# An Iterative Algorithm for the Generalized Reflexive Solution of the Matrix EquationsAXB=E, CXD=F

Authors: Deqin Chen; Feng Yin; Guang-Xin Huang
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

An iterative algorithm is constructed to solve the linear matrix equation pairAXB=E,CXD=F over generalized reflexive matrix X. When the matrix equation pair AXB=E,CXD=F is consistent over generalized reflexive matrix X, for any generalized reflexive initial iterative matrix X1, the generalized reflexive solution can be obtained by the iterative algorithm within finite iterative steps in the absence of round-off errors. The unique least-norm generalized reflexive iterative solution of the matrix equation pair can be derived when an appropriate initial iterative matrix is chosen. Furthermore, the optimal approximate solution of AXB=E,CXD=F for a given generalized reflexive matrix X0 can be derived by finding the least-norm generalized reflexive solution of a new corresponding matrix equation pair AX̃B=Ẽ,CX̃D=F̃ with Ẽ=E-AX0B,F̃=F-CX0D. Finally, several numerical examples are given to illustrate that our iterative algorithm is effective.


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## Body

## 1. Introduction

Letℛm×n denote the set of all m-by-n real matrices. In denotes the n order identity matrix. Let P∈ℛm×m and Q∈ℛn×n be two real generalized reflection matrices, that is,  PT=P,P2=Im,QT=Q,Q2=In. A matrix A∈ℛm×n is called generalized reflexive matrix with respect to the matrix pair (P,Q) if PAQ=A. For more properties and applications on generalized reflexive matrix, we refer to [1, 2]. The set of all m-by-n real generalized reflexive matrices with respect to matrix pair (P,Q) is denoted by ℛrm×n(P,Q).We denote by the superscripts T the transpose of a matrix. In matrix space ℛm×n, define inner product as tr(BTA)=trace(BTA) for all A,B∈ℛm×n;  ∥A∥ represents the Frobenius norm of A. ℛ(A) represents the column space of A. vec(·) represents the vector operator; that is, vec(A)=(a1T,a2T,…,anT)T∈ℛmn for the matrix A=(a1,a2,…,an)∈ℛm×n,ai∈Rm,i=1,2,…,n. A⊗B stands for the Kronecker product of matrices A and B.In this paper, we will consider the following two problems.Problem 1.
For given matricesA∈ℛp×m,B∈ℛn×q,C∈ℛs×m, D∈ℛn×t,E∈ℛp×q,F∈ℛs×t, find matrix X∈ℛrm×n(P,Q) such that
(1.1)AXB=E,CXD=F.Problem 2.
When Problem1 is consistent, let SE denote the set of the generalized reflexive solutions of Problem 1. For a given matrix X0∈ℛrm×n(P,Q), find X^∈SE such that
(1.2)‖X^-X0‖=min⁡X∈SE‖X-X0‖.The matrix equation pair (1.1) may arise in many areas of control and system theory. Dehghan and Hajarian [3] presented some examples to show a motivation for studying (1.1). Problem 2 occurs frequently in experiment design; see for instance [4]. In recent years, the matrix nearness problem has been studied extensively (e.g., [3, 5–19]).Research on solving the matrix equation pair (1.1) has been actively ongoing for last 40 or more years. For instance, Mitra [20, 21] gave conditions for the existence of a solution and a representation of the general common solution to the matrix equation pair (1.1). Shinozaki and Sibuya [22] and vander Woude[23] discussed conditions for the existence of a common solution to the matrix equation pair (1.1). Navarra et al. [11] derived sufficient and necessary conditions for the existence of a common solution to (1.1). Yuan [18] obtained an analytical expression of the least-squares solutions of (1.1) by using the generalized singular value decomposition (GSVD) of matrices. Recently, some finite iterative algorithms have also been developed to solve matrix