# The Structure of Symmetric Solutions of the Matrix EquationAX=B over a Principal Ideal Domain

Authors: V. M. Prokip
Journal: International Journal of Analysis
 (2017)

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## Abstract

We investigate the structure of symmetric solutions of the matrix equationAX=B, where A and B are m-by-n matrices over a principal ideal domain R and X is unknown n-by-n matrix over R. We prove that matrix equation AX=B over R has a symmetric solution if and only if equation AX=B has a solution over R and the matrix ABT is symmetric. If symmetric solution exists we propose the method for its construction.


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## Body

## 1. Introduction

LetR be a principal ideal domain with an identity element and let Rm×n denote the set of m×n matrices over R. In what follows 0m,k is the zero m×k matrix. The transpose matrix of matrix A will be denoted by AT. Matrix A∈Rn×n is called a symmetric matrix if A=AT.Consider the matrix equation(1)AX=B,where A,B∈Rm×n and X is unknown n×n matrix over R. The problem of solvability of (1) has drawn the attention of many mathematicians. Numerous works are devoted to the methods for its solution. This is explained not only by theoretical interest to this problem ([1–3]) but also by the existence of numerous applied problems connected with the necessity of solution of linear matrix equation (see also [4–6]). It may be noted that linear matrix equations have application in theory of differential equations, system theory, control theory, stability theory, and some other areas of pure and applied mathematics.Many authors addressed the question when (1) (over the field R of real numbers or the field C of complex numbers) has a solution which belongs to a special class of matrices. They are given necessary and sufficient conditions (using generalized inverses) for the existence of symmetric ([7–10]), symmetric with prescribed rank [11], Hermitian and skew-Hermitian ([12, 13]), reflexive and antireflexive [14], and general solutions which are described in terms of original elements or operators. Methods for constructed symmetric solutions of linear matrix equation AXB=C are proposed in [9, 13, 15, 16]. More details on this problem and many references to original literature can be found in [6, 17–21].LetR=F be a field, where F is R or C. It is well known (see [7–9]) that (1) over F has a symmetric solution if and only if the equation AX=B has a solution and ABT=BAT. In this article we affirm that this criterion is true for (1) over a principal ideal domain.The paper is organized as follows. In Section2 we prove preparatory results of this article. Necessary and sufficient conditions, under which matrix equation AX=B over a principal ideal domain has a symmetric solution, are proposed in Section 3. The method used for constructing of symmetric solutions of equation AX=B over a principal ideal domain is presented in Section 4. In this section numerical examples are also given.We note that obtained results are true for the matrix equationAX=B, where A and B are m-by-n matrices over an elementary divisor domain.


