# Iterative Methods to Solve the Generalized Coupled Sylvester-Conjugate Matrix Equations for Obtaining the Centrally Symmetric (Centrally Antisymmetric) Matrix Solutions

Authors: Yajun Xie; Changfeng Ma
Journal: Journal of Applied Mathematics
 (2014)

---

## Abstract

The iterative method is presented for obtaining the centrally symmetric (centrally antisymmetric) matrix pair(
X
,
Y
) solutions of the generalized coupled Sylvester-conjugate matrix equations A
1
X
+
B
1
Y
=
D
1
X
¯
E
1
+
F
1, A
2
Y
+
B
2
X
=
D
2
Y
¯
E
2
+
F
2. On the condition that the coupled matrix equations are consistent, we show that the solution pair (
X
*
,
Y
*
) can be obtained within finite iterative steps in the absence of round-off error for any initial value given centrally symmetric (centrally antisymmetric) matrix. Moreover, by choosing appropriate initial value, we can get the least Frobenius norm solution for the new generalized coupled Sylvester-conjugate linear matrix equations. Finally, some numerical examples are given to illustrate that the proposed iterative method is quite efficient.


---

## Body

## 1. Introduction

Many research papers are involved in the system of matrix equation ([1–33]). The following matrix equation
(1)
A
X
B
=
C
is a special case of coupled Sylvester linear matrix equations
(2)
∑
j
=
1
n
‍
A
i
j
X
j
B
i
j
=
C
,
(
i
=
1,2
,
…
,
m
)
.
In [34], an iterative algorithm was constructed to solve (1) for skew-symmetric matrix X. Navarra et al. studied a representation of the general solution for the matrix equations A
1
X
B
1
=
C
1, A
2
X
B
2
=
C
2 [35]. By Moore-Penrose generalized inverse, some necessary and sufficient conditions on the existence of the solution and the expressions for the matrix equation A
X
+
X
T
C
=
B were obtained in ([36]). Deng et al. give the consistent conditions and the general expressions of the Hermitian solutions for (1) [37]. In addition, by extending the well-known Jacobi and Gauss-Seidel iterations for A
x
=
b, Ding et al. gained iterative solutions for matrix equation (1) and the generalized Sylvester matrix equation A
X
B
+
C
X
D
=
F [38]. The closed form solutions to a family of generalized Sylvester matrix equations were given by utilizing the so-called Kronecker matrix polynomials in ([39]). In recent years, Dehghan and Hajarian considered the solution for the generalized coupled Sylvester matrix equations [40] A
X
B
+
C
Y
D
=
M, E
X
F
+
G
Y
H
=
N and presented a modified conjugate gradient method to solve the matrix equations over generalized bisymmetric matrix pair (
X
,
Y
). Liang and Liu proposed a modified conjugate gradient method to solve the following problem [41]:
(3)
A
1
X
B
1
+
C
1
X
T
D
1
=
F
1
,
A
2
X
B
2
+
C
2
X
T
D
2
=
F
2
.In the present paper, we conceive efficient algorithm to solve the following generalized coupled Sylvester-conjugate linear matrix equations for centrally symmetric (centrally antisymmetric) matrix pair(
X
,
Y
):
(4)
A
1
X
+
B
1
Y
=
D
1
X
¯
E
1
+
F
1
,
A
2
Y
+
B
2
X
=
D
2
Y
¯
E
2
+
F
2
,
where A
i
,
B
i
,
D
i
∈
C
p
×
m, E
i
∈
C
m
×
m, F
i
∈
C
p
×
m  (
i
=
1,2
) are given constant matrices, and X
,
Y
∈
C
m
×
m are unknown matrices to be solved. When A
2
=
B
2
=
D
2
=
0 and F
2
=
0, the problem (4) becomes the problem studied in [42]. When A
2
=
B
2
=
D
2
=
0, F
2
=
0, and A
1
=
I, this system becomes the Yakubovich-conjugate matrix equation investigated in [43]. When B
1
=
A
2
=
B
2
=
D
2
=
0, F
2
=
0, and A
1
=
I, the problem (4) becomes the equation considered in [44]. When A
2
=
B
2
=
D
2
=
0 and F
1
=
F
2
=
0, the problem (4) becomes the equation in [45]. When B
1
=
A
2
=
B
2
=
D
2
=
0, F
2
=
0, and D
1
=
I, (4) becomes the equation in [46].It is known that modified conjugate gradient (MCG) method is the most popular iterative method for solving the system of linear equation(5)
A
x
=
b
,
where x
∈
R
n is an unknown vector, A
∈
R
m
×
n is a given matrix, and b
∈
R
m is constant vector. By the definition of the Kronecker product, matrix equations can be transformed into the system (5). Then the MCG can be applied to various linear matrix equations [44, 45]. Based on this idea, in this paper, we propose a modified conjugate gradient method to solve the system (4) and show that a solution pair (
X
*
,
Y
*
) can be obtained within finite iterative steps in the absence of round-off error for any initial value given centrally symmetric (centrally antisymmetric) matrix. Furthermore, by choosing appropriate initial value matrix pair, we can obtain the least Frobenius norm solution for (4).As a matter of convenience, some terminology used throughout the paper follows.C
m
×
n is the set of m
×
n complex matrices and R
m
×
n is the set of m
×
n all real matrices. For A
∈
C
m
×
n, we write Re
(
A
), Im
⁡
(
A
), A
¯, A
T, A
H, A
-
1, ∥
A
∥
F, and R
(
A
) to denote the real part, the imaginary part, the conjugation, transpose, conjugate transpose, the inverse, the Frobenius norm, and the column space of the matrix A, respectively. Diag
{
A
1
,
A
2
,
…
,
A
n
} denotes the block diagonal matrix, where A
i
∈
R
m
×
m
(
i
=
1,2
,
…
,
n
). For any A
=
(
a
i
j
), B
=
(
b
i
j
), A
⊗
B denotes the Kronecker product defined as A
⊗
B
=
(
a
i
j
B
). For the matrix X
=
(
x
1
,
x
2
,
…
,
x
n
)
∈
C
m
×
n, vec
⁡
(
X
) denotes the vec
⁡ operator defined as vec
⁡
(
X
)
=
(
x
1
T
,
x
2
T
,
…
,
x
n
T
)
T. We use I to denote the identity matrix of size implied by context.Definition 1.
LetS
∈
R
m
×
m and S
=
(
e
m
,
e
m
-
1
,
…
,
e
1
), where e
j
(
j
=
1,2
,
…
,
m
) denotes the column vector whose j
th element is 1 and the other elements are zeros. An m
×
m complex matrix X is said to be a centrally symmetric (centrally antisymmetric) matrix if S
X
S
=
X
(
S
X
S
=
-
X
), denote the set of all centrally symmetric (centrally antisymmetric) matrices by C
S
C
m
×
m
(
C
A
S
C
m
×
m
).The rest of this paper is organized as follows. In Section2, we construct modified conjugate gradient (MCG) method for solving the system (4) and show that a solution pair (
X
*
,
Y
*
) for (4) can be obtained by the MCG method within finite iterative steps in the absence of round-off error for any initial value given centrally symmetric (centrally antisymmetric) matrix. Furthermore, we demonstrate that the least Frobenius norm solution can be obtained by choosing a special kind of initial matrix. Also we give some numerical examples which illustrate that the introduced iterative algorithm is efficient in Section 3. Conclusions are arranged in Section 4.