equations. Deng et al. [24] studied the consistent conditions and the general expressions about the Hermitian solutions of the matrix equations (AX,XB)=(C,D) and designed an iterative method for its Hermitian minimum norm solutions. Li and Wu [25] gave symmetric and skew-antisymmetric solutions to certain matrix equations A1X=C1,XB3=C3 over the real quaternion algebra H. Dehghan and Hajarian [26] proposed the necessary and sufficient conditions for the solvability of matrix equations A1XB1=D1,A1X=C1,XB2=C2 and A1X=C1,XB2=C2,A3X=C3,XB4=C4 over the reflexive or antireflexive matrix X and obtained the general expression of the solutions for a solvable case. Wang [27, 28] gave the centrosymmetric solution to the system of quaternion matrix equations A1X=C1,A3XB3=C3. Wang [29] also solved a system of matrix equations over arbitrary regular rings with identity. For more studies on iterative algorithms on coupled matrix equations, we refer to [6, 7, 15–17, 19, 30–34]. Peng et al. [13] presented iterative methods to obtain the symmetric solutions of (1.1). Sheng and Chen [14] presented a finite iterative method when (1.1) is consistent. Liao and Lei [9] presented an analytical expression of the least-squares solution and an algorithm for (1.1) with the minimum norm. Peng et al. [12] presented an efficient algorithm for the least-squares reflexive solution. Dehghan and Hajarian [3] presented an iterative algorithm for solving a pair of matrix equations (1.1) over generalized centrosymmetric matrices. Cai and Chen [35] presented an iterative algorithm for the least-squares bisymmetric solutions of the matrix equations (1.1). However, the problem of finding the generalized reflexive solutions of matrix equation pair (1.1) has not been solved. In this paper, we construct an iterative algorithm by which the solvability of Problem 1 can be determined automatically, the solution can be obtained within finite iterative steps when Problem 1 is consistent, and the solution of Problem 2 can be obtained by finding the least-norm generalized reflexive solution of a corresponding matrix equation pair.This paper is organized as follows. In Section2, we will solve Problem 1 by constructing an iterative algorithm; that is, if Problem 1 is consistent, then for an arbitrary initial matrix X1∈ℛrm×n(P,Q), we can obtain a solution X-∈ℛrm×n(P,Q) of Problem 1 within finite iterative steps in the absence of round-off errors. Let X1=ATHBT+CTH^DT+PATHBTQ+PCTH^DTQ, where H∈ℛp×q,H^∈ℛs×t are arbitrary matrices, or more especially, letting X1=0∈ℛrm×n(P,Q), we can obtain the unique least norm solution of Problem 1. Then in Section 3, we give the optimal approximate solution of Problem 2 by finding the least norm generalized reflexive solution of a corresponding new matrix equation pair. In Section 4, several numerical examples are given to illustrate the application of our iterative algorithm.


## 2. The Solution of Problem1

In this section, we will first introduce an iterative algorithm to solve Problem1 and then prove that it is convergent. The idea of the algorithm and it’s proof in this paper are originally inspired by those in [13]. The idea of our algorithm is also inspired by those in [3]. When P=Q,R=S,XT=X and YT=Y, the results in this paper reduce to those in [3].Algorithm 2.1.
Step1. Input matrices A∈ℛp×m,B∈ℛn×q,C∈ℛs×m,  D∈ℛn×t,E∈ℛp×q,F∈ℛs×t, and two generalized reflection matrix P∈ℛm×m,Q∈ℛn×n.
Step2. Choose an arbitrary matrix X1∈ℛrm×n(P,Q). Compute
(2.1)R1=(E-AX1B00F-CX1D),P1=12(AT(E-AX1B)BT+CT(F-CX1D)DT+PAT(E-AX1B)BTQ+PCT(F-CX1D)DTQ),k:=1.
Step3. If R1=0, then stop. Else go to Step 4.
