THE INVERSE PROBLEM OF CENTROSYMMETRIC MATRICES WITH A SUBMATRIX CONSTRAINT 1) 1. Introduction

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1 Journal of Computational Mathematics, Vol22, No4, 2004, THE INVERSE PROBLEM OF CENTROSYMMETRIC MATRICES WITH A SUBMATRIX CONSTRAINT 1 Zhen-yun Peng Department of Mathematics, Hunan University of Science and Technology, Xiangtan , China Xi-yan Hu Lei Zhang College of Mathematics and Econometrics, Hunan University, Changsha , China Abstract By using Moore-Penrose generalized inverse and the general singular value decomposition of matrices, this paper establishes the necessary and sufficient conditions for the existence of and the expressions for the centrosymmetric solutions with a submatrix constraint of matrix inverse problem AX = B In addition, in the solution set of corresponding problem, the expression of the optimal approximation solution to a given matrix is derived Mathematics subject classification: 65F15, 65H15 Key words: Matrix norm, Centrosymmetric matrix, Inverse problem, Optimal approximation 1 Introduction Inverse eigenvalue problem has widely used in control theory [1, 2], vibration theory [3, 4], structural design [5], molecular spectroscopy [6] In recent years, many authors have been devoted to the study of this kind of problem and a serial of good results have been made [7, 8, 9] Centrosymmetric matrices have practical application in information theory, linear system theory and numerical analysis theory However, inverse problems of centrosymmetric matrices, specifically centrosymmetric matrices with a submartix constraint, have not be concerned with In this paper, we will discuss this problem Denote the set of all n-by-m real matrices by R n m and the set of all n-by-n orthogonal matrices in R n n by OR n n Denote the column space, the null space, the Moore-Penrose generalized inverse and the Frobenius norm of a matrix A by RA,NA,A + and A, respectively I n denotes the n n unit matrix and S n denotes the n n reverse unit matrix We define the inner product in space R n m by A, B = traceb H A, A, B R n m Then R n m is a Hilbert inner product space The norm of a matrix generated by this inner product space is the Frobenius norm For A =a ij,b =b ij R n m, we using the notation A B =a ij b ij R n n denotes the Hadmard product of matrices A and B Definition 1 [10,15] A =a ij R n n is termed a Centrosymmetric matrix if a ij = a n+1 j,n+1 i i, j =1, 2,,n Received January 27, Research supported by National Natural Science Foundation of China , and by Hunan Province Educational Foundation 02C025

2 536 ZY PENG, XY HU AND L ZHANG The set of n n centrosymmetric matrices denoted by CSR n n In this paper, we consider the following two problems: Problem 1 Given X, B R n m,a 0 R r r, find A CSR n n such that AX = B, A 0 = A[1,r], where A[1,r] is a leading r r principal submatrix of matrix A Problem 2 Given A R n n, find  S E such that where S E is the solution set of Problem 1 A  = min A A, A S E In Section 2, we first discuss the structure of the set CSR n n, and then present the solvability conditions and provide the general solution formula for Problem 1 In Section 3, we first show the existence and uniqueness of the solution for Problem 2, and then derive an expression of the solution when the solution set S E is nonempty Finally, in section 4, we first give an algorithm to compute the solution to Problem 2, and then give a numerical example to illustrate the results obtained in this paper are correction 2 Solving Problem 1 We first characterize the set of all centrosymmetric matrices For all positive integers k, let D 2k = 1 Ik I k 2 S k S k Clearly, D n is orthogonal for all n Lemma 1 [10] A CSR n n if and only if A = S n AS n,d 2k+1 = 1 I k 0 I k S k 0 S k Lemma 2 A CSR n n if and only if A can be expressed as where A 1 R n k n k,a 2 R k k A = D n 1 D T n, 2 Proof We only prove the case for n =2k, thecaseforn =2k + 1 can be discussed similarly Partition the matrix A into the following form A11 A A = 12 A 21 A 22,A 11,A 22 R k k If A CSR 2k 2k,thenwehavefromLemma1that 0 Sk S k 0 A11 A 12 A 21 A 22 0 Sk S k 0 A11 A 12 = A 21 A 22,

