`Journal of Applied MathematicsVolume 2013 (2013), Article ID 230408, 6 pageshttp://dx.doi.org/10.1155/2013/230408`
Research Article

## Left and Right Inverse Eigenpairs Problem for -Hermitian Matrices

1Institute of Mathematics and Physics, School of Sciences, Central South University of Forestry and Technology, Changsha 410004, China
2College of Mathematics and Econometrics, Hunan University, Changsha 410082, China

Received 6 December 2012; Accepted 20 March 2013

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.

#### Abstract

Left and right inverse eigenpairs problem for -hermitian matrices and its optimal approximate problem are considered. Based on the special properties of -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 be the set of all complex matrices, , , denote the set of all unitary matrices, hermitian matrices, skew-hermitian matrices, respectively. Let , , and be the conjugate, conjugate transpose, and the Moore-Penrose generalized inverse of , respectively. For , , where denotes the real part of , the inner product of matrices and . The induced matrix norm is called Frobenius norm. That is, .

Left and right inverse eigenpairs problem is a special inverse eigenvalue problem. That is, giving partial left and right eigenpairs (eigenvalue and corresponding eigenvector), , ; , , a special matrix set , finding a matrix such that This problem, which usually arises in perturbation analysis of matrix eigenvalues and in recursive matters, has profound application background [16]. When the matrix set is different, it is easy to obtain different left and right inverse 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 -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 be a fixed product of disjoint transpositions, and let be the associated permutation matrix, that is, , , a matrix is said to be -hermitian matrices (skew -hermitian matrices) if and only if , . We denote the set of -hermitian matrices (skew -hermitian matrices) by .

From Definition 1, it is easy to see that hermitian matrices and perhermitian matrices are special cases of -hermitian matrices, with and , 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].

From Definition 1, it is also easy to prove the following conclusions. (1) if and only if . (2) if and only if . (3) If is a fixed permutation matrix, then and are the closed linear subspaces of and satisfy The notation stands for the orthogonal direct sum of linear subspace and . (4) if and only if there is a matrix such that . (5) if and only if there is a matrix such that .

Proof. (1) From Definition 1, if , then , this implies , for .
(2) With the same method, we can prove . So, the proof is omitted.
(3) (a) For any , there exist , such that where , .  (b) If there exist another , such that (3)-(4) yields Multiplying (5) on the left and on the right by , respectively, and according to and , we obtain Combining (5) and (6) gives , .  (c) For any , , we have This implies . Combining (a), (b), and (c) gives .
(4) Let , if , then . If , then and .
(5) With the same method, we can prove (5). So, the proof is omitted.

In this paper, we suppose that is a fixed permutation matrix and assume , , be right eigenpairs of ; , , be left eigenpairs of . If we let , ; , , then the problems studied in this paper can be described as follows.

Problem 2. Giving , ; , , find such that

Problem 3. Giving , find such that where 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 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

Lemma 4. Denoting , and , one has the following conclusions. (1)If , then . (2)If , then . (3)If , where , , then if and only if . In addition, one has .

Proof. (1) .
Hence, we have .
(2) .
Hence, we have .
(3) , we have , from (1) and (2). If , then , while . Therefore from the conclusion (3) of Definition 1, we have , that is, . On the contrary, if , it is clear that . The proof is completed.

Lemma 5. Let , if is a right eigenpair of , then is a left eigenpair of .

Proof. If is a right eigenpair of , then we have From the conclusion (1) of Definition 1, it follows that This implies So is a left eigenpair of .

From Lemma 5, without loss of the generality, we may assume that Problem 2 is as follows.

Combining (13) and the conclusion (4) of Definition 1, it is easy to derive the following lemma.

Lemma 6. If , , , are given by (13), then Problem 2 is equivalent to the following problem. If , , , are given by (13), find such that

Lemma 7 (see [11]). If giving , , then matrix equation has solution if and only if Moreover, the general solution can be expressed as

Theorem 8. If , , , are given by (13), then Problem 2 has a solution in if and only if Moreover, the general solution can be expressed as where

Proof. Necessity: If there is a matrix such that , , then from Lemma 6, there exists a matrix such that , and according to Lemma 7, we have It is easy to see that (20) is equivalent to (17).
Sufficiency: If (17) holds, then (20) holds. Hence, matrix equation has solution . Moreover, the general solution can be expressed as follows: Let This implies . Combining the definition of , and the first equation of (17), we have Hence, . Combining the definition of , , (13) and (17), we have Therefore, is a special solution of Problem 2. Combining the conclusion (4) of Definition 1, Lemma 4, and , it is easy to prove that if and only if . Hence, the solution set of Problem 2 can be expressed as (18).

#### 3. An Expression of the Solution of Problem 3

From (18), it is easy to prove that the solution set of Problem 2 is a nonempty closed convex set if Problem 2 has a solution in . We claim that for any given , there exists a unique optimal approximation for Problem 3.

Theorem 9. Giving , if the conditions of , , , are the same as those in Theorem 8, then Problem 3 has a unique solution . Moreover, can be expressed as where , are given by (19) and .

Proof. Denoting , it is easy to prove that matrices and are orthogonal projection matrices satisfying . It is clear that matrices and are also orthogonal projection matrices satisfying . According to the conclusion (3) of Definition 1, for any , there exists unique such that where Combining Theorem 8, for any , we have It is easy to prove that according to the definitions of , . So we have Obviously, is equivalent to Since , it is clear that , for any , is a solution of (31). Substituting this result to (18), we can obtain (25).

Algorithm 10. (1) Input , , , according to (13). (2) Compute , , , , if (17) holds, then continue; otherwise stop. (3) Compute according to (19), and compute according to (28). (4) According to (25) calculate .

Example 11 (, ).

From the first column to the fourth columnFrom the fifth column to the eighth columnIt is easy to see that matrices , , , satisfy (17). Hence, there exists the unique solution for Problem 3. Using the software “MATLAB”, we obtain the unique solution of Problem 3.
From the first column to the fourth columnFrom the fifth column to the eighth column

#### Conflict of Interests

There is no conflict of interests between the authors.

#### Acknowledgments

This research was supported by National Natural Science Foundation of China (31170532). The authors are very grateful to the referees for their valuable comments and also thank the editor for his helpful suggestions.

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