# Least-Norm of the General Solution to Some System of Quaternion Matrix Equations and Its Determinantal Representations

Authors: Abdur Rehman; Ivan Kyrchei; Muhammad Akram; Ilyas Ali; Abdul Shakoor
Journal: Abstract and Applied Analysis
 (2019)

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## Abstract

We constitute some necessary and sufficient conditions for the systemA1X1=C1, X1B1=C2, A2X2=C3, X2B2=C4, A3X1B3+A4X2B4=Cc, to have a solution over the quaternion skew field in this paper. A novel expression of general solution to this system is also established when it has a solution. The least norm of the solution to this system is also researched in this article. Some former consequences can be regarded as particular cases of this article. Finally, we give determinantal representations (analogs of Cramer’s rule) of the least norm solution to the system using row-column noncommutative determinants. An algorithm and numerical examples are given to elaborate our results.


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## Body

## 1. Introduction

In the whole article, the notationR is reserved for the real number field and Hm×n stands for the set of all m×n matrices over the quaternion skew field(1)H=b0+b1i+b2j+b3k∣i2=j2=k2=ijk=-1,b0,b1,b2,b3∈R.Hrm×n specifies its subset of matrices with a rank r. For A∈Hm×n, let A∗,R(A) and N(A) designate the conjugate transpose, the column right space and the left row space of A. dim R(A) illustrates the size of R(A) and dim R(A)=dim N(A) by [1], which is known as the rank of A denoted by r(A).Definition 1.
The Moore-Penrose inverse ofA∈Hm×n, denoted by A†, is defined to be the unique solution X to the following four matrix equations(2)1AXA=A,2XAX=X,3AX∗=AX,4XA∗=XA.Matrices satisfying (1) and (2) are known as reflexive inverses.Note that the reflexive inverse is denoted most often byAr- but sometimes by A+ (see, e.g., [2]) that is different from the denotation of the Moore-Penrose by A†. We will use the denotation A+ for the reflexive inverse.SupposeI refers an identity matrix with feasible size. In addition, RA=I-AA†, LA=I-A†A represent a pair of orthogonal projectors induced by A, respectively, and RA2=RA, RA∗=RA, LA2=LA, LA∗=LA, and RA∗=LA.Quaternions were invented by Hamilton in 1843. Zhang presented a detail survey on quaternion matrices in [3]. Quaternions provide a concise mathematical method for representing the automorphisms of three- and four-dimensional spaces. The representations by quaternions are more compact and quicker to compute than the representations by matrices [4]. For this reason, an increasing number of applications based on quaternions are found in various fields, such as color imaging, geometry, mechanics, linear adaptive filter, altitude control and computer science, signal processing, in particular as quaternion-valued neural networks, etc. [5–10].The research of matrix equations have both applied and theoretical importance. In particular, the Sylvester-type matrix equations have far reaching applications in singular system control [11], system design [12], robust control [13], feedback [14], perturbation theory [15], linear descriptor systems [16], neural networks [17], and theory of orbits [18].Some recent work on generalized Sylvester matrix equations and their systems can be observed in [19–31]. In 2014, Bao [32] examined the least-norm and extremal ranks of the least square solution to the quaternion matrix equations(3)A1X=C1,XB1=C2,A3XB3=Cc.Wang et al. [33] examined the expression of the general solution to the system(4)A1X1=C1,A2X2=C3,A3X1B3+A4X2B4=Cc,And, as an application, the P-symmetric and P-skew-symmetric solution to (5)AaX=Ca,AbXBb=Cbhas been established. Li et al. [34] established a novel expression to the general solution of system (4) and they computed the least-norm of general solution to (4). In 2009, Wang et al. [35] constituted the expression of the general solution to(6)A1X1=C1,X1B1=C2,A2X2=C3,X2B2=C4,A3X1B3+A4X2B4=Cc,and as an application they explored the (P,Q)-symmetric solution to the system (7)AaX=Ca,XBb=Cb,AcXBc=Cc.Some latest findings on the least-norm of matrix equations and(P,Q)-symmetric matrices can be consulted in [36–40]. Furthermore, our main system (6) is a special case of the following system:(8)A1X1=C1,X2B1=D1,A2X3=C2,X3B2=D2,A3X4=C3,X4B3=D3,A4X1+X2B4+C4X3D4+C5X4D5=Cc,which has been investigated by Zhang in 2014. But the expressions provided for the X1,X2,X3, and X4 in [41], we are in position to calculate the least-norm of the solutions with its determinantal representations. When some given matrices are zero in (8), then it becomes our system and we will give such kind of expressions in which the least-norm of the solutions can also be computed with its determinantal representations. It is worthy to note that Zhang examined (8) with complex settings and we will consider our system (6) with quaternion settings.According to our best of knowledge, the least-norm of the general solution to system (6) is not investigated by any one. Motivated by the vast application of quaternion matrices and the latest interest of least-norm of matrix equations, we construct a novel expression of the general solution to system (6) and apply this to investigate the least-norm of the general solution to system (6) over H in this paper. Observing that systems (3) and (4) are particular cases of our system (6), solving system (6) will encourage the least-norm to a wide class of problems in the collected work.Since the general solutions of considered systems are expressed in term of generalized inverses, another goal of the paper is to give determinantal representations of the least-norm of the general solution to system (6) based on determinantal representations of generalized inverses.Determinantal representation of a solution gives a direct method of its finding analogous to the classical Cramer’s rule that has important theoretical and practical significance. Through looking for their more applicable explicit expressions, there are various determinantal representations of generalized inverses even with the complex or real entries, in particular for the Moore-Penrose inverse (see, e.g., [42–44]). By virtue of noncommutativity of quaternions, the problem for determinantal representation of generalized quaternion inverses is more complicated, and only now it can be solved due to the theory of column-row determinants introduced in [45, 46]. Within the framework of the theory of noncommutative row-column determinants, determinantal representations of various generalized quaternion inverses and generalized inverse solutions to quaternion matrix equations have been derived by one of our authors (see, e.g.[47–54]) and by other researchers (see, e.g. [55–57]). Moreover, Song et al. [58] have just recently considered determinantal representations of general solution to the two-sided coupled generalized Sylvester matrix equation over H obtained using the theory of row-column determinants as well. But their proposed approach differs from our proposed. In [58], for determinantal representations of the general solution to the equation supplementary matrices have been used that not always easy to get. While, by proposed method only coefficient matrices of the equations are used. More detailed Cramer’s rule to solutions and (skew-)Hermitian solutions of some systems of matrix equations and generalized Sylvester matrix equation over H are recently explored in [59, 60] and [61, 62], respectively.The remainder of our article is directed as follows. In Section2, we commence with some needed known results about systems of matrix equations and determinantal representations of the Moore-Penrose inverse and of solutions to the quaternion matrix equations. In Section 3, we provide a new expression of the general solution to our system (6) and present an algorithm with an example. We discuss the least-norm of the general solution to (6) over H in Section 4. In Section 5, determinantal representations of the general solution to (6) are derived and other example to elaborate obtained Cramer’s Rule to system (6) with data from the example in Section 3 is given. As expected, we get the same solution. Finally, in Section 6, the conclusions are drawn.


## 2. Preliminaries

We commence with the following lemmas which have crucial function in the construction of the chief outcomes of the following sections.


