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

A?UDUH

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

AH?(UDUH)H?UDHUH?UDUH?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 (AB)H?(BA)H?AHBH?AB. Hence, AB is Hermitian. Conversely, if AB is Hermitian, then (AB)H?AB. Therefore, AB?BHAH?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?1,?2,?,?n. There exist unitary matrices U and V such that

UHAU?diag(?1,?2,?,?n), VHBV?diag(?1,?2,?,?n).

Thus,

UHAU?VHBV. (UH?U?1,VH?V?1)

That is (UVH)?1AUVH?B. Hence, A and B are similar.

Exercise 4

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

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

Method 1: (a) since AH??A, it follows that

(I?A)(I?A)?I?AA?I?AHA

For any x?0

xH(I?AHA)x?xHx?xHAHAx?xHx?(Ax)HAx?0

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:

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

(I?A)x?0. Thus, Ax?x,

xHAx?xHx. (1)

Since xHx is real, it follows that

(xHAx)H?xHx.

That is xHAHx?xHx. Since AH??A, it follows that

?xHAx?xHx (2)

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

Method 3:

Let ? be an eigenvalue of A and x be an associated eigenvector. Ax??x

xHAx(xHAx)HxHAHxxHAx??H. ???H??H???

xxxHxxxxxHence, ? 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 AH??A, A is normal. There exists a unitary matrix U such that UHAU?diag(?1,?2,?,?n)

(UHAU)H?(UHAHU)H??UHAU?diag(?1,?2,?,?n) diag(?1,?2,?,?n)??diag(?1,?2,?,?n) Each ?j is pure imaginary or zero.

I?A?U(I?diag(?1,?2,?,?n))UH

I?A?Udiag(1??1,1??2,?,1??n))UH

Since 1??i?0 for j?1,2,?,n, det(I?A)?0. 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

[(I?A)(I?A)?1]H(I?A)(I?A)?1

?(I?AH)?1(I?AH)(I?A)(I?A)?1 ( Note that (P?1)H?(PH)?1 if P is nonsingular.)

?(I?A)?1(I?A)(I?A)(I?A)?1 ?(I?A)?1(I?A)(I?A)(I?A)?1?I

Hence, (I?A)(I?A)?1 is a unitary matrix. Denote B?(I?A)(I?A)?1.

Since (?1)I?B?(?1)I?(I?A)(I?A)?1??(I?A?I?A)(I?A)?1??2(I?A)?1,

det(?I?B)?(?2)ndet[(I?A)?1]?0

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

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

I?A?Udiag(1??1,1??2,?,1??n))UH

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I?A?Udiag(1??1,1??2,?,1??n))UH

1??n1??11??2,,?,))UH (I?A)(I?A)?1?Udiag(1??11??21??nThe eigenvalues of (I?A)(I?A)?1 are

1??n1??11??2,,?,, which are all not equal to 1??11??21??n?1.

Method 3: Since (I?A)(I?A)?(I?A)(I?A), it follows that

(I?A)(I?A)?1?(I?A)?1(I?A)

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

(I?A)(I?A)?1x??x. That is (I?A)?1(I?A)x??x.

It follows that

(I?A)x??(I?A)x.

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

(I?A)(I?A)?1.

Exercise 6

If H is Hermitian, show that I?iH is invertible, and U?(I?iH)(I?iH)?1 is unitary. Proof Let A??iH. Then A is skew-Hermitian. By Exercises #4, I?A and I?A are invertible, and U?(I?A)(I?A)?1 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) f(x1,x2,x3)?ix1x2?x1x3?ix1x2?x1x3 Solution

f(x1,x2,x3)?x1?0i1??x1??0i1???????x3??i00??x2?, A???i00? ?100??x??100????3????x2? det?(I?A?)?3??2. Eigenvalues of A are ?1?2,?2??2, and ?3?0.

1i11i1Associated unit eigenvectors are u1?(,?,)T, u2?(,,?)T, and

2222221iu3?(0,,?)T, respectively. u1,u2,u3 form an orthonormal set.

22 Let U?(u1,u2,u3), and x?Uy. Then we obtain the canonical form

2y1y1?2y2y2

Exercise 9

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

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similar to a real diagonal matrix.

Proof Since A is positive definite, there exists an nonsingular Hermitian matrix P such that A?PPH AB?PPHB?P(PHBP)P?1

AB is similar to PHBP. Since PHBP 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 t0 such that tI?Ais positive definite.

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

?,?n, then the eigenvalues of tI?Aare t??1,t??2，?,t??n. Let ?1，?2，t?max{?1,?2，?,?n}

Then the eigenvalues of tI?A are all positive. And hence, tI?Ais positive definite.

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

Ir?Ar?)tr?terms involving lower powers in t. det(Hence, det(tIr?Ar) is positive for sufficiently large t.

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

That is, there exists a real number t0 such that det(tIr?Ar) is positive for t?t0 and for each r. Thus tI?A is positive definite for t?t0.

Exercise 11 Let

?AA??11H?A12A12?? be an Hermitian positive definite matrix. Show that A22?det(A)?det(A11)det(A22)

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

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

H A?PP

A?B?P(I?P?1B(P?1)H)PH det(A?B)?det(A)det(I?P?1B(P?1)H)

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

Thus

det(I?P?1B(P?1)H)?1

IO??A11??H?1??H??A12A11I??A12?1A12??I?A11A12?? ??A22??OI? 4

?A??11?O?1A12A12??A11??I?A11?????H?1A22?A12A11A12??OI??OO?? H?1A22?A12A11A12?H?1H?1A22?A12A11A12 is positive definite, and A12A11A12 is positive semi-definite, and

H?1A11A12) det(A)?det(A11)det(A22?A12H?1H?1H?1A11A12?A12A11A12)?det(A22?A12A11A12） Hence, det(A22)?det(A22?A12

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 A?(aij). Suppose that aij is of the largest norm, where i0?j0. Consider the

00?ai0i0principal minor ??aij?00ai0j0??. It must be positive definite since A is positive definite. (Recall that aj0j0??an Hermitian matrix is positive definite iff all its principal minors are positive.) ?ai0i0 Thus, det??aij?00ai0j0???0. aj0j0??ai0j0???ai0i0aj0j0?ai0j0aj0j0??2?ai0i0On the other hand, det??aij?00?0 since ai0j0is 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.

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