Left and Right Inverse Eigenpairs Problem for -Hermitian Matrices

Hindawi Publishing Corporation Journal of Applied Mathematics Volume 2013, Article ID 230408, 6 pages http://dx.doi.org/10.1155/2013/230408 Research Article Left and Right Inverse Eigenpairs Problem for 𝜅-Hermitian Matrices Fan-Liang Li,1 Xi-Yan Hu,2 and Lei Zhang2 1 Institute of Mathematics and Physics, School of Sciences, Central South University of Forestry and Technology, Changsha 410004, China 2 College of Mathematics and Econometrics, Hunan University, Changsha 410082, China Correspondence should be addressed to Fan-Liang Li; Received 6 December 2012; Accepted 20 March 2013 Academic Editor: Panayiotis J. Psarrakos Copyright © 2013 Fan-Liang Li et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Left and right inverse eigenpairs problem for 𝜅-hermitian matrices and its optimal approximate problem are considered. Based on the special properties of 𝜅-hermitian matrices, the equivalent problem is obtained. Combining a new inner product of matrices, the necessary and sufficient conditions for the solvability of the problem and its general solutions are derived. Furthermore, the optimal approximate solution and a calculation procedure to obtain the optimal approximate solution are provided. 1. Introduction Throughout this paper we use some notations as follows. Let 𝐶𝑛×𝑚 be the set of all 𝑛 × 𝑚 complex matrices, 𝑈𝐶𝑛×𝑛 , 𝐻𝐶𝑛×𝑛 , 𝑆𝐻𝐶𝑛×𝑛 denote the set of all 𝑛 × 𝑛 unitary matrices, hermitian matrices, skew-hermitian matrices, respectively. Let 𝐴, 𝐴𝐻, and 𝐴+ be the conjugate, conjugate transpose, and the MoorePenrose generalized inverse of 𝐴, respectively. For 𝐴, 𝐵 ∈ 𝐶𝑛×𝑚 , ⟨𝐴, 𝐵⟩ = re(tr(𝐵𝐻𝐴)), where re(tr(𝐵𝐻𝐴)) denotes the real part of tr(𝐵𝐻𝐴), the inner product of matrices 𝐴 and 𝐵. The induced matrix norm is called Frobenius norm. That is, ‖𝐴‖ = ⟨𝐴, 𝐴⟩1/2 = (tr(𝐴𝐻𝐴))1/2 . Left and right inverse eigenpairs problem is a special inverse eigenvalue problem. That is, giving partial left and right eigenpairs (eigenvalue and corresponding eigenvector), (𝜆 𝑖 , 𝑥𝑖 ), 𝑖 = 1, . . . , ℎ; (𝜇𝑗 , 𝑦𝑗 ), 𝑗 = 1, . . . , 𝑙, a special matrix set 𝑆, finding a matrix 𝐴 ∈ 𝑆 such that 𝐴𝑥𝑖 = 𝜆 𝑖 𝑥𝑖 , 𝑖 = 1, . . . , ℎ, 𝑦𝑗𝑇 𝐴 = 𝜇𝑗 𝑦𝑗𝑇 , 𝑗 = 1, . . . , 𝑙. eigenpairs problem. For example, we studied the left and