### 2.1. The General Solution to System (6)

Lemma 2 (see [63]).
LetA∈Hs×t,B∈Hs×k,andC∈Hl×t be given. Then
(1)r(A)+r(RAB)=r(B)+r(RBA)=rAB.
(2)r(A)+r(CLA)=r(C)+r(ALC)=rAC.
(3)r(B)+r(C)+r(RBALC)=rABC0.Lemma 3 (see [64]).
LetA, B, and C be known matrices over H with right sizes. Then
(1)A†=(A∗A)†A∗=A∗(AA∗)†.
(2)LA=LA2=LA∗,RA=RA2=RA∗.
(3)LA(BLA)†=(BLA)†,(RAC)†RA=(RAC)†.Lemma 4 (see [65]).
LetΦ,Ω be matrices over H and (9)Φ=Φ1Φ2,Ω=Ω1Ω2,F=Φ2LΦ1,T=RΩ1Ω2.Then(10)LΦ=LΦ1LF,LΩ=LΩ1-Ω1†Ω2LT0LT,RΩ=RTRΩ1,RΦ=RΦ10-RFΦ2Φ1†RF,where Φ1+, Ω1+ are any fixed reflexive inverses, LΦ1 and RΩ1 stand for the projectors LΦ1=I-Φ1+Φ1, RΩ1=I-Ω1Ω1+ induced by Φ1, Ω1, respectively.Remark 5.
Since Moore-Penrose inverses are reflexive inverses, this lemma can be used for Moore-Penrose inverses without any changes. It has taken place in ([64], Lemma 2.4). But for more credibility, we prove this lemma below for the Moore-Penroses inverse as well.Lemma 6 (see [66]).
Suppose that(11)B1XC1+B2YC2=Ais consistent linear matrix equation, where B1∈Hm×p, C1∈Hq×n, B2∈Hm×s, C2∈Ht×n and A∈Hm×n, respectively. Then (1)
The general solution of the homogeneous equation,(12)B1XC1+B2YC2=0,can be expressed by(13)X=X1X2+X3,Y=Y1Y2+Y3,whereX1-X3 and Y1-Y3 are general solution of the following four homogeneous matrix expressions (14)B1X1=-B2Y1,X2C1=Y2C2,B1X3C1=0,B2Y3C2=0.By computing the value of unknowns in the above equations and using them inX and Y, we have (15)X=S1LGURHT1+LB1V1+V2RC1,Y=S2LGURHT2+LB2W1+W2RC2,whereS1=[Ip,0], S2=[0,Is], T1=Iq0, T2=0It, G=[B1,B2], and H=C1-C2; the matrices U,V1,V2,W1andW2 are free to vary over H.(2)
Assume that the matrix expression (11) is solvable, then its general solution can be expressed as (16)X=X0+X1X2+X3,Y=Y0+Y1Y2+Y3,whereX0 and Y0 are any pair of particular solutions to (11).
It can also be written as(17)X=X0+S1LGURHT1+LB1V1+V2RC1,Y=Y0+S2LGURHT2+LB2W1+W2RC2.Lemma 7 (see [67]).
LetA1∈Hm1×n1, B1∈Hr1×s1, C1∈Hm1×r1, and C2∈Hn1×s1 be given and X1∈Hn1×r1 to be determined. Then the system(18)A1X1=C1,X1B1=C2,is consistent if and only if(19)RA1C1=0,C2LB1=0,A1C2=C1B1.Under these conditions, the general solution to (18) can be established as (20)X1=A1†C1+LA1C2B1†+LA1U1RB1,where U1 is a free matrix over H with accordant dimension.Lemma 8 (see [35]).
LetA1∈Hm1×p1, B1∈Hq1×n1, C1∈Hm1×q1, C2∈Hp1×n1, A2∈Hm2×p2, B2∈Hq2×n2, C3∈Hm2×q2, C4∈Hp2×n2,, A3∈Hs×p1, B3∈Hq1×t, A4∈Hs×p2, B4∈Hq2×t, Cc∈Hs×t be given and X1∈Hp1×q1, X2∈Hp2×q2 to be determined. Denote (21)A=A3LA1,B=RB1B3,C=A4LA2,D=RB2B4,N=DLB,M=RAC,S=CLM,E=Cc-A3A1†C1B3-AC2B1†B3-A4A2†C3B4-CC4B2†B4.Then the following conditions are tantamount: (1)
System (6) is resolvable.(2)
The conditions in (19) are met and(22)RA2C3=0,C4LB2=0,A2C4=C3B2,RMRAE=0,RAELD=0,ELBLN=0,RCELB=0.(3)
The equalities in (19) and (22) are satisfied and (23)MM†RAD†D=RAE,CC†ELBN†N=ELB.
In these conditions, the general solution to system (6) can be written as(24)X1=A1†C1+LA1C2B1†+LA1A†EB†RB1-LA1A†CM†EB†RB1-LA1A†SC†EN†DB†RB1-LA1A†SV1RNDB†RB1+LA1LAU1+Z1RBRB1,(25)X2=A2†C3+LA2C4B2†+LA2M†RAED†RB2+LA2LMbS†SC†EN†RB2+LA2LMV1-S†SV1NN†RB2+LA2W1RDRB2,where U1,V1,W1andZ1 are free matrices over H with agreeable dimensions.


### 2.2. Determinantal Representations of Solutions to the Quaternion Matrix Equations

Due to noncommutativity of quaternions there is a problem of a determinant of matrices with noncommutative entries (which are also defined as noncommutative determinants). There are several versions of defining of noncommutative determinants (e.g., see [68–70]). But any of the previous noncommutative determinants has not fully retained those properties which it owned for matrices with real settings. Moreover, if functional properties of a noncommutative determinant over a ring are satisfied, then it takes on a value in its commutative subset. This dilemma can be avoided due to the theory of row-column determinants.ForA∈Hn×n, we define n row determinants and n column determinants. Suppose Sn is the symmetric group on the set In={1,…,n}.Definition 9 (see [45]).
The i th row determinant of A=(aij)∈Hn×n is defined for all i=1,…,n by putting (26)rdetiA=∑σ∈Sn-1n-raiik1aik1ik1+1…aik1+l1i…aikrikr+1…aikr+lrikr,σ=iik1ik1+1…ik1+l1ik2ik2+1…ik2+l2…ikrikr+1…ikr+lr,where σ is the left-ordered permutation. It means that its first cycle from the left starts with i, other cycles start from the left with the minimal of all the integers which are contained in it,(27)ikt<ikt+sforallt=2,…,r,s=1,…,lt,and the order of disjoint cycles (except for the first one) is strictly conditioned by increase from left to right of their first elements, ik2<ik3<⋯<ikr.Definition 10 (see [45]).
The j th column determinant of A=(aij)∈Hn×n is defined for all j=1,…,n by putting (28)cdetjA=∑τ∈Sn-1n-rajkrjkr+lr…ajkr+1jkr…ajjk1+l1…ajk1+1jk1ajk1j,(29)τ=jkr+lr…jkr+1jkr…jk2+l2…jk2+1jk2jk1+l1…jk1+1jk1j,noindent where τ is the right-ordered permutation. It means that its first cycle from the right starts with j, other cycles start from the right with the minimal of all the integers which are contained in it,(30)jkt<jkt+sforallt=2,…,r,s=1,…,lt,and the order of disjoint cycles (except for the first one) is strictly conditioned by increase from right to left of their first elements, jk2<jk3<⋯<jkr.
Since [45] for Hermitian A we have(31)rdet1A=⋯=rdetnA=cdet1A=⋯=cdetnA∈R,the determinant of a Hermitian matrix is defined by putting(32)det⁡A≔rdetiA=cdetiAforalli=1,…,n.Its properties are similar to the properties of an usual (commutative) determinant and they have been completely explored in [46] by using row and column determinants that are so defined only by construction.For determinantal representations of the Moore-Penrose inverse, we shall use the following notations. Letα≔α1,…,αk⊆1,…,m and β≔β1,…,βk⊆1,…,n be subsets of the order 1≤k≤min⁡m,n. Let Aβα be a submatrix of A whose rows are indexed by α and the columns indexed by β. Similarly, let Aαα be a principal submatrix of A whose rows and columns indexed by α. If A∈Hn×n is Hermitian, then Aαα is the corresponding principal minor of det⁡A. For 1≤k≤n, the collection of strictly increasing sequences of k integers chosen from 1,…,n is denoted by Lk,n≔α:α=α1,…,αk,1≤α1<⋯<αk≤n. For fixed i∈α and j∈β, let Ir,mi≔α:α∈Lr,m,i∈α denotes the collection of sequences of row indexes that contain the index i, and Jr,nj≔β:β∈Lr,n,j∈β denotes the collection of sequences of column indexes that contain j.Leta.j be the jth column and ai. be the ith row of A, respectively. Suppose A.jb denotes the matrix obtained from A by replacing its jth column with the column-vector b, and Ai.b denotes the matrix obtained from A by replacing its ith row with the row-vector b. We denote the ith row and the jth column of A∗ by ai.∗ and a.j∗, respectively.Lemma 11 (see [47]).
IfA∈Hrm×n, then the Moore-Penrose inverse A†=aij†∈Hn×m have the following determinantal representations,(33)aij†=∑β∈Jr,nicdetiA∗A.ia.j∗ββ∑β∈Jr,nA∗Aββ,and(34)aij†=∑α∈Ir,mjrdetjAA∗j.ai.∗αα∑α∈Ir,mAA∗αα.Remark 12.
For an arbitrary full-rank matrixA∈Hrm×n, a row-vector b∈H1×m, and a column-vector c∈Hn×1, we put (35)irdetiAA∗i.b=∑α∈Im,mirdetiAA∗i.bαα,det⁡AA∗=∑α∈Im,mAA∗αα,whenr=m,iicdetjA∗A.jc=∑β∈Jn,njcdetjA∗A.jcββ,det⁡A∗A=∑β∈Jn,nA∗Aββ,whenr=n.Corollary 13.
IfA∈Hrm×n, then the projection matrix A†A=:QA=qijn×n has the determinantal representation (36)qij=∑β∈Jr,nicdetiA∗A.ia˙.jββ∑β∈Jr,nA∗Aββ,where a˙.j is the jth column of A∗A∈Hn×n.Corollary 14.
IfA∈Hrm×n, then the projection matrix AA†=:PA=pijm×m has the determinantal representation(37)pij=∑α∈Ir,mjrdetjAA∗j.a¨i.αα∑α∈Ir,mAA∗αα,where a¨i. is the ith row of AA∗∈Hm×m.Lemma 15 (see [2]).
LetA∈Hm×n, B∈Hr×s, C∈Hm×s be known and X∈Hn×r be unknown. Then the matrix equation(38)AXB=Cis consistent if and only if AA†CB†B=C. In this case, its general solution can be expressed as(39)X=A†CB†+LAV+WRB,where V,W are arbitrary matrices over H with appropriate dimensions.In [71], it’s proved that (39) is the least squares solution to (38), and its minimum norm least squares solution is XLS=A†CB†.Lemma 16 (see [48]).
LetA∈Hr1m×n, B∈Hr2r×s. Then the minimum norm least squares solution X=A†CB†=(xij)∈Hn×r to (38) have determinantal representations, (40)xij=∑β∈Jr1,nicdetiA∗A.id.jBββ∑β∈Jr1,nA∗Aββ∑α∈Ir2,rBB∗αα,or (41)xij=∑α∈Ir2,rjrdetjBB∗j.di.Aαα∑β∈Jr1,nA∗Aββ∑α∈Ir2,rBB∗αα,where (42)d.jB=∑α∈Ir2,rjrdetjBB∗j.c~k.αα∈Hn×1,k=1,…,n,di.A=∑β∈Jr1,nicdetiA∗A.ic~.lββ∈H1×r,l=1,…,r,are the column vector and the row vector, respectively. c~k. and c~.l are the kth row and the lth column of C~=A∗CB∗.Corollary 17.
LetA∈Hkm×n, C∈Hm×s be known and X∈Hn×s be unknown. Then the matrix equation AX=C is consistent if and only if AA†C=C. In this case, its general solution can be expressed as X=A†C+LAV, where V is an arbitrary matrix over H with appropriate dimensions. Its minimum norm least squares solution X=A†C has the following determinantal representation, (43)xij=∑β∈Jk,nicdetiA∗A.ic^.jββ∑β∈Jk,nA∗Aββ,where c^.j is the jth column of C^=A∗C.Corollary 18.
LetB∈Hkr×s, C∈Hn×s be given, and X∈Hn×r be unknown. Then the equation XB=C is solvable if and only if C=CB†B and its general solution is X=CB†+WRB, where W is a any matrix with conformable dimension. Moreover, its minimum norm least squares solution X=CB† has the determinantal representation, (44)xij=∑α∈Ik,rjrdetjBB∗j.c^i.αα∑α∈Ik,rBB∗αα,where c^i. is the ith row of C^=CB∗.


