5Eigenvalues and eigenvectors

IA Vectors and Matrices

5.2 Linearly independent eigenvectors

Theorem.

Suppose

n×n

matrix

A

has distinct eigenvalues

λ

1

, λ

2

, ··· , λ

n

. Then

the corresponding eigenvectors x

1

, x

2

, ··· , x

n

are linearly independent.

Proof.

Proof by contradiction: Suppose

x

1

, x

2

, ··· , x

n

are linearly dependent.

Then we can find non-zero constants d

i

for i = 1, 2, ··· , r, such that

d

1

x

1

+ d

2

x

2

+ ··· + d

r

x

r

= 0.

Suppose that this is the shortest non-trivial linear combination that gives

0

(we

may need to re-order x

i

).

Now apply (A − λ

1

I) to the whole equation to obtain

d

1

(λ

1

− λ

1

)x

1

+ d

2

(λ

2

− λ

1

)x

2

+ ··· + d

r

(λ

r

− λ

1

)x

r

= 0.

We know that the first term is

0

, while the others are not (since we assumed

λ

i

6= λ

j

for i 6= j). So

d

2

(λ

2

− λ

1

)x

2

+ ··· + d

r

(λ

r

− λ

1

)x

r

= 0,

and we have found a shorter linear combination that gives

0

. Contradiction.

Example.

(i) A =

0 1

−1 0

. Then p

A

(λ) = λ

2

+ 1 = 0. So λ

1

= i and λ

2

= −i.

To solve (A − λ

1

I)x = 0, we obtain

−i 1

−1 −i

x

1

x

2

= 0.

So we obtain

x

1

x

2

=

1

i

to be an eigenvector. Clearly any scalar multiple of

1

i

is also a solution,

but still in the same eigenspace E

i

= span

1

i

Solving (A − λ

2

I)x = 0 gives

x

1

x

2

=

1

−i

.

So E

−i

= span

1

−i

.

Note that

M

(

±i

) =

m

(

±i

) = 1, so ∆

±i

= 0. Also note that the two

eigenvectors are linearly independent and form a basis of C

2

.

(ii) Consider

A =

−2 2 −3

2 1 −6

−1 −2 0

Then

det

(

A − λI

) = 0 gives 45 + 21

λ − λ

2

− λ

3

. So

λ

1

= 5

, λ

2

=

λ

3

=

−

3.

The eigenvector with eigenvalue 5 is

x =

1

2

−1

We can find that the eigenvectors with eigenvalue −3 are

x =

−2x

2

+ 3x

3

x

2

x

3

for any

x

2

, x

3

. This gives two linearly independent eigenvectors, say

−2

1

0

,

3

0

1

.

So

M

(5) =

m

(5) = 1 and

M

(

−

3) =

m

(

−

3) = 2, and there is no defect for

both of them. Note that these three eigenvectors form a basis of C

3

.

(iii) Let

A =

−3 −1 1

−1 −3 1

−2 −2 0

Then 0 =

p

A

(

λ

) =

−

(

λ

+ 2)

4

. So

λ

=

−

2

, −

2

, −

2. To find the eigenvectors,

we have

(A + 2I)x =

−1 −1 1

−1 −1 1

−2 −2 2

x

1

x

2

x

3

= 0

The general solution is thus

x

1

+

x

2

− x

3

= 0, and the general solution is

thus x =

x

1

x

2

x

1

+ x

2

. The eigenspace E

−2

= span

1

0

1

,

0

1

1

.

Hence

M

(

−

2) = 3 and

m

(

−

2) = 2. Thus the defect ∆

−2

= 1. So the

eigenvectors do not form a basis of C

3

.

(iv)

Consider the reflection

R

in the plane with normal

n

. Clearly

Rn

=

−n

.

The eigenvalue is

−

1 and the eigenvector is

n

. Then

E

1

=

span{n}

. So

M(−1) = m(−1) = 1.

If

p

is any vector in the plane,

Rp

=

p

. So this has an eigenvalue of 1 and

eigenvectors being any vector in the plane. So M(1) = m(1) = 2.

So the eigenvectors form a basis of R

3

.

(v)

Consider a rotation

R

by

θ

about

n

. Since

Rn

=

n

, we have an eigenvalue

of 1 and eigenspace E

1

= span{n}.

We know that there are no other real eigenvalues since rotation changes

the direction of any other vector. The other eigenvalues turn out to be

e

±iθ

. If

θ 6

= 0, there are 3 distinct eigenvalues and the eigenvectors form a

basis of C

3

.

(vi) Consider a shear

A =

1 µ

0 1

The characteristic equation is (1

− λ

)

2

= 0 and

λ

= 1. The eigenvectors

corresponding to

λ

= 1 is

x

=

1

0

. We have

M

(1) = 2 and

m

(1) = 1. So

∆

1

= 1.

If

n × n

matrix

A

has

n

distinct eigenvalues, and hence has

n

linearly

independent eigenvectors

v

1

, v

2

, ···v

n

, then with respect to this eigenvector

basis, A is diagonal.

In this basis,

v

1

= (1

,

0

, ··· ,

0) etc. We know that

Av

i

=

λ

i

v

i

(no summation).

So the image of the

i

th basis vector is

λ

i

times the

i

th basis. Since the columns

of A are simply the images of the basis,

λ

1

0 ··· 0

0 λ

2

··· 0

.

.

.

.

.

.

.

.

.

.

.

.

0 0 ··· λ

n

The fact that

A

can be diagonalized by changing the basis is an important

observation. We will now look at how we can change bases and see how we can

make use of this.