigenvalue is obviously 2.
For λ = 2:
000
2200,
⎢
⎣
⎡
⎢
⎢
⎢
⎣
AI
⎡⎤
⎢⎥
−=
⎢⎥
and row reducing
[]
2AI−0
yields
1000
0100.
⎡
⎤
⎢
⎥
⎢
⎥
The general
18. The eigenvalues of A are given to be –2, -1 and 0.
For λ = –2:
422
2312,
220
AI
−−
⎡
⎤
⎢
⎥
+= − −
⎢
⎥
⎢
⎥
−
⎣
⎦
and row reducing
[]
2AI+0
yields
10 10
01 10.
00 0 0
−
⎡
⎤
⎢
⎥
−
⎢
⎥
⎢
⎥
⎣
⎦
The
⎡
⎢
⎢
⎢
⎣
⎡
⎢
⎢
⎢
⎣
⎡
⎢
⎢
⎢
⎣
general solution is
3
1
1/2 ,
1
x
⎡
⎤
⎢
⎥
⎢
⎥
⎢
⎥
⎣
⎦
and a basis for the eigenspace is
2
2
1
2
⎡
⎤
⎢
⎥
=
⎢
⎥
⎢
⎥
⎣
⎦
v
.
⎢
⎢
⎢
⎣
⎡
⎢
222
−−
⎡⎤
1100
−
⎡
⎤
5.3 • Solutions 295
⎢
⎢
⎢
⎣
12 1
⎡
⎤
200
−
⎡⎤
19. Since A is triangular, its eigenvalues are 2, 3, and 5.
For
= 2:
3309
0112
2,
0000
0000
−
⎡
⎤
⎢
⎥
−
⎢
⎥
−=
⎢
⎥
⎢
⎥
⎢
⎥
⎣
⎦
AI
and row reducing
[]
2 AI−0
yields
10 1 10
011 20
.
000 00
000 00
⎡⎤
⎢⎥
−
⎢⎥
⎢⎥
⎢⎥
⎢⎥
⎣⎦
12
11
12
{} .
10
01
⎧⎫−−
⎡⎤⎡⎤
⎪⎪
⎢⎥⎢⎥
−
⎪⎪
⎢⎥⎢⎥
,= ,
⎨⎬
⎢⎥⎢⎥
⎪⎪
⎢⎥⎢⎥
⎪⎪
⎢⎥⎢⎥
⎣⎦⎣⎦
⎩⎭
vv
⎡
⎢
⎢
⎢
⎢
⎢
⎣
⎡
⎢
⎢
⎢
⎢
⎢
⎣⎦
⎣
⎦
3
3
2.
0
0
⎡⎤
⎢⎥
⎢⎥
=⎢⎥
⎢⎥
⎢⎥
⎣⎦
v
⎡
⎢
⎢
⎢
⎢
⎢
⎣
⎡
⎢
⎢
⎢
⎢
⎢
⎣
296 CHAPTER 5 • Eigenvalues and Eigenvectors
From
123
,,vv v
and
4
v construct
1234
1131
1 220
.
1 000
0100
⎡⎤
⎣⎦
−−
⎡
⎤
⎢
⎥
−
⎢
⎥
==
⎢
⎥
⎢
⎥
⎢
⎥
⎣
⎦
vv vv
P
Then set
20. Since A is triangular, its eigenvalues are 2 and 3.
For λ = 2:
1000
0000
2,
0000
100 1
AI
⎡⎤
⎢⎥
⎢⎥
−=
⎢⎥
⎢⎥
⎣⎦
and row reducing
[]
2 AI−
0
yields
10000
00010
.
00000
00000
⎡
⎤
⎢
⎥
⎢
⎥
⎢
⎥
⎢
⎥
⎣
⎦
The
⎡
⎢
⎢
⎢
⎢
⎣
For λ = 3:
00 00
0100
3,
00 10
10 00
AI
⎡⎤
⎢⎥
−
⎢⎥
−=
⎢⎥
−
⎢⎥
⎣⎦
and row reducing
[]
3 AI−
0
yields
10000
01000
.
