Document Sample

NUMERICAL MATHEMATICS Vol. 15 No. 3, pp. 217-226 A Journal of Chinese Universities (English Series) August 2006 Least-Squares Mirrorsymmetric Solution for Matrix Equations (AX = B, XC = D) Fanliang Li1,2,∗, Xiyan Hu2 and Lei Zhang2 1 Department of Mathematics and Computational Science, Hunan City College, Yi Yang 413000, China. 2 College of Mathematics and Econometrics, Hunan University, Changsha 410082, China. Received June 9, 2005; Accepted (in revised version) October 24, 2005 Abstract. In this paper, least-squares mirrorsymmetric solution for matrix equations (AX = B, XC = D) and its optimal approximation is considered. With special expression of mir- rorsymmetric matrices, a general representation of solution for the least-squares problem is obtained. In addition, the optimal approximate solution and some algorithms to obtain the optimal approximation are provided. Key words: Mirrorsymmetric matrix; least-squares solution; optimal approximation. AMS subject classiﬁcations: 15A06; 15A57 1 Introduction Matrix equations (AX = B, XC = D) is a class of important matrix equations. The problem of the solution of the matrix equations (AX = B, XC = D) arise in engineering and in some special matrix inverse problems [1-3]. Many authors have been devoted to the study of this problem, and a series of useful results have been obtained. For example, Mitra [4] gave the common solution of minimum possible rank by using generalized inverse of matrix. Chu [5], Mitra [6] presented the necessary and suﬃcient conditions for the solvability and general solution by using the singular value decomposition (SVD) and generalized inverse of matrix, respectively. When the solution matrix X is constrained and the matrix equations are not consistent, it is necessary to study the least-squares constrained solution of the corresponding matrix equations. The purpose of this paper is to discuss the least-squares mirrorsymmetric solution of the matrix equations (AX = B, XC = D) by using the special structure of mirrorsymmetric ma- trices. The background for introducing the deﬁnition of mirrorsymmetric matrices is to study odd/even-mode decomposition of symmetric multiconductor transmission lines (MTL)[7]. We now introduce some notation. Let Rn×m be the set of all n × m real matrices; Rr n×m n×m n×n be the set of all matrices in R with rank r, R(A) denote the rank of A; OR denote ∗ Correspondenceto: Fanliang Li, Department of Mathematics and Computational Science, Hunan City College, Yi Yang 413000, China/College of Mathematics and Econometrics, Hunan University, Changsha 410082, China. E-mail: lfl302@tom.com