LINEAR ALGEBRA
W W L CHEN
c
W W L Chen, 1982, 2005.
This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990.
It is available free to all individuals, on the understanding that it is not to be used for financial gain,
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Chapter 7
EIGENVALUES AND EIGENVECTORS
7.1. Introduction
Example 7.1.1.
Consider a function
f
:
R
2
→
R
2
, defined for every (
x, y
)
∈
R
2
by
f
(
x, y
) = (
s, t
),
where
s
t
=
3
3
1
5
x
y
.
Note that
3
3
1
5
3
−
1
=
6
−
2
= 2
3
−
1
and
3
3
1
5
1
1
=
6
6
= 6
1
1
.
On the other hand, note that
v
1
=
3
−
1
and
v
2
=
1
1
form a basis for
R
2
. It follows that every
u
∈
R
2
can be written uniquely in the form
u
=
c
1
v
1
+
c
2
v
2
,
where
c
1
, c
2
∈
R
, so that
A
u
=
A
(
c
1
v
1
+
c
2
v
2
) =
c
1
A
v
1
+
c
2
A
v
2
= 2
c
1
v
1
+ 6
c
2
v
2
.
Note that in this case, the function
f
:
R
2
→
R
2
can be described easily in terms of the two special
vectors
v
1
and
v
2
and the two special numbers 2 and 6. Let us now examine how these special vectors
and numbers arise. We hope to find numbers
λ
∈
R
and nonzero vectors
v
∈
R
2
such that
3
3
1
5
v
=
λ
v
.
Chapter 7 : Eigenvalues and Eigenvectors
page 1 of 12
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Linear Algebra
c
W W L Chen, 1982, 2005
Since
λ
v
=
λ
1
0
0
1
v
=
λ
0
0
λ
v
,
we must have
3
3
1
5
−
λ
0
0
λ
v
=
0
.
In other words, we must have
(1)
3
−
λ
3
1
5
−
λ
v
=
0
.
In order to have nonzero
v
∈
R
2
, we must therefore ensure that
det
3
−
λ
3
1
5
−
λ
= 0
.
Hence (3
−
λ
)(5
−
λ
)
−
3 = 0, with roots
λ
1
= 2 and
λ
2
= 6. Substituting
λ
= 2 into (1), we obtain
1
3
1
3
v
=
0
,
with root
v
1
=
3
−
1
.
Substituting
λ
= 6 into (1), we obtain
−
3
3
1
−
1
v
=
0
,
with root
v
2
=
1
1
.
Definition.
Suppose that
(2)
A
=
a
11
. . .
a
1
n
.
.
.
.
.
.
a
n
1
. . .
a
nn
is an
n
×
n
matrix with entries in
R
. Suppose further that there exist a number
λ
∈
R
and a nonzero
vector
v
∈
R
n
such that
A
v
=
λ
v
. Then we say that
λ
is an eigenvalue of the matrix
A
, and that
v
is
an eigenvector corresponding to the eigenvalue
λ
.
Suppose that
λ
is an eigenvalue of the
n
×
n
matrix
A
, and that
v
is an eigenvector corresponding to
the eigenvalue
λ
. Then
A
v
=
λ
v
=
λI
v
, where
I
is the
n
×
n
identity matrix, so that (
A
−
λI
)
v
=
0
.
Since
v
∈
R
n
is nonzero, it follows that we must have
(3)
det(
A
−
λI
) = 0
.
In other words, we must have
det
a
11
−
λ
a
12
. . .
a
1
n
a
21
a
22
−
λ
a
2
n
.
.
.
.
.
.
.
.
.
a
n
1
a
n
2
. . .
a
nn
−
λ
= 0
.
Note that (3) is a polynomial equation. Solving this equation (3) gives the eigenvalues of the matrix
A
.
On the other hand, for any eigenvalue
λ
of the matrix
A
, the set
(4)
{
v
∈
R
n
: (
A
−
λI
)
v
=
0
}
is the nullspace of the matrix
A
−
λI
, a subspace of
R
n
.
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 Linear Algebra, Eigenvectors, Eigenvalues, W W L Chen

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