# Generalized Reflexive and Generalized Antireflexive Solutions to a System of Matrix Equations

Authors: Yong Lin; Qing-Wen Wang
Journal: Journal of Applied Mathematics
 (2014)

---

## Abstract

Two efficient iterative algorithms are presented to solve a system of matrix equationsA
1
X
1
B
1 + A
2
X
2
B
2
=
E, C
1
X
1
D
1 + C
2
X
2
D
2
=
F over generalized reflexive and generalized antireflexive matrices. By the algorithms, the least norm generalized reflexive (antireflexive) solutions and the unique optimal approximation generalized reflexive (antireflexive) solutions to the system can be obtained, too. For any initial value, it is proved that the iterative solutions obtained by the proposed algorithms converge to their true values. The given numerical examples demonstrate that the iterative algorithms are efficient.


---

## Body

## 1. Introduction

Throughout the paper, we denote the set of allm
×
n real matrices by R
m
×
n, the transpose matrix of A by A
T, the identity matrix of order n by I
n, the Kronecker product of A and B by A
⊗
B, the m
n
×
1 vector formed by the vertical concatenation of the respective columns of a matrix A
∈
R
m
×
n by vec
⁡
(
A
), the trace of a matrix A by tr
⁡
(
A
), and the Frobenius norm of a matrix A by ∥
A
∥, where ∥
A
∥
=
tr
⁡
(
A
T
A
). Let P
∈
R
m
×
m and Q
∈
R
n
×
n be two real generalized reflection matrices, that is; P
T
=
P, P
2
=
I
m and Q
T
=
Q, Q
2
=
I
n. A matrix A
∈
R
m
×
n is called generalized reflexive (antireflexive) matrix with respect to the matrix pair (
P
,
Q
) if A
=
P
A
Q
(
A
=
-
P
A
Q
). The set of all m
×
n real generalized reflexive (generalized antireflexive) matrices with respect to matrix pair (
P
,
Q
) is denoted by R
r
m
×
n
(
P
,
Q
) (R
a
m
×
n
(
P
,
Q
)). The generalized reflexive and generalized antireflexive matrices have been widely used in engineering, scientific computations, and various other fields.The linear matrix equation pairA
1
X
1
B
1
+
A
2
X
2
B
2
=
E, C
1
X
1
D
1
+
C
2
X
2
D
2
=
F is encountered in many systems and control applications. In this paper, we consider it with real generalized reflexive and generalized antireflexive constraint on the solutions. Let T
1
,
T
2
,
T
3, and T
4 be four real generalized reflection matrices; the following four problems are studied.Problem 1.
For given matricesA
1
∈
R
p
×
k, A
2
∈
R
p
×
m, B
1
∈
R
r
×
q, B
2
∈
R
n
×
q, C
1
∈
R
s
×
k, C
2
∈
R
s
×
m, D
1
∈
R
r
×
t, D
2
∈
R
n
×
t, E
∈
R
p
×
q, and F
∈
R
s
×
t, find X
1
∈
S
1
⊂
R
r
k
×
r
(
T
1
,
T
2
) and X
2
∈
S
2
⊂
R
r
m
×
n
(
T
3
,
T
4
) such that
(1)
A
1
X
1
B
1
+
A
2
X
2
B
2
=
E
,
C
1
X
1
D
1
+
C
2
X
2
D
2
=
F
.Problem 2.
When Problem1 is consistent, let S
r denote the set of its solutions. For the given matrices X
~
1
∈
R
k
×
r, X
~
2
∈
R
m
×
n, find {
X
^
1
,
X
^
2
}
∈
S
r such that
(2)
∥
X
^
1
-
X
~
1
∥
2
+
∥
X
^
2
-
X
~
2
∥
2
=
min
⁡
{
X
1
,
X
2
}
∈
S
r
(
∥
X
1
-
X
~
1
∥
2
+
∥
X
2
-
X
~
2
∥
2
)
.Problem 3.
For given matricesA
1
∈
R
p
×
k, A
2
∈
R
p
×
m, B
1
∈
R
r
×
q, B
2
∈
R
n
×
q, C
1
∈
R
s
×
k, C
2
∈
R
s
×
m, D
1
∈
R
r
×
t, D
2
∈
R
n
×
t, E
∈
R
p
×
q and F
∈
R
s
×
t, find X
1
∈
S
3
⊂
R
a
k
×
r
(
T
3
,
T
4
) and X
2
∈
S
4
⊂
R
a
m
×
n
(
T
3
,
T
4
) such that
(3)
A
1
X
1
B
1
+
A
2
X
2
B
2
=
E
,
C
1
X
1
D
1
+
C
2
X
2
D
2
=
F
.Problem 4.
When Problem3 is consistent, let S
a denote the set of its solutions. For the given matrices X
~
1
∈
R
k
×
r, X
~
2
∈
R
m
×
n, find {
X
^
1
,
X
^
2
}
∈
S
a such that
(4)
∥
X
^
1
-
X
~
1
∥
2
+
∥
X
^
2
-
X
~
2
∥
2
=
min
⁡
{
X
1
,
X
2
}
∈
S
a
(
∥
X
1
-
X
~
1
∥
2
+
∥
X
2
-
X
~
2
∥
2
)
.
Problem2 (4) is to find the optimal approximation generalized reflexive (antireflexive) solution to the given generalized reflexive (antireflexive) matrices X
~
1 and X
~
2 in the solution set of Problem 1 (3); it is occurs frequently in experiment design (see, e.g., [1]). In recent years, the matrix optimal approximation problem has been studied extensively (e.g., [2–12]).Research on solving matrix equation pair has been actively ongoing for the last 30 years or more. For instance, Mitra [13] gave conditions for the existence of a solution and a representation of the general common solution to A
X
B
=
E, C
X
D
=
F. Shinozaki and Sibuya [14] and van der Woude [15] discussed conditions for the existence of a common solution to A
X
B
=
E, C
X
D
=
F. Navarra et al. [5] derived sufficient and necessary conditions for the existence of a common solution to A
X
B
=
E, C
X
D
=
F. Yuan [11] obtained an analytical expression of the least-squares solutions of A
X
B
=
E, C
X
D
=
