AN ITERATIVE METHOD FOR THE HERMITIANGENERALIZED HAMILTONIAN SOLUTIONS TO THE INVERSE PROBLEM AX = B WITH A SUBMATRIX CONSTRAINT


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1 Bulletin of the Iranian Mathematical Society Vol. 39 No. 6 (2013), pp AN ITERATIVE METHOD FOR THE HERMITIANGENERALIZED HAMILTONIAN SOLUTIONS TO THE INVERSE PROBLEM AX = B WITH A SUBMATRIX CONSTRAINT J. CAI Communicated by Mohammad Asadzadeh Abstract. In this paper, an iterative method is proposed for solving the matrix inverse problem AX = B for Hermitiangeneralized Hamiltonian matrices with a submatrix constraint. By this iterative method, for any initial matrix A 0, a solution A can be obtained in finite iteration steps in the absence of roundoff errors, and the solution with least norm can be obtained by choosing a special kind of initial matrix. Furthermore, in the solution set of the above problem, the unique optimal approximation solution to a given matrix can also be obtained. A numerical example is presented to show the efficiency of the proposed algorithm. 1. Introduction Thought this paper, we adopt the following notation. Let C m n (R m n ) and HC n n denote the set of m n complex (real) matrices and n n Hermitian matrices, respectively. For a matrix A C m n, we denote its conjugate transpose, transpose, trace, column space, null space and Frobenius norm by A H, A T, tr(a), R(A), N(A) and A, respectively. In space C m n, we define inner product as: A, B = tr(b H A) for all MSC(2010): Primary: 15A29; Secondary: 65J22. Keywords: Inverse problem, Hermitiangeneralized Hamiltonian matrix, submatrix constraint, optimal approximation. Received: 7 August 2012, Accepted: 30 November c 2013 Iranian Mathematical Society. 129
2 1250 Cai A, B C m n, and the symbol Re A, B and A, B stand for its real part and conjugate number, respectively. Two matrices A and B are orthogonal if A, B = 0. Let Q s,n = {a = (a 1, a 2,, a s ) : 1 a 1 < a 2 < < a s n} denote the strictly increasing sequences of s elements from 1, 2,, n. For A C m n, p Q s,m and q Q t,n, let A[p q] stand for the s t submatrix of A determined by rows indexed by p and columns indexed by q. Let I n = (e 1, e 2,, e n ) be the n n unit matrix, where e i denotes its ith column. Let J n R n n be the orthogonal skewsymmetric matrix, i.e., Jn T J n = J n Jn T = I n and Jn T = J n. A matrix A C n n is called Hermitiangeneralized Hamiltonian if A H = A and (AJ n ) H = AJ n. The set of all n n Hermitiangeneralized ( Hamiltonian ) matrices is denoted 0 by HGH n n Ik. Particularly, if J n =, then the set HGH I k 0 n n reduces to the wellknown set of HermitianHamiltonian matrices, which have applications in many areas such as linearquadratic control problem [?,?], H optimization [?] and the related problem of solving algebraic Riccati equations [?]. Recently, there have been several papers considering solving the inverse problem AX = B for various matrices by direct methods based on different matrix decompositions. For instance, Xu and Li [?], Peng [?] and Zhou et al. [?] discuss its Hermitian reflexive, antireflexive solutions and leastsquare centrosymmetric solutions, respectively. Then Huang and Yin [?] and Huang et al. [?] generalize the results of the latter to the more general Rsymmetric and Rskew symmetric matrices, respectively. Li et al. [?] consider the inverse problem for symmetric P symmetric matrices with a submatrix constraint. Peng et al. [?,?] and Gong et al. [?] consider solving the inverse problem for centrosymmetric, bisymmetric and HermitianHamiltonian matrices, respectively, under the leading principal submatrix constraint. Zhao et al. [?] concerns the inverse problem for bisymmetric matrices under a central principal submatrix constraint. However, the inverse problem for the Hermitiangeneralized Hamiltonian matrices with general submatrix constraint has not been studied till now. Hence, in this paper, we consider solving the following problem and its associated best approximation problem which occurs frequently in experimental design ([?,?,?,?]) by iterative methods.
