# The HermitianR-Conjugate Generalized Procrustes Problem

Authors: Hai-Xia Chang; Xue-Feng Duan; Qing-Wen Wang
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We consider the HermitianR-conjugate generalized Procrustes problem to find Hermitian R-conjugate matrix X such that ∑k=1p∥AkX-Ck∥2+∑l=1q∥XBl-Dl∥2 is minimum, where Ak, Ck, Bl, and Dl (k=1,2,…,p, l=1,…,q) are given complex matrices, and p and q are positive integers. The expression of the solution to Hermitian R-conjugate generalized Procrustes problem is derived. And the optimal approximation solution in the solution set for Hermitian R-conjugate generalized Procrustes problem to a given matrix is also obtained. Furthermore, we establish necessary and sufficient conditions for the existence and the formula for Hermitian R-conjugate solution to the linear system of complex matrix equations A1X=C1, A2X=C2,…,ApX=Cp, XB1=D1,…,XBq=Dq (p and q are positive integers). The representation of the corresponding optimal approximation problem is presented. Finally, an algorithm for solving two problems above is proposed, and the numerical examples show its feasibility.


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## Body

## 1. Introduction

Throughout, let𝒞m×n denote the set of all complex m×n matrices, ℛm×n the set of all real m×n matrices, and ℛrm×n the set of all matrices in ℛm×n with rank r. The symbols I, A¯, AT, A*, A†, and ∥A∥, respectively, stand for the identity matrix with the appropriate size, the conjugate, the transpose, the conjugate transpose, the Moore-Penrose inverse, and the Frobenius norm of A∈𝒞m×n. For A=(aij), B=(bij)∈𝒞m×n; A*B=(aijbij)∈𝒞m×n represents the Hadamard product of A and B.A linear model of image restoration is a matrix-vector equation is(1)y=Kx+n,
where y represents the observed image, x the original true image, n additive noise, and K a blurring matrix. Image restoration is to minimize blur in an observed image, namely, recover a optimal approximation of x by given y and K, and get some statistical information of n. The process of the image restoration for the model (1) can be described as
(2)minx∥x∥2subjectto∥Kx-y∥2=ε,
where ε is a small positive parameter. It is known that x^=K†y is the least squares solution of (1) with minimal norm. However, for ε=0, the solution x^ε to (2), that is, n=0 in (1) is not feasible. We know if ε→0, then the solution x^ε converges to x^. In order to obtain a solution x^ε sufficiently near to x^, we usually take small ε such that 0<ε≪δ, where the error norm δ:=∥n∥ is given.Now we consider the generalized problem of the process of the image restoration.Problem 1.
Given𝒮⊆𝒞n×n, mk, hl, p, q being positive integers, Ak,Ck∈𝒞mk×n, Bl,Dl∈𝒞n×hl, E∈𝒞n×n, and k=1,…,p, l=1,…,q. Let
(3)C={X∈𝒮∣∑k=1p∥AkX-Ck∥2‍+∑l=1q∥XBl-Dl∥2‍=min}.
Find X^∈C such that
(4)∥X^-E∥=minX∈C∥X-E∥.
The constraint Procrustes problem associated with several kinds of sets𝒮, that is, p=1 and q=0 in (3) has been extensively studied, such as the orthogonal Procrustes problem with 𝒮 being the set of orthogonal matrices [1], the symmetric Procrustes problem [2], (M,N)-symmetric Procrustes problem [3], Hermitian, Hermitian R-symmetric and Hermitian R-skew-symmetric Procrustes problems [4], the Procrustes problems with 𝒮 constrained to the cone of symmetric positive semidefinite and symmetric elementwise matrices [5], and the generalized Procrustes analysis [6]. The optimal approximation problem (4) is initially proposed in the processes of testing or revising given data. A preliminary estimate E of the unknown matrix X in C can be obtained from experimental observation values and the information of statistical distribution.We characterize the caseAkX=Ck, XBl=Dl, k=1,…,p, l=1,…,q in Problem 1 and describe it as follows.Problem 2.
Given𝒮⊆𝒞n×n, m1,…,mp, h1,…,hq, p and q being positive integers, A1,C1∈𝒞m1×n, A2,C2∈𝒞m2×n,…,Ap,Cp∈𝒞mp×n, B1,D1∈𝒞n×h1,…,Bq,Dq∈𝒞n×hq, F∈𝒞n×n. Let
(5)L={X∈𝒮∣A1X=C1,A2X=C2,…,ApX=Cp,XB1=D1,…,XBq=Dq}.
When L is nonempty, find X~∈L such that
(6)∥X~-F∥=minX∈L∥X-F∥.For important results to solve Problem2 with different sets 𝒮, we refer to [7–14].Motivated by the work mentioned above, in this paper we mainly discuss the above two problems associated withS being the set of Hermitian R-conjugate matrices.Recall that ann×n complex matrix K is R-conjugate if K¯=RKR, where R∈ℛn×n is a nontrivial involution, that is, R2=I, R≠±I, which was defined in [15]. A matrix K∈𝒞n×n is Hermitian R-conjugate if K*=K and K¯=RKR, where RT=R-1=R≠±I. The Hermitian R-conjugate matrix is very useful in scientific computation and digital signal and image processing, its special case, for example, Hermitian Toeplitz matrix, have been studied by several authors, see [14, 16–21]. We denote the set of all n×n Hermitian R-conjugate matrices by HRc𝒞n×n. Let Rc𝒞n×n, Sℛn×n, and ASℛn×n denote the set of all n×n complex R-conjugate matrices, real symmetric matrices, real skew-symmetric matrices, respectively.This paper is organized as follows. In Section2, we give some preliminary lemmas. In Section 3, we derive the expression of the unique solution to the Problem 1 with 𝒮=HRc𝒞n×n. In Section 4, we establish the solvability conditions for existence and an expression of the solution for Problem 2 with 𝒮=HRc𝒞n×n. In Section 5, we give examples to illustrate the results obtained in this paper.


