The Symmetric and Antipersymmetric Solutions of
the Matrix Equation
A
1
X
1
B
1
+ A
2
X
2
B
2
+ . . . + A
l
X
l
B
l
= C and Its
Optimal Approximation
Ying Zhang, Member, IAENG
Abstract—A matrix A = (a
ij
) ∈ R
n×n
is said to be symmet-
ric and antipersymmetric matrix if a
ij
= a
ji
= −a
n−j+1,n−i+1
for all 1 ≤ i, j ≤ n. Peng gave the bisymmetric solutions of the
matrix equation A
1
X
1
B
1
+A
2
X
2
B
2
+. . .+A
l
X
l
B
l
= C, where
[X
1
, X
2
, . . . , X
l
] is a real matrices group. Based on this work,
an adjusted iterative method is proposed to find the symmetric
and antipersymmetric solutions of the above matrix equation.
When the matrix equation is consistent, for any initial symmet-
ric and antipersymmetric matrix group [X
(0)
1
, X
(0)
2
, . . . , X
(0)
l
],
the least norm symmetric and antipersymmetric solution group
can be obtained. In addition, for a given symmetric and
antipersymmetric matrix group [
¯
X
1
,
¯
X
2
, . . . ,
¯
X
l
], the optimal
approximation symmetric and antipersymmetric solution group
can be obtained. Given numerical examples show that the
iterative method is efficient.
Index Terms—Iterative method, Matrix equation, Symmetric
and antipersymmetric, Least-norm solution group, Optimal
approximation solution
I. INTRODUCTION
A
Matrix A = (a
ij
) ∈ R
n×n
is said to be sym-
metric and antipersymmetric matrix if a
ij
= a
ji
=
−a
n−j+1,n−i+1
for all 1 ≤ i, j ≤ n. Let R
m×n
, SR
n×n
and
SA
n
denote the set of m×n real matrices, n ×n real symmet-
ric matrices and n × n real symmetric and antipersymmetric
matrices respectively. S
n
(S
n
= (e
n
, e
n−1
, . . . , e
1
)) denotes
the n×n reverse unit matrix (e
i
denotes ith column of n×n
unit matrix). The superscripts T represents the transpose
of a matrix. In space R
m×n
, we define inner products as:
< A, B >= trace(B
T
A) for all A, B ∈ R
m×n
. Then the
norm of A generated by this inner product is Frobenius norm
and denoted by k A k.
Problem I. (See [1]) Given A
i
∈ R
p×n
i
, B
i
∈
R
n
i
×q
(i = 1, 2, . . . , l) and C ∈ R
p×q
, find matrix group
[X
1
, X
2
, . . . , X
l
] with X
i
∈ SA
n
i
, i = 1, 2, . . . , l, such that
A
1
X
1
B
1
+ A
2
X
2
B
2
+ . . . + A
l
X
l
B
l
= C. (1)
Problem II. (See [2]) When Problem I is consistent, let
S
E
denote its solution group set, for given matrix group
[
¯
X
1
,
¯
X
2
, . . . ,
¯
X
l
] with
¯
X
i
∈ SA
n
i
(i = 1, 2, . . . , l), find
Manuscript received 19 Oct, 2016; revised 25 Nov, 2016. This work was
supported by the National Natural Science Foundation of China under Grant
Nos. 61402071 and 61671099, Liaoning Province Nature Science Founda-
tion under Grant Nos. 2015020006 and 2015020011, and the Fundamental
Research Funds for the Central Universities under Grant Nos. 3132015230
and 3132016111.
Y. Zhang is with the Department of Mathematics, Dalian Maritime
University, Dalian, Liaoning, 116026 China, e-mail: zhgyg77@sina.com.
[
c
X
1
,
c
X
2
, . . . ,
c
X
l
] ∈ S
E
with
c
X
i
∈ SA
n
i
, such that
k
c
X
1
−
¯
X
1
k
2
+ k
c
X
2
−
¯
X
2
k
2
+ . . . + k
c
X
l
−
¯
X
l
k
2
= min
[X
1
,X
2
,...,X
l
]∈S
E
[k X
1
−
¯
X
1
k
2
+ k X
2
−
¯
X
2
k
2
+ . . . + k X
l
−
¯
X
l
k
2
].
(2)
In [3], an iterative method is constructed to find the bisym-
metric solutions of matrix equation A
1
X
1
B
1
+ A
2
X
2
B
2
+
. . . + A
l
X
l
B
l
= C, where [X
1
, X
2
, . . . , X
l
] is real matrices
group. In this paper, by adjusting the algorithm in [3], we can
find symmetric and antipersymmetric solutions of the above
matrix equation. In electricity, control theory, and processing
of digital signals, symmetric and antipersymmetric matrices
have wide application.
II. ITERATIVE METHODS FOR SOLVING PROBLEM I AND
II
Firstly we introduce some lemmas used to solve Problem
I.
Lemma 2.1.(see [4]) A matrix X ∈ SA
n
if and only if
X = X
T
= −S
n
XS
n
.
Lemma 2.2. Suppose that a matrix X ∈ SR
n×n
, then X −
S
n
XS
n
∈ SA
n
.
Lemma 2.3. Suppose that A ∈ R
n×n
, X ∈ SA
n
, then
< A, X >= h
1
4
[A + A
T
− S
n
(A + A
T
)S
n
], Xi.
Proof
h
1
4
[A + A
T
− S
n
(A + A
T
)S
n
], Xi
=
1
4
[hA, Xi + hA
T
, Xi − hS
n
AS
n
, Xi − hS
n
A
T
S
n
, Xi]
=
1
4
[hA, Xi + hA
T
, X
T
i − hA, S
n
XS
n
i − hA
T
, S
n
XS
n
i]
= hA, Xi.
(3)
Next, by adjusting the method in [3], an iterative method to
obtain the symmetric and antipersymmetric solution groups
of Problem I.
Algorithm 2.1.
Step 1. Input matrices A
i
∈ R
p×n
i
, B
i
∈ R
n
i
×q
, X
(0)
i
∈
SA
n
i
×n
i
(i = 1, 2, · · · , l) and C ∈ R
p×q
;
Step 2. Calculate R
0
= C −
P
l
i=1
A
i
X
(0)
i
B
i
;
Y
0,i
= A
T
i
R
0
B
T
i
, i = 1, 2, · · · , l;
P
0,i
=
1
4
[Y
0,i
+ Y
T
0,i
− S
n
i
Y
0,i
S
n
i
− S
n
i
Y
T
0,i
S
n
i
];
i = 1, 2, · · · , l;
k:=0;
Step 3. If R
k
= 0, then stop; else, k := k + 1;
Step 4. Calculate
X
(k)
i
= X
(k−1)
i
+
kR
k−1
k
2
P
l
j=1
kP
k−1,j
k
2
P
k−1,i
, i = 1, · · · , l;
Proceedings of the International MultiConference of Engineers and Computer Scientists 2017 Vol II,
IMECS 2017, March 15 - 17, 2017, Hong Kong
ISBN: 978-988-14047-7-0
ISSN: 2078-0958 (Print); ISSN: 2078-0966 (Online)