F by using the generalized singular value decomposition (GSVD) of matrices. Dehghan and Hajarian [32] presented some examples to show a motivation for studying the general coupled matrix equations ∑
j
=
1
l
‍
A
i
j
X
j
B
i
j
=
C
i, i
=
1,2
,
…
,
l, and they [17] constructed an iterative algorithm to solve the general coupled matrix equations ∑
j
=
1
p
‍
A
i
j
X
j
B
i
j
=
M
i, i
=
1,2
,
…
,
p. Wang [18, 19] gave the centrosymmetric solution to the system of quaternion matrix equations A
1
X
=
C
1, A
3
X
B
3
=
C
3. Wang [16] also solved a system of matrix equations over arbitrary regular rings with identity.Recently, some finite iterative algorithms have also been developed to solve matrix equations. Ding et al. [20–23] studied the iterative solutions of matrix equations A
X
B
=
F and A
i
X
B
i
=
F
i, generalized Sylvester matrix equations A
X
B
+
C
X
D
=
F and A
X
B
+
C
X
T
D
=
F. They presented gradient based and least-squares based iterative algorithms for the solution. Duan et al. [24–27] considered iterative method for some coupled linear matrix equations. Deng et al. [28] studied the consistent conditions and the general expressions about the Hermitian solutions of the matrix equations (
A
X
,
X
B
)
=
(
C
,
D
) and designed an iterative method for its Hermitian minimum norm solutions. Li and Wu [29] gave symmetric and skew-antisymmetric solutions to certain matrix equations A
1
X
=
C
1, X
B
3
=
C
3 over the real quaternion algebra H. For more studies on iterative algorithms on coupled matrix equations, we refer to [3, 8–10, 30–36]. Peng et al. [12] presented iterative methods to obtain the symmetric solutions of A
X
B
=
E, C
X
D
=
F. Sheng and Chen [7] presented a finite iterative method for solving A
X
B
=
E, C
X
D
=
F. Liao and Lei [37] presented an analytical expression of the least-squares solution and an algorithm for A
X
B
=
E, C
X
D
=
F with the minimum norm. Peng et al. [6] presented an algorithm for the least-squares reflexive solution. Dehghan and Hajarian [2] presented an iterative algorithm for solving a pair of matrix equations A
X
B
=
E, C
X
D
=
F over generalized matrices; they [31] also presented an algorithm for solving a pair of matrix equations A
X
B
+
C
Y
D
=
J, E
X
F
+
G
Y
H
=
K over generalized reflexive (antireflexive) matrices. Cai and Chen [38] presented an iterative algorithm for the least-squares bisymmetric solutions of the matrix equations A
X
B
=
E, C
X
D
=
F. Yin and Huang [39] presented an iterative algorithm to solve the least-squares generalized reflexive solutions of the matrix equations A
X
B
=
E, C
X
D
=
F. Lin and Wang [40] presented an iterative algorithm to solve a system of linear matrix equations A
1
X
1
B
1
+
A
2
X
2
B
2
=
E, C
1
X
1
D
1
+
C
2
X
2
D
2
=
F with real matrices X
1 and X
2.However, to our knowledge, there has been little information on finding the solutions to the Problems1–4 by iterative algorithm. In this paper, two efficient iterative algorithms are presented to solve the Problems 1–4. The suggested iterative algorithms automatically determine the solvability of equations pair (1) with the constraint. When the pair of equations is consistent, then, for any initial generalized reflexive (antireflexive) matrices X
1
1 and X
2
1, the solution can be obtained in the absence of round errors, and the least norm solution can be obtained by choosing a special kind of initial matrix. In addition, the unique optimal approximation solution pair X
^
1, X
^
2 to given matrix pair X
~
1, X
~
2 in Frobenius norm can be obtained by finding the least norm solution of a new pair of matrix equations A
1
X
¯
1
B
1
+
A
2
X
¯
2
B
2
=
E
¯, C
1
X
¯
1
D
1
+
C
2
X
¯
2
D
2
=
F
¯, where X
¯
i
=
X
i
-
X
~
i, (
i
=
1,2
), E
¯
=
E
-
A
1
X
~
1
B
1
+
A
2
X
~
2
B
2, F
¯
=
F
-
C
1
X
~
1
D
1
+
C
2
X
~
2
D
2. The given numerical examples demonstrate that our iterative algorithms are efficient. In particular, when the numbers of the parameter matrices A
1
,
A
2
,
B
1
,
B
2
,
C
1
,
C
2
,
D
1, and D
2 are large, our algorithms are efficient as well while the algorithm of [31] is not convergent. That is, our algorithms have merits of good numerical stability and ease to program.The rest of this paper is outlined as follows. In Section2, we first propose an efficient iterative algorithm for solving Problems 1 and 2; then we give some properties of this iterative algorithm. We show that the algorithm can obtain a solution group for any (special) initial generalized reflexive matrix group in the absence of round-off errors. In Section 3, Problems 3 and 4 are solved similarly. In Section 4 numerical examples are given to illustrate that our algorithms are quite efficient.