Step4. Compute
(2.2)Xk+1=Xk+‖Rk‖2‖Pk‖2Pk,Rk+1=(E-AXk+1B00F-CXk+1D)=Rk-‖Rk‖2‖Pk‖2(APkB00CPkD),Pk+1=12(AT(E-AXk+1B)BT+CT(F-CXk+1D)DT+PAT(E-AXk+1B)BTQ+PCT(F-CXk+1D)DTQ)+‖Rk+1‖2‖Rk‖2Pk.
Step5. If Rk+1=0, then stop. Else, letting k:=k+1, go to Step 4.Obviously, it can be seen thatPi∈Rrm×n(P,Q),Xi∈Rrm×n(P,Q), where i=1,2,….Lemma 2.2.
For the sequences{Ri}  and {Pi} generated in Algorithm 2.1, one has
(2.3)tr⁡(Ri+1TRj)=tr⁡(RiTRj)-‖Ri‖2‖Pi‖2tr⁡(PiTPj)+‖Ri‖2‖Rj‖2‖Pi‖2‖Rj-1‖2tr⁡(PiTPj-1),tr⁡(Pi+1TPj)=‖Pj‖2‖Rj‖2(tr⁡(Ri+1TRj)-tr⁡(Ri+1TRj+1))+‖Ri+1‖2‖Ri‖2tr⁡(PiTPj).Proof.
By Algorithm2.1, we have
(2.4)tr⁡(Ri+1TRj)=tr⁡((Ri-‖Ri‖2‖Pi‖2(APiB00CPiD))TRj)=tr⁡(RiTRj)-‖Ri‖2‖Pi‖2tr⁡((BTPiTAT00DTPiTCT)Rj)=tr⁡(RiTRj)-‖Ri‖2‖Pi‖2tr⁡((BTPiTAT00DTPiTCT)(E-AXjB00F-CXjD))=tr⁡(RiTRj)-‖Ri‖2‖Pi‖2tr⁡(BTPiTAT(E-AXjB)+DTPiTCT(F-CXjD))=tr⁡(RiTRj)-‖Ri‖2‖Pi‖2tr⁡(PiT(AT(E-AXjB)BT+CT(F-CXjD)DT))=tr⁡(RiTRj)-‖Ri‖2‖Pi‖2×tr⁡(PiT(AT(E-AXjB)BT+CT(F-CXjD)DT2+PAT(E-AXjB)BTQ+PCT(F-CXjD)DTQ2+AT(E-AXjB)BT+CT(F-CXjD)DT2+-PAT(E-AXjB)BTQ-PCT(F-CXjD)DTQ2))=tr⁡(RiTRj)-‖Ri‖2‖Pi‖2×tr⁡(PiT{AT(E-AXjB)BT+CT(F-CXjD)DT2+PAT(E-AXjB)BTQ+PCT(F-CXjD)DTQ2})=tr⁡(RiTRj)-‖Ri‖2‖Pi‖2tr⁡(PiT(Pj-‖Rj‖2‖Rj-1‖2Pj-1))=tr⁡(RiTRj)-‖Ri‖2‖Pi‖2tr⁡(PiTPj)+‖Ri‖2‖Rj‖2‖Pi‖2‖Rj-1‖2tr⁡(PiTPj-1),tr⁡(Pi+1TPj)=tr⁡((AT(E-AXi+1B)BT+CT(F-CXi+1D)DT2+PAT(E-AXi+1B)BTQ+PCT(F-CXi+1D)DTQ2+‖Ri+1‖2‖Ri‖2Pi)TPj)=tr⁡((AT(E-AXi+1B)BT+CT(F-CXi+1D)DT2+PAT(E-AXi+1B)BTQ+PCT(F-CXi+1D)DTQ2)TPj)+‖Ri+1‖2‖Ri‖2tr⁡(PiTPj)=tr⁡(PjT(AT(E-AXi+1B)BT+CT(F-CXi+1D)DT))+‖Ri+1‖2‖Ri‖2tr⁡(PiTPj)=tr⁡((E-AXi+1B)TAPjB+(F-CXi+1D)TCPjD)+‖Ri+1‖2‖Ri‖2tr⁡(PiTPj)=tr⁡(((E-AXi+1B)T00(F-CXi+1D)T)(APjB00CPjD))+‖Ri+1‖2‖Ri‖2tr⁡(PiTPj)=‖Pj‖2‖Rj‖2tr⁡(Ri+1T(Rj-Rj+1))+‖Ri+1‖2‖Ri‖2tr⁡(PiTPj)=‖Pj‖2‖Rj‖2(tr⁡(Ri+1TRj)-tr⁡(Ri+1TRj+1))+‖Ri+1‖2‖Ri‖2tr⁡(PiTPj).