3 The Inverse Problem of Centrosymmetric Matrices with a Submatrix Constraint 537 which is equivalent to Hence A 22 = S k A 11 S k, A 12 = S k A 21 S k Ik S k A11 S k A 21 S k Ik I k Let D T n AD n = 1 2 I k S k A 21 S k A 11 S k A11 + S k A 21 0 = 0 A 11 S k A 21 A 1 = A 11 + S k A 21,A 2 = A 11 S k A 21, and note that D n is a orthogonal matrix, we have 2 Conversely, for every A 1,A 2 R k k,wehave 0 Sk S k 0 D n 0 Dn T Sk S k 0 S k S k = D n D T n It follows from Lemma 1 that A = D n Dn T 0 A CSR2k 2k 2 Next, we give some lemmas According to [7], the following lemma is easy to verify Lemma 3 Given Z R m s,y R n s, then the matrix equation AZ = Y has a solution A R n m if and only if YZ + Z = Y In that case the general solution can be expressed as A = YZ + + GP T,whereP R n n t is an unit column-orthogonal matrix, RP =NZ T and t = rankz Let Γ={A CSR n n AX = B,X,B R n m } 3 Partitioning D T n X and D T n B into to the following form D T n X = X1 X 2,D T n B = B1 B 2 where X 1,B 1 R n k m,x 2,B 2 R k m We have the following lemma, 4 Lemma 4 Γ is nonempty if and only if B 1 X + 1 X 1 = B 1 and B 2 X + 2 X 2 = B 2 Furthermore, any matrix A Γ can be expressed as A = D n B1 X G 1U T B 2 X G 2P T 2 D T n, 5 where G 1 R n k n k r1,g 2 R k k r2 are arbitrary matrices, r 1 = rankx 1,r 2 = rankx 2, RU 2 =NX T 1, RP 2 =NX T 2 Proof If Γ is nonempty, we have from Lemma 2 and D n being an orthogonal matrix that A = D n D T n, 6

4 538 ZY PENG, XY HU AND L ZHANG with A 1 R n k n k,a 2 R k k satisfy It follows from Lemma 3 that and A 1 X 1 = B 1, A 2 X 2 = B 2 7 B 1 X + 1 X 1 = B 1,B 2 X + 2 X 2 = B 2 8 A 1 = B 1 X G 1U T 2,A 2 = B 2 X G 2P T 2, 9 where G 1 R n k n k r1, G 2 R k k r2 are arbitrary matrix, r 1 =rankx 1,r 2 =rankx 2, RU 2 =NX1 T, RP 2=NX2 T Substituting 9 into 6, we have 5 Conversely, if B 1 X 1 + X 1 = B 1 and B 2 X 2 + X 2 = B 2, then we have from Lemma 3 that there exist A 1 R n k n k and A 2 R k k such that which is equivalent to It in turn is equivalent to A 1 X 1 = B 1, A 2 X 2 = B 2, X1 X 2 = B1 B 2 D T n X = B D n Dn T Γ And this illustrate Γ is nonempty So the matrix A = D n Now, we investigate the consistency of Problem 1 with a centrosymmetric condition on the solution Obviously, solving Problem 1 is equivalent to find A Γ such that A 0 is a leading principal submatrix of A, ie, find A Γ such that I r, 0AI r, 0 T = A 0, 10 where I r, 0 R r n Note that any matrix A Γ can be expressed as 5, we partition the matrix I r, 0D n into the following form Let I r, 0D n =W 1,W 2,W 1 R r n k,w 2 R r k 11 Ẽ = A 0 W 1 B 1 X + 1 W T 1 W 2 B 2 X + 2 W T 2 12 Then equation 10 is equivalent to find G 1 and G 2 as in 5 such that W 1 G 1 N 1 + W 2 G 2 N 2 = Ẽ, 13 where N 1 = U T 2 W T 1,N 2 = P T 2 W T 2 Decomposing the matrix pairs W T 1,W T 2 andn 1,N 2 using the General Singular-Value Decomposition GSV D see[12, 13, 14]: W T 1 = ŨΣ 1M 1, W T 2 = Ṽ Σ 2M 1, N 1 = P Σ 3 M 2, N 2 = QΣ 4 M 2, 14

