Finite iterative Hamiltonian solutions of the generalized coupled Sylvester

Abstract

In this paper, we present an iterative algorithm to solve a generalized coupled Sylvester – conjugate matrix equations over Hamiltonian matrices. When the considered systems of matrix equations are consistent, it is proven that the solution can be obtained within finite iterative steps for any arbitrary initial generalized Hamiltonian matrices in the absence of round off errors. Two numerical examples are given to illustrate the effectiveness of the proposed method.

Keywords

Coupled Sylvester – conjugate matrix equation iterative algorithm inner product Frobenius norm Hamiltonian solutions

Introduction

In this paper, the symbols $A^{T}, \bar{A}, A^{H}$ and $tr (A)$ represent the transpose, conjugate, conjugate transpose and the trace of a matrix A, respectively. $Re (a)$ denote the real part of number $a \cdot C^{n \times n}$ represents the set of all $m \times n$ complex matrices. The matrix $A \in C^{n \times n}$ is a generalized Hamiltonian matrix if $JAJ = A^{H}$ where J is an $n \times n$ skew – Hermitian matrix, that is, $J = - J^{H}, J^{H} J = J J^{H} = I$ . Also, the set of all generalized Hamiltonian matrices is denoted by ${GHC}^{n \times n}$ , that is, ${GHC}^{n \times n} = {A : JAJ = A^{H}}$ where J is an $n \times n$ skew – Hermitian matrix.

Consider the generalized coupled Sylvester – conjugate matrix equations

\sum_{j = 1}^{q} A_{ij} X_{j} B_{ij} + C_{ij} \bar{X_{j}} D_{ij} = E_{i}, i = 1, 2, \dots, p

(1)

where $A_{ij}, C_{ij} \in C$ , $B_{ij}, D_{ij} \in C^{n \times r}$ and $E_{i} \in C^{m \times r}$ are given matrices, while $X_{j} \in {GH C}^{n \times n}, j = 1, 2, \dots, q$ are matrices to be determined. Matrix equations are often encountered in many areas of computational mathematics, control and system theory. By using the Kronecker product, the exact solutions can be readily obtained. However, this direct method may suffer from computing problems with the dimensions increasing. In Chang et al. (2010), the expression of $(R, S)$ – conjugate solution about $AX = C, XB = D$ by matrix decompositions is given. Trench (2004) defined R- conjugate matrix and defined $(R, S)$ - conjugate matrix in Trench (2005). Trench (2004) studied the system of linear equations $Az = w$ for R– conjugate matrices and $min ‖ Az - w ‖$ for $(R, S)$ – conjugate matrices in Trench (2005), respectively, where $z, w$ are known column vectors. Li (2017) constructed a finite iterative method to solve the generalized Hamiltonian coupled Sylvester matrix equations with conjugate transpose. Dehghan and Hajarian (2011a) introduced the reflexive solution of the Sylvester matrix equation $AY + YB = C$ . Madiseh and Dehghan (2014) extended the concept of generalized solution sets to the interval generalized Sylvester matrix equation $\sum_{i = 1}^{p} A_{i} X_{i} + \sum_{j = 1}^{q} Y_{j} B_{j} = C$ . The general expressions of the $(R, S)$ -symmetric and $(R, S)$ -skew symmetric solutions were given in Dehghan and Hajarian (2011b). (Dehghan and Hajarian, 2010, 2012a, 2012b) obtained the solutions, the generalized bisymmetric solutions, the generalized centro-symmetric and central anti-symmetric solutions by extending the CGNE iterative method and GI method. Hajarian (2014) solved general coupled matrix equations by using the matrix form of the CGS method. Beik and Salkuyeh (2017) developed a robust iterative algorithm to find the least-squares $(P, Q)$ -orthogonal symmetric and skew-symmetric solution sets of generalized coupled matrix equations. Beik and Moghadam (2014) proposed an algorithm to solve a class of general coupled linear matrix equations over the complex number field and the optimal approximately generalized reflexive and anti-reflexive solution groups are derived. Beik and Salkuyeh (2013) presented an iterative algorithm for solving coupled Sylvester-transpose matrix over generalized centro-symmetric matrices and derived the optimal approximate generalized centro-symmetric solution. Salkuyeh and Beik (2015) dealt with the problem of finding the minimum norm least-squares solution of a quite general class of coupled linear matrix equations defined over the complex number field and present the convergence properties of the algorithm.

This paper is organized as follows. First, in section 2, we introduce some definitions, lemmas and theorem that will be needed to develop this work. In section 3, we propose iterative methods to obtain numerical solutions to the generalized coupled Sylvester – conjugate matrix equation (1) over Hamiltonian matrices. In section 4, two numerical examples are given to explore the simplicity and the neatness of the presented methods.

Preliminaries

The following definition, lemmas and theorem will be used to develop the proposed work.

Definition 1. Inner product (Zhang, 2004): A real inner product space is a vector space V over the real field ℝ together with an inner product, that is, with a map

〈 ., . 〉 : V \times V \to R

Satisfying the following three axioms for all vectors $x, y, z \in V$ and all scalars $a \in R$

Symmetry: $〈 x, y 〉 = 〈 y, x 〉$ .

Linearity in the first argument $〈 ax, y 〉 = a 〈 x, y 〉, 〈 x + y, z 〉 = 〈 x, z 〉 + 〈 y, z 〉$ .

Positive definiteness: $〈 x, x 〉$ > 0 for all $x \neq 0$ .

The following theorem defines a real inner product on space $C^{m \times n}$ over the field ℝ.

Theorem 1 (Wu et al., 2011): In the space $C^{m \times n}$ over the field ℝ, an inner product can be defined as

〈 A, B 〉 = Re [tr (A^{H} B)]

(2)

The Frobenius norm of A is denoted by $‖ A ‖$ , that is, $‖ A ‖ = \sqrt{tr (A^{H} A)}$ . The matrices $A, B \in$ $C^{m \times n}$ are called orthogonal if $〈 A, B 〉 = 0$

Lemma 1 (Li, 2017): For any $X \in$ $C^{n \times n}$ , then $X + J X^{H} J \in$ ${GHC}^{n \times n}$ where J is an $n \times n$ skew – Hermitian matrix.