## 2. The Iterative Method for Solving the Matrix Equations (4)

In this section, we present the modified conjugate gradient method (MCG) for solving the system (4). Firstly, we recall that the definition of inner product came from [42].The inner product in spaceC
m
×
n is defined as
(6)
〈
A
,
B
〉
=
Re
[
tr
⁡
(
A
H
B
)
]
.
By Theorem 1 in [42], we know that the inner product defined by (6) satisfies the following three axioms:(1)
symmetry:〈
A
,
B
〉
=
〈
B
,
A
〉;
(2)
linearity in the first argument:(7)
〈
δ
1
A
1
+
δ
2
A
2
,
B
〉
=
δ
1
〈
A
1
,
B
〉
+
δ
2
〈
A
2
,
B
〉
,
where δ
1 and δ
2 are real constants;
(3)
positive definiteness:〈
A
,
A
〉
>
0, for all A
≠
0.For all real constantsδ
1
,
δ
2, by (1) and (2), we get
(8)
〈
A
,
δ
1
B
1
+
δ
2
B
2
〉
=
〈
δ
1
B
1
+
δ
2
B
2
,
A
〉
=
δ
1
〈
B
1
,
A
〉
+
δ
2
〈
B
2
,
A
〉
=
δ
1
〈
A
,
B
1
〉
+
δ
2
〈
A
,
B
2
〉
;
namely, the inner product defined by (6) is linear in the second argument.By the relation between the matrix trace and the conjugate operation, we get(9)
〈
A
,
B
〉
=
Re
[
tr
⁡
(
A
H
B
)
]
=
Re
[
tr
⁡
(
A
H
B
)
¯
]
=
Re
[
tr
⁡
(
A
H
B
¯
)
]
.The norm of a matrix generated by this inner product space is denoted by∥
·
∥. Then for A
∈
C
m
×
n, we obtain
(10)
∥
A
∥
2
=
〈
A
,
A
〉
=
Re
[
tr
⁡
(
A
H
A
)
]
.What is the relationship between this norm and the Frobenius norm? It is well known that∥
A
∥
F
2
=
tr
⁡
(
A
H
A
) and A
H
A is a Hermite matrix. Then by the knowledge of algebra, we know that tr
⁡
(
A
H
A
) is real; hence, tr
⁡
(
A
H
A
)
=
Re
[
tr
⁡
(
A
H
A
)
]. This shows that ∥
A
∥
F
=
∥
A
∥. Another interesting relationship is that
(11)
∥
A
∥
2
=
∥
Re
(
A
)
+
i
Im
⁡
(
A
)
∥
2
=
Re
{
tr
⁡
[
(
Re
(
A
)
+
i
Im
⁡
(
A
)
)
H
(
Re
(
A
)
+
i
Im
⁡
(
A
)
)
]
}
=
Re
{
tr
⁡
[
(
Re
(
A
)
T
-
i
Im
⁡
(
A
)
T
)
(
Re
(
A
)
+
i
Im
⁡
(
A
)
)
]
}
=
Re
[
tr
⁡
(
Re
(
A
)
T
Re
(
A
)
+
i
Re
(
A
)
T
Im
⁡
(
A
)
i
i
i
i
i
i
i
i
-
i
Im
⁡
(
A
)
T
Re
(
A
)
+
Im
⁡
(
A
)
T
Im
⁡
(
A
)
)
]
=
Re
[
tr
⁡
(
Re
(
A
)
T
Re
(
A
)
)
+
tr
⁡
(
Im
⁡
(
A
)
T
Im
⁡
(
A
)
)
i
i
i
i
i
i
i
+
i
tr
⁡
(
Re
(
A
)
T
Im
⁡
(
A
)
)
-
i
tr
⁡
(
Im
⁡
(
A
)
T
Re
(
A
)
)
]
=
tr
⁡
(
Re
(
A
)
T
Re
(
A
)
)
+
tr
⁡
(
Im
⁡
(
A
)
T
Im
⁡
(
A
)
)
=
∥
Re
(
A
)
∥
F
2
+
∥
Im
⁡
(
A
)
∥
F
2
.
That is, ∥
A
∥
2
=
∥
Re
(
A
)
∥
F
2
+
∥
Im
⁡
(
A
)
∥
F
2.In the following we present some algorithms. The ordinary conjugate gradient (CG) method to solve (5) is as follows [47].Algorithm 2 (CG method).
Consider the following steps.
Step  1. Input A, b. Choose the initial vectors x
0 and set k
:
=
0; calculate r
0
=
b
-
A
x
0, p
0
=
r
0.
Step  2. If r
k
=
0 or r
k
≠
0 and p
k
=
0, stop; otherwise, calculate
(12)
x
k
+
1
=
x
k
+
〈
r
k
,
r
k
〉
〈
p
k
,
p
k
〉
p
k
.
Step  3. Update the sequences
(13)
r
k
+
1
=
b
-
A
x
k
+
1
,
p
k
+
1
=
r
k
+
1
+
〈
r
k
+
1
,
r
k
+
1
〉
〈
r
k
,
r
k
〉
p
k
.
Step  4. Set k
∶
=
k
+
1; return to Step 2.It is known that the size of the linear equation (5) will be large, when (4) is transformed to a linear equation (5) by the Kronecker product. Therefore, the iterative Algorithm 2 will consume much more computer time and memory space once increasing the dimensionality of coefficient matrix.In view of these considerations, we construct the following so-called modified conjugate gradient (MCG) method to solve (4).Algorithm 3 (MCG method for centrally symmetric matrix version).
Consider the following steps.
Step  1. Input appropriate dimension matrices A
i, B
i, D
i, E
i, and F
i, (
i
=
1,2
). Choose the initial matrices X
1
∈
C
S
C
m
×
m and Y
1
∈
C
S
C
m
×
m, S
=
(
e
m
,
e
m
-
1
,
…
,
e
1
) in Definition 1. Compute
(14)
R
1
∶
=
(
R
1
(
1
)
0
0
R
1
(
2
)
)
,
R
1
(
1
)
=
D
1
X
1
¯
E
1
+
F
1
-
A
1
X
1
-
B
1
Y
1
,
R
1
(
2
)
=
D
2
Y
1
¯
E
2
+
F
2
-
A
2
Y
1
-
B
2
X
1
,
R
~
1
∶
=
(
R
~
1
(
1
)
0
0
R
~
1
(
2
)
)
,
R
~
1
(
1
)
=
A
1
H
R
1
(
1
)
+
B
2
H
R
1
(
2
)
-
D
1
T
R
1
(
1
)
¯
E
1
T
,
R
~
1
(
2
)
=
A
2
H
R
1
(
2
)
+
B
1
H
R
1
(
1
)
-
D
2
T
R
1
(
2
)
¯
E
2
T
,
M
1
=
1
2
(
R
~
1
(
1
)
+
S
R
~
1
(
1
)
S
)
,
N
1
=
1
2
(
R
~
1
(
2
)
+
S
R
~
1
(
2
)
S
)
;
set k
∶
=
1.
Step  2. If R
k
=
0 or R
k
≠
0, M
k
=
N
k
=
0, stop; otherwise, go to Step 3.
Step  3. Update the sequences
(15)
X
k
+
1
=
X
k
+
α
k
M
k
,
Y
k
+
1
=
Y
k
+
α
k
N
k
,
R
k
+
1
∶
=
(
R
k
+
1
(
1
)
0
0
R
k
+
1
(
2
)
)
,
R
k
+
1
(
1
)
=
D
1
X
k
+
1
¯
E
1
+
F
1
-
A
1
X
k
+
1
-
B
1
Y
k
+
1
,
R
k
+
1
(
2
)
=
D
2
Y
k
+
1
¯
E
2
+
F
2
-
A
2
Y
k
+
1
-
B
2
X
k
+
1
,
R
~
k
+
1
∶
=
(
R
~
k
+
1
(
1
)
0
0
R
~
k
+
1
(
2
)
)
,
R
~
k
+
1
(
1
)
=
A
1
H
R
k
+
1
(
1
)
+
B
2
H
R
k
+
1
(
2
)
-
D
1
T
R
k
+
1
(
1
)
¯
E
1
T
,
R
~
k
+
1
(
2
)
=
A
2
H
R
k
+
1
(
2
)
+
B
1
H
R
k
+
1
(
1
)
-
D
2
T
R
k
+
1
(
2
)
¯
E
2
T
,
M
k
+
1
=
1
2
(
R
~
k
+
1
(
1
)
+
S
R
~
k
+
1
(
1
)
S
)
+
β
k
M
k
,
N
k
+
1
=
1
2
(
R
~
k
+
1
(
2
)
+
S
R
~
k
+
1
(
2
)
S
)
+
β
k
N
k
,
where
(16)
α
k
∶
=
∥
R
k
∥
2
∥
M
k
∥
2
+
∥
N
k
∥
2
,
β
k
∶
=
∥
R
k
+
1
∥
2
∥
R
k
∥
2
.
Step  4. Set k
∶
=
k
+
1; return to Step 2.Algorithm 4 (MCG method for centrally antisymmetric matrix version).
Consider the following steps.
Step  1. Input matrices A
i, B
i, D
i, E
i, and F
i, (
i
=
1,2
). Choose the initial matrix X
1
∈
C
A
S
C
m
×
m and Y
1
∈
C
A
S
C
m
×
m, S
=
(
e
m
,
e
m
-
1
,
…
,
e
1
) in Definition 1. Compute
(17)
R
1
∶
=
(
R
1
(
1
)
0
0
R
1
(
2
)
)
,
R
1
(
1
)
=
D
1
X
1
¯
E
1
+
F
1
-
A
1
X
1
-
B
1
Y
1
,
R
1
(
2
)
=
D
2
Y
1
¯
E
2
+
F
2
-
A
2
Y
-
B
2
X
1
,
R
~
1
∶
=
(
R
~
1
(
1
)
0
0
R
~
1
(
2
)
)
,
R
~
1
(
1
)
=
A
1
H
R
1
(
1
)
+
B
2
H
R
1
(
2
)
-
D
1
T
R
1
(
1
)
¯
E
1
T
,
R
~
1
(
2
)
=
A
2
H
R
1
(
2
)
+
B
1
H
R
1
(
1
)
-
D
2
T
R
1
(
2
)
¯
E
2
T
,
M
1
=
1
2
(
R
~
1
(
1
)
-
S
R
~
1
(
1
)
S
)
,
N
1
=
1
2
(
R
~
1
(
2
)
-
S
R
~
1
(
2
)
S
)
;
set k
∶
=
1.
Step  2. If R
k
=
0 or R
k
≠
0, M
k
=
N
k
=
0, stop; otherwise, go to Step 3.
Step  3. Update the sequences
(18)
X
k
+
1
=
X
k
+
α
k
M
k
,
Y
k
+
1
=
Y
k
+
α
k
N
k
,
R
k
+
1
∶
=
(
R
k
+
1
(
1
)
0
0
R
k
+
1
(
2
)
)
,
R
k
+
1
(
1
)
=
D
1
X
k
+
1
¯
E
1
+
F
1
-
A
1
X
k
+
1
-
B
1
Y
k
+
1
,
R
k
+
1
(
2
)
=
D
2
Y
k
+
1
¯
E
2
+
F
2
-
A
2
Y
k
+
1
-
B
2
X
k
+
1
,
R
~
k
+
1
∶
=
(
R
~
k
+
1
(
1
)
0
0
R
~
k
+
1
(
2
)
)
,
R
~
k
+
1
(
1
)
=
A
1
H
R
k
+
1
(
1
)
+
B
2
H
R
k
+
1
(
2
)
-
D
1
T
R
k
+
1
(
1
)
¯
E
1
T
,
R
~
k
+
1
(
2
)
=
A
2
H
R
k
+
1
(
2
)
+
B
1
H
R
k
+
1
(
1
)
-
D
2
T
R
k
+
1
(
2
)
¯
E
2
T
,
(19)
M
k
+
1
=
1
2
(
R
~
k
+
1
(
1
)
-
S
R
~
k
+
1
(
1
)
S
)
+
β
k
M
k
,
N
k
+
1
=
1
2
(
R
~
k
+
1
(
2
)
-
S
R
~
k
+
1
(
2
)
S
)
+
β
k
N
k
,
where
(20)
α
k
:
=
∥
R
k
∥
2
∥
M
k
∥
2
+
∥
N
k
∥
2
,
β
k
:
=
∥
R
k
+
1
∥
2
∥
R
k
∥
2
.
Step  4. Set k
:
=
k
+
1; return to Step 2.Now, we will show that the sequence matrix pair{
X
k
,
Y
k
} generated by Algorithm 3 converges to the solution (
X
*
,
Y
*
) for (4) within finite iterative steps in the absence of round-off error for any initial value over centrally symmetric (centrally antisymmetric) matrix.Lemma 5.
Let the sequences{
R
k
}, {
M
k
}, {
N
k
}, {
R
~
j
(
1
)
}, {
R
~
j
(
2
)
}, and {
α
k
} generated by Algorithm 3; then have
(21)
〈
R
k
+
1
,
R
j
〉
=
〈
R
k
,
R
j
〉
-
α
k
(
〈
M
k
,
R
~
j
(
1
)
〉
+
〈
N
k
,
R
~
j
(
2
)
〉
)
,
〈
M
k
,
R
~
j
(
1
)
〉
k
,
j
=
1,2
,
…
.Proof.
By Algorithm3 and (20), we get
(22)
R
k
+
1
(
1
)
=
D
1
X
k
+
1
¯
E
1
+
F
1
-
A
1
X
k
+
1
-
B
1
Y
k
+
1
=
D
1
(
X
k
¯
+
α
k
M
k
¯
)
E
1
+
F
1
-
A
1
(
X
k
+
α
k
M
k
)
-
B
1
(
Y
k
+
α
k
N
k
)
=
D
1
X
k
¯
E
1
+
F
1
-
A
1
X
k
-
B
1
Y
k
+
α
k
(
D
1
M
k
¯
E
1
-
A
1
M
k
-
B
1
N
k
)
=
R
k
(
1
)
+
α
k
(
D
1
M
k
¯
E
1
-
A
1
M
k
-
B
1
N
k
)
.
In a similar way, we can get
(23)
R
k
+
1
(
2
)
=
R
k
(
2
)
+
α
k
(
D
2
N
k
¯
E
2
-
A
2
N
k
-
B
2
M
k
)
.
This together with the definition of inner product yields that
(24)
〈
R
k
+
1
(
1
)
,
R
j
(
1
)
〉
=
〈
R
k
(
1
)
,
R
j
(
1
)
〉
+
α
k
(
〈
D
1
M
k
¯
E
1
,
R
j
(
1
)
〉
-
〈
A
1
M
k
,
R
j
(
1
)
〉
+
α
-
〈
B
1
N
k
,
R
j
(
1
)
〉
)
=
〈
R
k
(
1
)
,
R
j
(
1
)
〉
+
α
k
×
{
Re
[
tr
⁡
(
E
1
H
M
k
T
D
1
H
R
j
(
1
)
)
]
-
Re
[
tr
⁡
(
M
k
H
A
1
H
R
j
(
1
)
)
]
+
α
-
Re
[
tr
⁡
(
N
k
H
B
1
H
R
j
(
1
)
)
]
}
=
〈
R
k
(
1
)
,
R
j
(
1
)
〉
+
α
k
×
{
Re
[
tr
⁡
(
E
1
H
M
k
T
D
1
H
R
j
(
1
)
¯
)
]
-
〈
M
k
,
A
1
H
R
j
(
1
)
〉
+
α
-
〈
N
k
,
B
1
H
R
j
(
1
)
〉
[
tr
⁡
(
E
1
H
M
k
T
D
1
H
R
j
(
1
)
¯
)
]
}
=
〈
R
k
(
1
)
,
R
j
(
1
)
〉
+
α
k
×
{
Re
[
tr
⁡
(
E
1
T
M
k
H
D
1
T
R
j
(
1
)
¯
)
]
-
〈
M
k
,
A
1
H
R
j
(
1
)
〉
+
α
-
〈
N
k
,
B
1
H
R
j
(
1
)
〉
[
tr
⁡
(
E
1
T
M
k
H
D
1
T
R
j
(
1
)
¯
)
]
}
=
〈
R
k
(
1
)
,
R
j
(
1
)
〉
+
α
k
×
{
Re
[
tr
⁡
(
M
k
H
D
1
T
R
j
(
1
)
¯
E
1
T
)
]
-
〈
M
k
,
A
1
H
R
j
(
1
)
〉
+
α
-
〈
N
k
,
B
1
H
R
j
(
1
)
〉
[
tr
⁡
(
M
k
H
D
1
T
R
j
(
1
)
¯
E
1
T
)
]
}
=
〈
R
k
(
1
)
,
R
j
(
1
)
〉
+
α
k