00100
00000
⎡⎤
⎢⎥
⎢⎥
⎢⎥
⎢⎥
⎣⎦
⎡
⎢
⎢
⎢
⎢
⎣
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.
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
23. A is diagonalizable because you know that five linearly independent eigenvectors exist: three in the
24. No, by Theorem 7(b). Here is an explanation that does not appeal to Theorem 7: Let
1
v and
2
v be
25. Let
1
{}v be a basis for the one-dimensional eigenspace, let
2
v and
3
v form a basis for the two-
26. Yes, if the third eigenspace is only one-dimensional. In this case, the sum of the dimensions of the
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
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,
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,
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
⎡
⎤
⎢
⎥
⎣
⎦ whose eigenvalues are 2 and 4. Unfortunately, a
⎡
⎢
⎣
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
⎡⎤
⎢⎥
⎣⎦
has the
33.
9424
56 32 28 44 ,
A
−−−
⎡⎤
⎢⎥
−−
⎢⎥
=⎢⎥
ev = eig(A)=(13,–12,-12, 13)
,
⎢
⎢
⎣
0.5000 0.3333
0 1.3333 .
−
⎡
⎤
⎢
⎥
−
⎢
⎥
⎪
⎢
⎪
⎢
⎨
⎢
⎪
⎢
⎪
⎣
11
⎧
⎫−
⎡
⎤⎡ ⎤
⎡
⎢
⎣
⎡
⎢
⎣
5.3 • Solutions 299
0.2857 0
⎡⎤
⎢⎥
⎪
⎢
⎨
⎢
⎪
⎢
⎪
⎣
20
⎧
⎫
⎡
⎤⎡ ⎤
⎪
⎪
⎢
⎥⎢ ⎥
1120
−
⎡⎤
⎢⎥
13000
⎡⎤
⎢⎥
order in P, and hence the eigenvalues are listed in a different order in D. Both answers are correct.
34.
49782
790714
,
5105510
A
−−
⎡⎤
⎢⎥
−−
⎢⎥
⎢⎥
=−−
ev = eig(A)= (5,– 2,-2,5,5)
,
⎢
⎢
⎢
⎢
⎢
⎣
2.0000 1.0000 2.0000
1.0000 1.0000 0
−
⎡
⎤
⎢
⎥
−
⎪
⎢
⎨
⎢
⎪
⎢
⎪
⎢
⎪
⎣
212
110
⎧
⎫
−
⎡
⎤⎡ ⎤⎡ ⎤
⎪
⎪
⎢
⎥⎢ ⎥⎢ ⎥
−
⎪
⎪
⎢
⎥⎢ ⎥⎢ ⎥
⎢
⎢
0.4000 0.6000
1.4000 1.4000
0 1.0000
−
⎡
⎤
⎢
⎥
⎢
⎥
⎢
⎥
⎢
⎥
⎣
⎦
300 CHAPTER 5 • Eigenvalues and Eigenvectors
⎪
⎢
⎪
⎢
⎨
⎢
⎪
⎢
⎪
⎢
⎪
23
77
⎧
⎫
−
⎡
⎤⎡ ⎤
⎪
⎪
⎢
⎥⎢ ⎥
⎣
⎦⎣ ⎦
⎩⎭
⎢
⎢
⎢
⎢
⎢
⎣
⎢
⎢
⎢
⎢
⎢
⎣
21223
11077
−−
⎡
⎤
⎢
⎥
−
500 0 0
050 0 0
⎡
⎤
⎢
⎥
35.
13129159
6 59159
,
6125 69
6129 89
61212 62
A
−−
⎡⎤
⎢⎥
−−
⎢⎥
⎢⎥
=−−
⎢⎥
−−
⎢⎥
⎢⎥
−−−
⎣⎦
ev = eig(A) −=(7, 14,-14,7,7)
,
⎢
⎢
⎢
⎢
⎢
⎣
2 0000 1 0000 1 5000
1 0000 0 0
..−.
⎡
⎤
⎢
⎥
.