## 2. Preliminaries

In order to study Problems1 and 2 with 𝒮=HRc𝒞n×n, we first give some preliminary lemmas in this section.For a nontrivial symmetric involutory matrixR∈ℛn×n, there exist positive integers r and s such that r+s=n and an n×n orthogonal matrix [P,Q] such that
(7)R=[P,Q][Ir00-Is][PTQT],
where P∈ℛn×r and Q∈ℛn×s. The columns of P(Q) form an orthogonal basis for the eigenspace of R associated with the eigenvalue 1(-1).Throughout this paper, we always suppose the nontrivial symmetric involutory matrixR is fixed which is given by (7).Lemma 3 (see Theorem 2.1 in [14]).
A matrixK∈Rc𝒞n×n if and only if there exists H1∈ℛn×n such that K=UH1U*, and K∈HRc𝒞n×n if and only if there exists H2∈Sℛn×n such that K=UH2U*, where
(8)U=[P,-iQ],
with P and Q being the same as (7).Lemma 4.
For any matrixA∈𝒞n×n, A=A1⊕A2⊕iA3, where
(9)B=A+RA¯R2,A1=B+B*2,A2=B-B*2,A3=A-RA¯R2i,
and ⊕ denotes the direct sum.Proof.
ForB∈𝒞n×n, it is obvious that B=A1⊕A2, where A1 are A2 are defined in (9). Hence, we just need to prove A=B⊕iA3.
For any matrixA∈𝒞n×n, it is obvious that A=B+iA3, where B and A3 are defined in (9).
We prove the uniqueness ofA=B+iA3. Note that
(10)RBR=B¯,RA3R=A3¯,
that is, B,A3∈Rc𝒞n×n. If there exist D and C3 satisfying
(11)RDR=D¯,RC3R=C3¯
such that A=D+iC3, then
(12)B-D=i(C3-A3).
Multiplying R on the left and right side and then taking the conjugate for (12), it yields
(13)B-D=i(A3-C3),
implying B=D and A3=C3.
SoA=A1⊕A2⊕iA3 holds, where A1, A2, and A3 are defined in (9).Lemma 5.
Let the symmetric involutionR∈ℛn×n be given in (7) and let U be defined as (8). Then(i)
a matrixA∈𝒞m×nsatisfiesAR=A¯if and only ifAU∈Rm×n,(ii)
a matrixB∈𝒞n×lsatisfiesRB=B¯if and only if  U*B∈Rn×l.Proof.
(i) It yields from (7) and (8) that
(14)R=[P,-iQ][Ir00-Is][PTiQT]=U[Ir00-Is]U*.
By (14)
(15)A¯=AR=AU[Ir00-Is]U*,
that is,
(16)A=AU¯[Ir00-Is]U¯*=AU¯U*,
implying AU=AU¯, that is, AU∈Rm×n.
Conversely, ifAU∈Rm×n, according to the proof of the necessity, we can get A¯=AR.
The proof of (ii) can be analogously completed according to the proof of (i).


