# On Hermitian Solutions of the Generalized Quaternion Matrix EquationAXB+CXD=E

Authors: Yong Tian; Xin Liu; Shi-Fang Yuan
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

The paper deals with the matrix equationAXB+CXD=E over the generalized quaternions. By the tools of the real representation of a generalized quaternion matrix, Kronecker product as well as vec-operator, the paper derives the necessary and sufficient conditions for the existence of a Hermitian solution and gives the explicit general expression of the solution when it is solvable and provides a numerical example to test our results. The paper proposes a unificated algebraic technique for finding Hermitian solutions to the mentioned matrix equation over the generalized quaternions, which includes many important quaternion algebras, such as the Hamilton quaternions and the split quaternions.


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## Body

## 1. Introduction

In 1843, Irish Mathematician William Rowan Hamilton introduced the Hamilton quaternions. It is a great event in the history of mathematics. The set of Hamilton quaternions can form a skew field [1, 2]. In 1849, James Cockle introduced the split quaternions. It can be used to express Lorentzian rotations, which is used in geometry and physics (see [3–5]). In this paper, we consider a more generalized case, that is, the generalized quaternions, which is in the form of [6](1)q=q1+q2i+q3j+q4k,q1,q2,q3,q4∈R.where i2=u,j2=v,k2=ijk=−uv,ij=−ji=k,jk=−kj=−vi,ik=−ki=uj. In the paper, we only focus on the cases of 0≠u,v∈R. We call the set QG by the generalized quaternions. Obviously, QG is a noncommutative 4-dimensional Clifford algebra. Specially, QG is the Hamilton quaternion ring Q if u=v=−1, QG is the split quaternion ring Qs if u=−1,v=1, QG is the nectarine ring Qn if u=1,v=−1, QG is the conectarine ring Qc if u=v=1.Throughout this paper, letRm×n, Cm×n, QGm×n, SRn×n, and ASRn×n denote the set of all m×n real matrices, the set of all m×n complex matrices, the set of all m×n generalized quaternion matrices, the set of all n×n real symmetric matrices, and the set of all n×n real antisymmetric matrices, respectively. The identity matrix of order n is denoted by In. The zero matrix with suitable size is denoted by 0. We define the conjugate of q∈QG as q¯=q1−q2i−q3j−q4k. For A=aij∈QGm×n; we use A¯=aij¯, AT to denote the conjugate matrix, the transpose matrix of A, respectively. AH=A1T−A2Ti−A3Tj−A4Tk is the conjugate transpose of A. We call a matrix A∈QGn×n is Hermitian if AH=A, which we denote it by A∈HQGn×n, where HQGn×n is the set of all Hermitian generalized quaternion matrices with the size of n×n.In recent decades, different kinds of matrix equations over some quaternion algebras had been studied, such as theAHX+XA=B, AX=B, AXB=C, AXB+CYD=E, and AXB+CXD=E, AX⋆−XB=CY+D, and X−AX⋆B=CY⋆+D over the real/complex fields or some quaternion algebras (see [7–29] and references cited therein). For now, only few papers explored some fundamental properties and matrix equation over the generalized quaternions, which one may refer to [9, 17, 30, 31].Hermitian matrix has attracted lots of attentions because of its great importance. There are some results about Hermitian solutions of matrix equations over several kinds of quaternion algebras (see [6, 26, 28, 32]). For example, Yu et al. [6] studied Hermitian solutions to the generalized quaternion matrix equation AXB+CX∗D=E by the real representation method; Yuan et al. [32] discussed Hermitian solutions to the split quaternion matrix equation AXB+CXD=E by using the complex representation method. Based on the work mentioned above, and inspired by the methods in ([28, 32]), we discuss the following problem:Problem I: givenA∈QGm×n,B∈QGn×s, C∈QGm×n,D∈QGn×s,andE∈QGm×s, find the solution set(2)HE=X|X∈HQGn×n,AXB+CXD=E.


