CHAPTER 6
EIGENVALUES AND EIGENVECTORS
SECTION 6.1
INTRODUCTION TO EIGENVALUES
In each of Problems 1–32 we first list the characteristic polynomial ()p
AI of the given
matrix A, and then the roots of ()
p
— which are the eigenvalues of A. All of the eigenvalues
that appear in Problems 1–26 are integers, so each characteristic polynomial factors readily. For
1. Characteristic polynomial: 2
() 5 6 ( 2)( 3)p
Eigenvalues: 12
2, 3
ab
1
2. Characteristic polynomial: 2
() 2 ( 1)( 2)p
Eigenvalues: 12
1, 2
ab
1
336 Chapter 6: Eigenvalues and Eigenvectors
3. Characteristic polynomial: 2
() 7 10 ( 2)( 5)p
Eigenvalues: 12
2, 5
ab
1
4. Characteristic polynomial: 2
() 3 2 ( 1)( 2)p
Eigenvalues: 12
2, 5
ab
1
5. Characteristic polynomial: 2
() 5 4 ( 1)( 4)p
Eigenvalues: 12
1, 4
ab
1
6. Characteristic polynomial: 2
() 5 6 ( 2)( 3)p
Eigenvalues: 12
1, 4
ab
1
7. Characteristic polynomial: 2
() 6 8 ( 2)( 4)p
Eigenvalues: 12
2, 4
Section 6.1: Introduction to Eigenvalues 337
ab
1
8. Characteristic polynomial: 2
() 3 2 ( 2)( 1)p
Eigenvalues: 12
2, 1
ab
2
9. Characteristic polynomial: 2
() 7 12 ( 3)( 4)p
Eigenvalues: 12
3, 4
ab
2
10. Characteristic polynomial: 2
() 9 20 ( 4)( 5)p
Eigenvalues: 12
4, 5
ab
2
11. Characteristic polynomial: 2
() 9 20 ( 4)( 5)p
Eigenvalues: 12
4, 5
ab
2
338 Chapter 6: Eigenvalues and Eigenvectors
12. Characteristic polynomial: 2
( ) 7 212 ( 3)( 4)p
Eigenvalues: 12
3, 4
ab
3
13. Characteristic polynomial: 32
() 3 2 ( 1)( 2)p
Eigenvalues: 123
0, 1, 2
With
10:
20
22 0
a
abc
1
0
1
v
14. Characteristic polynomial: 32
() 7 10 ( 2)( 5)p
Eigenvalues: 123
0, 2, 5
50
a
0
Section 6.1: Introduction to Eigenvalues 339
15. Characteristic polynomial: 32
() 3 2 ( 1)( 2)p
Eigenvalues: 123
0, 1, 2
22 0
ab
1
16. Characteristic polynomial: 32
() 4 3 ( 1)( 3)p
Eigenvalues: 123
0, 1, 3
0
ac
1
17. Characteristic polynomial: 32
() 6 11 6 ( 1)( 2)( 3)p
Eigenvalues: 12 3
1, 2, 3
252 0
abc
1
340 Chapter 6: Eigenvalues and Eigenvectors
18. Characteristic polynomial: 32
() 6 11 6 ( 1)( 2)( 3)p
Eigenvalues: 12 3
1, 2, 3
With
11:
00
672 0
abc
1
1
0
v
19. Characteristic polynomial: 32 2
() 5 7 3 ( 1)( 3)p
Eigenvalues: 12 3
1, 3
262 0
abc
1
3
20. Characteristic polynomial: 32 2
() 4 5 2 ( 1)( 2)p
Eigenvalues: 12 3
1, 2
Section 6.1: Introduction to Eigenvalues 341
00
1
3
21. Characteristic polynomial: 32 2
() 5 8 4 ( 1)( 2)p
Eigenvalues: 123
1, 2
33 0
abc
1
22. Characteristic polynomial: 32
() 3 2 ( 1)( 2)p
Eigenvalues: 12 3
1, 2
663 0
abc
1
1
342 Chapter 6: Eigenvalues and Eigenvectors
23. Characteristic polynomial: ( ) ( 1)( 2)( 3)( 4)p
Eigenvalues: 12 3 4
1, 2, 3, 4
With
11:
222 0
22 0
22 0
30
bcd
bcd
cd
d
1
1
0
0
0
v
00
1
24. Characteristic polynomial: 22
() ( 1)( 3)p
Eigenvalues: 12 34
1, 3
40
c
1
0
Section 6.1: Introduction to Eigenvalues 343
25. Characteristic polynomial: 22
() ( 1)( 2)p
Eigenvalues: 12 34
1, 2
0
c
1
0
26. Characteristic polynomial:
42 2 2
( ) 5 4 ( 1)( 4) ( 1)( 1)( 2)( 2)p
Eigenvalues: 1234
2, 1, 1, 2
With
12:
63 0
40
0
ad
b
c
1
1
0
0
v
344 Chapter 6: Eigenvalues and Eigenvectors
27. Characteristic polynomial: 2
() 1p
Eigenvalues: 12
,ii
ia b
i
28. Characteristic polynomial: 2
() 36p
Eigenvalues: 12
6, 6ii
21
29. Characteristic polynomial: 2
() 36p
Eigenvalues: 12
6, 6ii
30. Characteristic polynomial: 2
( ) 144p
Eigenvalues: 12
12 , 12ii
Section 6.1: Introduction to Eigenvalues 345
31. Characteristic polynomial: 2
( ) 144p
Eigenvalues: 12
12 , 12ii
32. Characteristic polynomial: 2
( ) 144p
Eigenvalues: 12
12 , 12ii
33. If
Av v and we assume that 11nn
Av v — meaning that 1n
is an eigenvalue
of 1n
A with associated eigenvector v, then multiplication by A yields
34. By the remark following Example 6, any eigenvalue of an invertible matrix is nonzero. If
0
is an eigenvalue of the invertible matrix A with associated eigenvalue v, then
35. (a) Note first that ( ) ( )
TT
AI A I because .
TII Since the determinant of a
square matrix equals the determinant of its transpose, it follows that
346 Chapter 6: Eigenvalues and Eigenvectors
(b) Consider the matrix 10
11
A with characteristic equation 2
(1)0
and the
36. If the nn matrix ij
a
A is either upper or lower triangular, then obviously its
characteristic equation is
37. If 1
110
(1)
nn n
n
ccc
AI , then substitution of 0
yields
00c AI A for the constant term in the characteristic polynomial.
39. If the characteristic equation of the nn matrix A with eigenvalues 12
,,,
n
(not
necessarily distinct) is written in the factored form
nnn
40. We find that trace A = 12 and det A = 60, so the characteristic polynomial of the given
matrix A is
Section 6.1: Introduction to Eigenvalues 347
Substitution of 1
and
so the other two eigenvalues are 23
4 and 5.
We proceed to find the eigenvectors
associated with these three eigenvalues.
With
13:
29 67 47 0 3 0
71713 0 2 0
7159 0 0 0
abc ac
abc bc
abc
1
3
2
1
v
41. We find that trace A = 8 and det A = –60, so the characteristic polynomial of the given
matrix A is
43 2
21
( ) 8 60.pcc
348 Chapter 6: Eigenvalues and Eigenvectors
21 21
43, 21cc cc
With 12:
24 9 8 8 0 (1/ 2) 0
10 5 14 2 0 0
10 10 10 0 (1/ 2) 0
29 9 3 13 0 0 0
abcd a d
ab cd b
acd c d
abc d
1
1
0
1
2
v
With 33:
19 9 8 8 0 (3/ 4) 0
10 10 14 2 0 (1/ 4) 0
10 5 10 0 (1/ 2) 0
29 9 3 18 0 0 0
abcd a d
abcd b d
ac d c d
abc d
3
3
1
2
4
v
Section 6.2: Diagonalization of Matrices 349
SECTION 6.2
DIAGONALIZATION OF MATRICES
In Problems 1–28 we first find the eigenvalues and associated eigenvectors of the given nn
matrix A. If A has n linearly independent eigenvectors, then we can proceed to set up the desired
diagonalizing matrix
12 n
Pvv v and diagonal matrix D such that P–1AP = D. If
you write the eigenvalues in a different order on the diagonal of D, then naturally the eigenvector
columns of P must be rearranged in the same order.