×

(
〈
M
k
,
D
1
T
R
j
(
1
)
¯
E
1
T
〉
-
〈
M
k
,
A
1
H
R
j
(
1
)
〉
-
〈
N
k
,
B
1
H
R
j
(
1
)
〉
)
=
〈
R
k
(
1
)
,
R
j
(
1
)
〉
+
α
k
×
(
〈
M
k
,
D
1
T
R
j
(
1
)
¯
E
1
T
-
A
1
H
R
j
(
1
)
〉
-
〈
N
k
,
B
1
H
R
j
(
1
)
〉
)
,
〈
R
k
+
1
(
2
)
,
R
j
(
2
)
〉
=
〈
R
k
(
2
)
,
R
j
(
2
)
〉
+
α
k
×
(
〈
D
2
N
k
¯
E
2
,
R
j
(
2
)
〉
-
〈
A
2
N
k
,
R
j
(
2
)
〉
-
〈
B
2
M
k
,
R
j
(
2
)
〉
)
=
〈
R
k
(
2
)
,
R
j
(
2
)
〉
+
α
k
×
{
Re
[
tr
⁡
(
E
2
H
N
k
T
D
2
H
R
j
(
2
)
)
]
-
Re
[
tr
⁡
(
N
k
H
A
2
H
R
j
(
2
)
)
]
×
-
Re
[
tr
⁡
(
M
k
H
B
2
H
R
j
(
2
)
)
]
}
=
〈
R
k
(
2
)
,
R
j
(
2
)
〉
+
α
k
×
{
Re
[
tr
⁡
(
E
2
H
N
k
T
D
2
H
R
j
(
2
)
¯
)
]
-
〈
N
k
,
A
2
H
R
j
(
2
)
〉
×
-
〈
M
k
,
B
2
H
R
j
(
2
)
〉
[
tr
⁡
(
E
2
H
N
k
T
D
2
H
R
j
(
2
)
¯
)
]
}
=
〈
R
k
(
2
)
,
R
j
(
2
)
〉
+
α
k
×
{
Re
[
tr
⁡
(
E
2
T
N
k
H
D
2
T
R
j
(
2
)
¯
)
]
-
〈
N
k
,
A
2
H
R
j
(
2
)
〉
×
-
〈
M
k
,
B
2
H
R
j
(
2
)
〉
[
tr
⁡
(
E
2
T
N
k
H
D
2
T
R
j
(
2
)
¯
)
]
}
=
〈
R
k
(
2
)
,
R
j
(
2
)
〉
+
α
k
×
{
Re
[
tr
⁡
(
N
k
H
D
2
T
R
j
(
2
)
¯
E
2
T
)
]
-
〈
N
k
,
A
2
H
R
j
(
2
)
〉
×
-
〈
M
k
,
B
2
H
R
j
(
2
)
〉
[
tr
⁡
(
N
k
H
D
2
T
R
j
(
2
)
¯
E
2
T
)
]
}
=
〈
R
k
(
2
)
,
R
j
(
2
)
〉
+
α
k

