Section 6.1
289
CHAPTER 6
EIGENVALUES AND EIGENVECTORS
SECTION 6.1
INTRODUCTION TO EIGENVALUES
In each of Problems 1–32 we first list the characteristic polynomial
( )
p
λ
λ
=
−
A
I
of the given
matrix
A
, and then the roots of
( )
p
λ
— 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
each eigenvalue
j
λ
of the matrix
A
,
we determine the associated eigenvector(s) by finding a basis
for the solution space of the linear system
(
)
.
j
λ
−
=
A
I v
0
We write this linear system in scalar
form in terms of the components of
[
]
.
T
a
b
=
v
In most cases an associated eigenvector is
then apparent.
If
A
is a
2
2
×
matrix, for instance, then our two scalar equations will be multiples
one of the other, so we can substitute a convenient numerical value for the first component
a
of
v
and then solve either equation for the second component
b
(or vice versa).
1.
Characteristic polynomial:
2
( )
5
6
(
2)(
3)
p
λ
λ
λ
λ
λ
=
−
+
=
−
−
Eigenvalues:
1
2
2,
3
λ
λ
=
=
With
1
2:
λ
=
2
2
0
0
a
b
a
b
−
=
−
=
1
1
1
=
v
With
2
3:
λ
=
2
0
2
0
a
b
a
b
−
=
−
=
2
2
1
=
v
2.
Characteristic polynomial:
2
( )
2
(
1)(
2)
p
λ
λ
λ
λ
λ
=
−
−
=
+
−
Eigenvalues:
1
2
1,
2
λ
λ
= −
=
With
1
1:
λ
= −
6
6
0
3
3
0
a
b
a
b
−
=
−
=
1
1
1
=
v
With
2
2:
λ
=
3
6
0
3
6
0
a
b
a
b
−
=
−
=
2
2
1
=
v
290
Chapter 6
3.
Characteristic polynomial:
2
( )
7
10
(
2)(
5)
p
λ
λ
λ
λ
λ
=
−
+
=
−
−
Eigenvalues:
1
2
2,
5
λ
λ
=
=
With
1
2:
λ
=
6
6
0
3
3
0
a
b
a
b
−
=
−
=
1
1
1
=
v
With
2
5:
λ
=
3
6
0
3
6
0
a
b
a
b
−
=
−
=
2
2
1
=
v
4.
Characteristic polynomial:
2
( )
3
2
(
1)(
2)
p
λ
λ
λ
λ
λ
=
−
+
=
−
−
Eigenvalues:
1
2
2,
5
λ
λ
=
=
With
1
1:
λ
=
3
3
0
2
2
0
a
b
a
b
−
=
−
=
1
1
1
=
v
With
2
2:
λ
=
2
3
0
2
3
0
a
b
a
b
−
=
−
=
2
3
2
=
v
5.
Characteristic polynomial:
2
( )
5
4
(
1)(
4)
p
λ
λ
λ
λ
λ
=
−
+
=
−
−
Eigenvalues:
1
2
1,
4
λ
λ
=
=
With
1
1:
λ
=
9
9
0
6
6
0
a
b
a
b
−
=
−
=
1
1
1
=
v
With
2
4:
λ
=
6
9
0
6
9
0
a
b
a
b
−
=
−
=
2
3
2
=
v
6.
Characteristic polynomial:
2
( )
5
6
(
2)(
3)
p
λ
λ
λ
λ
λ
=
−
+
=
−
−
Eigenvalues:
1
2
1,
4
λ
λ
=
=
With
1
2:
λ
=
4
4
0
3
3
0
a
b
a
b
−
=
−
=
1
1
1
=
v
With
2
3:
λ
=
3
4
0
3
4
0
a
b
a
b
−
=
−
=
2
4
3
=
v
7.
Characteristic polynomial:
2
( )
6
8
(
2)(
4)
p
λ
λ
λ
λ
λ
=
−
+
=
−
−
Eigenvalues:
1
2
2,
4
λ
λ
=
=
Section 6.1
291
With
1
2:
λ
=
8
8
0
6
6
0
a
b
a
b
−
=
−
=
1
1
1
=
v
With
2
4:
λ
=
6
8
0
6
8
0
a
b
a
b
−
=
−
=
2
4
3
=
v
8.
Characteristic polynomial:
2
( )
3
2
(
2)(
1)
p
λ
λ
λ
λ
λ
=
+
+
=
+
+
Eigenvalues:
1
2
2,
1
λ
λ
= −
= −
With
1
2:
λ
= −
9
6
0
12
8
0
a
b
a
b
−
=
−
=
1
2
3
=
v
With
2
1:
λ
= −
8
6
0
12
9
0
a
b
a
b
−
=
−
=
2
3
4
=
v
9.