## 2.1. The General Solution to System (6)

Lemma 2 (see [63]).
LetA∈Hs×t,B∈Hs×k,andC∈Hl×t be given. Then
(1)r(A)+r(RAB)=r(B)+r(RBA)=rAB.
(2)r(A)+r(CLA)=r(C)+r(ALC)=rAC.
(3)r(B)+r(C)+r(RBALC)=rABC0.Lemma 3 (see [64]).
LetA, B, and C be known matrices over H with right sizes. Then
(1)A†=(A∗A)†A∗=A∗(AA∗)†.
(2)LA=LA2=LA∗,RA=RA2=RA∗.
(3)LA(BLA)†=(BLA)†,(RAC)†RA=(RAC)†.Lemma 4 (see [65]).
LetΦ,Ω be matrices over H and (9)Φ=Φ1Φ2,Ω=Ω1Ω2,F=Φ2LΦ1,T=RΩ1Ω2.Then(10)LΦ=LΦ1LF,LΩ=LΩ1-Ω1†Ω2LT0LT,RΩ=RTRΩ1,RΦ=RΦ10-RFΦ2Φ1†RF,where Φ1+, Ω1+ are any fixed reflexive inverses, LΦ1 and RΩ1 stand for the projectors LΦ1=I-Φ1+Φ1, RΩ1=I-Ω1Ω1+ induced by Φ1, Ω1, respectively.Remark 5.
Since Moore-Penrose inverses are reflexive inverses, this lemma can be used for Moore-Penrose inverses without any changes. It has taken place in ([64], Lemma 2.4). But for more credibility, we prove this lemma below for the Moore-Penroses inverse as well.Lemma 6 (see [66]).
Suppose that(11)B1XC1+B2YC2=Ais consistent linear matrix equation, where B1∈Hm×p, C1∈Hq×n, B2∈Hm×s, C2∈Ht×n and A∈Hm×n, respectively. Then (1)
The general solution of the homogeneous equation,(12)B1XC1+B2YC2=0,can be expressed by(13)X=X1X2+X3,Y=Y1Y2+Y3,whereX1-X3 and Y1-Y3 are general solution of the following four homogeneous matrix expressions (14)B1X1=-B2Y1,X2C1=Y2C2,B1X3C1=0,B2Y3C2=0.By computing the value of unknowns in the above equations and using them inX and Y, we have (15)X=S1LGURHT1+LB1V1+V2RC1,Y=S2LGURHT2+LB2W1+W2RC2,whereS1=[Ip,0], S2=[0,Is], T1=Iq0, T2=0It, G=[B1,B2], and H=C1-C2; the matrices U,V1,V2,W1andW2 are free to vary over H.(2)
Assume that the matrix expression (11) is solvable, then its general solution can be expressed as (16)X=X0+X1X2+X3,Y=Y0+Y1Y2+Y3,whereX0 and Y0 are any pair of particular solutions to (11).
It can also be written as(17)X=X0+S1LGURHT1+LB1V1+V2RC1,Y=Y0+S2LGURHT2+LB2W1+W2RC2.Lemma 7 (see [67]).
LetA1∈Hm1×n1, B1∈Hr1×s1, C1∈Hm1×r1, and C2∈Hn1×s1 be given and X1∈Hn1×r1 to be determined. Then the system(18)A1X1=C1,X1B1=C2,is consistent if and only if(19)RA1C1=0,C2LB1=0,A1C2=C1B1.Under these conditions, the general solution to (18) can be established as (20)X1=A1†C1+LA1C2B1†+LA1U1RB1,where U1 is a free matrix over H with accordant dimension.Lemma 8 (see [35]).
LetA1∈Hm1×p1, B1∈Hq1×n1, C1∈Hm1×q1, C2∈Hp1×n1, A2∈Hm2×p2, B2∈Hq2×n2, C3∈Hm2×q2, C4∈Hp2×n2,, A3∈Hs×p1, B3∈Hq1×t, A4∈Hs×p2, B4∈Hq2×t, Cc∈Hs×t be given and X1∈Hp1×q1, X2∈Hp2×q2 to be determined. Denote (21)A=A3LA1,B=RB1B3,C=A4LA2,D=RB2B4,N=DLB,M=RAC,S=CLM,E=Cc-A3A1†C1B3-AC2B1†B3-A4A2†C3B4-CC4B2†B4.Then the following conditions are tantamount: (1)
System (6) is resolvable.(2)
The conditions in (19) are met and(22)RA2C3=0,C4LB2=0,A2C4=C3B2,RMRAE=0,RAELD=0,ELBLN=0,RCELB=0.(3)
The equalities in (19) and (22) are satisfied and (23)MM†RAD†D=RAE,CC†ELBN†N=ELB.
In these conditions, the general solution to system (6) can be written as(24)X1=A1†C1+LA1C2B1†+LA1A†EB†RB1-LA1A†CM†EB†RB1-LA1A†SC†EN†DB†RB1-LA1A†SV1RNDB†RB1+LA1LAU1+Z1RBRB1,(25)X2=A2†C3+LA2C4B2†+LA2M†RAED†RB2+LA2LMbS†SC†EN†RB2+LA2LMV1-S†SV1NN†RB2+LA2W1RDRB2,where U1,V1,W1andZ1 are free matrices over H with agreeable dimensions.