Lemma 2 (Li, 2017): For any $X \in$ $C^{n \times n}$ , $P \in {GHC}^{n \times n}$ , then $〈 P, \frac{X + J X^{H} J}{2} 〉 = 〈 P, X 〉$ .

Main results

In this section, we propose an iterative solution to the generalized coupled Sylvester – conjugate matrix equation (1).

Algorithm 1:

Input matrices $J \in C^{n \times n}$ and $A_{ij}, C_{ij} \in C^{m \times n}$ , $B_{ij}, D_{ij} \in C^{n \times r}$ and $E_{i} \in C^{m \times r}$ .

Chosen arbitrary generalized Hamiltonian matrices $X_{j} (1) \in$ ${GHC}^{n \times n}$ .

For $i = 1, 2, \dots, p$ $, j = 1, 2, \dots, q$ , compute

\begin{matrix} R_{i} (1) = E_{i} - \sum_{j = 1}^{q} [A_{ij} X_{j} (1) B_{ij} + C_{ij} \bar{X_{j} (1)} D_{ij}] . \\ P_{j} (1) = \frac{1}{2} (\sum_{i = 1}^{p} [A_{ij}^{H} R_{i} (1) B_{ij}^{H} + {\bar{C_{ij}}}^{H} \bar{R_{i} (1)} {\bar{D_{ij}}}^{H}] \\ + \sum_{i = 1}^{p} J [B_{ij} R_{i}^{H} (1) A_{ij} + \bar{D_{ij}} {\bar{R_{i} (1)}}^{H} \bar{C_{ij}}] J); \\ k : = 1 . \end{matrix}

If $R_{i} (1) = 0$ , $i = 1, 2, \dots, p$ , then stop and $X_{j} (k)$ , $j = 1, 2, \dots, q$ are the solution; else let $k : = k + 1$ go to Step 5.

Compute

\begin{matrix} X_{j} (k + 1) = X_{j} (k) + \frac{\sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (k) ‖}^{2}} P_{j} (k);, j = 1, 2, \dots, q \\ R_{i} (k + 1) = E_{i} - \sum_{j = 1}^{q} [A_{ij} X_{j} (k + 1) B_{ij} + C_{ij} \bar{X_{j} (k + 1)} D_{ij}]; \\ = R_{i} (k) - \frac{\sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (k) ‖}^{2}} \sum_{j = 1}^{q} [A_{ij} P_{j} (k) B_{ij} + C_{ij} \bar{P_{j} (k)} D_{ij}]; \\ P_{j} (k + 1) = \frac{1}{2} (\sum_{i = 1}^{p} [A_{ij}^{H} R_{i} (k + 1) B_{ij}^{H} + {\bar{C_{ij}}}^{H} \bar{R_{i} (k + 1)} {\bar{D_{ij}}}^{H}] \\ + \sum_{i = 1}^{p} J [B_{ij} R_{i}^{H} (k + 1) A_{ij} + \bar{D_{ij}} {\bar{R_{i} (k + 1)}}^{H} \bar{C_{ij}}] J) \\ + \frac{\sum_{i = 1}^{p} {‖ R_{i} (k + 1) ‖}^{2}}{\sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2}} P_{j} (k) . \end{matrix}

If $R_{i} (k + 1) = 0$ , $i = 1, 2, \dots, p$ , then stop and $X_{j} (k)$ , $j = 1, 2, \dots, q$ are the solution; else let $k = k + 1$ go to Step 5.

To prove the convergence property of Algorithm 1, we first establish the following basic properties.

Lemma 3: Assume the sequences ${R_{i} (k)}$ and ${P_{j} (k)}$ are generated by Algorithm 1 for $i, j = 1, 2, \dots$ , we have

\begin{array}{l} \sum_{i = 1}^{p} 〈 R_{i} (k + 1), R_{i} (1) 〉 \\ = \sum_{i = 1}^{p} 〈 R_{i} (k), R_{i} (1) 〉 - \frac{\sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (k) ‖}^{2}} \sum_{j = 1}^{q} 〈 P_{j} (k), P_{j} (1) 〉 \end{array}

When $l > 1$

\begin{matrix} \sum_{i = 1}^{p} 〈 R_{i} (k + 1), R_{i} (l) 〉 = \sum_{i = 1}^{p} 〈 R_{i} (k), R_{i} (l) 〉 - \frac{\sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (k) ‖}^{2}} \sum_{j = 1}^{q} 〈 P_{j} (k), P_{j} (l) 〉 + \frac{\sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2} \sum_{i = 1}^{p} {‖ R_{i} (l) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (k) ‖}^{2} \sum_{i = 1}^{p} {‖ R_{i} (l - 1) ‖}^{2}} \sum_{j = 1}^{q} 〈 P_{j} (k), P_{j} (l - 1) 〉 \end{matrix}

