Chapter 2.
Eigenvalues, Eigenvectors, Diagonalization
2.1.
Eigenvalues and Eigenvectors
Definition 2.1
Let
A
∈
M
n
(
R
)
. A nonzero vector
x
∈
M
n
×
1
(
R
)
is called an
eigenvector
of
A
if
A
x
is a scalar multiple of
x
, that is,
A
x
=
λ
x
(2.1)
for some
λ
∈
R
. The scalar
λ
is called an
eigenvalue
of
A
and
x
is said to be an eigenvector of
A
corresponding to
λ
.
In other words, the effect of multiplying
x
by the matrix
A
is equivalent to multiplying
x
by the
scalar
λ
. The geometric interpretation is that an eigenvector of a given matrix is a vector which
is rescaled by a constant (eigenvalue), possibly reverse the direction, when we apply the matrix to
the vector. For example, an eigenvalue
2
means that the eigenvector is doubled in length, while
an eigenvalue

1
means that the eigenvector stays the same in length but is reversed in direction.
Example 2.2
Let
A
=
6

1
16

4
!
and
x
=
1
8
!
. Then
A
x
=
6

1
16

4
!
·
1
8
!
=

2

16
!
=

2
1
8
!
=

2
x
.
Hence

2
is an eigenvalue of
A
and the vector
1
8
!
is an eigenvector of
A
corresponding to the
eigenvalue

2
.
2
Remark
Note that if
x
is an eigenvector of
A
corresponding to
λ
, then
k
x
,
k
6
= 0
, is also an
eigenvector of
A
corresponding to
λ
.
Equation (2.1) can be written as
(
A

λI
n
)
x
=
0
.
Let
B
=
A

λI
n
. If
λ
is an eigenvalue of
A
, then there exists a nonzero vector
x
such that
B
x
= 0
. Proposition 1.46 then implies that
B
is
not
invertible, that is,
B
is singular, or
det(
B
) = 0
.
Therefore, in order to find eigenvalues of
A
, we need to solve
det(
A

λI
n
) = 0
.
(2.2)
This equation is called the
characteristic equation
of
A
. The following gives the procedures of
finding eigenvalues and their corresponding eigenvectors.
Procedures to find eigenvalues and their corresponding eigenvectors
1. Find all the eigenvalues of
A
by solving the characteristic equation (2.2).
2. If
λ
0
is an eigenvalue of
A
, then the eigenvectors of
A
corresponding to
λ
0
are exactly
those NONTRIVIAL solutions of the system
(
A

λ
0
I
n
)
x
=
0
.
Example 2.3
Let
A
=
6

1
16

4
!
.
Find all the eigenvalues of
A
and their corresponding
eigenvectors.
Solution.
The characteristic equation of
A
is given by
6

λ

1
16

4

λ
= 0
.
Hence the eigenvalues of
A
are the roots of the equation
λ
2

2
λ

8
=
0
, that is,
λ
1
=

2
and
λ
2
= 4
. For
λ
1
=

2
, we have
A
+ 2
I
2
=
8

1
16

2
!
.
To find eigenvectors corresponding to
λ
1
=

2
, we solve
8

1
16

2
!
·
x
y
!
=
0
0
!
.
Calculation shows that the nontrivial solution to this system is given by
x
y
!
=
α
1
8
!
for any
α
∈
R
\ {
0
}
. Thus every vector of the form
α
1
8
!
is an eigenvector of
A
corresponding
to the eigenvalue
λ
1
=

2
. Similarly, for
λ
2
= 4
, we solve
2

1
16

8
!
·
x
y
!
=
0
0
!
,
which gives the nontrivial solution
x
y
!
=
β
1
2
!
for any
β
∈
R
\ {
0
}
, and so we have obtained eigenvectors of
A
corresponding to the eigenvalue
λ
2
= 4
.
2
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Example 2.4
Let
A
=
5

4
1
1
!
.
Find all the eigenvalues of
A
and their corresponding
eigenvectors.
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