This completes the proof.Lemma 2.3.
For the sequences{Ri} and {Pi} generated by Algorithm 2.1, and k≥2, one has
(2.5)tr⁡(RiTRj)=0,tr⁡(PiTPj)=0,i,j=1,2,…,k,i≠j.Proof.
Sincetr(RiTRj)=tr(RjTRi) and tr(PiTPj)=tr(PjTPi) for all i,j=1,2,…,k, we only need to prove that tr(RiTRj)=0,tr(PiTPj)=0 for all 1≤j<i≤k. We prove the conclusion by induction, and two steps are required.
Step1. We will show that
(2.6)tr⁡(Ri+1TRi)=0,tr⁡(Pi+1TPi)=0,i=1,2,…,k-1.
To prove this conclusion, we also use induction.
Fori=1, by Algorithm 2.1 and the proof of Lemma 2.2, we have that
(2.7)tr⁡(R2TR1)=tr⁡((R1-‖R1‖2‖P1‖2(AP1B00CP1D))TR1)=tr⁡(R1TR1)-‖R1‖2‖P1‖2×tr⁡(P1T{AT(E-AX1B)BT+CT(F-CX1D)DT2+PAT(E-AX1B)BTQ+PCT(F-CX1D)DTQ2})=‖R1‖2-‖R1‖2‖P1‖2tr⁡(P1TP1)=0,tr⁡(P2TP1)=‖P1‖2‖R1‖2(tr⁡(R2TR1)-tr⁡(R2TR2))+‖R2‖2‖R1‖2‖P1‖2=0.
Assume (2.6) holds for i=s-1, that is,  tr(RsTRs-1)=0,tr(PsTPs-1)=0. When i=s, by Lemma 2.2, we have that
(2.8)tr⁡(Rs+1TRs)=tr⁡(RsTRs)-‖Rs‖2‖Ps‖2tr⁡(PsTPs)+‖Rs‖4‖Ps‖2‖Rs-1‖2tr⁡(PsTPs-1)=‖Rs‖2-‖Rs‖2+‖Rs‖4‖Ps‖2‖Rs-1‖2tr⁡(PsTPs-1)=0,tr⁡(Ps+1TPs)=‖Ps‖2‖Rs‖2(tr⁡(Rs+1TRs)-tr⁡(Rs+1TRs+1))+‖Rs+1‖2‖Rs‖2tr⁡(PsTPs)=-‖Ps‖2‖Rs‖2‖Rs+1‖2+‖Rs+1‖2‖Rs‖2‖Ps‖2=0.
Hence, (2.6) holds for i=s. Therefor, (2.6) holds by the principle of induction.
Step2. Assuming that tr(RsTRj)=0,tr(PsTPj)=0,j=1,2,…,s-1, then we show that
(2.9)tr⁡(Rs+1TRj)=0,tr⁡(Ps+1TPj)=0,j=1,2,…,s.
In fact, by Lemma 2.2 we have
(2.10)tr⁡(Rs+1TRj)=tr⁡(RsTRj)-‖Rs‖2‖Ps‖2tr⁡(PsTPj)+‖Rs‖2‖Rj‖2‖Ps‖2‖Rj-1‖2tr⁡(PsTPj-1)=0.
