The Hermitian R-symmetric Solutions of the Matrix Equation AXA = B

International Journal of Algebra, Vol. 6, 0, no. 9, 903-9 The Hermitian R-symmetric Solutions of the Matrix Equation AXA = B Qingfeng Xiao Department of Basic Dongguan olytechnic Dongguan 53808, China qfxiao@hnu.edu.cn Abstract In this paper, we mainly discuss solving the following problems. roblem I. Given matrices A C m n and B C m m, find X HRS n n such that AXA = B, where HRS n n = {X H n n RXR = X, for given R C n n satisfying R = R = R I n }. roblem II. Given a matrix X C n n, find ˆX S E such that X ˆX F = inf X X F, X S E where is the Frobenius norm, and S E is the solution set of roblem I. Expressions for the general solution of roblem I are derived. Necessary and sufficient conditions for the solvability of roblem I are provided. For roblem II, an expression for the solution is given as well. Mathematics Subject Classification: 65F5, 65F0 Keywords: Matrix equation, Hermitian R-symmetric, Inverse problem, The Optimal approximation Introduction Throughout this paper, let C n m be the set of all n m complex matrices, H n n denote the class of n n hermitian matrices, U n n be the set of all n n unitary matrices. Denote by I n the identity matrix with order n. Let J = (e n,e n,...,e ), where e i is the ith column of I n. For matrix A, A, A +, A F and r(a) represent its conjugate transpose, Moore-enrose inverse, Frobenius norm and rank, respectively. For matrices A, B C n m, the expression A B will be the Hadamard product of A and B. Defining the inner product (A, B) =

904 Qingfeng Xiao tr(b A) for matrices A, B C n m, C n m becomes a Hilbert space. The norm of a matrix generated by this inner product is the Frobenius norm. Consider the linear matrix equation AXA = B, () where A C m n, B C m m are given, and X C n n is unknown(matrix). Such a matrix equation with symmetric pattern appears frequently in matrix theory and applications such as in statistics, vibration theory, etc. For important results on the inverse problem () associated with several kinds of different sets S, for instance, symmetric matrices, symmetric nonnegative definite matrices, bisymmetric (same as persymmetric) matrices, bisymmetric nonnegative definite matrices and so on, we refer the reader to [-8. In this paper we discuss the above inverse problem () associated with a new S. Definition. Let R C n n be a nontrivial unitary involution; i.e., R = R = R I n. We say that X C n n is a Hermitian R-symmetric matrix, if X = X, RXR = X. We denoted by HRS n n the set of all n n Hermitian R-symmetric matrices. The Hermitian R-symmetric is a class of important matrices and have engineering and scientific applications. For the case the unknown A is Hermitian R-symmetric, [9 has discussed the inverse problems AX = B. However, for this case, the inverse problem () has not been dealt with yet. This problem will be considered here. The problem studied in this paper can now be described as follows. roblem I. Given matrices A C m n and B C m m, find a Hermitian R-symmetric matrix X such that AXA = B. In this paper, we discuss the solvability of this problem and an expression for its solution is presented. The optimal approximation problem of a matrix with the above-given matrix restriction comes up in the processes of test or recovery of a linear system due to incomplete data or revising given data. A preliminary estimate X of the unknown matrix X can be obtained by the experimental observation values and the information of statistical distribution. The optimal estimate of X is a matrix ˆX that satisfies the given matrix restriction for X and is the best approximation of X, see [0-. In this paper, we will also consider the so-called optimal approximation problem associated with (). It reads as follows.

Hermitian R-symmetric solutions 905 roblem II. Given matrix X C n n, find ˆX S E such that X ˆX F = inf X S E X X F, where S E is the solution set of roblem I. We point out that if roblem I is solvable, then roblem II has a unique solution, and in this case an expression for the solution can be derived. The paper is organized as follows. In Section we first discuss some properties of Hermitian R-symmetric matrices. Then we establish the solvability conditions for roblem I and give an expression of the general solution of roblem I. In Section 3 we prove the existence and uniqueness of the solution to roblem II and derive an expression. The expression of the general solution of problem I We first know that hermitian R-symmetric matrices have the following properties. Since R = R = R I n, the only possible eigenvalues of R are + and. Let r and s be respectively the dimensions of the eigenspaces of R associated with the eigenvalues λ =,andλ = ; thus r, s > and r+s = n. Let =[p p r, and Q =[q q s, () where {p,...,p r } and {q,...,q s } are orthonormal bases for the eigenspaces. and Q can be found by applying the Gram-Schmidt process to the columns of I + R and I R, respectively. If R = J the standard choices for and Q are = [ Im J m, and Q = [ Im J m if n =m, or = I m 0 0, and Q = J m 0, I m 0 J m, if n =m +. In general, and R = and RQ = Q, (3) [ [ Q = Q.

