Eigenvalues, Eigenvectors, and Eigenspaces

DEFINITION: Let A be a square matrix of size n.If a NONZERO vector ~x ∈ R

n and a scalar λ satisfy

A~x = λ~x, or, equivalently, (A − λIn)~x = 0,

scalar λ is called an eigenvalue of A,vector ~x 6= 0 is called an eigenvector of A associated with eigenvalue λ,and the null space of A − λI

nis called the eigenspace of A associated with eigenvalue λ.

HOW TO COMPUTE?

The eigenvalues of A are given by the roots of the polynomial

det(A − λIn) = 0.

The corresponding eigenvectors are the nonzero solutions of the linear system

(A − λIn)~x = 0.

Collecting all solutions of this system, we get the corresponding eigenspace.

EXERCISES:

For each given matrix, find the eigenvalues, and for each eigenvalue give a basis of thecorresponding eigenspace.

(a)

[

1 01 2

]

, (b)

[

2 01 2

]

, (c)

[

1 3−1 5

]

,

(d)

2 0 00 3 00 4 3

, (e)

−1 4 −2−3 4 0−3 1 3

, (f)

0 0 −21 2 11 0 3

, (g)

2 4 3−4 −6 −33 3 1

,

(h)

−1 1 1 −2−1 1 3 21 1 −1 −20 −1 −1 1

.

Turn over for the answers

Answers:

(a) Eigenvalues: λ1 = 1, λ2 = 2

The eigenspace associated to λ1 = 1, which is Ker(A − I): v1 =

[

−11

]

gives a basis.

The eigenspace associated to λ2 = 2, which is Ker(A − 2I): v2 =

[

01

]

gives a basis.

(b) Eigenvalues: λ1 = λ2 = 2

Ker(A − 2I), the eigenspace associated to λ1 = λ2 = 2: v1 =

[

01

]

gives a basis.

(c) Eigenvalues: λ1 = 2, λ2 = 4

Ker(A − 2I), the eigenspace associated to λ1 = 2: v1 =

[

31

]

gives a basis.

Ker(A − 4I), the eigenspace associated to λ2 = 4: v2 =

[

11

]

gives a basis.

(d) Eigenvalues: λ1 = 2, λ2 = λ3 = 3

Ker(A − 2I), the eigenspace associated to λ1 = 2: v1 =

100

gives a basis.

Ker(A − 3I), the eigenspace associated to λ2 = λ3 = 3: v2 =

001

gives a basis.

(e) Eigenvalues: λ1 = 1, λ2 = 2, λ3 = 3

The eigenspace associated to λ1 = 1: v1 =

111

gives a basis.

The eigenspace associated to λ2 = 2: v2 =

2/311

gives a basis.

The eigenspace associated to λ3 = 3: v3 =

1/43/41

gives a basis.

(f) Eigenvalues: λ1 = 1, λ2 = λ3 = 2

The eigenspace associated to λ1 = 1: v1 =

−211

gives a basis.

The eigenspace associated to λ2 = λ3 = 2: v2 =

010

, v3 =

−101

form a basis.

(g) Eigenvalues: λ1 = 1, λ2 = λ3 = −2

The eigenspace associated to λ1 = 1: v1 =

1−11

gives a basis.

The eigenspace associated to λ2 = λ3 = −2: v2 =

−110

gives a basis.

(h) Eigenvalues: λ1 = −1, λ2 = 1, λ3 = −2, λ4 = 2

The eigenspace associated to λ1 = −1: v1 =

1−110

gives a basis.

The eigenspace associated to λ2 = 1: v2 =

−11−11

gives a basis.

The eigenspace associated to λ3 = −2: v3 =

0−110

gives a basis.

The eigenspace associated to λ4 = 2: v4 =

−10−11

gives a basis.