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 A1X+B1Y=D1X¯E1+F1, A2Y+B2X=D2Y¯E2+F2. 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.

1. Introduction

Many research papers are involved in the system of matrix equation ([1–33]). The following matrix equation
(1)AXB=C
is a special case of coupled Sylvester linear matrix equations
(2)∑j=1nAijXjBij=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 A1XB1=C1, A2XB2=C2 [35]. By Moore-Penrose generalized inverse, some necessary and sufficient conditions on the existence of the solution and the expressions for the matrix equation AX+XTC=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 Ax=b, Ding et al. gained iterative solutions for matrix equation (1) and the generalized Sylvester matrix equation AXB+CXD=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] AXB+CYD=M, EXF+GYH=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)A1XB1+C1XTD1=F1,A2XB2+C2XTD2=F2.

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)A1X+B1Y=D1X¯E1+F1,A2Y+B2X=D2Y¯E2+F2,
where Ai,Bi,Di∈Cp×m, Ei∈Cm×m, Fi∈Cp×m(i=1,2) are given constant matrices, and X,Y∈Cm×m are unknown matrices to be solved. When A2=B2=D2=0 and F2=0, the problem (4) becomes the problem studied in [42]. When A2=B2=D2=0, F2=0, and A1=I, this system becomes the Yakubovich-conjugate matrix equation investigated in [43]. When B1=A2=B2=D2=0, F2=0, and A1=I, the problem (4) becomes the equation considered in [44]. When A2=B2=D2=0 and F1=F2=0, the problem (4) becomes the equation in [45]. When B1=A2=B2=D2=0, F2=0, and D1=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)Ax=b,
where x∈Rn is an unknown vector, A∈Rm×n is a given matrix, and b∈Rm 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.

Cm×n is the set of m×n complex matrices and Rm×n is the set of m×n all real matrices. For A∈Cm×n, we write Re(A), Im(A), A¯, AT, AH, 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{A1,A2,…,An} denotes the block diagonal matrix, where Ai∈Rm×m(i=1,2,…,n). For any A=(aij), B=(bij), A⊗B denotes the Kronecker product defined as A⊗B=(aijB). For the matrix X=(x1,x2,…,xn)∈Cm×n, vec(X) denotes the vec operator defined as vec(X)=(x1T,x2T,…,xnT)T. We use I to denote the identity matrix of size implied by context.

Definition 1.

Let S∈Rm×m and S=(em,em-1,…,e1), where ej(j=1,2,…,m) denotes the column vector whose jth 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 SXS=X(SXS=-X), denote the set of all centrally symmetric (centrally antisymmetric) matrices by CSCm×m(CASCm×m).

The rest of this paper is organized as follows. In Section 2, 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 (<xref ref-type="disp-formula" rid="EEq1.4">4</xref>)

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 space Cm×n is defined as
(6)〈A,B〉=Re[tr(AHB)].
By Theorem 1 in [42], we know that the inner product defined by (6) satisfies the following three axioms:

symmetry: 〈A,B〉=〈B,A〉;

linearity in the first argument:
(7)〈δ1A1+δ2A2,B〉=δ1〈A1,B〉+δ2〈A2,B〉,
where δ1 and δ2 are real constants;

positive definiteness: 〈A,A〉>0, for all A≠0.

For all real constants δ1,δ2, by (1) and (2), we get
(8)〈A,δ1B1+δ2B2〉=〈δ1B1+δ2B2,A〉=δ1〈B1,A〉+δ2〈B2,A〉=δ1〈A,B1〉+δ2〈A,B2〉;
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(AHB)]=Re[tr(AHB)¯]=Re[tr(AHB¯)].

The norm of a matrix generated by this inner product space is denoted by ∥·∥. Then for A∈Cm×n, we obtain
(10)∥A∥2=〈A,A〉=Re[tr(AHA)].

What is the relationship between this norm and the Frobenius norm? It is well known that ∥A∥F2=tr(AHA) and AHA is a Hermite matrix. Then by the knowledge of algebra, we know that tr(AHA) is real; hence, tr(AHA)=Re[tr(AHA)]. This shows that ∥A∥F=∥A∥. Another interesting relationship is that
(11)∥A∥2=∥Re(A)+iIm(A)∥2=Re{tr[(Re(A)+iIm(A))H(Re(A)+iIm(A))]}=Re{tr[(Re(A)T-iIm(A)T)(Re(A)+iIm(A))]}=Re[tr(Re(A)TRe(A)+iRe(A)TIm(A)iiiiiiii-iIm(A)TRe(A)+Im(A)TIm(A))]=Re[tr(Re(A)TRe(A))+tr(Im(A)TIm(A))iiiiiii+itr(Re(A)TIm(A))-itr(Im(A)TRe(A))]=tr(Re(A)TRe(A))+tr(Im(A)TIm(A))=∥Re(A)∥F2+∥Im(A)∥F2.
That is, ∥A∥2=∥Re(A)∥F2+∥Im(A)∥F2.