5 The Inverse Problem of Centrosymmetric Matrices with a Submatrix Constraint 539 where Ũ,Ṽ, P and Q are orthogonal matrices, and M 1 and M 2 are nonsingular matrices of order r, Σ 1 R n k r,σ 2 R k r,σ 3 R n k r1 r,σ 4 R k r2 r,with Σ 1 = I 1 S Σ 2 = 0 2 S 2 0 I 2 15 r 3 r 4 p 1 r 3 r 4 r p 1 r 3 r 4 p 1 r 3 r 4 r p 1 Σ 3 = I 3 S Σ 4 = 0 4 S 4 0 I 4 16 r 5 r 6 p 2 r 5 r 6 r p 2 r 5 r 6 p 2 r 5 r 6 r p 2 here I 1,I 2,I 3 and I 4 are identity matrices, 0 1, 0 2, 0 3 and 0 4 are zero matrices, p 1 =rankc 1 =rankw 1,W 2, p 2 =rankc 2 =rankn 1,N 2, and S 1 = diagα 1,,α r4,s 2 = diagβ 1,,β r4, 17 S 3 = diagσ 1,,σ r6,s 4 = diagδ 1,,δ r6, 18 with 1 >α 1 α r4 > 0, 0 <β 1 β r4 < 1, 1 >σ 1 σ r6 > 0, 0 <δ 1 δ r6 < 1, and α 2 i + β2 i =1, i =1,,r 4, σi 2 + δ2 i =1, i =1,,r 6 Some submatrices in equations 15 and 16 may disappear, depending on the structure of the matrices W 1,W 2,N 1 and N 2 Define Y = Ũ T G 1 P,Z = Ṽ T G 2 Q and E = M T 1 ẼM 1 2 Equation 13 now reads Σ T 1 Y Σ 3 +Σ T 2 ZΣ 4 = E 19 Note that transforming equation 13 to 19 does not change the equation s consistency Partitioning the matrices Y,Z and E according to the Σ s, equation 19 is equivalent to Y 11 Y 12 S S 1 Y 21 S 1 Y 22 S 3 + S 2 Z 22 S 4 S 2 Ỹ Z 32 S 4 Z = From the above discussion, we can prove the following theorem: Theorem 1 Problem 1 is solvable if and only if a B i X + i X i = B i i =1, 2 E 11 E 12 E 13 E 14 E 21 E 22 E 23 E 24 E 31 E 32 E 33 E 34 E 41 E 42 E 43 E 44 b E 13 =0,E 31 =0,E 14 =0,E 24 =0,E 34 =0,E 41 =0,E 42 =0,E 43 =0,E 44 =0 When the conditions a and b are satisfied, the general solution can be expressed as 20 A = D n B1 X G 1U T B 2 X G 2P T 2 D T n, 21