⎢
⎨
⎢
⎪
⎢
⎪
⎢
⎪
⎣
21 3
10 0
⎧
⎫
−
⎡
⎤⎡⎤⎡ ⎤
⎪
⎪
⎢
⎥⎢⎥⎢ ⎥
⎪
⎪
⎢
⎥⎢⎥⎢ ⎥
⎪
⎪
nulbasis(A-ev(2)*eye(5))
10
10
01
10
01
⎡
⎤
⎢
⎥
⎢
⎥
⎢
⎥
=,
−
⎢
⎥
⎢
⎥
⎢
⎥
⎣
⎦
A basis for the eigenspace of
10
10
14 is , .
01
⎪
⎢
⎪
⎢
⎪
⎣
⎧
⎫
⎡
⎤⎡ ⎤
⎪
⎪
⎢
⎥⎢ ⎥
⎪
⎪
⎢
⎥⎢ ⎥
⎪
⎪
⎢
⎥⎢ ⎥
λ=− −
⎨
⎬
⎢
⎥⎢ ⎥
5.3 • Solutions 301
⎢
⎢
⎢
⎢
⎢
⎣
⎢
⎢
⎢
⎢
⎢
⎣
21 31 0
10 0 1 0
−
⎡
⎤
⎢
⎥
700 0 0
070 0 0
⎡
⎤
⎢
⎥
36.
24 6 2 6 2
72 51 9 99 9
,
063156363
A
−
⎡⎤
⎢⎥
−
⎢⎥
⎢⎥
=−
ev = eig(A)=(24,-48,36,-48,36)
,
⎢
⎢
⎢
1
1
0
⎡
⎤
⎢
⎥
⎢
⎥
⎢
⎥
⎣
⎦
⎢
⎢
⎢
⎢
⎢
⎣
1
1
⎡
⎤
⎢
⎥
nulbasis(A-ev(2)*eye(5)),
⎪
⎢
⎪
⎢
⎨
⎢
⎪
⎢
⎪
⎢
⎪
⎣
00
10
⎧
⎫
⎡
⎤⎡ ⎤
⎪
⎪
⎢
⎥⎢ ⎥
⎢
⎢
⎢
⎢
⎣
1 0000 0 3333
0 0000 1 0000
.−.
⎡
⎤
⎢
⎥
..
⎢
⎥
⎪
⎢
⎪
⎢
⎨
⎢
⎪
⎢
⎪
⎢
⎪
⎣
11
03
⎧
⎫
−
⎡
⎤⎡ ⎤
⎪
⎪
⎢
⎥⎢ ⎥
302 CHAPTER 5 • Eigenvalues and Eigenvectors
Thus we construct
10 0 1 1
11 00 3
00 13 0
⎢
⎢
⎢
⎣
⎢
⎢
⎢
⎣
P
−
⎡
⎤
⎢
⎥
⎢
⎥
⎢
⎥
=−
and
24 0 0 0 0
048000
00 4800
D
⎡
⎤
⎢
⎥
−
⎢
⎥
⎢
⎥
=−
. Notice that
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,
3 7332,
−.
and
6 0345.−.
68530
73530
37535
−−
⎡⎤
⎢⎥
−−
⎢⎥
⎢⎥
−− −
5.4 SOLUTIONS
1. Since
1121
3
()3 5 [()] .
5
⎡⎤
=−, =
⎢⎥
−
⎣
⎡
⎢
⎣
bddb
D
TT
Likewise
212
() 6T=− +bdd
implies that
2
1
[( )] 6
D
T−
⎡
⎤
=
⎢
⎥
b
2.Since
1121
3
()3 3 [()] .
3
B
TT
⎡⎤
=−, =
⎢⎥
−
⎣⎦
dbbd Likewise
212
() 2 5T=− +dbb
implies that
2
2
[( )] .
5
B
T−
⎡
⎤
=
⎢
⎥
⎣
⎦
d
⎡
⎢
⎣
5.4 • Solutions 303
3. a.
1 1232 1233 123
() 0 0 ( ) 2 0 () 2 0 3TT T=−+, =−−+, =++ebbbe bbbebbb
b.