From the previous results, we have tr(Rs+1TRj+1)=0. By Lemma 2.2 we have that
(2.11)tr⁡(Ps+1TPj)=‖Pj‖2‖Rj‖2(tr⁡(Rs+1TRj)-tr⁡(Rs+1TRj+1))+‖Rs+1‖2‖Rs‖2tr⁡(PsTPj)=‖Pj‖2‖Rj‖2(tr⁡(Rs+1TRj)-tr⁡(Rs+1TRj+1))=0.
By the principle of induction, (2.9) holds. Note that (2.5) is implied in Steps 1 and 2 by the principle of induction. This completes the proof.Lemma 2.4.
SupposingX- is an arbitrary solution of Problem 1, that is, AX-B=E and CX-D=F, then
(2.12)tr⁡((X--Xk)TPk)=‖Rk‖2,k=1,2,…,
where the sequences {Xk}, {Rk}, and {Pk} are generated by Algorithm 2.1.Proof.
We proof the conclusion by induction.
Fork=1, we have that
(2.13)tr⁡((X--X1)TP1)=tr⁡((X--X1)T12(AT(E-AX1B)BT+CT(F-CX1D)DT+PAT(E-AX1B)BTQ+PCT(F-CX1D)DTQ)12)=tr⁡((X--X1)T(AT(E-AX1B)BT+CT(F-CX1D)DT))=tr⁡((X--X1)TAT(E-AX1B)BT+(X--X1)TCT(F-CX1D)DT)=tr⁡((E-AX1B)TA(X--X1)B+(F-CX1D)TC(X--X1)D)=tr⁡(((E-AX1B)T00(F-CX1D)T)(A(X--X1)B00C(X--X1)D))=tr⁡(((E-AX1B)T00(F-CX1D)T)(E-AX1B00F-CX1D))=tr⁡(R1TR1)=‖R1‖2.
Assume (2.12) holds for k=s. By Algorithm 2.1, we have that
(2.14)tr⁡((X--Xs+1)TPs+1)=tr⁡((X--Xs+1)T×((AT(E-AXs+1B)BT+CT(F-CXs+1D)DT2+PAT(E-AXs+1B)BTQ+PCT(F-CXs+1D)DTQ2)+‖Rs+1‖2‖Rs‖2Ps))=tr⁡((X--Xs+1)T(AT(E-AXs+1B)BT+CT(F-CXs+1D)DT+‖Rs+1‖2‖Rs‖2Ps))=tr⁡(((E-AXs+1B)T00(F-CXs+1D)T)(E-AXs+1B00F-CXs+1D))+‖Rs+1‖2‖Rs‖2tr⁡((X--Xs+1)TPs)=tr⁡(Rs+1TRs+1)+‖Rs+1‖2‖Rs‖2tr⁡((X--Xs)TPs)-‖Rs+1‖2‖Rs‖2‖Rs‖2‖Ps‖2tr⁡(PsTPs)=‖Rs+1‖2+‖Rs+1‖2‖Rs‖2‖Rs‖2-‖Rs+1‖2‖Rs‖2‖Rs‖2‖Ps‖2‖Ps‖2=‖Rs+1‖2.
Therefore, (2.12) holds for k=s+1. By the principle of induction, (2.12) holds. This completes the proof.Theorem 2.5.
Supposing that Problem1 is consistent, then for an arbitrary initial matrix X1∈ℛrm×n(P,Q), a solution of Problem 1 can be obtained with finite iteration steps in the absence of round-off errors.Proof.
IfRi≠0,i=1,2,…,pq+st, by Lemma 2.4 we have Pi≠0,i=1,2,…,pq+st, then we can compute Xpq+st+1,Rpq+st+1 by Algorithm 2.1.
By Lemma2.3, we have
(2.15)tr⁡(Rpq+st+1TRi)=0,i=1,2,…,pq+st,tr⁡(RiTRj)=0,i,j=1,2,…,pq+st,i≠j.