×

(
〈
N
k
,
D
2
T
R
j
(
2
)
¯
E
2
T
〉
-
〈
N
k
,
A
2
H
R
j
(
2
)
〉
-
〈
M
k
,
B
2
H
R
j
(
2
)
〉
)
=
〈
R
k
(
2
)
,
R
j
(
2
)
〉
+
α
k
×
(
〈
N
k
,
D
2
T
R
j
(
2
)
¯
E
2
T
-
A
2
H
R
j
(
2
)
〉
-
〈
M
k
,
B
2
H
R
j
(
2
)
〉
)
.
Then by the updated formulas of R
k, R
~
j
(
1
), and R
~
j
(
2
), we obtain
(25)
〈
R
k
+
1
,
R
j
〉
=
〈
R
k
+
1
(
1
)
,
R
j
(
1
)
〉
+
〈
R
k
+
1
(
2
)
,
R
j
(
2
)
〉
=
〈
R
k
(
1
)
,
R
j
(
1
)
〉
+
α
k
×
(
〈
M
k
,
D
1
T
R
j
(
1
)
¯
E
1
T
-
A
1
H
R
j
(
1
)
〉
-
〈
N
k
,
B
1
H
R
j
(
1
)
〉
)
+
〈
R
k
(
2
)
,
R
j
(
2
)
〉
+
α
k
×
(
〈
N
k
,
D
2
T
R
j
(
2
)
¯
E
2
T
-
A
2
H
R
j
(
2
)
〉
-
〈
M
k
,
B
2
H
R
j
(
2
)
〉
)
=
〈
R
k
,
R
j
〉
+
α
k
×
(
〈
M
k
,
D
1
T
R
j
(
1
)
¯
E
1
T
-
A
1
H
R
j
(
1
)
-
B
2
H
R
j
(
2
)
〉
+
〈
N
k
,
D
2
T
R
j
(
2
)
¯
E
2
T
-
A
2
H
R
j
(
2
)
-
B
1
H
R
j
(
1
)
〉
)
=
〈
R
k
,
R
j
〉
-
α
k
(
〈
M
k
,
R
~
j
(
1
)
〉
+
〈
N
k
,
R
~
j
(
2
)
〉
)
,
which completes the proof.Lemma 6.
Let the sequences,{
R
k
}, {
M
k
}, and {
N
k
}, be generated by Algorithm 3; one has
(26)
〈
R
i
,
R
j
〉
=
0
,
〈
M
i
,
M
j
〉
+
〈
N
i
,
N
j
〉
=
0
,
i
i
i
i
i
i
,
j
=
1,2
,
…
,
k
,
i
≠
j
.Proof.
Firstly, we prove(27)
〈
R
i
,
R
j
〉
=
0
,
〈
M
i
,
M
j
〉
+
〈
N
i
,
N
j
〉
=
0
,
1
≤
j
<
i
≤
k
.
By mathematical induction, for k
=
2, by Lemma 5, and noticing M
j
∈
C
S
C
m
×
m, N
j
∈
C
S
C
m
×
m
(
j
=
1,2
,
…
,
k
) generated by Algorithm 3, we get
(28)
〈
R
2
,
R
1
〉
=
〈
R
1
,
R
1
〉
-
α
1
(
〈
M
1
,
R
~
1
(
1
)
〉
+
〈
N
1
,
R
~
1
(
2
)
〉
)
=
∥
R
1
∥
2
-
α
1
(
〈
M
1
,
R
~
1
(
1
)
+
S
R
~
1
(
1
)
S
2
〉
=
∥
R
1
∥
2
-
α
1
+
〈
N
1
,
R
~
1
(
2
)
+
S
R
~
1
(
2
)
S
2
〉
)
=
∥
R
1
∥
2
-
∥
R
1
∥
2
∥
M
1
∥
2
+
∥
N
1
∥
2
(
〈
M
1
,
M
1
〉
+
〈
N
1
,
N
1
〉
)
=
∥
R
1
∥
2
-
∥
R
1
∥
2
=
0
,
where the second equality is from the fact
(29)
〈
M
1
,
S
R
~
1
(
1
)
S
〉
=
〈
M
1
,
R
~
1
(
1
)
〉
,
〈
N
1
,
S
R
~
1
(
2
)
S
〉
=
〈
N
1
,
R
~
1
(
2
)
〉
.
In addition, by (19), (20), and Lemma 5, we have
(30)
〈
M
2
,
M
1
〉
+
〈
N
2
,
N
1
〉
=
〈
R
~
2
(
1
)
+
S
R
~
2
(
1
)
S
2
+
β
1
M
1
,
M
1
〉
+
〈
R
~
2
(
2
)
+
S
R
~
2
(
2
)
S
2
+
β
1
N
1
,
N
1
〉
=
〈
R
~
2
(
1
)
,
M
1
〉
+
〈
R
~
2
(
2
)
,
N
1
〉
+
β
1
(
∥
M
1
∥
2
+
∥
N
1
∥
2
)
=
1
α
1
(
〈
R
1
,
R
2
〉
-
〈
R
2
,
R
2
〉
)
+
β
1
(
∥
M
1
∥
2
+
∥
N
1
∥
2
)
=
0
,
where the second equality is from (6) and the fact
(31)
〈
M
1
,
S
R
~
2
(
1
)
S
〉
=
〈
M
1
,
R
~
2
(
1
)
〉
,
〈
N
1
,
S
R
~
2
(
2
)
S
〉
=
〈
N
1
,
R
~
2
(
2
)
〉
.
Therefore, (27) holds for k
=
2.
Suppose that (27) holds, for k
=
l
(
l
≥
2
). For k
=
l
+
1, it follows from Lemma 5, (9) that
(32)
〈
R
l
+
1
,
R
l
〉
=
〈
R
l
,
R
l
〉
-
α
l
(
〈
M
l
,
R
~
l
(
1
)
〉
+
〈
N
l
,
R
~
l
(
2
)
〉
)
=
∥
R
l
∥
2
-
α
l
×
(
〈
M
l
,
R
~
l
(
1
)
+
S
R
~
l
(
1
)
S
2
〉
+
〈
N
l
,
R
~
l
(
2
)
+
S
R
~
l
(
2
)
S
2
〉
)
=
∥
R
l
∥
2
-
α
l
×
(
〈
M
l
,
M
l
-
β
l
-
1
M
l
-
1
〉
+
〈
N
l
,
N
l
-
β
l
-
1
N
l
-
1
〉
)
=
∥
R
l
∥
2
-
α
l
(
∥
M
l
∥
2
+
∥
N
l
∥
2
)
=
0
,
where the fourth equality holds by the induction assumption. Combining (19) and (20) and by induction with the above result, we obtain
(33)
〈
M
l
+
1
,
M
l
〉
+
〈
N
l
+
1
,
N
l
〉
=
〈
R
~
l
+
1
(
1
)
+
S
R
~
l
+
1
(
1
)
S
2
+
β
l
M
l
,
M
l
〉
+
〈
R
~
l
+
1
(
2
)
+
S
R
~
l
+
1
(
2
)
S
2
+
β
l
N
l
,
N
l
〉
=
〈
R
~
l
+
1
(
1
)
,
M
l
〉
+
〈
R
~
l
+
1
(
2
)
,
N
l
〉
+
β
l
(
∥
M
l
∥
2
+
∥
N
l
∥
2
)
=
1
α
l
(
〈
R
l
,
R
l
+
1
〉
-
〈
R
l
+
1
,
R
l
+
1
〉
)
+
β
l
(
∥
M
l
∥
2
+
∥
N
l
∥
2
)
=
0
,
where the third equality is from Lemma 5.
Forj
=
1, by Lemma 5 and the induction, we have
(34)
〈
R
l
+
1
,
R
1
〉
=
〈
R
l
,
R
1
〉
-
α
l
(
〈
M
l
,
R
~
1
(
1
)
〉
+
〈
N
l
,
R
~
1
(
2
)
〉
)
=
-
α
l
(
〈
M
l
,
R
~
1
(
1
)
+
S
R
~
1
(
1
)
S
2
〉
+
〈
N
l
,
R
~
1
(
2
)
+
S
R
~
1
(
2
)
S
2
〉
)
=
-
α
l
(
〈
M
l
,
M
1
〉
+
〈
N
l
,
N
1
〉
)
=
0
.
Analogously, for j
=
2,3
,
…
,
l
-
1, then we obtain
(35)
〈
R
l
+
1
,
R
j
〉
=
〈
R
l
,
R
j
〉
-
α
l
(
〈
M
l
,
R
~
j
(
1
)
〉
+
〈
N
l
,
R
~
j
(
2
)
〉
)
=
-
α
l
(
〈
M
l
,
R
~
j
(
1
)
+
S
R
~
j
(
1
)
S
2
〉
=
-
α
l
+
〈
N
l
,
R
~
j
(
2
)
+
S
R
~
j
(
2
)
S
2
〉
)
=
-
α
l
(
〈
M
l
,
M
j
-
β
j
-
1
M
j
-
1
〉
M
,
M
+
〈
N
l
,
N
j
-
β
j
-
1
N
j
-
1
〉
)
=
-
α
l
(
〈
M
l
,
M
j
〉
+
〈
N
l
,
N
j
〉
-
β
j
-
1
(
〈
M
l
,
M
j
-
1
〉
+
〈
N
l
,
N
j
-
1
〉
)
-
α
l
×
(
〈
M
l
,
M
j
-
1
〉
+
〈
N
l
,
N
j
-
1
〉
)
)
=
0
.
In addition, from Lemma 5 and the induction, for j
=
1,2
,
…
,
l
-
1, we get
(36)
〈
M
l
+
1
,
M
j
〉
+
〈
N
l
+
1
,
N
j
〉
=
〈
R
~
l
+
1
(
1
)
+
S
R
~
l
+
1
(
1
)
S
2
,
M
j
〉
+
〈
R
~
l
+
1
(
2
)
+
S
R
~
l
+
1
(
2
)
S
2
,
N
j
〉
+
β
l
(
〈
M
l
,
M
j
〉
+
〈
N
l
,
N
j
〉
)
=
〈
R
~
l
+
1
(
1
)
,
M
j
〉
+
〈
R
~
l
+
1
(
2
)
,
N
j
〉
=
1
α
l
(
〈
R
j
,
R
l
+
1
〉
-
〈
R
j
+
1
,
R
l
+
1
〉
)
=
0
.
So (27) holds, for k
=
l
+
1. By induction principle, (27) holds, for all 1
≤
j
<
i
≤
k. For j
>
i, we obtain
(37)
〈
R
i
,
R
j
〉
=
〈
R
j
,
R
i
〉
=
0
,
〈
M
i
,
M
j
〉
+
〈
N
i
,
N
j
〉
=
〈
M
j
,
M
i
〉
+
〈
N
j
,
N
i
〉
=
0
,
which completes the proof.Lemma 7.
Suppose that the system of matrix equations (4) is consistent; let (
X
*
,
Y
*
) be an arbitrary solution pair of (4). Then for any initial matrices X
1
∈
C
M
C
m
×
m, Y
1
∈
C
M
C
m
×
m, the sequences {
X
k
}, {
Y
k
}, {
R
k
}, {
M
k
}, and {
N
k
} generated by Algorithm 3 satisfy
(38)
〈
X
*
-
X
k
,
M
k
〉
+
〈
Y
*
-
Y
k
,
N
k
〉
=
∥
R
k
∥
2
,
k
=
1,2
,
…
,
n
.Proof.
The conclusion is accomplished by mathematical induction.
Firstly, we notice that the sequences pair(
X
k
,
Y
k
), (
k
=
1,2
,
…
) generated by Algorithm 3 are all central symmetric matrices since initial matrix pair (
X
1
,
Y
1
) is centrally symmetric matrix. Then for k
=
1, it follows from Algorithm 3 that
(39)
〈
X
*
-
X
1
,
M
1
〉
=
〈
X
*
-
X
1
,
R
~
1
(
1
)
+
S
R
~
1
(
1
)
S
2
〉
=
〈
X
*
-
X
1
,
R
~
1
(
1
)
〉
=
〈
X
*
-
X
1
,
A
1
H
R
1
(
1
)
+
B
2
H
R
1
(
2
)
-
D
1
T
R
1
(
1
)
¯
E
1
T
〉
=
〈
X
*
-
X
1
,
A
1
H
R
1
(
1
)
〉
+
〈
X
*
-
X
1
,
B
2
H
R
1
(
2
)
〉
-
〈
X
*
-
X
1
,
D
1
T
R
1
(
1
)
¯
E
1
T
〉
=
Re
[
tr
⁡
(
(
X
*
-
X
1
)
H
A
1
H
R
1
(
1
)
)
]
+
Re
[
tr
⁡
(
(
X
*
-
X
1
)
H
B
2
H
R
1
(
2
)
)
]
-
Re
[
tr
⁡
(
(
X
*
-
X
1
)
H
D
1
T
R
1
(
1
)
¯
E
1
T
)
]
=
〈
A
1
(
X
*
-
X
1
)
,
R
1
(
1
)
〉
+
〈
B
2
(
X
*
-
X
1
)
,
R
1
(
2
)
〉
-
Re
[
tr
⁡
(
(
X
*
-
X
1
)
H
D
1
T
R
1
(
1
)
¯
E
1
T
¯
)
]
=
〈
A
1
(
X
*
-
X
1
)
,
R
1
(
1
)
〉
+
〈
B
2
(
X
*
-
X
1
)
,
R
1
(
2
)
〉
-
Re
[
tr
⁡
(
(
X
*
-
X
1
)
T
D
1
H
R
1
(
1
)
E
1
H
)
]
=
〈
A
1
(
X
*
-
X
1
)
,
R
1
(
1
)
〉
+
〈
B
2
(
X
*
-
X
1
)
,
R
1
(
2
)
〉
-
Re
[
tr
⁡
(
E
1
H
(
X
*
-
X
1
)
T
D
1
H
R
1
(
1
)
)
]
=
〈
A
1
(
X
*
-
X
1
)
,
R
1
(
1
)
〉
+
〈
B
2
(
X
*
-
X
1
)
,
R
1
(
2
)
〉
-
〈
D
1
(
X
*
-
X
1
¯
)
E
1
,
R
1
(
1
)
〉
=
〈
A
1
(
X
*
-
X
1
)
-
D
1
(
X
*
-
X
1
¯
)
E
1
,
R
1
(
1
)
〉
+
〈
B
2
(
X
*
-
X
1
)
,
R
1
(
2
)
〉
.
In the same way, we can get
(40)
〈
Y
*
-
Y
1
,
N
1
〉
=
〈
A
2
(
Y
*
-
Y
1
)
-
D
2
(
Y
*
-
Y
1
¯
)
E
2
,
R
1
(
2
)
〉
+
〈
B
1
(
Y
*
-
Y
1
)
,
R
1
(
1
)
〉
.
This shows that
(41)
〈
X
*
-
X
1
,
M
1
〉
+
〈
Y
*
-
Y
1
,
N
1
〉
=
〈
A
1
(
X
*
-
X
1
)
-
D
1
(
X
*
-
X
1
¯
)
E
1
,
R
1
(
1
)
〉
+
〈
B
2
(
X
*
-
X
1
)
,
R
1
(
2
)
〉
+
〈
A
2
(
Y
*
-
Y
1
)
-
D
2
(
Y
*
-
Y
1
¯
)
E
2
,
R
1
(
2
)
〉
+
〈
B
1
(
Y
*
-
Y
1
)
,
R
1
(
1
)
〉
=
〈
A
1
(
X
*
-
X
1
)
+
B
1
(
Y
*
-
Y
1
)
R
1
(
1
)
-
D
1
(
X
*
-
X
1
¯
)
E
1
,
R
1
(
1
)
〉
+
〈