## 2.2. Determinantal Representations of Solutions to the Quaternion Matrix Equations

Due to noncommutativity of quaternions there is a problem of a determinant of matrices with noncommutative entries (which are also defined as noncommutative determinants). There are several versions of defining of noncommutative determinants (e.g., see [68–70]). But any of the previous noncommutative determinants has not fully retained those properties which it owned for matrices with real settings. Moreover, if functional properties of a noncommutative determinant over a ring are satisfied, then it takes on a value in its commutative subset. This dilemma can be avoided due to the theory of row-column determinants.ForA∈Hn×n, we define n row determinants and n column determinants. Suppose Sn is the symmetric group on the set In={1,…,n}.Definition 9 (see [45]).
The i th row determinant of A=(aij)∈Hn×n is defined for all i=1,…,n by putting (26)rdetiA=∑σ∈Sn-1n-raiik1aik1ik1+1…aik1+l1i…aikrikr+1…aikr+lrikr,σ=iik1ik1+1…ik1+l1ik2ik2+1…ik2+l2…ikrikr+1…ikr+lr,where σ is the left-ordered permutation. It means that its first cycle from the left starts with i, other cycles start from the left with the minimal of all the integers which are contained in it,(27)ikt<ikt+sforallt=2,…,r,s=1,…,lt,and the order of disjoint cycles (except for the first one) is strictly conditioned by increase from left to right of their first elements, ik2<ik3<⋯<ikr.Definition 10 (see [45]).
The j th column determinant of A=(aij)∈Hn×n is defined for all j=1,…,n by putting (28)cdetjA=∑τ∈Sn-1n-rajkrjkr+lr…ajkr+1jkr…ajjk1+l1…ajk1+1jk1ajk1j,(29)τ=jkr+lr…jkr+1jkr…jk2+l2…jk2+1jk2jk1+l1…jk1+1jk1j,noindent where τ is the right-ordered permutation. It means that its first cycle from the right starts with j, other cycles start from the right with the minimal of all the integers which are contained in it,(30)jkt<jkt+sforallt=2,…,r,s=1,…,lt,and the order of disjoint cycles (except for the first one) is strictly conditioned by increase from right to left of their first elements, jk2<jk3<⋯<jkr.
Since [45] for Hermitian A we have(31)rdet1A=⋯=rdetnA=cdet1A=⋯=cdetnA∈R,the determinant of a Hermitian matrix is defined by putting(32)det⁡A≔rdetiA=cdetiAforalli=1,…,n.Its properties are similar to the properties of an usual (commutative) determinant and they have been completely explored in [46] by using row and column determinants that are so defined only by construction.For determinantal representations of the Moore-Penrose inverse, we shall use the following notations. Letα≔α1,…,αk⊆1,…,m and β≔β1,…,βk⊆1,…,n be subsets of the order 1≤k≤min⁡m,n. Let Aβα be a submatrix of A whose rows are indexed by α and the columns indexed by β. Similarly, let Aαα be a principal submatrix of A whose rows and columns indexed by α. If A∈Hn×n is Hermitian, then Aαα is the corresponding principal minor of det⁡A. For 1≤k≤n, the collection of strictly increasing sequences of k integers chosen from 1,…,n is denoted by Lk,n≔α:α=α1,…,αk,1≤α1<⋯<αk≤n. For fixed i∈α and j∈β, let Ir,mi≔α:α∈Lr,m,i∈α denotes the collection of sequences of row indexes that contain the index i, and Jr,nj≔β:β∈Lr,n,j∈β denotes the collection of sequences of column indexes that contain j.Leta.j be the jth column and ai. be the ith row of A, respectively. Suppose A.jb denotes the matrix obtained from A by replacing its jth column with the column-vector b, and Ai.b denotes the matrix obtained from A by replacing its ith row with the row-vector b. We denote the ith row and the jth column of A∗ by ai.∗ and a.j∗, respectively.Lemma 11 (see [47]).
IfA∈Hrm×n, then the Moore-Penrose inverse A†=aij†∈Hn×m have the following determinantal representations,(33)aij†=∑β∈Jr,nicdetiA∗A.ia.j∗ββ∑β∈Jr,nA∗Aββ,and(34)aij†=∑α∈Ir,mjrdetjAA∗j.ai.∗αα∑α∈Ir,mAA∗αα.Remark 12.
For an arbitrary full-rank matrixA∈Hrm×n, a row-vector b∈H1×m, and a column-vector c∈Hn×1, we put (35)irdetiAA∗i.b=∑α∈Im,mirdetiAA∗i.bαα,det⁡AA∗=∑α∈Im,mAA∗αα,whenr=m,iicdetjA∗A.jc=∑β∈Jn,njcdetjA∗A.jcββ,det⁡A∗A=∑β∈Jn,nA∗Aββ,whenr=n.Corollary 13.
IfA∈Hrm×n, then the projection matrix A†A=:QA=qijn×n has the determinantal representation (36)qij=∑β∈Jr,nicdetiA∗A.ia˙.jββ∑β∈Jr,nA∗Aββ,where a˙.j is the jth column of A∗A∈Hn×n.Corollary 14.
IfA∈Hrm×n, then the projection matrix AA†=:PA=pijm×m has the determinantal representation(37)pij=∑α∈Ir,mjrdetjAA∗j.a¨i.αα∑α∈Ir,mAA∗αα,where a¨i. is the ith row of AA∗∈Hm×m.Lemma 15 (see [2]).
LetA∈Hm×n, B∈Hr×s, C∈Hm×s be known and X∈Hn×r be unknown. Then the matrix equation(38)AXB=Cis consistent if and only if AA†CB†B=C. In this case, its general solution can be expressed as(39)X=A†CB†+LAV+WRB,where V,W are arbitrary matrices over H with appropriate dimensions.In [71], it’s proved that (39) is the least squares solution to (38), and its minimum norm least squares solution is XLS=A†CB†.Lemma 16 (see [48]).
LetA∈Hr1m×n, B∈Hr2r×s. Then the minimum norm least squares solution X=A†CB†=(xij)∈Hn×r to (38) have determinantal representations, (40)xij=∑β∈Jr1,nicdetiA∗A.id.jBββ∑β∈Jr1,nA∗Aββ∑α∈Ir2,rBB∗αα,or (41)xij=∑α∈Ir2,rjrdetjBB∗j.di.Aαα∑β∈Jr1,nA∗Aββ∑α∈Ir2,rBB∗αα,where (42)d.jB=∑α∈Ir2,rjrdetjBB∗j.c~k.αα∈Hn×1,k=1,…,n,di.A=∑β∈Jr1,nicdetiA∗A.ic~.lββ∈H1×r,l=1,…,r,are the column vector and the row vector, respectively. c~k. and c~.l are the kth row and the lth column of C~=A∗CB∗.Corollary 17.
LetA∈Hkm×n, C∈Hm×s be known and X∈Hn×s be unknown. Then the matrix equation AX=C is consistent if and only if AA†C=C. In this case, its general solution can be expressed as X=A†C+LAV, where V is an arbitrary matrix over H with appropriate dimensions. Its minimum norm least squares solution X=A†C has the following determinantal representation, (43)xij=∑β∈Jk,nicdetiA∗A.ic^.jββ∑β∈Jk,nA∗Aββ,where c^.j is the jth column of C^=A∗C.Corollary 18.
LetB∈Hkr×s, C∈Hn×s be given, and X∈Hn×r be unknown. Then the equation XB=C is solvable if and only if C=CB†B and its general solution is X=CB†+WRB, where W is a any matrix with conformable dimension. Moreover, its minimum norm least squares solution X=CB† has the determinantal representation, (44)xij=∑α∈Ik,rjrdetjBB∗j.c^i.αα∑α∈Ik,rBB∗αα,where c^i. is the ith row of C^=CB∗.