## 2. Properties of the Generalized Quaternion Matrices

For anyA=A1+A2i+A3j+A4k∈QGm×n with A1,A2,A3,A4∈Rm×n, we define(3)ΦA=A1,A2,A3,A4.Obviously, the mapΦA is an isomorphism of A, we denote by ΦA≅A. Next, we propose a real matrix representation for the generalized quaternion matrix A∈QGm×n:(4)AR=A1uA2vA3−uvA4A2A1vA4−vA3A3−uA4A1uA2A4−A3A2A1∈R4m×4n.ForA=aij∈QGm×n,B∈QGp×q, the Kronecker product of A and B is defined as A⊗B=aijB∈QGmp×nq. For the generalized quaternion matrices A,B,C,D,E,F,G,andH with suitable sizes and the real number k, we have(5)kA⊗C=A⊗kC=kA⊗C,A,B,C⊗D=A⊗D,B⊗D,C⊗D,EGFH⊗K=E⊗KG⊗KF⊗KH⊗K.For the matrixA=aij∈QGm×n, let aj=a1j,a2j,…,amj with j=1,2,…,n, we denote the vector vecA by(6)vecA=a1,a2,…,anT.Throughout the paper, we denote(7)Nn=In0000uIn0000vIn0000−uvIn,Qn=0In00In0000001vIn00−1vIn0,Ln=00In0000−1uInIn00001uIn00,Sn=000In00−In00In00In000.The following are some properties of generalized quaternion matrices.Proposition 1.
Letq∈QG,A∈QGm×n,B∈QGn×t, andC∈QGn×n. Then,(i)
qq¯≥0 does not hold in general(ii)
ABT≠BTAT in general(iii)
AB¯≠A¯B¯ in general(iv)
AHR≠ARH in general(v)
C¯−1≠C−1¯ in general(vi)
CT−1≠C−1T in generalProof.
Whenu=v=−1 and u=−1,v=1, QG is the quaternion ring and the split quaternion ring, we can refer to [26, 32]), and the other cases can be easily obtained by direct calculation.
Some important properties ofAR and ΦA are as follows.Proposition 2.
Letk∈R, A,B∈QGm×n, C∈QGn×t, and D∈QGn×n. Then,(i)
A=B if and only if AR=BR, A=B if and only if ΦA=ΦB(ii)
A+BR=AR+BR, kAR=kAR, ΦA+B=ΦA+ΦB, ΦkA=kΦA(iii)
ΦAC=ΦANnCRNt−1, ACR=ARCR(iv)
IfD is invertible, then D−1R=DR−1(v)
InR=I4nProof.
Since the proofs of (i), (ii), (iv), and (v) are easy, we only prove (iii). By direct calculation, we have(8)AC=A1+A2i+A3j+A4kC1+C2i+C3j+C4k=A1C1+uA2C2+vA3C3−uvA4C4+A1C2+A2C1−vA3C4+vA4C3i+A1C3+uA2C4+A3C1−uA4C2j+A1C4+A2C3−A3C2+A4C1k=Z1+Z2i+Z3j+Z4k.
Thus,(9)AC≅ΦAC=Z1Z2Z3Z4,where(10)Z1=A1C1+uA2C2+vA3C3−uvA4C4,Z2=A1C2+A2C1−vA3C4+vA4C3,Z3=A1C3+uA2C4+A3C1−uA4C2,Z4=A1C4+A2C3−A3C2+A4C1.
Now, it is easy to verifyAC≅ΦAC=Z1Z2Z3Z4=ΦANnCRNt−1.


