MATH 2040 COURSE NOTES
ROB NOBLE
The following course notes are based closely on [Poole2006].
4.
Eigenvalues and Eigenvectors
4.1.
Introduction to Eigenvalues and Eigenvectors.
Definition 1
(Eigenvalues and Eigenvectors)
.
A scalar
λ
2
R
is called an
eigenvalue
of the matrix
A
2
R
n
⇥
n
provided
A
x
=
λ
x
for at least one nonzero vector
x
2
R
n
. Any such vector
x
2
R
n
is
referred to as an
eigenvector
associated to
λ
.
Example 1.
Show that
v
= [2
,

1
,
1]
T
is an eigenvector of the matrix
A
=
2
6
4
3
0
0
0
1

2
1
0
1
3
7
5
and
determine the associated eigenvalue.
Solution.
We need to verify that
A
v
is a scalar multiple of
v
, and determine the multiple. We
compute
A
v
=
2
6
4
3
0
0
0
1

2
1
0
1
3
7
5
2
6
4
2

1
1
3
7
5
=
2
6
4
6

3
3
3
7
5
= 3
2
6
4
2

1
1
3
7
5
= 3
v
.
Therefore
A
v
is indeed a scalar multiple of
v
, and the multiple is 3. Thus
v
is an eigenvector of
A
with associated eigenvalue
λ
= 3
.
For
A
2
R
n
⇥
n
, and
λ
2
R
, we have the following chain of equivalences:
x
is an eigenvector of
A
associated to
λ
()
x
6
=
0
and
A
x
=
λ
x
()
x
6
=
0
and
A
x

λ
x
=
0
()
x
6
=
0
and (
A

λ
I
)
x
=
0
()
x
is a nonzero vector in the null space of
A

λ
I
.
Date
: March 19, 2014.
1
2
ROB NOBLE
The eigenvectors of
A
associated to
λ
are therefore the nonzero vectors in the null space of
A

λ
I
.
This particular null space associated to
λ
is of great importance.
Definition 2
(Eigenspace)
.
Let
λ
2
R
be an eigenvalue of
A
2
R
n
⇥
n
. By the
eigenspace
associated
to
λ
, denoted
E
λ
, we mean the null space of
A

λ
I
.
Therefore, the eigenspace
E
λ
consists of all vectors on which
A
acts as scalar multiplication by
λ
.
That is
E
λ
=
{
x
2
R
n

A
x
=
λ
x
}
consists of the zero vector together with all the eigenvectors of
A
associated to
λ
. In particular, if
we know an eigenvalue
λ
of
A
2
R
n
⇥
n
, we can find the associated eigenvectors by calculating the
null space of the matrix
A

λ
I
.
Example 2.
Show that
λ
=

1 is an eigenvalue of the matrix
A
=
2
6
4
1
0
2

1
1
1
2
0
1
3
7
5
and determine
the associated eigenspace.
Solution.
We need to show that
A
v
= (

1)
v
for at least one nonzero vector
v
and then determine
the eigenspace
E

1
consisting of all vectors
v
(including
0
) in
R
3
for which
A
v
= (

1)
v
. Since
it will be easier, we will first compute the null space of
A

(

1)
I
in order to find
E

1
, and then
verify that

1
is indeed an eigenvalue of
A
by choosing a nonzero vector in
E

1
and demonstrating
that
A
transforms this vector into its negative. We compute
A

(

1)
I
=
2
6
4
2
0
2

1
2
1
2
0
2
3
7
5
⇠
2
6
4
2
0
2

1
2
1
0
0
0
3
7
5
⇠
2
6
4
2
0
2
0
2
2
0
0
0
3
7
5
⇠
2
6
4
1
0
1
0
2
2
0
0
0
3
7
5
⇠
2
6
4
1
0
1
0
1
1
0
0
0
3
7
5
.
The null space in question is therefore
E

1
= span
⇣
[

1
,

1
,
1]
T
⌘
.
All that remains is to verify that

1
is in fact an eigenvalue of
A
. We need to produce a nonzero
vector that
A
sends to its negative.
Any nonzero vector in
E

1
will do, and so we choose
v
=
[

1
,

1
,
1]
and complete the solution with the computation
A
v
=
2
6
4
1
0
2

1
1
1
2
0
1
3
7
5
2
6
4

1

1
1
3
7
5
=
2
6
4
1
1

1
3
7
5
=

v
.
MATH 2040 COURSE NOTES
3
The question arises: “How do we determine the eigenvalues of a matrix?” Well, when
A
is a matrix
whose action on vectors is known (e.g. a rotation matrix or reflection matrix), then the eigenvectors