The skew-symmetric orthogonal solutions of the matrix equation AX = B

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1 Linear Algebra and its Applications 402 (2005) The skew-symmetric orthogonal solutions of the matrix equation AX = B Chunjun Meng, Xiyan Hu, Lei Zhang College of Mathematics and Econometrics, Hunan University, Changsha , PR China Received 15 January 2004; accepted 18 January 2005 Available online 16 March 2005 Submitted by R. Byers Abstract An n n real matrix X is said to be a skew-symmetric orthogonal matrix if X T = X and X T X = I. Using the special form of the C S decomposition of an orthogonal matrix with skew-symmetric k k leading principal submatrix, this paper establishes the necessary and sufficient conditions for the existence of and the expressions for the skew-symmetric orthogonal solutions of the matrix equation AX = B. In addition, in corresponding solution set of the equation, the explicit expression of the nearest matrix to a given matrix in the Frobenius norm have been provided. Furthermore, the Procrustes problem of skew-symmetric orthogonal matrices is considered and the formula solutions are provided. Finally an algorithm is proposed for solving the first and third problems. Numerical experiments show that it is feasible Elsevier Inc. All rights reserved. AMS classification: 65F15; 65F20 Keywords: Skew-symmetric orthogonal matrix; Leading principal submatrix; C S decomposition; The matrix nearness problem; The least-square solutions The work was supported by the National Natural Science Foundation of China ( ) and by Science Fund of Hunan University ( ). Corresponding author. address: chunjunm@hnu.cn (C. Meng) /$ - see front matter 2005 Elsevier Inc. All rights reserved. doi: /j.laa

2 304 C. Meng et al. / Linear Algebra and its Applications 402 (2005) Introduction We first introduce some notations to be used. Let R m n denote the set of real m n matrices, and OR n n, SR n n, ASR n n denote the sets of orthogonal n n matrices, real n n symmetric matrices, real n n skew-symmetric matrices respectively, I n the identity matrix of order n. A T means the transpose of the real matrix A. We define the inner product in space R m n, (A, B) = trace(b T A) = mi=1 nj=1 a ij b ij A,B R n m.thenr m n is a Hilbert inner product space. The norm of a matrix generated by the inner product is Frobenius norm, denoted by. Skew-symmetric orthogonal matrices play an important role in numerical analysis and numerical solution of partial differential equation. For example, the symplectic transformation can preserve the Hamiltonian structure, ( and an) matrix S is symplectic matrix if and only if SJS T 0 In = J,whereJ =. There the I n 0 matrix J is a skew-symmetric orthogonal matrix. It is pointed out that the order of a skew-symmetric orthogonal matrix is necessarily even number. So we denote the set of skew-symmetric orthogonal matrices by SSOR 2m 2m. As another example for skew-symmetric orthogonal matrices, we can see that if P OR m m, then the 0 P matrix P T SSOR 0 2m 2m. In this paper we consider the skew-symmetric orthogonal solutions of the matrix equation AX = B, (1) where A and B are given matrices in R n 2m. This problem has important geometry meanings. If X SSOR 2m 2m satisfies Eq. (1), then there exists an orthogonal transformation with skew-symmetric structure that can transform the row space of the matrix A onto that of B. And an orthogonal transformation is widely applied in numerical analysis because of its numerical stability, hence the problem we will discuss here is useful both theoretically and practically. We also consider the matrix nearness problem min X X, (2) X S X where X is a given matrix in R 2m 2m and S X is the solution set of Eq. (1). In Section 3 we will prove that problem (1) is solvable whenever the given matrices A and B satisfy two equations. However, the data matrices A and B are often perturbed by lots of means, such as observing error, model error, rounding error etc. In many cases the changed A and B do not meet the solvability conditions, which makes problem (1) have no solution. Hence we consider the least-square solutions of problem (1). That means to find a X SSOR 2m 2m to minimize the distance between AX and B,i.e. min AX B. (3) X SSOR2m 2m