3 Iterative Hermitiangeneralized Hamiltonian solutions 1251 Problem I. Given X, B C n m, A S HC s t, p = (p 1, p 2,, p s ) Q s,n, and q = (q 1, q 2,, q t ) Q t,n, find A HGH n n, such that (1.1) AX = B and A[p q] = A S. Problem II. Let S E denote the set of solutions of Problem I, for given Ā Cn n, find Â S E, such that (1.2) Â Ā = min A Ā. A S E The rest of this paper is organized as follows. In Section 2, we propose an iterative algorithm for solving Problem I and present some basic properties of this algorithm. In Section 3, we consider the iterative method for solving Problem II. A numerical example is given in Section to show the efficiency of the proposed algorithm. Conclusions will be put in Section Iterative algorithm for solving Problem I Firstly, we present several basic properties of Hermitiangeneralized Hamiltonian matrices in the following lemmas. Lemma 2.1. Consider a matrix Y C n n. Then Y + Y H + J n (Y + Y H )J n HGH n n. Proof. The proof is easy, thus is omitted. Lemma 2.2. Suppose a matrix Y C n n and a matrix D HGH n n. Then Re Y, D = Y + Y H + J n (Y + Y H )J n, D. Proof. Since Y H, D = tr(d H Y H ) = tr((y D) H ) = Y, D H = Y, D, we have Y + Y H, D = 2Re Y, D. Then we get J n (Y + Y H )J n, D = tr(d H J n (Y + Y H )J n ) = tr(j n D H J n (Y + Y H )) = tr(d H (Y + Y H )) = Y + Y H, D = 2Re Y, D. Hence we have Y + Y H + J n (Y + Y H )J n, D = Re Y, D. Next we propose an iterative algorithm for solving Problem I.
4 1252 Cai Algorithm 1. Step 1. Input X, B C n m, A S HC s t, p = (p 1, p 2,, p s ) Q s,n, q = (q 1, q 2,, q t ) Q t,n and an arbitrary A 0 HGH n n ; Step 2. Compute R 0 = B A 0 X; S 0 = A S (e p1, e p2,, e ps ) T A 0 (e q1, e q2,, e qt ); E 0 = R 0 X H + (e p1, e p2,, e ps )S 0 (e q1, e q2,, e qt ) T ; F 0 = 1 [E 0 + E H 0 + J n (E 0 + E H 0 )J n ]; P 0 = F 0 ; k := 0; Step 3. If R k = S k = 0 then stop; else, k := k + 1; Step. Compute α k 1 = R k S k 1 2 P k 1 2 ; A k = A k 1 + α k 1 P k 1 ; R k = B A k X; S k = A S (e p1, e p2,, e ps ) T A k (e q1, e q2,, e qt ); E k = R k X H + (e p1, e p2,, e ps )S k (e q1, e q2,, e qt ) T ; F k = 1 [E k + E H k β k 1 = tr(f kp k 1 ) P k 1 2 ; P k = F k β k 1 P k 1 ; Step 5. Go to Step 3. + J n(e k + E H k )J n]; Remark 2.3. By Lemma 2.1, one can easily see that the matrix sequences {A k },{P k } and {F k } generated by Algorithm 1 are all the Hermitiangeneralized Hamiltonian matrices. And Algorithm 1 implies that if R k = S k = 0, then A k is the solution of Problem I. We list some basic properties of Algorithm 1 as follows. Theorem 2.. Assume that A is a solution of Problem I. Then the sequences {A i }, {P i }, {R i } and {S i } generated by Algorithm 2.1 satisfy the following equality: (2.1) P i, A A i = R i 2 + S i 2, i = 0, 1, 2,. Proof. From Remark 2.3, it follows that A A i HGH n n, i = 0, 1, 2,. Then according to Lemma 2.2 and Algorithm 1, for i = 0, we have P 0, A A 0 = 1 (E 0 + E H 0 + J n (E 0 + E H 0 )J n ), A A 0 = Re E 0, A A 0