Therefore, R1,R2,…,Rpq+st is an orthogonal basis of the matrix space
(2.16)S={W∣W=(W100W4),W1∈Rp×q,W4∈Rs×t},
which implies that Rpq+st+1=0; that is, Xpq+st+1 is a solution of Problem 1. This completes the proof.To show the least norm generalized reflexive solution of Problem1, we first introduce the following result.Lemma 2.6 (see [8, Lemma  2.4]).
Supposing that the consistent system of linear equationMy=b has a solution y0∈R(MT), then y0 is the least norm solution of the system of linear equations.By Lemma2.6, the following result can be obtained.Theorem 2.7.
Suppose that Problem1 is consistent. If one chooses the initial iterative matrix X1=ATHBT+CTH^DT+PATHBTQ+PCTH^DTQ, where H∈ℛp×q,H^∈ℛs×t are arbitrary matrices, especially, let X1=0∈ℛrm×n, one can obtain the unique least norm generalized reflexive solution of Problem 1 within finite iterative steps in the absence of round-off errors by using Algorithm 2.1.Proof.
By Algorithm2.1 and Theorem 2.5, if we let X1=ATHBT+CTH^DT+PATHBTQ+PCTH^DTQ, where H∈ℛp×q,H^∈ℛs×t are arbitrary matrices, we can obtain the solution X* of Problem 1 within finite iterative steps in the absence of round-off errors, and the solution X* can be represented that X*=ATGBT+CTG^DT+PATGBTQ+PCTG^DTQ.
In the sequel, we will prove thatX* is just the least norm solution of Problem 1.
Consider the following system of matrix equations:(2.17)AXB=E,CXD=F,APXQB=E,CPXQD=F.
If Problem1 has a solution X0∈ℛrm×n(P,Q), then
(2.18)PX0Q=X0,AX0B=E,CX0D=F.
Thus
(2.19)APX0QB=E,CPX0QD=F.
Hence, the systems of matrix equations (2.17) also have a solution X0.
Conversely, if the systems of matrix equations (2.17) have a solution X¯∈Rm×n, let X0=(X¯+PX¯Q)/2, then X0∈Rrm×n(P,Q), and
(2.20)AX0B=12A(X¯+PX¯Q)B=12(AX¯B+APX¯QB)=12(E+E)=E,CX0D=12C(X¯+PX¯Q)D=12(CX¯D+CPX¯QD)=12(F+F)=F.
Therefore, X0 is a solution of Problem 1.
So the solvability of Problem1 is equivalent to that of the systems of matrix equations (2.17), and the solution of Problem 1 must be the solution of the systems of matrix equations (2.17).
LettingSE′ denote the set of all solutions of the systems of matrix equations (2.17), then we know that SE⊂SE′, where SE is the set of all solutions of Problem 1. In order to prove that X* is the least-norm solution of Problem 1, it is enough to prove that X* is the least-norm solution of the systems of matrix equations (2.21). Denoting vec(X)=x,vec(X*)=x*,vec(G)=g1,vec(G^)=g2,vec(E)=e,vec(F)=f, then the systems of matrix equations (2.17) are equivalent to the systems of linear equations
(2.21)(BT⊗ADT⊗CBTQ⊗APDTQ⊗CP)x=(efef).
Noting that
(2.22)x*=vec⁡(ATGBT+CTG^DT+PATGBTQ+PCTG^DTQ)=(B⊗AT)g1+(D⊗CT)y2+(QB⊗PAT)g1+(QD⊗PCT)g2=(B⊗ATD⊗CTQB⊗PATQD⊗PCT)(g1g2g1g2)=(BT⊗ADT⊗CBTQ⊗APDTQ⊗CP)T(g1g2g1g2)∈R((BT⊗ADT⊗CBTQ⊗APDTQ⊗CP)T),
by Lemma 2.6 we know that X* is the least norm solution of the systems of linear equations (2.21). Since vector operator is isomorphic and X* is the unique least norm solution of the systems of matrix equations (2.17), then X* is the unique least norm solution of Problem 1.