Proof: From Algorithm 1, one has

\begin{matrix} \sum_{i = 1}^{p} 〈 R_{i} (k + 1), R_{i} (l) 〉 = \sum_{i = 1}^{p} Re {tr [R_{i}^{H} (k + 1), R_{i} (l)]} \\ = \sum_{i = 1}^{p} Re {tr [(R_{i} (k) - \frac{\sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (k) ‖}^{2}} \sum_{j = 1}^{q} [A_{ij} P_{j} (k) B_{ij} + C_{ij} \bar{P_{j} (k)} D_{ij}])^{H} R_{i} (l)]} \\ = \sum_{i = 1}^{p} Re {tr [R_{i}^{H} (k) R_{i} (l)]} - \frac{\sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (k) ‖}^{2}} \sum_{i = 1}^{p} Re {tr [\sum_{j = 1}^{q} [A_{ij} P_{j} (k) B_{ij} + C_{ij} \bar{P_{j} (k)} D_{ij}]^{H} R_{i} (l)]} \\ = \sum_{i = 1}^{p} Re {tr [R_{i}^{H} (k) R_{i} (l)]} - \frac{\sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (k) ‖}^{2}} \sum_{i = 1}^{p} Re {tr [\sum_{j = 1}^{q} B_{ij}^{H} P_{j}^{H} (k) A_{ij}^{H} R_{i} (l) + D_{ij}^{H} {\bar{P_{j} (k)}}^{H} C_{ij}^{H} R_{i} (l)]} \\ = \sum_{i = 1}^{p} Re {tr [R_{i}^{H} (k) R_{i} (l)]} - \frac{\sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (k) ‖}^{2}} \sum_{i = 1}^{p} Re {tr [\sum_{j = 1}^{q} P_{j}^{H} (k) A_{ij}^{H} R_{i} (l) B_{ij}^{H} + P_{j}^{H} (k) {\bar{C}}_{ij}^{H} \bar{R_{i}} (l) {\bar{D}}_{ij}^{H}]} \\ = \sum_{i = 1}^{p} Re {tr [R_{i}^{H} (k) R_{i} (l)]} - \frac{\sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (k) ‖}^{2}} \sum_{i = 1}^{p} Re {tr [\sum_{j = 1}^{q} P_{j}^{H} (k) (A_{ij}^{H} R_{i} (l) B_{ij}^{H} + {\bar{C}}_{ij}^{H} \bar{R_{i}} (l) {\bar{D}}_{ij}^{H})]} \\ = \sum_{i = 1}^{p} Re {tr [R_{i}^{H} (k) R_{i} (l)]} - \frac{\sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (k) ‖}^{2}} . \\ \sum_{j = 1}^{q} Re {tr [\sum_{i = 1}^{p} P_{j}^{H} (k) (\frac{1}{2} (A_{ij}^{H} R_{i} (l) B_{ij}^{H} + {\bar{C}}_{ij}^{H} \bar{R_{i}} (l) {\bar{D}}_{ij}^{H}) + \frac{1}{2} (J [B_{ij} R_{i}^{H} (l) A_{ij} + \bar{D_{ij}} {\bar{R_{i} (l)}}^{H} \bar{C_{ij}}] J))]} \\ = \sum_{i = 1}^{p} Re {tr [R_{i}^{H} (k) R_{i} (l)]} - \frac{\sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (k) ‖}^{2}} . \sum_{j = 1}^{q} Re {tr [P_{j}^{H} (k) (P_{j} (l) - \frac{\sum_{i = 1}^{p} {‖ R_{i} (l) ‖}^{2}}{\sum_{j = 1}^{q} {‖ R_{i} (l - 1) ‖}^{2}} P_{j} (l - 1))]} \\ = \sum_{i = 1}^{p} Re {tr [R_{i}^{H} (k) R_{i} (l)]} - \frac{\sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (k) ‖}^{2}} . \sum_{j = 1}^{q} Re {tr [P_{j}^{H} (k) P_{j} (l)]} \\ + \frac{\sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2} \sum_{i = 1}^{p} {‖ R_{i} (l) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (k) ‖}^{2} \sum_{j = 1}^{q} {‖ R_{i} (l - 1) ‖}^{2}} . \sum_{j = 1}^{q} Re {tr [P_{j}^{H} (k) P_{j} (l - 1)]} \\ \sum_{i = 1}^{p} 〈 R_{i} (k + 1), R_{i} (l) 〉 = \sum_{i = 1}^{p} 〈 R_{i} (k), R_{i} (l) 〉 - \frac{\sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (k) ‖}^{2}} \sum_{j = 1}^{q} 〈 P_{j} (k), P_{j} (l) 〉 + \frac{\sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2} \sum_{i = 1}^{p} {‖ R_{i} (l) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (k) ‖}^{2} \sum_{i = 1}^{p} {‖ R_{i} (l - 1) ‖}^{2}} \sum_{j = 1}^{q} 〈 P_{j} (k), P_{j} (l - 1) 〉 \end{matrix}

Using this proof, we have

\begin{matrix} \sum_{i = 1}^{p} 〈 R_{i} (k + 1), R_{i} (1) 〉 = \sum_{i = 1}^{p} Re {tr [R_{i}^{H} (k + 1) R_{i} (1)]} \\ = \sum_{i = 1}^{p} Re {tr [R_{i}^{H} (k) R_{i} (1)]} - \frac{\sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (k) ‖}^{2}} . \sum_{j = 1}^{q} Re {tr [\sum_{i = 1}^{p} P_{j}^{H} (k) (\frac{1}{2} (A_{ij}^{H} R_{i} (1) B_{ij}^{H} + {\bar{C}}_{ij}^{H} \bar{R_{i}} (1) {\bar{D}}_{ij}^{H}) + \frac{1}{2} (J [B_{ij} R_{i}^{H} (1) A_{ij} + \bar{D_{ij}} {\bar{R_{i} (1)}}^{H} \bar{C_{ij}}] J))]} \\ = \sum_{i = 1}^{p} Re {tr [R_{i}^{H} (k) R_{i} (1)]} - \frac{\sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (k) ‖}^{2}} . \sum_{j = 1}^{q} Re {tr [P_{j}^{H} (k) P_{j} (1)]} \\ \sum_{i = 1}^{p} 〈 R_{i} (k + 1), R_{i} (1) 〉 = \sum_{i = 1}^{p} 〈 R_{i} (k), R_{i} (1) 〉 - \frac{\sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (k) ‖}^{2}} \sum_{j = 1}^{q} 〈 P_{j} (k), P_{j} (1) 〉 \end{matrix}

Lemma 4: Assume the sequences ${R_{i} (k)}$ and ${P_{j} (k)}$ are generated by Algorithm 1 for $i, j = 1, 2, \dots$ , we have

\begin{array}{l} \sum_{j = 1}^{q} 〈 P_{j} (k + 1), P_{j} (l) 〉 = \frac{\sum_{i = 1}^{p} {‖ R_{i} (k + 1) ‖}^{2}}{\sum_{j = 1}^{q} {‖ R_{i} (k) ‖}^{2}} \\ \sum_{j = 1}^{q} 〈 P_{j} (k), P_{j} (l) 〉 + \frac{\sum_{j = 1}^{q} {‖ P_{j} (l) ‖}^{2}}{\sum_{i = 1}^{p} {‖ R_{i} (l) ‖}^{2}} \sum_{i = 1}^{p} 〈 R_{i} (k + 1), R_{i} (l) - R_{i} (l + 1) 〉 \end{array}

