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2024年4月16日发(作者：openstack创建内部网络)
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
AUDU
H
where D is a diagonal matrix and the diagonal elements of
D
are eigenvalues
of
A
.
If eigenvalues of
A
are all real, then
A
H
(UDU
H
)
H
UD
H
U
H
UDU
H
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
ABBA
.
Proof
1
If
HHHH
ABBA
, then
(AB)(BA)ABAB
. Hence,
AB
is Hermitian.
Conversely, if
AB
is Hermitian, then
(AB)
H
AB
. Therefore,
ABB
H
A
H
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
U
H
AUdiag(
1
,
2
,,
n
)
,
V
H
BVdiag(
1
,
2
,,
n
)
.
Thus,
U
H
H1H1
AUV
H
BV
. (
UU,VV
)
That is
(UV
H
)
1
AUV
H
B
. Hence,
A
and
B
are similar.
Exercise 4
Let
A
be a skew-Hermitian matrix, i.e.,
A
H
A
, show that
(a)
IA
and
IA
are invertible.
2
(b)
(IA)(IA)
1
is a unitary matrix with eigenvalues not equal to
1
.
Proof of Part (a)
Method 1: (a) since
A
H
A
, it follows that
(IA)(IA)IAAIA
H
A
For any
x0
x
H
(IA
H
A)xx
H
xx
H
A
H
Axx
H
x(Ax)
H
Ax0
Hence,
(IA)(IA)
is positive definite. It follows that
(IA)(IA)
Hence, both
IA
and
IA
are invertible.
Method 2:
If
IA
is singular, then there exists a nonzero vector x such that
(IA)x0
. Thus,
Axx
,
x
H
Axx
H
x
. (1)
Since
x
H
x
is real, it follows that
(x
H
Ax)
H
x
H
x
.
invertible.
3
is
That is
x
H
A
H
xx
H
x
. Since
A
H
A
, it follows that
x
H
Axx
H
x
(2)
Equation (1) and (2) implies that
is nonzero.
x
H
x0
. This contradicts the assumption that x
Therefore,
IA
is invertible.
Method 3:
Let
be an eigenvalue of
A
and x be an associated eigenvector.
Ax
x
x
H
Ax(x
H
Ax)
H
x
H
A
H
xx
H
Ax
H
H
H
xx
.
x
H
xxxxx
Hence,
is either zero or pure imaginary. 1 and
1
can not be eigenvalues of
A
. Hence,
IA
and
IA
are invertible.
Method 4: Since
A
H
A
,
A
is normal. There exists a unitary matrix
U
such that
U
H
AUdiag(
1
,
2
,,
n
)
(U
H
AU)
H
(U
H
A
H
U)
H
U
H
AUdiag(
1
,
2
,,
n
)
4
Each
j
diag(
1
,
2
,,
n
)
diag(
1
,
2
,,
n
)
is pure imaginary or zero.
IAU(Idiag(
1
,
2
,,
n
))U
H
IAUdiag(1
1
,1
2
,,1
n
))U
H
Since
1
i
0
for
can prove that
j1,2,,n
, det
(IA)0
. Hence,
IA
is invertible. Similarly, we
IA
is invertible.
Proof of Part (b)
Method 1:
Since
(IA)(IA)(IA)(IA)
, it follows that
[(IA)(IA)
1
]
H
(IA)(IA)
1
( Note that
(P
1
)
H
(P
H
)
1
(IA
H
)
1
(IA
H
)(IA)(IA)
1
if
P
is nonsingular.)
(IA)
1
(IA)(IA)(IA)
1
(IA)
1
(IA)(IA)(IA)
1
I
5
Hence,
(IA)(IA)
1
is a unitary matrix. Denote
B(IA)(IA)
1
.
Since
(1)IB(1)I(IA)(IA)
1
(IAIA)(IA)
1
2(IA)
1
,
det(IB)(2)
n
det[(IA)
1
]0
Hence,
1
1
can not be an eigenvalue of
(IA)(IA)
.
Method 2:
By method 4 of the Proof of Part (a),
IAUdiag(1
1
,1
2
,,1
n
))U
H
IAUdiag(1
1
,1
2
,,1
n
))U
H
1
n
))U
H
1
n
(IA)(IA)
1
1
1
1
2
Udiag(,,
1
1
1
2
,
The eigenvalues of
1
.
(IA)(IA)
1
are
1
1
1
2
,,
1
1
1
2
,
1
n
1
n
, which are all not equal to
Method 3: Since
(IA)(IA)(IA)(IA)
, it follows that
(IA)(IA)
1
(IA)
1
(IA)
6
If
1
1
is an eigenvalue of
(IA)(IA)
, then there is a nonzero vector x, such that
(IA)(IA)
1
xx
. That is
(IA)
1
(IA)xx
.