## 3. The Structure ofvecAXB

In the section, we investigate the structure ofvecAXB. For A∈Cm×n, B∈Cn×s, and C∈Cs×t, it is well known that(11)vecABC=CT⊗AvecB.However, (11) cannot hold in the generalized quaternions for the noncommutative multiplication of the generalized quaternions. Thus, we need to study the structure of vecΦABC.Theorem 1.
LetA=A1+A2i+A3j+A4k∈QGm×n, B=B1+B2i+B3j+B4k∈QGn×s, and C=C1+C2i+C3j+C4k∈QGs×t, where Ai∈Rm×n,Bi∈Rn×s, and Ci∈Rs×ti=1,2,3,4. Then,(12)vecΦABC=NsCRNt−1T⊗A1+NsCRQtT⊗A2+NsCRLtT⊗A3+NsCRStT⊗A4vecΦB.Proof.
By (iii) in Proposition2,(13)ΦABC=ΦANnBCRNt−1=ΦANnBRCRNt−1=A1A2A3A4In0000uIn0000vIn0000−uvInB1uB2vB3−uvB4B2B1vB4−vB3B3−uB4B1uB2B4−B3B2B1×C1uC2vC3−uvC4C2C1vC4−vC3C3−uC4C1uC2C4−C3C2C1It00001uIt00001vIt0000−1uvIt=γ1γ2γ3γ4,where(14)γ1=A1B1C1+uA2B2C1+vA3B3C1−uvA4B4C1+uA1B2C2+uA2B1C2−uvA3B4C2+uvA4B3C2+vA1B3C3+uvA2B4C3+vA3B1C3−uvA4B2C3−uvA1B4C4−uvA2B3C4+uvA3B2C4−uvA4B1C4,γ2=A1B1C2+uA2B2C2+vA3B3C2−uvA4B4C2+A1B2C1+A2B1C2−vA3B4C1+vA4B3C1−vA1B3C4−uvA2B4C4−vA3B1C4+uvA4B2C4+vA1B4C3+vA2B3C3−vA3B2C3+vA4B1C3,γ3=A1B1C3+uA2B2C3+vA3B3C3−uvA4B4C3+uA1B2C4+uA2B1C4−uvA3B4C4+uvA4B3C4+A1B3C1+uA2B4C1+A3B1C1−uA4B2C1−uA1B4C2−uA2B3C2+uA3B2C2−uA4B1C2,γ4=A1B1C4+uA2B2C4+vA3B3C4−uvA4B4C4+A1B2C3+A2B1C3−vA3B4C3+vA4B3C3−A1B3C2−uA2B4C2−A3B1C2+uA4B2C2+A1B4C1+A2B3C1−A3B2C1+A4B1C1.
It follows from (11) that(15)vecγ1=C1T⊗A1vecB1+uC1T⊗A2vecB2+vC1T⊗A3vecB3−uvC1T⊗A4vecB4+uC2T⊗A1vecB2+uC2T⊗A2vecB1−uvC2T⊗A3vecB4+uvC2T⊗A4vecB3+vC3T⊗A1vecB3+uvC3T⊗A2vecB4+vC3T⊗A3vecB1−uvC3T⊗A4vecB2−uvC4T⊗A1vecB4−uvC4T⊗A2vecB3+uvC4T⊗A3vecB2−uvC4T⊗A4vecB1,vecγ2=C2T⊗A1vecB1+uC2T⊗A2vecB2+vC2T⊗A3vecB3−uvC2T⊗A4vecB4+C1T⊗A1vecB2+C1T⊗A2vecB1−vC1T⊗A3vecB4+vC1T⊗A4vecB3−vC4T⊗A1vecB3−uvC4T⊗A2vecB4−vC4T⊗A3vecB1+uvC4T⊗A4vecB2+vC3T⊗A1vecB4+vC3T⊗A2vecB3−vC3T⊗A3vecB2+vC3T⊗A4vecB1,vecγ3=C3T⊗A1vecB1+uC3T⊗A2vecB2+vC3T⊗A3vecB3−uvC3T⊗A4vecB4+uC4T⊗A1vecB2+uC4T⊗A2vecB1−uvC4T⊗A3vecB4+uvC4T⊗A4vecB3+C1T⊗A1vecB3+uC1T⊗A2vecB4+C1T⊗A3vecB1−uC1T⊗A4vecB2−uC2T⊗A1vecB4−uC2T⊗A2vecB3+uC2T⊗A3vecB2−uC2T⊗A4vecB1,vecγ4=C4T⊗A1vecB1+uC4T⊗A2vecB2+vC4T⊗A3vecB3−uvC4T⊗A4vecB4+C3T⊗A1vecB2+C3T⊗A2vecB1−vC3T⊗A3vecB4+vC3T⊗A4vecB3−C2T⊗A1vecB3−uC2T⊗A2vecB4−C2T⊗A3vecB1+uC2T⊗A4vecB2+C1T⊗A1vecB4+C1T⊗A2vecB3−C1T⊗A3vecB2+C1T⊗A4vecB1.
Thus,(16)vecΦABC=vecγ1vecγ2vecγ3vecγ4=C1C2C3C4uC2C1uC4C3vC3−vC4C1−C2−uvC4vC3−uC2C1T⊗A1vecΦB+uC2C1uC4C3uC1uC2uC3uC4−uvC4vC3−uC2C1uvC3−uvC4uC1−uC2T⊗A2vecΦB+vC3−vC4C1−C2uvC4−vC3uC2−C1vC1vC2vC3vC4−uvC2−vC1−uvC4−vC3T⊗A3vecΦB+−uvC4vC3−uC2C1−uvC3uvC4−uC1uC2uvC2vC1uvC4vC3−uvC1−uvC2−uvC3−uvC4T⊗A4vecΦB=NsCRNt−1T⊗A1+NsCRQtT⊗A2+NsCRLtT⊗A3+NsCRStT⊗A4vecΦB,which completed our proof.