Proof: From Algorithm 1 and Lemma 2, one has

\begin{matrix} \sum_{j = 1}^{q} 〈 P_{j} (k + 1), P_{j} (l) 〉 = \sum_{j = 1}^{q} Re {tr [P_{j}^{H} (k + 1) P_{j} (l)]} \\ = \sum_{j = 1}^{q} Re {tr [(\frac{1}{2} (\sum_{i = 1}^{p} [A_{ij}^{H} R_{i} (k + 1) B_{ij}^{H} + {\bar{C_{ij}}}^{H} \bar{R_{i} (k + 1)} {\bar{D_{ij}}}^{H}] + \sum_{i = 1}^{p} J [B_{ij} R_{i}^{H} (k + 1) A_{ij} + \bar{D_{ij}} {\bar{R_{i} (k + 1)}}^{H} \bar{C_{ij}}] J) + \frac{\sum_{i = 1}^{p} {‖ R_{i} (k + 1) ‖}^{2}}{\sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2}} P_{j} (k))^{H} P_{j} (l)]} \\ = \frac{\sum_{i = 1}^{p} {‖ R_{i} (k + 1) ‖}^{2}}{\sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2}} \sum_{j = 1}^{q} Re {tr [P_{j}^{H} (k) P_{j} (l)]} + \sum_{j = 1}^{q} Re {tr [(\frac{1}{2} (\sum_{i = 1}^{p} [A_{ij}^{H} R_{i} (k + 1) B_{ij}^{H} + {\bar{C_{ij}}}^{H} \bar{R_{i} (k + 1)} {\bar{D_{ij}}}^{H}] \\ + \sum_{i = 1}^{p} J [B_{ij} R_{i}^{H} (k + 1) A_{ij} + \bar{D_{ij}} {\bar{R_{i} (k + 1)}}^{H} \bar{C_{ij}}] J))^{H} P_{j} (l)]} \\ = \frac{\sum_{i = 1}^{p} {‖ R_{i} (k + 1) ‖}^{2}}{\sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2}} \sum_{j = 1}^{q} Re {tr [P_{j}^{H} (k) P_{j} (l)]} + \sum_{j = 1}^{q} Re {tr [(\sum_{i = 1}^{p} [A_{ij}^{H} R_{i} (k + 1) B_{ij}^{H} + {\bar{C_{ij}}}^{H} \bar{R_{i} (k + 1)} {\bar{D_{ij}}}^{H}])^{H} P_{j} (l)]} \\ = \frac{\sum_{i = 1}^{p} {‖ R_{i} (k + 1) ‖}^{2}}{\sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2}} \sum_{j = 1}^{q} Re {tr [P_{j}^{H} (k) P_{j} (l)]} + \sum_{j = 1}^{q} Re {tr [\sum_{i = 1}^{p} B_{ij} R_{i}^{H} (k + 1) A_{ij} + \bar{D_{ij}} {\bar{R_{i} (k + 1)}}^{H} \bar{C_{ij}}] P_{j} (l)]} \\ = \frac{\sum_{i = 1}^{p} {‖ R_{i} (k + 1) ‖}^{2}}{\sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2}} \sum_{j = 1}^{q} Re {tr [P_{j}^{H} (k) P_{j} (l)]} + \sum_{j = 1}^{q} Re {tr [\sum_{i = 1}^{p} R_{i}^{H} (k + 1) A_{ij} P_{j} (l) B_{ij} + R_{i}^{H} (k + 1) C_{ij} \bar{P_{j} (l)} D_{ij}]} \\ = \frac{\sum_{i = 1}^{p} {‖ R_{i} (k + 1) ‖}^{2}}{\sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2}} \sum_{j = 1}^{q} Re {tr [P_{j}^{H} (k) P_{j} (l)]} + \sum_{i = 1}^{p} Re {tr [\sum_{j = 1}^{q} R_{i}^{H} (k + 1) [A_{ij} P_{j} (l) B_{ij} + C_{ij} \bar{P_{j} (l)} D_{ij}]]} \\ = \frac{\sum_{i = 1}^{p} {‖ R_{i} (k + 1) ‖}^{2}}{\sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2}} \sum_{j = 1}^{q} Re {tr [P_{j}^{H} (k) P_{j} (l)]} + \frac{\sum_{j = 1}^{q} {‖ P_{j} (l) ‖}^{2}}{\sum_{i = 1}^{p} {‖ R_{i} (l) ‖}^{2}} \sum_{i = 1}^{p} Re {tr [R_{i}^{H} (k + 1) [R_{i} (l) - R_{i} (l + 1)]]} \\ \sum_{j = 1}^{q} 〈 P_{j} (k + 1), P_{j} (l) 〉 = \frac{\sum_{i = 1}^{p} {‖ R_{i} (k + 1) ‖}^{2}}{\sum_{j = 1}^{q} {‖ R_{i} (k) ‖}^{2}} \sum_{j = 1}^{q} 〈 P_{j} (k), P_{j} (l) 〉 + \frac{\sum_{j = 1}^{q} {‖ P_{j} (l) ‖}^{2}}{\sum_{i = 1}^{p} {‖ R_{i} (l) ‖}^{2}} \sum_{i = 1}^{p} 〈 R_{i} (k + 1), R_{i} (l) - R_{i} (l + 1) 〉 \end{matrix}

Lemma 5: Assume the sequences ${R_{i} (k)}$ and ${P_{j} (k)}$ are generated by Algorithm 1 for $i, j = 1, 2, \dots$ , we have

\sum_{i = 1}^{p} 〈 R_{i} (k), R_{i} (l) 〉 = 0

(3)

and

\sum_{j = 1}^{q} 〈 P_{j} (k), P_{j} (l) 〉 = 0, for k, l = 1, 2, \dots, t, k \neq l

(4)

Proof: We apply mathematical induction.

