# The Least Squares Hermitian (Anti)reflexive Solution with the Least Norm to Matrix EquationAXB=C

Authors: Xin Liu; Qing-Wen Wang
Journal: Mathematical Problems in Engineering
 (2017)

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## Abstract

For a given generalized reflection matrixJ, that is, JH=J, J2=I, where JH is the conjugate transpose matrix of J, a matrix A∈Cn×n is called a Hermitian (anti)reflexive matrix with respect to J if AH=A and A=±JAJ. By using the Kronecker product, we derive the explicit expression of least squares Hermitian (anti)reflexive solution with the least norm to matrix equation AXB=C over complex field.


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## Body

## 1. Introduction

Throughout this paper, we denote the set of allm×n matrices over the real field and complex field by Rm×n and Cm×n, respectively. Let the symbols HCn×n, SRn×n, ASRn×n, In, and 0 stand for the set of all n×n complex Hermitian matrices, the set of all n×n real symmetric matrices, the set of all n×n real skew symmetric matrices, the n×n identity matrix, and the zero matrix with appropriate size, respectively. We also denote the transpose, the conjugate transpose, the Moore-Penrose inverse, and the Frobenius norm of A by AT, AH, A†, and A, respectively. Let A∈Cm×n; we define vec(A)=a1T,a2T,…,anTT, where ai is the i-column of A,  i=1,2,…,n.In this paper, we consider the Hermitian (anti)reflexive solution problem of matrix equationAXB=C. Let us first recall the definition of those two kinds of matrices.Definition 1.
Given a generalized reflection matrixJ, that is, JH=J, J2=I, a matrix A∈Cn×n is called a Hermitian reflexive matrix with respect to J if AH=A and A=JAJ. The set of all n×n Hermitian reflexive matrices with respect to J is denoted by HCrn×n(J).Definition 2.
Given a generalized reflection matrixJ, that is, JH=J, J2=I, a matrix A∈Cn×n is called a Hermitian antireflexive matrix with respect to J if AH=A and A=-JAJ. The set of all n×n Hermitian antireflexive matrices with respect to J is denoted by HCan×n(J).As an important kind of matrices, the reflexive matrices with different constraints were studied by many authors in recent years in literatures and references therein [1–29]. For instance, Zhao et al. [1] derived the general Hermitian reflexive solution to AX=B, while Yu and Shen [2] considered a more general case, that is, the Hermitian {J,k+1}-(anti)-reflexive solution to AX=B. As an extension of matrix equation AX=B, the matrix system AX=B,XC=D was considered by Zhou