## 2. Preparatory Results

It is well known that every matrixA with entries in a principal ideal domain R admits diagonal reduction; that is, for the matrix A∈Rm×n there exist matrices U∈GL(m,R) and V∈GL(n,R) such that (2)SA=UAV=diaga1,a2,…,ar,0,…,0∈Rm×nis a diagonal matrix, where ai≠0 for all i=1,2,…,r and aj-1∣aj (divides) for all j=2,…,r. The elements a1,a2,…,ar are called invariant factors of A and the diagonal matrix SA is called the Smith normal form of matrix A (see [1], Chapter 15).Consider the following matrix equation:(3)AX=B,where A∈Rm×n, B∈Rm×k, and X is unknown element in Rn×k. We denote by A~=AB∈Rm×(n+k) the augmented matrix of (3); this is the matrix obtained from A by adding matrix B. The problem of solvability of (3) over a principal ideal domain has a long history (see, e.g., [22–25] and references therein). In more general form, this theory is presented in [1, 3, 26, 27].Thus, matrix equationAX=B over a principal ideal domain is solvable if and only if (4)rankA=rankA~=r,ai=a~i,∀i=1,2,…,r,where a~i are invariant factors of the matrix A~ (see [26], Chapter 1).In [25] the following statement was proved. Matrix equation (3) over a principal ideal domain is solvable if and only if the Hermite normal forms of the matrices A0m,k and A~ coincide; that is, matrices A0m,k and A~ are right equivalent.In this part we present new necessary and sufficient conditions for the solvability of the equationAX=B over a principal ideal domain.Lemma 1.
LetA∈Rm×n, B∈Rm×k, rank(A)=r, and let (5)SA=UAV=diaga1,a2,…,ar,0,…,0be the Smith normal form of matrix A, where U∈GL(m,R) and V∈GL(n,R). The matrix equation AX=B is solvable over R if and only if (6)UB=B10m-r,kand the matrix B1∈Rr×k admits the representation (7)B1=diaga1,a2,…,arG,whereG∈Rr×k.
If matrix equationAX=B is solvable, then for arbitrary matrix P∈R(n-r)×k matrix Xp=VGP is a general solution of the equation AX=B.Proof.
LetX0∈Rn×k be the solution of the equation AX=B and matrices U∈GL(m,R) and V∈GL(n,R) such that (8)UAV=SA=diaga1,a2,…ar,0,…,0is the Smith normal form of the matrix A. From equality AX0=B it follows that(9)UAVV-1X0=UB.Put S(A)=diaga1,a2,…ar∈Rr×r, V-1X0=Y1Y2, and UB=B1B2, where Y1∈Rr×k and B1∈Rr×k. We rewrite equality (9) in the form (10)SA0r,n-r0m-r,r0m-r,n-rY1Y2=B1B2.From this we have B1=S(A)Y1 and B2=0m-r,k.
Conversely, letU∈GL(m,R) and V∈GL(n,R) be invertible matrices such that (11)UAV=SA=SA0r,n-r0m-r,r0m-r,n-r,UB=B10m-r,k,where B1∈Rr×k and B1=S(A)G, where S(A)=diaga1,a2,…,ar∈Rr×r and G∈Rr×k. So, we have (12)SA0r,n-r0m-r,r0m-r,n-rG0n-r,k=B10m-r,k.Let P∈R(n-r)×k be an arbitrary matrix. It is obvious that next equality holds (13)SA0r,n-r0m-r,r0m-r,n-r0r,kP=0m,k.Thus(14)SA0r,n-r0m-r,r0m-r,n-rGP=B10m-r,k.From the above it follows that the matrix Xp=VGP∈Rn×k is a general solution of equation AX=B. This completes the proof.LetR=F be a field. From Lemma 1 it follows.Corollary 2.
LetA∈Fm×n and B∈Fm×k be nonzero matrices and let U∈Fm×m and V∈Fn×n be nonsingular matrices such that (15)UAV=Ir0r,m-r0m-r,r0m-r,n-r,wherer=rankA.The equation AX=B is solvable over F if and only if UB=B10m-r,k, where B1∈Fr×k. If equation AX=B is solvable, then for arbitrary matrix P∈F(n-r)×k the matrix Xp=VB1P is a general solution of equation AX=B.