5 Iterative Hermitiangeneralized Hamiltonian solutions 1253 = Re R 0 X H + (e p1, e p2,, e ps )S 0 (e q1, e q2,, e qt ) T, A A 0 = Re (e p1, e p2,, e ps )S 0 (e q1, e q2,, e qt ) T, A A 0 +Re R 0 X H, A A 0 = Retr((e q1, e q2,, e qt ) T (A A 0 )(e p1, e p2,, e ps )S 0 ) +Retr(X H (A A 0 )R 0 ) = Retr(R H 0 R 0) + Retr(S H 0 S 0) = tr(r H 0 R 0) + tr(s H 0 S 0) = R S 0 2. Assume that the conclusion holds for i = k(k > 0), i.e., P k, A A k = R k 2 + S k 2, then for i = k + 1, we have P k+1, A A k+1 = F k+1, A A k+1 β k P k, A A k+1 = 1 E k+1 + E H k+1 + J n(e k+1 + E H k+1 )J n, A A k+1 β k P k, A A k α k P k = Re E k+1, A A k+1 β k P k, A A k + β k α k P k 2 = R k+1 X H + (e p1, e p2,, e ps )S k+1 (e q1, e q2,, e qt ) T, A A k+1 β k ( R k 2 + S k 2 ) + β k R k 2 + S k 2 P k 2 P k 2 = R k S k+1 2. This completes the proof by the principle of induction. Remark 2.5. Theorem 2. implies that if Problem I is consistent, then R i 2 + S i 2 0 implies that P i 0. On the other hand, if there exists a positive number k such that R k 2 + S k 2 0 but P k = 0, then Problem I must be inconsistent. Lemma 2.6. For the ( sequences ) {R i }, {S i (}, {P i }) and {F i } generated by Algorithm 1, let ˆR Ri Si i = and Ŝi =. Then it follows that R H i (2.2) ˆR i+1, ˆR j + Ŝi+1, Ŝj = ˆR i, ˆR j + Ŝi, Ŝj 2α i F j, P i. Proof. By Algorithm 1, Remark 2.3 and Lemma 2.2, we have ˆR i+1, ˆR j + Ŝi+1, Ŝj = tr(r H j R i+1 +R j R H i+1)+ tr(s H j S i+1 + S j S H i+1) = tr(r H j (R i α i P i X) + R j (R i α i P i X) H ) + tr(s H j (S i α i (e p1, e p2,, e ps ) T P i (e q1, e q2,, e qt )) + S j (S i α i (e q1, e q2,, e qt ) T P i (e p1, e p2,, e ps ))) S H i
6 125 Cai = tr(r H j R i + R j R H i ) + tr(sh j S i + S j S H i ) α i tr(r H j P ix + R j (P i X) H +S H j (e p 1, e p2,, e ps ) T P i (e q1, e q2,, e qt ) + S j (e q1, e q2,, e qt ) T P i (e p1, e p2,, e ps )) = ˆR i, ˆR j + Ŝi, Ŝj α i E i + E H i, P i = ˆR i, ˆR j + Ŝi, Ŝj α i 2 E i + E H i + J n (E i + E H i )J n, P i = ˆR i, ˆR j + Ŝi, Ŝj 2α i F i, P i. Theorem 2.7. For the sequences { ˆR i }, {Ŝi} and {P i } generated by Algorithm 1, if there exists a positive number k such that ˆR i 0 for all i = 0, 1, 2,..., k, then we have (2.3) ˆR i, ˆR j + Ŝi, Ŝj = 0, (i, j = 0, 1, 2,..., k, i