characteristic values and vectors (of a matrix)

characteristic values and vectors (of a matrix)

Over the

spectrum

σ

⁢

(

A

)

of a matrix

A

, its

eigenvalues

λ

1

,

λ

2

,

…

,

λ

s

possess

multiplicities

n

1

,

n

2

,

…

,

n

s

, respectively, with

∑

k

=

1

s

n

k

=

n

. Its associated

characteristic polynomial

is then factored as

Δ

⁢

(

λ

)

≡

|

λ

⁢

I

-

A

|

=

Π

k

=

1

s

⁢

(

λ

-

λ

k

)

n

k

.

(1)

Let us set

mult

⁢

(

λ

k

)

=

n

k

for multiplicity of

λ

k

(

k

=

1

,

…

,

s

). We will now prove the following theorem.

Theorem 1

.

If

σ

⁢

(

A

)

=

{

λ

k

}

k

=

1

s

,

mult

⁢

(

λ

k

)

=

n

k

, and

g

⁢

(

μ

)

is a scalar

polynomial

, then

σ

⁢

(

g

⁢

(

A

)

)

=

{

g

⁢

(

λ

k

)

}

k

=

1

s

,

mult

⁢

(

g

⁢

(

λ

k

)

)

=

n

k

.

Proof.

Let

g

⁢

(

μ

)

be an arbitrary scalar polynomial. We want to find the

characteristic values

of

g

⁢

(

A

)

. For this purpose we split

g

⁢

(

μ

)

into linear factors

g

⁢

(

μ

)

=

a

0

⁢

Π

i

=

1

t

⁢

(

μ

-

μ

i

)

l

i

,

a

0

≠

0

,

∑

i

=

1

t

l

i

=

l

.

(2)

On substitution

μ

↦

A

, we have

g

⁢

(

A

)

=

a

0

⁢

Π

i

=

1

t

⁢

(

A

-

μ

i

⁢

I

)

l

i

,

(3)

being

I

the

identity matrix

. Let us compute the

determinant

of

g

⁢

(

A

)

. (Coefficient

a

0

will be powered to

n

, the order of the

square matrix

A

).

|

g

⁢

(

A

)

|

=

a

0

n

⁢

Π

i

=

1

t

⁢

|

(

-

1

)

⁢

(

μ

i

⁢

I

-

A

)

|

l

i

=

a

0

n

⁢

Π

i

=

1

t

⁢

(

-

1

)

n

⁢

l

i

⁢

|

μ

i

⁢

I

-

A

|

l

i

=

a

0

n

⁢

(

-

1

)

n

⁢

∑

i

=

1

t

l

i

⁢

Π

i

=

1

t

⁢

|

μ

i

⁢

I

-

A

|

l

i

=

a

0

n

⁢

(

-

1

)

n

⁢

l

⁢

Π

i

=

1

t

⁢

Δ

⁢

(

μ

i

)

l

i

=

a

0

n

⁢

(

-

1

)

n

⁢

l

⁢

Π

i

=

1

t

⁢

[

Π

k

=

1

s

⁢

(

μ

i

-

λ

k

)

n

k

]

l

i

,

because on substitution

λ

↦

μ

i

in (1). Next we commute the binomial by introducing

(

-

1

)

n

⁢

l

into the product signs and also we note that

a

0

n

=

a

0

∑

k

=

1

s

n

k

=

Π

k

=

1

s

⁢

a

0

n

k

, so that

|

g

⁢

(

A

)

|

=

Π

k

=

1

s

⁢

[

a

0

⁢

Π

i

=

1

t

⁢

(

λ

k

-

μ

i

)

l

i

]

n

k

,

and we may use (2) for

μ

=

λ

k

to obtain

|

g

⁢

(

A

)

|

=

Π

k

=

1

s

⁢

g

⁢

(

λ

k

)

n

k

.

(4)

Finally we substitute the polynomial

g

⁢

(

μ

)

by

λ

-

g

⁢

(

μ

)

, where

λ

is an arbitrary

parameter

, getting for (4)

Δ

⁢

(

g

⁢

(

A

)

)

≡

|

λ

⁢

I

-

g

⁢

(

A

)

|

=

Π

k

=

1

s

⁢

[

λ

-

g

⁢

(

λ

k

)

]

n

k

.

(5)

This proves the theorem.
∎

As an important particular case we have:

σ

⁢

(

A

m

)

=

{

λ

k

m

}

k

=

1

s

, (

m

=

0

,

1

,

⋯

),

mult

⁢

(

λ

k

)

=

n

k

.