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 x0 and set k:=0; calculate r0=b-Ax0, p0=r0.

Step 2. If rk=0 or rk≠0 and pk=0, stop; otherwise, calculate
(12)xk+1=xk+〈rk,rk〉〈pk,pk〉pk.

Step 3. Update the sequences
(13)rk+1=b-Axk+1,pk+1=rk+1+〈rk+1,rk+1〉〈rk,rk〉pk.

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 Ai, Bi, Di, Ei, and Fi, (i=1,2). Choose the initial matrices X1∈CSCm×m and Y1∈CSCm×m, S=(em,em-1,…,e1) in Definition 1. Compute
(14)R1∶=(R1(1)00R1(2)),R1(1)=D1X1¯E1+F1-A1X1-B1Y1,R1(2)=D2Y1¯E2+F2-A2Y1-B2X1,R~1∶=(R~1(1)00R~1(2)),R~1(1)=A1HR1(1)+B2HR1(2)-D1TR1(1)¯E1T,R~1(2)=A2HR1(2)+B1HR1(1)-D2TR1(2)¯E2T,M1=12(R~1(1)+SR~1(1)S),N1=12(R~1(2)+SR~1(2)S);
set k∶=1.

Step 2. If Rk=0 or Rk≠0, Mk=Nk=0, stop; otherwise, go to Step 3.

Step 3. Update the sequences
(15)Xk+1=Xk+αkMk,Yk+1=Yk+αkNk,Rk+1∶=(Rk+1(1)00Rk+1(2)),Rk+1(1)=D1Xk+1¯E1+F1-A1Xk+1-B1Yk+1,Rk+1(2)=D2Yk+1¯E2+F2-A2Yk+1-B2Xk+1,R~k+1∶=(R~k+1(1)00R~k+1(2)),R~k+1(1)=A1HRk+1(1)+B2HRk+1(2)-D1TRk+1(1)¯E1T,R~k+1(2)=A2HRk+1(2)+B1HRk+1(1)-D2TRk+1(2)¯E2T,Mk+1=12(R~k+1(1)+SR~k+1(1)S)+βkMk,Nk+1=12(R~k+1(2)+SR~k+1(2)S)+βkNk,
where
(16)αk∶=∥Rk∥2∥Mk∥2+∥Nk∥2,βk∶=∥Rk+1∥2∥Rk∥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 Ai, Bi, Di, Ei, and Fi, (i=1,2). Choose the initial matrix X1∈CASCm×m and Y1∈CASCm×m, S=(em,em-1,…,e1) in Definition 1. Compute
(17)R1∶=(R1(1)00R1(2)),R1(1)=D1X1¯E1+F1-A1X1-B1Y1,R1(2)=D2Y1¯E2+F2-A2Y-B2X1,R~1∶=(R~1(1)00R~1(2)),R~1(1)=A1HR1(1)+B2HR1(2)-D1TR1(1)¯E1T,R~1(2)=A2HR1(2)+B1HR1(1)-D2TR1(2)¯E2T,M1=12(R~1(1)-SR~1(1)S),N1=12(R~1(2)-SR~1(2)S);
set k∶=1.

Step 2. If Rk=0 or Rk≠0, Mk=Nk=0, stop; otherwise, go to Step 3.

Step 3. Update the sequences
(18)Xk+1=Xk+αkMk,Yk+1=Yk+αkNk,Rk+1∶=(Rk+1(1)00Rk+1(2)),Rk+1(1)=D1Xk+1¯E1+F1-A1Xk+1-B1Yk+1,Rk+1(2)=D2Yk+1¯E2+F2-A2Yk+1-B2Xk+1,R~k+1∶=(R~k+1(1)00R~k+1(2)),R~k+1(1)=A1HRk+1(1)+B2HRk+1(2)-D1TRk+1(1)¯E1T,R~k+1(2)=A2HRk+1(2)+B1HRk+1(1)-D2TRk+1(2)¯E2T,(19)Mk+1=12(R~k+1(1)-SR~k+1(1)S)+βkMk,Nk+1=12(R~k+1(2)-SR~k+1(2)S)+βkNk,
where
(20)αk:=∥Rk∥2∥Mk∥2+∥Nk∥2,βk:=∥Rk+1∥2∥Rk∥2.Step 4. Set k:=k+1; return to Step 2.

Now, we will show that the sequence matrix pair {Xk,Yk} 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 {Rk}, {Mk}, {Nk}, {R~j(1)}, {R~j(2)}, and {αk} generated by Algorithm 3; then have
(21)〈Rk+1,Rj〉=〈Rk,Rj〉-αk(〈Mk,R~j(1)〉+〈Nk,R~j(2)〉),〈Mk,R~j(1)〉k,j=1,2,….

Proof.