3 C. Meng et al. / Linear Algebra and its Applications 402 (2005) The well-known equation (1) with the unknown matrix X being symmetric, skewsymmetric, symmetric positive semidefinite, bisymmetric and orthogonal were studied (see, for instance, [1 8]). All in this paper, the necessary and sufficient conditions for the existence of and the expressions for the solution of the equation were given by using the structure properties of matrices in required subset and the singular value decomposition of the matrix. But the skew-symmetric orthogonal solutions of Eq. (1) have not been considered yet, which motivates us to study the problem. Another motivation to solve the skew-symmetric orthogonal solutions of Eq. (1) is that it will be probably helpful for checking if the matrix A is generalized Hamiltonian matrix. The matrix A is called to be generalized Hamiltonian iff there exists a skew-symmetric orthogonal matrix Q such that AQ is symmetric. If the skew-symmetric orthogonal matrix X solves Eq. (1) whereb is symmetric matrix then we can claim that the matrix A is generalized Hamiltonian. To see the applications of generalized Hamiltonian matrix, we refer the readers to the paper [9]. The matrix nearness problem (2), that is, finding the nearest matrix in the solution setofeq.(1) to a given matrix in Frobenius norm, was initially proposed in the process of test or recovery of linear systems due to incomplete dates or revising given dates. A preliminary estimate X of the unknown matrix X can be obtained by the experimental observation values and the information of statistical distribution. The other form of the matrix nearness problem (2), min A A A subject to AX = B, where X and B are given matrices, A is given matrix and A is a symmetric, bisymmetric or symmetric positive semidefinite matrix were discussed (see [10,11] and references therein). Problem (3) is a special case of the Procrustes problem. Schonemann studied the orthogonal Procrustes problem in the paper [12] and obtained a explicit solution. The symmetric Procrustes problem was researched by Higham in the paper [13], where the general solutions of the problem and the algorithm to compute it are presented. Problem (3) we will discuss here may be called the skew-symmetric orthogonal Procrustes problem and be the extension of the research of Procrustes problem. The paper is organized as follows: in Section 2, the special form of C S decomposition of an orthogonal matrix with a skew-symmetric k k leading principal submatrix will be introduced. In Section 3, the necessary and sufficient conditions for the existence of and the expressions for the skew-symmetric orthogonal solutions of Eq. (1) will be derived. Section 4 will provide the expressions of the solution of the matrix nearness problem (2). In Section 5, the formula of solutions of problem (3)

4 306 C. Meng et al. / Linear Algebra and its Applications 402 (2005) is presented and in Section 6 some examples are given to illustrate the main results obtained in the paper. 2. The C S decomposition of an orthogonal matrix with a skew-symmetric k k leading principal submatrix Let X OR 2m 2m, partition X as X11 X X = 12, (4) X 21 X 22 where X11 T = X 11,i.e.X 11 ASR k k. Using the skew-symmetric structure of k k leading principal submatrix of the orthogonal matrix X, we prove that the C S decomposition of X has special form described as the following theorem. Theorem 1. If X OR 2m 2m, and partitioned as (4), X 11 ASR k k, then the C S decomposition of X can be expressed as T D1 0 X11 X 12 D1 0 Σ11 Σ = 12, (5) 0 D 2 X 21 X 22 0 R 2 Σ 21 Σ 22 here, D 1 OR k k,d 2,R 2 OR (2m k) (2m k), I 0 Σ 11 = 0 C 0, Σ 12 = 0 S 0, 0 I 0 I 0 0 Σ 21 = 0 S 0, Σ 22 = 0 C T 0, I 0 where I = diag( I 1,..., I r ), I 1 = = I 0 1 r =, C = diag(c 1 0 1,...,C l ), 0 ci si 0 C i =,i= 1,...,l. S = diag(s c i 0 1,...,S l ), S i =,S 0 s i TS i+ Ci TC i = i I 2,i= 1,...,l. 2r + 2l = rank(x 11 ), Σ 21 = Σ T 12,k 2r = rank(x 21). Proof. Suppose X OR 2m 2m, and partitioned as (4), X 11 ASR k k.asx 11 is skew-symmetric, and the k k leading principle submatrix of the orthogonal matrix X, thenifλ is one of eigenvalue of X 11,wemusthavethatλ is a pure imaginary number and the absolute value of λ, denoted by λ, is not larger than 1, i.e. λ 1. Assume the spectral decomposition of the real skew-symmetric matrix X 11 is written as I D1 T X 11D 1 = Σ 11 = 0 C 0, (6) 0