A
2
(
Y
*
-
Y
1
)
+
B
2
(
X
*
-
X
1
)
R
1
(
1
)
-
D
2
(
Y
*
-
Y
1
¯
)
E
2
,
R
1
(
2
)
〉
=
〈
F
1
-
A
1
X
1
-
B
1
Y
1
+
D
1
X
1
¯
E
1
,
R
1
(
1
)
〉
+
〈
F
2
-
A
2
Y
1
-
B
2
X
1
+
D
2
Y
1
¯
E
2
,
R
1
(
2
)
〉
=
∥
R
1
(
1
)
∥
2
+
∥
R
1
(
2
)
∥
2
=
∥
R
1
∥
2
.
That is, (38) holds, for k
=
1.
Assume (38) holds, for k
=
l. For k
=
l
+
1, it follows from the updated formulas of X
l
+
1, Y
l
+
1 that
(42)
〈
X
*
-
X
l
+
1
,
M
l
〉
=
〈
X
*
-
X
l
-
α
l
M
l
,
M
l
〉
=
〈
X
*
-
X
l
,
M
l
〉
-
α
l
∥
M
l
∥
2
,
〈
Y
*
-
Y
l
+
1
,
N
l
〉
=
〈
Y
*
-
Y
l
-
α
l
N
l
,
N
l
〉
=
〈
Y
*
-
Y
l
,
N
l
〉
-
α
l
∥
N
l
∥
2
.
Then
(43)
〈
X
*
-
X
l
+
1
,
M
l
+
1
〉
=
〈
X
*
-
X
l
+
1
,
R
~
l
+
1
(
1
)
+
S
R
~
l
+
1
(
1
)
S
2
+
β
l
M
l
〉
=
〈
X
*
-
X
l
+
1
,
R
~
l
+
1
(
1
)
〉
+
β
l
〈
X
*
-
X
l
+
1
,
M
l
〉
=
〈
X
*
-
X
l
+
1
,
R
~
l
+
1
(
1
)
〉
+
β
l
(
〈
X
*
-
X
l
,
M
l
〉
-
α
l
∥
M
l
∥
2
)
,
〈
Y
*
-
Y
l
+
1
,
N
l
+
1
〉
=
〈
Y
*
-
Y
l
+
1
,
R
~
l
+
1
(
2
)
+
S
R
~
l
+
1
(
2
)
S
2
+
β
l
N
l
〉
=
〈
Y
*
-
Y
l
+
1
,
R
~
l
+
1
(
2
)
〉
+
β
l
〈
Y
*
-
Y
l
+
1
,
N
l
〉
=
〈
Y
*
-
Y
l
+
1
,
R
~
l
+
1
(
2
)
〉
+
β
l
(
〈
Y
*
-
Y
l
,
N
l
〉
-
α
l
∥
N
l
∥
2
)
.
On the other hand, we have
(44)
〈
X
*
-
X
l
+
1
,
R
~
l
+
1
(
1
)
〉
+
〈
Y
*
-
Y
l
+
1
,
R
~
l
+
1
(
2
)
〉
=
〈
X
*
-
X
l
+
1
,
A
1
H
R
l
+
1
(
1
)
+
B
2
H
R
l
+
1
(
2
)
-
D
1
T
R
l
+
1
(
1
)
¯
E
1
T
〉
+
〈
Y
*
-
Y
l
+
1
,
A
2
H
R
l
+
1
(
2
)
+
B
1
H
R
l
+
1
(
1
)
-
D
2
T
R
l
+
1
(
2
)
¯
E
2
T
〉
=
〈
A
1
(
X
*
-
X
l
+
1
)
,
R
l
+
1
(
1
)
〉
+
〈
B
2
(
X
*
-
X
l
+
1
)
,
R
l
+
1
(
2
)
〉
-
Re
[
tr
⁡
(
(
X
*
-
X
l
+
1
)
H
D
1
T
R
l
+
1
(
1
)
¯
E
1
T
)
]
+
〈
A
2
(
Y
*
-
Y
l
+
1
)
,
R
l
+
1
(
2
)
〉
+
〈
B
1
(
Y
*
-
Y
l
+
1
)
,
R
l
+
1
(
1
)
〉
-
Re
[
tr
⁡
(
(
Y
*
-
Y
l
+
1
)
H
D
2
T
R
l
+
1
(
2
)
¯
E
2
T
)
]
=
〈
A
1
(
X
*
-
X
l
+
1
)
+
B
1
(
Y
*
-
Y
l
+
1
)
,
R
l
+
1
(
1
)
〉
-
Re
[
tr
⁡
(
(
X
*
-
X
l
+
1
)
T
D
1
H
R
l
+
1
(
1
)
E
1
H
)
]
+
〈
A
2
(
Y
*
-
Y
l
+
1
)
+
B
2
(
X
*
-
X
l
+
1
)
,
R
l
+
1
(
2
)
〉
-
Re
[
tr
⁡
(
(
Y
*
-
Y
l
+
1
)
T
D
2
H
R
l
+
1
(
2
)
E
2
H
)
]
=
〈
A
1
(
X
*
-
X
l
+
1
)
+
B
1
(
Y
*
-
Y
l
+
1
)
,
R
l
+
1
(
1
)
〉
-
Re
[
tr
⁡
(
E
1
H
(
X
*
-
X
l
+
1
)
T
D
1
H
R
l
+
1
(
1
)
)
]
+
〈
A
2
(
Y
*
-
Y
l
+
1
)
+
B
2
(
X
*
-
X
l
+
1
)
,
R
l
+
1
(
2
)
〉
-
Re
[
tr
⁡
(
E
2
H
(
Y
*
-
Y
l
+
1
)
T
D
2
H
R
l
+
1
(
2
)
)
]
=
〈
A
1
(
X
*
-
X
l
+
1
)
+
B
1
(
Y
*
-
Y
l
+
1
)
R
l
+
1
(
1
)
-
D
1
(
X
*
-
X
l
+
1
¯
)
E
1
,
R
l
+
1
(
1
)
〉
+
〈
A
2
(
Y
*
-
Y
l
+
1
)
+
B
2
(
X
*
-
X
l
+
1
)
R
l
+
1
(
2
)
-
D
2
(
Y
*
-
Y
l
+
1
¯
)
E
2
,
R
l
+
1
(
2
)
〉
=
〈
F
1
-
A
1
X
l
+
1
-
B
1
Y
l
+
1
+
D
1
X
l
+
1
¯
E
1
,
R
l
+
1
(
1
)
〉
+
〈
F
2
-
A
2
Y
l
+
1
-
B
2
X
l
+
1
+
D
2
Y
l
+
1
¯
E
2
,
R
l
+
1
(
2
)
〉
=
∥
R
l
+
1
(
1
)
∥
2
+
∥
R
l
+
1
(
2
)
∥
2
=
∥
R
l
+
1
∥
2
.
Therefore, by (20) we get
(45)
〈
X
*
-
X
l
+
1
,
M
l
+
1
〉
+
〈
Y
*
-
Y
l
+
1
,
N
l
+
1
〉
=
〈
X
*
-
X
l
+
1
,
R
~
l
+
1
(
1
)
+
S
R
~
l
+
1
(
1
)
S
2
+
β
l
M
l
〉
+
〈
Y
*
-
Y
l
+
1
,
R
~
l
+
1
(
2
)
+
S
R
~
l
+
1
(
2
)
S
2
+
β
l
N
l
〉
=
〈
X
*
-
X
l
+
1
,
R
~
l
+
1
(
1
)
〉
+
β
l
(
〈
X
*
-
X
l
,
M
l
〉
-
α
l
∥
M
l
∥
2
)
+
〈
Y
*
-
Y
l
+
1
,
R
~
l
+
1
(
2
)
〉
+
β
l
(
〈
Y
*
-
Y
l
,
N
l
〉
-
α
l
∥
N
l
∥
2
)
=
〈
X
*
-
X
l
+
1
,
R
~
l
+
1
(
1
)
〉
+
〈
Y
*
-
Y
l
+
1
,
R
~
l
+
1
(
2
)
〉
+
β
l
[
〈
X
*
-
X
l
,
M
l
〉
+
〈
Y
*
-
Y
l
,
N
l
〉
(
∥
M
l
∥
2
+
∥
N
l
∥
2
)
〈
X
*
X
l
,
M
l
〉
-
α
l
(
∥
M
l
∥
2
+
∥
N
l
∥
2
)
]
=
〈
X
*
-
X
l
+
1
,
R
~
l
+
1
(
1
)
〉
+
〈
Y
*
-
Y
l
+
1
,
R
~
l
+
1
(
2
)
〉
+
β
l
[
∥
R
l
∥
2
-
∥
R
l
∥
2
∥
M
l
∥
2
+
∥
N
l
∥
2
(
∥
M
l
∥
2
+
∥
N
l
∥
2
)
]
=
〈
X
*
-
X
l
+
1
,
R
~
l
+
1
(
1
)
〉
+
〈
Y
*
-
Y
l
+
1
,
R
~
l
+
1
(
2
)
〉
=
∥
R
l
+
1
∥
2
.
Hence, the proof is completed.Remark 8.
Lemma7 implies that if (4) is consistent, then ∥
M
k
∥
2
+
∥
N
k
∥
2
≠
0 when R
k
≠
0. Conversely, if there exists a positive integer k
0 such that R
k
0
≠
0 and ∥
M
k
0
∥
2
+
∥
N
k
0
∥
2
=
0 in the iteration process of Algorithm 3, then (4) is inconsistent; we will study this condition with other papers in the future.Remark 9.
The above lemmas are achieved under the assumption that initial value is centrally symmetric matrix. Similarly, if the initial matrix is centrally antisymmetric matrix, we can get the same conclusions easily (see Definition1). Hence, we need not show these results in detail; in the following content, we only discuss the version when X
,
Y
∈
C
M
C
m
×
m.Theorem 10.
Suppose the system (4) is consistent; then, for any initial matrix X
1
∈
C
M
C
m
×
m, Y
1
∈
C
M
C
m
×
m, an exact solution of (4) can be derived at most 2
p
m
+
1 iteration steps by Algorithm 3.Proof.
AssumeR
k
≠
0, for k
=
1,2
,
…
,
2
p
m. It follows from Lemma 7 that ∥
M
k
∥
2
+
∥
N
k
∥
2
≠
0 for k
=
1,2
,
…
,
2
p
m. Then R
2
p
m
+
1 can be derived by Algorithm 3. According to Lemma 6, we know 〈
R
i
,
R
j
〉
=
0, for i
,
j
=
1,2
,
…
,
2
p
m
+
1, i
≠
j. Then the matrix sequence of R
1
,
R
2
,
…
,
R
2
p
m is an orthogonal basis of the linear space
(46)
H
=
{
H
∣
H
=
(
H
1
0
0
H
2
)
}
,
where H
1
,
H
2
∈
R
p
×
m. Since R
2
p
m
+
1
∈
H and 〈
R
2
p
m
+
1
,
R
k
〉
=
0, for k
=
1,2
,
…
,
2
p
m, hence R
2
p
m
+
1
=
0, which completes the proof.Although the proof is trivial, the consequences of this result are of major importance.When (4) is consistent, the solution of (4) is not unique. Then we need to find the unique least Frobenius norm solution of (4). Next, we introduce the following lemma.Lemma 11.
SupposeA
∈
R
m
×
n, b
∈
R
m, and the linear matrix equation A
x
=
b has a solution x
*
∈
R
(
A
T
); then x
* is the unique least Frobenius norm solution of A
x
=
b.For a rigorous proof of this lemma above the reader is referred to [46, 48].Lemma 12.
SupposeA
∈
C
m
×
n, b
∈
C
m, and the linear matrix equation A
x
=
b has a solution x
*
∈
R
n. If x
*
∈
R
[
(
Re
(
A
)
T
,
Im
⁡
(
A
)
T
)
], then x
* is the unique least Frobenius norm solution of A
x
=
b.Proof.
LetA
=
Re
(
A
)
+
i
Im
⁡
(
A
) and b
=
Re
(
b
)
+
i
Im
⁡
(
b
). Then A
x
=
b can be written as
(47)
(
Re
(
A
)
+
i
Im
⁡
(
A
)
)
x
=
Re
(
A
)
x
+
i
Im
⁡
(
A
)
x
=
Re
(
b
)
+
i
Im
⁡
(
b
)
.
It shows that
(48)
Re
(
A
)
x
=
Re
(
b
)
,
Im
⁡
(
A
)
x
=
Im
⁡
(
b
)
,
or
(49)
(
Re
(
A
)
Im
⁡
(
A
)
)
x
=
(
Re
(
b
)
Im
⁡
(
b
)
)
.
Since
(50)
x
*
∈
R
[
(
Re
(
A
)
T
,
Im
⁡
(
A
)
T
)
]
=
R
[
(
Re
(
A
)
Im
⁡
(
A
)
)
T
]
,
this together with Lemma 11 completes the result.In order to get the least Frobenius norm solution of (4), we need to transform the problem (4).LetX
=
Re
(
X
)
+
i
Im
⁡
(
X
), Y
=
Re
(
Y
)
+
i
Im
⁡
(
Y
). Then the problem (4) can be equivalently written as
(51)
A
1
(
Re
(
X
)
+
i
Im
⁡
(
X
)
)
+
B
1
(
Re
(
Y
)
+
i
Im
⁡
(
Y
)
)
=
D
1
(
Re
(
X
)
-
i
Im
⁡
(
X
)
)
E
1
+
F
1
,
A
2
(
Re
(
Y
)
+
i
Im
⁡
(
Y
)
)
+
B
2
(
Re
(
X
)
+
i
Im
⁡
(
X
)
)
=
D
2
(
Re
(
Y
)
-
i
Im
⁡
(
Y
)
)
E
2
+
F
2
,
or
(52)
A
1
Re
(
X
)
-
D
1
Re
(
X
)
E
1
+
i
A
1
Im
⁡
(
X
)
+
i
D
1
Im
⁡
(
X
)
E
1
+
B
1
Re
(
Y
)
+
i
B
1
Im
⁡
(
Y
)
=
F
1
,
A
2
Re
(
Y
)
-
D
2
Re
(
Y
)
E
2
+
i
A
2
Im
⁡
(
Y
)
+
i
D
2
Im
⁡
(
Y
)
E
2
+
B
2
Re
(
X
)
+
i
B
2
Im
⁡
(
X
)
=
F
2
.
This together with the definition of the Kronecker product yields
(53)
W
(
vec
⁡
[
Re
(
X
)
]
vec
⁡
[
Im
⁡
(