12 3
012
[( )] 0 [( )] 2 [( )] 0
BB B
TT T
−
⎡⎤ ⎡ ⎤ ⎡⎤
⎢⎥ ⎢ ⎥ ⎢⎥
=, =−, =
ee e
.
4. Let
12
{}=,eeE be the standard basis for . Since
11 2 2
23
[ ( )] ( ) [ ( )] ( ) ,
20
TT T T
−
⎡
⎤⎡⎤
==, ==
⎢
⎥⎢⎥
−
⎣
⎦⎣⎦
bb b b
EE
⎡
⎢
⎣
5. a.
223
( ) ( 3)(3 2 ) 9 3T t tt tt t=+ − + =−+ +p
b. Let
p
and
q
be polynomials in
2
, and let
c
be any scalar. Then
( () ()) ( 3)[ () ()] ( 3) () ( 3) ()
(()) (())
Tttt tttttt
Tt Tt
+=+ +=+ ++
=+
pq pq p q
pq
c. Let
2
{1 }Btt=,,
and
23
{1 } .=,,,Cttt
Since
11
3
1
( ) (1) ( 3)(1) 3 [ ( )] .
0
0
C
TTt tT
⎡⎤
⎢⎥
⎢⎥
==+ =+, =
⎢⎥
⎢⎥
⎣⎦
bb
⎢
⎢
⎢
⎢
⎣
0
⎡
⎤
304 CHAPTER 5 • Eigenvalues and Eigenvectors
2232
33
0
0
( ) ( ) ( 3)( ) 3 [ ( )] .
3
1
C
TTtttttT
⎡
⎤
⎢
⎥
⎢
⎥
==+ =+, =
⎢
⎥
⎢
⎥
⎣
⎡
⎢
⎢
⎢
⎢
⎣
bb
Thus the matrix for T relative to B and
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
(() ()) [() ()] 2 [() ()]
Tt t t t t t t
+=++ +
pq pq pq
c. Let
2
{1 }Btt=,,
and
234
{1 } .=,,,,Ctttt
Since
22
11
1
0
() (1)12(1)12 [()] .
2
0
0
C
TT t tT
⎡
⎤
⎢
⎥
⎢
⎥
⎢
⎥
==+ =+, =
⎢
⎥
⎢
⎥
⎢
⎥
⎣
⎦
bb
⎢
⎢
⎢
⎢
0
0
⎡
⎤
⎢
⎥
⎢
⎥
⎣
⎦
⎢
⎢
⎢
0
0
2
⎡
⎤
⎢
⎥
⎢
⎥
⎢
⎥
⎣
⎦
⎢
⎢
⎢
⎣
100
010
⎡
⎤
⎢
⎥
⎢
⎥
5.4 • Solutions 305
7. Since
11
3
() (1)35[()] 5.
0
⎡
⎤
⎢
⎥
==+, =
⎢
⎥
⎢
⎥
⎣
bb
B
TT tT
Likewise since
matrix representation of T relative to the basis
B
is
123
300
[ ( )] [ ( )] [ ( )] 5 2 0 .
041
⎡⎤
⎣⎦
⎡
⎤
⎢
⎥
=−
⎢
⎥
⎢
⎥
⎣
⎦
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:
The matrix that implements the multiplication along the bottom of the figure is easily filled in by
8. Since
12
4
[4 3 ] 3 ,
0
B
⎡
⎤
⎢
⎥
−=−
⎢
⎥
⎢
⎥
⎣
⎦
bb
12 12
00 1 4 0
[(4 3 )] [][4 3 ] 2 1 2 3 5
13 1 0 5
BB B
TT
⎡
⎤⎡ ⎤ ⎡ ⎤
⎢
⎥⎢ ⎥ ⎢ ⎥
−= −= −−=
⎢
⎥⎢ ⎥ ⎢ ⎥
⎢
⎥⎢ ⎥ ⎢ ⎥
−
⎣
⎦⎣ ⎦ ⎣ ⎦
bb bb
53(1) 2
53(1) 8
+−
⎡⎤⎡⎤
⎢⎥⎢⎥
+
⎣⎦⎣⎦
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) ()()
+− −+− − −
⎡ ⎤⎡ ⎤⎡⎤⎡⎤
⎢ ⎥⎢ ⎥⎢⎥⎢⎥
+= + = + = + = +
⎢ ⎥⎢ ⎥⎢⎥⎢⎥
pq p q p q
pq pq p q p q p q
TTT
306 CHAPTER 5 • Eigenvalues and Eigenvectors
and T is a linear transformation.