right inverse eigenpairs problem of skew-centrosymmetric matrices and generalized centrosymmetric matrices, respectively [5, 6]. Based on the special properties of left and right eigenpairs of these matrices, we derived the solvability conditions of the problem and its general solutions. In this paper, combining the special properties of 𝜅-hermitian matrices and a new inner product of matrices, we first obtain the equivalent problem, then derive the necessary and sufficient conditions for the solvability of the problem and its general solutions. Hill and Waters [7] introduced the following matrices. Definition 1. Let 𝜅 be a fixed product of disjoint transpositions, and let 𝐾 be the associated permutation matrix, that is, 𝐾 = 𝐾𝐻 = 𝐾, 𝐾2 = 𝐼𝑛 , a matrix 𝐴 ∈ 𝐶𝑛×𝑛 is said to be 𝜅hermitian matrices (skew 𝜅-hermitian matrices) if and only if 𝑎𝑖𝑗 = 𝑎𝑘(𝑗)𝑘(𝑖) (𝑎𝑖𝑗 = −𝑎𝑘(𝑗)𝑘(𝑖) ), 𝑖, 𝑗 = 1, . . . , 𝑛. We denote the set of 𝜅-hermitian matrices (skew 𝜅-hermitian matrices) by 𝐾𝐻𝐶𝑛×𝑛 (𝑆𝐾𝐻𝐶𝑛×𝑛 ). (1) This problem, which usually arises in perturbation analysis of matrix eigenvalues and in recursive matters, has profound application background [1–6]. When the matrix set 𝑆 is different, it is easy to obtain different left and right inverse From Definition 1, it is easy to see that hermitian matrices and perhermitian matrices are special cases of 𝜅-hermitian matrices, with 𝑘(𝑖) = 𝑖 and 𝑘(𝑖) = 𝑛 − 𝑖 + 1, respectively. Hermitian matrices and perhermitian matrices, which are one of twelve symmetry patterns of matrices [8], are applied in engineering, statistics, and so on [9, 10]. 2 Journal of Applied Mathematics From Definition 1, it is also easy to prove the following conclusions. (1) 𝐴 ∈ 𝐾𝐻𝐶𝑛×𝑛 if and only if 𝐴 = 𝐾𝐴𝐻𝐾. (2) 𝐴 ∈ 𝑆𝐾𝐻𝐶𝑛×𝑛 if and only if 𝐴 = −𝐾𝐴𝐻𝐾. (3) If 𝐾 is a fixed permutation matrix, then 𝐾𝐻𝐶𝑛×𝑛 and 𝑆𝐾𝐻𝐶𝑛×𝑛 are the closed linear subspaces of 𝐶𝑛×𝑛 and satisfy 𝐶𝑛×𝑛 = 𝐾𝐻𝐶𝑛×𝑛 ⨁ 𝑆𝐾𝐻𝐶𝑛×𝑛 . (2) The notation 𝑉1 ⊕ 𝑉2 stands for the orthogonal direct sum of linear subspace 𝑉1 and 𝑉2 . ̃ ∈ (4) 𝐴 ∈ 𝐾𝐻𝐶𝑛×𝑛 if and only if there is a matrix 𝐴 ̃ = 𝐾𝐴. 𝐻𝐶𝑛×𝑛 such that 𝐴 ̃ ∈ (5) 𝐴 ∈ 𝑆𝐾𝐻𝐶𝑛×𝑛 if and only if there is a matrix 𝐴 𝑛×𝑛 ̃ = 𝐾𝐴. 