By Algorithm 3 and (20), we get
(22)Rk+1(1)=D1Xk+1¯E1+F1-A1Xk+1-B1Yk+1=D1(Xk¯+αkMk¯)E1+F1-A1(Xk+αkMk)-B1(Yk+αkNk)=D1Xk¯E1+F1-A1Xk-B1Yk+αk(D1Mk¯E1-A1Mk-B1Nk)=Rk(1)+αk(D1Mk¯E1-A1Mk-B1Nk).
In a similar way, we can get
(23)Rk+1(2)=Rk(2)+αk(D2Nk¯E2-A2Nk-B2Mk).
This together with the definition of inner product yields that
(24)〈Rk+1(1),Rj(1)〉=〈Rk(1),Rj(1)〉+αk(〈D1Mk¯E1,Rj(1)〉-〈A1Mk,Rj(1)〉+α-〈B1Nk,Rj(1)〉)=〈Rk(1),Rj(1)〉+αk×{Re[tr(E1HMkTD1HRj(1))]-Re[tr(MkHA1HRj(1))]+α-Re[tr(NkHB1HRj(1))]}=〈Rk(1),Rj(1)〉+αk×{Re[tr(E1HMkTD1HRj(1)¯)]-〈Mk,A1HRj(1)〉+α-〈Nk,B1HRj(1)〉[tr(E1HMkTD1HRj(1)¯)]}=〈Rk(1),Rj(1)〉+αk×{Re[tr(E1TMkHD1TRj(1)¯)]-〈Mk,A1HRj(1)〉+α-〈Nk,B1HRj(1)〉[tr(E1TMkHD1TRj(1)¯)]}=〈Rk(1),Rj(1)〉+αk×{Re[tr(MkHD1TRj(1)¯E1T)]-〈Mk,A1HRj(1)〉+α-〈Nk,B1HRj(1)〉[tr(MkHD1TRj(1)¯E1T)]}=〈Rk(1),Rj(1)〉+αk×(〈Mk,D1TRj(1)¯E1T〉-〈Mk,A1HRj(1)〉-〈Nk,B1HRj(1)〉)=〈Rk(1),Rj(1)〉+αk×(〈Mk,D1TRj(1)¯E1T-A1HRj(1)〉-〈Nk,B1HRj(1)〉),〈Rk+1(2),Rj(2)〉=〈Rk(2),Rj(2)〉+αk×(〈D2Nk¯E2,Rj(2)〉-〈A2Nk,Rj(2)〉-〈B2Mk,Rj(2)〉)=〈Rk(2),Rj(2)〉+αk×{Re[tr(E2HNkTD2HRj(2))]-Re[tr(NkHA2HRj(2))]×-Re[tr(MkHB2HRj(2))]}=〈Rk(2),Rj(2)〉+αk×{Re[tr(E2HNkTD2HRj(2)¯)]-〈Nk,A2HRj(2)〉×-〈Mk,B2HRj(2)〉[tr(E2HNkTD2HRj(2)¯)]}=〈Rk(2),Rj(2)〉+αk×{Re[tr(E2TNkHD2TRj(2)¯)]-〈Nk,A2HRj(2)〉×-〈Mk,B2HRj(2)〉[tr(E2TNkHD2TRj(2)¯)]}=〈Rk(2),Rj(2)〉+αk×{Re[tr(NkHD2TRj(2)¯E2T)]-〈Nk,A2HRj(2)〉×-〈Mk,B2HRj(2)〉[tr(NkHD2TRj(2)¯E2T)]}=〈Rk(2),Rj(2)〉+αk×(〈Nk,D2TRj(2)¯E2T〉-〈Nk,A2HRj(2)〉-〈Mk,B2HRj(2)〉)=〈Rk(2),Rj(2)〉+αk×(〈Nk,D2TRj(2)¯E2T-A2HRj(2)〉-〈Mk,B2HRj(2)〉).
Then by the updated formulas of Rk, R~j(1), and R~j(2), we obtain
(25)〈Rk+1,Rj〉=〈Rk+1(1),Rj(1)〉+〈Rk+1(2),Rj(2)〉=〈Rk(1),Rj(1)〉+αk×(〈Mk,D1TRj(1)¯E1T-A1HRj(1)〉-〈Nk,B1HRj(1)〉)+〈Rk(2),Rj(2)〉+αk×(〈Nk,D2TRj(2)¯E2T-A2HRj(2)〉-〈Mk,B2HRj(2)〉)=〈Rk,Rj〉+αk×(〈Mk,D1TRj(1)¯E1T-A1HRj(1)-B2HRj(2)〉+〈Nk,D2TRj(2)¯E2T-A2HRj(2)-B1HRj(1)〉)=〈Rk,Rj〉-αk(〈Mk,R~j(1)〉+〈Nk,R~j(2)〉),
which completes the proof.

Lemma 6.