c. Let
2
{1 }=,,Btt
and
123
{}=,,ee eE be the standard basis for
3
. Since
11
−
⎡⎤ ⎡ ⎤
⎣⎦
111
−
⎡⎤
⎢⎥
10. a. Let p and q be polynomials in
3
, and let c be any scalar. Then
⎢
⎢
⎣
( )(2) (2) (2)
( )(3) (3) (3)
+− −+−
⎡
⎤⎡⎤
⎢
⎥⎢⎥
++
⎢
⎥⎢⎥
pq p q
pq p q
(2) (2)
(3) (3) () ()
−−
⎡⎤⎡⎤
⎢⎥⎢⎥
⎢⎥⎢⎥
pq
pq pq
( )(2) ((2)) (2)
( )(3) ( (3)) (3)
cc
cc
⋅− ⋅− −
⎡⎤⎡⎤⎡⎤
⎢⎥⎢⎥⎢⎥
⋅⋅
ppp
ppp
b. Let
23
{1 }=,,,Bttt
and
1234
{}=,,,ee eeE be the standard basis for
4
. Since
124
100
−
⎡⎤ ⎡ ⎤ ⎡ ⎤
⎢⎥ ⎢ ⎥ ⎢ ⎥
⎣⎦ ⎣ ⎦ ⎣ ⎦
8
0
−
⎡⎤
⎢⎥
⎣⎦
⎢
⎢
⎢
1248
1000
−−
⎡
⎤
⎢
⎥
⎣
⎦
5.4 • Solutions 307
12. Following Example 4, if
12
01
,
12
P
⎡⎤
⎢⎥
⎣⎦
−
⎡
⎤
==
⎢
⎥
⎣
⎦
bb then the B-matrix is
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
−
⎡⎤
−=
⎢⎥
−
⎣⎦
AI The equation
()AI−=x0
amounts to
12
0,xx−+ = so
12
xx= with
2
x
⎡
⎢
⎣
31
−
⎣⎦
with
2
x free. A nice basis vector for the eigenspace is thus
2
1.
3
⎡
⎤
=
⎢
⎥
⎣
⎦
v
⎡
⎢
⎣
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 −
⎡⎤
−=
⎢⎥
−
⎣⎦
The equation
(5)AI−=x0
amounts to
12
0,xx−=
so
12
xx= with
⎡
⎢
⎣
free. A nice basis vector for the eigenspace is thus
2
1.
1
−
⎡
⎤
=
⎢
⎥
⎣
⎦
v
⎡
⎢
⎣
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
⎡⎤
+=
⎢⎥
⎣⎦
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
2
2.
1
⎡
⎤
=
⎢
⎥
⎣
⎦
v
⎡
⎢
⎣
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 −
⎡⎤
−=
⎢⎥
−
⎡
⎢
⎣
The equation
(3)AI−=x0
amounts to
12
20,xx−=
so
12
2xx=
2
x free. A basis vector for the eigenspace is thus
2
1.
1
−
⎡
⎤
=
⎢
⎥
⎣
⎡
⎢
⎣
v
17. a. We compute that
11
411 3 3
12 1 3
A⎡⎤⎡⎤⎡⎤
===
⎢⎥⎢⎥⎢⎥
−−−
⎣⎦⎣⎦⎣⎦
bb
so
1
b is an eigenvector of A corresponding to the eigenvalue 3. The characteristic polynomial of
5.4 • Solutions 309
b. Following Example 4, if
12
,
⎡⎤
⎢⎥
⎣⎦
=bbP
then the B-matrix for T is
18. If there is a basis
B
such that []
B
T is diagonal, then A is similar to a diagonal matrix, by the second
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,
20. If
1
,
−
=APBP
then
211 11 121
()()() .
