# The Solutions to Matrix EquationA

Authors: Chang-Zhou Dong; Yu-Ping Zhang
Journal: Journal of Applied Mathematics
 (2014)

---

## Abstract

LetP be a given Hermitian matrix satisfying P
2
=
I. Using the eigenvalue decomposition of P, we consider the least squares solutions to the matrix equation A
X
=
B with the constraints P
X
=
X
P and X
*
=
X. A similar problem of this matrix equation with generalized constrained is also discussed.


---

## Body

## 1. Introduction

Throughout we denote the complexm
×
n matrix space by ℂ
m
×
n. The symbols I,  A
*,  A
-
1, and ∥
A
∥ stand for the identity matrix with the appropriate size, the conjugate transpose, the inverse, and the Frobenius norm of A
∈
ℂ
m
×
n, respectively.It is a very active research topic to study solutions to various matrix equations [1–4]. There are many authors who have investigated the classical matrix equation
(1)
A
X
=
B
with different constraints such as symmetric, reflexive, Hermitian-generalized Hamiltonian, and repositive definite [5–9]. By special matrix decompositions such as singular value decompositions (SVDs) and CS decompositions [10–12], Hu and his collaborators [13–15], Dai [16], and Don [17] have presented the existence conditions and detailed representations of constrained solutions for (1) with corresponding constraints, respectively. For instance, Peng and Hu [18] presented the eigenvectors-involved solutions to (1) with reflexive and antireflexive constraints; Wang and Yu [19] derived the bi(skew-)symmetric solutions and the bi(skew-)symmetric least squares solutions with the minimum norm to this matrix equation; Qiu and Wang [20] proposed an eigenvectors-free method to (1) with P
X
=
X
P and X
*
=
s
X constraints, where P is a Hermitian involutory matrix and s
=
±
1.Inspired by the work mentioned above, we focus on the matrix equation (1) with P
X
=
X
P and X
*
=
X constraints, which can be described as follows: find X such that
(2)
{
∥
A
X
-
B
∥
2
=
min
⁡
,
P
X
=
X
P
,
X
*
=
X
}
.Moreover, we also discuss the least squares solutions of (1) with P
X
=
X
G
P
G
* and X
*
=
X constraints, where G is a given unitary matrix of order n.In Section2, we present the least squares solutions to the matrix equation (1) with the constraints P
X
=
X
P and X
*
=
X. In Section 3, we derive the least squares solutions to the matrix equation (1) with the constraints P
X
=
X
G
P
G
* and X
*
=
X. In Section 4, we give an algorithm and a numerical example to illustrate our results.