## 2. Iterative Algorithm for Solving Problems1 and 2

In this section, we present an iterative algorithm for solving Problems1 and 2.Algorithm 5.
(1) Input matricesA
1
∈
R
p
×
k, A
2
∈
R
p
×
m, B
1
∈
R
r
×
q, B
2
∈
R
n
×
q, C
1
∈
R
s
×
k, C
2
∈
R
s
×
m, D
1
∈
R
r
×
t, D
2
∈
R
n
×
t, E
∈
R
p
×
q, F
∈
R
s
×
t, X
1
1
∈
R
r
k
×
r, and X
2
1
∈
R
r
m
×
n (where X
1
1
,
X
2
1 are any initial generalized reflexive matrices).
(2) Calculate(5)
E
1
=
E
;
F
1
=
F
;
Q
1
1
=
A
1
T
E
1
B
1
T
+
C
1
T
F
1
D
1
T
;
Q
2
1
=
A
2
T
E
1
B
2
T
+
C
2
T
F
1
D
2
T
;
P
1
1
=
Q
1
1
+
T
1
Q
1
1
T
2
;
P
2
1
=
Q
2
1
+
T
3
Q
2
1
T
4
;
β
1
=
(
tr
⁡
[
(
E
1
)
T
(
A
1
P
1
1
B
1
+
A
2
P
2
1
B
2
)
]
x
x
k
+
tr
⁡
[
(
F
1
)
T
(
C
1
P
1
1
D
1
+
C
2
P
2
1
D
2
)
]
)
×
(
∥
A
1
P
1
1
B
1
+
A
2
P
2
1
B
2
∥
2
+
∥
C
1
P
1
1
D
1
+
C
2
P
2
1
D
2
∥
2
)
-
1
;
Δ
X
1
1
=
β
1
P
1
1
;
Δ
X
2
1
=
β
1
P
2
1
;
k
=
1
.
(3) IfΔ
X
=
diag
⁡
(
Δ
X
1
k
,
Δ
X
2
k
)
=
0 (k
=
1,2
,
…), then stop. Otherwise,
(6)
X
1
k
+
1
=
X
1
k
+
Δ
X
1
k
;
X
2
k
+
1
=
X
2
k
+
Δ
X
2
k
.
(4) Calculate(7)
E
k
+
1
=
E
k
-
(
A
1
Δ
X
1
k
B
1
+
A
2
Δ
X
2
k
B
2
)
;
F
k
+
1
=
F
k
-
(
C
1
Δ
X
1
k
D
1
+
C
2
Δ
X
2
k
D
2
)
;
Q
1
k
+
1
=
A
1
T
E
k
+
1
B
1
T
+
C
1
T
F
k
+
1
D
1
T
;
Q
2
k
+
1
=
A
2
T
E
k
+
1
B
2
T
+
C
2
T
F
k
+
1
D
2
T
;
P
1
k
+
1
=
Q
1
k
+
1
+
T
1
Q
1
k
+
1
T
2
;
P
2
k
+
1
=
Q
2
k
+
1
+
T
3
Q
2
k
+
1
T
4
;
β
k
+
1
=
(
tr
⁡
[
(
E
k
+
1
)
T
(
A
1
P
1
k
+
1
B
1
+
A
2
P
2
k
+
1
B
2
)
]
ccci
+
tr
⁡
[
(
F
k
+
1
)
T
(
C
1
P
1
k
+
1
D
1
+
C
2
P
2
k
+
1
D
2
)
]
)
×
(
∥
C
1
P
1
k
+
1
D
1
+
C
2
P
2
k
+
1
D
2
∥
2
∥
A
1
P
1
k
+
1
B
1
+
A
2
P
2
k
+
1
B
2
∥
2
c
c
c
c
c
c
+
∥
C
1
P
1
k
+
1
D
1
+
C
2
P
2
k
+
1
D
2
∥
2
)
-
1
;
Δ
X
1
k
+
1
=
β
k
+
1
P
1
k
+
1
;
Δ
X
2
k
+
1
=
β
k
+
1
P
2
k
+
1
;
k
=
k
+
1
.
Go to (3).Lemma 6.
In Algorithm5, the choice of β
k makes ∥
diag
⁡
(
E
k
+
1
,
F
k
+
1
)
∥ reach a minimum and diag
⁡
(
E
k
+
1
,
F
k
+
1
) and diag
⁡
(
A
1
Δ
X
1
k
B
1
+
A
2
Δ
X
2
k
B
2
,
C
1
Δ
X
1
k
D
1
+
C
2
Δ
X
2
k
D
2
) orthogonal to each other.Proof.
From Algorithm5, we have
(8)
∥
diag
⁡
(
E
k
+
1
,
F
k
+
1
)
∥
2
=
∥
diag
⁡
(
(
C
1
Δ
X
1
k
D
1
+
C
2
Δ
X
2
k
D
2
)
E
k
-
(
A
1
Δ
X
1
k
B
1
+
A
2
Δ
X
2
k
B
2
)
,
c
c
c
c
c
c
i
c
c
c
c
c
F
k
-
(
C
1
Δ
X
1
k
D
1
+
C
2
Δ
X
2
k
D
2
)
)
∥
2
=
∥
diag
⁡
(
E
k
-
(
A
1
β
k
P
1
k
B
1
+
A
2
β
k
P
2
k
B
2
)
,
c
c
c
c
c
c
c
i
c
c
i
F
k
-
(
C
1
β
k
P
1
k
D
1
+
C
2
β
k
P
2
k
D
2
)
)
∥
2
=
∥
E
k
-
(
A
1
β
k
P
1
k
B
1
+
A
2
β
k
P
2
k
B
2
)
∥
2
+
∥
F
k
-
(
C
1
β
k
P
1
k
D
1
+
C
2
β
k
P
2
k
D
2
)
∥
2
=
∥
E
k
∥
2
+
∥
F
k
∥
2
-
2
[
tr
⁡
(
(
E
k
)
T
(
A
1
P
1
k
B
1
+
A
2
P
2
k
B
2
)
)
cccccic
+
tr
⁡
(
(
F
k
)
T
(
C
1
P
1
k
D
1
+
C
2
P
2
k
D
2
)
)
]
β
k
+
[
∥
A
1
P
1
k
B
1
+
A
2
P
2
k
B
2
∥
2
ccccc
+
∥
C
1
P
1
k
D
1
+
C
2
P
2
k
D
2
∥
2
]
(
β
k
)
2
.
From the above, the condition of ∥
diag
⁡
(
E
k
+
1
,
F
k
+
1
)
∥ reaching a minimum is
(9)
β
k
=
(
tr
⁡
[
(
E
k
)
T
(
A
1
P
1
k
B
1
+
A
2
P
2
k
B
2
)
]
ccii
+
tr
⁡
[
(
F
k
)
T
(
C
1
P
1
k
D
1
+
C
2
P
2
k
D
2
)
]
)
×
(
∥
A
1
P
1
k
B
1
+
A
2
P
2
k
B
2
∥
2
+
∥
C
1
P
1
k
D
1
+
C
2
P
2
k
D
2
∥
2
)
-
1
.
On the other side, if the choice ofβ
k makes diag
⁡
(
E
k
+
1
,
F
k
+
1
) and diag
⁡
(
A
1
Δ
X
1
k
B
1
+
A
2
Δ
X
2
k
B
2
,
C
1
Δ
X
1
k
D
1
+
C
2
Δ
X
2
k
D
2
) orthogonal to each other, that is, tr
⁡
[
diag
⁡
(
E
k
+
1
,
F
k
+
1
)
T
diag
⁡
(
A
1
Δ
X
1
k
B
1
+
A
2
Δ
X
2
k
B
2
,
C
1
Δ
X
1
k
D
1
+
C
2
Δ
X
2
k
D
2
)
]
=
0, we can have the same β
k as (3).Theorem 7.
Algorithm5 is bound to be convergent.Proof.
From Algorithm5 and Lemma 6 we have
(10)
diag
⁡
(
E
k
,
F
k
)
=
diag
⁡
(
E
k
+
1
,
F
k
+
1
)