5 C. Meng et al. / Linear Algebra and its Applications 402 (2005) where D 1 OR k k, I = diag( I 1,..., I r ), I 1 = = I 0 1 r =. C = 1 0 diag(c 1,...,C l ), C i = rank(x 11 ). ( 0 ci c i 0 ), 0 < c i < 1, i = 1,...,l and 2(r + l) = T D1 0 X11 Consider the matrix D 0 I X 1, it remains column orthogonal. Let 21 C X 21 D 1 = (W 1,W 2 ), W 1 R (2m k) 2r, W 2 R (2m k) (k 2r) 0, Ĉ = R (k 2r) (k 2r),then Therefore, T D1 0 X11 D 0 I X 1 = 21 D T 1 X 11 D 1 = X 21 D 1 I 0 0 Ĉ W 1 W 2. (7) W 1 = 0, W2 T W 2 = I k 2r Ĉ T Ĉ. (8) Write s i = 1 ci 2, i = r + 1,...,r + l. s 2r+2l+1 = =s k = 1, and let then ( 1 Z 2 = W 2 diag, s r+1 Z T 2 Z 2 = I. 1 s r+1,..., 1 s r+l, 1 s r+l,..., 1 s k ), (9) And we see that Z 2 R (2m k) (k 2r) has orthonormal columns, hence the matrix Z 2 can be expanded into an orthogonal matrix D 2, i.e. D 2 = (Z 1,Z 2 ) OR (2m k) (2m k). From (8) and(9), we obtain Z T 1 W 2 = Z T 1 Z 2 diag(s r+1,...,s k ) = 0 (10) and ( 1 Z2 T W 2 = diag,..., 1 ) W T S 0 2 s r+1 s W 2 =. (11) k 0 I here, S = diag(s r+1,s r+1,...,s r+l,s r+l ), s i = 1 ci 2, i = r + 1,...,r + l. Therefore, we get T Z T D 2 X21 D 1 = 1 (0 ) 0 W2 = Z2 T 0 S 0. (12) I

6 308 C. Meng et al. / Linear Algebra and its Applications 402 (2005) Applying the same method, we can also find an orthogonal matrix R 2 OR (2m k) (2m k) such that 0 D1 T X 12R 2 = 0 S 0. (13) I Now we consider the matrix D T 1 0 X11 X 12 D1 0 0 D 2 T, X 21 X 22 0 R 2 it is still orthogonal matrix, and we have, from (6), (12)and(13), that I 0 0 C S 0 D1 0 X11 X 12 D1 0 = 0 I 0 D 2 X 21 X 22 0 R 2 0 X 44 X 45 X S 0 X 54 X 55 X 56 I X 64 X 65 X 66 (14) It can deduce that X 64 = X 65 = X 66 = X 56 = X 46 = 0. And the CS T + SX55 T = 0 implies that X 55 = (S 1 CS T ) T = C T, X 45 = X 54 = 0, X 44 OR 2r 2r. (15) Let D 2 = D 2 diag(x 44,I), the expression (5) can be obtained. Thus the proof is completed. 3. The solvable conditions for problem (1) At first, we assume that there exists a matrix X SSOR 2m 2m such that Eq. (1) holds, then we get and BB T = AXX T A T = AA T (16) AB T = AX T A T = AXA T = BA T. (17) Therefore Eqs. (16) and(17) are necessary for the solvability of problem (1), we will prove in the following paper that they is also sufficient for the solvability.