## 2. Least Squares Solutions to the Matrix Equation (1) with the Constraints P
X
=
X
P  and X
*
=
X

It is required to transform the constrained problem to unconstrained one. To this end, let(3)
P
=
U
diag
⁡
(
I
k
,
-
I
n
-
k
)
U
*
be the eigenvalue decomposition of the Hermitian matrix P with unitary matrix U. Obviously, P
X
=
X
P holds if and only if
(4)
diag
⁡
(
I
k
,
-
I
n
-
k
)
X
¯
=
X
¯
diag
⁡
(
I
k
,
-
I
n
-
k
)
,
where X
¯
=
U
*
X
U. Partitioning
(5)
X
¯
=
(
X
11
X
12
X
21
X
22
)
,
X
11
∈
ℂ
k
×
k
,
X
22
∈
ℂ
(
n
-
k
)
×
(
n
-
k
)
,
(4) is equivalent to
(6)
X
12
=
-
X
12
,
X
21
=
-
X
21
.
Therefore,
(7)
X
=
U
diag
⁡
(
X
11
,
X
22
)
U
*
,
X
11
∈
ℂ
k
×
k
,
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
X
22
∈
ℂ
(
n
-
k
)
×
(
n
-
k
)
.
The constraint X
*
=
X is equivalent to
(8)
X
=
U
diag
⁡
(
X
1
,
X
2
)
U
*
,
X
i
*
=
X
i
,
i
=
1,2
,
with X
1
∈
ℂ
k
×
k
,
X
2
∈
ℂ
(
n
-
k
)
×
(
n
-
k
).PartitionU
=
(
U
1
,
U
2
) and denote
(9)
A
1
=
A
U
1
,
A
2
=
A
U
2
,
B
1
=
B
U
1
,
B
2
=
B
U
2
;
then assume that the singular value decomposition of A
1 and A
2 is as follows:
(10)
A
1
=
M
1
[
Σ
1
0
0
0
]
N
1
*
,
A
2
=
M
2
[
Σ
2
0
0
0
]
N
2
*
,
where M
1
,
M
2
,
N
1, and N
2 are unitary matrices, Σ
1
=
diag
⁡
(
α
1
,
…
,
α
r
),  α
i
>
0
(
i
=
1
,
…
,
r
),  r
=
rank
⁡
(
A
1
),  Σ
2
=
diag
⁡
(
β
1
,
…
,
β
l
),  β
j
>
0
(
j
=
1
,
⋯
,
l
),   and  l
=
rank
⁡
(
A
2
).Theorem 1.
GivenA
,
B
∈
ℂ
m
×
n. Then the least squares solutions to the matrix equation (1) with the constraints P
X
=
X
P and X
*
=
X can be expressed as(11)
X
=
U
diag
⁡
(
N
1
(
Σ
1
-
1
B
11
+
B
11
*
Σ
1
-
1
2
Σ
1
-
1
B
12
B
12
*
Σ
1
-
1
X
14
)
N
1
*
0
0
N
2
(
Σ
2
-
1
B
21
+
B
21
*
Σ
2
-
1
2
Σ
2
-
1
B
22
B
22
*
Σ
2
-
1
X
24
)
N
2
*
)
U
*
, where X
14
=
X
14
* and X
24
=
X
24
* are arbitrary matrix.Proof.
According to (8) and the unitary invariance of Frobenius norm
(12)
∥
A
X
-
B
∥
=
∥
A
U
diag
⁡
(
X
1
,
X
2
)
U
*
-
B
∥
=
∥
A
U
diag
⁡
(
X
1
,
X
2
)
-
B
U
∥
.
By (9), the least squares problem is equivalent to
(13)
∥
A
X
-
B
∥
=
∥
(
A
1
X
1
-
B
1
,
A
2
X
2
-
B
2
)
∥
.
We get
(14)
∥
A
X
-
B
∥
2
=
∥
A
1
X
1
-
B
1
∥
2
+
∥
A
2
X
2
-
B
2
∥
2
.
According to (10), the least squares problem is equivalent to
(15)
∥
A
X
-
B
∥
2
=
∥
M
1
[
Σ
1
0
0
0
]
N
1
*
X
1
-
B
1
∥
2
+
∥
M
2
[
Σ
2
0
0
0
]
N
2
*
X
2
-
B
2
∥
2
=
∥
[
Σ
1
0
0
0
]
N
1
*
X
1
N
1
-
M
1
*
B
1
N
1
∥
2
+
∥
[
Σ
2
0
0
0
]
N
2
*
X
2
N
2
-
M
2
*
B
2
N
2
∥
2
.
Assume that
(16)
N
1
*
X
1
N
1
=
[
X
11
X
12
X
13
X
14
]
,
N
2
*
X
2
N
2
=
[
X
21
X
22
X
23
X
24
]
,
M
1
*
B
1
N
1
=
[
B
11
B
12
B
13
B
14
]
,
M
2
*
B
2
N
2
=
[
B
21
B
22
B
23
B
24
]
.
Then we have
(17)
∥
A
X
-
B
∥
2
=
∥
[
Σ
1
0
0
0
]
[
X
11
X
12
X
13
X
14
]
-
[
B
11
B
12
B
13
B
14
]
∥
2
+
∥
[
Σ
2
0
0
0
]
[
X
21
X
22
X
23
X
24
]
-
[
B
21
B
22
B
23
B
24
]
∥
2
=
∥
Σ
1
X
11
-
B
11
∥
2
+
∥
Σ
2
X
21
-
B
21
∥
2
+
∥
Σ
1
X
12
-
B
12
∥
2
+
∥
Σ
2
X
22
-
B
22
∥
2
+
∥
B
13
∥
2
+
∥
B
14
∥
2
+
∥
B
23
∥
2
+
∥
B
24
∥
2
.
Hence
(18)
∥
A
X
-
B
∥
2
=
min
⁡
is solvable if and only if there exist X
11
,
X
12
,
X
21
,
X
22 such that
(19)
∥
Σ
1
X
11
-
B
11
∥
2
=
min
⁡
,
∥
Σ
1
X
12
-
B
12
∥
2
=
min
⁡
,
∥
Σ
2
X
21
-
B
21
∥
2
=
min
⁡
,
∥
Σ
2
X
22
-
B
22
∥
2
=
min
⁡
.
It follows from (19) that
(20)
X
11
=
Σ
1
-
1
B
11
+
B
11
*
Σ
1
-
1
2
,
X
12
=
Σ
1
-
1
B
12
,
X
21
=
Σ
2
-
1
B
21
+
B
21
*
Σ
2
-
1
2
,
X
22
=
Σ
2
-
1
B
22
.
Substituting (20) into (16) and then into (8), we can get that the form of X is (11).