+
diag
⁡
(
A
1
Δ
X
1
k
B
1

+
A
2
Δ
X
2
k
B
2
,
C
1
Δ
X
1
k
D
1

+
C
2
Δ
X
2
k
D
2
)
,
⟺
∥
diag
⁡
(
E
k
,
F
k
)
∥
2
=
∥
diag
⁡
(
E
k
+
1
,
F
k
+
1
)
+
diag
⁡
(
A
1
Δ
X
1
k
B
1
+
A
2
Δ
X
2
k
B
2
,
cccccccc
C
1
Δ
X
1
k
D
1
+
C
2
Δ
X
2
k
D
2
)
∥
2
=
∥
diag
⁡
(
E
k
+
1
,
F
k
+
1
)
∥
2
+
∥
diag
⁡
(
A
1
Δ
X
1
k
B
1
+
A
2
Δ
X
2
k
B
2
,
cccccccccc
C
1
Δ
X
1
k
D
1
+
C
2
Δ
X
2
k
D
2
)
∥
2
such that
(11)
∥
diag
⁡
(
E
k
+
1
,
F
k
+
1
)
∥
2
=
∥
diag
⁡
(
E
k
,
F
k
)
∥
2
-
∥
diag
⁡
(
A
1
Δ
X
1
k
B
1
+
A
2
Δ
X
2
k
B
2
,
ccccccccci
C
1
Δ
X
1
k
D
1
+
C
2
Δ
X
2
k
D
2
)
∥
2
≤
∥
diag
⁡
(
E
k
,
F
k
)
∥
2
.
From (11), we know that Algorithm 5 is convergent.Lemma 8 (see [41]).
Suppose that the consistent system of linear equationsM
y
=
b has a solution y
0
∈
R
(
M
T
); then y
0 is the least Frobenius norm solution of the system of linear equations.Theorem 9.
Assume that the system (1) is consistent. Let X
1
1
=
A
1
T
Y
B
1
T +
C
1
T
Z
D
1
T
+ T
1
(
A
1
T
Y
B
1
T
+
C
1
T
Z
D
1
T
)
T
2, X
2
1
=
A
2
T
Y
B
2
T +
C
2
T
Z
D
2
T
+ T
3
(
A
2
T
Y
B
2
T
+
C
2
T
Z
D
2
T
)
T
4 be initial matrices, where Y
∈
R
p
×
q, Z
∈
R
s
×
t are any initial matrices, or, especially, X
1
1
=
0, X
2
1
=
0; then the solution generated by Algorithm 5 is the least Frobenius norm solution to (1).Proof.
If (1) is consistent, from X
1
1
=
A
1
T
Y
B
1
T
+
C
1
T
Z
D
1
T
+
T
1
(
A
1
T
Y
B
1
T
+
C
1
T
Z
D
1
T
)
T
2, X
2
1
=
A
2
T
Y
B
2
T
+ C
2
T
Z
D
2
T
+
T
3
(
A
2
T
Y
B
2
T
+
C
2
T
Z
D
2
T
)
T
4, using Algorithm 5, we have the generalized reflexive iterative solution pair X
1
k, X
2
k of (1) as the following:
(12)
X
1
k
=
X
1
k
-
1
+
Δ
X
1
k
-
1
=
X
1
1
+
Δ
X
1
1
+
⋯
+
Δ
X
1
k
-
1
=
A
1
T
Y
B
1
T
+
C
1
T
Z
D
1
T
+
T
1
(
A
1
T
Y
B
1
T
+
C
1
T
Z
D
1
T
)
T
2
+
A
1
T
[
β
1
E
1
+
⋯
+
β
k
-
1
E
k
-
1
]
B
1
T
+
C
1
T
[
β
1
F
1
+
⋯
+
β
k
-
1
F
k
-
1
]
D
1
T
+
T
1
A
1
T
[
β
1
E
1
+
⋯
+
β
k
-
1
E
k
-
1
]
B
1
T
T
2
+
T
1
C
1
T
[
β
1
F
1
+
⋯
+
β
k
-
1
F
k
-
1
]
D
1
T
T
2
=
A
1
T
M
B
1
T
+
C
1
T
N
D
1
T
+
T
1
(
A
1
T
M
B
1
T
+
C
1
T
N
D
1
T
)
T
2
,
X
2
k
=
X
2
k
-
1
+
Δ
X
2
k
-
1
=
X
2
1
+
Δ
X
2
1
+
⋯
+
Δ
X
2
k
-
1
=
A
2
T
Y
B
2
T
+
C
2
T
Z
D
2
T
+
T
3
(
A
2
T
Y
B
2
T
+
C
2
T
Z
D
2
T
)
T
4
+
A
2
T
[
β
1
E
1
+
⋯
+
β
k
-
1
E
k
-
1
]
B
2
T
+
C
2
T
[
β
1
F
1
+
⋯
+
β
k
-
1
F
k
-
1
]
D
2
T
+
T
3
A
2
T
[
β
1
E
1
+
⋯
+
β
k
-
1
E
k
-
1
]
B
2
T
T
4
+
T
3
C
2
T
[
β
1
F
1
+
⋯
+
β
k
-
1
F
k
-
1
]
D
2
T
T
4
=
A
2
T
M
B
2
T
+
C
2
T
N
D
2
T
+
T
3
(
A
2
T
M
B
2
T
+
C
2
T
N
D
2
T
)
T
4
.
We know that (1) is equivalent to the system
(13)
(
B
1
T
⊗
A
1
B
2
T
⊗
A
2
D
1
T
⊗
C
1
D
2
T
⊗
C
2
(
T
2
B
1
)
T
⊗
(
A
1
T
1
)
(
T
4
B
2
)
T
⊗
(
A
2
T
3
)
(
T
2
D
1
)
T
⊗
(
C
1
T
1
)
(
D
2
T
4
)
T
⊗
(
C
2
T
3
)
)
(
vec
⁡
(
X
1
)
vec
⁡
(
X
2
)
)
=
(
vec
⁡
(
E
)
vec
⁡
(
F
)
vec
⁡
(
E
)