and Yang [3]. They gave the necessary and sufficient conditions for the existence of a Hermitian reflexive solution to the above matrix system. As we can see, many important associated works were widely studied. But finding the least squares Hermitian (anti)reflexive solution with the least norm to matrix equation AXB=C is still a problem. Therefore, we in this paper will fill this gap.We consider the following two problems.Problem 3.
Given matricesA∈Cm×n,  B∈Cn×k, and C∈Cm×k, let (1)S=X∣X∈HCrn×nJ,AXB-C=minX^∈HCrn×nJ⁡AX^B-C. Find X0∈S such that X0=minX∈S⁡X.Problem 4.
Given matricesA∈Cm×n,  B∈Cn×k, and C∈Cm×k, let (2)SA=X∣X∈HCan×nJ,AXB-C=minX^∈HCan×nJ⁡AX^B-C. Find X0∈SA such that X0=minX∈SAX.Now, we start with some useful and fundamental results in the following.Lemma 5.
The least squares solution to the matrix equationAx=b with A∈Cm×n and b∈Cm is given by x=A†b+(I-A†A)y, where y∈Cn is an arbitrary vector. And the least squares solution with the least norm is x=A†b.Kronecker product method plays an important role in this paper. We use this method to get some important vectorizing formulas.For anyA∈Cm×n, A=A1+A2i, where A1, A2∈Rm×n. The real representation matrix of A=A1+A2i∈Cm×n is given by f(A)=A1A2-A2A1∈C2m×2n, which has the following properties:(1)f(A+B)=f(A)+f(B).(2)f(AB)=f(A)f(B).(3) ForA=A1+A2i∈Cm×n, we define g(A)=(A1,A2). Then(3)vec⁡A=vec⁡gA=vec⁡A1vec⁡A2,gAB=gAfB.Analogously to the method of Yuan et al. (see [30]), we can get a fundamental result as below.Lemma 6.
LetA, B, and C be complex matrices with appropriate sizes; then vec⁡(g(ABC))=f(CH⊗A¯)vec(g(B)).Proof.
We denoteA=A1+A2i, B=B1+B2i, and C=C1+C2i.g(ABC)=g(A)f(BC)=g(A)f(B)f(C). Then(4)gAfBfC=A1,A2B1B2-B2B1C1C2-C2C1=A1B1C1-A2B2C1-A1B2C2-A2B1C2,A1B1C2-A2B2C2+A1B2C1+A2B1C1. By the rule of (5)vec⁡XYZ=ZT⊗Xvec⁡Y, we have (6)vec⁡gABC=C1T⊗A1-C2T⊗A2-C2T⊗A1-C1T⊗A2C2T⊗A1+C1T⊗A2C1T⊗A1-C2T⊗A2vec⁡B1vec⁡B2=fCH⊗A¯vec⁡B1vec⁡B2=fCH⊗A¯vec⁡gB.The structure of Hermitian (anti)reflexive matrix promises the following decompositions, which would simplify our problems.Lemma 7.
LetJ be the n×n generalized reflection matrix. Then there exists an n×n unitary matrix U such that(7)J=UIr00-In-rUH.Lemma 8 (see [18]).
(1) LetA∈Cn×n and let the spectral decomposition of the n×n reflection matrix J be given as (7). Then A∈HCrn×n(J) if and only if(8)A=UF100F2UH,where F1∈HCr×r and F2∈HC(n-r)×(n-r).
(2) See [4]. Let A∈Cn×n and let the spectral decomposition of the n×n reflection matrix J be given as (7). Then A∈HCan×n(J) if and only if(9)A=U0FFH0UH,where F∈Cr×(n-r).