7 C. Meng et al. / Linear Algebra and its Applications 402 (2005) Theorem 2. Given A, B R n 2m, there is a X SSOR 2m 2m such that Eq. (1) holds if and only if (a) BB T = AA T ; (b) AB T = BA T. When these conditions are satisfied, the solutions of problem (1) can be expressed as I 0 X = Ṽ Q T, (18) 0 G where Ṽ, Q T OR 2m 2m have relation to given matrices A and B. G SSOR 2r 2r is arbitrary. In order to prove the sufficiency and obtain the expression of solutions, we introduce some important lemmas. Lemma 1. Given A, B R n 2m, if A, B satisfy the condition (a) of theorem 2, then the singular value decomposition of A, B can be described as Σ 0 A = U V T Σ 0, B = U Q T, (19) where U = U 1 U 2 OR n n,v=(v 1 V 2 ), Q = (Q 1 Q 2 ) OR 2m 2m,U 1 R n k,p 1,Q 1 R 2m k,k= rank(a), and Σ R k k > 0 is diagonal, its diagonal elements are nonzero singular value of A or B. Proof. Suppose the singular value decomposition of A be written as the fore part of (19), from condition (a), we have BB T Σ 2 0 = U U T. It implies that B T U 2 = 0, Σ 1 U1 T BBT U 1 Σ 1 = I k. (20) Hence we may find Q 2 such that Q = (B T U 1 Σ 1 Q 2 ) OR 2m 2m.Using(20), we obtain that U T Σ 0 BQ =, that ends our proof. Lemma 2. Given A, B R n 2m, there exists an orthogonal matrix X such that Eq. (1) holds if and only if (a) BB T = AA T.

8 310 C. Meng et al. / Linear Algebra and its Applications 402 (2005) Let the singular value decomposition of A, B be (19), then the orthogonal solutions of Eq. (1) can be described as Ik 0 X = V Q T, (21) 0 P where P OR (2m k) (2m k) is arbitrary. Proof. Substitute (19) into Eq. (1), we get Σ 0 V T Σ 0 XQ =. (22) Write V T XQ = X = Using (22) and(23), it can be deduced that Ik 0 X =. 0 P Therefore we have proved the lemma. X 11 X 12, X 11 R X 21 X k k. (23) 22 Proof of Theorem 2. Since the given matrices A and B satisfy the condition (a), we can assume, from the Lemma 1, that the singular value decomposition of A, B be (19). Morever A and B satisfy the condition (b), then Eq. (1) has orthogonal solutions with the form (21). Substitute (19) into the condition (b), we get V1 T Q 1 = Q T 1 V 1. (24) Note that X T = X is equivalent to V T I 0 I 0 Q = 0 P 0 P T Q T V. Comparing the corresponding blocks, with V = ( V 1 ) ( V 2, Q = Q1 ) Q 2,we have Q T 1 V 1 = V1 T Q 1, (25) V T 1 Q 2P = Q T 1 V 2, (26) V2 T Q 2P = P T Q T 2 V 2. (27) And (25) already stands from (24). Then we intend to find the common orthogonal solutions of (26) and(27). Firstly, consider (26), we observe that ( Q T 1 V 2)( Q T 1 V 2) T = V1 T Q 2(V1 T Q 2) T. Then we know, from the Lemma, that Eq. (26) has orthogonal solutions P.

9 C. Meng et al. / Linear Algebra and its Applications 402 (2005) Consider the orthogonal matrix V T Q,thep p leading principal submatrix of which is skew-symmetric from (24). Applying Theorem 1, weget Q T 1 V 2 = D 1 Σ 12 D T 2, VT 1 Q 2 = D 1 Σ 12 R T 2. (28) And the orthogonal solutions P of (26) can be expressed as G P = R 2 0 I 0 D2 T, (29) I where, G OR 2r 2r is arbitrary. Since V T 2 Q 2 = D 2 Σ 22 R T 2. (30) Using (27), (29) and(30), we obtain that G T = G, (31) that means G SSOR 2r 2r,let ˆQ = Q diag(i, R 2 ), ˆV = U diag(i, D 2 ),weget the skew-symmetric orthogonal solutions of Eq. (1) can be described as I 0 0 X = ˆV 0 G 0 ˆQ. I I 0 0 Let E be the permutation matrix and E = I, partitioned with accordance 0 I 0 I to the 0 G 0, write Ṽ = ˆVE, Q = ˆQE. Therefore, the solutions of I problem (1)be(18), so the proof is finished. 4. The solutions of problem (2) We first consider the following matrix nearness problem. Given C R 2m 2m, seek a matrix P SSOR 2m 2m such that C P =min. (32) For 2 is a continuous function, and the subset SSOR 2m 2m is closed and bounded in R 2m 2m, the problem (32) must have solutions. But the set SSOR 2m 2m is not convex set, then problem (32) has possibly not only one solution.