Let the sequences, {Rk}, {Mk}, and {Nk}, be generated by Algorithm 3; one has
(26)〈Ri,Rj〉=0,〈Mi,Mj〉+〈Ni,Nj〉=0,iiiiii,j=1,2,…,k,i≠j.

Proof.

Firstly, we prove
(27)〈Ri,Rj〉=0,〈Mi,Mj〉+〈Ni,Nj〉=0,1≤j<i≤k.
By mathematical induction, for k=2, by Lemma 5, and noticing Mj∈CSCm×m, Nj∈CSCm×m(j=1,2,…,k) generated by Algorithm 3, we get
(28)〈R2,R1〉=〈R1,R1〉-α1(〈M1,R~1(1)〉+〈N1,R~1(2)〉)=∥R1∥2-α1(〈M1,R~1(1)+SR~1(1)S2〉=∥R1∥2-α1+〈N1,R~1(2)+SR~1(2)S2〉)=∥R1∥2-∥R1∥2∥M1∥2+∥N1∥2(〈M1,M1〉+〈N1,N1〉)=∥R1∥2-∥R1∥2=0,
where the second equality is from the fact
(29)〈M1,SR~1(1)S〉=〈M1,R~1(1)〉,〈N1,SR~1(2)S〉=〈N1,R~1(2)〉.
In addition, by (19), (20), and Lemma 5, we have
(30)〈M2,M1〉+〈N2,N1〉=〈R~2(1)+SR~2(1)S2+β1M1,M1〉+〈R~2(2)+SR~2(2)S2+β1N1,N1〉=〈R~2(1),M1〉+〈R~2(2),N1〉+β1(∥M1∥2+∥N1∥2)=1α1(〈R1,R2〉-〈R2,R2〉)+β1(∥M1∥2+∥N1∥2)=0,
where the second equality is from (6) and the fact
(31)〈M1,SR~2(1)S〉=〈M1,R~2(1)〉,〈N1,SR~2(2)S〉=〈N1,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)〈Rl+1,Rl〉=〈Rl,Rl〉-αl(〈Ml,R~l(1)〉+〈Nl,R~l(2)〉)=∥Rl∥2-αl×(〈Ml,R~l(1)+SR~l(1)S2〉+〈Nl,R~l(2)+SR~l(2)S2〉)=∥Rl∥2-αl×(〈Ml,Ml-βl-1Ml-1〉+〈Nl,Nl-βl-1Nl-1〉)=∥Rl∥2-αl(∥Ml∥2+∥Nl∥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)〈Ml+1,Ml〉+〈Nl+1,Nl〉=〈R~l+1(1)+SR~l+1(1)S2+βlMl,Ml〉+〈R~l+1(2)+SR~l+1(2)S2+βlNl,Nl〉=〈R~l+1(1),Ml〉+〈R~l+1(2),Nl〉+βl(∥Ml∥2+∥Nl∥2)=1αl(〈Rl,Rl+1〉-〈Rl+1,Rl+1〉)+βl(∥Ml∥2+∥Nl∥2)=0,
where the third equality is from Lemma 5.

For j=1, by Lemma 5 and the induction, we have
(34)〈Rl+1,R1〉=〈Rl,R1〉-αl(〈Ml,R~1(1)〉+〈Nl,R~1(2)〉)=-αl(〈Ml,R~1(1)+SR~1(1)S2〉+〈Nl,R~1(2)+SR~1(2)S2〉)=-αl(〈Ml,M1〉+〈Nl,N1〉)=0.
Analogously, for j=2,3,…,l-1, then we obtain
(35)〈Rl+1,Rj〉=〈Rl,Rj〉-αl(〈Ml,R~j(1)〉+〈Nl,R~j(2)〉)=-αl(〈Ml,R~j(1)+SR~j(1)S2〉=-αl+〈Nl,R~j(2)+SR~j(2)S2〉)=-αl(〈Ml,Mj-βj-1Mj-1〉M,M+〈Nl,Nj-βj-1Nj-1〉)=-αl(〈Ml,Mj〉+〈Nl,Nj〉-βj-1(〈Ml,Mj-1〉+〈Nl,Nj-1〉)-αl×(〈Ml,Mj-1〉+〈Nl,Nj-1〉))=0.
In addition, from Lemma 5 and the induction, for j=1,2,…,l-1, we get
(36)〈Ml+1,Mj〉+〈Nl+1,Nj〉=〈R~l+1(1)+SR~l+1(1)S2,Mj〉+〈R~l+1(2)+SR~l+1(2)S2,Nj〉+βl(〈Ml,Mj〉+〈Nl,Nj〉)=〈R~l+1(1),Mj〉+〈R~l+1(2),Nj〉=1αl(〈Rj,Rl+1〉-〈Rj+1,Rl+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)〈Ri,Rj〉=〈Rj,Ri〉=0,〈Mi,Mj〉+〈Ni,Nj〉=〈Mj,Mi〉+〈Nj,Ni〉=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 X1∈CMCm×m, Y1∈CMCm×m, the sequences {Xk}, {Yk}, {Rk}, {Mk}, and {Nk} generated by Algorithm 3 satisfy
(38)〈X*-Xk,Mk〉+〈Y*-Yk,Nk〉=∥Rk∥2,k=1,2,…,n.

