# A Study of Inverse Problems Based on Two Kinds of Special Matrix Equations in Euclidean Space

Authors: Rui Huang; Xiaodong Wu; Ruihe Wang; Hui Li
Journal: Algebra
 (2014)

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## Abstract

Two special classes of symmetric coefficient matrices were defined based on characteristics matrix; meanwhile, the expressions of the solution to inverse problems are given and the conditions for the solvability of these problems are studied relying on researching. Finally, the optimal approximation solution of these problems is provided.


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## Body

## 1. Introduction

In recent years, a lot of matrix problems have been used widely in the fields of structural design, automatic control, physical, electrical, nonlinear programming and numerical calculation, for example, a matrix Eigen value problem was applied for mixed convection stability analysis in the Darcy media by Serebriiskii et al. [1] and some of the problems based on the nonskew symmetric orthogonal matrices were studied by Hamed and Bennacer in 2008 [2], but some of the matrix inverse problems still need further research in order to make it easier to discuss relevant issues. Therefore, in this paper, we studied the inverse problems of two kinds of special matrix equations based on the existing research achievements, moreover, the expressions and conditions of the matrix solutions are given by related matrix-calculation methods. Some definitions and assumptions of the inverse problem for two forms of special matrices are given in Section 2. In Sections 3 and 5 we discuss the existence and expressions of general solution based on the two classes of matrices, and in Sections 4 and 6 we prove the uniqueness of matrices for researching related inverse problems.


## 2. Definitions and Assumptions of Inverse Problems for Two Forms of Special Matrices

In order to research some inverse problems of related matrices, we give the following definitions and assumptions.Definition 1.
WhenP∈Rn×n, P=PT=P-1, A=(aij)∈Rn×n, i,j=1,2,3,…,n, (PA)T=-PA, and aij=-aji, A will be called the first-class special symmetric matrix and the set of these special symmetric matrices is denoted by A1Rn×n. The corresponding problems are as follows.
Problem
1. When X,B∈Rn×m, A∈A1Rn×n can be obtained, so that AX=B.
Problem
2. When A˙∈Rn×n, A^∈SE can be obtained, so that ∥A˙-A^∥=min⁡A∈SE∥A˙-A∥, where SE is the solution set of the first problem.Definition 2.
WhenA=(aij)∈Rn×n, i,j=1,2,3,…,n and aij=-an+1-i,n+1-j, A will be called the second-class special symmetric matrix and the set of these special symmetric matrices is denoted by A2Rn×n. The corresponding problems are as follows.
Problem
1. When X,B∈Rn×m, A∈A2Rn×n can be found, so that AX=B.
Problem
2. When A˙∈Rn×n, A^∈S~E can be found, so that ∥A˙-A^∥=min⁡A∈S~E∥A˙-A∥, where S~E is the solution set of the first problem.