## 3. A New Expression of the General Solution to System (6)

First, we show that Lemma4 is true for the Moore-Penrose inverses.Lemma 19.
LetΦ,Ω be matrices over H and (45)Φ=Φ1Φ2,Ω=Ω1Ω2,F=Φ2LΦ1,T=RΩ1Ω2.Then(46)LΦ=LΦ1LF,LΩ=LΩ1-Ω1†Ω2LT0LT,(47)RΩ=RTRΩ1,RΦ=RΦ10-RFΦ2Φ1†RF.where Ω1†, Φ1† are the Moore-Penrose inverses, and LΦ1, RΩ1, LT, RF, LΩ, and RΦ are projectors with respect to the corresponding Moore-Penrose inverses.Proof.
In ([65], Lemma 2.4), it is proved that for fixed reflexive inverses Ω1+ and T+, the reflexive inverse Ω+ can be expressed as follows, (48)Ω+=Ω1+-Ω1+Ω2T+RΩ1T+RΩ1.We choose Ω1†, T† as the Moore-Penrose inverses, and RΩ1 as the projector with respect to the Moore-Penrose inverse Ω1† and show that the obtained matrix(49)Ω†=Ω1†-Ω1†Ω2T†RΩ1T†RΩ1is the Moore-Penrose inverse of Ω. For this, it is enough to proof that Ω† satisfies the conditions (3) and (4) in Definition 1.
Since by Lemma3, T†RΩ1=RΩ1Ω2†RΩ1=T†, then Ω† can be expressed as (50)Ω†=Ω1†-Ω1†Ω2T†T†.So, (51)ΩΩ†=Ω1Ω2Ω1†-Ω1†Ω2T†T†=Ω1Ω1†-Ω1Ω1†Ω2T†+Ω2T†=Ω1Ω1†+RΩ1Ω2T†=Ω1Ω1†+RΩ1Ω2RΩ1Ω2†Since condition (3) is satisfied by components, namely, (52)Ω1Ω1†∗=Ω1Ω1†,(53)RΩ1Ω2RΩ1Ω2†∗=RΩ1Ω2RΩ1Ω2†it follows that Ω† satisfies condition (3) as well; i.e., ΩΩ†∗=ΩΩ†.
Similar, it can be shown thatΩ† satisfies condition (4). Hence, the Moore-Penrose inverse of Ω can be expressed by (49). From this (46) immediately follow.
The equations (47) can be proved similarly.Now we demonstrate the principal theorem of this section.Theorem 20.
Assume thatS1=[Ip10], S2=[0Ip2], T1=Iq10, T2=Iq20, G=[AC], H=B-D, H1=LA1LA, H2=LA1S1LG, H3=RHT1RB1, H4=LA2LC, H5=LA2S2LG, H6=RHT2RB2 and system (6) is solvable, then the general solution to our system can be formed as(54)X1=A1†C1+LA1C2B1†+LA1A†EB†RB1-LA1A†CM†EB†RB1-LA1A†SC†EN†DB†RB1+H1V1RB1+H2UH3+LA1V2RBRB1,(55)X2=A2†C3+LA2C4B2†+LA2M†RAED†RB2+LA2LMS†SC†EN†RB2+H4W1RB2+H5UH6+LA2W2RDRB2,where U,V1,V2,W1,andW2 are free matrices over H with allowable dimensions.Proof.
Our proof contains three parts. At the first step, we show that the matricesX1 and X2 have the forms of(56)X1=ϕ0+H1V1RB1+LA1V2RBRB1+H2UH3,(57)X2=ψ0+H4W1RB2+LA2W2RDRB2+H5UH6,where ϕ0 and ψ0 are any pair of particular solution to system (6), V1, V2, W1, W2 and U are free matrices of able shapes over H, are solutions to system (6). At the second step, we display that any couple of solutions μ0 and ν0 to system (6) can be established as (56) and (57), respectively. At the end, we confirm that (58)μ=A1†C1+LA1C2B1†+A†EB†-A†CM†EB†-A†SC†EN†DB†and(59)ν=A2†C3+LA2C4B2†+LA2M†RAED†+LA2LMS†SC†EN†RB2 are a couple of particular solutions to system (6).
Now we prove that a couple of matricesX1 and X2 having the shape of (56) and (57), respectively, are solutions to system (6). Observe that (60)A1†C1B1+LA1C2B1†B1=A1†A1C2+LA1C2=C2,A2†C3B2+LA2C4B2†B2=A2†A2C4+LA2C4=C4.It is evident that X1 having the form (56) is a solution of A1X1=C1, and X1B1=C2 and X2 having the form (57) is a solution to A2X2=C3,X2B2=C4. Now we are left to show that A3X1B3+A4X2B4=Cc is satisfied by X1 and X2 given in (56) and (57). By Lemma 4, we have(61)AS1LG=AIp10LA-A†CLM0LM=ALA-A†CLM=0-AA†CLM=0-C-MLM=0-CLM=-0S=-CS2LG,
and(62)RHT1B=RB0RNDB†RNIq10B=RBRNDB†B=0RNDB†B=0RNDI-LB=0RND=RHT2D.Observe that ALA=0 and by using (61) and (62), we arrive that (63)A3X1B3+A4X2B4=Cc.Conversely, assume that μ0 and ν0 are any couple of solutions to our system (6). By Lemma 7, we have (64)A1A1†C1=C1,C2B1†B1=C2,A2A2†C3=C3,C4B2†B2=C4,A1C2=C1B1,A2C4=C3B2.Observe that (65)LA1μ0RB1=I-A1†A1μ0I-B1B1†=μ0-μ0B1B1†-A1†A1μ0+A1†A1μ0B1B1†=μ0-C2B1†-A1†C1+A1†A1C2B1†=μ0-LA1C2B1†-A1†C1produces(66)μ0=LA1C2B1†+A1†C1+LA1μ0RB1.On the same lines, we can get(67)ν0=LA2C4B2†+A2†C3+LA2ν0RB2.It is manifest that μ0 and ν0 defined in (66)-(67) are also solution pair of(68)AX1B+CX2D=E.Since (69)AX1B+CX2D=A3LA1μ0RB1B3+A4LA2ν0RB2B4=A3μ0-LA1C2B1†-A1†C1B3+A4ν0-LA2C4B2†-A2†C3B4=A3μ0B3-A3LA1C2B1†B3-A1†C1B3+A4ν0B4-A4LA2C4B2†B4-A4A2†C3B4=A3μ0B3+A4ν0B4-AC2B1†B3-A1†C1B3-CC4B2†B4-A4A2†C3B4=Cc-AC2B1†B3-A1†C1B3-CC4B2†B4-A4A2†C3B4=E. Hence by Lemma 6, μ0 and ν0 can be written as(70)μ0=X01+S1LGURHT1+LAV1+V2RB,(71)ν0=X02+S2LGURHT2+LCW1+W2RD,where X01 and X02 are a couple of special solutions to (68) and U,V1,V2,W1 and W2 are free matrices with agreeable dimensions. Using (70) and (71) in (66) and (67), respectively, we get (72)μ0=X10+H2UH3+H1V1RB1+LA1V2RBRB1,ν0=X20+H5UH6+H4W1RB2+LA2W2RDRB2,where X10=A1†C1+LA1C2B1†+LA1X01RB1 and X20=A2†C3+LA2C4B2†+LA2X02RB2. It is evident that X10 and X20 are a couple of solutions to system (6). It is clear that μ0 and ν0 can be represented by (56) and (57), respectively. Lastly, by putting U1,V1,W1, and Z1 equal to zero in (24) and (25), we conclude that μ and ν are special solutions to system (6). Hence the expressions (54) and (55) represent the general solution to system (6) and the theorem is completed.Remark 21.
Due to Lemma3 and taking into account LA2LM=LMLA2, we have the following simplification of the solution pair to system (6) that is identical for (24)-(25) and (54)-(55) when U,U1,V1,V2,Z1,W1, and W2 disappear, (73)X1=A1†C1+LA1C2B1†+A†EB†-A†A4M†EB†-A†SC†EN†B4B†,X2=A2†C3+LA2C4B2†+M†ED†+S†SC†EN†.Comment 1.
We have established a novel expression of the general solution to system (6) in Theorem 20 which is different from one created in [35]. With the help of this novel expression, we can explore the least-norm of the general solution which can not be studied with the help of the expression given in [35], which is one of the advantage of our new expression.Now we discuss some special cases of our system.IfB1,B2,C2 and C4 disappear in Theorem 20, then we gain the following conclusion.Corollary 22.
DenoteS1=[Ip10], S2=[0Ip2], T1=Iq10, T2=Iq20, G=[AC],H=B3-B4, H1=LA1LA,H2=LA1S1LG,H3=RHT1,H4=LA2LC,H5=LA2S2LG,H6=RHT2 and system (4) is solvable, then the general solution to system (4) can be formed as (74)X1=A1†C1+A†EB3†-A†A4M†EB3†-A†SC†EN†B4B3†-H1Y1+H2VH3+LA1Y2RB3,X2=A2†C3+M†EB4†+S†SC†EN†+H4Z1+H5VH6+LA2Z2RB4,where A,C,N,M,S are the same as in Lemma 6, E=Cc-A3A1†C1B3-A4A2†C3B4, V,Y1,Y2,Z1, and Z2 are free matrices over H obeying agreeable dimensions.Comment 2.
The above consequence is a chief result of [64].
IfA2,B2,C3,A4,B4 and C4 vanish in our system (6), then we get the following outcome.Corollary 23.
Suppose thatA1,B1,C1,C2,A3,B3 and Cc are given. Then the general solution to system (3) is established by (75)X1=A1†C1+LA1C2B1†+A3LA1†Cc-A3A1†C1B3-A3LA1C2B1†B3RB1B3†+LA1LA3LA1W1RB1+LA1W2RRB1B3RB1, where W1 and W2 are arbitrary matrices over H with appropriate sizes.Comment 3.
Corollary23 is the rudimentary result of [32].Comment 4.
WhenA1,B1,A4 and B4 become zero in (8), then we will get the least-norm of the solution of (8) with the help of Theorem 20 quite smoothly. This is one of the advantage of the our expressions over the expressions given in [41].An algorithm and numerical example is provided to obtain the general solution of (6) with the help of Theorem 20.Algorithm 24.
(1) InputA1, B1, C1, A2, B2, C2, D2, A3, B3, C3, D3, A4, B4, C4 with viable dimensions over H.
(2) EvaluateX1 and X2 by (54)-(55).Example 25.
For given matrices(76)A1=1-ijk,B1=-jki1i-k,C1=ikj-1,C2=-ij1-k-1j,A2=i-kj,C3=i-k1jj1,A3=i1-1ikj,A4=-10j,B2=jk,B3=i1k-j,B4=j-kk-j,C4=-i+j,Cc=-1-j+k-2+i-j+k-i-1-1+j+k-i+j+k.By these given matrices, the consistency conditions of (6) from Lemma 3 are fulfilled. So, system (6) is resolvable. Now we compute the partial solution to system (6) when U,U1,V1,V2,Z1,W1, and W2 disappear. Using determinantal representations (33)-(34) for computing Moore-Penrose inverses, we find that (77)A1†=141-ji-k,LA1=121i-i1,B1†=16j1-k-i-ik,RB1=121j-j1,B2†=12-j-k,A2†=13-ik-j,A=0000kj,B=i1k-j,A†=1200-k00-j,B†=14-i-k1j,E=k-j-1ijk,RB2=121i-i1.
SinceLA2=0 and D=0, then C,S,M,N are zero-matrices. Hence the general solution to our system (6) is (78)X1=A1†C1+LA1C2B1†+A†EB†=1125+i-2j-k-2-i+7j+5k-5+i-j+2k-1+2i-j-k,X2=A2†C3=132+k-i-2j.


