Section 6.2: Diagonalization of Matrices 355
With
33:
352 0
462 0
abc
abc
3
1
1
v
19. Characteristic polynomial: 32
() 6 11 6 ( 1)( 2)( 3)p
Eigenvalues: 12 3
1, 2, 3
With
11:
0
23 0
44 0
bc
abc
ab
1
1
1
1
v
20. Characteristic polynomial: 32
( ) 13 52 60 ( 2)( 5)( 6)p
Eigenvalues: 123
2, 5, 6
00
1
356 Chapter 6: Eigenvalues and Eigenvectors
21. Characteristic polynomial: 32 3
() 3 3 1 ( 1)p
Eigenvalues: 123
1, 1, 1
22. Characteristic polynomial: 32 3
() 3 3 1 ( 1)p
Eigenvalues: 123
1, 1, 1
20
abc
1
23. Characteristic polynomial: 32 2
() 4 5 2 ( 1)( 2)p
Eigenvalues: 123
1, 1, 2
Section 6.2: Diagonalization of Matrices 357
24. Characteristic polynomial: 32 2
() 5 8 4 ( 1)( 2)p
Eigenvalues: 12 3
1, 2, 2
22 0
abc
1
25. Characteristic polynomial: 22
() ( 1)( 1)p
Eigenvalues: 1234
1, 1, 1, 1
With
11:
22 0
22 0
00
00
ac
bc
12
01
01
,
01
10
vv
358 Chapter 6: Eigenvalues and Eigenvectors
26. Characteristic polynomial: 3
() ( 1)( 2)p
Eigenvalues: 1234
1, 1, 1, 2
With
11:
0
0
0
0
d
d
d
d
123
001
010
,,
100
000
vvv
27. Characteristic polynomial: 3
() ( 1)( 2)p
Eigenvalues: 1234
1, 1, 1, 2
0
b
1
Section 6.2: Diagonalization of Matrices 359
28. Characteristic polynomial: 22
() ( 1)( 2)p
Eigenvalues: 123 4
1, 1, 2, 2
0
0
bd
cd
1
0
29. If A is similar to B and B is similar to C, so A = P–1BP and B = Q–1CQ, then
30. If A is similar to B so A = P–1BP then
111 11
()()()()()
n
A P BP P BP P BP P BP P BP
360 Chapter 6: Eigenvalues and Eigenvectors
31. If A is similar to B so A = P–1BP then A–1 = (P–1BP)–1 = P–1B–1P, so A–1 is similar
to B–1.
33. If A and B are similar with A = P–1BP, then
1
11 .
APBPPBPPBPB
Moreover, by Problem 32 the two matrices have the same eigenvalues, and by Problem 39
in Section 6.1, the trace of a square matrix with real eigenvalues is equal to the sum of those
eigenvalues. Therefore trace A = (eigenvalue sum) = trace B.
34. The characteristic equation of the 2 2 matrix ab
cd
A is
2()( )0,a d ad bc
and the discriminant of this quadratic equation is
35. Three eigenvectors associated with three distinct eigenvalues can be arranged in six different
orders as the column vectors of the diagonalizing matrix
123
.
T
Pvvv
Section 6.3: Applications Involving Powers of Matrices 361
36. The fact that the matrices A and B have the same eigenvalues (with the same
37. If A = PDP–1 with P the eigenvector matrix of A and D its diagonal matrix of
38. If the nn matrix A has n linearly independent eigenvectors associated with the single
eigenvalue
, then A = PDP–1 with D =
I, so 11
() .
APIP PP ID
39. Let the nn matrix A have k n distinct eigenvalues 12
,,,.
k
Then the definition
of algebraic multiplicity and the fact that all solutions of the nth degree polynomial equation
SECTION 6.3
APPLICATIONS INVOLVING
POWERS OF MATRICES
In Problems 1–10 we first find the eigensystem of the given matrix A so as to determine its
eigenvector matrix P and its diagonal eigenvalue matrix D. Then we calculate the matrix
power A5 = PD5P–1.