Numer. Math. J. Chinese Univ. (English Ser.) 217 http://www.global-sci.org/nm 218 Least-Squares Mirrorsymmetric Solution for Matrix Equations the set of all n × n orthogonal matrices; The identity matrix of order n by In ; AT and A+ be the transpose and the Moore-Penrose generalized inverse of A, respectively. For A = (aij ), B = (bij ) ∈ Rn×m , A ∗ B = (aij bij ) denote the Hadamard product of matrices A and B; (A, B) = trB T A denote the inner product of matrices A and B. The induced matrix norm is 1 1 Frobenius norm, i.e ||A|| = (A, A) 2 = (tr(AT A)) 2 . Deﬁnition 1.1. The (k, p)-mirror matrix W(k,p) is deﬁned by Jk W(k,p) = Ip , (1) Jk where Jk is the k-square backward identity matrix with ones along the secondary diagonal and zero elsewhere. The dimension of the (k, p)-mirror matrix is n = 2k + p, where k ≥ 1, p ≥ 0. The (k, p)-mirror matrix W(k,p) is orthogonal and symmetric, i.e. W −1 = W T = W . When p = 0 or 1, mirror matrix W(k,p) is backward identity matrix Jn . Deﬁnition 1.2. A matrix A ∈ R(2k+p)×(2k+p) is called the (k, p)-mirrorsymmetric matrix if and only if A = W(k,p) AW(k,p) . (2) We denote the set of all (k, p)-mirrorsymmetric matrices by M S(k,p) . A matrix A ∈ R(2k+p)×(2k+p) is called the (k, p)-mirrorskewsymmetric matrix if and only if A = −W(k,p) AW(k,p) . (3) We denote the set of all (k, p)-mirrorskewsymmetric matrices by M SS(k,p) . From Deﬁnition 1.2, it is easy to see that the (k, 1)-mirrorsymmetric matrices and the (k, 0)- mirrorsymmetric matrices are centrosymmetric matrices. That is to say, all centrosymmetric matrices are the special cases of mirrorsymmetric matrices, i.e. when p = 0 or 1, mirror matrix W(k,p) is the backward identity matrix Jn . Then (2) becomes A = Jn AJn ,which is the deﬁnition of centrosymmetric matrices [8]. We study the following problems in this paper. Problem I. Given A ∈ Rh×(2k+p) , B ∈ Rh×(2k+p) , C ∈ R(2k+p)×l , D ∈ R(2k+p)×l , ﬁnding ∆ X ∈ M S(k,p) such that f (X) = ||AX − B||2 + ||XC − D||2 = min. Problem II. ˜ Given X ∗ ∈ R(2k+p)×(2k+p) ﬁnding X ∈ SE such that ˜ X ∗ − X = min X∗ − X , ∀X∈SE where SE is the solution set of Problem I. The paper is organized as follows. In Section 2, we at ﬁrst discuss the structure of mirrorsym- metric matrices. Then with the special structure of mirrorsymmetric matrices and the singular value decomposition (SVD) of matrix, we obtain the solution set of Problem I. In Section 3, the unique approximation solution of Problem II is presented by applying the decomposition of space. Finally, some algorithms to obtain the optimal approximate solution and numerical experiment to illustrate the results