Step 1: We prove

\sum_{i = 1}^{p} 〈 R_{i} (k + 1), R_{i} (k) 〉 = 0

(5)

and

\sum_{j = 1}^{q} 〈 P_{j} (k + 1), P_{j} (k) 〉 = 0

(6)

for $k, l = 1, 2, \dots, t, k \neq l .$

First, from Lemma 3 and Lemma 4 we have

\begin{matrix} \sum_{i = 1}^{p} 〈 R_{i} (2), R_{i} (1) 〉 \\ = \sum_{i = 1}^{p} 〈 R_{i} (1), R_{i} (1) 〉 - \frac{\sum_{i = 1}^{p} {‖ R_{i} (1) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (1) ‖}^{2}} \sum_{j = 1}^{q} 〈 P_{j} (1), P_{j} (1) 〉 \\ = \sum_{i = 1}^{p} {‖ R_{i} (1) ‖}^{2} - \sum_{i = 1}^{p} {‖ R_{i} (1) ‖}^{2} = 0 \end{matrix}

Thus, equation (5) is satisfied for $k = 1$ , and

\begin{matrix} \sum_{j = 1}^{q} 〈 P_{j} (2), P_{j} (1) 〉 \\ = \frac{\sum_{i = 1}^{p} {‖ R_{i} (2) ‖}^{2}}{\sum_{j = 1}^{q} {‖ R_{i} (1) ‖}^{2}} \sum_{j = 1}^{q} 〈 P_{j} (1), P_{j} (1) 〉 \\ + \frac{\sum_{j = 1}^{q} {‖ P_{j} (1) ‖}^{2}}{\sum_{i = 1}^{p} {‖ R_{i} (1) ‖}^{2}} \sum_{i = 1}^{p} 〈 R_{i} (2), R_{i} (1) - R_{i} (2) 〉 \\ = \frac{\sum_{i = 1}^{p} {‖ R_{i} (2) ‖}^{2}}{\sum_{j = 1}^{q} {‖ R_{i} (1) ‖}^{2}} \sum_{j = 1}^{q} 〈 P_{j} (1), P_{j} (1) 〉 \\ + \frac{\sum_{j = 1}^{q} {‖ P_{j} (1) ‖}^{2}}{\sum_{i = 1}^{p} {‖ R_{i} (1) ‖}^{2}} \sum_{i = 1}^{p} 〈 R_{i} (2), - R_{i} (2) 〉 \\ = \frac{\sum_{i = 1}^{p} {‖ R_{i} (2) ‖}^{2}}{\sum_{j = 1}^{q} {‖ R_{i} (1) ‖}^{2}} \sum_{j = 1}^{q} {‖ P_{j} (1) ‖}^{2} - \frac{\sum_{j = 1}^{q} {‖ P_{j} (1) ‖}^{2}}{\sum_{i = 1}^{p} {‖ R_{i} (1) ‖}^{2}} \sum_{i = 1}^{p} {‖ R_{i} (2) ‖}^{2} = 0 \end{matrix}

Thus, equation (6) is satisfied for $k = 1$ .

Now, assume equation (5) and equation (6) hold for $k = t - 1$ .

That is $\sum_{i = 1}^{p} 〈 R_{i} (t), R_{i} (t - 1) 〉 = 0$

and

\sum_{j = 1}^{q} 〈 P_{j} (t), P_{j} (t - 1) 〉 = 0

By Lemma 3 it follows that

\begin{matrix} \sum_{i = 1}^{p} 〈 R_{i} (t + 1), R_{i} (t) 〉 \\ = \sum_{i = 1}^{p} 〈 R_{i} (t), R_{i} (t) 〉 - \frac{\sum_{i = 1}^{p} {‖ R_{i} (t) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (t) ‖}^{2}} \sum_{j = 1}^{q} 〈 P_{j} (t), P_{j} (t) 〉 \\ + \frac{\sum_{i = 1}^{p} {‖ R_{i} (t) ‖}^{2} \sum_{i = 1}^{p} {‖ R_{i} (t) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (t) ‖}^{2} \sum_{i = 1}^{p} {‖ R_{i} (t - 1) ‖}^{2}} \sum_{j = 1}^{q} 〈 P_{j} (t), P_{j} (t - 1) 〉 \\ = \sum_{i = 1}^{p} {‖ R_{i} (t) ‖}^{2} - \frac{\sum_{i = 1}^{p} {‖ R_{i} (t) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (t) ‖}^{2}} \sum_{j = 1}^{q} {‖ P_{j} (t) ‖}^{2} = 0 \end{matrix}

Thus, equation (5) holds for $k = t$ .

Also, by Lemma 4 it follows that

\begin{matrix} \sum_{j = 1}^{q} 〈 P_{j} (t + 1), P_{j} (t) 〉 = \frac{\sum_{i = 1}^{p} {‖ R_{i} (t + 1) ‖}^{2}}{\sum_{j = 1}^{q} {‖ R_{i} (t) ‖}^{2}} \sum_{j = 1}^{q} 〈 P_{j} (t), P_{j} (t) 〉 \\ + \frac{\sum_{j = 1}^{q} {‖ P_{j} (t) ‖}^{2}}{\sum_{i = 1}^{p} {‖ R_{i} (t) ‖}^{2}} \sum_{i = 1}^{p} 〈 R_{i} (t + 1), R_{i} (t) - R_{i} (t + 1) 〉 \\ = \frac{\sum_{i = 1}^{p} {‖ R_{i} (t + 1) ‖}^{2}}{\sum_{j = 1}^{q} {‖ R_{i} (t) ‖}^{2}} \sum_{j = 1}^{q} {‖ P_{j} (t) ‖}^{2} \\ - \frac{\sum_{j = 1}^{q} {‖ P_{j} (t) ‖}^{2}}{\sum_{i = 1}^{p} {‖ R_{i} (t) ‖}^{2}} \sum_{i = 1}^{p} {‖ R_{i} (t + 1) ‖}^{2} = 0 \end{matrix}

Thus, equation (6) holds for $k = t$ .