## 3. Existence and Expression of General Solutions Based on the First-Class Special Symmetric Matrix for Problem 1

To research the structure and properties of the special symmetric matrixA∈A1Rn×n, first of all, we have the following conclusion from Definition 1.Conclusion 1.
ConsiderA=(aij)∈Rn×n, and the sufficient and necessary conditions for A∈A1Rn×n are
(1)AT=-A,ATP+PA=0.Theorem 3.
ConsiderA=(aij)∈Rn×n, and the sufficient and necessary conditions for A∈A1Rn×n are
(2)A=U[A1100A22]UT,
where A11∈A1Rk×k, A22∈A1R(n-k)×(n-k), and U is an orthogonal matrix. Consider U∈Rn×n and UUT=In.Proof.
Rely on the decomposition theorem of symmetric orthogonal matrix [3, 4]. When P is a symmetric orthogonal matrix, P∈Rn×n, the P can be represented as the following equation by an orthogonal matrix U,  U∈Rn×n:
(3)P=U[Ik00-In-k]UT,
where I is identity matrix [5].
WhenA∈A1Rn×n, we can derive from (1) and (3) the following:
(4)UTATU[Ik00-In-k]+[Ik00-In-k]UTAU=0.
Based onAT=-A and UTAU∈A1Rn×n, there will be
(5)UTAU=[A11A12-A12TA22]
and A=U[A11A12-A12TA22]UT can be derived from (4) and (5).
Conversely, when(6)A=U[A11A12-A12TA22]UT,AT=-A, ATP+PA=0 can be obtained, and, relying on Conclusion 1,  A∈A1Rn×n can be obtained.Theorem 4.
WhenX,B∈Rn×m and the X singular value decomposition is
(7)X=U1[Σ1000]V1T,
where U1=[U(1)U(2)]∈ORn×n, V1=[V(1)V(2)]∈ORm×m, and Σ1=diag⁡(δ1,…,δr)>0, r=rank⁡(X), U(1)∈Rn×r, U(2)∈Rn×(n-r), V(1)∈Rm×r, V(2)∈Rm×(m-r), ORn×n is the set of orthogonal matrix, and X+ is Moore-Penrose generalized inverse matrix [6–8].
The sufficient and necessary conditions for the existence of solutionAX=B are
(8)BTX=-XTB,BX+X=B.
And the general solution is as follows:(9)A=BX+-(BX+)T(In-XX+)+U(2)G(U(2))T,
where G∈A1R(n-r)×(n-r).Proof.
Define(10)U1TAU1=[A11A12A21A22],U1TBV1=[B11B12B21B22],
where A11∈Rr×r and Bij=(U(1))TBV(1),  i,j=1,2.
The equationAX=B can be represented as the following equation by (7) and (10):
(11)[A11A12A21A22][Σ1000]=[B11B12B21B22].
According to (7) and (8), there will be
(12)B11TΣ1=-Σ1B11,B12=0,B22=0.
Then (11) is equivalent to the following equation:
(13)A11Σ1=B11,A21Σ1=B21,
and A11=B11Σ1-1∈A1Rr×r and A21=B21Σ1-1can be obtained from (12) and (13), so the equation AX=B has solutions, A∈A1Rn×n. And the general solution can be represented as the following equation:
(14)A=U1[B11Σ1-1-Σ1-1B21TB21Σ1-1A22]U1T=BX+-(BX+)T(In-XX+)+U(2)G(U(2))T,
where G∈A1R(n-r)×(n-r).
Conversely, when the equationAX=B has solutions and A∈A1Rn×n, BX+X=B and XTB=XTAX can be obtained relying on the Penrose theorem [9, 10].
There will be(15)BTX=(XTB)T=(XTAX)T=-XTAX=-XTB.
The equations BTX=-XTB and BX+X=B are provided.
Finally the proof of Theorems3 and 4 is completed.From Theorems3 and 4 we have a corollary as follows.Corollary 5.
Consider thatX,B∈Rn×m, Xi and Bi(i=1,2), X2,B2∈R(n-k)×m, and X1,B1∈Rk×m are given by these equations:
(16)UTX=[X1X2],UTB=[B1B2].
TheXi singular value decompositions are
(17)X1=U1[Σ1000]V1T,X2=U2[Σ2000]V2T,
where U1=[U1(1)U1(2)]∈ORk×k, V1=[V1(1)V1(2)]∈ORm×m, U2=[U2(1)U2(2)]∈OR(n-k)×(n-k), V2=[V2(1)V2(2)]∈ORm×m, Σ1=diag⁡(δ11,…,δr11)>0, Σ2=diag⁡(δ12,…,δr22)>0, r1=rank⁡(X1), r2=rank⁡(X2), V1(1)∈Rm×r1, U2(1)∈R(n-k)×r2, U1(1)∈Rk×r1, V2(1)∈Rm×r2.
The sufficient and necessary conditions for the existence of solutionAX=B are
(18)BiTXi=-XiTBi,BiXi+Xi=Bi.
And the general solution is as follows:(19)A=U[A1100A22]UT+U[U1(2)G1(U1(2))T00U2(2)G2(U2(2))T]UT,A11=B1X1+-(B1X1+)T(Ik-X1X1+)A22=B2X2+-(B2X2+)T(In-k-X2X2+),
where G1∈A1R(k-r1)×(k-r1) and G2∈A1R(n-k-r2)×(n-k-r2).The proof is completed.