j). Proof. Since ˆR i, ˆR j = ˆR j, ˆR i and Ŝi, Ŝj = Ŝj, Ŝi, we only need to prove that (2.3) holds for all 0 j < i k. For k = 1, it follows from Lemma 2.6 that ˆR 1, ˆR 0 + Ŝ1, Ŝ0 = ˆR 0, ˆR 0 + Ŝ0, Ŝ0 2α 0 F 0, P 0 = tr(r H 0 R 0 + R 0 R H 0 ) + tr(sh 0 S 0 + S 0 S H 0 ) 2 R S 0 2 P 0 2 P 0, P 0 = 2( R S 0 2 ) 2 R S 0 2 P 0 2 P 0 2 = 0. and P 1, P 0 = F 1 tr(f 1P 0 ) P 0 2 P 0, P 0 = 0, Assume that ˆR m, ˆR j + Ŝm, Ŝj = 0 and P m, P j = 0 hold for all 0 j < m, 0 < m k. We shall show that ˆR m+1, ˆR j + Ŝm+1, Ŝj = 0 and P m+1, P j = 0 hold for all 0 j < m + 1, 0 < m + 1 k. For 0 j < m, by Lemma 2.6, we have ˆR m+1, ˆR j + Ŝm+1, Ŝj = ˆR m, ˆR j + Ŝm, Ŝj 2α m F j, P m = ˆR m, ˆR j + Ŝm, Ŝj α m P j + β j 1 P j 1, P m = α m P j, P m = 0, and P m+1, P j = F m+1, P j β m P m, P j = F m+1, P j = ˆR j, ˆR m+1 + Ŝj, Ŝm+1 + ˆR j+1, ˆR m+1 + Ŝj+1, Ŝm+1 2α m+1 = 0.
7 Iterative Hermitiangeneralized Hamiltonian solutions 1255 For j = m, it follows from Lemma 2.6 and the hypothesis that ˆR m+1, ˆR m + Ŝm+1, Ŝm = ˆR m, ˆR m + Ŝm, Ŝm 2α s F m, P m = 2( R m 2 + S m 2 ) 2α s P m + β m 1 P m 1, P m = 2( R m 2 + S m 2 ) 2 R m 2 + S m 2 P m 2 P m, P m = 0, and P m+1, P m = F m+1 β m P m, P m = F m+1, P m tr(f m+1p m ) P m 2 P m 2 = 0. Hence ˆR m+1, ˆR j + Ŝm+1, Ŝj = 0 and P m+1, P j = 0 hold for all 0 j < m + 1, 0 < m + 1 k. This completes the proof by the principle of induction. Remark 2.8. Based on Theorem 2.7, we can further ( demonstrate ) the ˆRk finite termination property of Algorithm 1. Let Z k =. Theorem 2.7 implies that the matrix sequences Z 0, Z 1, are orthogonal to each other in the finite dimension matrix subspace. Hence there exists a positive integer t 0 such that Z t0 = 0. Then we have R t0 = S t0 = 0. Thus the iteration will be terminated in finite steps in the absence of roundoff errors. Next we consider the least Frobenius norm solution of Problem I. Lemma 2.9. [?] Suppose that the consistent system of linear equations Ax = b has a solution x R(A H ), then x is the unique least norm solution of the system of linear equations. Theorem Suppose that Problem I is consistent. If we choose the initial bisymmetric matrix as follows: (2.) A 0 = Y 1 X H + XY1 H + J n (Y 1 X H + XY H Ŝ k 1 )J n + (e p1, e p2,, e ps )Y 2 (e q1, e q2,, e qt ) T + (e q1, e q2,, e qt )Y H 2 (e p 1, e p2,, e ps ) T +J n ((e p1, e p2,, e ps )Y 2 (e q1, e q2,, e qt ) T + (e q1, e q2,, e qt )Y H 2 (e p1, e p2,, e ps ) T )J n, where Y 1, Y 2 are arbitrary n n complex matrices, or more especially, if A 0 = 0, then the solution obtained by Algorithm 1 is the least Frobenius norm solution.