Step 2: We want to show that

\sum_{i = 1}^{p} 〈 R_{i} (t + l), R_{i} (t) 〉 = 0

(7)

and

\sum_{j = 1}^{q} 〈 P_{j} (t + l), P_{j} (t) 〉 = 0

(8)

hold for integer $l \geq 1$ . We will prove this conclusion by induction. The case of $l = 1$ has been proven in Step 1. Now, we assume that (7) and (8) hold for $l \leq q, q \geq 1$ the aim is to show

\sum_{i = 1}^{p} 〈 R_{i} (t + q + 1), R_{i} (t) 〉 = 0

(9)

and

\sum_{j = 1}^{q} 〈 P_{j} (t + q + 1), P_{j} (t) 〉 = 0

(10)

First, we prove the following

\sum_{i = 1}^{p} 〈 R_{i} (q + 2), R_{i} (1) 〉 = 0

(11)

and

\sum_{j = 1}^{q} 〈 P_{j} (q + 2), P_{j} (1) 〉 = 0

(12)

By Lemma 3 and Lemma 4 we have

\begin{array}{l} \sum_{i = 1}^{p} 〈 R_{i} (q + 2), R_{i} (1) 〉 = \sum_{i = 1}^{p} 〈 R_{i} (q + 1), R_{i} (1) 〉 \\ - \frac{\sum_{i = 1}^{p} {‖ R_{i} (q + 1) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (q + 1) ‖}^{2}} \sum_{j = 1}^{q} 〈 P_{j} (q + 1), P_{j} (1) 〉 = 0 \end{array}

and

\begin{array}{l} \sum_{j = 1}^{q} 〈 P_{j} (q + 2), P_{j} (1) 〉 \\ = \frac{\sum_{i = 1}^{p} {‖ R_{i} (q + 2) ‖}^{2}}{\sum_{j = 1}^{q} {‖ R_{i} (q + 1) ‖}^{2}} \sum_{j = 1}^{q} 〈 P_{j} (q + 1), P_{j} (1) 〉 \\ + \frac{\sum_{j = 1}^{q} {‖ P_{j} (l) ‖}^{2}}{\sum_{i = 1}^{p} {‖ R_{i} (l) ‖}^{2}} \sum_{i = 1}^{p} 〈 R_{i} (q + 2), R_{i} (1) - R_{i} (2) 〉 = 0 \end{array}

This implies that (11) and (12) holds.

By Lemma 3 and Lemma 4 it follows that

\begin{matrix} \sum_{j = 1}^{q} 〈 P_{j} (t + q + 1), P_{j} (t) 〉 \\ = \frac{\sum_{i = 1}^{p} {‖ R_{i} (t + q + 1) ‖}^{2}}{\sum_{j = 1}^{q} {‖ R_{i} (t + q) ‖}^{2}} \sum_{j = 1}^{q} 〈 P_{j} (t + q), P_{j} (t) 〉 \\ + \frac{\sum_{j = 1}^{q} {‖ P_{j} (t) ‖}^{2}}{\sum_{i = 1}^{p} {‖ R_{i} (t) ‖}^{2}} \sum_{i = 1}^{p} 〈 R_{i} (t + q + 1), R_{i} (t) - R_{i} (t + 1) 〉 \\ = \frac{\sum_{j = 1}^{q} {‖ P_{j} (t) ‖}^{2}}{\sum_{i = 1}^{p} {‖ R_{i} (t) ‖}^{2}} \sum_{i = 1}^{p} 〈 R_{i} (t + q + 1), R_{i} (t) 〉 \end{matrix}

(13)

and

\begin{matrix} \sum_{i = 1}^{p} 〈 R_{i} (t + q + 1), R_{i} (t) 〉 \\ = \sum_{i = 1}^{p} 〈 R_{i} (t + q), R_{i} (t) 〉 - \frac{\sum_{i = 1}^{p} {‖ R_{i} (t + q) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (t + q) ‖}^{2}} \sum_{j = 1}^{q} 〈 P_{j} (t + q), P_{j} (t) 〉 \\ + \frac{\sum_{i = 1}^{p} {‖ R_{i} (t + q) ‖}^{2} \sum_{i = 1}^{p} {‖ R_{i} (t) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (t + q) ‖}^{2} \sum_{i = 1}^{p} {‖ R_{i} (t - 1) ‖}^{2}} \sum_{j = 1}^{q} 〈 P_{j} (t + q), P_{j} (t - 1) 〉 \\ = \frac{\sum_{i = 1}^{p} {‖ R_{i} (t + q) ‖}^{2} \sum_{i = 1}^{p} {‖ R_{i} (t) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (t + q) ‖}^{2} \sum_{i = 1}^{p} {‖ R_{i} (t - 1) ‖}^{2}} \sum_{j = 1}^{q} 〈 P_{j} (t + q), P_{j} (t - 1) 〉 \end{matrix}

(14)

Repeating (13) and (14), one can easily obtain for certain $α$ and $β$

\sum_{j = 1}^{q} 〈 P_{j} (t + q + 1), P_{j} (t) 〉 = α (\sum_{j = 1}^{q} 〈 P_{j} (q + 2), P_{j} (1) 〉)

and

\sum_{i = 1}^{p} 〈 R_{i} (t + q + 1), R_{i} (t) 〉 = β (\sum_{i = 1}^{p} 〈 R_{i} (q + 2), R_{i} (1) 〉)

This implies that (7) and (8) holds for $l = q + 1$ .

Thus, from Steps (1) and (2), the conclusion holds by the principle of induction.

Lemma 6: Suppose that the system of matrix equations (1) is consistent and let $X_{j}^{*} \in {GHC}^{n \times n}$ , $j = 1, 2, \dots, q$ be its generalized Hamiltonian solutions. Then for any initial matrices $X_{j} (1) \in {GHC}^{n \times n}$ , $j = 1, 2, \dots, q$ . We have

\sum_{j = 1}^{q} 〈 X_{j}^{*} - X_{j} (k), P_{j} (k) 〉 = \sum_{i = 1}^{p} {‖ R_{i} (k) ‖}^{2}

(15)

where the sequences ${X_{j} (k)},$ ${R_{i} (k)}$ and ${P_{j} (k)}$ are generated by Algorithm 1 for $i, j = 1, 2, \dots$ .