## 4. The Unique Solutions on the First-Class Special Symmetric Matrix for Problem 2

Consider the following theorem.Theorem 6.
WhenX,B∈Rn×m, A∈Rn×n, and Xi and Bi(i=1,2) are given by (16) and obey (18), Problem 2 has the unique solution A^∈SE.
Define(20)UTA˙U=[A˙11A˙12A˙21A˙22],A˙11∈Rk×k.
Then the unique solutionA^ can be represented as follows:
(21)A^=U[A˙1100A˙22]UT,
where
(22)A^11=A110+0.5·(Ik-X1X1+)(A˙11-A˙11T)(Ik-X1X1+),A^22=A220+0.5·(In-k-X2X2+)(A˙22-A˙22T)(In-k-X2X2+).
A
0 is defined by the following equation:
(23)A0=U[A11000A220]UT,(24)A110=B1X1+-(B1X1+)T(Ik-X1X1+),A220=B2X2+-(B2X2+)T(In-k-X2X2+).
FinallySE can represent the following set by Theorem 4:
(25)SE=A0+U[U1(2)G2(U1(2))T00U2(2)G2(U2(2))T]UT.
Consider G1∈A1R(k-r1)×(k-r1) and G2∈A1R(n-k-r2)×(n-k-r2).Proof.
Define these following equations:(26)U1T(A˙11-A110)U1=[A˙111A˙121A˙211A˙221],U2T(A˙22-A220)U2=[A˙112A˙122A˙212A˙222],
where A˙111∈Rr1×r1 and A˙221∈Rr2×r2.
Because ofA∈SE we can obtain
(27)A=A0+U[U1(2)G1(U1(2))T00U2(2)G2(U2(2))T]UT=A0+U[U1[000G1]U1T00U2[000G2]U2T]UT,(28)∥A˙-A∥2=∥[U1(2)G2(U1(2))T00U2(2)G2(U2(2))T]-UT(A˙-A0)U∥2=∥[U1(2)G2(U1(2))T00U2(2)G2(U2(2))T]-[A˙11-A110A˙12A˙21A˙22-A220]∥2=∥A˙12∥2+∥A˙21∥2+∥[000G1]-U1T(A˙11-A110)U1∥2+∥[000G2]-U2T(A˙22-A220)U2∥2.
Finally,(29)∥A˙-A∥2=∥A˙12∥2+∥A˙21∥2+∥A˙111∥2+∥A˙121∥2+∥A˙211∥2+∥A˙112∥2+∥A˙122∥2+∥A˙212∥2+∥G1-A˙221∥2+∥G2-A˙222∥2.
WhenH=-HT, H∈Rn×m and E∈Rn×m, we can obtain:
(30)∥E-E-ET2∥≤∥E-H∥.
When G1=0.5·(A˙221-(A˙221)T) and G2=0.5·(A˙222-(A˙222)T), so that ∥A˙-A∥=min⁡A∈SE,
(31)G1=0.5·(U1(2))T(A˙11-A˙11T-2A110)U1(2),G2=0.5·(U2(2))T(A˙22-A˙22T-2A220)U2(2)
can be derived from (26) and
(32)(U1(2))TA110U1(2)=0,(U2(2))TA220U2(2)=0
can be obtained according to (17), (18), and (24).
Therefore,(33)G1=0.5·(U1(2))T(A˙11-A˙11T)U1(2),G2=0.5·(U2(2))T(A˙22-A˙22T)U2(2).
Relying on (27) the solution A^ can be provided as follows:
(34)A^=A0+12UCUT,
where(35)C=[U1(2)(U1(2))T(A˙11-A˙11T)U1(2)(U1(2))T00U2(2)(U2(2))T(A˙22-A˙22T)U2(2)(U2(2))T]and A^=U[A˙1100A˙22]UT can be obtained because of U1(2)(U1(2))T=In-k-X1X1+ and U2(2)(U2(2))T=Ik-X2X2+.
Finally, rely on the optimal approximation [11, 12], and SE is a closed convex set; Problem 2 has the unique solution A^∈SE, SE∈Rn×n, so that ∥A˙-A^∥=min⁡A∈S~E∥A˙-A∥.
The proof is completed.


