Lecture 30
Andrei Antonenko
April 25, 2003
1 Eigenvectors. Eigenvalues
Last lecture we saw, that in order to ﬁnd vectors, “stretched” by the operator with matrix
A
,
we need to solve the characteristic equation
det(
A

λI
) = 0
,
(1)
which will give us diﬀerent
λ
i
’s — coeﬃcients, showing, how the vectors are changed after
applying the operator. Now we will give the following deﬁnition.
Deﬁnition 1.1.
Let
V
be a vector space, and let
A
be a linear operator in vector space
V
.
Then the vector
x
is called
eigenvector
of the operator
A
is there exist a number
λ
, which is
called
eigenvalue
such that
A
(
x
) =
λx.
So, our goal is to ﬁnd eigenvectors, since the following proposition holds:
Proposition 1.2.
Let
V
be an
n
dimensional vector space, and
A
be a linear operator. Then
if there are
n
linearly independent eigenvectors, then the matrix of
A
is diagonal in the basis,
consisting of eigenvectors.
So far we know how to ﬁnd
λ
i
’s — eigenvalues of the operator. In order to ﬁnd eigenvectors,
we need to solve the system
(
A

λ
i
I
)
x
= 0
(2)
for every found eigenvalue
λ
i
.
We will give an example of computing eigenvalues and eigenvectors.
Example 1.3.
Let
A
=
ˆ
1

3
1
5
!
. Let’s compute its eigenvalues and eigenvectors.
A

λI
=
ˆ
1

λ

3
1
5

λ
!
1
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View Full Documentdet(
A

λI
) = (1

λ
)(5

λ
) + 3
= 5

6
λ
+
λ
2
+ 3
=
λ
2

6
λ
+ 8
.
The roots of this equation are
λ
1
= 2
and
λ
2
= 4
. Now we’ll ﬁnd eigenvectors corresponding to
these eigenvalues.
λ
= 2
.
Let’s subtract
λ
’s from the diagonal. We’ll get the following matrix, and the system:
ˆ

1

3
1
3
!
,
(

x

3
y
= 0
x
+ 3
y
= 0
From this system it follows that
x
=

3
y
, so each vector of the form
(

3
c,c
)
, i.e.
(

3
,
1)
is an eigenvector corresponding to the eigenvalue
λ
= 2
.
λ
= 4
.
Let’s subtract
λ
’s from the diagonal. We’ll get the following matrix, and the system:
ˆ

3

3
1
1
!
,
(

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 Spring '03
 Andant
 Linear Algebra, Eigenvectors, Characteristic polynomial, λ, Det

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