10 312 C. Meng et al. / Linear Algebra and its Applications 402 (2005) Next we will derive the expression of solutions for problem (32). From the properties of Frobenius norm, one know that C P 2 = (C + C T )/2 2 + (C C T )/2 P 2. Hence C P =min is equivalent to (C C T )/2 P 2 = min. Let the spectral decomposition of the skew-symmetric matrix Ĉ = (C C T )/2 be Λ 0 Ĉ = D D T, (33) 0 λi here Λ = diag(λ 1,...,Λ k ), Λ i =,λ λ i 0 i > 0, i = 1,...,k,2k = rank(ĉ), D T D = DD T = I 2m. With the help of (33) and the definition of Frobenius norm, we have ( ) (C C T )/2 P 2 = 2m + Ĉ 2 Λ 0 2 trace D T PD. As D T PDis still skew-symmetric orthogonal matrix, we obtain that the solutions of problem (32) can be expressed as I 0 E 0 (Ĉ) = D D T, (34) 0 H where I = diag( I 1,..., I k ), is arbitrary. I 1 = = I k = 0 1, H SSOR 1 0 (2m 2k) (2m 2k) Theorem 3. Suppose the spectral decomposition of the skew-symmetric matrix Ĉ = (C C T )/2 be (33), then the solution set of problem (32) can be written as (34). Because the solution set of problem (1) with expression (18) is closed and bounded but not convex, the solution of (2) exists and is not unique if the conditions of Theorem 1 are satisfied. Given X R 2m 2m,using(18), we get X X 2 = I 0 Ṽ T X 2 Q 0 G. Write Ṽ = Ṽ 1 Ṽ 2 ), Q = Q 1 Q 2, partitioned according to the matrix I 0, then we observe that X X 0 G =min is equivalent to min G SSOR 2r 2r G Ṽ T 2 X Q 2 2.

11 C. Meng et al. / Linear Algebra and its Applications 402 (2005) Let C = Ṽ T 2 X Q 2 and Ĉ = (C C T )/2, from Theorem 3, the solution set of can be expressed as (34). Theorem 4. Given X R 2m 2m, and A, B R n 2m, satisfying the conditions of Theorem 1, the solution set of problem (1) be (18), write Ṽ = Ṽ 1 Ṽ 2, Q = I 0 Q 1 Q 2, partitioned according to the matrix, and let C = Ṽ 0 G 2 TX Q 2, Ĉ = (C C T )/2, then if and only if I 0 X = Ṽ Q T, G E 0 G 0 (Ĉ), the minimum of (2) can be reached. 5. The solutions of problem (3) Consider (3), from the definition of Frobenius norm and X SSOR 2m 2m,itis obvious to see that AX B 2 = A 2 + B 2 2 trace(b T AX). (35) Furthermore, 2 trace(b T AX) = trace((b T A A T B)X). Denote the skew-symmetric matrix (B T A A T B) by Ĉ. Assume the spectral decomposition of Ĉ is the form with (33), then we know that 2 trace(b T AX) 2 k i=1 λ i and the equality holds if and only if X E 0 (Ĉ). Therefore we obtain the following theorem. Theorem 5. Given A, B R n 2m, denote (A T B B T A) by Ĉ. Suppose the spectral decomposition of Ĉ be (33), then the solution set of problem (3) can be described as (34). 6. Numerical algorithm and several examples The following algorithm can be used to solve problem (1) and problem (3).

12 314 C. Meng et al. / Linear Algebra and its Applications 402 (2005) Algorithm Step 1. Input A, B R n 2m, compute the matrix Ĉ = B T A A T B. Step 2. Compute the spectral decomposition of Ĉ with the form (33). Step 3. Compute the solutions of problem (1) orproblem(3) according to (34). Example 1. Suppose A, B R 6 6 and A = , B = We can easily see that the given A and B do not satisfy the solvable conditions of problem (1). Hence we tryto seek an orthogonalskew-symmetric matrixtominimize AX B F. and Solution: Applying the algorithm, we get the following the result: X = min AX B = , X SSOR6 6 e1 = X T X I 6 = , e2 = X T + X = Remark 1 (1) Problem (3) has a unique solution if and only if the matrix (B T A A T B) is nonsingular matrix. Example 2 is just the case.