vec
⁡
(
F
)
)
.
From (12) and (13) we have(14)
(
vec
⁡
(
A
1
T
M
B
1
T
+
C
1
T
N
D
1
T
+
T
1
(
A
1
T
M
B
1
T
+
C
1
T
N
D
1
T
)
T
2
)
vec
⁡
(
A
2
T
M
B
2
T
+
C
2
T
N
D
2
T
+
T
3
(
A
2
T
M
B
2
T
+
C
2
T
N
D
2
T
)
T
4
)
)
=
(
B
1
⊗
A
1
T
D
1
⊗
C
1
T
(
T
2
B
1
)
⊗
(
T
1
A
1
T
)
(
T
2
D
1
)
⊗
(
T
1
C
1
T
)
B
2
⊗
A
2
T
D
2
⊗
C
2
T
(
T
4
B
2
)
⊗
(
T
3
A
2
T
)
(
T
4
D
2
)
⊗
(
T
3
C
2
T
)
)
(
vec
⁡
(
M
)
vec
⁡
(
N
)
vec
⁡
(
M
)
vec
⁡
(
N
)
)
=
(
B
1
T
⊗
A
1
B
2
T
⊗
A
2
D
1
T
⊗
C
1
D
2
T
⊗
C
2
(
B
1
T
T
2
)
⊗
(
A
1
T
1
)
(
B
2
T
T
4
)
⊗
(
A
2
T
3
)
(
D
1
T
T
2
)
⊗
(
C
1
T
1
)
(
D
2
T
T
4
)
⊗
(
C
2
T
3
)
)
T
(
vec
⁡
(
M
)
vec
⁡
(
N
)
vec
⁡
(
M
)
vec
⁡
(
N
)
)
∈
R
(
(
B
1
T
⊗
A
1
B
2
T
⊗
A
2
D
1
T
⊗
C
1
D
2
T
⊗
C
2
(
B
1
T
T
2
)
⊗
(
A
1
T
1
)
(
B
2
T
T
4
)
⊗
(
A
2
T
3
)
(
D
1
T
T
2
)
⊗
(
C
1
T
1
)
(
D
2
T
T
4
)
⊗
(
C
2
T
3
)
)
T
)
,where R
(
*
) is the column space of matrix *.
Considering Lemma8, with the initial matrices X
1
1
=
A
1
T
Y
B
1
T
+
C
1
T
Z
D
1
T
+
T
1
(
A
1
T
Y
B
1
T
+
C
1
T
Z
D
1
T
)
T
2, X
2
1
=
A
2
T
Y
B
2
T
+
C
2
T
Z
D
2
T
+
T
3
(
A
2
T
Y
B
2
T
+
C
2
T
Z
D
2
T
)
T
4, where Y
∈
R
p
×
q, Z
∈
R
s
×
t are arbitrary, or, especially, X
1
1
=
0 and X
2
1
=
0, then the solution pair X
1
k
,
X
2
k, generated by Algorithm 5, is the least Frobenius norm solution of the matrix equations (1).Suppose that Problem1 is consistent. Obviously the solution set S
r of (1) is nonempty. For given matrices pair X
~
1
∈
R
k
×
r, X
~
2
∈
R
m
×
n, we consider two case:Case 1.
ConsiderX
~
1
∈
R
r
k
×
r
(
T
1
,
T
2
)
,
X
~
2
∈
R
r
m
×
n
(
T
3
,
T
4
).
We have(15)
A
1
X
1
B
1
+
A
2
X
2
B
2
=
E
,
C
1
X
1
D
1
+
C
2
X
2
D
2
=
F
,
(16)
⟺
{
A
1
(
X
1
-
X
~
1
)
B
1
+
A
2
(
X
2
-
X
~
2
)
B
2
=
E
-
A
1
X
~
1
B
1
-
A
2
X
~
2
B
2
,
C
1
(
X
1
-
X
~
1
)
D
1
+
C
2
(
X
2
-
X
~
2
)
D
2
=
F
-
C
1
X
~
1
D
1
-
C
2
X
~
2
D
2
.
Let X
¯
1
=
X
1
-
X
~
1, X
¯
2
=
X
2
-
X
~
2, E
¯
=
E
-
A
1
X
~
1
B
1
-
A
2
X
~
2
B
2, and F
¯
=
F
-
C
1
X
~
1
D
1
-
C
2
X
~
2
D
2; then Problem 2 is equivalent to finding the least Frobenius norm solution pair of the system
(17)
A
1
X
¯
1
B
1
+
A
2
X
¯
2
B
2
=
E
¯
,
C
1
X
¯
1
D
1
+
C
2
X
¯
2
D
2
=
F
¯
,
which can be obtained using Algorithm 5 with the initial matrix pair X
1
1
=
A
1
T
Y
B
1
T +
C
1
T
Z
D
1
T
+
T
1
(
A
1
T
Y
B
1
T
+
C
1
T
Z
D
1
T
)
T
2, X
2
1
=
A
2
T
Y
B
2
T +
C
2
T
Z
D
2
T
+ T
3
(
A
2
T
Y
B
2
T
+
C
2
T
Z
D
2
T
)
T
4, where Y
∈
R
p
×
q, Z
∈
R
s
×
t are arbitrary, or, especially, X
¯
1
1
=
0, X
¯
2
1
=
0, and the solution of the matrix optimal approximation Problem 2 can be represented as X
^
1
=
X
¯
1
k
+
X
~
1, X
^
2
=
X
¯
2
k
+
X
~
2.Case 2.
X
~
1
∈
R
k
×
r, X
~
2
∈
R
m
×
n, and X
~
1
,
X
~
2 are not generalized reflexive matrices.Lemma 10.
SupposeH
1
∈
R
r
k
×
r
(
T
1
,
T
2
), H