Proof.

The conclusion is accomplished by mathematical induction.

Firstly, we notice that the sequences pair (Xk,Yk), (k=1,2,…) generated by Algorithm 3 are all central symmetric matrices since initial matrix pair (X1,Y1) is centrally symmetric matrix. Then for k=1, it follows from Algorithm 3 that
(39)〈X*-X1,M1〉=〈X*-X1,R~1(1)+SR~1(1)S2〉=〈X*-X1,R~1(1)〉=〈X*-X1,A1HR1(1)+B2HR1(2)-D1TR1(1)¯E1T〉=〈X*-X1,A1HR1(1)〉+〈X*-X1,B2HR1(2)〉-〈X*-X1,D1TR1(1)¯E1T〉=Re[tr((X*-X1)HA1HR1(1))]+Re[tr((X*-X1)HB2HR1(2))]-Re[tr((X*-X1)HD1TR1(1)¯E1T)]=〈A1(X*-X1),R1(1)〉+〈B2(X*-X1),R1(2)〉-Re[tr((X*-X1)HD1TR1(1)¯E1T¯)]=〈A1(X*-X1),R1(1)〉+〈B2(X*-X1),R1(2)〉-Re[tr((X*-X1)TD1HR1(1)E1H)]=〈A1(X*-X1),R1(1)〉+〈B2(X*-X1),R1(2)〉-Re[tr(E1H(X*-X1)TD1HR1(1))]=〈A1(X*-X1),R1(1)〉+〈B2(X*-X1),R1(2)〉-〈D1(X*-X1¯)E1,R1(1)〉=〈A1(X*-X1)-D1(X*-X1¯)E1,R1(1)〉+〈B2(X*-X1),R1(2)〉.
In the same way, we can get
(40)〈Y*-Y1,N1〉=〈A2(Y*-Y1)-D2(Y*-Y1¯)E2,R1(2)〉+〈B1(Y*-Y1),R1(1)〉.
This shows that
(41)〈X*-X1,M1〉+〈Y*-Y1,N1〉=〈A1(X*-X1)-D1(X*-X1¯)E1,R1(1)〉+〈B2(X*-X1),R1(2)〉+〈A2(Y*-Y1)-D2(Y*-Y1¯)E2,R1(2)〉+〈B1(Y*-Y1),R1(1)〉=〈A1(X*-X1)+B1(Y*-Y1)R1(1)-D1(X*-X1¯)E1,R1(1)〉+〈A2(Y*-Y1)+B2(X*-X1)R1(1)-D2(Y*-Y1¯)E2,R1(2)〉=〈F1-A1X1-B1Y1+D1X1¯E1,R1(1)〉+〈F2-A2Y1-B2X1+D2Y1¯E2,R1(2)〉=∥R1(1)∥2+∥R1(2)∥2=∥R1∥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 Xl+1, Yl+1 that
(42)〈X*-Xl+1,Ml〉=〈X*-Xl-αlMl,Ml〉=〈X*-Xl,Ml〉-αl∥Ml∥2,〈Y*-Yl+1,Nl〉=〈Y*-Yl-αlNl,Nl〉=〈Y*-Yl,Nl〉-αl∥Nl∥2.
Then
(43)〈X*-Xl+1,Ml+1〉=〈X*-Xl+1,R~l+1(1)+SR~l+1(1)S2+βlMl〉=〈X*-Xl+1,R~l+1(1)〉+βl〈X*-Xl+1,Ml〉=〈X*-Xl+1,R~l+1(1)〉+βl(〈X*-Xl,Ml〉-αl∥Ml∥2),〈Y*-Yl+1,Nl+1〉=〈Y*-Yl+1,R~l+1(2)+SR~l+1(2)S2+βlNl〉=〈Y*-Yl+1,R~l+1(2)〉+βl〈Y*-Yl+1,Nl〉=〈Y*-Yl+1,R~l+1(2)〉+βl(〈Y*-Yl,Nl〉-αl∥Nl∥2).
On the other hand, we have
(44)〈X*-Xl+1,R~l+1(1)〉+〈Y*-Yl+1,R~l+1(2)〉=〈X*-Xl+1,A1HRl+1(1)+B2HRl+1(2)-D1TRl+1(1)¯E1T〉+〈Y*-Yl+1,A2HRl+1(2)+B1HRl+1(1)-D2TRl+1(2)¯E2T〉=〈A1(X*-Xl+1),Rl+1(1)〉+〈B2(X*-Xl+1),Rl+1(2)〉-Re[tr((X*-Xl+1)HD1TRl+1(1)¯E1T)]+〈A2(Y*-Yl+1),Rl+1(2)〉+〈B1(Y*-Yl+1),Rl+1(1)〉-Re[tr((Y*-Yl+1)HD2TRl+1(2)¯E2T)]=〈A1(X*-Xl+1)+B1(Y*-Yl+1),Rl+1(1)〉-Re[tr((X*-Xl+1)TD1HRl+1(1)E1H)]+〈A2(Y*-Yl+1)+B2(X*-Xl+1),Rl+1(2)〉-Re[tr((Y*-Yl+1)TD2HRl+1(2)E2H)]=〈A1(X*-Xl+1)+B1(Y*-Yl+1),Rl+1(1)〉-Re[tr(E1H(X*-Xl+1)TD1HRl+1(1))]+〈A2(Y*-Yl+1)+B2(X*-Xl+1),Rl+1(2)〉-Re[tr(E2H(Y*-Yl+1)TD2HRl+1(2))]=〈A1(X*-Xl+1)+B1(Y*-Yl+1)Rl+1(1)-D1(X*-Xl+1¯)E1,Rl+1(1)〉+〈A2(Y*-Yl+1)+B2(X*-Xl+1)Rl+1(2)-D2(Y*-Yl+1¯)E2,Rl+1(2)〉=〈F1-A1Xl+1-B1Yl+1+D1Xl+1¯E1,Rl+1(1)〉+〈F2-A2Yl+1-B2Xl+1+D2Yl+1¯E2,Rl+1(2)〉=∥Rl+1(1)∥2+∥Rl+1(2)∥2=∥Rl+1∥2.
Therefore, by (20) we get
(45)〈X*-Xl+1,Ml+1〉+〈Y*-Yl+1,Nl+1〉=〈X*-Xl+1,R~l+1(1)+SR~l+1(1)S2+βlMl〉+〈Y*-Yl+1,R~l+1(2)+SR~l+1(2)S2+βlNl〉=〈X*-Xl+1,R~l+1(1)〉+βl(〈X*-Xl,Ml〉-αl∥Ml∥2)+〈Y*-Yl+1,R~l+1(2)〉+βl(〈Y*-Yl,Nl〉-αl∥Nl∥2)=〈X*-Xl+1,R~l+1(1)〉+〈Y*-Yl+1,R~l+1(2)〉+βl[〈X*-Xl,Ml〉+〈Y*-Yl,Nl〉(∥Ml∥2+∥Nl∥2)〈X*Xl,Ml〉-αl(∥Ml∥2+∥Nl∥2)]=〈X*-Xl+1,R~l+1(1)〉+〈Y*-Yl+1,R~l+1(2)〉+βl[∥Rl∥2-∥Rl∥2∥Ml∥2+∥Nl∥2(∥Ml∥2+∥Nl∥2)]=〈X*-Xl+1,R~l+1(1)〉+〈Y*-Yl+1,R~l+1(2)〉=∥Rl+1∥2.
Hence, the proof is completed.