## 3. The Main Result

In this section we establish conditions under which a symmetric solution over a principal ideal domainR exists for matrix equation AX=B.Lemma 3.
LetA∈Rm×n and rank(A)=r<n. The homogeneous equation AX=0m,n has a nontrivial symmetric solution.Proof.
For the matrixA there exist matrices U∈GL(m,R) and V∈GL(n,R) such that UAV=SA=diag(a1,a2,…,ar,0,…,0) is the Smith normal form of A. Let P∈R(n-r)×(n-r) be a nonzero symmetric matrix. For the symmetric matrix X0=V0r,r0r,n-r0n-r,rPVT we have AX0=0m,n and the proof of Lemma 3 is complete.The main result of this article is the following theorem.Theorem 4.
LetA,B∈Rm×n. The matrix equation AX=B has a symmetric solution over a principal ideal domain R if and only if the equation AX=B is solvable over R and ABT=BAT.Proof.
Necessity. Let X0∈Rn×n be a symmetric solution of matrix equation AX=B; that is, X0=X0T and AX0=B. Thus, X0AT=BT. From the last equality we have AX0AT=BAT=ABT. The necessity is proved.
Sufficiency. Let X0∈Rn×n be a solution of equation AX=B; that is, AX0=B and ABT=BAT. This implies that X0TAT=BT. Further, let rank(A)=r. For the matrix A there exist matrices U∈GL(m,R) and V∈GL(n,R) such that (16)UAV=SA=diaga1,a2,…,ar,0,…,0is the Smith normal form of A. Let us introduce for V∈GL(n,R) the following notation (V-1)T=V-T.
From equalitiesAX0=B and X0TAT=BT we have(17)UAVV-1X0V-T=UBV-T,V-1X0TV-TVTATUT=V-1BTUT.Put V-1X0V-T=G and UBV-T=B1. Thus, V-1X0TV-T=GT and V-1BTUT=B1T. We rewrite system (17) as(18)SAG=B1,GTSAT=B1T.
We write matricesG and B1 in block forms (19)G=G11G12G21G22,B1=B11B12B21B22,where G11∈Rr×r, G12∈Rr×(n-r), G21∈R(n-r)×r, B11∈Rr×r, B12∈Rr×(n-r), and B21∈R(m-r)×r. From (18) we obtain (20)SA0r,n-r0m-r,r0m-r,n-rG11G12G21G22=B11B12B21B22,(21)G11TG21TG12TG22TSA0r,m-r0n-r,r0n-r,m-r=B11TB21TB12TB22T,where S(A)=diag(a1,a2,…,ar)∈Rr×r.
By Lemma1, from (20) it follows that B21=0m-r,r and B22=0m-r,n-r. Similarly, from equality (21) it follows that B21T=0r,m-r and B22T=0n-r,m-r. Thus, (22)SAG11G12G21G22=B11B120m-r,r0m-r,n-r,G11TG21TG12TG22TSAT=B11T0r,m-rB12T0n-r,m-r.
From equalityABT=BAT we have UABTUT=UBATUT. Hence, (23)UABTUT=UAVV-1BTUT=SAGTSAT=SAG11T0r,n-r0n-r,r0n-r,n-rSAT,UBATUT=UBV-TVTATUT=SAGSAT=SAG110r,n-r0n-r,r0n-r,n-rSAT.From the last two equalities we have (24)SAG11SA=SAG11TSA.This implies that G11=G11T. Thus, matrix G11 is symmetric.
It is easily verified that the symmetric matrix(25)X0=VG11G12G12T0n-r,n-rVT∈Rn×nis the common solution of equations AX=B and XAT=BT. So, for equation AX=B there exists a symmetric solution over R.
Now we prove that for arbitrary symmetric matrixP∈R(n-r)×(n-r) the matrix (26)Xp=VG11G12G12TPVTis a symmetric solution of equation AX=B. By Lemma 3, the symmetric matrix Xs=V0r,r0r,n-r0n-r,rPVT is the solution of the homogenous equation AX=0m,n. Put Xp=X0+Xs. It is easy to see that (27)AXp=AX0+Xs=AX0+AXs=B.
Thus, for arbitrary symmetric matrixP∈R(n-r)×(n-r) the matrix (28)Xp=VG11G12G12TPVT∈Rn×nis a general symmetric solution of matrix equations AX=B. The proof of Theorem 4 is complete.