## 3. Least Squares Solutions to the Matrix Equation (1) with the Constraints P
X
=
X
G
P
G
* and X
*
=
X

In this section, we generalize the constraintsP
X
=
X
P to P
X
=
X
G
P
G
*, where G is a given unitary matrix of order n. Obviously, the constraint is equal to
(21)
P
X
G
=
X
G
P
.
Notice that (1) can be equivalently rewritten in
(22)
A
X
G
=
B
G
.
Denoting by Y
=
X
G and setting C
=
B
G, the equation becomes
(23)
A
Y
=
C
,
with the constraints P
Y
=
Y
P and Y
*
=
Y.Therefore, the least squares solutions to matrix equation (1) with the constraints P
X
=
X
G
P
G
* and X
*
=
X can be solved similar to Theorem 1.Theorem 2.
GivenA
,
B
∈
ℂ
m
×
n. Then the least squares solutions to the matrix equation (1) with the constraints P
X
=
X
G
P
G
* and X
*
=
X can be expressed as(24)
X
=
U
diag
⁡
(
N
1
(
Σ
1
-
1
C
11
+
C
11
*
Σ
1
-
1
2
Σ
1
-
1
C
12
C
12
*
Σ
1
-
1
Y
14
)
N
1
*
0
0
N
2
(
Σ
2
-
1
C
21
+
C
21
*
Σ
2
-
1
2
Σ
2
-
1
C
22
C
22
*
Σ
2
-
1
Y
24
)
N
2
*
)
U
*
G
*
,where Y
14
=
Y
14
* and Y
24
=
Y
24
* are arbitrary matrix.


## 4. An Algorithm and Numerical Examples

Based on the main results of this paper, we in this section propose an algorithm for finding the least squares solutions to the matrix equationA
X
=
B with the constraints P
X
=
X
P and X
*
=
X. All the tests are performed by MATLAB 6.5 which has a machine precision of around 1
0
-
16.Algorithm 3.
(
1
) Input A
,
B
∈
ℂ
m
×
n
,
P
∈
ℂ
n
×
n and compute U
∈
ℂ
n
×
n,  I
k
∈
ℂ
k
×
k,  -
I
n
-
k
∈
ℂ
(
n
-
k
)
×
(
n
-
k
) by the eigenvalue decomposition to P.
(
2
) Compute A
1
,
A
2
,
B
1
,
B
2 according to (9).
(
3
) Compute N
1
,
N
2
,
M
1
,
M
2
,
Σ
1
,
Σ
2 by the singular value decomposition of A
1
,
A
2.
(
4
) Compute B
11
,
B
12
,
B
21
,
B
22 according to (16).
(
5
) Compute X by Theorem 1.Example 4.
Suppose(25)
A
=
[
0
0
0
0
0
1.2
i
0
0
0
0
0
0.8
i
]
,
B
=
[
-
3
-
0.8
i
-
1
-
3
i
-
1
-
1
-
i
-
1
9
i
-
7
-
2
-
2
2
i
-
2
]
,
P
=
[
1
0
0
0
0
1
0
0
0
0
-
1
0
0
0
0
-
1
]
.
Applying Algorithm3, we obtain the following:
(26)
U
=
[
0
-
i
0
0
i
0
0
0
0
0
0
1
0
0
1
0
]
,
A
1
=
[
0
0
1.2
0
0
0
]
,
A
2
=
[
0
0
0
0
0.8
i
0
]
,
B
1
=
[
0.8
3
.
i
-
i
-
1
+
i
-
2
i
2
i
]
,
B
2
=
[
-
1
-
1
-
3
i
-
7
9
i
-
2
2
i
]
,
M
1
=
[
0
i
0
-
i
0
0
0
0
1
]
,
M
2
=
[
0
1
0
0
0
i
1
0
0
]
,
N
1
=
[
i
0
0
1
]
,
N
2
=
[
-
i
0
0
i
]
,
Σ
1
=
[
1.2
]
,
Σ
2
=
[
0.8
]
,
B
11
=
[
i
]
,
B
12
=
[
-
1
-
i
]
,
B
21
=
[
2
i
]
,
B
22
=
[
-
2
]
,
X
=
[
3
-
0.83
+
0.83
i
0
0
-
0.83
-
0.83
i
0
0
0
0
0
-
2
2.5
0
0
2.5
0
]
.



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*Source: 412094-2014-04-07.xml*