−− −− − −
===⋅⋅=A PBP PBP PB P P BP PB I BP PB P
So
2
A
is
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
22. If A is diagonalizable, then
1
APDP
−
=
for some P. Also, if B is similar to A, then
1
BQAQ
−
=
23. If
0,=,≠xxxA
then
11
.PA P
−−
=xx
If
1
,
−
=BPAP
then
1111 1
() ()BP P APP P A P
−−−− −
===xxxx
(*)
24. If
1
,
−
=APBP
then
11
rank rank ( ) rank ,
−−
==A P BP 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.
25. If
1
,
−
=APBP
then
11
tr( ) tr(( ) ) tr( ( )) By the trace property
APBP PPB
−−
==
26. If
1
APDP
−
=
for some P, then the general trace property from Exercise 25 shows that
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
E
is simply
28. For each
() ,,=bb
jj
jI
and
[( )] [ ].=bb
jC jC
I
By formula (4), the matrix for I relative to the bases
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
…
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
⎡⎤
⎢⎥
⎣⎦
−− −
⎡⎤ ⎡⎤
⎢⎥ ⎢⎥
=−,= =−,
⎢⎥ ⎢⎥
⎢⎥ ⎢⎥
−
⎣⎦ ⎣⎦
bbb
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
⎡⎤
⎢⎥
⎣⎦
−− − −−
⎡⎤ ⎡⎤
⎢⎥ ⎢⎥
=,==−,
⎢⎥ ⎢⎥
⎢⎥ ⎢⎥
−− − −−
⎣⎦ ⎣⎦
bbbAP
32. [M]
6409
3016
,
1210
A
−
⎡⎤
⎢⎥
−
⎢⎥
=⎢⎥
−−
5.4 • Solutions 311
ev
=
eig(A)
=
(5, 1, -2, -2)
⎢
⎢
⎢
1 0000
1 0000
.
⎡
⎤
⎢
⎥
.
⎣
⎦
⎢
⎢
⎢
2
2
⎡
⎤
⎢
⎥
⎣
⎦
⎢
⎢
⎢
1 0000
1.0000
.
⎡
⎤
⎢
⎥
⎣
⎦
⎢
⎢
2
7
2
⎡
⎤
⎢
⎥
−
⎢
⎥
⎣
⎦
⎢
⎢
⎢
1 0000 1.5000
0 1.0000
.
⎡
⎤
⎢
⎥
⎣
⎦
A basis for the eigenspace of
2=−
is
{}
34
16
13
,,.
10
04
⎧
⎫
⎡
⎤⎡ ⎤
⎪
⎪
⎢
⎥⎢ ⎥
−
⎪
⎪
⎢
⎥⎢ ⎥
=
⎨
⎬
⎢
⎥⎢ ⎥
⎪
⎪
⎢
⎥⎢ ⎥
⎪
⎪
⎣
⎦⎣ ⎦
⎩⎭
bb
The basis
1234
{}B=,,,bb bb is a basis for
4
with the property that []
B
T is diagonal.
5.5 SOLUTIONS _____________________________________________
1.12 1 2
13 13
AAI
λ
λλ
−−−
⎡⎤ ⎡ ⎤
=,−=
⎢⎥ ⎢ ⎥
−
⎣⎦ ⎣ ⎦
2
det( )(1 )(3 )(2) 4 5AI−=− −−−=−+
For λ = 2 + i: 12
(2 ) .
11
−− −
⎡⎤
−+ =
⎢⎥
−
⎣⎦
i
AiI i The equation
( )AI−=x0
gives
As in Example 2, the two equations are equivalent—each determines the same relation between
1
x
and
.x So use the second equation to obtain
(1 ) ,=− −xix
with
x free. The general solution is
For λ = 2 – i: Let
1
2
1.
1
−−
⎡⎤
==
⎢⎥
⎣⎦
vv i The remark prior to Example 5 shows that
2
v is
2. 33
.
33
A−
⎡⎤
=⎢⎥
⎣⎦
The characteristic polynomial is
2
618,−+
so the eigenvalues of A are
63672
33.
2i
±−
==±