X
)
]
vec
⁡
[
Re
(
Y
)
]
vec
⁡
[
Im
⁡
(
Y
)
]
)
=
(
vec
⁡
(
F
1
)
vec
⁡
(
F
2
)
)
,
where(54)
W
=
(
I
⊗
A
1
-
E
1
T
⊗
D
1
i
(
I
⊗
A
1
+
E
1
T
⊗
D
1
)
I
⊗
B
1
i
(
I
⊗
B
1
)
I
⊗
B
2
i
(
I
⊗
B
2
)
I
⊗
A
2
-
E
2
T
⊗
D
2
i
(
I
⊗
A
2
+
E
2
T
⊗
D
2
)
)
.
By some simple calculating, we have(55)
W
=
Re
(
W
)
+
i
Im
⁡
(
W
)
,
where
(56)
Re
(
W
)
=
(
N
11
N
12
N
13
N
14
N
21
N
22
N
23
N
24
)
,
(57)
Im
⁡
(
W
)
=
(
L
11
L
12
L
13
L
14
L
21
L
22
L
23
L
24
)
,
(58)
N
11
=
I
⊗
Re
(
A
1
)
-
Re
(
E
1
)
T
⊗
Re
(
D
1
)
+
Im
⁡
(
E
1
)
T
⊗
Im
⁡
(
D
1
)
,
N
12
=
-
I
⊗
Im
⁡
(
A
1
)
-
Im
⁡
(
E
1
)
T
⊗
Re
(
D
1
)
-
Re
(
E
1
)
T
⊗
Im
⁡
(
D
1
)
,
N
13
=
I
⊗
Re
(
B
1
)
,
N
14
=
-
I
⊗
Im
⁡
(
B
1
)
,
N
21
=
I
⊗
Re
(
B
2
)
,
N
22
=
-
I
⊗
Im
⁡
(
B
2
)
,
N
23
=
I
⊗
Re
(
A
2
)
-
Re
(
E
2
)
T
⊗
Re
(
D
2
)
+
Im
⁡
(
E
2
)
T
⊗
Im
⁡
(
D
2
)
,
N
24
=
-
I
⊗
Im
⁡
(
A
2
)
-
Im
⁡
(
E
2
)
T
⊗
Re
(
D
2
)
-
Re
(
E
2
)
T
⊗
Im
⁡
(
D
2
)
,
L
11
=
I
⊗
Im
⁡
(
A
1
)
-
Im
⁡
(
E
1
)
T
⊗
Re
(
D
1
)
-
Re
(
E
1
)
T
⊗
Im
⁡
(
D
1
)
,
L
12
=
I
⊗
Re
(
A
1
)
+
Re
(
E
1
)
T
⊗
Re
(
D
1
)
-
Im
⁡
(
E
1
)
T
⊗
Im
⁡
(
D
1
)
,
L
13
=
I
⊗
Im
⁡
(
B
1
)
,
L
14
=
I
⊗
Re
(
B
1
)
,
L
21
=
I
⊗
Im
⁡
(
B
2
)
,
L
22
=
I
⊗
Re
(
B
2
)
,
L
23
=
I
⊗
Im
⁡
(
A
2
)
-
Im
⁡
(
E
2
)
T
⊗
Re
(
D
2
)
-
Re
(
E
2
)
T
⊗
Im
⁡
(
D
2
)
,
L
24
=
I
⊗
Re
(
A
2
)
+
Re
(
E
2
)
T
⊗
Re
(
D
2
)
-
Im
⁡
(
E
2
)
T
⊗
Im
⁡
(
D
2
)
.
Particularly, when X
∈
C
S
C
m
×
m, X can be substituted with (
X
+
S
X
S
)
/
2 in (51); then we obtain
(59)
W
¯
·
(
vec
⁡
[
Re
(
X
)
]
vec
⁡
[
Im
⁡
(
X
)
]
vec
⁡
[
Re
(
Y
)
]
vec
⁡
[
Im
⁡
(
Y
)
]
)
=
(
vec
⁡
(
F
1
)
vec
⁡
(
F
2
)
)
,
where W
¯
∶
=
(
1
/
2
)
[
(
I
+
K
)
(
Re
(
W
)
T
,
(
I
+
K
)
(
Im
⁡
(
W
)
T
)
]
T, K
:
=
Diag
{
S
¯
,
S
¯
,
S
¯
,
S
¯
}, S
¯
∶
=
S
T
⊗
S
=
S
⊗
S
∈
R
m
2
×
m
2 (see Definition 1).Obviously, ifx
∈
R
(
Z
), then x
∈
R
(
(
1
/
2
)
Z
), where Z is a matrix. So, from the above analysis, we can get the result.Lemma 13.
LetW
¯
∈
C
4
m
2
×
4
m
2, b
∈
C
4
m
2, and the linear matrix equation W
¯
x
=
b has a solution x
*
∈
R
4
m
2, where x
=
(
vec
⁡
[
Re
(
X
)
]
T
,
vec
⁡
[
Re
(
Y
)
]
T
,
vec
⁡
[
Im
⁡
(
X
)
]
T
,
vec
⁡
[
Im
⁡
(
Y
)
]
T
)
T. If x
*
∈
R
[
(
I
+
K
)
(
Re
(
W
)
T
,
(
I
+
K
)
Im
⁡
(
W
)
T
)
], then x
* is the unique least Frobenius norm solution of W
¯
x
=
b in (59).Theorem 14.
Suppose the system (4) is consistent; if one chooses the initial matrix pair
(60)
X
1
=
A
1
H
U
1
+
B
2
H
V
1
-
D
1
T
U
1
¯
E
1
T
+
S
A
1
H
U
1
S
+
S
B
2
H
V
1
S
-
S
D
1
T
U
1
¯
E
1
T
S
,
(61)
Y
1
=
A
2
H
V
1
+
B
1
H
U
1
-
D
2
T
V
1
¯
E
2
T
+
S
A
2
H
V
1
S
+
S
B
1
H
U
1
S
-
S
D
2
T
V
1
¯
E
2
T
S
,
where U
1
,
V
1
∈
C
p
×
m are two arbitrary matrices, especially, taking X
1
=
Y
1
=
0
∈
R
m
×
m, the solution (
X
*
,
Y
*
) given by Algorithm 3 is the unique least Frobenius norm solution of (4).Proof.
IfX
1
,
Y
1 has the form of (60), (61), respectively, by Step
2 of Algorithm 3, we have
(62)
X
2
=
X
1
+
α
1
M
1
=
X
1
+
α
1
(
1
2
(
R
~
1
(
1
)
+
S
R
~
1
(
1
)
S
)
)
=
A
1
H
U
1
+
B
2
H
V
1
-
D
1
T
U
1
¯
E
1
T
+
S
A
1
H
U
1
S
+
S
B
2
H
V
1
S
-
S
D
1
T
U
1
¯
E
1
T
S
+
α
1
[
1
2
(
A
1
H
R
1
(
1
)
+
B
2
H
R
1
(
2
)
-
D
1
T
R
1
(
1
)
¯
E
1
T
+
S
A
1
H
R
1
(
1
)
S
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
+
S
B
2
H
R
1
(
2
)
S
-
S
D
1
T
R
1
(
1
)
¯
E
1
T
S
)
1
2
]
=
A
1
H
(
U
1
+
1
2
α
1
R
1
(
1
)
)
+
B
2
H
(
V
1
+
1
2
α
1
R
1
(
2
)
)
-
D
1
T
(
U
1
+
1
2
α
1
R
1
(
1
)
¯
)
E
1
T
+
S
A
1
H
(
U
1
+
1
2
α
1
R
1
(
1
)
)
S
+
S
B
2
H
(
V
1
+
1
2
α
1
R
1
(
2
)
)
S
-
S
D
1
T
(
U
1
+
1
2
α
1
R
1
(
1
)
¯
)
E
1
T
S
∶
=
A
1
H
U
2
+
B
2
H
V
2
-
D
1
T
U
2
¯
E
1
T
+
S
A
1
H
U
2
S
+
S
B
2
H
V
2
S
-
S
D
1
T
U
2
¯
E
1
T
S
,
Y
2
=
Y
1
+
α
1
N
1
=
Y
1
+
α
1
(
1
2
(
R
~
1
(
2
)
+
S
R
~
1
(
2
)
S
)
)
=
A
2
H
V
1
+
B
1
H
U
1
-
D
2
T
V
1
¯
E
2
T
+
S
A
2
H
V
1
S
+
S
B
1
H
U
1
S
-
S
D
2
T
V
1
¯
E
2
T
S
+
α
1
[
1
2
(
A
2
H
R
1
(
2
)
+
B
1
H
R
1
(
1
)
-
D
2
T
R
1
(
2
)
¯
E
2
T
+
S
A
2
H
R
1
(
2
)
S
+
S
B
1
H
R
1
(
1
)
S
-
S
D
2
T
R
1
(
2
)
¯
E
2
T
S
)
1
2
]
=
A
2
H
(
V
1
+
1
2
α
1
R
1
(
2
)
)
+
B
1
H
(
U
1
+
1
2
α
1
R
1
(
1
)
)
-
D
2
T
(
V
1
+
1
2
α
1
R
1
(
2
)
¯
)
E
2
T
+
S
A
2
H
(
V
1
+
1
2
α
1
R
1
(
2
)
)
S
+
S
B
1
H
(
U
1
+
1
2
α
1
R
1
(
1
)
)
S
-
S
D
2
T
(
V
1
+
1
2
α
1
R
1
(
2
)
¯
)
E
2
T
S
∶
=
A
2
H
V
2
+
B
1
H
U
2
-
D
2
T
V
2
¯
E
2
T
+
S
A
2
H
V
2
S
+
S
B
1
H
U
2
S
-
S
D
2
T
V
2
¯
E
2
T
S
,
where U
2
∶
=
U
1
+
(
1
/
2
)
α
1
R
1
(
1
) and V
2
∶
=
V
1
+
(
1
/
2
)
α
1
R
1
(
2
). Since
(63)
M
2
=
1
2
(
R
~
2
(
1
)
+
S
R
~
2
(
1
)
S
)
+
β
1
M
1
=
1
2
(
A
1
H
R
2
(
1
)
+
B
2
H
R
2
(
2
)
-
D
1
T
R
2
(
1
)
¯
E
1
T
+
S
A
1
H
R
2
(
1
)
S
+
S
B
2
H
R
2
(
2
)
S
-
S
D
1
T
R
2
(
1
)
¯
E
1
T
S
)
+
β
1
2
(
A
1
H
R
1
(
1
)
+
B
2
H
R
1
(
2
)
-
D
1
T
R
1
(
1
)
¯
E
1
T
+
S
A
1
H
R
1
(
1
)
S
+
S
B
2
H
R
1
(
2
)
S
-
S
D
1
T
R
1
(
1
)
¯
E
1
T
S
)
,
N
2
=
1
2
(
R
~
2
(
2
)
+
S
R
~
2
(
2
)
S
)
+
β
1
N
1
=
1
2
(
A
1
H
R
2
(
2
)
+
B
2
H
R
2
(
1
)
-
D
1
T
R
2
(
2
)
¯
E
1
T
+
S
A
1
H
R
2
(
2
)
S
+
S
B
2
H
R
2
(
1
)
S
-
S
D
1
T
R
2
(
2
)
¯
E
1
T
S
)
+
β
1
2
(
A
1
H
R
1
(
2
)
+
B
2
H
R
1
(
1
)
-
D
1
T
R
1
(
2
)
¯
E
1
T
+
S
A
1
H
R
1
(
2
)
S
i
i
i
i
i
i
i
i
i
i
i
+
S
B
2
H
R
1
(
1
)
S
-
S
D
1
T
R
1
(
2
)
¯
E
1
T
S
)
,
we have
(64)
X
3
=
X
2
+
α
2
M
2
=
A
1
H
U
3
+
B
2
H
V
3
-
D
1
T
U
3
¯
E
1
T
+
S
A
1
H
U
3
S
+
S
B
2
H
V
3
S
-
S
D
1
T
U
3
¯
E
1
T
S
,
Y
3
=
Y
2
+
α
2
N
2
=
A
2
H
V
3
+
B
1
H
U
3
-
D
2
T
V
3
¯
E
2
T
+
S
A
2
H
V
3
S
+
S
B
1
H
U
3
S
-
S
D
2
T
V
3
¯
E
2
T
S
,
where U
3
∶
=
U
2
+
(
α
2
/
2
)
R
2
(
1
)
+
(
β
1
α
2
/
2
)
R
1
(
1
) and V
3
∶
=
V
2
+
(
α
2
/
2
)
R
2
(
2
)
+
(
β
1
α
2
/
2
)
R
1
(
2
). By parity of reasoning, we can prove that
(65)
X
k
+
1
=
A
1
H
U
k
+
1
+
B
2
H
V
k
+
1
-
D
1
T
U
k
+
1
¯
E
1
T
+
S
A
1
H
U
k
+
1
S
+
S
B
2
H
V
k
+
1
S
-
S
D
1
T
U
k
+
1
¯
E
1
T
S
,
Y
k
+
1
=
A
2
H
V
k
+
1
+
B
1
H
U
k
+
1
-
D
2
T
V
k
+
1
¯
E
2
T
+
S
A
2
H
V
k
+
1
S
+
S
B
1
H
U
k
+
1
S
-
S
D
2
T
V
k
+
1
¯
E
2
T
S
,
where U
k
+
1
,
V
k
+
1
∈
C
p
×
m. This together with Theorem 10 yields that
(66)
X
k
+
1
⟶
X
*
=
A
1
H
U
^
+
B
2
H
V
^
-
D
1
T
U
^
¯
E
1
T
+
S
A
1
H
U
^
S
+
S
B
2
H
V
^
S
-
S
D
1
T
U
^
¯
E
1
T
S
(
k
⟶
∞
)
,
Y
k
+
1
⟶
Y
*
=
A
2
H
V
^
+
B
1
H
U
^
-
D
2
T
V
^
¯
E
2
T
+
S
A
2
H
V
^
S
+
S
B
1
H
U
^
S
-
S
D
2
T
V
^
¯
E
2
T
S
(
k
⟶
∞
)
,
where U
^
,
V
^
∈
R
p
×
m. Since
(67)
X
*
=
A
1
H
U
^
+
B
2
H
V
^
-
D
1
T
U
^
¯
E
1
T
+
S
A
1
H
U
^
S
+
S
B
2
H
V
^
S
-
S
D
1
T
U
^
¯
E
1
T
S
=
[
Re
(
A
1
)
+
i
Im
⁡
(
A
1
)
]
H
[
Re
(
U
^
)
+
i
Im
⁡
(
U
^
)
]
+
[
Re
(
B
2
)
+
i
Im
⁡
(
B
2
)
]
H
[
Re
(
V
^
)
+
i
Im
⁡
(
V
^
)
]
-
[
Re
(
D
1
)
+
i
Im
⁡
(
D
1
)
]
T
[
Re
(
U
^
)
+
i
Im
⁡
(
U
^
)
¯
]
×
[
Re
(
E
1
)
+
i
Im
⁡
(
E
1
)
]
T
+
S
[
Re
(
A
1
)
+
i
Im
⁡
(
A
1
)
]
H
×
[
Re
(
U
^
)
+
i
Im
⁡
(
U
^
)
]
S
+
S
[
Re
(
B
2
)
+
i
Im
⁡
(
B
2
)
]
H
×
[
Re
(
V
^
)
+
i
Im
⁡
(
V
^
)
]
S
-
S
[
Re
(
D
1
)
+
i
Im
⁡
(
D
1
)
]
T
×
[
Re
(
U
^
)
+
i
Im
⁡
(
U
^
)
¯
]
[
Re
(
E
1
)
+
i
Im
⁡
(
E
1
)
]
T
S
=
[
Re
(
A
1
)
T
-
i
Im
⁡
(
A
1
)
T
]
[
Re
(
U
^
)
+
i
Im
⁡
(
U
^
)
]
+
[
Re
(
B
2
)
T
-
i
Im
⁡
(
B
2
)
T
]
[
Re
(
V
^
)
+
i
Im
⁡
(
V
^
)
]
-
[
Re
(
D
1
)
T
+
i
Im
⁡
(
D
1
)
T
]
[
Re
(
U
^
)
-
i
Im
⁡
(
U
^
)
]