2
∈
R
a
m
×
n
(
T
1
,
T
2
); then t
r
(
H
1
T
H
2
)
=
0.Proof.
SinceH
1
=
T
1
H
1
T
2, H
2
=
-
T
1
H
2
T
2, we have
(18)
tr
⁡
(
H
1
T
H
2
)
=
tr
⁡
[
(
T
1
H
1
T
2
)
T
(
-
T
1
H
2
T
2
)
]
=
-
tr
⁡
(
T
2
T
H
1
T
T
1
T
T
1
H
2
T
2
)
=
-
tr
⁡
(
T
2
T
H
1
T
T
1
T
T
1
H
2
T
2
)
=
-
tr
⁡
(
T
2
T
H
1
T
H
2
T
2
)
=
-
tr
⁡
(
H
1
T
H
2
T
2
T
2
T
)
=
-
tr
⁡
(
H
1
T
H
2
)
.
So, tr
⁡
(
H
1
T
H
2
)
=
0.For{
X
1
,
X
2
}
∈
S
r, by Lemma 10, we have
(19)
∥
X
1
-
X
~
1
∥
2
=
∥
X
1
-
(
X
~
1
+
T
1
X
~
1
T
2
2
+
X
~
1
-
T
1
X
~
1
T
2
2
)
∥
2
=
∥
(
X
1
-
X
~
1
+
T
1
X
~
1
T
2
2
)
-
X
~
1
-
T
1
X
~
1
T
2
2
∥
2
=
∥
X
1
-
X
~
1
+
T
1
X
~
1
T
2
2
∥
2
+
∥
X
~
1
-
T
1
X
~
1
T
2
2
∥
2
,
∥
X
2
-
X
~
2
∥
2
=
∥
X
2
-
(
X
~
2
+
T
3
X
~
2
T
4
2
+
X
~
2
-
T
3
X
~
2
T
4
2
)
∥
2
=
∥
(
X
2
-
X
~
2
+
T
3
X
~
2
T
4
2
)
-
X
~
2
-
T
3
X
~
2
T
4
2
∥
2
=
∥
X
2
-
X
~
2
+
T
3
X
~
2
T
4
2
∥
2
+
∥
X
~
2
-
T
3
X
~
2
T
4
2
∥
2
.
For any {
X
1
,
X
2
}
∈
S
r, we have
(20)
∥
X
^
1
-
X
~
1
∥
2
+
∥
X
^
2
-
X
~
2
∥
2
=
min
⁡
{
X
1
,
X
2
}
∈
S
r
(
∥
X
1
-
X
~
1
∥
2
+
∥
X
2
-
X
~
2
∥
2
)
=
∥
X
~
1
-
T
1
X
~
1
T
2
2
∥
2
+
∥
X
~
2
-
T
3
X
~
2
T
4
2
∥
2
+
min
⁡
{
X
1
,
X
2
}
∈
S
r
(
∥
X
1
-
X
~
1
+
T
1
X
~
1
T
2
2
∥
2
c
c
c
c
c
c
c
c
c
c
c
c
c
+
∥
X
2
-
X
~
2
+
T
3
X
~
2
T
4
2
∥
2
)
.
That is, Problem 2 is equivalent to finding {
X
^
1
,
X
^
2
}
∈
S
r such that ∥
X
1
-
(
X
~
1
+
T
1
X
~
1
T
2
)
/
2
∥
2
+ ∥
X
2
-
(
X
~
1
+
T
3
X
~
2
T
4
)
/
2
∥
2 reach a minimum.For any{
X
1
,
X
2
}
∈
S
r, we have
(21)
A
1
X
1
B
1
+
A
2
X
2
B
2
=
E
,
C
1
X
1
D
1
+
C
2
X
2
D
2
=
F
,
(22)
⟺
{
A
1
(
X
1
-
X
~
1
+
T
1
X
~
1
T
2
2
)
B
1
+
A
2
(
X
2
-
X
~
2
+
T
3
X
~
2
T
4
2
)
B
2
=
E
-
A
1
X
~
1
+
T
1
X
~
1
T
2
2
B
1
-
A
2
X
~
2
+
T
3
X
~
2
T
4
2
B
2
,
C
1
(
X
1
-
X
~
1
+
T
1
X
~
1
T
2
2
)
D
1
+
C
2
(
X
2
-
X
~
2
+
T
3
X
~
2
T
4
2
)
D
2
=
F
-
C
1
X
~
1
+
T
1
X
~
1
T
2
2
D
1
-
C
2
X
~
2
+
T
3
X
~
2
T
4
2
D
2
.
Let X
¯
1
=
X
1
-
(
X
~
1
+
T
1
X
~
1
T
2
)
/
2, X
¯
2
=
X
2
- (
X
~
2
+
T
3
X
~
2
T
4
)
/
2, E
¯
=
E
-
A
1
(
(
X
~
1
+
T
1
X
~
1
T
2
)
/
2
)
B
1 -
A
2
(
(
X
~
2
+
T
3
X
~
2
T
4
)
/
2
)
B
2, and F
¯
=
F
-
C
1
(
(
X
~
1
+
T
1
X
~
1
T
2
)
/
2
)
D
1 -
C
2
(
(
X
~
2
+
T
3
X
~
2
T
4
)
/
2
)
D
2 and then to finding {
X
^
1
,
X
^
2
}
∈
S such that ∥
X
1
-
(
X
~
1
+
T
1
X
~
1
T
2
)
/
2
∥
2
+
∥
X
2
-
(
X
~
2
+
T
3
X
~
2
T
4
)
/
2
∥
2 reach a minimum is equivalent to finding the least Frobenius norm generalized reflexive solution pair X
¯
1
k
,
X
¯
2
k of the system
(23)
A
1
X
¯
1
B
1
+
A
2
X
¯
2
B
2
=
E
¯
,
C
1
X
¯
1
D
1
+
C
2
X
¯
2
D
2
=
F
¯
,
which can be obtained using Algorithm 5.So we can have the least Frobenius norm generalized reflexive solution of Problem2 as X
^
1
=
X
¯
1
k
+
(
X
~
1
+
T
1
X
~
1
T
2
)
/
2, X
^
2
=
X
¯
2
k
+
(
X
~
2
+
T
3
X
~
2
T
4
)
/
2.