## 3. The Solution of Problem2

In this section, we will show that the optimal approximate solution of Problem2 for a given generalized reflexive matrix can be derived by finding the least norm generalized reflexive solution of a new corresponding matrix equation pair AX~B=E~, CX~D=F~.When Problem1 is consistent, the set of solutions of Problem 1 denoted by SE is not empty. For a given matrix X0∈ℛrm×n(P,Q) and X∈SE, we have that the matrix equation pair (1.1) is equivalent to the following equation pair:
(3.1)AX~B=E~,CX~D=F~,
where X~=X-X0,E~=E-AX0B,F~=F-CX0D. Then Problem 2 is equivalent to finding the least norm generalized reflexive solution X~* of the matrix equation pair (3.1).By using Algorithm2.1, let initially iterative matrix  X~1=ATHBT+CTH^DT+PATHBTQ+PCTH^DTQ, or more especially, letting X~1=0∈Rrm×n(P,Q), we can obtain the unique least norm generalized reflexive solution X~* of the matrix equation pair (3.1); then we can obtain the generalized reflexive solution X^ of Problem 2, and X^ can be represented that X^=X~*+X0.


## 4. Examples for the Iterative Methods

In this section, we will show several numerical examples to illustrate our results. All the tests are performed by MATLAB 7.8.Example 4.1.
Consider the generalized reflexive solution of the equation pair (1.1), where
(4.1)A=(13-57-92046-10-296-836227-13-55-22-1-1184-6-9-19),B=(408-54-150-234-10250392-6-27-8111),C=(632-57-921046-119-1293-8136427-15-515-22-13-1129-6-9-19),D=(718-614-450-233-1208251694-6-58-2917),E=(592-11911216-244-13313054311234-518221814-4071668-11765371434-1794083-1374-808242-3150-13621104-2848423-29091441-182-3326),F=(-288228302992291-48494096701090-783-7933363-1262979-385124626321734553-3709-100-1774-4534-45481256-6896864-2512-1136-1633-5412).
Let
(4.2)P=(000100000100-1001000001000),Q=(0000-10001000-10001000-10000).
We will find the generalized reflexive solution of the matrix equation pairAXB=E,CXD=F by using Algorithm 2.1. It can be verified that the matrix equation pair is consistent over generalized reflexive matrix and has a solution with respect to P,Q as follows:
(4.3)X*=(53-612-5-118-19713-4-841351263-5-791811)∈Rr5×5(P,Q).
Because of the influence of the error of calculation, the residualRi is usually unequal to zero in the process of the iteration, where i=1,2,…. For any chosen positive number ε, however small enough, for example, ε=1.0000e-010, whenever ∥Rk∥<ε, stop the iteration, and Xk is regarded to be a generalized reflexive solution of the matrix equation pair AXB=E,CXD=F. Choose an initially iterative matrix X1∈ℛr5×5(P,Q), such as
(4.4)X1=(110-612-5-68-114913-4-8413512610-1-914186).
By Algorithm 2.1, we have
(4.5)X17=(5.00003.0000-6.000012.0000-5.0000-11.00008.0000-1.00009.00007.000013.0000-4.0000-8.00004.000013.00005.000012.00006.00003.0000-5.0000-7.00009.00001.00008.000011.0000),‖R17‖=3.2286e-011<ε.
So we obtain a generalized reflexive solution of the matrix equation pair AXB=E,CXD=F as follows:
(4.6)X-=(5.00003.0000-6.000012.0000-5.0000-11.00008.0000-1.00009.00007.000013.0000-4.0000-8.00004.000013.00005.000012.00006.00003.0000-5.0000-7.00009.00001.00008.000011.0000).