Proof: We apply mathematical induction.

For $i = 1$ , from Algorithm 1 one has

\begin{matrix} \sum_{j = 1}^{q} 〈 X_{j}^{*} - X_{j} (1), P_{j} (1) 〉 \\ = \sum_{j = 1}^{q} Re {tr [{(X_{j}^{*} - X_{j} (1))}^{H} P_{j} (1)]} \\ = \sum_{j = 1}^{q} Re {tr [{(X_{j}^{*} - X_{j} (1))}^{H} . \frac{1}{2} (\sum_{i = 1}^{p} [A_{ij}^{H} R_{i} (1) B_{ij}^{H} + {\bar{C_{ij}}}^{H} \bar{R_{i} (1)} {\bar{D_{ij}}}^{H}] \\ + \sum_{i = 1}^{p} J [B_{ij} R_{i}^{H} (1) A_{ij} + \bar{D_{ij}} {\bar{R_{i} (1)}}^{H} \bar{C_{ij}}] J)]} \\ = \sum_{j = 1}^{q} Re {tr [{(X_{j}^{*} - X_{j} (1))}^{H} . (\sum_{i = 1}^{p} [A_{ij}^{H} R_{i} (1) B_{ij}^{H} + {\bar{C_{ij}}}^{H} \bar{R_{i} (1)} {\bar{D_{ij}}}^{H}])]} \\ = \sum_{j = 1}^{q} Re {tr {[(\sum_{i = 1}^{p} [A_{ij}^{H} R_{i} (1) B_{ij}^{H} + {\bar{C_{ij}}}^{H} \bar{R_{i} (1)} {\bar{D_{ij}}}^{H}])}^{H} (X_{j}^{*} - X_{j} (1))]} \\ = \sum_{j = 1}^{q} Re {tr [\sum_{i = 1}^{p} [B_{ij} R_{i}^{H} (1) A_{ij} (X_{j}^{*} - X_{j} (1)) + \bar{D_{ij}} {\bar{R_{i} (1)}}^{H} \bar{C_{ij}} (X_{j}^{*} - X_{j} (1))]} \\ = \sum_{j = 1}^{q} Re {tr [\sum_{i = 1}^{p} [R_{i}^{H} (1) A_{ij} (X_{j}^{*} - X_{j} (1)) B_{ij} + R_{i}^{H} (1) C_{ij} \bar{(X_{j}^{*} - X_{j} (1))} D_{ij}]} \\ = \sum_{i = 1}^{p} Re {tr [R_{i}^{H} (1) (\sum_{j = 1}^{q} (A_{ij} (X_{j}^{*} - X_{j} (1)) B_{ij} + C_{ij} \bar{(X_{j}^{*} - X_{j} (1))} D_{ij}))]} \end{matrix}

In view that $X_{j}^{*}$ are solutions of the generalized coupled Sylvester – conjugate matrix equation (1), it is easy to obtain from the above relation

\begin{matrix} \sum_{j = 1}^{q} 〈 X_{j}^{*} - X_{j} (1), P_{j} (1) 〉 \\ = \sum_{i = 1}^{p} Re {tr [R_{i}^{H} (1) (E_{i} - \sum_{j = 1}^{q} [A_{ij} X_{j} (1) B_{ij} + C_{ij} \bar{X_{j} (1)} D_{ij}])]} \\ = \sum_{i = 1}^{p} Re {tr [R_{i}^{H} (1) R_{i} (1)]} = \sum_{i = 1}^{p} {‖ R_{i} (1) ‖}^{2} \end{matrix}

This implies that (15) holds for $k = 1$ .

Now, assume that (15) holds for $k = t$ . That is

\sum_{j = 1}^{q} 〈 X_{j}^{*} - X_{j} (t), P_{j} (t) 〉 = \sum_{i = 1}^{p} {‖ R_{i} (t) ‖}^{2}

Then we have to prove that the conclusion holds for $k = t + 1$ .

Since

\begin{matrix} \sum_{j = 1}^{q} 〈 X_{j}^{*} - X_{j} (t + 1), P_{j} (t) 〉 \\ = \sum_{j = 1}^{q} Re {tr [{(X_{j}^{*} - X_{j} (t + 1))}^{H} P_{j} (t)]} \\ = \sum_{j = 1}^{q} Re {tr [(X_{j}^{*} - X_{j} (t) - \frac{\sum_{i = 1}^{p} {‖ R_{i} (t) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (t) ‖}^{2}} P_{j} (t))^{H} P_{j} (t)]} \\ = \sum_{j = 1}^{q} Re {tr [(X_{j}^{*} - X_{j} (t)) P_{j} (t)]} - \frac{\sum_{i = 1}^{p} {‖ R_{i} (t) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (t) ‖}^{2}} \sum_{j = 1}^{q} Re {tr [P_{j}^{H} (t)) P_{j} (t)]} \\ = \sum_{j = 1}^{q} Re {tr [(X_{j}^{*} - X_{j} (t)) P_{j} (t)]} - \frac{\sum_{i = 1}^{p} {‖ R_{i} (t) ‖}^{2}}{\sum_{j = 1}^{q} {‖ P_{j} (t) ‖}^{2}} \sum_{j = 1}^{q} {‖ P_{j} (t) ‖}^{2} = 0 \end{matrix}

