We consider the Hermitian R-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.
1. Introduction
Throughout, let 𝒞m×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.
Given 𝒮⊆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 sets 𝒮, 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 case AkX=Ck, XBl=Dl, k=1,…,p, l=1,…,q in Problem 1 and describe it as follows.
Problem 2.
Given 𝒮⊆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 Problem 2 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 with S being the set of Hermitian R-conjugate matrices.
Recall that an n×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 Section 2, 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 Problems 1 and 2 with 𝒮=HRc𝒞n×n, we first give some preliminary lemmas in this section.
For a nontrivial symmetric involutory matrix R∈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 matrix R is fixed which is given by (7).
Lemma 3 (see Theorem 2.1 in [<xref ref-type="bibr" rid="B21">14</xref>]).
A matrix K∈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 matrix A∈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.
For B∈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 matrix A∈𝒞n×n, it is obvious that A=B+iA3, where B and A3 are defined in (9).
We prove the uniqueness of A=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.
So A=A1⊕A2⊕iA3 holds, where A1, A2, and A3 are defined in (9).
Lemma 5.
Let the symmetric involution R∈ℛn×n be given in (7) and let U be defined as (8). Then
a matrix A∈𝒞m×n satisfies AR=A¯ if and only if AU∈Rm×n,
a matrix B∈𝒞n×l satisfies RB=B¯ if and only ifU*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, if AU∈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 Problem <xref ref-type="statement" rid="problem1">1</xref> with <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M198"><mml:mi>𝒮</mml:mi><mml:mo mathvariant="bold">=</mml:mo><mml:mi>H</mml:mi><mml:msub><mml:mrow><mml:mi>R</mml:mi></mml:mrow><mml:mrow><mml:mi>c</mml:mi></mml:mrow></mml:msub><mml:msup><mml:mrow><mml:mi>𝒞</mml:mi></mml:mrow><mml:mrow><mml:mi>n</mml:mi><mml:mo mathvariant="bold">×</mml:mo><mml:mi>n</mml:mi></mml:mrow></mml:msup></mml:math></inline-formula>
We, in this section, give the explicit expression of the solution to Problem 1 with 𝒮=HRc𝒞n×n. In the following, we refer to the 𝒮=HRc𝒞n×n in C.
According to Lemma 3, 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.
Given S,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.
Given Ak,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.
Given E∈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.
∥Xmin∥=minX∈C∥X∥ if and only if
(36)Xmin=UV×[Φ*(Σr1W1TGV1+V1TGTW1Σr1)Σr1-1W1TGV2V2TGTW1Σr1-10]×VTU*.
Remark 10.
when p=1 and q=1 in Theorem 8, we can derive a result of Theorem 4.1 in [14].
4. The Solution to Problem <xref ref-type="statement" rid="problem2">2</xref> with <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M269"><mml:mi>𝒮</mml:mi><mml:mo mathvariant="bold">=</mml:mo><mml:mi>H</mml:mi><mml:msub><mml:mrow><mml:mi>R</mml:mi></mml:mrow><mml:mrow><mml:mi>c</mml:mi></mml:mrow></mml:msub><mml:msup><mml:mrow><mml:mi>𝒞</mml:mi></mml:mrow><mml:mrow><mml:mi>n</mml:mi><mml:mo mathvariant="bold">×</mml:mo><mml:mi>n</mml:mi></mml:mrow></mml:msup></mml:math></inline-formula>
We refer to 𝒮=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 [<xref ref-type="bibr" rid="B8">7</xref>]).
Given S,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.
Given A1,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 set L 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.
Given F∈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 Lemma 4, 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.
When p=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 Problems 1 and 2 with 𝒮=HRc𝒞n×n and give illustrative numerical examples.
Let the symmetric involutory matrix R=[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].
(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 of p=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 Problem 2 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.
Let Wa∈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.