Remark 8.

Lemma 7 implies that if (4) is consistent, then ∥Mk∥2+∥Nk∥2≠0 when Rk≠0. Conversely, if there exists a positive integer k0 such that Rk0≠0 and ∥Mk0∥2+∥Nk0∥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 Definition 1). Hence, we need not show these results in detail; in the following content, we only discuss the version when X,Y∈CMCm×m.

Theorem 10.

Suppose the system (4) is consistent; then, for any initial matrix X1∈CMCm×m, Y1∈CMCm×m, an exact solution of (4) can be derived at most 2pm+1 iteration steps by Algorithm 3.

Proof.

Assume Rk≠0, for k=1,2,…,2pm. It follows from Lemma 7 that ∥Mk∥2+∥Nk∥2≠0 for k=1,2,…,2pm. Then R2pm+1 can be derived by Algorithm 3. According to Lemma 6, we know 〈Ri,Rj〉=0, for i,j=1,2,…,2pm+1, i≠j. Then the matrix sequence of R1,R2,…,R2pm is an orthogonal basis of the linear space
(46)H={H∣H=(H100H2)},
where H1,H2∈Rp×m. Since R2pm+1∈H and 〈R2pm+1,Rk〉=0, for k=1,2,…,2pm, hence R2pm+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.

Suppose A∈Rm×n, b∈Rm, and the linear matrix equation Ax=b has a solution x*∈R(AT); then x* is the unique least Frobenius norm solution of Ax=b.

For a rigorous proof of this lemma above the reader is referred to [46, 48].

Lemma 12.

Suppose A∈Cm×n, b∈Cm, and the linear matrix equation Ax=b has a solution x*∈Rn. If x*∈R[(Re(A)T,Im(A)T)], then x* is the unique least Frobenius norm solution of Ax=b.

Proof.