## 3. The Solution to Problem1 with 𝒮=HRc𝒞n×n

We, in this section, give the explicit expression of the solution to Problem1 with 𝒮=HRc𝒞n×n. In the following, we refer to the 𝒮=HRc𝒞n×n in C.According to Lemma3, if X∈HRc𝒞n×n, then
(17)X=UYU*,
where U is defined as (8) and Y∈Sℛn×n. Let
(18)Ak=Ak1+iAk2∈𝒞mk×n,Ak1=Ak+Ak¯R2,Ak2=Ak-Ak¯R2i,(19)Bl=Bl1+iBl2∈𝒞n×hl,Bl1=Bl+RBl¯2,Bl2=Bl-RBl¯2i,(20)CkU=Ck1+iCk2,Ck1,Ck2∈ℛmk×n,U*Dl=Dl1+iDl2,Dl1,Dl2∈ℛn×hl.
By Lemma 5, it is easy to verify Ak1U,Ak2U∈ℛmk×n, U*Bl1,U*Bl2∈ℛn×hl. We obtain
(21)AkX-Ck=(Ak1+iAk2)UYU*-Ck=[(Ak1UY-Ck1)+i(Ak2UY-Ck2)]U*,(22)XBl-Dl=UYU*(Bl1+iBl2)-Dl=U[YU*(Bl1-Dl1)+iYU*(Bl2-Dl2)].
It follows from the unitary invariance of Frobenius norm, (21), (22), and Y∈Sℛn×n that
(23)∑k=1p‍∥AkX-Ck∥2+∑l=1q‍∥XBl-Dl∥2=∑k=1p‍(∥Ak1UY-Ck1∥2+∥Ak2UY-Ck2∥2)+∑l=1q‍(∥YU*Bl1-Dl1∥2+∥YU*Bl2-Dl2∥2)=∑k=1p‍∥[Ak1UAk2U]Y-[Ck1Ck2]∥2+∑l=1q‍∥Y[U*Bl1,U*Bl2]-[Dl1,Dl2]∥2=∑k=1p‍∥[Ak1UAk2U]Y-[Ck1Ck2]∥2+∑l=1q‍∥[(U*Bl1)T(U*Bl2)T]Y-[Dl1TDl2T]∥2.
Suppose
(24)S=[(A11U)T,(A12U)T,…,(Ap1U)T,(Ap2U)T,U*B11,U*B12,…,U*Bq1,U*Bq2]T,(25)G=[C11T,C12T,…,Cp1T,Cp2T,D11,D12,…,Dq1,Dq2]T,
then
(26)∑k=1p‍∥AkX-Ck∥2+∑l=1q‍∥XBl-Dl∥2=∥SY-G∥2.We first give the following lemma which can be obtained by contrast with Lemma 2.1 in [22].Lemma 6.
GivenS,G∈ℛ(m1+⋯+mp+h1+⋯+hq)×n; therefore, the singular value decomposition (SVD) of S∈ℛr1(m1+⋯+mp+h1+⋯+hq)×n can be described as
(27)S=W[Σr1000]VT=W1Dr1V1T,
where
(28)W=[W1,W2],V=[V1,V2],
Let W1∈ℛ(m1+⋯+mp+h1+⋯+hq)×r1, W2∈ℛ(m1+⋯+mp+h1+⋯+hq)×(m1+⋯+mp+h1+⋯+hq-r1), V1∈ℛn×r1, V2∈ℛn×(n-r1), and Σr1=diag(d1,…,dr1) with di>0, i=1,…,r1; Φ=(1/(di2+dj2))r1×r1. Then minY∈Sℛn×n∥SY-G∥ is consistent if and only if
(29)Y=V[Φ*(Σr1W1TGV1+V1TGTW1Σr1)Σr1-1W1TGV2V2TGTW1Σr1-1L]VT,
where L∈Sℛ(n-r1)×(n-r1) is arbitrary.Theorem 7.