## 5. Conditions for the Existence and Expression of General Solutions Based on the Second-Class Special Symmetric Matrix for Problem 1

Consider the following theorem.Theorem 7.
Consider thatA=(aij)∈Rn×n, and the sufficient and necessary conditions for A∈A2Rn×n are
(36)A=U[0F1F20]UT,
where U is an orthogonal matrix. U∈Rn×n and UUT=In.
Whenn=2k+1, define
(37)U=12[Ik0Ik020Hk0-Hk].
Whenn=2k, define
(38)U=12[IkIkHk-Hk],Hk=[0⋯010⋯10⋯⋯⋯⋯1⋯00],HkTHk=Ik,HkT=Hk,F1∈A2R(n-k)×k, F2∈A2Rk×(n-k).Proof.
First relying on the definition and property of the matrixA∈A2Rn×n, A2Rn×n can represent the following set.Whenn=2k+1,
(39)A=[C1uC2Hk-vT0vTHk-HkC2-Hku-HkC1Hk],C1,C2∈Rk×k  and  u,v∈Rk.Whenn=2k,
(40)A=[C1C2Hk-HkC2-HkC1Hk],C1,C2∈Rk×k.
We first discuss the topicn=2k+1.
From (37) and (39),
(41)UTAU=12[Ik0Hk020Ik0-Hk]C[Ik0Ik020Hk0-Hk],C=[C1uC2Hk-vT0vTHk-HkC2-Hku-HkC1Hk],(42)UTAU=[00C1-C200-2vTC1+C22u0].
DefineF1=[C1-C2-2vT], F2=[C1+C22u].
According toC1,C2∈Rk×k and u,v∈Rk, we have F1∈R(n-k)×k and F2∈Rk×(n-k), and
(43)A=UUTAUUT=U[0F1F20]UT
can be obtained from (42).
Conversely, whenA=U[0F1F20]UT, F1∈R(n-k)×k, F2∈Rk×(n-k), A=-HnAHn can be obtained and A∈A2Rn×n will be provided.
Whenn=2k, in the same way, we can prove the above theorem.
The proof is completed.Theorem 8.
WhenX,B∈Rn×m, A∈A2Rn×n, and Xi and Bi  (i=1,2) are given by (16), the sufficient and necessary conditions for the existence of solution AX=B are
(44)BiXi+Xi=Bi.
And the general solution is as follows:(45)A=U[0A12A210]UT,A12=B1X2++M1(Ik-X2X2+),A21=B2X1++M2(In-k-X1X1+),
where
(46)U=12[Ik0Ik020Hk0-Hk],n=2k+1,U=12[IkIkHk-Hk],n=2k.
Meanwhile M1∈R(n-k)×k, M2∈Rk×(n-k), and according to (17) we can obtain Xi singular value decompositions:
(47)X1=U1[Σ1000]V1T,X2=U2[Σ2000]V2T.Proof.
Relying on Theorem7, if there is A∈A2Rn×n, it will have
(48)A=U[0F1F20]UT,F1∈A2R(n-k)×k,F2∈A2Rk×(n-k).
BecauseU is an orthogonal matrix and UUT=In,
(49)∥AX-B∥2=∥UTAUUTX-UTB∥2.
We can derive from (16), (48), and (49)
(50)∥AX-B∥2=∥[0F1F20][X1X2]-[B1B2]∥2=∥F1X2-B1∥2+∥F2X1-B2∥2.
It will be known that, whenX∈Rn×k, B∈Rn×k, and rank⁡(X)=r [13], the sufficient and necessary conditions for the existence of solution AX=B are BX+X=B and the general solution is
(51)A=BX++M(Im-XX+),M∈Rn×m.
There will be(52)F1=B1X2++M1(Ik-X2X2+),M1∈R(n-k)×k;F2=B2X1++M2(In-k-X1X1+),M2∈Rk×(n-k).
The general solution can obtain(53)A=U[0A12A210]UT,A12=B1X2++M1(Ik-X2X2+),A21=B2X1++M2(In-k-X1X1+).
The sufficient and necessary conditions for the existence of solutionAX=B are BiXi+Xi=Bi.
The proof is completed.