It follows that
(IA)x(IA)x
.
This implies that
eigenvalue of
x0
. This contradiction shows that
1
can not be an
(IA)(IA)
1
.
Exercise 6
If
H
is Hermitian, show that
1
IiH
is invertible, and
U(IiH)(IiH)
is unitary.
Proof Let
AiH
. Then
A
is skew-Hermitian. By Exercises #4,
IA
and
IA
1
are invertible, and
U(IA)(IA)
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(x
1
,x
2
,x
3
)ix
1
x
2
x
1
x
3
ix
1
x
2
x
1
x
3
Solution
7
f(x
1
,x
2
,x
3
)x
1
x
2
0i1
x
1
0i1
x
3
i00
x
2
A
i00
100
x
100
3
,
det(
IA)
3
2
. Eigenvalues of
A
are
1
2
,
2
2
, and
3
0
.
Associated unit eigenvectors are
u
1
(
1i1
1i1
T
,,)
T
u
2
(,,)
2
22
,
2
22
, and
u
3
(0,
1
2
,
i
2
)
T
, respectively.
u
1
,u
2
,u
3
form an orthonormal set.
Let
U(u
1
,u
2
,u
3
)
, and
xUy
. Then we obtain the canonical form
2y
1
y
1
2y
2
y
2
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
APP
H
ABPP
H
BP(P
H
BP)P
1
8
AB
is similar to
P
H
BP
. Since
P
H
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
t
0
such that
tIA
is positive definite.
Proof 1: The matrix
of
A
are
tIA
is Hermitian for real values of
t
. If the eigenvalues
1
，
2
，,
n
, then the eigenvalues of
tIA
are
t
1
,t
2
，,t
n
. Let
tmax{
1
,
2
，,
n
}
tIA
are Then the eigenvalues of
definite.
all positive. And hence,
tIA
is positive
Proof 2: The matrix
tIA
is Hermitian for real values of
t
. Let
A
r
be the
leading principle minor of
A
of order
r
.
det(tI
r
A
r
)t
r
terms involving lower powers in
t
.
Hence,
det(tI
r
A
r
)
is positive for sufficiently large
t
.
9
Thus, if
t
is sufficiently large, all leading principal minors of
positive.
tIA
will be
That is, there exists a real number
t
0
such that
and for each
r
. Thus
tIA
is positive definite for
tt
0
.
det(tI
r
A
r
)
is positive for
tt
0
Exercise 11
A
A
11
H
A
12
Let
det(A)det(A
11
)det(A
22
)
A
12
A
22
be an Hermitian positive definite matrix. Show that
Proof We first prove that if
A
is Hermitian positive definite and
B
is Hermitian
semi-positive definite, then
det(AB)det(A)
.
Since
A
is positive definite, there exists a nonsingular hermitian matrix
P
such that
APP
H
ABP(IP
1
B(P
1
)
H
)P
H
det(AB)det(A)det(IP
1
B(P
1
)
H
)
IP
1
B(P
1
)
H
is positive semi- definite. Its eigenvalues are all greater than or
equal to 1. Thus
det(IP
1
B(P
1
)
H
)1
10
IO
A
11
H1
H
A
12
A
11
I
A
12
1
A
12
IA
11
A
12
A
22
OI
O
H1
A
22
A
12
A
11
A
12
A
11
O
1
A
12
A
12
A
11
IA
11
H1
A
22
A
12
A
11
A
12
OI
O
is positive semi-definite, and
H1
A
22
A
12
A
11
A
12
is positive definite, and
H1
A
12
A
11
A
12
H1
det(A)det(A
11
)det(A
22
A
12
A
11
A
12
)
Hence,
H1H1H1
det(A
22
)det(A
22
A
12
A
11
A
12
A
12
A
11
A
12
)det(A
22
A
12
A
11
A
12
）
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.
a
i
0
j
0
Proof Let
A(a
ij
)
. Suppose that
a
i
0
i
0
a
ij
00
is of the largest norm, where
i
0
j
0
.
Consider the principal minor
a
i
0
j
0
a
j
0
j
0
. 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,
a
i
0
i
0
det
a
ij
00
a
i
0
j
0
0
a
j
0
j
0
.
11
On the other hand,
a
i
0
i
0
det
a
ij
00
a
i
0
j
0
a
i
0
i
0
a
j
0
j
0
a
i
0
j
0
a
j
0
j
0
2
0
since
a
i
0
j
0
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.
12
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