第七章部分习题参考答案
Exerci 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
where D is a diagonal matrix and the diagonal elements of D are eigenvalues of 大鱼海棠讲的什么故事A.
If eigenvalues of A are all real, then
Therefore, A is Hermitian.
Exerci 2
Let A and B be Hermitian matrices of the same order. Show that AB is Hermitian if and only if .
Proof
If , then . Hence, AB is Hermitian.
Converly, if AB is Hermitian, then . Therefore, .
Exerci 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.
现代女子医院 Proof Since matrix A and B have the same characteristic polynomial, they have the same eigenvalues. There exist unitary matrices U and V such that
, .
Thus,
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That is . Hence, A and B are similar.
Exerci 4
Let A be a skew-Hermitian matrix, i.e., , show that
(a) and are invertible.
(b) is a unitary matrix with eigenvalues not equal to .
Proof of Part (a)
Method 1: (a) since , it follows that
For any
Hence, is positive definite. It follows that is invertible. Hence, both and are invertible.
Method 2:
If is singular, then there exists a nonzero vector x such that
. Thus, ,
. (1)
Since is real, it follows that
.
That is . Since , it follows that
(2)
Equation (1) and (2) implies that . This contradicts the assumption that x is nonzero.
视觉设计师Therefore, is invertible.
Method 3:
Let be an eigenvalue of A and x be an associated eigenvector.
.
Hence, is either zero or pure imaginary. 1 and can not be eigenvalues of A. Hence, and are invertible.
Method 4: Since , A is normal. There exists a unitary matrix U such that
Each is pure imaginary or zero.
Since for , det. Hence, is invertible. Similarly, we can prove that is invertible.
高脚杯炸弹Proof of Part (b)
Method 1:
Since , it follows that
( Note that if P is nonsingular.)
Hence, is a unitary matrix. 成本项目Denote .
Since ,
Hence, can not be an eigenvalue of .
Method 2:
By method 4 of the Proof of Part (a),
The eigenvalues of are , which are all not equal to .
Method 3: Since , it follows that
If is an eigenvalue of , then there is a nonzero vector x学校日记, such that 草莓蛋糕图片
. That is .
It follows that
.
This implies that . This contradiction shows that can not be an eigenvalue of .
Exerci 6
If H is Hermitian, show that is invertible, and is unitary.
Proof Let . Then A is skew-Hermitian. By Exercis #4, and are invertible, and is unitary. This finishes the proof.
Exerci 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)
Solution
,
. Eigenvalues of A are ,, and .
Associated unit eigenvectors are , , and
, respectively. form an orthonormal t.
Let , and . Then we obtain the canonical form
Exerci 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
AB is similar to . Since is Hermitian, it is similar to a real diagonal matrix. Hence, AB is similar to a real diagonal matrix.
Exerci 10
Let A be an Hermitian matrix of order n. Show that there exists a real number such that
is positive definite.
Proof 1: The matrix is Hermitian for real values of t. If the eigenvalues of A are
, then the eigenvalues of are . Let
Then the eigenvalues of are all positive. And hence, is positive definite.
Proof 2: The matrix is Hermitian for real values of t. Let be the leading principle minor of A of order r.
terms involving lower powers in t.
Hence, is positive for sufficiently large t.
Thus, if t is sufficiently large, all leading principal minors of will be positive.
That is, there exists a real number such that is positive for and for each r. Thus is positive definite for .
Exerci 11
Let be an Hermitian positive definite matrix. Show that
Proof We first prove that if A is Hermitian positive definite and B is Hermitian mi-positive definite, then .