The relative error of the solution and the residual are shown in Figure 1, where the relative error rek=∥Xk-X*∥/∥X*∥ and the residual rk=∥Rk∥.Figure 1
The relative error of the solution and the residual for Example4.1 with X1≠0.Letting(4.7)X1=(0000000000000000000000000),
by Algorithm 2.1, we have
(4.8)X17=(5.00003.0000-6.000012.0000-5.0000-11.00008.0000-1.00009.00007.000013.0000-4.0000-8.00004.000013.00005.000012.00006.00003.0000-5.0000-7.00009.00001.00008.000011.0000),‖R17‖=3.1999e-011<ε.
So we obtain a generalized reflexive solution of the matrix equation pair AXB=E,CXD=F as follows:
(4.9)X-=(5.00003.0000-6.000012.0000-5.0000-11.00008.0000-1.00009.00007.000013.0000-4.0000-8.00004.000013.00005.000012.00006.00003.0000-5.0000-7.00009.00001.00008.000011.0000).
The relative error of the solution and the residual are shown in Figure 2.Figure 2
The relative error of the solution and the residual for Example4.1 with X1=0.Example 4.2.
Consider the least norm generalized reflexive solution of the matrix equation pair in Example4.1. Let
(4.10)H=(101020-10101-10012010-301210-10-2-10),H^=(-11-100010-131-10-202010-301210-10-212),X1=ATHBT+CTH^DT+PATHBTQ+PCTH^DTQ.
By using Algorithm 2.1, we have
(4.11)X19=(5.00003.0000-6.000012.0000-5.0000-11.00008.0000-1.00009.00007.000013.0000-4.0000-8.00004.000013.00005.000012.00006.00003.0000-5.0000-7.00009.00001.00008.000011.0000),‖R19‖=6.3115e-011<ε.
So we obtain the least norm generalized reflexive solution of the matrix equation pair AXB=E,CXD=F as follows:
(4.12)X*=(5.00003.0000-6.000012.0000-5.0000-11.00008.0000-1.00009.00007.000013.0000-4.0000-8.00004.000013.00005.000012.00006.00003.0000-5.0000-7.00009.00001.00008.000011.0000).
The relative error of the solution and the residual are shown in Figure 3.Figure 3
The relative error of the solution and the residual for Example4.2.Example 4.3.
LetSE denote the set of all generalized reflexive solutions of the matrix equation pair in Example 4.1. For a given matrix,
(4.13)X0=(-331110-7161010-90910-11-133-106-1-70)∈Rr5×5(P,Q),
we will find X^∈SE, such that
(4.14)‖X^-X0‖=min⁡X∈SE‖X-X0‖.
That is, find the optimal approximate solution to the matrix X0 in SE.LettingX~=X-X0,E~=E-AX0B,F~=F-CX0D, by the method mentioned in Section 3, we can obtain the least norm generalized reflexive solution X~* of the matrix equation pair AX~B=E~,CX~D=F~ by choosing the initial iteration matrix X~1=0, and X~* is that
(4.15)X~17*=(8.0000-0.0000-7.000011.0000-6.0000-11.000015.0000-2.00003.0000-3.00003.00005.0000-8.0000-5.00003.00006.000011.00007.0000-0.0000-8.00003.00003.00002.000015.000011.0000),‖R17‖=3.0690e-011<ε=1.0000e-010,X^=X~17*+X0=(5.00003.0000-6.000012.0000-5.0000-11.00008.0000-1.00009.00007.000013.0000-4.0000-8.00004.000013.00005.000012.00006.00003.0000-5.0000-7.00009.00001.00008.000011.0000).
The relative error of the solution and the residual are shown in Figure 4, where the relative error rek=∥X~k+X0-X*∥/∥X*∥ and the residual rk=∥Rk∥.Figure 4
The relative error of the solution and the residual for Example4.3.



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*Source: 492951-2012-07-15.xml*