GivenAk,Ck∈𝒞mk×n, Bl,Dl∈𝒞n×hl, and positive integers mk, hl, p, and q, where k=1,…,p, l=1,…,q, the notations Ak1, Ak2, Bl1, Bl2, Ck1, Ck2, Dl1, Dl2, S, G are defined as (18), (19), (20), (24), and (25), respectively. Let the SVD of S∈ℛr1(m1+⋯+mp+h1+⋯+hq)×n be of the form (27) with (28). Then
(30)C={[Φ*(Σr1W1TGV1+V1TGTW1Σr1)Σr1-1W1TGV2V2TGTW1Σr1-1L]L∈Sℛ(n-r1)×(n-r1)∣UV×[Φ*(Σr1W1TGV1+V1TGTW1Σr1)Σr1-1W1TGV2V2TGTW1Σr1-1L]×VTU*[Φ*(Σr1W1TGV1+V1TGTW1Σr1)Σr1-1W1TGV2V2TGTW1Σr1-1L]}.Proof.
It yields from (26) that
(31)minX∈HRc𝒞n×n∑k=1p‍∥AkX-Ck∥2+∑l=1q‍∥XBl-Dl∥2=minY∈Sℛn×n∥SY-G∥2.
By Lemma 6, minY∈Sℛn×n∥SY-G∥ is consistent if and only if Y has the expression of (29). Taking (29) into (17), we obtain the solution set C is (30).Theorem 8.
GivenE∈𝒞n×n, the equation (4) is consistent if and only if
(32)X^=UV×[Φ*(Σr1W1TGV1+V1TGTW1Σr1)Σr1-1W1TGV2V2TGTW1Σr1-1V2TU*(E+E*)UV22]×VTU*.Proof.
Obviously,C is a closed convex set. Hence, there exists the unique element X^∈L such that (4) holds. By applying Theorem 7 and the unitary invariance of Frobenius norm, for X∈C, we get
(33)∥X-E∥2=∥UV[Φ*(Σr1W1TGV1+V1TGTW1Σr1)Σr1-1W1TGV2V2TGTW1Σr1-1L]×VTU*-E[Φ*(Σr1W1TGV1+V1TGTW1Σr1)Σr1-1W1TGV2V2TGTW1Σr1-1L]∥2=∥[Φ*(Σr1W1TGV1+V1TGTW1Σr1)Σr1-1W1TGV2V2TGTW1Σr1-1L]-VTU*EUV[Φ*(Σr1W1TGV1+V1TGTW1Σr1)Σr1-1W1TGV2V2TGTW1Σr1-1L]∥2=∥Φ*(Σr1W1TGV1+V1TGTW1Σr1)-V1TU*EUV1∥2+∥Σr1-1W1TGV2-V1TU*EUV2∥2+∥V2TGTW1Σr1-1-V2TU*EUV1∥2+∥L-V2TU*EUV2∥2.
Then minX∈C∥X-E∥ is equivalent to
(34)minL∈Sℛ(n-r1)×(n-r1)∥L-V2TU*EUV2∥.
(34) holds if and only if
(35)L=V2TU*(E+E*)UV22.
Substituting (35) into X∈C, we get X^ is (32).Corollary 9.
∥
X
min
∥
=
min
X
∈
C
∥
X
∥ if and only if
(36)Xmin=UV×[Φ*(Σr1W1TGV1+V1TGTW1Σr1)Σr1-1W1TGV2V2TGTW1Σr1-10]×VTU*.Remark 10.
whenp=1 and q=1 in Theorem 8, we can derive a result of Theorem 4.1 in [14].