## 4. The Least-Norm of the General Solution to System (6)

We experience the least-norm to system (6) in this section. We first modify the description of quaternionic inner product space defined in [72] as follows:A rightH-vector space Vr is a quaternionic inner product space if there is a mapping 〈·,·〉:V×V→H such that for all q1,q2∈H and ξ,ξ1,ξ2∈Vr:(1)
〈ξ,ξ1q1+ξ2q2〉=q1¯〈ξ,ξ1〉+q2¯〈ξ,ξ2〉; 〈ξ1q1+ξ2q2,ξ〉=〈ξ1,ξ〉q1+〈ξ2,ξ〉q2;(2)
〈ξ1,ξ2〉=〈ξ2,ξ1〉¯;(3)
〈ξ,ξ〉≥0,and〈ξ,ξ〉=0ifandonlyifξ=0.It can be achieved by putting〈ξ,η〉=∑iηi¯ξi for ξ=(ξi)i=1n,η=(ηi)i=1n∈Vr. ξ=〈ξ,ξ〉 is referred as the norm of ξ. It is routine to verify that H is a quaternionic inner product space under the inner product defined by 〈C,D〉=tr(D∗C) where C,D∈Hm×n. A=(tr(A∗A))1/2 is the matrix norm defined by A. The real part of a quaternion q is denoted by re[q].By the definition and [73], we can get the following result easily.Lemma 26.
LetA∈Hm×n,B∈Hn×m. Then we have
(1)A+B2=A2+B2+2Re[tr(B∗A)].
(2)Re[tr(AB)]=Re[tr(BA)].Theorem 27.
Assume that system (6) is solvable, then the least-norm of the solution pair X1 and X2 to system (6) can be extracted as follows:(79)X1min=A1†C1+LA1C2B1†+A†EB†-A†A4M†EB†-A†SC†EN†B4B†,(80)X2min=A2†C3+LA2C4B2†+M†ED†+S†SC†EN†.Proof.
With the help of Theorem20 and Remark 21, the general solution to system (6) can be formed as (81)X1=A1†C1+LA1C2B1†+A†EB†-A†A4M†EB†-A†SC†EN†B4B†-H1V1RB1+H2UH3+LA1V2RBRB1,X2=A2†C3+LA2C4B2†+M†ED†+S†SC†EN†+H4W1RB2+H5UH6+LA2W2RDRB2,where U,V1,V2,W1, and W2 are free matrices over H having executable dimensions. By Lemma 26, the norm of X1 can be established as(82)X12=A1†C1+LA1C2B1†+A†EB†-A†A4M†EB†-A†SC†EN†B4B†-H1V1RB1+H2UH3+LA1V2RBRB12=A1†C1+LA1C2B1†+A†EB†-A†A4M†EB†-A†SC†EN†B4B†2+H1V1RB1+H2UH3+LA1V2RBRB12+J,where(83)J=2Re⁡tr⁡H1V1RB1+H2UH3+LA1V2RBRB1∗A1†C1+LA1C2B1†+A†EB†-A†A4M†EB†-A†SC†EN†B4B†.Now we want to show that J=0. Applying Lemmas 3, 4, and 26, we have(84)RetrH1V1RB1∗A1†C1+LA1C2B1†+A†EB†-A†A4M†EB†-A†SC†EN†B4B†=RetrRB1V1∗H1∗A1†C1+LA1C2B1†+A†EB†-A†A4M†EB†-A†SC†EN†B4B†=RetrRB1V1∗LALA1A1†C1+LA1C2B1†+A†EB†-A†A4M†EB†-A†SC†EN†B4B†=RetrRB1V1∗LALA1LA1C2B1†=RetrV1∗LALA1LA1C2B1†RB1=0,(85)RetrLA1V2RBRB1∗A1†C1+LA1C2B1†+A†EB†-A†A4M†EB†-A†SC†EN†B4B†=RetrRB1RBV2∗LA1∗A1†C1+LA1C2B1†+A†EB†-A†A4M†EB†-A†SC†EN†B4B†=RetrRB1RBV2∗LA1LA1C2B1†+A†EB†-A†A4M†EB†-A†SC†EN†B4B†=RetrV2∗LA1LA1C2B1†+A†EB†-A†A4M†EB†-A†SC†EN†B4B†RB1RB=RetrV2∗LA1A†EB†-A†A4M†EB†-A†SC†EN†B4B†RB=0,(86)RetrH2UH3∗A1†C1+LA1C2B1†+A†EB†-A†A4M†EB†-A†SC†EN†B4B†=RetrH3∗U∗H2∗A1†C1+LA1C2B1†+A†EB†-A†A4M†EB†-A†SC†EN†B4B†=RetrH3∗U∗LGS1∗LA1LA1C2B1†+A†EB†-A†A4M†EB†-A†SC†EN†B4B†=RetrH3∗U∗LA-A†CLM0LMI0LA1C2B1†+A†EB†-A†A4M†EB†-A†SC†EN†B4B†=RetrH3∗U∗LAA†EB†-A†A4M†EB†-A†SC†EN†B4B†=RetrH3∗U∗LALA1C2B1†=RetrRB1T1∗RHU∗LALA1C2B1†=RetrT1∗RHU∗LALA1C2B1†RB1=0.By using (84)-(86) in (83) produces J=0. Since X1 is arbitrary, we get (79) from (82). On the same way, we can prove that (80) hold.A special cases of our system (6) are given below.IfB1,B2,C2 and C4 become zero matrices in Theorem 27, then again we get the principal result of [30].Corollary 28.
Assume that system (4) is solvable, then the least-norm of the solution pair X1 and X2 to system (4) can be furnished as (87)X1min=A1†C1+A†EB3†-A†A4M†EB3†-A†SC†EN†B4B3†,X2min=A2†C3+M†EB4†+S†SC†EN†.IfA2, B2, C3, A4, B4, and C4 vanish in our system, then we get the next consequence.Corollary 29.
Suppose thatA1,B1,C1,C2,A3,B3, and Cc are given. Then the least-norm of the least square solution to system (3) is launched by (88)X1min=A1†C1+LA1C2B1†+A3LA1†Cc-A3A1†C1B3-A3LA1C2B1†B3RB1B3†.Comment 5.
Corollary29 is the key result of [32].