Comparison of results by Algorithm 15 and the MATLAB procedure A∖C.
t
e1
e2
e3
e4
te1
te2
te3
te4
-13
2·10-15
1·10-15
0
2·10-15
3·10-15
3·10-2
5·10-2
4·10-2
-12
2·10-15
1·10-15
0
1·10-15
3·10-15
3·10-3
5·10-3
4·10-3
-11
2·10-15
2·10-15
0
1·10-15
3·10-15
3·10-4
4·10-4
5·10-4
-10
2·10-15
1·10-15
0
2·10-15
3·10-15
3·10-5
3·10-5
4·10-5
-9
3·10-15
2·10-15
0
2·10-15
2·10-15
3·10-6
5·10-6
4·10-6
-8
2·10-15
1·10-15
0
2·10-15
2·10-15
3·10-7
4·10-7
3·10-7
-7
3·10-15
2·10-15
0
1·10-15
2·10-15
4·10-8
5·10-8
5·10-8
-6
2·10-15
2·10-15
0
2·10-15
2·10-15
2·10-9
4·10-9
4·10-9
-5
2·10-15
1·10-15
0
2·10-15
2·10-15
2·10-10
3·10-10
3·10-10
-4
2·10-15
1·10-15
0
1·10-15
2·10-15
3·10-11
3·10-11
4·10-11
-3
2·10-15
1·10-15
0
2·10-15
3·10-15
3·10-12
5·10-12
5·10-12
-2
2·10-15
1·10-15
0
2·10-15
2·10-15
2·10-13
3·10-13
3·10-13
-1
2·10-15
1·10-15
0
2·10-15
3·10-15
3·10-14
5·10-14
5·10-14
0
2·10-15
2·10-15
0
2·10-15
4·10-15
4·10-15
6·10-15
6·10-15
1
9·10-15
2·10-15
0
2·10-15
1·10-14
5·10-15
6·10-15
8·10-15
6. Conclusion
In this paper, we converted the Hermitian R-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.
Acknowledgments
The authors are very much indebted to the anonymous referees for their constructive and valuable comments and suggestions which greatly improved the original paper. This research was supported by the Innovation Foundation of Shanghai Municipal Education Commission (13ZZ080), the National Natural Science Foundation of China (11391240185), the Shanghai University Scientific Selection and Cultivation for Outstanding Young Teachers in special fund (sjr10009), and the Natural Science Foundation of Guangxi Province (no. 2012GXNSFBA053006).
GreenB. F.The orthogonal approximation of an oblique structure in factor analysisHighamN. J.The symmetric Procrustes problemPengJ.HuX.-Y.ZhangL.The (M,N)-symmetric Procrustes problemTrenchW. F.Hermitian, Hermitian R-symmetric, and Hermitian R-skew symmetric Procrustes problemsAnderssonL.-E.ElfvingT.A constrained Procrustes problemGowerJ. C.Generalized Procrustes analysisDaiH.On the symmetric solutions of linear matrix equationsPengZ.-Y.HuX.-Y.The reflexive and anti-reflexive solutions of the matrix equation AX=BHornR. A.SergeichukV. V.Shaked-MondererN.Solution of linear matrix equations in a ^{∗}congruence classHuaG. K.HuX.ZhangL.A new iterative method for the matrix equation AX=BPorterA. D.Solvability of the matrix equation AX=BWangQ.-W.YuJ.On the generalized bi (skew-) symmetric solutions of a linear matrix equation and its procrust problemsHannaY. S.On the solutions of tridiagonal linear systemsDongC.-Z.WangQ.-W.ZhangY.-P.On the Hermitian R-conjugate solution of a system of matrix equationsTrenchW. F.Characterization and properties of matrices with generalized symmetry or skew symmetryHuckleT.Serra-CapizzanoS.Tablino-PossioC.Preconditioning strategies for Hermitian indefinite Toeplitz linear systemsChanR. H.YipA. M.NgM. K.The best circulant preconditioners for Hermitian Toeplitz systemsLeeA.Centro-Hermitian and skew-centro-Hermitian matricesOppenheimA. V.NgM. K.PlemmonsR. J.PimentelF.A new approach to constrained total least squares image restorationKouassiR.GoutonP.PaindavoineM.Approximation of the Karhunen-Loève tranformation and its application to
colour imagesDengY.-B.HuX.-Y.ZhangL.Least squares solution of BXAT=T over symmetric, skew-symmetric, and positive semidefinite X