## 4. The Solution to Problem2 with 𝒮=HRc𝒞n×n

We refer to𝒮=HRc𝒞n×n in L in the following text. In this section, we establish necessary and sufficient conditions for the existence and the expression of L. When L is nonempty, we present the expression of the unique solution to (6).It follows from (21) that the system A1X=C1, A2X=C2,…,ApX=Cp, XB1=D1,…,XBq=Dq with unknown X∈HRc𝒞n×n is consistent if and only if there exists Y∈Sℛn×n such that SY=G. We first give the following lemma.Lemma 11 (see Theorem 1 in [7]).
GivenS,G∈R(m1+⋯+mp+h1+⋯+hq)×n. Let the SVD of S∈Rr1(m1+⋯+mp+h1+⋯+hq)×n be (27) with (28). Then the matrix equation SY=G has a symmetric solution if and only if
(37)SS†G=G,GST=SGT,
and the symmetric solution can be expressed as
(38)Y=S†G+(I-S†S)(S†G)T+V2ZV2T,
where Z∈Sℛ(n-r1)×(n-r1) is arbitrary.Theorem 12.
GivenA1,C1∈𝒞m1×n, A2,C2∈𝒞m2×n,…,Ap,Cp∈𝒞mp×n, B1,D1∈𝒞n×h1,…,Bq,Dq∈𝒞n×hq, and positive integers p and q. The notations Ak1, Ak2, Bl1, Bl2, Ck1, Ck2, Dl1, Dl2, S, G are defined as (18), (19), (20), (24), and (25), respectively. Let the SVD of S∈ℛr1(m1+⋯+mp+h1+⋯+hq)×n be of the form (27) with (28). Then the solution set L is nonempty if and only if (37) holds, in which case,
(39)L={[S†G+(I-S†S)(S†G)T+V2ZV2T]Z∈Sℛ(n-r1)×(n-r1)∣U[S†G+(I-S†S)(S†G)T+V2ZV2T]U*}.Proof.
If the solution setL is nonempty, then there exists a matrix X∈HRc𝒞n×n such that A1X=C1, A2X=C2,…,ApX=Cp, XB1=D1,…,XBq=Dq. We know that A1X=C1, A2X=C2,…,ApX=Cp, XB1=D1,…,XBq=Dq with X∈HRc𝒞n×n is consistent if and only if there exists Y∈Sℛn×n such that SY=G. By Lemma 11, SY=G is solvable for Y∈Sℛn×n if and only if (37) holds, and the expression of the solution is (38). Insert (38) into (17), then we obtain L is of the form (39).
Conversely, assume (37) holds, according to the proof of the necessity, A1X=C1, A2X=C2,…,ApX=Cp, XB1=D1,…,XBq=Dq is solvable for X∈HRc𝒞n×n. For Z∈Sℛ(n-r1)×(n-r1), by Lemma 3, U[S†G+(I-S†S)(S†G)T+V2ZV2T]U*∈HRc𝒞n×n.Theorem 13.
GivenF∈𝒞n×n and L is nonempty. Let F1=(1/4)[F+RF¯R+(F+RF¯R)*]. Then (6) is consistent for X~∈L if and only if
(40)X~=U[S†G+(I-S†S)(S†G)T+V2V2TUF1U*V2V2T]U*.
In particular, if ∥Xinf∥=minX∈L∥X∥, then
(41)Xinf=U[S†G+(I-S†S)(S†G)T]U*.Proof.
By Lemma4, F=F1⊕F2⊕iF3, where
(42)F1=14[F+RF¯R+(F+RF¯R)*],F2=14[F+RF¯R-(F+RF¯R)*],F3=F-RF¯R2i.
When L is nonempty, for X∈L, we get
(43)∥X-F∥2=∥U[S†G+(I-S†S)(S†G)T+V2ZV2T]U*-(F1+F2+iF3)[S†G+(I-S†S)(S†G)T+V2ZV2T]∥2=∥S†G+(I-S†S)(S†G)T+V2ZV2T-U*F1U-U*F2U-iU*F3U(I-S†S)(S†G)T∥2.
Since F1∈HRc𝒞n×n, F2,F3∈Rc𝒞n×n, by Lemma 3, we obtain U*F1U∈Sℛn×n, U*F2U,U*F3U∈ℛn×n. Note that U*F2U∈ASℛn×n, then
(44)∥X-F∥2=∥UF1U*-S†G-(I-S†S)(S†G)T-V2ZV2T∥2+∥UF2U*∥2+∥UF3U*∥2.
Hence, minX∈L∥X-F∥ is equivalent to
(45)minZ∈Sℛ(n-r1)×(n-r1)∥UF1U*-S†G-(I-S†S)(S†G)T-V2ZV2T∥.
For the orthogonal matrix V, we get
(46)∥UF1U*-S†G-(I-S†S)(S†G)T-V2ZV2T∥=∥VT[UF1U*-S†G-(I-S†S)(S†G)T-V2ZV2T]V∥=∥V1T[UF1U*-S†G-(I-S†S)(S†G)T]V1∥+∥V1T[UF1U*-S†G-(I-S†S)(S†G)T]V2∥+∥V2T[UF1U*-S†G-(I-S†S)(S†G)T]V1∥+∥V2T[UF1U*-S†G-(I-S†S)(S†G)T]V2-Z∥.
Then minZ∈Sℛ(n-r1)×(n-r1)∥UF1U*-S†G-(I-S†S)(S†G)T-V2ZV2T∥ is solvable if and only if
(47)Z=V2T[UF1U*-S†G-(I-S†S)(S†G)T]V2.
It follows from (27) that
(48)S†=V1Dr1-1W1T.
By (48) and (47) we get
(49)Z=V2TUF1U*V2.
Hence, from (39) and (49), we obtain X~ which can be expressed as (40).Remark 14.
Whenp=1 and q=1 in Theorem 13, we can get a conclusion of Theorem 3.1 in [14].