6 540 ZY PENG, XY HU AND L ZHANG where G 1 = Ũ E 11 E 12 S3 1 Y 13 S S S 2 Z 22 S 4 S3 1 Y 23 Y 31 Y 32 Y 33 Z 11 Z 12 Z 13 G 2 = Ṽ Z 21 Z 22 S2 1 E 23 Z 31 E 32 S4 1 E 33 with Y 13,Y 23,Y 31,Y 32,Y 33,Z 11,Z 12,Z 13,Z 21,Z 22,Z 31 are arbitrary matrices P T, 22 QT, 23 Proof Problem 1 having a solution A CSR n n is equivalent to there exists A Γsuch that 10 holds Condition a is the necessary and sufficient conditions for Γ is nonempty Condition b is the necessary and sufficient conditions for equation 10 has centrosymmetric solution with the given matrix as its submatrix The general solution can be obtained by the definition of Y and Z and the Equations 5 and 20 Letusfirstintroducealemma 3 Solving Problem 2 Lemma 5 [11] Given E,F R n n, Ω 1 =diaga 1,,a n > 0, Ω 2 =diagb 1,,b n > 0, Φ= 1 R n n, then there exists an unique matrix 1+a Ŝ 2 Rn n such that S E 2 + Ω 1 SΩ 2 i b2 j F 2 = min, andŝ can be expressed as Ŝ =Φ E +Ω 1F Ω 2 Partition the matrix Dn T A D n intoa2 2 block matrix A ij 2 2 with A 11 R n k n k, A 22 Rk k Partition matrices Ũ T A 11 U 2 P and Ṽ T A 22 P 2 Q into 3 3 block matrices Ỹij 3 3 with Ỹ11 R r3 r5, Ỹ22 R r4 r6, Ỹ33 R n k r3 r4 n k r1 r5 r6, and Z ij 3 3 with Z 11 R k+r3 p1 k r2+r3 p2, Z 22 R r4 r6, Z 33 R p1 r3 r4 p2 r5 r6, respectively Then we have the following theorem Theorem 2 Given X, B R n m,a 0 R r r and A R n n X, B, A 0 satisfy conditions of Theorem 1 Then Problem 2 has an unique optimal approximate solution which can be expressed as B 1 X 1 + Â = D + Ĝ1U2 T 0 n 0 B 2 X Ĝ2P2 T Dn T 24 where Ĝ 1 = Ũ E 11 E 12 S3 1 Ỹ 13 S1 1 E 21 S1 1 E 22 S 2 Ẑ 22 S 4 S3 1 Ỹ 23 Ỹ 31 Ỹ 32 Ỹ 33 Ĝ 2 = Ṽ Z 11 Z12 Z13 Z 21 Ẑ 22 S2 1 E 23 Z 31 E 32 S4 1 E 33 P T, 25 QT, 26 Z 22 =Φ [ Z 22 + S 1 1 S 2Ỹ22 S 1 1 E 22S 1 3 S 4S 1 3 ] 27 with Φ=ϕ ij R r4 r6,ϕ ij = the same as 14 α 2 i δ2 j α 2 i δ2 j +β2 i σ2 j, U 2,P 2 are the same as 5, and Ũ,Ṽ, P and Q are

7 The Inverse Problem of Centrosymmetric Matrices with a Submatrix Constraint 541 Proof Because X, B and A 0 satisfy the conditions of the Theorem 1, the solution set S E of Problem 1 is nonempty According to the proof of theorem 74 in [7], it is easy to verify that S E is a closed convex cone Hence, the corresponding Problem 2 has an unique optimal approximate solution Choose U 1 and P 1 such that U =U 1,U 2 andp =P 1,P 2 are orthogonal matrices Attention to U, P, D, Ũ,Ṽ, P and Q are orthogonal matrices, and RU 2 =NX1 T,RP 2=NX2 T, we have from that A A 2 = B 1 X G 1U T 2 A B 2 X G 2P T 2 A A A 21 2 = G 1 A 11U G 2 A 22P B 1 X + 1 A 11U B 2 X 2 + A 22 P A A 21 2 E 11 E 12 S 1 3 Y 13 = S1 1 E 21 S1 1 E 22 S 2 Z 22 S 4 S3 1 Y 23 Ũ T A 11U 2 P Y 31 Y 32 Y 33 2 Z 11 Z 12 Z 13 + Z 21 Z 22 S2 1 E 23 Ṽ T A 22P 2 Q Z 31 E 32 S4 1 E 33 + B 1 X + 1 U 1 A 11 U B 2 X + 2 P 1 A 22 P A A Hence, there exists A S E such that A A = min is equivalent to Y i3 Ỹi3 =min, Y 3j Ỹ3j =min, Z i1 Z i1 =min, i =1, 2, 3,j =1, 2, Z 12 Z 12 =min, Z 22 Z S1 1 E 22 S 2 Z 22 S 4 S3 1 Ỹ22 2 =min 28 It follow from Lemma 5 that Y i3 = Ỹi3,Y 3j = Ỹ3j,Z i1 = Z i1, i =1, 2, 3,j =1, 2,Z 12 = Z and Z 22 =Φ Z 22 + S1 1 S 2Ỹ22 S1 1 E 22S3 1 S 4S Substituting 29 and 30 into 21-23, we obtain The Algorithm Description and Numerical Example According to discuss in section 2 and 3, we now give a method for solving Problem 2 as following steps: step 1 According to 4 calculate X i,b i i =1, 2, furthermore calculate B i X i + X i i =1, 2 step 2 If B i X i + X i = B i i =1, 2, then the set Γ is nonempty and we continue Otherwise we stop step 3 According to 11 calculate W 1,W 2 According to 12 calculate Ẽ step 4 Find unit column orthogonal matrices U 2,P 2 basis of linear equations X1 T u =0 and X2 T v =0 ChosenU 1,P 1 such that U =U 1,U 2,P =P 1,P 2 are orthogonal matrices step 5 According to 14 decomposing the matrix pairs W1 T,WT 2 andn 1,N 2 usingthe GSV D step 6 Partitioning the matrix E according to the right side of 20