1. Characteristic polynomial: 2
() 3 2 ( 1)( 2)p
Eigenvalues: 12
1, 2
362 Chapter 6: Eigenvalues and Eigenvectors
2. Characteristic polynomial: 2
() 2 ( 1)( 2)p
Eigenvalues: 12
1, 2
With
11:
66 0
33 0
ab
ab
1
1
1
v
3. Characteristic polynomial: 2
() 2 ( 2)p
Eigenvalues: 12
0, 2
With
66 0
ab
1
v
Section 6.3: Applications Involving Powers of Matrices 363
4. Characteristic polynomial: 2
() 3 2 ( 1)( 2)p
Eigenvalues: 12
1, 2
With
11:
33 0
22 0
ab
ab
1
1
1
v
5. Characteristic polynomial: 2
() 3 2 ( 1)( 2)p
Eigenvalues: 12
1, 2
6. Characteristic polynomial: 2
() 3 2 ( 1)( 2)p
Eigenvalues: 12
1, 2
364 Chapter 6: Eigenvalues and Eigenvectors
7. Characteristic polynomial: 2
() ( 1)( 2)p
Eigenvalues: 12 3
1, 2, 2
8. Characteristic polynomial: 32 2
() 4 5 2 ( 1)( 2)p
Eigenvalues: 123
1, 1, 2
20
cb
01
Section 6.3: Applications Involving Powers of Matrices 365
9. Characteristic polynomial: 2
() ( 1)( 2)p
Eigenvalues: 12 3
1, 2, 2
With
11:
30
0
cb
b
1
1
0
v
10. Characteristic polynomial: 2
() ( 1)( 2)p
Eigenvalues: 12 3
1, 2, 2
33 0
abc
1
366 Chapter 6: Eigenvalues and Eigenvectors
11. Characteristic polynomial: 3
() ( 1)( 1)p
Eigenvalues: 123
1, 0, 1
With
11:
20
662 0
21 15 5 0
a
abc
abc
1
0
1
3
v
12. Characteristic polynomial: 22
( ) (1 )( 1) ( 1)( 1)p
Eigenvalues: 123
1, 1, 1
Section 6.3: Applications Involving Powers of Matrices 367
13. Characteristic polynomial: 3
() ( 1)( 1)p
Eigenvalues: 123
1, 0, 1
With
11:
20
20
442 0
abc
abc
abc
1
1
0
2
v
14. Characteristic polynomial: 3
() ( 1)( 1)p
Eigenvalues: 123
1, 0, 1
368 Chapter 6: Eigenvalues and Eigenvectors
With
11:
653 0
20
442 0
abc
abc
abc
1
1
0
2
v
15. 22
() 3 2so 3 2 0p
AAI
2
5 4 1 0 13 12
32 3 2
32 01 9 8
AAI
16. 22
() 3 2so 3 2 0p
AAI
Section 6.3: Applications Involving Powers of Matrices 369
17. 32 3 2
() 5 8 4so 5 8 4p
AAAI0
232
190 1210
040, 5 8 4 0 8 0
004 0 0 8
AAAAI
18. 32 3 2
() 4 5 2so 4 5 2p
AAAI0
232
163 1147
010, 4 5 2 0 1 0
064 0148
AAAAI
19. 32 3 2
() 5 8 4so 5 8 4p
AAAI0
232
193 1217
040, 5 8 4 0 8 0
004 0 0 8
AAAAI
370 Chapter 6: Eigenvalues and Eigenvectors
20. 32 3 2
() 5 8 4so 5 8 4p
AAAI0
232
10 9 3 22 21 7
653, 5 84 14137
004 0 08
AAAAI
21. 33
() sop
AA 0
23
100 100
78 5 2 , 6 5 2
195 15 6 21 15 6
AAA
22. 32 3 2
() 1sop
AAAI 0
232
100 11 6 2
0 1 0 , 20 11 4
001 0 0 1
AIAAAI A
Section 6.3: Applications Involving Powers of Matrices 371
23. 33
() sop
AA 0
23
331 111
221, 221
001 4 41
AAA
24. 33
() sop
AA 0
23
331 553
221, 221
AAA
In Problems 25–30 we first find the eigensystem of the given transition matrix A so as to
determine its eigenvector matrix P and its diagonal eigenvalue matrix D. Then we determine how
the matrix power Ak = PDkP–1 behaves as .k For simpler calculations of eigenvalues and
eigenvectors, we write the entries of A in fractional rather than decimal form.
25. Characteristic polynomial: 2941
() ( 1)(5 4)
555
p
Eigenvalues: 12
4
1, 5
372 Chapter 6: Eigenvalues and Eigenvectors
1
00
0
000
0
11 10 11
1
,,
11 04/5 11
2
1110 11
1
11 04/5 11
2
1110 11 11 1/2
11
1 1 0 0 11 11 1/2
22
k
k
k
CCS
S
PD P
xAx x
x
as .k Thus the long-term distribution of population is 50% city, 50% suburban.
26. Characteristic polynomial: 2941
() ( 1)(5 4)
555
p
Eigenvalues: 12
4
1, 5
20 20
1
00
11 10 11
1
,,
31 04/5 31
4
1110 11
1
31 04/5 31
4
k
k
k
PD P
xAx x
Section 6.3: Applications Involving Powers of Matrices 373
as .k Thus the long-term distribution of population is 25% city, 75% suburban.
27. Characteristic polynomial: 2831
( ) ( 1)(5 3)
555
p
Eigenvalues: 12
3
1, 5
1
31 10 11
1
,,
51 03/5 53
8
PD P
00
3110 11
1
51 03/5 53
8
k
k
k
xAx x
28. Characteristic polynomial: 217 7 1
() ( 1)(10 7)
10 10 10
p
Eigenvalues: 12
7
1, 10
374 Chapter 6: Eigenvalues and Eigenvectors
1
11 1 0 11
1
,,
21 07/10 21
3
PD P
29. Characteristic polynomial: 237 17 1
( ) ( 1)(20 17)
20 20 20
p
Eigenvalues: 12
17
1, 20
With
11:
11 0
10 20
11 0
10 20
ab
ab
1
1
2
v
00
111 0 11
1
2 1 017/20 21
3
k
k
k
xAx x
30. Characteristic polynomial: 233 13 1
( ) ( 1)(20 13)
20 20 20
p