南航双语矩阵论matrix theory第7章部分习题参考答案

第七章部分习题参考答案

Exercise 1

Show that a normal matrix A is Hermitian if its eigenvalues are all real.

Proof If A is a normal matrix, then there is a unitary matrix that diagonalizes A . That is, there is a unitary matrix U such that

H A UDU =

where D is a diagonal matrix and the diagonal elements of D are eigenvalues of A . If eigenvalues of A are all real, then

()H H H H H H A UDU UD U UDU A ====

Therefore, A is Hermitian.

Exercise 2

Let A and B be Hermitian matrices of the same order. Show that AB is Hermitian if and only if AB BA =. Proof

If AB BA =, then ()()H H H H AB BA A B AB ===. Hence, AB is Hermitian. Conversely, if AB is Hermitian, then ()H AB AB =. Therefore, H H AB B A BA ==.

Exercise 3

Let A and B be Hermitian matrices of the same order. Show that A and B are similar if they have the same characteristic polynomial.

Proof Since matrix A and B have the same characteristic polynomial, they have the same eigenvalues 12,,,n λλλ . There exist unitary matrices U and V such that

12diag(,,,)H n U AU λλλ= , 12diag(,,,)H n V BV μμμ= .

Thus,

H H U AU V BV =. (11,H H U U V V --==)

That is 1()H H UV AUV B -=. Hence, A and B are similar.

Exercise 4

Let A be a skew-Hermitian matrix, i.e., H A A =-, show that (a) I A - and I A + are invertible.

(b) 1()()I A I A --+ is a unitary matrix with eigenvalues not equal to 1-. Proof of Part (a)

Method 1: (a) since H A A =-, it follows that

()()H I A I A I AA I A A -+=-=+

For any x 0≠

()()0x x x x x x x x x x H H H H H H H I A A A A A A +=+=+>

Hence, ()()I A I A -+ is positive definite. It follows that ()()I A I A -+ is invertible. Hence, both I A - and I A + are invertible.

Method 2:

If I A - is singular, then there exists a nonzero vector x such that

()x 0I A -=. Thus, x x A =,

x x x x H H A =. (1)

Since x x H is real, it follows that

()x x x x H H H A =.

That is x x x x H H H A =. Since H A A =-, it follows that

x x x x H H A -= (2)

Equation (1) and (2) implies that 0x x H =. This contradicts the assumption that x is nonzero. Therefore, I A - is invertible.

Method 3:

Let λ be an eigenvalue of A and x be an associated eigenvector. x x A λ=

x x x x H H A λ=. ()x x x x x x x x x x x x

H H H H H H H H A A A λλ===-=-

Hence, λ is either zero or pure imaginary. 1 and 1- can not be eigenvalues of A . Hence, I A -

and I A + are invertible.

Method 4: Since H A A =-, A is normal. There exists a unitary matrix U such that 12diag(,,,)H n U AU λλλ=

12()()diag(,,,)H H H H H H n U AU U A U U AU ==-= 12diag(,,,)n λλλ= 12diag(,,,)n λλλ- Each j λ is pure imaginary or zero.

12(diag(,,,))H n I A U I U λλλ-=-

12diag(1,1,,1))H n I A U U λλλ-=---

Since 10i λ-≠ for 1,2,,j n = , det ()0I A -≠. Hence, I A - is invertible. Similarly, we can prove that I A + is invertible.

Proof of Part (b) Method 1:

Since ()()()()I A I A I A I A +-=-+, it follows that

11[()()]()()H I A I A I A I A ---+-+

11()()()()H H I A I A I A I A --=+--+ ( Note that 11()()H H P P --= if P is nonsingular.)

11()()()()I A I A I A I A --=-+-+ 11()()()()I A I A I A I A I --=--++=

Hence, 1()()I A I A --+ is a unitary matrix. Denote 1()()B I A I A -=-+.

Since 111(1)(1)()()()()2()I B I I A I A I A I A I A I A -----=---+=-++-+=-+,

1det()(2)det[()]0n I B I A ---=-+≠

Hence, 1- can not be an eigenvalue of 1()()I A I A --+. Method 2:

By method 4 of the Proof of Part (a),

12diag(1,1,,1))H n I A U U λλλ-=---

12diag(1,1,,1))H n I A U U λλλ+=+++

1()()I A I A --+12

12111diag(

,,,))111H n n

U U λλλλλλ---=+++ The eigenvalues of 1()()I A I A --+ are

12

12111,,,111n n

λλλλλλ---+++ , which are all not equal to 1-.

Method 3: Since ()()()()I A I A I A I A +-=-+, it follows that

11()()()()I A I A I A I A ---+=+-

If 1- is an eigenvalue of 1()()I A I A --+, then there is a nonzero vector x , such that

1()()x x I A I A --+=-. That is 1()()x x I A I A -+-=-.