×
[
Re
(
E
1
)
T
+
i
Im
⁡
(
E
1
)
T
]

+
S
[
Re
(
A
1
)
T
-
i
Im
⁡
(
A
1
)
T
]
×
[
Re
(
U
^
)
+
i
Im
⁡
(
U
^
)
]
S
+
S
[
Re
(
B
2
)
T
-
i
Im
⁡
(
B
2
)
T
]
×
[
Re
(
V
^
)
+
i
Im
⁡
(
V
^
)
]
S
-
S
[
Re
(
D
1
)
T
+
i
Im
⁡
(
D
1
)
T
]
×
[
Re
(
U
^
)
-
i
Im
⁡
(
U
^
)
]
[
Re
(
E
1
)
T
+
i
Im
⁡
(
E
1
)
T
]
S
=
Re
(
A
1
)
T
Re
(
U
^
)
+
i
Re
(
A
1
)
T
Im
⁡
(
U
^
)
-
i
Im
⁡
(
A
1
)
T
Re
(
U
^
)
+
Im
⁡
(
A
1
)
T
Im
⁡
(
U
^
)
+
Re
(
B
2
)
T
Re
(
V
^
)
+
i
Re
(
B
2
)
T
Im
⁡
(
V
^
)
-
i
Im
⁡
(
B
2
)
T
Re
(
V
^
)
+
Im
⁡
(
B
2
)
T
Im
⁡
(
V
^
)
-
Re
(
D
1
)
T
Re
(
U
^
)
Re
(
E
1
)
T
-
i
Re
(
D
1
)
T
Re
(
U
^
)
Im
⁡
(
E
1
)
T
+
i
Re
(
D
1
)
T
Im
⁡
(
U
^
)
Re
(
E
1
)
T
-
Re
(
D
1
)
T
Im
⁡
(
U
^
)
Im
⁡
(
E
1
)
T
-
i
Im
⁡
(
D
1
)
T
Re
(
U
^
)
Re
(
E
1
)
T
+
Im
⁡
(
D
1
)
T
Re
(
U
^
)
Im
⁡
(
E
1
)
T
-
Im
⁡
(
D
1
)
T
Im
⁡
(
U
^
)
Re
(
E
1
)
T
-
i
Im
⁡
(
D
1
)
T
Im
⁡
(
U
^
)
Im
⁡
(
E
1
)
T
+
S
Re
(
A
1
)
T
Re
(
U
^
)
S
+
i
S
Re
(
A
1
)
T
Im
⁡
(
U
^
)
S
-
i
S
Im
⁡
(
A
1
)
T
Re
(
U
^
)
S
+
S
Im
⁡
(
A
1
)
T
Im
⁡
(
U
^
)
S
+
S
Re
(
B
2
)
T
Re
(
V
^
)
S
+
i
S
Re
(
B
2
)
T
Im
⁡
(
V
^
)
S
-
i
S
Im
⁡
(
B
2
)
T
Re
(
V
^
)
S
+
S
Im
⁡
(
B
2
)
T
Im
⁡
(
V
^
)
S
-
S
Re
(
D
1
)
T
Re
(
U
^
)
Re
(
E
1
)
T
S
-
i
S
Re
(
D
1
)
T
Re
(
U
^
)
Im
⁡
(
E
1
)
T
S
+
i
S
Re
(
D
1
)
T
Im
⁡
(
U
^
)
Re
(
E
1
)
T
S
-
S
Re
(
D
1
)
T
Im
⁡
(
U
^
)
Im
⁡
(
E
1
)
T
S
-
i
S
Im
⁡
(
D
1
)
T
Re
(
U
^
)
Re
(
E
1
)
T
S
+
S
Im
⁡
(
D
1
)
T
Re
(
U
^
)
Im
⁡
(
E
1
)
T
S
-
S
Im
⁡
(
D
1
)
T
Im
⁡
(
U
^
)
Re
(
E
1
)
T
S
-
i
S
Im
⁡
(
D
1
)
T
Im
⁡
(
U
^
)
Im
⁡
(
E
1
)
T
S
,
we have
(68)
Re
(
X
*
)
=
Re
(
A
1
)
T
Re
(
U
^
)
+
Im
⁡
(
A
1
)
T
Im
⁡
(
U
^
)
+
Re
(
B
2
)
T
Re
(
V
^
)
+
Im
⁡
(
B
2
)
T
Im
⁡
(
V
^
)
-
Re
(
D
1
)
T
Re
(
U
^
)
Re
(
E
1
)
T
-
Re
(
D
1
)
T
Im
⁡
(
U
^
)
Im
⁡
(
E
1
)
T
+
Im
⁡
(
D
1
)
T
Re
(
U
^
)
Im
⁡
(
E
1
)
T
-
Im
⁡
(
D
1
)
T
Im
⁡
(
U
^
)
Re
(
E
1
)
T
+
S
Re
(
A
1
)
T
Re
(
U
^
)
S
+
S
Im
⁡
(
A
1
)
T
Im
⁡
(
U
^
)
S
+
S
Re
(
B
2
)
T
Re
(
V
^
)
S
+
S
Im
⁡
(
B
2
)
T
Im
⁡
(
V
^
)
S
-
S
Re
(
D
1
)
T
Re
(
U
^
)
Re
(
E
1
)
T
S
-
S
Re
(
D
1
)
T
Im
⁡
(
U
^
)
Im
⁡
(
E
1
)
T
S
+
S
Im
⁡
(
D
1
)
T
Re
(
U
^
)
Im
⁡
(
E
1
)
T
S
-
S
Im
⁡
(
D
1
)
T
Im
⁡
(
U
^
)
Re
(
E
1
)
T
S
,
Im
⁡
(
X
*
)
=
Re
(
A
1
)
T
Im
⁡
(
U
^
)
-
Im
⁡
(
A
1
)
T
Re
(
U
^
)
+
Re
(
B
2
)
T
Im
⁡
(
V
^
)
-
Im
⁡
(
B
2
)
T
Re
(
V
^
)
-
Re
(
D
1
)
T
Re
(
U
^
)
Im
⁡
(
E
1
)
T
+
Re
(
D
1
)
T
Im
⁡
(
U
^
)
Re
(
E
1
)
T
-
Im
⁡
(
D
1
)
T
Re
(
U
^
)
Re
(
E
1
)
T
-
Im
⁡
(
D
1
)
T
Im
⁡
(
U
^
)
Im
⁡
(
E
1
)
T
+
S
Re
(
A
1
)
T
Im
⁡
(
U
^
)
S
-
S
Im
⁡
(
A
1
)
T
Re
(
U
^
)
S
+
S
Re
(
B
2
)
T
Im
⁡
(
V
^
)
S
-
S
Im
⁡
(
B
2
)
T
Re
(
V
^
)
S
-
S
Re
(
D
1
)
T
Re
(
U
^
)
Im
⁡
(
E
1
)
T
S
+
S
Re
(
D
1
)
T
Im
⁡
(
U
^
)
Re
(
E
1
)
T
S
-
S
Im
⁡
(
D
1
)
T
Re
(
U
^
)
Re
(
E
1
)
T
S
-
S
Im
⁡
(
D
1
)
T
Im
⁡
(
U
^
)
Im
⁡
(
E
1
)
T
S
.
Let γ
=
(
vec
⁡
[
Re
(
U
^
)
]
T, vec
⁡
[
Re
(
V
^
)
]
T, vec
⁡
[
Im
⁡
(
U
^
)
]
T, vec
⁡
[
Im
⁡
(
V
^
)
]
T
)
T and K in (59); using Kronecker product, we get
(69)
vec
⁡
[
Re
(
X
*
)
]
=
(
I
+
K
)
×
(
I
⊗
Re
(
A
1
)
-
Re
(
E
1
)
T
⊗
Re
(
D
1
)
+
Im
⁡
(
E
1
)
T
⊗
Im
⁡
(
D
1
)
I
⊗
Re
(
B
2
)
I
⊗
Im
⁡
(
A
1
)
-
Im
⁡
(
E
1
)
T
⊗
Re
(
D
1
)
-
Re
(
E
1
)
T
⊗
Im
⁡
(
D
1
)
I
⊗
Im
⁡
(
B
2
)
)
T
·
(
vec
⁡
[
Re
(
U
^
)
]
vec
⁡
[
Re
(
V
^
)
]
vec
⁡
[
Im
⁡
(
U
^
)
]
vec
⁡
[
Im
⁡
(
V
^
)
]
)
=
(
I
+
K
)
·
(
N
11
T
,
N
21
T
,
L
11
T
,
L
21
T
)
γ
,
vec
⁡
[
Im
⁡
(
X
*
)
]
=
(
I
+
K
)
×
(
-
I
⊗
Im
⁡
(
A
1
)
-
Im
⁡
(
E
1
)
T
⊗
Re
(
D
1
)
-
Re
(
E
1
)
T
⊗
Im
⁡
(
D
1
)
-
I
⊗
Im
⁡
(
B
2
)
I
⊗
Re
(
A
1
)
+
Re
(
E
1
)
T
⊗
Re
(
D
1
)
-
Im
⁡
(
E
1
)
T
⊗
Im
⁡
(
D
1
)
I
⊗
Re
(
B
2
)
)
T
·
(
vec
⁡
[
Re
(
U
^
)
]
vec
⁡
[
Re
(
V
^
)
]
vec
⁡
[
Im
⁡
(
U
^
)
]
vec
⁡
[
Im
⁡
(
V
^
)
]
)
=
(
I
+
K
)
·
(
N
12
T
,
N
22
T
,
L
12
T
,
L
22
T
)
γ
.
In the same way, we can prove that
(70)
vec
⁡
[
Re
(
Y
*
)
]
=
(
I
+
K
)
×
(
I
⊗
Re
(
B
1
)
I
⊗
Re
(
A
2
)
-
Re
(
E
2
)
T
⊗
Re
(
D
2
)
+
Im
⁡
(
E
2
)
T
⊗
Im
⁡
(
D
2
)
I
⊗
Im
⁡
(
B
1
)
I
⊗
Im
⁡
(
A
2
)
-
Im
⁡
(
E
2
)
T
⊗
Re
(
D
2
)
-
Re
(
E
2
)
T
⊗
Im
⁡
(
D
2
)
)
T
·
(
vec
⁡
[
Re
(
U
^
)
]
vec
⁡
[
Re
(
V
^
)
]
vec
⁡
[
Im
⁡
(
U
^
)
]
vec
⁡
[
Im
⁡
(
V
^
)
]
)
=
(
I
+
K
)
·
(
N
13
T
,
N
23
T
,
L
13
T
,
L
23
T
)
γ
,
vec
⁡
[
Im
⁡
(
Y
*
)
]
=
(
I
+
K
)
×
(
-
I
⊗
Im
⁡
(
B
1
)
-
I
⊗
Im
⁡
(
A
2
)
-
Im
⁡
(
E
2
)
T
⊗
Re
(
D
2
)
-
Re
(
E
2
)
T
⊗
Im
⁡
(
D
2
)
I
⊗
Re
(
B
1
)
I
⊗
Re
(
A
2
)
+
Re
(
E
2
)
T
⊗
Re
(
D
2
)
-
Im
⁡
(
E
2
)
T
⊗
Im
⁡
(
D
2
)
)
T
·
(
vec
⁡
[
Re
(
U
^
)
]
vec
⁡
[
Re
(
V
^
)
]
vec
⁡
[
Im
⁡
(
U
^
)
]
vec
⁡
[
Im
⁡
(
V
^
)
]
)
=
(
I
+
K
)
·
(
N
14
T
,
N
24
T
,
L
14
T
,
L
24
T
)
γ
.
This shows that
(71)
(
vec
⁡
[
Re
(
X
*
)
]
vec
⁡
[
Im
⁡
(
X
*
)
]
vec
⁡
[
Re
(
Y
*
)
]
vec
⁡
[
Im
⁡
(
Y
*
)
]
)
=
(
I
+
K
)
(
N
11
N
12
N
13
N
14
N
21
N
22
N
23
N
24
L
11
L
12
L
13
L
14
L
21
L
22
L
23
L
24
)
T
γ
=
(
I
+
K
)
(
Re
(
W
)
Im
⁡
(
W
)
)
T
γ
=
(
Re
(
W
)
(
I
+
K
)
Im
⁡
(
W
)
(
I
+
K
)
)
T
γ
.
Notice that X
*
,
Y
*
∈
C
M
C
m
×
m and K
T
=
K in (59). By (56) and (57), we have
(72)
(
vec
⁡
[
Re
(
X
*
)
]
vec
⁡
[
Im
⁡
(
X
*
)
]
vec
⁡
[
Re
(
Y
*
)
]
vec
⁡
[
Im
⁡
(
Y
*
)
]
)
∈
R
[
(
I
+
K
)
Re
(
W
)
T
,
(
I
+
K
)
Im
⁡
(
W
)
T
]
.
It follows from Lemma 11, Lemma 12, and (59) that
(73)
∥
vec
⁡
[
Re
(
X
*
)
]
∥
2
+
∥
vec
⁡
[
Im
⁡
(
X
*
)
]
∥
2
+
∥
vec
⁡
[
Re
(
Y
*
)
]
∥
2
+
∥
vec
⁡
[
Im
⁡
(
Y
*
)
]
∥
2
≤
∥
ξ
¯
∥
2
,
where
(74)
ξ
¯
=
(
vec
⁡
[
Re
(
X
¯
)
]
vec
⁡
[
Im
⁡
(
X
¯
)
]
vec
⁡
[
Re
(
Y
¯
)
]
vec
⁡
[
Im
⁡
(
Y
¯
)
]
)
∈
R
4
m
2
for all solutions (
X
¯
,
Y
¯
) of (4). Since
(75)
∥
vec
⁡
[
Re
(
X
*
)
]
∥
2
+
∥
vec
⁡
[
Im
⁡
(
X
*
)
]
∥
2
=
∥
Re
(
X
*
)
∥
F
2
+
∥
Im
⁡
(
X
*
)
∥
F
2
=
∥
X
*
∥
2
,
∥
vec
⁡
[
Re
(
Y
*
)
]
∥
2
+
∥
vec
⁡
[
Im
⁡
(
Y
*
)
]
∥
2
=
∥
Re
(
Y
*
)
∥
F
2
+
∥
Im
⁡
(
Y
*
)
∥
F
2
=
∥
Y
*
∥
2
,
we obtain
(76)
∥
vec
⁡
[
Re
(
X
*
)
]
∥
2
+
∥
vec
⁡
[
Im
⁡
(
X
*
)
]
∥
2
+
∥
vec
⁡
[
Re
(
Y
*
)
]
∥
2
+
∥
vec
⁡
[
Im
⁡
(
Y
*
)
]
∥
2
=
∥
X
*
∥
2
+
∥
Y
*
∥
2
≤
∥
X
¯
∥
2
+
∥
Y
¯
∥
2
,
for all solutions (
X
¯
,
Y
¯
) of (4). Hence (
X
*
,
Y
*
) is the unique least Frobenius norm solution of (4).