𝑆𝐻𝐶 such that 𝐴 Proof. (1) From Definition 1, if 𝐴 = (𝑎𝑖𝑗 ) ∈ 𝐾𝐻𝐶𝑛×𝑛 , then 𝑎𝑖𝑗 = 𝑎𝑘(𝑗)𝑘(𝑖) , this implies 𝐴 = 𝐾𝐴𝐻𝐾, for 𝐾𝐴𝐻𝐾 = (𝑎𝑘(𝑗)𝑘(𝑖) ). (2) With the same method, we can prove (2). So, the proof is omitted. (3) (a) For any 𝐴 ∈ 𝐶𝑛×𝑛 , there exist 𝐴 1 ∈ 𝐾𝐻𝐶𝑛×𝑛 , 𝐴 2 ∈ 𝑆𝐾𝐻𝐶𝑛×𝑛 such that 𝐴 = 𝐴 1 + 𝐴 2, (3) where 𝐴 1 = (1/2)(A + 𝐾𝐴𝐻𝐾), 𝐴 2 = (1/2)(𝐴 − 𝐾𝐴𝐻𝐾). (b) If there exist another 𝐴1 ∈ 𝐾𝐻𝐶𝑛×𝑛 , 𝐴2 ∈ 𝑆𝐾𝐻𝐶𝑛×𝑛 such that 𝐴 = 𝐴1 + 𝐴2 , (4) (5) Multiplying (5) on the left and on the right by 𝐾, respectively, and according to (1) and (2), we obtain 𝐴 1 − 𝐴1 = 𝐴 2 − 𝐴2 . Problem 2. Giving 𝑋 ∈ 𝐶𝑛×ℎ , Λ = diag(𝜆 1 , . . . , 𝜆 ℎ ) ∈ 𝐶ℎ×ℎ ; 𝑌 ∈ 𝐶𝑛×𝑙 , Γ = diag(𝜇1 , . . . , 𝜇𝑙 ) ∈ 𝐶𝑙×𝑙 , find 𝐴 ∈ 𝐾𝐻𝐶𝑛×𝑛 such that 𝐴𝑋 = 𝑋Λ, 𝑌𝑇 𝐴 = Γ𝑌𝑇 . ̂ ∈ 𝑆𝐸 such that Problem 3. Giving 𝐵 ∈ 𝐶𝑛×𝑛 , find 𝐴 󵄩󵄩 ̂󵄩󵄩󵄩󵄩 = min ‖𝐵 − 𝐴‖ , 󵄩󵄩𝐵 − 𝐴 󵄩 ∀𝐴∈𝑆𝐸 󵄩 (8) (9) where 𝑆𝐸 is the solution set of Problem 2. This paper is organized as follows. In Section 2, we first obtain the equivalent problem with the properties of 𝐾𝐻𝐶𝑛×𝑛 and then derive the solvability conditions of Problem 2 and its general solution’s expression. In Section 3, we first attest the existence and uniqueness theorem of Problem 3 then present the unique approximation solution. Finally, we provide a calculation procedure to compute the unique approximation solution and numerical experiment to illustrate the results obtained in this paper correction. 2. Solvability Conditions of Problem 2 We first discuss the properties of 𝐾𝐻𝐶𝑛×𝑛 Lemma 4. Denoting 𝑀 = 𝐾𝐸𝐾𝐺𝐸, and 𝐸 ∈ 𝐻𝐶𝑛×𝑛 , one has the following conclusions. (3)-(4) yields 𝐴 1 − 𝐴1 = − (𝐴 2 − 𝐴2 ) . In this paper, we suppose that 𝐾 is a fixed permutation matrix and assume (𝜆 𝑖 , 𝑥𝑖 ), 𝑖 = 1, . . . , ℎ, be right eigenpairs of 𝐴; (𝜇𝑗 , 𝑦𝑗 ), 𝑗 = 1, . . . , 𝑙, be left eigenpairs of 𝐴. If we let 𝑋 = (𝑥1 , . . . , 𝑥ℎ ) ∈ 𝐶𝑛×ℎ , Λ = diag (𝜆 1 , . . . , 𝜆 ℎ ) ∈ 𝐶ℎ×ℎ ; 𝑌 = (𝑦1 , . . . , 𝑦𝑙 ) ∈ 𝐶𝑛×𝑙 , Γ = diag(𝜇1 , . . . , 𝜇𝑙 ) ∈ 𝐶𝑙×𝑙 , then the problems studied in this paper can be described as follows. (6) Combining (5) and (6) gives 𝐴 1 = 𝐴1 , 𝐴 2 = 𝐴2 . (c) For any 𝐴 1 ∈ 𝐾𝐻𝐶𝑛×𝑛 , 𝐴 2 ∈ 𝑆𝐾𝐻𝐶𝑛×𝑛 , we have 𝐻 ⟨𝐴 1 , 𝐴 2 ⟩ = re (tr (𝐴𝐻 2 𝐴 1 )) = re (tr (𝐾𝐴 2 𝐾𝐾𝐴 1 𝐾)) = re (tr (−𝐴𝐻 2 𝐴 1 )) = − ⟨𝐴 1 , 𝐴 2 ⟩ . (7) This implies ⟨𝐴 1 , 𝐴 2 ⟩ = 0. Combining (a), (b), and (c) gives (3). ̃𝐻 = 𝐴 ̃ ∈ 𝐻𝐶𝑛×𝑛 . ̃ = 𝐾𝐴, if 𝐴 ∈ 𝐾𝐻𝐶𝑛×𝑛 , then 𝐴 (4) Let 𝐴 𝐻 𝑛×n 𝐻 ̃ ∈ 𝐻𝐶 , then 𝐴 = 𝐾𝐴 ̃ and 𝐾𝐴 𝐾 = 𝐾𝐴 ̃𝐻𝐾𝐾 = ̃ =𝐴 If 𝐴 ̃ = 𝐴 ∈ (...truncated)