906 Qingfeng Xiao Therefore any X C n n can be written conformably in block form as [ [ X C X =[ Q, (4) C with X = X H r r, X Q = Q XQ H s s, and C = XQ C r s, From (3) and (4), [ [ X C RXR =[ Q C, (5) X Q Lemma. [9 X C n n is Hermitian R-symmetric if and only if [ [ X 0 X =[ Q 0 X Q Q, (6) X Q with X = X H r r, X Q = Q XQ H s s. Let A C m n and B C m m,, Q with the form of (). Let A[ Q=[A A, A C m r, A C m s. (7) By making generalized singular value decomposition (see [3) to the matrix pair [A, A, we have Q Q A = MΣ U, A = MΣ V, (8) where M C m m is a nonsingular matrix, U U r r, V U s s and I 0 0 l k 0 0 0 l k 0 S Σ = 0 k 0 S, Σ 0 0 0 t r = 0 k, (9) 0 0 I t r 0 0 0 n t 0 0 0 n t t = r([a,a ), l = r(a ), k = r(a )+r(a ) r([a,a ), S = diag(α,...,α k ), S = diag(β,...,β k ), 0 <α k α <, 0 <β β k <, α i + β i =,i =,...,k. Now we can establish the existence theorems as follows. Theorem. Given A C m n and B C m m,, Q defined in(), and A[ Q has the partition form of (7), the generalized singular value decomposition of the matrix pair [A,A as (8). artition the matrix M BM as B B B 3 B 4 l k M BM B = B B 3 B 4 k, (0) B 3 B 3 B 33 B 34 t r B 4 B 4 B 43 B 44 m t

Hermitian R-symmetric solutions 907 then the problem I has a solution X HRS n n if and only if B = B,B 3 = O, B 4 = O, B 4 = O, B 43 = O, B 44 = O. () In that case it has the general solution [ X 0 X =[ Q 0 X Q [ Q, () where X = U B B S X 3 S B X X 3 X 3 X 3 X 33 Y Y Y 3 X Q = V Y S (B S X S )S S B 3 Y3 B 3 S B 33 U, (3) V, (4) with X 3 C (l k) (r l), X H k k, X 3 C k (r l), X 33 C (r l) (r l), Y H (s+l k t) (s+l k t), Y C (s+l k t) k, Y 3 C (s+l k t) (t l) are arbitrary matrices. roof The necessity. Assume Eq.() has a solution X HRS n n. By the definition of HRS n n, it is easy to verify that B = B, and we have from Lemma that X can be expressed as [ [ X 0 X =[ Q 0 X Q Q, (5) where X H r r, X Q H s s. Note that [ Q is an unitary matrix, and the definition of A i (i =, ), Eq.() is equivalent to Substituting (8) into (6), then we have A X A + A X Q A = B. (6) Σ U X UΣ +Σ V X Q V Σ = M BM. (7) artition the matrices U X U, V X Q V as follows: U X U = X X X 3 X X X 3 X 3 X 3 X 33, V X Q V = Y Y Y 3 Y Y Y 3 Y 3 Y 3 Y 33, (8) where X C (l k) (l k), X C k k, X 33 C (r l) (r l), and Y C (s+l k t) (s+l k t), Y C k k, Y 33 C (t l) (t l).