## 2. The Solution of Problem3

In this section, we provide the least squares Hermitian reflexive solution with the least norm to matrix equationAXB=C. There are the vectorizing formulas of Hermitian matrix, which are very useful in what comes below.Definition 9.
LetX=(xij)∈Rn×n; we define(10)vecS⁡X=x1,x2,…,xn-1,xnT∈Rnn+1/2,where x1=(x11,2x21,…,2xn1), x2=(x22,2x32,…,2xn2),…,xn-1=(x(n-1)(n-1),  2xn(n-1)), and xn=xnn.Definition 10.
LetX=(xij)∈Rn×n; we define(11)vecA⁡X=2x1,x2,…,xn-2,xn-1T∈Rnn-1/2,where x1=(x21,x31,…,xn1), x2=(x32,x42,…,xn2),…,xn-2=(x(n-1)(n-2),xn(n-2)), and xn-1=xn(n-1).Lemma 11 (see [30]).
Suppose thatX∈Rn×n. Then X∈SRn×n if and only if vec⁡(X)=KSvecS⁡(X), where vecS⁡(X) is given by (10), where(12)KS=122e1e2e3⋯en-1en00⋯00⋯0000e10⋯002e2e3⋯en-1en⋯00000e1⋯000e2⋯00⋯000⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮000⋯e1000⋯e20⋯2en-1en0000⋯0e100⋯0e2⋯0en-12en.Obviously, ei is the i-column of In, KS∈Rn2×n(n+1)/2, and KSTKS=In(n+1)/2.Lemma 12.
Suppose thatX∈Rn×n. Then X∈ASRn×n if and only if vec⁡(X)=KAvecA⁡(X), where vecA⁡(X) is given by (11), where(13)KA=12e2e3⋯en-1en0⋯00⋯0-e10⋯00e3⋯en-1en⋯00-e1⋯00-e2⋯00⋯0⋮⋮⋮⋮⋮00000⋯-e100⋯-e20⋯en00⋯0-e10⋯0-e2⋯-en-1.Obviously, ei is the i-column of In, KA∈Rn2×n(n-1)/2, and KATKA=In(n-1)/2.LetX=X1+X2i∈HCn×n, X1∈Rn×n, X2∈Rn×n. Then X1=X1T and X2=-X2T. By Lemmas 11 and 12, one has(14)vec⁡X1vec⁡X2=KS00KAvecS⁡X1vecA⁡X2.IfX∈HCrn×n(J), then, by Lemma 8, there exists a unitary matrix U such that(15)X=UY00ZUH,where Y=Y1+Y2i∈HCr×r, and Z=Z1+Z2i∈HC(n-r)×(n-r).Set(16)AU=D,E,UHB=FG,(17)M=fFH⊗D¯,fGH⊗E¯KS10000KA10000KS20000KA2,where U is given by (7). KS1, KS2 are given by (12) with the sizes r2×r(r+1)/2, (n-r)2×(n-r)(n-r+1)/2, respectively. KA1, KA2 are given by (13) with the sizes r2×r(r-1)/2, (n-r)2×(n-r)(n-r-1)/2, respectively. Then one has the following result.Theorem 13.
LetA∈Cm×n, B∈Cn×k, and C∈Cm×k. Then the solution X∈S to matrix equation AXB=C is given by(18)X=UY00ZUH,(19)vec⁡Y=KS1,KA1i,0,0M†c+I-M†My,vec⁡Z=0,0,KS2,KA2iM†c+I-M†My, where U is given by (7), c=vec⁡(g(C))∈R2km, and y∈Rr2+(n-r)2 is an arbitrary vector. And the solution X∈S with the least norm is given by(20)X=UY000Z0UH,(21)vec⁡Y0=KS1,KA1i,0,0M†c,vec⁡Z0=0,0,KS2,KA2iM†c.Proof.
Since(22)Y1∈SRr×r,Y2∈ASRr×r,Z1∈SRn-r×n-r,Z2∈ASRn-r×n-r,it follows from (14) that(23)vec⁡Y1vec⁡Y2vec⁡Z1vec⁡Z2=KS10000KA10000KS20000KA2vecS⁡Y1vecA⁡Y2vecS⁡Z1vecA⁡Z2.Now, by Lemmas 6 and 11 and (23), we have(24)AXB-C=AUY00ZUHB-C=DYF+EZG-C=gDYF+gEZG-gC=vec⁡gDYF+vec⁡gEZG-vec⁡gC=fFH⊗D¯vec⁡Y1vec⁡Y2+fGH⊗E¯vec⁡Z1vec⁡Z2-vec⁡gC=fFH⊗D¯,fGH⊗E¯vec⁡Y1vec⁡Y2vec⁡Z1vec⁡Z2-vec⁡gC=Mx-c, where M is given by (17); x=vecS⁡(Y1)vecA⁡(Y2)vecS⁡(Z1)vecA⁡(Z2). Then, by Lemma 5,(25)x=M†c+I-M†My is the required least squares solution, where y∈Rr2+(n-r)2 is an arbitrary vector. Recall that (26)vec⁡Y1vec⁡Y2vec⁡Z1vec⁡Z2=KS10000KA10000KS20000KA2x. Therefore (27)vec⁡Y=vec⁡Y1+vec⁡Y2i=KS1KA1i00x,vec⁡Z=vec⁡Z1+vec⁡Z2i=00KS2KA2ix. Then the least squares Hermitian reflexive solution is given by (18). Obviously, the least squares Hermitian reflexive solution with the least norm is given by (20).