## 6. The Unique Solutions on the Second-Class Special Symmetric Matrix for Problem 2

Consider the following theorem.Theorem 9.
When,B∈Rn×m,  A∈Rn×n, and Xi and Bi  (i=1,2) are given by (16) and (17) and obey (18), Problem 2 has the unique solution A^∈S~E.
The unique solutionA^ can be represented as follows:
(54)A^=UCUT,C=[0B1X2++A12(Ik-X2X2+)B2X1++A21(In-k-X1X1+)0],
where
(55)A0=U[0B1X2+B2X1+0]UT;
n=2k+1, A˙2∈A2Rn×n:
(56)A12=0.5·[Ik0Hk020](A˙2-A0)[Ik0-Hk],(57)A21=0.5·[Ik0-Hk](A˙2-A0)[Ik002Hk0];
n=2k, A˙2∈A2Rn×n:
(58)A12=0.5·[IkHk](A˙2-A0)[Ik-Hk],A21=0.5·[Ik-Hk](A˙2-A0)[IkHk].From Theorem9  S~E can be represented as follows:
(59)S~E=A0+U[0M1(Ik-X2X2+)M2(In-k-X1X1+)0]UThhhhhhhhhhhhhhhM1∈R(n-k)×k,M2∈Rk×(n-k).Proof.
Defining this following equation:(60)UT(A˙2-A0)U=[A11A12A21A22],
where n=2k+1, A˙2∈A2Rn×n,
(61)A11=0.5·[Ik0Hk020](A˙2-A0),A12=0.5·[Ik0Hk020](A˙2-A0)[Ik0-Hk],A21=0.5·[Ik0-Hk](A˙2-A0)[Ik002Hk0],A22=0.5·[Ik0-Hk](A˙2-A0)[Ik0-Hk];n=2k, A˙2∈A2Rn×n:
(62)A11=0.5·[IkHk](A˙2-A0)[IkHk],A12=0.5·[IkHk](A˙2-A0)[Ik-Hk],A21=0.5·[Ik-Hk](A˙2-A0)[IkHk],A22=0.5·[Ik-Hk](A˙2-A0)[Ik-Hk].
ForA∈S~E the following equation can be obtained:
(63)A=A0+UCUT,C=[0M1(Ik-X2X2+)M2(In-k-X1X1+)0].
WhenA˙∈Rn×n is known and relies on the definition of A∈A2Rn×n, there will be the unique set A˙1=(a-ij)∈Rn×n(a-ij=a-(n+1-i),(n+1-j)) and A˙2∈A2Rn×n to make the following equations true:
(64)A˙=A˙1+A˙2,(A˙1,A˙2)=0,A˙2=0.5·(A˙+HnA˙Hn),A˙1=0.5·(A˙-HnA˙Hn),(65)∥A˙-A∥2=∥A˙1+A˙2-A∥2=∥A˙1∥2+∥A˙2-A∥2.
From (60), (63), and (65) we obtain
(66)∥A˙-A∥2=∥A˙1∥2+∥U[A11A12A21A22]UT-UCUT∥2=∥A˙1∥2+∥[A11A12A21A22]-C∥2,C=[0M1(Ik-X2X2+)M2(In-k-X1X1+)0].
Equation (67) can be known from (17) and (64) as follows:
(67)U2(2)(U2(2))T=Ik-X2X2+,U1(2)(U1(2))T=In-k-X1X1+,(U2(2))TU2(2)=Ik-r2,(U1(2))TU1(2)=In-k-r1,(U2(2))TU2(1)=0,(U1(2))TU1(1)=0,(68)∥A˙-A∥2=∥A˙1∥2+∥A11∥2+∥A22∥2+∥A12-M1U2(2)(U2(2))T∥2+∥A12-M2U1(2)(U1(2))T∥2=∥A˙1∥2+∥A11∥2+∥A22∥2+∥A12U2(1)∥2+∥A21U1(1)∥2+∥A12U2(2)-M1U2(2)∥2+∥A21U1(2)-M2U1(2)∥2.
Therefore,∥A˙-A∥=min⁡A∈S~E is equivalent to
(69){∥A12U2(2)-M1U2(2)∥=min⁡M1∈R(n-k)×k∥A12U1(2)-M2U1(2)∥=min⁡M2∈Rk×(n-k)⟹{A12U2(2)-M1U2(2)=0A12U1(2)-M2U1(2)=0⟹{A12U2(2)=M1U2(2)A12U1(2)=M2U1(2),
and A12(Ik-X2X2+)=M1(Ik-X2X2+), A21(In-k-X1X1+)=M2(In-k-X1X1+) can be obtained.
From the above results and (63), the solution A^(A^∈S~E) can be represented as follows:
(70)A^=UCUT,C=[0B1X2++A12(Ik-X2X2+)B2X1++A21(In-k-X1X1+)0].
Finally, becauseS~E is a closed convex set, Problem 2 has the unique solution A^∈S~E, S~E∈Rn×n, so that ∥A˙-A^∥=min⁡A∈S~E∥A˙-A∥.
The proof is completed.



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*Source: 392467-2014-05-26.xml*