obtained in this paper correction are provided. F. Li, X. Hu and L. Zhang 219 2 The solution of problem I Lemma 2.1. A ∈ M S(k,p) if and only if Ak×k Bk×p Ck×k Jk A = Dp×k Ep×p Dp×k Jk . (4) Jk Ck×k Jk Bk×p Jk Ak×k Jk The above result follows directly from Deﬁnition 1.2. Hence the proof is omitted. From Deﬁnition 1.1, 1.2 and Lemma 2.1, it is easy to derive the following lemma. Lemma 2.2. A ∈ M S(k,p) if and only if A1 0 A=K KT , (5) 0 A2 where √ Ak×k + Ck×k 2Bk×p A1 = √ ∈ R(k+p)×(k+p) , A2 = Jk (Ak×k − Ck×k )Jk ∈ Rk×k , 2Dp×k Ep×p 1 Ik √0 −Jk K = √ 0 2Ip 0 . (6) 2 J 0 Ik k Lemma 2.3. R(2k+p)×(2k+p) = M S(k,p) M SS(k,p) . To prove Lemma 2.3, we only prove that for any A ∈ R(2k+p)×(2k+p) , there exist a unique A1 ∈ M S(k,p) , and a unique A2 ∈ M SS(k,p) such that A1 + A2 = A and (A1 , A2 ) = 0. Proof 1) For any A ∈ R(2k+p)×(2k+p) , there exist A1 ∈ M S(k,p) , A2 ∈ M SS(k,p) such that A = A1 + A2 , (7) 1 where A1 = 2 (A + W(k,p) AW(k,p) ), A2 = 1 (A − W(k,p) AW(k,p) ). 2 ¯ ¯ 2) If there exist another A1 ∈ M S(k,p) , A2 ∈ M SS(k,p) such that ¯ ¯ A = A1 + A2 . (8) Combining (7) and (8) gives ¯ ¯ A1 − A1 = −(A2 − A2 ). (9) Multiplying (9) on the left and the right by W(k,p) , respectively, and using the deﬁnition of A1 , ¯ ¯ A2 , A1 , A2 , we obtain ¯ ¯ A1 − A1 = (A2 − A2 ). (10) ¯ ¯ It follows from (9) and (10) that A1 = A1 , A2 = A2 . 3) For any A1 ∈ M S(k,p) , A2 ∈ M SS(k,p) , (A1 , A2 ) = tr(AT A1 ) = tr(AT W(k,p) W(k,p) A1 ) 2 2 = tr(W(k,p) AT W(k,p) W(k,p) A1 W(k,p) ) 2 = tr(−AT A1 ) = −tr(AT A1 ). 2 2 220 Least-Squares Mirrorsymmetric Solution for Matrix Equations Hence (A1 , A2 ) = tr(AT A1 ) = 0. Combining 1), 2) and 3), we have 2 R(2k+p)×(2k+p) = M S(k,p) M SS(k,p) . Let A ∈ Rh×(2k+p) , B ∈ Rh×(2k+p) , C ∈ R(2k+p)×l , D ∈ R(2k+p)×l , K be given by (6). Denote C1 D1 AK = (A1 , A2 ), BK = (B1 , B2 ), KT C = , KT D = , (11) C2 D2 A1 ∈ Rh×(k+p) , B1 ∈ Rh×(k+p) , C1 ∈ R(k+p)×l , D1 ∈ R(k+p)×l , A2 ∈ Rh×k , B2 ∈ Rh×k , C2 ∈ Rk×l , D2 ∈ Rk×l . Denote the SVDs of A1 , A2 , C1 , C2 as follows, respectively. T Σ1 0 Σ1 0 V11 A1 = U1 V T = (U11 , U12 ) T = U11 Σ1 V11 , 0 0 1 0 0 T V12 U1 = (U11 , U12 ) ∈ ORh×h , V1 = (V11 , V12 ) ∈ OR(k+p)×(k+p) , (12) h×r1 (k+p)×r1 U11 ∈ R , V11 ∈ R , Σ1 = diag(σ1 , · · · , σr1 ), σi > 0, i = 1, · · · , r1 , r1 = R(A1 ). T Σ2 0 Σ2 0 V21 A2 = U2 V T = (U21 , U22 ) T = U21 Σ2 V21 , 0 0 2 0 0 T V22 U2 = (U21 , U22 ) ∈ ORh×h , V2 = (V21 , V22 ) ∈ ORk×k , (13) U21 ∈ Rh×r2 , V21 ∈ Rk×r2 , Σ2 = diag(δ1 , · · · , δr2 ), δi > 0, i = 1, · · · , r2 , r2 = R(A2 ). τ1 0 τ 0 QT C1 = P1 QT = (P11 , P12 ) 1 1 11 = P11 τ1 QT , 11 0 0 0 0 QT 12 P1 = (P11 , P12 ) ∈ OR(k+p)×(k+p) , Q1 = (Q11 , Q12 ) ∈ ORl×l , (14) (k+p)×s1 l×s1 P11 ∈ R , Q11 ∈ R , τ1 = diag(α1 , · · · , αs1 ), αi > 0, i = 1, · · · , s1 , s1 = R(C1 ). τ2 0 τ 0 QT C2 = P2 QT = (P21 , P22 ) 2 2 21 = P21 τ2 QT , 21 0 0 0 0 QT 22 P2 = (P21 , P22 ) ∈ ORk×k , Q2 = (Q21 , Q22 ) ∈ ORl×l , (15) P21 ∈ Rk×s2 , Q21 ∈ Rl×s2 , τ2 = diag(β1 , · · · , βs2 ), βi > 0, i = 1, · · · , s2 , s2 = R(C2 ). Lemma 2.4 ([9]). If B ∈ Rn×n , C ∈ Rn×n , Σ = diag(α1 , · · · , αn ) ∈ Rn×n , τ = diag(β1 , · · · , βn ) ∈ Rn×n , then the solution for problem XΣ − B 2 + τ X − C 2 = minX∈Rn×n can be expressed as 1 X = Φ ∗ (BΣ + τ C), where Φ = (ϕij ), ϕij = 2. α2 + βi j F. Li, X. Hu and L. Zhang 221 Lemma 2.5. If A1 , B1 , C1 , D1 are given by (11), and the SV Ds of A1 , C1 be (12) and (14) respectively. Then the general representation of solution for problem A1 X − B1 2 + XC1 − D1 2 = minX∈R(k+p)×(k+p) can be expressed as Φ ∗ (Σ1 U11 B1 P11 + V11 D1 Q11 τ1 ) Σ−1 U11 B1 P12 T T 1 T T X = V1 T −1 P1 , (16) V12 D1 Q11 τ1 X22 1 ∀ X22 ∈ R(k+p−r1 )×(k+p−s1 ) , Φ = (ϕij ), ϕij = , 1 ≤ i ≤ r1 , 1 ≤ j ≤ s1 . 2 σi + α2 j Proof From (12) and (14), we have 2 2 A1 X − B1 + XC1 − D1 2 2 Σ1 0 τ1 0 = U1 V T X − B1 + XP1 QT − D1 0 0 1 0 0 1 2 2 Σ1 0 τ1 0 = T V T XP1 − U1 B1 P1 + V1T XP1 − V1T D1 Q1 . 0 0 1 0 0 Denote X11 X12 B11 B12 D11 D12 V1T XP1 = , T U1 B1 P1 = , V1T D1 Q1 = . X21 X22 B21 B22 D21 D22 Hence 2 2 A1 X − B1 + XC1 − D1 2 2 Σ1 0 X11 X12 B11 B12 X11 X12 τ1 0 D11 D12 = − + − 0 0 X21 X22 B21 B22 X21 X22 0 0 D21 D22 = Σ1 X11 − B11 2 + X11 τ1 − D11 2 + Σ1 X12 − B12 2 + X21 τ1 − D21 2 + B21 2 + B22 2 + D12 2 + D22 2 . 2 2 So A1 X − B1 + XC1 − D1 = minX∈R(k+p)×(k+p) if and only if 2 2 Σ1 X11 − B11 + X11 τ1 − D11 = min , X11 ∈Rr1 ×s1 2 Σ1 X12 − B12 = min, 2 X21 τ1 − D21 = min . From Lemma 2.4, we have T T X11 = Φ ∗ (Σ1 B11 + D11 τ1 ) = Φ ∗ (Σ1 U11 B1 P11 + V11 D1 Q11 τ1 ), 1 Φ = (ϕij ), ϕij = , 1 ≤ i ≤ r1 , 1 ≤ j ≤ s1 . 2 σi + α2 j X12 = Σ−1 B12 = Σ−1 U11 B1 P12 , X21 = D21 τ1 = V12 D1 Q11 τ1 . 1 1 T T This completes the proof of Lemma 2.5. 222 Least-Squares Mirrorsymmetric Solution for Matrix Equations Theorem 2.1. If A, B, C, D are given by (11). Let the SVDs of A1 , A2 , C1 , C2 be (12)-(15), respectively. Then Problem I has a solution in M S(k,p) . Moreover, the general solution can be expressed as X1 0 X=K KT , (17) 0 X2 where Φ ∗ (Σ1 U11 B1 P11 + V11 D1 Q11 τ1 ) Σ−1 U11 B1 P12 T T 1 T T X1 = V1 T −1 P1 , (18) V12 D1 Q11 τ1 X22 1 ∀ X22 ∈ R(k+p−r1 )×(k+p−s1 ) , Φ = (ϕij ), ϕij = 2 , 1 ≤ i ≤ r1 , 1 ≤ j ≤ s1 , σi + α2 j Φ′ ∗ (Σ2 U21 B2 P21 + V21 D2 Q21 τ2 ) Σ−1 U21 B2 P22 T T 2 T T X2 = V2 T −1 ′ P2 , (19) V22 D2 Q21 τ2 X22 1 ∀ X22 ∈ R(k−r2 )×(k−s2 ) , ′ Φ′ = (ϕ′ ), ij ϕ′ = ij 2 2, 1 ≤ i ≤ r2 , 1 ≤ j ≤ s2 . δi + βj Proof For any X ∈ M S(k,p) , from Lemma 2.2, X can be expressed as X1 0 X=K KT . (20) 0 X2 From (11), (20), we have 2 2 f (X) = AX − B + XC − D 2 2 X1 0 X1 0 = AK KT − B + K KT C − D 0 X2 0 X2 2 2 X1 0 X1 0 = AK − BK + KT C − KT D 0 X2 0 X2 2 2 X1 0 X1 0 C1 D1 = (A1 , A2 ) − (B1 , B2 ) + − 0 X2 0 X2 C2 D2 2 2 2 = A1 X1 − B1 + A2 X2 − B2 + X1 C1 − D1 + X2 C2 − D2 2 . Hence, there exists X ∈ M S(k,p) such that f (X) = minX∈MS(k,p) is equivalent to the fact that there exist X1 ∈ R(k+p)×(k+p) and X2 ∈ Rk×k satisfying 2 2 A1 X1 − B1 + X1 C1 − D1 = min , X1 ∈R(k+p)×(k+p) 2 2 A2 X2 − B2 + X2 C2 − D2 = min . X2 ∈Rk×k According to Lemma 2.5, the general solution of X1 can be expressed by (18), and the general solution of X2 can be expressed by (19). Substituting (18) and (19) into (20) gives (17). 3 The solution of Problem II Denote SE the solution set of Problem I. From (16), it is easy to see that the SE is a nonempty closed convex set. We can claim that for any given X ∗ ∈ R(2k+p)×(2k+p) , there exists a unique optimal approximation for Problem II. F. Li, X. Hu and L. Zhang 223 Theorem 3.1. Let X ∗ ∈ R(2k+p)×(2k+p) , A, B, C and D are given by (11). Problem II has a ˜ ˜ unique solution X ∈ SE . Moreover X can be expressed as ˜ X1 0 ˜ X=K T (21) 0 ˜ K , X2 where ˜ Φ ∗ (Σ1 U11 B1 P11 + V11 D1 Q11 τ1 ) Σ−1 U11 B1 P12 T T 1 T T X1 = V1 T −1 T ∗ P1 , (22) V12 D1 Q11 τ1 V12 X11 P12 ˜ Φ′ ∗ (Σ2 U21 B2 P21 + V21 D2 Q21 τ2 ) Σ−1 U21 B2 P22 T T 2 T T X2 = V2 T −1 T ∗ P2 , (23) V22 D2 Q21 τ2 V22 X22 P22 Ik √ 0 X ∗ + W(k,p) X ∗ W(k,p) ∗ 1 Ik √ 0 Jk ∗ ∗ X11 = X1 0 2Ip , X1 = , 2 0 2Ip 0 2 Jk 0 −Jk ∗ 1 ∗ X22 = (−Jk , 0, Ik )X1 0 , (24) 2 Ik where Φ and Φ′ are given by Theorem 2.1. Proof For any X ∗ ∈ R(2k+p)×(2k+p) , from Lemma 2.3, there exist only X1 ∈ M S(k,p) , only ∗ 1 ∗ ∗ ∗ X2 ∈ M SS(k,p) such that X = X1 + X2 , (X1 , X2 ) = 0, where X1 = 2 (X + W(k,p) X ∗ W(k,p) ). ∗ ∗ ∗ ∗ ∗ Therefore, for any X ∈ SE , from Theorem 2.1, we have ||X ∗ − X||2 = ||X1 + X2 − X||2 = ||X1 − X||2 + ||X2 ||2 ∗ ∗ ∗ ∗ 2 X1 = ∗ X1 − K KT + ||X2 ||2 ∗ 0 X2 2 X1 0 = K T X1 K − ∗ + ||X2 ||2 . ∗ 0 X2 ∗ ∗ X11 X12 Denote K T X1 K = ∗ ∗ ∗ , where X21 X22 1 Ik √0 1 Ik √ 0 −Jk ∗ X11 = Ik √0 Jk ∗ X1 0 ∗ 2Ip , X12 = Jk ∗ X1 0 , 2 0 2Ip 0 2 0 2Ip 0 Jk 0 Ik ∗ 1 ∗ Ik √0 ∗ 1 ∗ −Jk X21 = (−Jk , 0, Ik )X1 0 2Ip , X22 = (−Jk , 0, Ik )X1 0 . 