## 3. Numerical Experiments

In this section, we report some numerical results to support our Algorithm3. The iterations have been carried out by MATLAB R2011b (7.13), Intel(R) Core(TM) i7-2670QM, CPU 2.20 GHZ, RAM 8.GB PC Environment.Example 1.
We consider (4) with the following matrices:(77)
A
1
=
(
1
-
i
1
+
i
2
-
2
i
-
3
-
i
4
+
i
3
+
2
i
4
-
3
i
2
-
i
2
+
i
1
-
2
i
0
+
i
4
+
i
7
-
i
2
-
i
4
+
2
i
-
1
+
i
-
1
-
i
-
1
+
i
2
-
i
4
+
i
4
+
i
4
-
i
3
+
i
2
-
i
1
+
2
i
)
,
A
2
=
(
2
-
i
2
+
i
3
-
i
1
+
i
1
+
i
0
-
2
i
5
-
i
4
+
i
-
2
+
i
-
2
+
i
2
+
i
3
-
i
4
+
i
1
+
i
1
+
i
2
-
i
0
+
i
2
-
2
i
0
-
i
1
+
i
-
3
-
i
-
3
+
i
1
+
i
2
-
i
2
+
i
)
,
B
1
=
(
4
+
i
3
+
i
4
-
i
4
-
i
1
+
i
-
2
-
i
-
2
+
i
3
+
i
4
-
3
i
4
-
i
5
+
i
6
-
i
5
+
i
0
+
i
1
-
i
5
-
i
4
-
i
5
+
i
3
-
i
3
-
3
i
1
-
2
i
2
-
2
i
0
+
i
0
-
i
1
-
i
)
,
B
2
=
(
-
2
-
i
-
i
-
2
+
i
3
+
i
2
-
i
6
-
i
5
-
i
4
+
i
4
-
i
3
-
i
2
-
i
3
+
i
2
-
i
1
+
i
1
-
i
1
+
3
i
1
-
2
i
2
-
i
4
+
i
1
-
3
i
0
-
i
0
+
i
2
-
i
3
-
i
2
+
i
)
,
D
1
=
(
1
-
i
2
+
i
3
-
i
1
-
i
2
-
i
0
+
i
1
+
i
2
-
i
3
+
i
1
-
3
i
4
-
i
4
-
i
2
-
i
1
+
i
3
+
i
1
-
i
0
-
i
0
-
i
-
1
-
i
i
2
-
i
4
+
i
5
+
i
3
+
i
2
-
i
)
,
D
2
=
(
1
+
i
2
-
i
3
+
i
-
1
+
i
-
1
+
i
3
-
i
0
-
i
0
+
i
3
-
i
3
-
i
1
-
i
2
+
i
3
-
3
i
0
+
2
i
-
2
-
3
i
-
1
+
i
-
1
-
i
2
+
2
i
2
-
2
i
3
+
i
5
-
i
4
-
i
5
+
i
4
+
i
4
-
4
i
)
,
E
1
=
(
1
-
i
2
-
i
1
-
i
2
+
i
1
-
i
3
+
i
3
-
i
3
+
i
1
-
i
2
+
i
1
-
i
2
-
i
3
-
i
-
4
-
i
-
4
+
i
5
-
i
5
-
i
5
+
i
4
+
i
4
-
i
2
-
i
-
2
+
i
-
2
+
i
-
2
+
i
1
-
i
)
,
E
2
=
(
2
+
i
4
+
i
3
+
i
2
-
i
1
-
i
0
+
i
-
3
+
i
-
3
+
i
-
3
-
i
2
-
i
1
+
i
1
+
i
1
-
i
0
-
i
0
-
i
1
-
i
1
+
i
2
+
i
3
-
i
4
-
i
2
-
i
3
+
i
4
+
i
5
-
i
2
+
i
)
,
F
1
=
1
0
2
·
(
-
1.2100
+
1.1500
i
-
1.1100
+
0.8000
i
-
1.5500
+
0.2300
i
-
0.2800
+
0.2000
i
-
0.4300
+
0.9500
i
-
0.9400
+
1.5300
i
-
1.1200
+
0.6000
i
-
1.5900
-
0.1400
i
-
0.8200
+
0.4400
i
-
0.3700
+
1.0300
i
-
2.5200
+
1.5700
i
-
2.1900
+
1.4200
i
-
2.6600
-
0.1500
i
-
0.4200
+
0.2800
i
-
0.5900
+
1.3200
i
0.5100
+
0.1500
i
0.6800
+
0.2500
i
0.3100
+
0.3300
i
0.6700
-
0.1500
i
0.4300
-
0.0500
i
-
2.5700
+
1.7800
i
-
2.2700
+
1.1300
i
-
2.6400
-
0.5200
i
-
1.0900
+
0.5600
i
-
0.8500
+
1.6700
i
)
,
F
2
=
1
0
2
·
(
-
0.0900
-
0.0500
i
-
0.4400
-
0.2700
i
-
0.6500
-
0.2700
i
-
1.1700
+
0.1100
i
-
0.8100
-
0.3600
i
-
0.6700
-
0.0200
i
-
0.0300
-
0.1700
i
-
0.1200
-
0.2400
i
0.2800
+
0.7800
i
-
0.7800
+
0.9300
i
-
0.1700
+
0.7100
i
-
0.4400
+
0.2500
i
-
0.6800
+
0.6800
i
0.0100
+
1.3100
i
-
0.7000
+
1.0900
i
-
0.3000
-
0.2400
i
0.3100
-
0.1300
i
0.2300
-
0.5800
i
-
0.0600
-
0.2400
i
-
0.0700
-
0.3600
i
-
1.5800
+
0.2400
i
-
1.4700
-
0.8400
i
-
2.0700
-
0.2300
i
-
2.2000
+
2.6500
i
-
3.6200
+
0.9700
i
)
.We can show that the matrix equation (4) is consistent. Choose the initial matrix pair X
1
=
0, Y
1
=
0; by Algorithm 3, we obtain its exact solution:(78)
X
116
=
(
2.0000
+
1.0000
i
3.0000
-
1.0000
i
1.0000
+
1.0000
i
3.0000
+
1.0000
i
1.0000
-
1.0000
i
-
1.0000
+
2.0000
i
0.0000
+
1.0000
i
2.0000
-
1.0000
i
3.0000
+
2.0000
i
1.0000
+
2.0000
i
1.0000
+
1.0000
i
1.0000
-
1.0000
i
-
0.0000
-
1.0000
i
1.0000
-
1.0000
i
1.0000
+
1.0000
i
1.0000
+
2.0000
i
3.0000
+
2.0000
i
2.0000
-
1.0000
i
0.0000
+
1.0000
i
-
1.0000
+
2.0000
i
1.0000
-
1.0000
i
3.0000
+
1.0000
i
1.0000
+
1.0000
i
3.0000
-
1.0000
i
2.0000
+
1.0000
i
)
,
Y
116
=
(
2.0000
-
1.0000
i
1.0000
-
1.0000
i
1.0000
-
1.0000
i
3.0000
-
1.0000
i
1.0000
+
1.0000
i
-
1.0000
-
2.0000
i
0.0000
+
1.0000
i
2.0000
+
1.0000
i
3.0000
-
2.0000
i
2.0000
+
2.0000
i
2.0000
+
1.0000
i
1.0000
-
1.0000
i
0.0000
-
1.0000
i
1.0000
-
1.0000
i
2.0000
+
1.0000
i
2.0000
+
2.0000
i
3.0000
-
2.0000
i
2.0000
+
1.0000
i
0.0000
+
1.0000
i
-
1.0000
-
2.0000
i
1.0000
+
1.0000
i
3.0000
-
1.0000
i
1.0000
-
1.0000
i
1.0000
-
1.0000
i
2.0000
-
1.0000
i
)
.
The corresponding residual and least Frobenius norm are(79)
∥
R
116
∥
F
=
4.6203
×
1
0
-
11
,
∥
X
116
∥
F
=
9.1726
,
∥
Y
116
∥
F
=
7.5890
.
The results above are presented in Figure 1, where r
k
=
∥
R
k
∥, δ
k
=
(
∥
X
k
-
X
*
∥
+
∥
Y
k
-
Y
*
∥
)
/
(
∥
X
*
∥
+
∥
Y
*
∥
).Figure 1
The relative error of solution and the residual for Example1.Example 2.
Consider (4) with the following matrices:(80)
A
1
=
(
5
+
6
i
3
+
i
3
-
5
i
3
-
6
i
2
-
6
i
7
+
3
i
0
+
6
i
1
+
7
i
3
+
6
i
)
,
A
2
=
(
5
-
6
*
i
2
-
3
i
3
+
2
i
6
+
2
*
i
3
-
i
3
+
5
i
3
-
6
*
i
2
+
3
i
5
+
4
i
0
+
3
*
i
3
-
4
i
3
+
2
i
)
,
B
1
=
(
3
+
2
i
2
i
4
+
2
i
5
-
3
i
3
+
i
2
+
5
i
3
+
6
i
2
-
6
i
6
+
3
i
0
+
6
i
1
+
7
i
3
+
6
i
)
,
B
2
=
(
3
+
2
i
2
+
3
i
4
+
2
i
3
-
3
i
3
-
1
i
4
+
5
i
3
+
2
i
2
-
6
i
7
+
3
i
0
-
6
i
1
+
7
i
3
+
6
i
)
,
D
1
=
(
5
+
6
i
2
+
3
i
6
+
2
i
2
+
3
i
3
+
1
i
3
+
5
i
3
-
2
i
2
-
6
i
5
-
3
i
0
+
6
i
1
+
i
3
+
6
i
)
,
D
2
=
(
5
+
3
i
2
+
3
i
6
+
2
i
5
+
3
i
3
-
1
i
3
-
5
i
3
+
4
i
2
-
3
i
4
+
3
i
0
-
6
i
1
+
6
i
3
+
6
i
)
,
E
1
=
(
5
+
6
i
2
+
3
i
6
+
1
i
5
-
3
i
3
-
1
i
2
-
5
i
2
+
6
i
2
-
5
i
8
-
3
i
)
,
E
2
=
(
8
-
6
i
1
+
3
i
4
+
1
i
5
-
3
i
4
-
1
i
2
+
2
i
2
+
8
i
1
+
3
i
9
-
2
i
)
,
F
1
=
1
0
2
·
(
0.4500
-
5.6000
i
-
1.0300
-
1.6100
i
-
4.4300
-
2.9100
i
0.9100
-
3.3300
i
-
0.3400
-
0.3800
i
-
2.2400
-
1.8000
i
-
3.6300
+
0.4600
i
0.5000
+
0.8100
i
-
0.6500
+
4.2400
i
2.8300
-
2.5900
i
0.4400
-
0.6200
i
-
1.2100
-
2.8100
i
)
,
F
2
=
1
0
2
·
(
-
4.8800
-
3.5100
i
-
0.3100
-
4.0900
i
-
4.4000
-
3.8400
i
-
2.5500
+
2.3800
i
-
1.6500
-
1.8100
i
-
4.2900
-
0.3900
i
-
3.0200
-
2.8300
i
0.5300
-
2.9700
i
-
3.5600
-
3.7000
i
-
3.4400
-
3.8200
i
0.0100
-
1.3000
i
-
0.2200
+
0.2600
i
)
.We can show that the matrix equation (4) is also consistent. Choose the initial matrix pair X
1
=
0, Y
1
=
0; from Algorithm 3, we obtain its exact solution only the 29th iterative step:
(81)
X
29
=
(
1.0000
+
1.0000
i
-
0.0000
-
2.0000
i
2.0000
-
1.0000
i
3.0000
+
1.0000
i
2.0000
+
2.0000
i
3.0000
+
1.0000
i
2.0000
-
1.0000
i
-
0.0000
-
2.0000
i
1.0000
+
1.0000
i
)
,
Y
29
=
(
2.0000
+
1.0000
i
2.0000
-
1.0000
i
5.0000
-
1.0000
i
1.0000
-
1.0000
i
3.0000
-
1.0000
i
1.0000
-
1.0000
i
5.0000
-
1.0000
i
2.0000
-
1.0000
i
2.0000
+
1.0000
i
)
.
The corresponding residual and least Frobenius norm are
(82)
∥
R
29
∥
F
=
9.9651
×
1
0
-
13
,
∥
X
29
∥
F
=
5.9876
,
∥
Y
29
∥
F
=
8.2249
.
The above results are presented in Figure 2, where r
k
=
∥
R
k
∥, δ
k
=
(
∥
X
k
-
X
*
∥
+
∥
Y
k
-
Y
*
∥
)
/
(
∥
X
*
∥
+
∥
Y
*
∥
).Figure 2
The relative error of solution and the residual for Example2.


## 4. Conclusion

Iterative method is proposed to solve the generalized coupled Sylvester-conjugate linear matrix equationsA
1
X
+
B
1
Y
=
D
1
X
¯
E
1
+
F
1, A
2
Y
+
B
2
X
=
D
2
Y
¯
E
2
+
F
2 for center-symmetry (center-antisymmetry) matrix pair (
X
,
Y
). When (4) is consistent, utilizing the Kronecker product, it has been revealed that an exact solution can be obtained by the proposed algorithm within finite iterative steps in the absence of round-off error for any initial value chosen center-symmetry (center-antisymmetry) matrix. Furthermore, we show that the least Frobenius norm solution is obtained by choosing a special kind of initial matrix. Finally, some numerical examples were given to show the efficiency for the proposed method.



---

*Source: 515816-2014-07-08.xml*