It follows that

()()x x I A I A -=-+.

This implies that x 0=. This contradiction shows that 1- can not be an eigenvalue of

1()()I A I A --+.

Exercise 6

If H is Hermitian, show that i I H - is invertible, and 1(i )(i )U I H I H -=+- is unitary. Proof Let i A H =-. Then A is skew-Hermitian. By Exercises #4, I A - and I A + are invertible, and 1()()U I A I A -=-+ is unitary. This finishes the proof.

Exercise 7

Find the Hermitian matrix for each of the following quadratic forms. And reduce each quadratic form to its canonical form by a unitary transformation (a) 12312131213(,,)i i f x x x x x x x x x x x =+-+ Solution

(

)

11231

2

3230i 1(,,)i 00100x f x x x x x x x x ???? ???=- ??? ???????, 0i 1i 00100A ?? ?=- ? ???

3d e t ()2I A λλλ-=-. Eigenvalues of A

are 1λ=

2λ=30λ=.

Associated unit eigenvectors are

1i 1,)22u T =-

, 2i 1

,)22

u T =-, and

3u T =, respectively. 123,,u u u form an orthonormal set.

Let 123(,,)u u u U =, and x y U =. Then we obtain the canonical form

1122y y

Exercise 9

Let A and B be Hermitian matrices of order n , and A be positive definite. Show that AB is

similar to a real diagonal matrix.

Proof Since A is positive definite, there exists an nonsingular Hermitian matrix P such that H A PP = 1()H H AB PP B P P BP P -==

AB is similar to H P BP . Since H P BP is Hermitian, it is similar to a real diagonal matrix. Hence, AB is similar to a real diagonal matrix.

Exercise 10

Let A be an Hermitian matrix of order n . Show that there exists a real number 0t such that t I A +is positive definite.

Proof 1: The matrix t I A + is Hermitian for real values of t . If the eigenvalues of A are

12,n λλλ ，，

, then the eigenvalues of t I A +are 12,,n t t t λλλ+++ ，. Let 12max{,,}n t λλλ> ，

Then the eigenvalues of t I A + are all positive. And hence, tI A +is positive definite.

Proof 2: The matrix t I A + is Hermitian for real values of t . Let r A be the leading principle minor of A of order r .

d e t (

)r r r I A t +=+terms involving lower powers in t . Hence, det()r r t I A + is positive for sufficiently large t .

Thus, if t is sufficiently large, all leading principal minors of t I A + will be positive.

That is, there exists a real number 0t such that det()r r t I A + is positive for 0t t > and for each r . Thus t I A + is positive definite for 0t t >.

Exercise 11 Let

11

121222H A A A A A ??

= ???

be an Hermitian positive definite matrix. Show that 1122det()det()det()A A A ≤

Proof We first prove that if A is Hermitian positive definite and B is Hermitian semi-positive

definite, then det()det()A B A +≥. Since A is positive definite, there exists a nonsingular hermitian matrix P such that

H

A P P =

11(())H H A B P I P B P P --+=+ 11det()det()det(())H A B A I P B P --+=+

11()H I P B P --+ is positive semi- definite. Its eigenvalues are all greater than or equal to 1.

Thus

11det(())1H I P B P --+≥

1

11121112112111222H H I O A A I A A A A I A A O I --??

-?

???

? ???-??????

1

11121111

121

1

2212111222121112H H A A A O I A A O A A A A O A A A A O I ---??-????

== ? ? ?--????

?? 122121112H A A A A -- is positive definite, and 1

121112H A A A - is positive semi-definite, and

1

112212

1112det()det()det()H A A A A A A -=- Hence, 111

222212111212111222121112det()det()det(H H H A A A A A A A A A A A A ---=-+≥-）

This finishes the proof.

Exercise 12

Let A be a positive definite Hermitian matrix of order n . Show that the element in A with the largest norm must be in the main diagonal.

Proof Let ()ij A a =. Suppose that 00

i j a is of the largest norm, where 00i j ≠. Consider the

principal minor 00

000000i i i j i j j j a a a a ??

? ???

. It must be positive definite since A is positive definite. (Recall that an Hermitian matrix is positive definite iff all its principal minors are positive.) Thus, 00

0000

00det 0i i i j i j j j a a a a ??

?> ???

. On the other hand, 00

00000000

00

002

det 0i i i j i i j j i j i j j j a a a a a a a ??

?=-≤ ???

since 00i j a is of the largest norm.

(Remark: The diagonal elements in an Hermitian matrix must be real.)

This contradiction implies that the element in A with the largest norm must be in the main diagonal.