uY u d¥ £ ¡ ¨ ¢¤¥

n ¦¢¥

¤¤§ ¥ §¢

¡

£ ¡© ¥ ¡ © ¥ ¡ ¥ © ¥ ¥

¥ ¨

0#¤0 0

matrix. Let be the Schur canonical form of where is unitary and is

5 5

I

upper triangular. By the normality of ,

I #5¤0 I 0 I 5¤0

0 0 I ¤0 I #5¤0

0

5 I

or,

I 0#5 I ¨0 I 0 I 5 ¨0

5 5

0 0

Upon multiplication by on the left and on the right, this leads to the equality 5I5

I

which means that is normal, and according to the previous lemma this is only

5 H5 5

I

possible if is diagonal. 5

Thus, any normal matrix is diagonalizable and admits an orthonormal basis of eigenvectors,

0

namely, the column vectors of .

The following result will be used in a later chapter. The question that is asked is:

Assuming that any eigenvector of a matrix is also an eigenvector of , is normal? P

I

¨

If had a full set of eigenvectors, then the result is true and easy to prove. Indeed, if

¨ 0¨I 6 0 7 ¨

˜ 6 ™˜ 0 7 67

¨

is the matrix of common eigenvectors, then and , with

6 7 0 7 ¨ ¨IHX DI

¨

7 )¨

and diagonal. Then, and and, therefore,

7 7

DI I X

. It turns out that the result is true in general, i.e., independently of the

number of eigenvectors that admits.

§ ¥“£¢§

U

T

A matrix is normal if and only if each of its eigenvectors is also an

IH

eigenvector of .

§ § ¥¦£

T

If is normal, then its left and right eigenvectors are identical, so the suf¬cient

d

IH

condition is trivial. Assume now that a matrix is such that each of its eigenvectors ,

I B

˜# 111)‘

, with is an eigenvector of . For each eigenvector of , , @

B I Q IH B B

and since is also an eigenvector of , then . Observe that ¦

C B Q

C C I C

and because , it follows that . Next, it @ @

¦ ¦

C

is proved by contradiction that there are no elementary divisors. Assume that the contrary

is true for . Then, the ¬rst principal vector associated with is de¬ned by

@ V @

P x C B

@3 V

Taking the inner product of the above relation with , we obtain

t §¦¥

C T B — C P EB C T B ¨

&

W

V @ V

On the other hand, it is also true that

B C Q % C I B C B C t §¦¥

B ¨

¢&

(

W

V V V @ @ V

C B

7

A result of (1.26) and (1.27) is that which is a contradiction. Therefore, has

a full set of eigenvectors. This leads to the situation discussed just before the lemma, from

which it is concluded that must be normal.

Clearly, Hermitian matrices are a particular case of normal matrices. Since a normal

I 0 7)0 0

matrix satis¬es the relation , with diagonal and unitary, the eigenvalues

7

I

of are the diagonal entries of . Therefore, if these entries are real it is clear that

7

. This is restated in the following corollary.

nd¥ ¥ ¡ £ ¥ £ ”˜

uu d

u

¥ ¢ ¡ ¥ ¥ ¥ § ¥ ¤© ¡

¡

¥

§

©

¡ ¥ ©

„ ¡b£ £ G

¦

¦¢

`Q

Q U

T

A normal matrix whose eigenvalues are real is Hermitian.

As will be seen shortly, the converse is also true, i.e., a Hermitian matrix has real eigenval-

ues.

An eigenvalue of any matrix satis¬es the relation

@

C B B

W

V V

@

VV

C

where is an associated eigenvector. Generally, one might consider the complex scalars

V

C xx B B C

©¦

¦