2 2 Jk 0 Ik Then, we have ∗ ∗ 2 X11 − X1 X12 ||X ∗ − X||2 = ∗ ∗ ∗ + X2 2 X21 X22 − X2 = ||X11 − X1 ||2 + ||X22 − X2 ||2 + ||X12 ||2 + ||X21 ||2 + ||X2 ||2 . ∗ ∗ ∗ ∗ ∗ 224 Least-Squares Mirrorsymmetric Solution for Matrix Equations From this, it is obvious that ∗ X11 − X1 = min, X ∗ − X = min is equivalent to ∗ (25) X∈SE X22 − X2 = min . From (18),we have ∗ 2 X11 − X1 2 ∗ T T Φ ∗ (Σ1 U11 B1 P11 + V11 D1 Q11 τ1 ) Σ−1 U11 B1 P12 1 T T = X11 − V1 T −1 P1 V11 D1 Q12 τ1 X22 2 Φ ∗ (Σ1 U11 B1 P11 + V11 D1 Q11 τ1 ) Σ−1 U11 B1 P12 T T T = V1T X11 P1 ∗ − T −1 1 V11 D1 Q12 τ1 X22 2 V11 X11 P11 − Φ ∗ (Σ1 U11 B1 P11 + V11 D1 Q11 τ1 ) V11 X11 P12 − Σ−1 U11 B1 P12 T ∗ T T T ∗ 1 T = T ∗ T −1 T ∗ V12 X11 P11 − V11 D1 Q12 τ1 V12 X11 P12 − X22 2 2 = T ∗ T T V11 X11 P11 − Φ ∗ (Σ1 U11 B1 P11 + V11 D1 Q11 τ1 ) + V11 X11 P12 − Σ−1 U11 B1 P12 T ∗ 1 T 2 T ∗ T −1 + V12 X11 P11 − V11 D1 Q12 τ1 + V12 X11 P12 − X22 2 . T ∗ ∗ T ∗ Hence, X11 − X1 = min is equivalent to V12 X11 P12 − X22 = min∀X22 ∈R(k+p−r1 )×(k+p−s1 ) . Then, we have T ∗ X22 = V12 X11 P12 . (26) ∗ Similarly, from X22 − X2 = min we can also obtain ′ T ∗ X22 = V22 X11 P22 . (27) Substituting (26) and (27) into (17) yields the conclusion. Algorithm 1. Input A, B, C, D, X ∗ . 2. Compute A1 , A2 , B1 , B2 , C1 , C2 , D1 , D2 according to (11). ∗ ∗ ∗ 3. Compute X1 , X11 , X22 according to (24). ˜ ˜ 4. Compute X1 , X2 according to (22), (23), respectively. ˜ 5. According to (21) calculate X. Example. Let k = 2, p = 3, n = 2k + p = 7, h = 5, l = 6. Giving 12.1 8.3 −5.6 6.3 9.4 10.7 7.6 11.8 2.9 8.6 6.9 9.5 −7.8 3.5 A= 10.6 2.7 11.5 7.8 6.7 8.9 4.8 , 3.6 7.8 4.9 11.9 9.3 5.9 −2.7 4.5 6.3 7.8 3.1 5.6 11.6 10.8 8.5 9.3 3.6 7.8 6.3 4.7 −4.3 2.5 3.6 7.9 9.4 5.6 7.8 1.9 B = 3.7 6.7 8.6 9.8 3.1 2.9 8.7 , −4.3 6.1 5.7 7.4 5.4 9.5 6.3 2.9 3.9 −5.1 6.3 7.8 4.6 6.4 F. Li, X. Hu and L. Zhang 225 5.4 6.4 3.7 5.6 9.5 7.5 3.5 4.2 7.8 7.8 −6.3 6.1 6.7 3.5 −4.6 2.9 2.7 −3.8 C= −2.7 7.1 10.8 3.7 3.8 8.5 , 1.9 3.9 8.1 5.6 11.2 7.8 8.9 7.8 9.3 7.9 5.6 2.8 9.5 5.1 4.7 8.5 1.7 5.6 7.9 9.5 5.3 2.8 8.6 1.9 8.6 2.6 6.7 8.4 8.1 2.8 5.7 3.9 −2.9 5.1 1.9 3.6 D= 4.8 5.8 1.8 −7.1 5.8 −4.3 . 2.7 7.9 4.5 6.7 9.6 5.8 9.5 4.1 3.4 9.7 3.9 9.7 −6.3 6.7 3.8 7.6 5.7 5.8 2.9 1.9 5.6 1.6 3.8 1.5 3.3 3.5 3.6 1.9 1.4 5.4 2.5 4.1 1.8 −2.7 4.7 4.6 1.7 3.6 5.2 X∗ = 4.3 1.7 5.1 5.3 4.6 −2.3 1.6 . 1.9 1.4 2.1 2.9 3.8 1.7 4.5 −1.8 5.3 3.7 3.5 2.5 5.4 6.7 3.6 2.8 1.7 4.6 1.8 2.9 1.8 ˜ Using MATLAB, we can obtain the unique solution X of Problem II given by 0.1929 −0.1608 −0.0367 0.3664 −0.0732 −0.2489 0.0367 0.3204 −0.1139 −0.1980 −0.2026 0.3445 0.2523 0.2614 0.1639 0.1307 −0.4161 −0.0133 0.0820 0.1307 0.1639 ˜ X = −0.5199 −0.6649 0.0779 0.7340 0.0054 −0.6649 −0.5199 . 