## 3. Iterative Algorithm for Solving Problems3 and 4

In this section, we present an iterative algorithm for solving Problems3 and 4.Algorithm 11.
(1) Input matricesA
1
∈
R
p
×
k, A
2
∈
R
p
×
m, B
1
∈
R
r
×
q, B
2
∈
R
n
×
q, C
1
∈
R
s
×
k, C
2
∈
R
s
×
m, D
1
∈
R
r
×
t, D
2
∈
R
n
×
t, E
∈
R
p
×
q, F
∈
R
s
×
t, X
1
1
∈
R
a
k
×
r, and X
2
1
∈
R
a
m
×
n (where X
1
1
,
X
2
1 are any initial generalized antireflexive matrices).
(2) Calculate(24)
E
1
=
E
;
F
1
=
F
;
Q
1
1
=
A
1
T
E
1
B
1
T
+
C
1
T
F
1
D
1
T
;
Q
2
1
=
A
2
T
E
1
B
2
T
+
C
2
T
F
1
D
2
T
;
P
1
1
=
Q
1
1
-
T
1
Q
1
1
T
2
;
P
2
1
=
Q
2
1
-
T
3
Q
2
1
T
4
;
β
1
=
(
tr
⁡
[
(
E
1
)
T
(
A
1
P
1
1
B
1
+
A
2
P
2
1
B
2
)
]
xcx
+
tr
⁡
[
(
F
1
)
T
(
C
1
P
1
1
D
1
+
C
2
P
2
1
D
2
)
]
)
×
(
∥
A
1
P
1
1
B
1
+
A
2
P
2
1
B
2
∥
2
+
∥
C
1
P
1
1
D
1
+
C
2
P
2
1
D
2
∥
2
)
-
1
;
Δ
X
1
1
=
β
1
P
1
1
;
Δ
X
2
1
=
β
1
P
2
1
;
k
=
1
.
(3) IfΔ
X
=
diag
⁡
(
Δ
X
1
k
,
Δ
X
2
k
)
=
0 (k
=
1,2
,
…), then stop. Otherwise,
(25)
X
1
k
+
1
=
X
1
k
+
Δ
X
1
k
;
X
2
k
+
1
=
X
2
k
+
Δ
X
2
k
.
(4) Calculate(26)
E
k
+
1
=
E
k
-
(
A
1
Δ
X
1
k
B
1
+
A
2
Δ
X
2
k
B
2
)
;
F
k
+
1
=
F
k
-
(
C
1
Δ
X
1
k
D
1
+
C
2
Δ
X
2
k
D
2
)
;
Q
1
k
+
1
=
A
1
T
E
k
+
1
B
1
T
+
C
1
T
F
k
+
1
D
1
T
;
Q
2
k
+
1
=
A
2
T
E
k
+
1
B
2
T
+
C
2
T
F
k
+
1
D
2
T
;
P
1
k
+
1
=
Q
1
k
+
1
-
T
1
Q
1
k
+
1
T
2
;
P
2
k
+
1
=
Q
2
k
+
1
-
T
3
Q
2
k
+
1
T
4
;
β
k
+
1
=
(
tr
⁡
[
(
E
k
+
1
)
T
(
A
1
P
1
k
+
1
B
1
+
A
2
P
2
k
+
1
B
2
)
]
ccc
+
tr
⁡
[
(
F
k
+
1
)
T
(
C
1
P
1
k
+
1
D
1
+
C
2
P
2
k
+
1
D
2
)
]
)
×
(
∥
A
1
P
1
k
+
1
B
1
+
A
2
P
2
k
+
1
B
2
∥
2
c
c
c
c
c
c
+
∥
C
1
P
1
k
+
1
D
1
+
C
2
P
2
k
+
1
D
2
∥
2
)
-
1
;
Δ
X
1
k
+
1
=
β
k
+
1
P
1
k
+
1
;
Δ
X
2
k
+
1
=
β
k
+
1
P
2
k
+
1
;
k
=
k
+
1
.
Go to (3).The properties of Algorithm11 can be proposed similarly to Algorithm 5.