Yuan et al. [26] studied the vecΦABC over Q, while Theorem 1 extends it to the result over QG. As we can see that Theorem 1 maps the product of generalized quaternion matrices into the product of real matrices by using the real representation method, by this way, we can convert a generalized quaternion matrix equation into a real one.
In the following, we introduce some definitions and useful lemmas.Definition 1.
For the matrixA=aij∈QGn×n, let a1=a11,2a21,…,2an1, a2=a22,2a32,…,2an2,…, an−1=an−1n−1,2ann−1, an=ann, we denote(17)vecSA=a1,a2,…,an−1,anT∈QGnn+1/2.Definition 2.
For the matrixB=bij∈QGn×n, let b1=b21,b31,…,bn1, b2=b32,b42,…,bn2,\\…, bn−2=bn−1n−2,bnn−2, bn−1=bnn−1, we denote(18)vecAB=2b1,b2,…,bn−2,bn−1T∈QGnn−1/2.Lemma 1 (see [27]).
SupposeX∈Rn×n, then(19)iIfX∈SRn×n,then vecX=KSvecSX,where vecSX is represented as (4), and the matrix KS∈Rn2×nn+1/2 is of the following form:(20)KS=122e1e2⋯en−1en00⋯0⋯0000e1⋯002e2e3⋯en⋯00000⋯000e2⋯0⋯000⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮00⋯e1000⋯0⋯2en−1en000⋯0e100⋯e2⋯0en−12en,where ei is the i-th column of the identity matrix of order n.(21)iiIfX∈ASRn×n,then vecX=KAvecAX,where vecAX is represented as (5), and the matrix KA∈Rn2×nn−1/2 is of the following form:(22)KA=12e2e3⋯en−1en0⋯00⋯0−e10⋯00e3⋯en−1en⋯00−e1⋯00−e2⋯00⋯0⋮⋮⋮⋮⋮⋮⋮⋮00⋯−e100⋯−e20⋯en00⋯0−e10⋯0−e2⋯−en−1,where ei is the i-th column of In. Obviously, KSTKS=Inn+1/2, KATKA=Inn−1/2.
By Lemma1, we have the following.Theorem 2.
ForX=X1+X2i+X3j+X4k∈HQGn×n, then(23)vecΦX=WvecSX1vecAX2vecAX3vecAX4,in which(24)W=KS0000KA0000KA0000KA.Proof.
For anyX=X1+X2i+X3j+X4k∈HQGn×n,Xi∈Rn×ni=1,2,3,4, it is easy to see that(25)X∈HQGn×n⟺X1T=X1,X2T=−X2,X3T=−X3,X4T=−X4.
Obviously,X1 is symmetric, and X2, X3, and X4 are antisymmetric. By Lemma 1, we have(26)vecΦX=vecX1vecX2vecX3vecX4=KSvecSX1KAvecAX2KAvecAX3KAvecAX4=KS0000KA0000KA0000KAvecSX1vecAX2vecAX3vecAX4=WvecSX1vecAX2vecAX3vecAX4.
Combining Theorems1 and 2, we yield the following result.Theorem 3.
LetA=A1+A2i+A3j+A4k∈QGm×n, X=X1+X2i+X3j+X4k∈HQGn×n, and B=B1+B2i+B3j+B4k∈QGn×s, where Ai∈Rm×n,Xi∈Rn×n, and Bi∈Rn×si=1,2,3,4. Then,(27)vecΦAXB=NnBRNs−1T⊗A1+NnBRQsT⊗A2+NnBRLsT⊗A3+NnBRSsT⊗A4WvecSX3vecAX3vecAX3vecAX3.