Let A=Re(A)+iIm(A) and b=Re(b)+iIm(b). Then Ax=b can be written as
(47)(Re(A)+iIm(A))x=Re(A)x+iIm(A)x=Re(b)+iIm(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).

Let X=Re(X)+iIm(X), Y=Re(Y)+iIm(Y). Then the problem (4) can be equivalently written as
(51)A1(Re(X)+iIm(X))+B1(Re(Y)+iIm(Y))=D1(Re(X)-iIm(X))E1+F1,A2(Re(Y)+iIm(Y))+B2(Re(X)+iIm(X))=D2(Re(Y)-iIm(Y))E2+F2,
or
(52)A1Re(X)-D1Re(X)E1+iA1Im(X)+iD1Im(X)E1+B1Re(Y)+iB1Im(Y)=F1,A2Re(Y)-D2Re(Y)E2+iA2Im(Y)+iD2Im(Y)E2+B2Re(X)+iB2Im(X)=F2.
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(F1)vec(F2)),
where(54)W=(I⊗A1-E1T⊗D1i(I⊗A1+E1T⊗D1)I⊗B1i(I⊗B1)I⊗B2i(I⊗B2)I⊗A2-E2T⊗D2i(I⊗A2+E2T⊗D2)).
By some simple calculating, we have(55)W=Re(W)+iIm(W),
where
(56)Re(W)=(N11N12N13N14N21N22N23N24),(57)Im(W)=(L11L12L13L14L21L22L23L24),(58)N11=I⊗Re(A1)-Re(E1)T⊗Re(D1)+Im(E1)T⊗Im(D1),N12=-I⊗Im(A1)-Im(E1)T⊗Re(D1)-Re(E1)T⊗Im(D1),N13=I⊗Re(B1),N14=-I⊗Im(B1),N21=I⊗Re(B2),N22=-I⊗Im(B2),N23=I⊗Re(A2)-Re(E2)T⊗Re(D2)+Im(E2)T⊗Im(D2),N24=-I⊗Im(A2)-Im(E2)T⊗Re(D2)-Re(E2)T⊗Im(D2),L11=I⊗Im(A1)-Im(E1)T⊗Re(D1)-Re(E1)T⊗Im(D1),L12=I⊗Re(A1)+Re(E1)T⊗Re(D1)-Im(E1)T⊗Im(D1),L13=I⊗Im(B1),L14=I⊗Re(B1),L21=I⊗Im(B2),L22=I⊗Re(B2),L23=I⊗Im(A2)-Im(E2)T⊗Re(D2)-Re(E2)T⊗Im(D2),L24=I⊗Re(A2)+Re(E2)T⊗Re(D2)-Im(E2)T⊗Im(D2).
Particularly, when X∈CSCm×m, X can be substituted with (X+SXS)/2 in (51); then we obtain
(59)W¯·(vec[Re(X)]vec[Im(X)]vec[Re(Y)]vec[Im(Y)])=(vec(F1)vec(F2)),
where W¯∶=(1/2)[(I+K)(Re(W)T,(I+K)(Im(W)T)]T, K:=Diag{S¯,S¯,S¯,S¯}, S¯∶=ST⊗S=S⊗S∈Rm2×m2 (see Definition 1).

Obviously, if x∈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.

Let W¯∈C4m2×4m2, b∈C4m2, and the linear matrix equation W¯x=b has a solution x*∈R4m2, 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)X1=A1HU1+B2HV1-D1TU1¯E1T+SA1HU1S+SB2HV1S-SD1TU1¯E1TS,(61)Y1=A2HV1+B1HU1-D2TV1¯E2T+SA2HV1S+SB1HU1S-SD2TV1¯E2TS,
where U1,V1∈Cp×m are two arbitrary matrices, especially, taking X1=Y1=0∈Rm×m, the solution (X*,Y*) given by Algorithm 3 is the unique least Frobenius norm solution of (4).

Proof.

