proof of theorem for normal matrices

proof of theorem for normal matrices

1) (

A

H

=

g

⁢

(

A

)

→

A

is normal)

Keeping in mind that every matrix commutes with its own powers, let’s compute

A

⁢

A

H

=

A

⁢

g

⁢

(

A

)

=

A

⁢

∑

i

=

0

n

-

1

a

i

⁢

A

i

=

∑

i

=

0

n

-

1

a

i

⁢

A

⁢

A

i

=

∑

i

=

0

n

-

1

a

i

⁢

A

i

⁢

A

=

(

∑

i

=

0

n

-

1

a

i

⁢

A

i

)

⁢

A

=

g

⁢

(

A

)

⁢

A

=

A

H

⁢

A

which shows

A

to be normal.

2) (

A

is normal

→

A

H

=

g

⁢

(

A

)

)

Let

λ

1

,

λ

2

,

…

,

λ

r

,

1

≤

r

≤

n

be the distinct

eigenvalues

of A, and let

Λ

=

d

⁢

i

⁢

a

⁢

g

⁢

{

λ

1

,

λ

2

,

…

,

λ

n

}

; then it’s possible to
find a

(

r

-

1

)

-degree polynomial

g

⁢

(

t

)

such that

g

⁢

(

λ

i

)

=

λ

i

∗

1

≤

i

≤

r

, solving the

r

×

r

linear Vandermonde
system:

[

1

λ

1

λ

1

2

⋯

λ

1

r

-

1

1

λ

2

λ

2

2

⋯

λ

2

r

-

1

⋮

⋮

⋮

⋮

⋮

1

λ

r

-

1

λ

r

-

1

2

⋯

λ

r

-

1

r

-

1

1

λ

r

λ

r

2

⋯

λ

r

r

-

1

]

⁢

[

a

0

a

1

a

2

⋮

a

r

-

1

]

=

[

λ

1

∗

λ

2

∗

λ

3

∗

⋮

λ

r

∗

]

Since these

r

eigenvalues are distinct, the

Vandermonde matrix

is full
rank, and the

linear system

admits a unique solution; so a

(

r

-

1

)

-degree
polynomial

g

⁢

(

t

)

can be found such that

g

⁢

(

λ

i

)

=

λ

i

∗

1

≤

i

≤

r

and therefore

g

⁢

(

λ

i

)

=

λ

i

∗

1

≤

i

≤

n

. Writing these equations in matrix form, we have

g

⁢

(

Λ

)

=

Λ

∗

By Schur’s

decomposition

theorem

, a

unitary matrix

U

and an

upper
triangular matrix

T

exist such that

A

=

U

⁢

T

⁢

U

H

and since

A

is normal we have

T

=

Λ

.

Let’s evaluate

g

⁢

(

A

)

.

g

⁢

(

A

)

=

g

⁢

(

U

⁢

Λ

⁢

U

H

)

=

∑

i

=

0

r

-

1

a

i

⁢

(

U

⁢

Λ

⁢

U

H

)

i

But, keeping in mind that

U

H

⁢

U

=

I

,

(

U

⁢

Λ

⁢

U

H

)

i

=

U

⁢

Λ

⁢

U

H

⁢

U

⁢

Λ

⁢

U

H

⁢

U

⁢

Λ

⁢

U

H

⁢

⋯

⁢

U

⁢

Λ

⁢

U

H

⏞

i

⁢

t

⁢

i

⁢

m

⁢

e

⁢

s

=

U

⁢

Λ

i

⁢

U

H

and so

g

⁢

(

A

)

=

∑

i

=

0

r

-

1

a

i

⁢

(

U

⁢

Λ

i

⁢

U

H

)

=

U

⁢

(

∑

i

=

0

r

-

1

a

i

⁢

Λ

i

)

⁢

U

H

=

U

⁢

g

⁢

(

Λ

)

⁢

U

H

=

U

⁢

Λ

∗

⁢

U

H

=

U

⁢

Λ

H

⁢

U

H

=

(

U

⁢

Λ

⁢

U

H

)

H

=

A

H

which is the thesis.

Remark: note that this is a

constructive proof

, giving explicitly a way to
find

g

⁢

(

t

)

polynomial by solving Vandermonde system in the eigenvalues.

Example:

Let

A

=

1

2

⁢

[

1

+

j

-

1

-

j

1

+

j

1

+

j

]

(which is easily checked to be normal),

with

U

=

1

2

⁢

[

1

-

j

j

-

1

]

. Then

σ

⁢

(

A

)

=

{

1

,

j

}

and the Vandermonde system is

[

1

1

1

j

]

⁢

[

a

0

a

1

]

=

[

1

-

j

]

from which we find

g

⁢

(

t

)

=

(

1

-

j

)

+

j

⁢

t

A simple calculation yields

g

⁢

(

A

)

=

(

1

-

j

)

⁢

I

+

j

⁢

A

=

1

2

⁢

[

1

-

j

1

-

j

-

1

+

j

1

-

j

]

=

A

H

Title

proof of theorem for normal matrices

Canonical name

ProofOfTheoremForNormalMatrices

Date of creation

2013-03-22 15:36:36

Last modified on

2013-03-22 15:36:36

Owner

Andrea Ambrosio (7332)

Last modified by

Andrea Ambrosio (7332)

Numerical id

16

Author

Andrea Ambrosio (7332)

Entry type

Proof

Classification

msc 15A21

Generated on Fri Feb 9 18:26:45 2018 by

LaTeXML