Connection between the characteristic polynomial

Δ

⁢

(

λ

)

and the

adjugate matrix

B

⁢

(

λ

)

of

A

.

As it is well known, the adjugate matrix

B

of a matrix

A

there corresponds to the

algebraic complement

or cofactor matrix of the

transpose

of

A

. From this definition we have

B

⁢

(

λ

)

⁢

(

λ

⁢

I

-

A

)

=

Δ

⁢

(

λ

)

⁢

I

and

(

λ

⁢

I

-

A

)

⁢

B

⁢

(

λ

)

=

Δ

⁢

(

λ

)

⁢

I

.

(6)

Let us suppose

Δ

⁢

(

λ

)

is given by

Δ

⁢

(

λ

)

=

λ

n

-

∑

k

=

1

n

c

k

⁢

λ

n

-

k

.

(7)

It is clear that the

difference

Δ

⁢

(

λ

)

-

Δ

⁢

(

μ

)

is

divisible

by

λ

-

μ

without

remainder

, hence

δ

⁢

(

λ

,

μ

)

≡

Δ

⁢

(

λ

)

-

Δ

⁢

(

μ

)

λ

-

μ

=

λ

n

-

1

+

(

μ

-

c

1

)

⁢

λ

n

-

2

+

(

μ

2

-

c

1

⁢

μ

-

c

2

)

⁢

λ

n

-

3

+

⋯

(8)

is a polynomial in

λ

,

μ

. If we replace in (8)

(

λ

,

μ

)

by the

permutable

matrices

(

λ

⁢

I

,

A

)

and recalling that from

Cayley-Hamilton theorem

Δ

⁢

(

A

)

=

0

, then

δ

⁢

(

λ

⁢

I

,

A

)

⁢

(

λ

⁢

I

-

A

)

=

Δ

⁢

(

λ

)

⁢

I

,

(9)

which by comparing it with (6)

1

we conclude that

B

⁢

(

λ

)

=

δ

⁢

(

λ

⁢

I

,

A

)

(10)

is the desired formula by virtue of the uniqueness of the

quotient

. Therefore (10) and (8) let to write the

adjugate

B

⁢

(

λ

)

as the matrix polynomial

B

⁢

(

λ

)

=

I

⁢

λ

n

-

1

+

∑

k

=

1

n

-

1

B

k

⁢

λ

n

-

k

-

1

,

(11)

where (

μ

↦

A

in (8))

B

k

=

A

k

-

∑

i

=

1

k

c

i

A

k

-

i

,

(

k

=

1

,

…

,

n

-

1

)

,

(12)

which can also be obtained from the recurrence equation

B

k

=

A

B

k

-

1

-

c

k

I

,

(

k

=

1

,

…

,

n

-

1

;

B

0

=

I

)

.

(13)

What is more,

A

⁢

B

n

-

1

-

c

n

⁢

I

=

0

≡

B

n

.

(14)

(13) as well as (14) follow inmediately from (6)

2

if we equate the coefficients of equal powers of

λ

on both sides. Also, if we substitute

B

n

-

1

from (12), into (14), we get

Δ

⁢

(

A

)

=

0

(Cayley-Hamilton), an implicit consequence of generalized Bézout theorem. On the other hand, by setting

λ

=

0

in (7) we obtain

c

n

=

Δ

⁢

(

0

)

/

(

-

1

)

=

|

-

A

|

/

(

-

1

)

=

(

-

1

)

n

-

1

⁢

|

A

|

≠

0

, whenever

A

be non-

singular

. From this and from (14) follow that

A

-

1

=

1

c

n

⁢

B

n

-

1

.

(15)

Let now

λ

c

be a characteristic value of

A

, then

Δ

⁢

(

λ

c

)

=

0

and (6)

2

becomes

(

λ

c

⁢

I

-

A

)

⁢

B

⁢

(

λ

c

)

=

0

.

(16)

Let us assume that

B

⁢

(

λ

c

)

≠

0

and denote by

𝐛

an arbitrary non-zero column of this matrix. From (16) we have

(

λ

c

⁢

I

-

A

)

⁢

𝐛

=

𝟎

. That is,

A

⁢

𝐛

=

λ

c

⁢

𝐛

.

(17)

Therefore every non-zero column of

B

⁢

(

λ

c

)

determines a characteristic vector corresponding to the characteristic value

λ

c

. Moreover, if to the characteristic value

λ

c

there correspond

l

linearly independent

characteristic vectors,

n

-

l

will be the rank of

λ

c

⁢

I

-

A

and so the rank of

B

⁢

(

λ

c

)

does not exceed

l

. In particular, if only one characteristic vector there corresponds to

λ

c

, then in

B

⁢

(

λ

c

)

the

elements of

any two columns will be proportional (In such a case

l

=

1

, hence the rank of

λ

c

⁢

I

-

A

will be

n

-

1

).