13 C. Meng et al. / Linear Algebra and its Applications 402 (2005) (2) When we use the stable algorithms to do the steps 1 3, we find, by computing, that if the perturbations A, B are small, the solution perturbation of X generally remains small. We can also apply the algorithm to solve problem (1). The following example is in the case. We will compare the solutions computed by our algorithm with that computed by MATLAB procedure X = A \ B. Example 2. Let U OR 10 10,V OR 6 6,and U = , V = S1 Assume S 1 = diag(2, 1,e/1,e/2,e/3,e/4), S = with e = 10,e = 1,e = ,e = 10 2,...,e = 10 13, A = USV T. Let X SSOR 6 6, X = and B = AX. Clearly, these given matrices A and B are consistent with a skewsymmetric orthogonal solution X. For such matrices A and B, we first seek the skewsymmetric orthogonal solution of AX = B applying our algorithm, then we compute the solution of AX = B using MATLAB procedure X = A\B. The entries in the following table are with exponent and with the mantissas which absolute value are less than one. And we only report their exponents.

14 316 C. Meng et al. / Linear Algebra and its Applications 402 (2005) The results 1 of Example 2 e e 1 e 2 e 3 e 4 ee 1 ee 2 ee 3 ee here, X i s denote the solutions computed by our algorithm and X i s are the solutions by MATLAB procedure X = A\B. And e 1 = AX i B, e 3 = X i + X T i, ee 1 = A X i B, e 2 = X i X, e 4 = X T i X i + I, ee 2 = X i X, ee 3 = X i + X T i, ee 4 = X T i X i + I. Remark 2 (1) The results show that the matrices X i computed by our algorithm can solve well problem (1) in 14 correct significant digits. At the same time the MATLAB procedure X = A\B can also solve well the equation AX = B with the solutions X i. (2) When e = 10, 1,...,10 4, both Ĉ = B T A A T B and A are full rank and the solution of AX = B is unique, so X i,x i,x are very nearness. In this case we suggest that one should prefer to choose MATLAB procedure X = A\B to solve problem (1) for it is simple. But when some of eigenvalues of the matrix A become nearer and nearer to zero, we find that the solution X i s have gradually lost the skewsymmetric and orthogonality. While the solution X i computed by our algorithm still preserve the property of skew-symmetry and orthogonality well. Therefore when A

15 C. Meng et al. / Linear Algebra and its Applications 402 (2005) has small eigenvalues near to zero, the MATLAB procedure X = A\B cannot settle problem (1) any longer while our algorithm can do it well. (3) It is pointed out that AX = B may be ill-conditioned linear equation or least squares problem but a well conditioned skew-symmetric-orthogonal-solutionproblem. An extreme example is A = ( 1 0 ), B = ( 0 1 ). The linear equation or least squares ( problem ) has condition. However, since the matrix Ĉ = B T A A T 0 1 B =, it is well-conditioned for skew-symmetric-orthogonalsolution-problem Conclusion In this paper, we have discussed the following three problems: Problem 1. Given matrices A, B R n 2m, find the skew-symmetric orthogonal solutions of the matrix equation AX = B. Problem 2. Given X R 2m 2m, find a matrix X S X, which denotes the solution set of problem 1, such that min X S X X X. Problem 3. Given matrices A, B R n 2m, find a skew-symmetric orthogonal matrix (SSOR 2m 2m )X such that min AX B. X SSOR2m 2m By applying the special form of the C S decomposition of an orthogonal matrix with skew-symmetric k k leading principal submatrix, we have obtained necessary and sufficient conditions for the solvability of problem 1 and the general forms of solutions for these three problems. Furthermore, we propose an algorithm to solve problems (1) and(3), which can preserve both orthogonality and skew-symmetry in the presence of rounding error, and compute the solutions of both problem (1) and problem (3). Numerical examples show that it is feasible. Acknowledgments We would like to thank the referee for his comments and suggestions that significantly improve the presentation of the work.

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