## 5. Numerical Examples

We, in this section, propose an algorithm for finding the solution of Problems1 and 2 with 𝒮=HRc𝒞n×n and give illustrative numerical examples.Let the symmetric involutory matrixR=[00V20I20V200]∈ℛ6×6, where V2=[0110]. Applying the spectral decomposition of R, we obtain the orthogonal matrix [P,Q] satisfying (7) and then by (8), we get
(50)U=12[00100-i0001i02000000200000001-i000100i].Algorithm 15.
(
1
) Input A1,A2,…,Ap, C1,C2,…,Cp, B1,…,Bq, D1,…,Dq;
(
2
) compute Ak1, Ak2, Bl1, Bl2, Ck1, Ck2, Dl1, Dl2, S, G by (18), (19), (20), (24), and (25);
(
3
) make the SVD of S with the form of (27) and compute W1, W2, V1, V2 by (28);
(
4
) if (37) holds, continue, or go to step 6;
(
5
) input E∈𝒞n×n, compute the solution to Problem 2 with 𝒮=HRc𝒞n×n by (40);
(
6
) input F∈𝒞n×n, compute the solution of Problem 1 with 𝒮=HRc𝒞n×n by (32).Example 1.
We consider the case ofp=1 and q=1. Suppose A,C∈𝒞5×6, B,D∈𝒞6×4, and(51)A=[-2.2163+1.2010i2.2524-1.1328i0-0.0955+1.3420i4.3215-2.3145i1.2120+1.9804i-1.16253.2510i02.1800-0.3125i-0.2235+1.4140i5.70711.0500-2.0120i2.4020+1.2500i03.2106i-1.60802.25600000003.2000-4.1955+1.2000i01.1025-0.2400i-0.8902-0.2408i1.2010+0.7071i],C=[4.1365+2.6614i1.5828-0.2259i2.2091+1.5678i2.4154i-1.8939+0.4692i2.5155-0.7604i1.0118-0.2844i-3.4235+1.5798i1.3798-0.7942i1.3050+2.9900i-2.7833+0.0592i1.8744-0.7889i0000002.1991-1.0648i-1.5298i5.6808+1.6029i-4.0150+2.1200i2.4387-1.9600i3.4200-2.6288i2.98832.4329-0.6405i1.4611-1.0503i3.7680-1.4980i-2.0266-0.9334i1.5756i],B=[0.5200+0.7200i-0.20000.4500+0.3600i0-0.3394-0.5091i0.1414-0.3536i-0.3200-0.6500i00.5940+0.2970i0.0707+0.0354i0.070000.0200i0.25000.3500+0.8400i0-0.3960-0.5091i0.1414+0.3536i-0.3200+0.2400i00.5940+0.2970i0.2121+0.0354i0.07070],D=[-0.3140+0.1069i0.4954-0.0615i0.7469+0.4876i-0.0662+0.0817i0.4255-0.5754i0.0348+0.5182i-0.2522+0.5751i-0.0105-0.0928i0.0883+0.0649i0.2201+0.0182i0.3133+0.3271i0.0076-0.0138i0.8675+0.7972i-0.3105-0.0837i-0.1461+0.2541i0.2620+0.0983i-0.4591+0.6268i-0.0405-0.6447i0.1517-0.6198i0.0256+0.2057i0.1439-0.2759i0.4722+0.1091i0.4952+0.3680i-0.0414-0.1178i].We can easily verify the solvability condition (37) does not holds. For given E∈𝒞6×6,(52)E=[0.4424+0.6274i-0.1765-0.0108i0.4006+0.3308i0.3069-0.1505+0.1262i1.200+0.0045i0.6374-0.0108i0.2246-0.2738i0.1509-0.4327i-0.0434-0.0915i0.1385-0.1800i0-0.4950-0.4172i0.1414-0.0354i-0.3818-0.1750i-0.2475+0.0619i-0.1630-0.0540i0.2032i-0.5129-0.4025i0.1956-0.0157i0.0868+0.5565i-0.1227-0.2382-0.0911i00.1878+0.3187i0.1789+0.3334i0.0271+0.7668i-0.0434+0.0915i-0.1024-0.0002+0.0150i-0.2404-0.6010i0.1414+0.0354i-0.2546-0.3341i-0.2475-0.0619i-0.0002-0.0150i0.0256],applying Algorithm 15, we get the following results:(53)X^=[-0.1949+0.0000i-0.0395+0.3097i0.1067+0.0353i0.6067-0.2888i0.0380-0.6272i-0.1218+0.1242i-0.0395-0.3097i-0.7354+0.0000i0.0606-0.1460i-0.1095+0.4086i0.6954-0.0593i0.0380-0.6272i0.1067-0.0353i0.0606+0.1460i0.00090.39570.0606-0.1460i0.1067+0.0353i0.6067+0.2888i-0.1095-0.4086i0.3957-0.1562-0.1095+0.4086i0.6067-0.2888i0.0380+0.6272i0.6954+0.0593i0.0606+0.1460i-0.1095-0.4086i-0.7354-0.0000i-0.0395+0.3097i-0.1218-0.1242i0.0380+0.6272i0.1067-0.0353i0.6067+0.2888i-0.0395-0.3097i-0.1949-0.0000i],(54)∥AX^-C∥2+∥X^B-D∥2=minX∈HRcℂ6×6∥AX-C∥2+∥XB-D∥2=160.0357,∥X^-X^*∥=4.1192×10-16,∥X¯-RXR∥=0,∥X^-E∥=3.7397.The following example is about Problem2 with 𝒮=HRc𝒞6×6, p=1 and q=0. We list results of comparison of the solutions computed by Algorithm 15 and MATLAB procedure X=A∖C.Example 2.