1. Characteristic polynomial: 2
() 4 3 ( 1)( 3)p
Eigenvalues: 12
1, 3
ab
1
2. Characteristic polynomial: 2
() 2 ( 2)p
Eigenvalues: 12
0, 2
3. Characteristic polynomial: 2
() 5 6 ( 2)( 3)p
Eigenvalues: 12
2, 3
350 Chapter 6: Eigenvalues and Eigenvectors
4. Characteristic polynomial: 2
() 3 2 ( 1)( 2)p
Eigenvalues: 12
1, 2
With
44 0
ab
1
v
5. Characteristic polynomial: 2
() 4 3 ( 1)( 3)p
Eigenvalues: 12
1, 3
With
11:
88 0
66 0
ab
ab
1
1
1
v
6. Characteristic polynomial: 2
() 3 2 ( 1)( 2)p
Eigenvalues: 12
1, 2
With
11:
96 0
12 8 0
ab
ab
1
2
3
v
Section 6.2: Diagonalization of Matrices 351
7. Characteristic polynomial: 2
() 3 2 ( 1)( 2)p
Eigenvalues: 12
1, 2
8. Characteristic polynomial: 2
() 3 2 ( 1)( 2)p
Eigenvalues: 12
1, 2
With
11:
10 15 0
69 0
ab
ab
3
2
v
9. Characteristic polynomial: 22
() 2 1 ( 1)p
Eigenvalues: 12
1, 1
10. Characteristic polynomial: 22
() 4 4 ( 2)p
Eigenvalues: 12
2, 2
352 Chapter 6: Eigenvalues and Eigenvectors
11. Characteristic polynomial: 22
() 4 4 ( 2)p
Eigenvalues: 12
2, 2
12. Characteristic polynomial: 22
() 2 1 ( 1)p
Eigenvalues: 12
1, 1
13. Characteristic polynomial: 32 2
() 5 8 4 ( 1)( 2)p
Eigenvalues: 12 3
1, 2, 2
With
11:
30
0
b
b
1
1
0
v
14. Characteristic polynomial: 32 2
() ( 1)p
Eigenvalues: 123
0, 0, 1
22 0
abc
11
Section 6.2: Diagonalization of Matrices 353
The eigenspace of 10
is 2-dimensional. We get the eigenvector v with
0, 2,bc and the eigenvector v2 with 1, 0.bc
15. Characteristic polynomial: 32 2
() 2 ( 1)p
Eigenvalues: 123
0, 1, 1
33 0
abc
1
The eigenspace of 22
is 2-dimensional. We get the eigenvector v2 with
0, 2,bc and the eigenvector v3 with 2, 0.bc
16. Characteristic polynomial: 32 2
() 5 7 3 ( 1)( 3)p
Eigenvalues: 123
1, 1, 3
22 0
ab
01
354 Chapter 6: Eigenvalues and Eigenvectors
10 2 003
17. Characteristic polynomial: 32
() 2 2 ( 1)( 1)( 2)p
Eigenvalues: 123
1, 1, 2
With
11:
883 0
663 0
223 0
abc
abc
abc
1
1
1
0
v
18. Characteristic polynomial: 32
() 6 11 6 ( 1)( 2)( 3)p
Eigenvalues: 12 3
1, 2, 3
552 0
abc
1