## 3. The Solution of Problem4

In this section, we provide the least squares Hermitian antireflexive solution with the least norm to matrix equationAXB=C.We first derive the relation betweenvec⁡(g(Y)) and vec⁡(g(YH)).Lemma 14 (see [31]).
LetY∈Rm×n be any matrix. Then(28)vec⁡YT=Pm,nvec⁡Y, where(29)Pm,n=E11TE12T⋯E1nTE21TE22T⋯E2nT⋮⋮⋱⋮Em1TEm2T⋯EmnT∈Rmn×mn,with Eij, i=1,2,…,m, j=1,2,…,n, being an m×n matrix with the element at position (i,j) being 1 and the others being 0. Moreover, P(m,n) is an orthogonal matrix, which is uniquely determined by the integers m,n.LetX∈HCan×n(J); then, by Lemma 8, there exists a unitary matrix U such that(30)X=U0YYH0UH, where Y∈Cr×(n-r) and U is given by (7). Since YH=Y1T-Y2Ti,(31)gYH=Y1T,-Y2T.By Lemma14, one has(32)vec⁡gYH=vec⁡Y1Tvec⁡-Y2T=Pr,n-rvec⁡Y1-Pr,n-rvec⁡Y2=Pr,n-r00-Pr,n-rvec⁡gY.Set (33)AU=D,E,UHB=FG,N=fFH⊗E¯Pr,n-r00-Pr,n-r+fGH⊗D¯, and P(r,n-r) is given by (29). Then one has the following results.Theorem 15.
LetA∈Cm×n, B∈Cn×k, and C∈Cm×k. Then on has the following:
(1) The solutionX∈SA to matrix equation AXB=C is given by(34)X=U0YYH0UH,where (35)vec⁡Y=Irn-r,iIrn-rN†c+I-N†Ny,U is given by (7), c=vec⁡(g(C))∈R2km, and y∈R2r(n-r) is an arbitrary vector.
(2) The solutionX∈SA with the least norm is given by(36)X=U0Y0Y0H0UH,where U is given by (7) and vec⁡(Y0)=[Irn-r,iIrn-r]N†c, c=vec⁡(g(C)).Proof.
It follows from Lemma6 that(37)vec⁡gEYHF=fFH⊗E¯vec⁡gYH,(38)vec⁡gDYG=fGH⊗D¯vec⁡gY. Combining (32) and (37), we obtain (39)vec⁡gEYHF=fFH⊗E¯vec⁡gYH=fFH⊗E¯Pr,n-r00-Pr,n-rvec⁡gY. Then (40)AXB-C=AU0YYH0UHB-C=EYHF+DYG-C=gEYHF+gDYG-gC=vec⁡gEYHF+vec⁡gDYG-vec⁡gC=fFH⊗E¯Pr,n-r00-Pr,n-rvec⁡gY+fGH⊗D¯vec⁡gY-vec⁡gC=Nx-c,
wherec=vec⁡(g(C)) and x=vec⁡(g(Y)).
By Lemma5, the least squares solution is given by(41)x=vec⁡Y1vec⁡Y2=N†c+I-N†Ny,
wherey∈R2r(n-r) is an arbitrary vector.
The least squares solution with least norm isx0=N†c.
So(42)vec⁡Y1=Irn-r,0x,vec⁡Y2=0,Irn-rx,vec⁡Y10=Irn-r,0x0,vec⁡Y20=0,Irn-rx0. Now, it is easy to see that our required solutions are given by (34) and (36).


## 4. Conclusion

By exploiting the decomposition of Hermitian (anti)reflexive matrix and the Kronecker product, we derive the expression of the least squares Hermitian (anti)reflexive solution with the least norm toAXB=C. This method overcomes the difficulty of finding the structured least squares solution with the least norm but creates a large-scale linear system A^x=b^. Therefore, it fits well with the small-scale coefficient matrix, but, for the large-scale case, it may not work so well.



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