0.2299 −0.0157 0.1387 −0.1238 0.5227 −0.0157 0.2299 0.2614 0.2523 −0.1980 −0.2026 0.3445 −0.1139 0.3204 0.0367 −0.2489 −0.0367 0.3664 0.0732 −0.1608 0.1929 Acknowledgments This research was supported by National Natural Science Foundation of China (1057,1047). References [1] Xie D X, Hu X Y, Zhang Z. The solvability conditions for the inverse eigenproblems of symmetric and anti-persymmetric matrices and its approximation. Numer. Linear Algebra Appl., 2003, 10: 223-234. [2] Xie D X, Sheng Y. Inverse eigenproblem of anti-symmetric and persymmetric matrices and its approximation. Inverse Problems, 2003, 19: 217-225. [3] Trench W F. Hermitian, hermitian R-symmetric, and hermitian R-skew symmetric Procrustes prob- lems. Linear Algebra Appl., 2004, 387: 83-98. [4] Mitra S K. The matrix equations AX=C, XB=D. Linear Algebra Appl., 1984, 59: 171-181. [5] Chu. K-W E. Singular value and generalized singular value decomposition and the solution of linear matrix equations. Linear Algebra Appl., 1987, 88/89: 83-98. 226 Least-Squares Mirrorsymmetric Solution for Matrix Equations [6] Mitra S K. A pair of simultaneous linear matrix equations A1 XB1 =C1 A2 XB2 =C2 and a matrix programming problem. Linear Algebra Appl., 1990, 131: 107-123. [7] Li G L, Feng Z H. Mirrorsymmetric matrices, their basic properties, and an application on odd/even- mode decomposition of symmetric multiconductor transmission line. SIAM J. Matrix Anal. Appl., 2002, 24(1): 78-90. [8] Weaver J R. Centrosymmetric (cross-symmetric) matrices, their properties, eigenvalues, and eigen- vectors. Amer. Math. Monthly, 1985, 92: 711-717. [9] Zhang L, Xie D X. A class of inverse eigenvalue problems. Math. Sci. Acta., 1993, 13(1): 94-99.

DOCUMENT INFO

Shared By:

Categories:

Tags:
numerical mathematics, matrix equations, lei zhang, yi wu, l. zhang, ting wu, hong li, xian xu, qing tan, chinese universities, optimal approximation, english series, order system, global approximation, partition of unity

Stats:

views: | 7 |

posted: | 6/16/2010 |

language: | English |

pages: | 10 |

OTHER DOCS BY pkj12584

How are you planning on using Docstoc?
BUSINESS
PERSONAL

By registering with docstoc.com you agree to our
privacy policy and
terms of service, and to receive content and offer notifications.

Docstoc is the premier online destination to start and grow small businesses. It hosts the best quality and widest selection of professional documents (over 20 million) and resources including expert videos, articles and productivity tools to make every small business better.

Search or Browse for any specific document or resource you need for your business. Or explore our curated resources for Starting a Business, Growing a Business or for Professional Development.

Feel free to Contact Us with any questions you might have.