## 4. Examples

In this section, we show two numerical examples to illustrate the efficiency of Algorithms5 and 11. All computations are performed by MATLAB 7. For the influence of the error of calculation, we consider the matrix R as a zero matrix if ∥
R
∥
<
1
0
-
10.Example 12.
Consider the generalized reflexive solution of the linear matrix equations:(27)
A
1
X
1
B
1
+
A
2
X
2
B
2
=
E
,
C
1
X
1
D
1
+
C
2
X
2
D
2
=
F
,
where
(28)
A
1
=
(
139
105
54
124
176
50
159
35
175
191
196
147
)
,
B
1
=
(
13
117
103
87
116
198
85
67
45
152
)
,
A
2
=
(
27
60
2
132
179
57
40
94
)
,
B
2
=
(
106
76
92
12
83
128
157
114
121
61
42
136
159
10
175
)
,
C
1
=
(
3
88
192
154
100
145
194
43
82
198
129
149
158
64
54
)
,
D
1
=
(
88
31
140
71
135
146
)
,
C
2
=
(
4
63
115
3
90
77
9
137
5
19
)
,
D
2
=
(
96
90
55
111
143
51
24
179
173
)
,
T
1
=
(
-
1
0
0
0
0
1
0
1
0
)
,
T
2
=
(
0
-
1
-
1
0
)
,
T
3
=
(
0
1
1
0
)
,
T
4
=
(
0
1
0
1
0
0
0
0
-
1
)
,
E

=

(
4079684
5931341
5257546
3760503
5978636
4507138
7309406
6800596
4204046
7209376
5983467
8290427
6916092
5829089
5964332
4470189
8583167
7694045
5535093
8046508
)
,
F
=
(
1148659
-
415508
175592
5221721
3459626
5372914
6867542
6499561
7947258
6828512
5933355
8858444
3684631
3115265
5873214
)
.Let(29)
X
1
1
=
(
0
0
0
0
0
0
)
,
X
2
1
=
(
0
0
0
0
0
0
)
.Using Algorithm5 and iterating 1214 steps, we obtain the least Frobenius norm solution pair of the matrix equation in Example 12 as follows:
(30)
X
1
=
(
122.0000
122.0000
86.0000
-
29.0000
29.0000
-
86.0000
)
,
X
2
=
(
57.0000
126.0000
-
35.0000
126.0000
57.0000
35.0000
)
.Example 13.
Consider the generalized antireflexive solution of the linear matrix equations:(31)
A
1
X
1
B
1
+
A
2
X
2
B
2
=
E
,
C
1
X
1
D
1
+
C
2
X
2
D
2
=
F
,
where A
1
,
B
1
,
A
2
,
B
2
,
C
1
,
D
1
,
C
2
,
D
2
,
T
1
,
T
2
,
T
3, and T
4 are the same matrices of Example 12:
(32)
E
=
(
-
743633
6710933
7426284
2563311
8338860
1535166
7945314
9163408
1986230
12338200
982308
14732018
17074320
3775802
18238112
1999993
12662336
13364050
5402942
16219108
)
,
F
=
(
5228109
6926269
12710902
7508081
7671994
14810466
4201528
4853203
12104012
5302290
5127440
15650036
2867500
-
410942
5039552
)
.Let(33)
X
1
1
=
(
0
0
0
0
0
0
)
,
X
2
1
=
(
0
0
0
0
0
0
)
.Using Algorithm11 and iterating 240 steps, we obtain the least Frobenius norm solution pair of the matrix equation in Example 13 as follows:
(34)
X
1
=
(
226.0000
-
226.0000
59.0000
191.0000
191.0000
59.0000
)
,
X
2
=
(
189.0000
-
63.0000
268.0000
63.0000
-
189.0000
268.0000
)
.The numbers of the parameter matricesA
1
,
A
2
,
B
1
,
B
2
,
C
1
,
C
2
,
D
1, and D
2 in our examples are larger than the numbers of the parameter matrices in the example of [31]. To our examples, the algorithm of [31] is not convergent. The numerical examples demonstrate that our algorithm has merits of good numerical stability and ease to program.



---

*Source: 352327-2014-02-23.xml*