## 4. The Hermitian Solutions

Based on our earlier discussion, we now pay our attention to Problem I. The following notation is necessary for deriving a solution to Problem I. LetA=A1+A2i+A3j+A4k∈QGm×n, B=B1+B2i+B3j+B4k∈QGn×s, C=C1+C2i+C3j+C4k∈QGm×n, D=D1+D2i+D3j+D4k∈QGn×s, and E=E1+E2i+E3j+E4k∈QGm×s. In the remaining of the paper, we set(28)P=NnBRNs−1T⊗A1+NnBRQsT⊗A2+NnBRLsT⊗A3+NnBRSsT⊗A4W+NnDRNs−1T⊗C1+NnDRQsT⊗C2+NnDRLsT⊗C3+NnDRSsT⊗C4W,e=vecE1vecE2vecE3vecE4.We also need the following lemma.Lemma 2 (see [33]).
The matrix equationAx=b, with A∈Rm×n and b∈Rm, has a solution x∈Rn if and only if(29)AA+b=b,where A+ is the Moore–Penrose inverse of the matrix A. In this case, it has the general solution(30)x=A+b+In−A+Ay,where y∈Rn is an arbitrary vector, and it has the unique solution x=A+b for the case when rankA=n. The solution of the matrix equation Ax=b with the least norm is x=A+b.Theorem 4.
LetA∈QGm×n,B∈QGn×s, C∈QGm×n,D∈QGn×s, and E∈QGm×s. Then, Problem I has a solution X∈HE if and only if(31)PP+e=e.If this condition satisfies, then(32)HE=X|vecΦX=WP+e+I2n2−n−P+Py,where y∈R2n2−n is an arbitrary vector.Furthermore, if (31) holds, then the generalized quaternion matrix equation (2) has a unique solution X∈HE if and only if(33)rankP=2n2−n.In this case,(34)HE=X|vecΦX=WP+e.Proof.
By (ii) in Proposition2 and Theorem 3, we have(35)AXB+CXD=E⟺ΦAXB+ΦCXD=ΦE⟺vecΦAXB+vecΦCXD=vecΦE⟺PvecSX1vecAX2vecAX3vecAX4=e.
By Lemma2, Problem I has a solution X∈HE if and only if (31) holds. If this condition satisfies, then(36)vecSX1vecAX2vecAX3vecAX4=P+e+I2n2−n−P+Py.
Also by (23),(37)vecΦX=WvecSX1vecAX2vecAX3vecAX4=WP+e+I2n2−n−P+Py,where y∈R2n2−n is an arbitrary vector. We can draw the conclusion (32). Furthermore, if (31) holds, Problem I has a unique solution X∈HE if and only if(38)P+P=I2n2−n.
That is, (33) holds. In the case, we obtain (34).


## 5. Example

In this section, we give two examples to illustrate our results.Example 1.
Consider the Hamilton quaternion matrix equationAXB+CXD=E, where(39)A=i1+j,B=−2+ij−2k,C=k1+j,D=3k1−2i+j,E=1+2i+3j−k.
Obviously, the Hamilton quaternions meanu=v=−1. By (4) and (28), we easily get(40)BR=−2−10000−121−200002100−2−11−200001−2−2−100,DR=000−312−1000−30−21010300101230000−1−21,P=N2BRN1−1T⊗A1+N2BRQ1T⊗A2+N2BRL1T⊗A3+N2BRS1T⊗A4W+N2DRN1−1T⊗C1+N2DRQ1T⊗C2+N2DRL1T⊗C3+N2DRS1T⊗C4W,e=123−1.
By Theorem4 and MATLAB, calculating the formula vecΦX=WP+e gives a solution(41)X=−0.0778−0.3899−0.38990.1907+00.5091−0.50910i+0−0.31800.31800j+00.2478−0.24780k.Example 2.
Consider the generalized quaternion matrix equationAXB+CXD=E with u=2,v=3, where(42)A=j1−i,B=−1−ij+2k,C=−k1−k+2j,D=−k1−j,E=1−2i+j−3k.
By (4) and (28), we easily get(43)BR=−1−200003−12−1−100006−300−1−21−40000−1−12−100,DR=000610−3000−3001030200−1010−10000101,P=N2BRN1−1T⊗A1+N2BRQ1T⊗A2+N2BRL1T⊗A3+N2BRS1T⊗A4W+N2DRN1−1T⊗C1+N2DRQ1T⊗C2+N2DRL1T⊗C3+N2DRS1T⊗C4W,e=1−21−3.
By Theorem4 and MATLAB, calculating the formula vecΦX=WP+e gives a solution(44)X=0.8970−0.1826−0.1826−0.2064+00.2769−0.27690i+0−0.32280.32280j+0−0.01110.01110k.


## 6. Conclusion

In this paper, we provide a direct method to find Hermitian solutions of the generalized quaternion matrix equationAXB+CXD=E by using the real representation of generalized quaternion matrices, Kronecker product and vec-operator. We give the necessary and sufficient conditions for the existence of a Hermitian solution and also derive the general solution when the matrix equation is consistent. The paper proposes an algebraic technique for finding the Hermitian solutions to the above matrix equation over the generalized quaternions. The generalized quaternions include many important quaternion algebras, for instance, Q, Qs, Qn, and Qc, thus the paper actually proposes a unified technique to solve the Hermitian solution problems over the several quaternion algebras.



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