If X1,Y1 has the form of (60), (61), respectively, by Step2 of Algorithm 3, we have
(62)X2=X1+α1M1=X1+α1(12(R~1(1)+SR~1(1)S))=A1HU1+B2HV1-D1TU1¯E1T+SA1HU1S+SB2HV1S-SD1TU1¯E1TS+α1[12(A1HR1(1)+B2HR1(2)-D1TR1(1)¯E1T+SA1HR1(1)Siiiiiiiiiiiiiii+SB2HR1(2)S-SD1TR1(1)¯E1TS)12]=A1H(U1+12α1R1(1))+B2H(V1+12α1R1(2))-D1T(U1+12α1R1(1)¯)E1T+SA1H(U1+12α1R1(1))S+SB2H(V1+12α1R1(2))S-SD1T(U1+12α1R1(1)¯)E1TS∶=A1HU2+B2HV2-D1TU2¯E1T+SA1HU2S+SB2HV2S-SD1TU2¯E1TS,Y2=Y1+α1N1=Y1+α1(12(R~1(2)+SR~1(2)S))=A2HV1+B1HU1-D2TV1¯E2T+SA2HV1S+SB1HU1S-SD2TV1¯E2TS+α1[12(A2HR1(2)+B1HR1(1)-D2TR1(2)¯E2T+SA2HR1(2)S+SB1HR1(1)S-SD2TR1(2)¯E2TS)12]=A2H(V1+12α1R1(2))+B1H(U1+12α1R1(1))-D2T(V1+12α1R1(2)¯)E2T+SA2H(V1+12α1R1(2))S+SB1H(U1+12α1R1(1))S-SD2T(V1+12α1R1(2)¯)E2TS∶=A2HV2+B1HU2-D2TV2¯E2T+SA2HV2S+SB1HU2S-SD2TV2¯E2TS,
where U2∶=U1+(1/2)α1R1(1) and V2∶=V1+(1/2)α1R1(2). Since
(63)M2=12(R~2(1)+SR~2(1)S)+β1M1=12(A1HR2(1)+B2HR2(2)-D1TR2(1)¯E1T+SA1HR2(1)S+SB2HR2(2)S-SD1TR2(1)¯E1TS)+β12(A1HR1(1)+B2HR1(2)-D1TR1(1)¯E1T+SA1HR1(1)S+SB2HR1(2)S-SD1TR1(1)¯E1TS),N2=12(R~2(2)+SR~2(2)S)+β1N1=12(A1HR2(2)+B2HR2(1)-D1TR2(2)¯E1T+SA1HR2(2)S+SB2HR2(1)S-SD1TR2(2)¯E1TS)+β12(A1HR1(2)+B2HR1(1)-D1TR1(2)¯E1T+SA1HR1(2)Siiiiiiiiiii+SB2HR1(1)S-SD1TR1(2)¯E1TS),
we have
(64)X3=X2+α2M2=A1HU3+B2HV3-D1TU3¯E1T+SA1HU3S+SB2HV3S-SD1TU3¯E1TS,Y3=Y2+α2N2=A2HV3+B1HU3-D2TV3¯E2T+SA2HV3S+SB1HU3S-SD2TV3¯E2TS,
where U3∶=U2+(α2/2)R2(1)+(β1α2/2)R1(1) and V3∶=V2+(α2/2)R2(2)+(β1α2/2)R1(2). By parity of reasoning, we can prove that
(65)Xk+1=A1HUk+1+B2HVk+1-D1TUk+1¯E1T+SA1HUk+1S+SB2HVk+1S-SD1TUk+1¯E1TS,Yk+1=A2HVk+1+B1HUk+1-D2TVk+1¯E2T+SA2HVk+1S+SB1HUk+1S-SD2TVk+1¯E2TS,
where Uk+1,Vk+1∈Cp×m. This together with Theorem 10 yields that
(66)Xk+1⟶X*=A1HU^+B2HV^-D1TU^¯E1T+SA1HU^S+SB2HV^S-SD1TU^¯E1TS(k⟶∞),Yk+1⟶Y*=A2HV^+B1HU^-D2TV^¯E2T+SA2HV^S+SB1HU^S-SD2TV^¯E2TS(k⟶∞),
where U^,V^∈Rp×m. Since
(67)X*=A1HU^+B2HV^-D1TU^¯E1T+SA1HU^S+SB2HV^S-SD1TU^¯E1TS=[Re(A1)+iIm(A1)]H[Re(U^)+iIm(U^)]+[Re(B2)+iIm(B2)]H[Re(V^)+iIm(V^)]-[Re(D1)+iIm(D1)]T[Re(U^)+iIm(U^)¯]×[Re(E1)+iIm(E1)]T+S[Re(A1)+iIm(A1)]H×[Re(U^)+iIm(U^)]S+S[Re(B2)+iIm(B2)]H×[Re(V^)+iIm(V^)]S-S[Re(D1)+iIm(D1)]T×[Re(U^)+iIm(U^)¯][Re(E1)+iIm(E1)]TS=[Re(A1)T-iIm(A1)T][Re(U^)+iIm(U^)]+[Re(B2)T-iIm(B2)T][Re(V^)+iIm(V^)]-[Re(D1)T+iIm(D1)T][Re(U^)-iIm(U^)]×[Re(E1)T+iIm(E1)T]+S[Re(A1)T-iIm(A1)T]×[Re(U^)+iIm(U^)]S+S[Re(B2)T-iIm(B2)T]×[Re(V^)+iIm(V^)]S-S[Re(D1)T+iIm(D1)T]×[Re(U^)-iIm(U^)][Re(E1)T+iIm(E1)T]S=Re(A1)TRe(U^)+iRe(A1)TIm(U^)-iIm(A1)TRe(U^)+Im(A1)TIm(U^)+Re(B2)TRe(V^)+iRe(B2)