## 5. Determinantal Representations of the Least-Norm Solution to System (6)

In this section, we give determinantal representations of the least-norm solution to system (6). Let A1∈Hr1m×n, B1∈Hr2r×s, A2∈Hr3k×p, B2∈Hr4q×l, A3∈Hr5t×n, B3∈Hr6r×h, A4∈Hr7t×pB4∈Hr8q×h, r(A)=r9, r(B)=r10, r(C)=r11, r(D)=r12, r(M)=r13, r(N)=r14, and r(S)=r15.First, consider each term of (79) separately.(i) DenoteC11≔A1∗C1. Due to Corollary 17 for the first term of (79), X11=A1†C1=(xij(11)), we have(89)xij11=∑β∈Jr1,nicdetiA1∗A1.ic.j11ββ∑β∈Jr1,nA1∗A1ββ,where c.j(11) is the jth column of C11.(ii) For the second term of (79) we have, X12=xij(12)≔LA1C2B1†=C2B1†-QA1C2B1†. So, due to Corollaries 18 and 13,(90)xij12=∑α∈Ir2,rjrdetjB1B1∗j.ci.12αα∑α∈Ir2,rB1B1∗αα-∑f∑β∈Jr1,nicdetiA1∗A1.ia˙.f1ββ∑α∈Ir2,rjrdetjB1B1∗j.cf.12αα∑β∈Jr1,nA1∗A1αα∑α∈Ir2,rB1B1∗αα,where ci.(12) is the ith row of C12≔C2B1∗ and a˙.f(1) is the fth column of A1∗A1.Construct the matrixΨ1=ψif(1), where(91)ψif1=∑β∈Jr1,nicdetiA1∗A1.ia˙.f1ββ,and denoteΨ~1=Ψ1C2B1∗. Then, from (90), it follows that(92)xij12=∑α∈Ir2,rjrdetjB1B1∗j.ci.12αα∑α∈Ir2,rB1B1∗αα-∑α∈Ir2,rjrdetjB1B1∗j.ψ~i.1αα∑β∈Jr1,nA1∗A1αα∑α∈Ir2,rB1B1∗αα, where ψ~i.(1) is the ith row of the matrix Ψ~1.If we construct the matrixΨ2=ψif(2), where (93)ψfj2=∑α∈Ir2,rjrdetjB1B1∗j.cf.12αα,and denote Ψ~2≔A1∗A1Ψ2, and then, from (90), we obtain(94)xij12=∑α∈Ir2,rjrdetjB1B1∗j.ci.12αα∑α∈Ir2,rB1B1∗αα-∑β∈Jr1,nicdetiA1∗A1.iψ~.j2ββ∑β∈Jr1,nA1∗A1αα∑α∈Ir2,rB1B1∗αα,where ψ~.j(2) is the jth column of the matrix Ψ~2.(iii) Due to Theorem 2.15 for the third termA†EB†=:X13=(xij(13)), we obtain(95)xij13=∑β∈Jr9,micdetiA∗A.id.jBββ∑β∈Jr9,mA∗Aββ∑α∈Ir10,rBB∗αα,or(96)xij13=∑α∈Ir10,rjrdetjBB∗j.di.Aαα∑β∈Jr9,mA∗Aββ∑α∈Ir10,rBB∗αα,where(97)d.jB=∑α∈Ir10,rjrdetjBB∗j.eu.1αα∈Hp×1,u=1,…,m,di.A=∑β∈Jr9,micdetiA∗A.ie.v1ββ∈H1×r,v=1,…,r,are the column vector and the row vector, respectively. eu.(1) and e.v(1) are the uth row and the vth column of E1≔A∗EB∗.(iv) For the fourth term of (79), A†A4M†EB†≔X14=xij(14), using the determinantal representation (33) for A† and by Theorem 2.15, we have(98)xij14=∑f∑β∈Jr9,micdetiA∗A.ia.f14ββϕfj∑β∈Jr9,mA∗Aββ∑β∈Jr13,pM∗Mββ∑α∈Ir10,rBB∗αα,where(99)ϕfj=∑β∈Jr13,pfcdetfM∗M.fφ.jBββ=∑α∈Ir10,rjrdetjBB∗j.φf.Mαα,and (100)φ.jB=∑α∈Ir10,rjrdetjBB∗j.eu.2αα∈Hp×1,u=1,…,p,φf.M=∑β∈Jr13,pfcdetfM∗M.fe.v2ββ∈H1×r,v=1,…,r,are the column vector and the row vector, respectively. a.f(14) is the fth column of A14≔A∗A4, eu.(2), and e.v(2) are the uth row and the vth column of E2≔M∗EB∗, respectively. Construct the matrix Φ=ϕij, where ϕij is given by (99) and denote A∗AΦ=:Φ~=ϕ~ij. Then, from (98), we get the following final determinantal representation of the fourth term of (79),(101)xij14=∑β∈Jr9,micdetiA∗A.iϕ~.jββ∑β∈Jr9,mA∗Aββ∑β∈Jr13,pM∗Mββ∑α∈Ir10,rBB∗αα,where ϕ~.j is the jth column of Φ~.(v) For the fifth term of (79), A†SC†EN†B4B†≔X15=xij(15), due to Corollary 17 to A†S, by Theorem 2.15 to C†EN†, and Corollary 18 to B4B†, we obtain (102)xij15=∑f∑l∑β∈Jr9,nicdetiA∗A.is.l1ββωlf∑α∈Ir10,rjrdetjBB∗j.bf.15αα∑β∈Jr9,nA∗Aββ∑β∈Jr11,pC∗Cββ∑β∈Ir14,qNN∗αα∑β∈Ir10,rBB∗αα,where s.l(1) is the lth column of S1≔A∗S, bf.(15) is the fth row of B15≔B4B∗,(103)ωlf=∑β∈Jr11,plcdetlC∗C.lζ.fNββ=∑α∈Ir14,qfrdetfNN∗f.ζl.Cαα,and(104)ζ.fN=∑α∈Ir14,qfrdetfNN∗f.eu.3αα∈Hm×1,u=1,…,p,ζl.C=∑β∈Jr11,picdetlC∗C.le.v3ββ∈H1×t,v=1,…,q,are the column vector and the row vector, respectively. eu.(3) and e.v(3) are the uth row and the vth column of E3=C∗EN∗. Construct the matrix Ω=(ωlf), where ωlf is determined by (103), and denote Ω^≔A∗SΩB4B∗. Then, from (102), it follows that (105)xij15=∑f∑l∑β∈Jr9,nicdetiA∗A.ie.lββω^lf∑α∈Ir10,rjrdetjBB∗j.ef.αα∑β∈Jr9,nA∗Aββ∑β∈Ir11,pC∗Cββ∑α∈Ir14,qNN∗αα∑β∈Ir10,rBB∗αα,where ef. and e.l are the unit row-vector and the unit column-vector whose components are 0 except the fth or lth components which are 1, respectively.If we denote(106)ωif1≔∑l∑β∈Jr9,nicdetiA∗A.ie.lββω^lf=∑β∈Jr9,nicdetiA∗A.iω^.fββ,then, from (102), it follows the determinantal representation(107)xij15=∑α∈Ir10,rjrdetjBB∗j.ωi.1αα∑β∈Jr9,nA∗Aββ∑β∈Jr11,pC∗Cββ∑α∈Ir14,qNN∗αα∑β∈Ir10,rBB∗αα,where ωi.(1) is the ith row of the matrix Ω(1)=(ωif(1)) that is determined by (106).If we denote(108)ωlj2≔∑fω^lf∑α∈Ir10,rjrdetjBB∗j.ef.αα=∑α∈Ir10,rjrdetjBB∗j.ω^l.αα,then, from (102), it follows the determinantal representation(109)xij15=∑β∈Jr9,nicdetiA∗A.iω.j2ββ∑β∈Jr9,nA∗Aββ∑β∈Jr11,pC∗Cββ∑α∈Ir14,qNN∗αα∑β∈Ir10,rBB∗αα,where ω.j(2) is the jth column of the matrix Ω(2)=(ωlj(2)) that is determined by (108).Similarly, consider each term of (80) separately.