Math311-102 June14, 2005: slide #2
Eigenvectors and eigenvalues
A non-zero vector
vis an eigenvector of a matrix Aif A
vis a
scalar multiple of
v.
Example. If A
3
1 1
1 1
1
1
1 4
and
v
1
1
0
,
then A
v
2
2
0
2
v.
Thus
vis an eigenvector of Awith eigenvalue equal to 2.
How can you find eigenvectors?
The strategy is to find the eigenvalues first.
Math311-102 June14, 2005: slide #3
Finding eigenvalues
Given a matrix A, the goal is to find a non-zero vector
vand a
scalar λsuch that A
v
λ
v. An equivalent equation is
A
λI
v
0, where Iis the identity matrix.
There can be such a non-zero vector
vonly if the matrix
A
λIfails to be invertible.
The equation det
A
λI
0determines the eigenvalues; it is
called the characteristic equation of the matrix A.
Example. Find the eigenvalues of
13 5
30 12
.
Solution. Solve det
13
λ5
30 12
λ
0or
13
λ
12
λ
150
0or λ2
λ
6
0or
λ
3
λ
2
0. The eigenvalues are 2and
3.
Math311-102 June14, 2005: slide #4
Finding eigenvectors
Continuing the example: the matrix A
13 5
30 12
has
eigenvalues 2and
3. Find corresponding eigenvectors.
Solution. The eigenvector
vcorresponding to eigenvalue 2
satisfies
A
2I
v
0or
15 5
30 10
v1
v2
0. Row
reducing gives
3 1
0 0
v1
v2
0, so
v1
v2
1
3
.
Check:
13 5
30 12
1
3
2
6
2
1
3
.
Similarly solving
A
3I
v
0or
10 5
30 15
v1
v2
0gives
1
2
as an eigenvector with eigenvalue
3.