908 Qingfeng Xiao Substituting (9), (0) and (8) into (7), we have X X S 0 0 B B B 3 B 4 S X S X S + S Y S S Y 3 0 0 Y 3 S Y 33 0 = B B B 3 B 4 B 3 B 3 B 33 B 34 0 0 0 0 B 4 B 4 B 43 B 44. (9) Therefore (9) holds if and only if () holds and X = B, X = X = B S, Y 3 = Y3 = S B 3, Y 33 = B 33, Y = S (B S X S )S where X H k k is arbitrary matrix. Substituting the above into (8) and (5), thus we have formulation (), (3) and (4). The sufficiency. Assume () holds. Let X = U B B S X 3 S B X X 3 X 3 X 3 X 33 U, Y Y Y 3 X Q = V Y S (B S X S )S S B 3 V, Y3 B 3 S B 33 where X 3 C (l k) (r l), X H k k, X 3 C k (r l), X 33 C (r l) (r l), Y H (s+l k t) (s+l k t), Y C (s+l k t) k, Y 3 C (s+l k t) (t l) are arbitrary matrices. Obviously, X H r r, XQ H s s. By Lemma. and X =[ Q [ X 0 0 XQ [ we have X HRS n n. Hence A XA = A[ [ [ X 0 Q 0 XQ Q A = A X A + A XQ A = B B 0 0 0 0 0 0 B M S X S 0 0 0 0 0 0 + 0 B S X S B 3 0 0 B3 B 33 0 0 0 0 0 0 0 0 0 = B B 0 0 B M B B 3 0 0 B3 B 33 0 M = B. 0 0 0 0 [ [ This implies that X X 0 =[ Q 0 XQ Q is the Hermitian R- symmetric solution of Eq.(). The proof is completed. Q,

Hermitian R-symmetric solutions 909 3 The expression of the solution of roblem II To prepare for an explicit expression for the solution of the matrix nearness problem II, we first verify the following lemma. Lemma 3. [6 Suppose that G, H C n n, S σ = diag(σ,,σ n ), σ i > 0(i =,,,n), Then there exist a unique Ŝ Hn n such that S G F + S σss σ H F = min (0) and where Ŝ =Φ [G + G + S σ (H + H )S σ, Φ=(ψ ij ) C n n,ψ ij =, i, j n. ( + σi σj ) Theorem 3. Let the matrix X C n n, the other relevant [ conditions and symbols be the same as those stated in Theorem.. artition X[ Q, U Z U, V Z V as follows: U Z U = [ Q X[ X X X3 X X X3 X 3 X3 X33 Q= [ Z Z Z, V Z V = Z Q, () Ỹ Ỹ Ỹ 3 Ỹ Ỹ Ỹ 3 Ỹ 3 Ỹ 3 Ỹ 33. () When the solution set S E of roblem I is nonempty, then roblem II has a unique solution ˆX, which can be expressed as where ˆX = U ˆX =[ Q [ ˆX 0 0 ˆXQ [ Q B B S ( X 3 + X 3 ) S B K ( X 3 + X 3 ) ( X 3 + X 3) ( X 3 + X 3) ( X 33 + X 33), (3) U, (4) ˆX (Ỹ + Ỹ ) (Ỹ + Ỹ ) (Ỹ3 + Ỹ 3) = V (Ỹ + Ỹ ) S (B S KS )S S B 3 V, (5) (Ỹ3 + Ỹ 3 ) B 3S B 33

90 Qingfeng Xiao with K =Φ [ X + X + S S (S B S Ỹ Ỹ )S S, (6) Φ=(ψ ij ) R k k, ψ ij = βi β j, i, j k. (αi αj + βi βj ) roof When S m is nonempty, it is easy to verify from () that S m is a closed convex set, there exists a unique solution for roblem II. Using the invariance of the Frobenius norm under unitary transformations, from (), () and () we have X X F = X Z F + X Z F + Z F + Z F = + U Z U + Z F + Z F F Y Y Y 3 Y S (B S X S )S S B 3 V Z V. Y3 B 3 S B 33 F B B S X 3 S B X X 3 X3 X3 X 33 Thus X ˆX F = inf X S E X X F is equivalent to X 3 X 3 F + X 3 X 3 F = min, X 3 X 3 F + X 3 X 3 F = min, X 33 X 33 F = min, Y Ỹ F = min, Y Ỹ F + Y +Ỹ F = min, Y 3 Ỹ3 F + Y 3 Ỹ3 F = min, X X F + S S X S S (S B S Ỹ) F = min. From Lemma 3. we have X 3 = ( X 3 + X 3 ), X 3 = ( X 3 + X 3 ), X 33 = ( X 33 + X 33 ),Y = (Ỹ + Ỹ ),Y = (Ỹ + Ỹ ),Y 3 = (Ỹ3 + Ỹ 3 ) and X =Φ [ X + X + S S (S B S Ỹ Ỹ )S S. Taking X i,y i3,i =,, 3 and X into (), (3), (4), we obtain that the solution of (the matrix nearness) roblem II can be expressed as (3). The proof is completed. ACKNOWLEDGEMENTS. This research was supported by Scientific Research Fund of Dongguan olytechnic (0a5).

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