Chapter 4: Eigenvalues and Eigenvectors
April 17, 2009
Week 1112
1 Eigenvalues and eigenvectors
If a linear transformation
T
:
R
n
→
R
n
has the form
T
x
1
x
2
.
.
.
x
n
=
λ
1
0
···
0
0
λ
2
···
0
.
.
.
.
.
.
.
.
.
.
.
.
0
0
···
λ
n
x
1
x
2
.
.
.
x
n
,
then we can easily see that
T
(
e
i
) =
λ
i
e
i
(
i
= 1
,
2
,...,n
), i.e., each
e
i
is dilated
λ
i
times by
T
.
Example 1.1.
The linear transformation
T
:
R
2
→
R
2
, deﬁned by
T
(
x
) =
•
2 0
0 3
‚•
x
1
x
2
‚
=
•
2
x
1
3
x
2
‚
,
is to dilate the ﬁrst coordinate two times and the second coordinate three times.
Example 1.2.
Let
T
:
R
2
→
R
2
be a linear transformation deﬁned by
T
(
x
) =
•
4

2
3

3
‚•
x
1
x
2
‚
.
What can we say geometrically about
T
? Consider the basis
B
=
{
u
1
,
u
2
}
of
R
2
, where
u
1
=
•
2
1
‚
,
u
2
=
•
1
3
‚
.
Then
T
(
u
1
) =
•
6
3
‚
= 3
•
2
1
‚
= 3
u
1
,
T
(
u
2
) =
•

2

6
‚
=

2
•
1
3
‚
=

2
u
2
.
For any vector
v
=
c
1
u
1
+
c
2
u
2
, we have [
v
]
B
=
•
c
1
c
2
‚
, and
T
(
v
) =
c
1
T
(
u
1
) +
c
2
T
(
u
2
) = 3
c
1
u
1

2
c
2
u
2
,
1
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[
T
(
v
)]
B
=
•
3
c
1

2
c
2
‚
=
•
3
0
0

2
‚•
c
1
c
2
‚
.
If one uses the basis
B
to describe vector
v
with coordinate vector [
v
]
B
, then the coordinate vector
of
T
(
v
) under the basis
B
simply described as
£
T
(
v
)
/
B
=
•
3
0
0

2
‚
£
v
/
B
.
This means that the matrix of
T
relative to the basis
B
is as simple as a diagonal matrix.
The above discussion demonstrates that for a linear transformation
T
:
V
→
V
, the nonzero
vectors
v
satisfying the condition
T
(
v
) =
λ
v
(1.1)
for some scalar
λ
is important in simplifying a linear transformation
T
.
Deﬁnition 1.1.
Given a linear transformation
T
:
R
n
→
R
n
with
T
(
x
) =
A
x
. A nonzero vector
v
in
R
n
is called an
eigenvector
of
T
(the matrix
A
) if there exists a scalar
λ
such that
T
(
v
) =
A
v
=
λ
v
.
(1.2)
The scalar
λ
is called an
eigenvalue
of
T
(the matrix
A
) and the nonzero vector
v
is called an
eigenvector of
T
(the matrix
A
) corresponding to the eigenvalue
λ
.
Example 1.3.
Let
A
=
•
1 6
5 2
‚
. Then
u
=
•
6

5
‚
is an eigenvector of
A
. However, but
v
=
•
3

2
‚
is not an eigenvector of
A
.
Proposition 1.2.
For any
n
×
n
matrix
A
, the value
0
is an eigenvalue of
A
⇐⇒
det
A
= 0
.
Proof.
Note that det
A
= 0
⇐⇒
A
is not invertible
⇐⇒
Nul
A
6
=
{
0
}
. The set of eigenvectors of
A
corresponding to the zero eigenvalue is the set Nul
A
 {
0
}
.
Theorem 1.3.
Let
v
1
,...,
v
p
be eigenvectors of a matrix
A
corresponding to distinct eigenvalues
λ
1
,...,λ
p
, respectively. Then
v
1
,...,
v
p
are linearly independent.
Proof.
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 Fall '08
 BeifangChan
 Calculus, Linear Algebra, Eigenvectors, Vectors, λ, .

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