LetWa∈𝒞8×8, Va∈𝒞6×6 be unitary matrices,(55)Wa=[-0.1590+0.2812i-0.1803-0.1822i0.4873+0.0557i0.1272+0.2856i:0.1159-0.1077i0.4983-0.2783i0.4127-0.2359i0.3195+0.2342i:0.2197-0.1364i0.1146-0.1466i0.2900+0.0440i-0.5309-0.0423i:0.0016-0.0047i0.0108-0.0047i-0.0257+0.0018i-0.0181+0.0126i:0.2576-0.0271i0.2935+0.2509i-0.3014+0.0304i0.3562+0.0778i:0.4283+0.1012i0.0858-0.1579i-0.1758-0.3913i-0.5198+0.0316i:0.3884-0.2103i0.2167+0.2285i0.1236-0.0447i0.1348-0.0089i:0.5625+0.1774i-0.2720-0.4682i-0.0047+0.3918i0.1951+0.0273i:-0.5270-0.1794i0.1734-0.2471i0.2466+0.1275i0.0437+0.1077i0.0562-0.0163i-0.2264+0.4255i-0.1461+0.0547i-0.0662-0.0426i-0.1748+0.4297i0.4788+0.1119i-0.1543-0.2114i0.0422-0.0388i-0.0011-0.0201i0.0249+0.0695i0.0506+0.2945i0.7638-0.5658i-0.3766-0.0582i0.3605-0.2131i-0.4447+0.1694i-0.0868-0.0677i-0.1860-0.3130i-0.2654-0.1257i0.0594+0.2924i-0.0976+0.0376i0.3112-0.2036i0.3628-0.1215i0.5860-0.0319i0.1016+0.1755i0.2087-0.1071i-0.0638-0.1313i-0.1736-0.2252i0.0555-0.0788i],Va=[0.46210.1052-0.6218-0.1836-0.58970.0852-0.2474-0.0088i-0.1896-0.5726i0.1951-0.1974i-0.5870-0.1096i-0.2811+0.1327i-0.2113-0.0036i0.4418-0.0309i-0.2001-0.2357i0.0004+0.2178i0.0696+0.1631i0.1926-0.3664i-0.6629-0.1362i-0.1299-0.4383i0.1856+0.1319i0.1214-0.4193i0.0858+0.5057i-0.2366-0.1016i-0.0917-0.4592i-0.2390+0.2258i-0.1938-0.3803i0.0533+0.3571i0.3007+0.3102i-0.3167-0.3755i0.3810-0.0794i0.4263+0.1739i0.3926-0.3760i0.3391-0.2301i0.2360+0.2699i-0.0153+0.2763i0.0807+0.3359i],and D(t)=[Dzeros(2,6)], where D=diag(3,2,1,t/1,t/2,t/3) and zeros(2,6) is a 2×6 zeros matrix. Let A(t)=WaD(t)Va* and X∈HRc𝒞6×6,(56)X=[0.2543-0.3625-1.1647i-0.1609-0.2793i0.8839-0.4914i0.5675-0.0043i-0.4143-0.4378i-0.3625+1.1647i-1.38860.0672-0.6470i-0.2316+0.2397i0.0524-0.4418i0.5675-0.0043i-0.1609+0.2793i0.0672+0.6470i-0.4312-0.25000.0672-0.6470i-0.1609-0.2793i0.8839+0.4914i-0.2316-0.2397i-0.2500-0.0380-0.2316+0.2397i0.8839-0.4914i0.5675+0.0043i0.0524+0.4418i0.0672+0.6470i-0.2316-0.2397i-1.3886-0.3625-1.1647i-0.4143+0.4378i0.5675+0.0043i-0.1609+0.2793i0.8839+0.4914i-0.3625+1.1647i0.2543].Then compute C(t)=A(t)X for t=10,1,10-1,10-2,…,10-13. Obviously, Problem 2 with 𝒮=HRc𝒞n×n is consistent for each value of t. For matrices A, C obtained above, we first use Algorithm 15 to obtain the Hermitian R-conjugate solutions approximate to X, then compute the solutions of AX=C by MATLAB procedure X=A∖C. Let X~is denote the solutions computed by Algorithm 15 and Xis the solutions by MATLAB procedure X=A∖C. Let