(i) DenoteC21≔A2∗C3. Due to Corollary 17 for the first term of (80), X21=A2†C3=(xgf(21)), we have(110)xgf21=∑β∈Jr3,pgcdetgA2∗A2.ic.f21ββ∑β∈Jr3,pA2∗A2ββ,where c.f(21) is the fth column of C21.(ii) For the second term of (80) we have, X22=xij(22)≔LA2C4B2†=C4B2†-PA2C4B2†. So, due to Corollaries 18 and 13,(111)xgf22=∑α∈Ir4,qfrdetfB2B2∗f.cg.22αα∑α∈Ir4,qB2B2∗αα-∑j∑β∈Jr3,pgcdetgA2∗A2.ga˙.j2ββ∑α∈Ir4,qfrdetfB2B2∗f.cj.22αα∑β∈Jr3,pA2∗A2αα∑α∈Ir4,qB2B2∗αα,where cg.(22) is the gth row of C22≔C4B2∗ and a˙.j(2) is the jth column of A2∗A2.Construct the matrixΥ1=υgj(1), where(112)υgj1=∑β∈Jr3,pgcdetgA2∗A2.ga˙.j2ββ,and denote Υ~1=Υ1C4B2∗. Then, from (111), it follows that(113)xgf22=∑α∈Ir4,qfrdetfB2B2∗f.cg.22αα∑α∈Ir4,qB2B2∗αα-∑α∈Ir4,qfrdetfB2B2∗f.υ~g.1αα∑β∈Jr3,pA2∗A2αα∑α∈Ir4,qB2B2∗αα,where υ~g.(1) is the gth row of Υ~1.If we construct the matrixΥ2=υif(2), where(114)υjf2=∑α∈Ir4,qfrdetfB2B2∗f.cj.22αα,and denote Υ~2=A2∗A2Υ2, then, from (111), we obtain(115)xgf22=∑α∈Ir4,qfrdetfB2B2∗f.cg.22αα∑α∈Ir4,qB2B2∗αα-∑β∈Jr3,pgcdetgA2∗A2.gυ~.f2ββ∑β∈Jr3,pA2∗A2αα∑α∈Ir4,qB2B2∗αα,where υ~.f(2) is the gth row of Υ~2.(iii) Due to Theorem 2.15 for the third termM†ED†=:X23=(xgf(23)), we obtain(116)xgf23=∑β∈Jr13,pgcdetgM∗M.gd.fDββ∑β∈Jr13,pM∗Mββ∑α∈Ir10,rBB∗αα,or(117)xij13=∑α∈Ir10,rjrdetjBB∗j.di.Aαα∑β∈Jr9,mA∗Aββ∑α∈Ir10,rBB∗αα,where (118)d.jB=∑α∈Ir10,rjrdetjBB∗j.eu.4αα∈Hp×1,u=1,…,m,di.A=∑β∈Jr9,micdetiA∗A.ie.v4ββ∈H1×r,v=1,…,r,are the column vector and the row vector, respectively. eu.(1) and e.v(1) are the uth row and the vth column of E4≔M∗ED∗.(iv) Using Corollary13 to S†S and by Theorem 2.15 to C†EN†, we obtain the the following representation of the fourth term, X24=(xgf(24))≔S†SC†EN†, of (80)(119)xgf24=∑l∑β∈Jr15,pgcdetgS∗S.gs˙.lββωlf∑β∈Jr15,pS∗Sββ∑α∈Ir11,tCC∗αα∑β∈Ir14,qNN∗αα,where ωlf is determined by (103). Construct the matrix Ω=(ωlf) and denote Ω~=S∗SΩ. Then, from (119) finally, we have(120)xgf24=∑β∈Jr15,pgcdetgS∗S.gω~.fββ∑β∈Jr15,pS∗Sββ∑α∈Ir11,tCC∗αα∑β∈Ir14,qNN∗αα,where ω~.f is the fth column of Ω~.So, we prove the following theorem.Theorem 30.
LetA1∈Hr1m×n, B1∈Hr2r×s, A2∈Hr3k×p, B2∈Hr4q×l, A3∈Hr5t×n, B3∈Hr6r×h, A4∈Hr7t×pB4∈Hr8q×h, r(A)=r9, r(B)=r10, r(C)=r11, r(D)=r12, r(M)=r13, r(N)=r14, and r(S)=r15. The least-norm solution (79)-(80) to system (8), X1=xij(1)∈Hn×r, X2=xgf(2)∈Hp×q, by components(121)xij1=xij11+xij12+xij13-xij14-xij15,xgf2=xgf21+xgf22+xgf23+xgf24,has determinantal representations, where the term xij(11) is (89), xij(12) is (92) or (94), xij(13) is (95) or (96), xij(14) is (101), and xij(15) is (107) or (109); similarly, xgf(21) is (110), xgf(22) is (113) or (115), xgf(23) is (116) or (117), and xgf(24) is (120).A numerical example is provided to obtain the least norm of the general solution of (6) with the help of Theorem 30.Example 31.
We use the given matrices from the Example25. Since r(A1)=1 and(122)C11=A1∗C1=1+ij+k-1+i-j+k,A1∗A1=2-2i2i2,and then, by (89), (123)x1111=14+14i,x1211=14j+14k,x2111=-14+14i,x2211=-14j+14k. Now, by (92), we find xij(12) for all i,j=1,2. So,(124)C12=C2B1∗=-2i-k-i+2ki+2k2i-k,B1B1∗=3-3j3j3,Similarly, by (91), Ψ1=A1∗A1. So(125)Ψ~1=Ψ1C2B1∗=2-4i+4j-2k4-2i-2j+4k4+2i+2j+4k2+4i-4j-2k
Sincer(B1)=1, then by (92), (126)x1112=16-2i-k-1242-4i+4j-2k=-112-16i-16j-112k,x1212=16-i+2k-1244-2i-2j+4k=-16-112i+112j+16k,x2112=16i+2k-1244+2i+2j+4k=-16+112i-112j+16k,x2212=162i-k-1242+4i-4j-2k=-112+16i+16j-112k.Since r(A)=r(B)=1 and(127)E1=A∗EB∗=22j-2i-2k,A∗A=1i-i1,BB∗=22j-2j2,and then, by (95), (128)d.1B=2-2i,d.2B=2j-2k,and (129)x1113=14,x1213=14j,x2113=-14i,x2213=-14k. Further, due to Example 25, xij(14)=xij(15)=0 for all i,j=1,2. So, (130)x111=512+112i-16j-112k,x121=-16-112i+712j+512k,x211=-512+112i-112j+16k,x221=-112+16i-112j-112k.
Since,r(A2)=1 and(131)C21=A2∗C3=2+k-i-2j,A2∗A2=3,and, due to Example 25, x1j(22)=x1j(23)=x1j(24)=0 for all j=1,2; then by (110) and (121) (132)x112=x1121=23+13k,x122=x1221=-13i-23j.Hence, the least norm solution of (6) obtained by Cramer’s Rule and the matrix method in Example 25 are the same as expected.
Note that we used Maple with the package CLIFFORD in the calculations.


## 6. Conclusion

We have constructed a novel expression of the general solution to system (6) over H and used this result to explore the least-norm of the general solution to this system when it is solvable. Some particular cases of our system are also discussed. Our results carry the principal results of [32, 64]. Finally, we give determinantal representations (analogous of Cramer’s Rule) of the least norm solutions to the systems using row-column noncommutative determinants. Numerical examples are also provided to interpret the results.



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