8 1256 Cai Proof. Consider the matrix equations as follows: AX = B, X H A = B H, (e p1, e p2,, e ps ) T A(e q1, e q2,, e qt ) = A S, (e (2.5) q1, e q2,, e qt ) T A(e p1, e p2,, e ps ) = A H S, J n AJ n X = B, X H J n AJ n = B H, (e p1, e p2,, e ps ) T J n AJ n (e q1, e q2,, e qt ) = A S, (e q1, e q2,, e qt ) T J n AJ n (e p1, e p2,, e ps ) = A H S. If A is a solution of Problem I, then it must be a solution of (2.5). Conversely, if (2.5) has a solution A, let Ã = A + AH + J n (A + A H )J n, then it is easy to verify that Ã is a solution of Problem I. Therefore, the consistency of Problem I is equivalent to that of (2.5). By using Kronecker products, (2.5) can be equivalently written as (2.6) X T A I X H (e q1, e q2,, e qt ) T (e p1, e p2,, e ps ) T (e p1, e p2,, e ps ) T (e q1, e q2,, e qt ) T X T J T n J n J T n X H J n (e q1, e q2,, e qt ) T J T n (e p1, e p2,, e ps ) T J n (e p1, e p2,, e ps ) T J T n (e q1, e q2,, e qt ) T J n vec(a) = vec B B H A S A H S B B H Let the initial matrix A 0 be of the form (2.), then by Algorithm 1 and Remark 2.5, we can obtain the solution A of Problem I within finite iteration steps, which can be represented in the same form. Hence we have vec(a ) R X T A I X H (e q1, e q2,, e qt ) T (e p1, e p2,, e ps ) T (e p1, e p2,, e ps ) T (e q1, e q2,, e qt ) T X T J T n J n J T n X H J n (e q1, e q2,, e qt ) T J T n (e p1, e p2,, e ps ) T J n (e p1, e p2,, e ps ) T J T n (e q1, e q2,, e qt ) T J n A S A H S H..
9 Iterative Hermitiangeneralized Hamiltonian solutions 1257 According to Lemma 2.9, vec(a ) is the least norm solution of (2.6), i.e., A is the least norm solution of the (2.5). Since the solution set of Problem I is a subset of that of (2.5), A also is the least norm solution of Problem I. This completes the proof. 3. Iterative algorithm for solving Problem II In this section, we consider iterative algorithm for solving Problem II. For given Ā Cn n and arbitrary A S E, we have A Ā 2 = A Ā+ĀH Ā ĀH 2 2 = A Ā+ĀH +J n(ā+āh )J n 2 + Ā+ĀH J n(ā+āh )J n 2 + Ā ĀH 2 2, which implies that min A Ā is equivalent to A S E min A Ā + ĀH + J n (Ā + ĀH )J n. A S E When Problem I is consistent, for A S E, it follows that { (A Ā+ĀH +J n(ā+āh )J n )X = B (Ā+ĀH +J n(ā+āh )J n)x, (A Ā+ĀH +J n(ā+āh )J n )[p q] = A S ( Ā+ĀH +J n(ā+āh )J n )[p q]. Hence Problem II is equivalent to finding the least norm Hermitiangeneralized Hamiltonian solution of the following problem: ÃX = B, Ã[p q] = ÃS, where Ã = A Ā+ĀH +J n(ā+āh )J n, B = B (Ā+ĀH +J n(ā+āh )J n)x and Ã S = A S ( Ā+ĀH +J n(ā+āh )J n )[p q]. Once the unique least norm solution Ã of the above problem is obtained by applying Algorithm 1 with the initial matrix A 0 having the representation assumed in Theorem 2.10, the unique solution Â of Problem II can then be obtained, where Â = Ã + Ā+ĀH +J n(ā+āh )J n.. A numerical example In this section, we give a numerical example to illustrate the efficiency of the proposed iterative algorithm. All the tests are performed by MATLAB 7. with machine precision around Let zeros(n) denote
10 1258 Cai the n n matrix whose all elements are zero. Because of the influence of the error of calculation, we shall regard a matrix X as zero matrix if X < 1.0e 010. Example.1. Given matrices X and B as follows: X = , B = Let p = (1, 3, 5), q = (1, 2, 6) Q 3,6 and A S = Consider the least Frobenius norm Hermitiangeneralized Hamiltonian solution of the following inverse problem with submatrix constraint: (.1) AX = B and A[1, 3, 5 1, 2, 6] = A S. If we choose the initial matrix A 0 = zeros(6), then by Algorithm 1 and iterating 11 steps, we obtain the least Frobenius norm solution of the problem (.1) as follows: A 11 = with R S 11 2 = e Conclusions In this paper, we construct an iterative method to solve the inverse problem AX = B of the Hermitiangeneralized Hamiltonian matrices with general submatrix constraint. In the solution set of the matrix equations, the optimal approximation solution to a given matrix can.