8 542 ZY PENG, XY HU AND L ZHANG step 7 If E 13 =0,E 31 =0,E 14 =0,E 24 =0,E 34 =0,E 41 =0,E 42 =0,E 43 =0,E 44 =0, then the solution set S E to Problem 1 is nonempty and we continue Otherwise we stop step 8 Partition the matrices Dn T A D n, Ũ T A 11U 2 P and Ṽ T A 22P 2 Q into block matrices according to Dn T AD n in 21, Ũ T G 1 P in 22 and Ṽ T G 2 Q in 23, respectively step 9 According to calculate  Example 1 Taking A 0 = X = A = we obtain X + 1,X+ 2,B 1 and B 2 are,respectively,,b = , By a direct computing, we know that B i X + i X i = B i i =1, 2 According to step 4, we get two orthogonal matrices U =U 1,U 2,P =P 1,P 2 as follow U 1 = U 2 =

9 The Inverse Problem of Centrosymmetric Matrices with a Submatrix Constraint 543 P 1 = P 2 = After calculating according to step 5 and 6, the condition b of Theorem 1 is satisfied Hence, the conditions of Theorem 1 are satisfied, and the corresponding problem 2 has an unique solution  After calculating by 24-27, we have  as follow: References [1] LR Fletcher, An Inverse Eigenvalue Problem from Control Theory, in Numerical Treatment of Inverse problems for Differential and Integral Equations, eds P Deuflhard and E Hairer, Birkhauser, Beston, 1983, [2] WM Wonham, Linear Multivariable Control: A Geometric Approach, Springer-Verlag, 1979 [3] V Barcilon, Inverse Problem for a Vibrating Beam, J Appl Math Phys, , [4] GML Gladwell, Inverse Problems in Vibration, Martinus Nijhoff, Dordrecht, The Netherlands, Boston, MA, 1986 [5] KT Jeseph, Inverse Eigenvalue Problem in Structural design, AIAA J, , [6] S Friedland, The Reconstruction of a Symmetric Matrix from the spectral data, J Math Anal Applic, , [7] SQ Zhou, HDai, The Algebraic Inverse Eigenvalue Problem in Chinese, Henan Science and Technology Press, Zhengzhou, China, 1991 [8] JG Sun, Some New Result on Algebraic Inverse Eigenvalue Problems, in Proceedings of the First Chinese National Conference on Inverse Eigenvalue Problems, 1986 [9] SF Xu, An Introduction to Inverse Eigenvalue Problems, Peking University Press, Beijing, China, 1998 [10] JL Zhen, XH Zhen, Special Matrices in Chinese, Qinghua University Press, Beijing, China, 2001 [11] AP Liao, A Generalization of a Class of Inverse Eigenvalue Problems, Journal of Hunan University, , 7-10 [12] CC Paige and MA Saunders, Towards a Generalized Singular Value Decomposition, SIAMJ Numer Anal, 18:1 1981, [13] GH Golub and CF Van Loan, Matrix Computations, The Johns Hopkins University Press, American, 1983 [14] GW Stewart, Computing the CS-decomposition of a partitional orthogonal matrix, Numer Math, ,

10 544 ZY PENG, XY HU AND L ZHANG [15] Z Xu, KY Zhang and Q Lu, The fast algorithm for solving Toeplitz matrices in Chinese, Xibei Industry University Press, 1999