In conclusion:

If the coefficients of the characteristic polynomial are known, then the adjugate matrix may be found by (10). In

addition

, if the given matrix

A

is non-singular, then the inverse matrix

A

-

1

can be found from (15). Also if

λ

c

is a characteristic value of

A

, the non-zero columns of

B

⁢

(

λ

c

)

are characteristc vectors of A for

λ

=

λ

c

.

Example.

We find out the characteristic values and vectors from the matrix

A

=

[

3

-

3

2

-

1

5

-

2

-

1

3

0

]

.

From (1),

Δ

⁢

(

λ

)

=

|

λ

⁢

I

-

A

|

=

|

λ

-

3

3

-

2

1

λ

-

5

2

1

-

3

λ

|

=

λ

3

-

8

⁢

λ

2

+

20

⁢

λ

-

16

.

Comparing with (7), we have

c

1

=

8

,

c

2

=

-

20

,

c

3

=

16

.

Next we use (8),

δ

⁢

(

λ

,

μ

)

=

Δ

⁢

(

λ

)

-

Δ

⁢

(

μ

)

λ

-

μ

=

λ

2

+

(

μ

-

8

)

⁢

λ

+

μ

2

-

8

⁢

μ

+

20

,

so that from (11)

B

⁢

(

λ

)

=

δ

⁢

(

λ

⁢

I

,

A

)

=

λ

2

⁢

I

+

(

A

-

8

⁢

I

⏟

B

1

)

⁢

λ

+

A

2

-

8

⁢

A

+

20

⁢

I

⏟

B

2

.

We will now evaluate

B

1

and

B

2

by using (12) and (13), respectively.

B

1

=

A

-

8

⁢

I

=

[

-

5

-

3

2

-

1

-

3

-

2

-

1

3

-

8

]

,

B

2

=

A

⁢

B

1

+

20

⁢

I

=

[

6

6

-

4

2

2

4

2

-

6

12

]

,

thus

B

⁢

(

λ

)

is

B

⁢

(

λ

)

=

[

λ

2

-

5

⁢

λ

+

6

-

3

⁢

λ

+

6

2

⁢

λ

-

4

-

λ

+

2

λ

2

-

3

⁢

λ

+

2

-

2

⁢

λ

+

4

-

λ

+

2

3

⁢

λ

-

6

λ

2

-

8

⁢

λ

+

12

]

.

Also

|

A

|

=

16

and

A

-

1

is obtained from (15), i.e.

A

-

1

=

1

16

⁢

B

2

=

1

8

⁢

[

3

3

-

2

1

1

2

1

-

3

6

]

.

Furthermore,

Δ

⁢

(

λ

)

=

(

λ

-

2

)

2

⁢

(

λ

-

4

)

.

We notice the eigenvalue

λ

=

2

possesses multiplicity

2

and also that all the entries of the adjugate

B

⁢

(

λ

)

are divisible by the binomial

λ

-

2

(

|

B

⁢

(

2

)

|

=

0

, i.e.

λ

=

2

annihilates it), therefore it can be reduced which makes instructive this problem. Thus,

C

⁢

(

λ

)

=

[

λ

-

3

-

3

2

-

1

λ

-

1

-

2

-

1

3

λ

-

6

]

,

which for

λ

=

2

it becomes

C

⁢

(

2

)

=

[

-

1

-

3

2

-

1

1

-

2

-

1

3

-

4

]

.

From this we get the charactreristic vectors

(

1

,

1

,

1

)

by multiplying the first colum by

-

1

, and also

(

-

3

,

1

,

3

)

, both correponding to

λ

=

2

. Third column is a

linear combination

of the first two (subtract it). Likewise we find for the another characteristic value

λ

=

4

C

⁢

(

4

)

=

[

1

-

3

2

-

1

3

-

2

-

1

3

-

2

]

,

whence we get the

eigenvector

(

1

,

-

1

,

-

1

)

, being the remaining two columns clearly proportional to the first one.

Title

characteristic values and vectors (of a matrix)

Canonical name

CharacteristicValuesAndVectorsofAMatrix

Date of creation

2013-03-22 17:43:58

Last modified on

2013-03-22 17:43:58

Owner

perucho (2192)

Last modified by

perucho (2192)

Numerical id

6

Author

perucho (2192)

Entry type

Topic

Classification

msc 15A18

Synonym

eigenvalues

Synonym

eigenvectors

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