11 Iterative Hermitiangeneralized Hamiltonian solutions 1259 also be found by this iterative method. The given numerical example show that the proposed iterative method is quite efficient. Acknowledgments Research supported by National Natural Science Foundation of China (No ), Natural Science Foundation of Zhejiang Province (No.Y611003) and The University Natural Science Research key Project of Anhui Province (No. KJ2013A20) References [1] A. BenIsrael and T. N. E. Greville, Generalized Inverse: Theory and Applications, Second Ed., John Wiley & Sons, New York, [2] L. S. Gong, X. Y. Hu and L. Zhang, An inverse problem for Hermitian Hamiltonian matrices with a submatrix constraint, Acta Math. Sci. Ser. A Chin. Ed. 28 (2008), no., [3] N. J. Higham, Computing a nearest symmetric positive semidefinite matrix, Linear Algebra Appl. 103 (1988) [] G. X. Huang and F. Yin, Matrix inverse problem and its optimal approximation problem for Rsymmetric matrices, Appl. Math. Comput. 189 (2007), no. 1, [5] G. X. Huang, F. Yin, H. F. Chen, L. Chen and K. Guo, Matrix inverse problem and its optimal approximation problem for Rskew symmetric matrices, Appl. Math. Comput. 216 (2010), no. 12, [6] M. Jamshidi, An overview on the solutions of the algebra matrix Riccati equation and related problems, Large Scale Systems: Theory and Appl. 1 (1980), no. 3, [7] Z. Jiang and Q. Lu, On optimal approximation of a matrix under a spectral restriction, Math. Numer. Sinica 8 (1986), no. 1, [8] J. F. Li, X. Y. Hu and L. Zhang, Inverse problem for symmetric P symmetric matrices with a submatrix constraint, Bull. Belg. Math. Soc. Simon Stevin 17 (2010), no., [9] V. L. Mehrmann, The Autonomous Linear Quadratic Control Problem, Theory and Numerical Solution, SpringerVerlag, Berlin, [10] Z. Y. Peng, The inverse eigenvalue problem for Hermitian antireflexive matrices and its approximation, Appl. Math. Comput. 162 (2005), no. 3, [11] Z. Y. Peng, X. Y. Hu and L. Zhang, The inverse problem of centrosymmetric matrices with a submatrix constraint, J. Comput. Math. 22 (200), no., [12] Z. Y. Peng, X. Y. Hu and L. Zhang, The inverse problem of bisymmetric matrices with a submatrix constraint, Numer. Linear Algebra Appl. 11 (200), no. 1,
12 1260 Cai [13] R. Penrose, On best approximation solutions of linear matrix equations, Proc. Cambridge Philos. Soc. 52 (1956) [1] A. J. Pritchard and D. Salamon, The linear quadratic control problem for retarded systems with delays in control and observation, IMA J. Math. Control Information 2 (1985) [15] W. W. Xu and W. Li, The Hermitian reflexive solutions to the matrix inverse problem AX = B, Appl. Math. Comput. 211 (2009), no. 1, [16] Y. X. Yuan, Two classes of best approximation problems of matrices, Math. Numer. Sin. 23 (2001), no., [17] L. J. Zhao, X. Y. Hu and L. Zhang, Least squares solutions to AX = B for bisymmetric matrices under a central principal submatrix constraint and the optimal approximation, Linear Algebra Appl. 28 (2008), no., [18] K. M. Zhou, J. Doyle and K. Glover, Robust and Optimal Control, Prentice Hall, Upper Saddle River, New Jersey, [19] F. Z. Zhou, L. Zhang and X. Y. Hu, Leastsquare solutions for inverse problems of centrosymmetric matrices, Comput. Math. Appl. 5 (2003), no , J. Cai School of Science, Huzhou Teachers College, Huzhou, P.R. China
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