## 4. The Representation of Symmetric Solutions

In this section we give new conditions under which for the equationAX=B there exists a symmetric solution. If a symmetric solution exists we propose the method for its construction.Theorem 5.
LetA,B∈Rm×n and rank(A)=r and let U∈GL(m,R) and V∈GL(n,R) such that (29)UAV=SA=diaga1,a2,…,ar,0,…,0is the Smith normal form of matrix A. The equation AX=B has a symmetric solution if and only if (30)UBV-T=B11B120m-r,n,B11B12=diaga1,a2,…,arD11D12,where D11∈Rr×r is a symmetric matrix and D12∈Rr×(n-r).
If a symmetric solution of equationAX=B exists, then for arbitrary symmetric matrix P∈R(n-r)×(n-r) the matrix (31)Xs=VD11D12D12TPVTis a general solution of AX=B.Proof.
Necessity. Let the symmetric matrix X0∈Rn×n be the solution of equation AX=B. Further, let U∈GL(m,R) and V∈GL(n,R) such that (32)UAV=SA=SA0r,n-r0m-r,r0m-r,n-r,where S(A)=diaga1,a2,…,ar∈Rr×r. By Lemma 1, from equality AX0=B we have (33)SA0r,n-r0m-r,r0m-r,n-rV-1X0=UB=SAB10m-r,n,where B1∈Rr×n. Put V-T=(V-1)T. Multiplying the last equality from the right by V-T, we obtain (34)SA0r,n-r0m-r,r0m-r,n-rV-1X0V-T=SAB10m-r,nV-T.We write matrices V-1X0V-T and UBV-T=B10m-r,nV-T in block forms (35)V-1X0V-T=Y0=Y11Y12Y21Y22,UBV-T=SAB10m-r,nV-T=SAD11SAD120m-r,r0m-r,n-r,where Y11,D11∈Rr×r, Y22∈R(n-r)×(n-r), and D12∈Rr×(n-r).
It is obvious thatY0 is a symmetric matrix. Thus, Y11 and Y22 are symmetric matrices and Y21=Y12T. From equality(36)SA0r,n-r0m-r,r0m-r,n-rY0=SAD11SAD120m-r,r0m-r,n-rwe obtain that Y11=D11; that is, D11 is the symmetric matrix and Y12=D12 and Y21=D12T. It is easily verified that for arbitrary symmetric matrix P∈R(n-r)×(n-r) the following equality holds: (37)SA0r,n-r0m-r,r0m-r,n-rSAD11SAD12D12TP=SAD11SAD120m-r,r0m-r,n-r.The necessity is proved.
Sufficiency. Let U∈GL(m,R) and V∈GL(n,R) and let (38)UAV=SA=SA0r,n-r0m-r,r0m-r,n-rbe the Smith normal form of matrix A, where S(A)=diaga1,a2,…,ar∈Rr×r. Further, let (39)UBV-T=SAD11SAD120m-r,r0m-r,n-r,where D11∈Rr×r is symmetric matrix and D12∈Rr×(n-r). For the symmetric matrix Y0=D11D12D12T0n-r,n-r∈Rn×n we have(40)SA0r,n-r0m-r,r0m-r,n-rY0=SAD11SAD120m-r,r0m-r,n-r,(41)Y0SA0r,m-r0n-r,r0n-r,m-r=D11SA0r,m-rD12TSA0n-r,m-r.
From equality (40) it follows that(42)U-1SA0r,n-r0m-r,r0m-r,n-rV-1VY0VT=AX0=U-1SAD11SAD120m-r,r0m-r,n-rVT=B.Since Y0 is the symmetric matrix we finally obtain that X0=VY0VT is also the symmetric matrix. From equality (41) we have that X0AT=BT. Giving the above, equation AX=B has a symmetric solution.
LetP∈R(n-r)×(n-r) be a symmetric matrix. By Lemma 3, matrix Xp=V0r,r0r,n-r0m-r,rPVT∈Rn×n is the common symmetric solution of the homogeneous matrix equations AX=0m,n and XAT=0n,m. Hence, the matrix (43)Xs=VD11D12D12TPVTis the general symmetric solution of the equation AX=B and the proof is complete.Consider the following example.Example 6.
LetR=Z be the ring of integers and A=2-8-21213-6-1∈M3,3(Z) and B=0-6-26316-3-1∈M3,3(Z). Consider the matrix equation AX=B. For invertible matrices(44)U=010-12011-1,V=1-1100101-3over Z we have (45)UAV=SA=100040000,UB=63112124000=diag1,4,0·631331000.Thus, equation AX=B is solvable. Moreover, matrix UBV-T may be represented in the form (46)UBV-T=diag1,4,01310310103000,where D11=13101010 is a symmetric matrix. Thus, equation AX=B has a symmetric solution.
For arbitraryp∈Z the matrix Yp=131031010333p is a symmetric solution of the equation SAY=UB(V-T), and the matrix (47)Xs=VYpVT=3+pp3-3ppp3-3p3-3p3-3p-8+9pis the general symmetric solution of the matrix equation AX=B.LetR=F be a field. Using this terminology, we have the following corollary to Theorem 5 (see also [7–9, 28]).Corollary 7.
LetA,B∈Fm×n and let U∈GL(m,F) and V∈GL(n,F) such that UAV=Ir0r,n-r0m-r,r0m-r,n-r, where r=rank(A). The equation AX=B has a symmetric solution if and only if UBV-T=B11B120m-r,n, where B11∈Fr×r is a symmetric matrix and B12∈Fr×(n-r).
If a symmetric solution of the equationAX=B exists, then for arbitrary symmetric matrix P∈F(n-r)×(n-r) the matrix (48)Xs=VD11D12D12TPVTis a general solution of AX=B.Consider the following example.Example 8.
LetF=Q be the field of rational numbers and (49)A=-32-11-74-1310-1-110-1-1∈M4,4Q,B=-1-102-3-3-2411201120∈M4,4Q.
Consider the matrix equationAX=B. For invertible matrices (50)U=125-300-1100-424-200-11,V=11012310001-100-12over Q we have (51)UAV=1000010000000000,UB=446-2-2-2-2200000000.Thus, equation AX=B is solvable.
It is easily verified thatUBV-T=6-6104-64-2000000000. This means that equation AX=B has a symmetric solution over Q and (52)Xs=3558+0110P01102031+0110P1-1-122301+1-1-12P01101-1-12P1-1-12is the general symmetric solution of matrix equation AX=B for arbitrary symmetric matrix P∈M2,2(Q).Open Question. Let A and B be m-by-n matrices over a commutative ring R with an identity element. Further, let the matrix equation AX=B be solvable over D and the matrix ABT be symmetric. When are these conditions sufficient for the existence of symmetric solution of the equation AX=B over R?



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