(57)e1=∥AX~i-C∥,e2=∥X~i-X∥,e3=∥X~i¯-RX~iR∥,e4=∥X~iT-X~i∥,te1=∥AXi-C∥,te2=∥Xi-X∥,te3=∥Xi¯-RXiR∥,te4=∥XiT-Xi∥.Analysis of Results. As a general observation from Table 1, we find that the performance of Algorithm 15 to solve Problem 2 is very good and that of the MATLAB procedure X=A∖C is quite sensitive to the conditioning of matrix A. For t=10,1,…,10-4, both methods behave well. In this case we should choose MATLAB procedure X=A∖C to solve Problem 2 with 𝒮=HRc𝒞n×n for it is simple. However, as some singular values of A are close to zero, the solutions Xi computed by MATLAB procedure X=A∖C do not satisfy AX=C and gradually lose the property of Hermitian R-conjugate matrix, while Algorithm 15 does it well. Hence, when A has small singular values close to zero, the Algorithm 15 predominates over MATLAB procedure X=A∖C.Table 1
Comparison of results by Algorithm15 and the MATLAB procedure A∖C.
t
e
1
e
2
e
3
e
4
t
e
1
t
e
2
t
e
3
t
e
4
-
13
2
·
1
0
-
15
1
·
1
0
-
15
0
2
·
1
0
-
15
3
·
1
0
-
15
3
·
1
0
-
2
5
·
1
0
-
2
4
·
1
0
-
2
-
12
2
·
1
0
-
15
1
·
1
0
-
15
0
1
·
1
0
-
15
3
·
1
0
-
15
3
·
1
0
-
3
5
·
1
0
-
3
4
·
1
0
-
3
-
11
2
·
1
0
-
15
2
·
1
0
-
15
0
1
·
1
0
-
15
3
·
1
0
-
15
3
·
1
0
-
4
4
·
1
0
-
4
5
·
1
0
-
4
-
10
2
·
1
0
-
15
1
·
1
0
-
15
0
2
·
1
0
-
15
3
·
1
0
-
15
3
·
1
0
-
5
3
·
1
0
-
5
4
·
1
0
-
5
-
9
3
·
1
0
-
15
2
·
1
0
-
15
0
2
·
1
0
-
15
2
·
1
0
-
15
3
·
1
0
-
6
5
·
1
0
-
6
4
·
1
0
-
6
-
8
2
·
1
0
-
15
1
·
1
0
-
15
0
2
·
1
0
-
15
2
·
1
0
-
15
3
·
1
0
-
7
4
·
1
0
-
7
3
·
1
0
-
7
-
7
3
·
1
0
-
15
2
·
1
0
-
15
0
1
·
1
0
-
15
2
·
1
0
-
15
4
·
1
0
-
8
5
·
1
0
-
8
5
·
1
0
-
8
-
6
2
·
1
0
-
15
2
·
1
0
-
15
0
2
·
1
0
-
15
2
·
1
0
-
15
2
·
1
0
-
9
4
·
1
0
-
9
4
·
1
0
-
9
-
5
2
·
1
0
-
15
1
·
1
0
-
15
0
2
·
1
0
-
15
2
·
1
0
-
15
2
·
1
0
-
10
3
·
1
0
-
10
3
·
1
0
-
10
-
4
2
·
1
0
-
15
1
·
1
0
-
15
0
1
·
1
0
-
15
2
·
1
0
-
15
3
·
1
0
-
11
3
·
1
0
-
11
4
·
1
0
-
11
-
3
2
·
1
0
-
15
1
·
1
0
-
15
0
2
·
1
0
-
15
3
·
1
0
-
15
3
·
1
0
-
12
5
·
1
0
-
12
5
·
1
0
-
12
-
2
2
·
1
0
-
15
1
·
1
0
-
15
0
2
·
1
0
-
15
2
·
1
0
-
15
2
·
1
0
-
13
3
·
1
0
-
13
3
·
1
0
-
13
-
1
2
·
1
0
-
15
1
·
1
0
-
15
0
2
·
1
0
-
15
3
·
1
0
-
15
3
·
1
0
-
14
5
·
1
0
-
14
5
·
1
0
-
14
0
2
·
1
0
-
15
2
·
1
0
-
15
0
2
·
1
0
-
15
4
·
1
0
-
15
4
·
1
0
-
15
6
·
1
0
-
15
6
·
1
0
-
15
1
9
·
1
0
-
15
2
·
1
0
-
15
0
2
·
1
0
-
15
1
·
1
0
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14
5
·
1
0
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15
6
·
1
0
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15
8
·
1
0
-
15


## 6. Conclusion

In this paper, we converted the HermitianR-conjugate generalized Procrustes problem to real symmetric Procrustes problem trickly and obtained its solution set. We also investigated the Hermitian R-conjugate solution to the linear system of complex matrix equations A1X=C1, A2X=C2,…,ApX=Cp, XB1=D1,…,XBq=Dq and established solvable conditions and the formula for the Hermitian R-conjugate solution. Moreover, we showed the optimal approximation solution to a given matrix in the above two corresponding solution set is unique, respectively. As applications, a numerical algorithm has been given and the examples have illustrated the feasibility of the algorithm. Additionally, we can further consider the least square Hermitian R-conjugate solution of the system AkXAk*=Ck, k=1,2,…,n (n is positive integer) and the corresponding optimal approximation problem.



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*Source: 423605-2013-09-28.xml*