We get

\begin{matrix} \sum_{j = 1}^{q} 〈 X_{j}^{*} - X_{j} (t + 1), P_{j} (t + 1) 〉 = \sum_{j = 1}^{q} Re {tr [{(X_{j}^{*} - X_{j} (t + 1))}^{H} P_{j} (t + 1)]} \\ = \sum_{j = 1}^{q} Re {tr [{(X_{j}^{*} - X_{j} (t + 1))}^{H} ((\frac{1}{2} (\sum_{i = 1}^{p} [A_{ij}^{H} R_{i} (t + 1) B_{ij}^{H} + {\bar{C_{ij}}}^{H} \bar{R_{i} (t + 1)} {\bar{D_{ij}}}^{H}] \\ + \sum_{i = 1}^{p} J [B_{ij} R_{i}^{H} (t + 1) A_{ij} + \bar{D_{ij}} {\bar{R_{i} (t + 1)}}^{H} \bar{C_{ij}}] J) + \frac{\sum_{i = 1}^{p} {‖ R_{i} (t + 1) ‖}^{2}}{\sum_{i = 1}^{p} {‖ R_{i} (t) ‖}^{2}} P_{j} (t))]} \\ = \sum_{j = 1}^{q} Re {tr [{(X_{j}^{*} - X_{j} (t + 1))}^{H} (\sum_{i = 1}^{p} [A_{ij}^{H} R_{i} (k + 1) B_{ij}^{H} + {\bar{C_{ij}}}^{H} \bar{R_{i} (k + 1)} {\bar{D_{ij}}}^{H}])]} \\ = \sum_{j = 1}^{q} Re {tr [(\sum_{i = 1}^{p} [A_{ij}^{H} R_{i} (t + 1) B_{ij}^{H} + {\bar{C_{ij}}}^{H} \bar{R_{i} (t + 1)} {\bar{D_{ij}}}^{H} {])}^{H} (X_{j}^{*} - X_{j} (t + 1))]} \\ = \sum_{j = 1}^{q} Re {tr [\sum_{i = 1}^{p} [B_{ij} R_{i}^{H} (t + 1) A_{ij} (X_{j}^{*} - X_{j} (t + 1)) + \bar{D_{ij}} {\bar{R_{i} (t + 1)}}^{H} \bar{C_{ij}} (X_{j}^{*} - X_{j} (t + 1))]} \\ = \sum_{j = 1}^{q} Re {tr [\sum_{i = 1}^{p} [R_{i}^{H} (t + 1) A_{ij} (X_{j}^{*} - X_{j} (t + 1)) B_{ij} + R_{i}^{H} (t + 1) C_{ij} \bar{(X_{j}^{*} - X_{j} (t + 1))} D_{ij}]} \\ = \sum_{i = 1}^{p} Re {tr [R_{i}^{H} (t + 1) (\sum_{j = 1}^{q} [A_{ij} (X_{j}^{*} - X_{j} (t + 1)) B_{ij} + C_{ij} \bar{(X_{j}^{*} - X_{j} (t + 1))} D_{ij}])]} \end{matrix}

In view that $X_{j}^{*}$ are solutions of the generalized coupled Sylvester – conjugate matrix equation (1), it is easy to obtain from the above relation

\begin{matrix} \sum_{j = 1}^{q} 〈 X_{j}^{*} - X_{j} (t + 1), P_{j} (t + 1) 〉 \\ = \sum_{i = 1}^{p} Re {tr [R_{i}^{H} (t + 1) (E_{i} - \sum_{j = 1}^{q} [A_{ij} X_{j} (t + 1) B_{ij} + C_{ij} \bar{X_{j} (t + 1)} D_{ij}])]} \\ = \sum_{i = 1}^{p} Re {tr [R_{i}^{H} (t + 1) R_{i} (t + 1)]} = \sum_{i = 1}^{p} {‖ R_{i} (t + 1) ‖}^{2} \end{matrix}

This implies that (15) holds for $k = t + 1$ . Hence, relation (15) holds by the principle of induction.

Remark: Lemma 6 implies that if there exist a positive number j such that $P_{j} (k) = 0$ but, then the system of matrix equation (1) is inconsistent.

With the above two lemmas, one has the following theorem that guarantees the algorithm will converge after a finite number of iterations.

Theorem 2: If the system of matrix equation (1) is consistent, then a solution can be obtained within a finite number of iteration steps in the absence of round off errors by using Algorithm 1 for any arbitrary initial generalized Hamiltonian matrices $X_{j} (1) \in {GHC}^{n \times n}$ , $j = 1, 2, \dots, q$ .

Proof:

Let $\begin{array}{l} a R (k) = (\begin{matrix} R_{1} (k) & 0 & 0 & 0 \\ 0 & R_{2} (k) & 0 & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & R_{p} (k) \end{matrix}), \\ P (k) = (\begin{matrix} P_{1} (k) & 0 & 0 & 0 \\ 0 & P_{2} (k) & 0 & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & P_{q} (k) \end{matrix}) \end{array}$ ,

Suppose that $R (k) \neq 0$ for $k = 1, 2, 3, \dots, 2 pmr$ , we get $P (k) \neq 0$ from Lemma 6 and remark. Then we can compute $X_{j} (2 pmr + 1), R_{i} (2 pmr + 1)$ , $i = 1, 2, \dots, p$ $, j = 1, 2, \dots, q$ by Algorithm 1. Also, from Lemma 5, we have

〈 R (l), R (t) 〉 = 0 for l, t = 1, 2, 3, \dots, 2 pmr, t \neq l

So, the set of $R (1), R (2), \dots, R (2 pmr)$ is an orthogonal basis of the linear space $Ω$ of dimension $2 pmr + 1$ where $Ω = {U | U = diag (K_{1}, K_{2}, \dots, K_{p}) where K_{i} \in$ $C^{m \times r}$ }

This implies that

$R (2 pmr + 1) = 0$ , that is, $X_{j} (2 pmr + 1)$ $, j = 1, 2, \dots, q$ are the generalized Hamiltonian solution of system of matrix equation (1).

Numerical example

In this section, we present two numerical examples to illustrate the application of our proposed algorithm.

Example 1: Consider the generalized coupled Sylvester – conjugate matrix equations

\begin{matrix} A_{11} X_{1} B_{11} + C_{11} \bar{X_{1}} D_{11} + A_{12} X_{2} B_{12} = E_{1} \\ A_{21} X_{1} B_{21} + C_{21} \bar{X_{1}} D_{21} + A_{22} X_{2} B_{22} = E_{2} \end{matrix}}

(16)

where

\begin{matrix} A_{11} = [\begin{matrix} 3 + i & - i & - 3 i \\ 1 + 2 i & 4 + i & 0 \\ 2 i & 1 + i & 4 \end{matrix}], A_{21} = [\begin{matrix} 2 + 3 i & 1 + 2 i & 0 \\ - 1 + 3 i & 3 - i & 2 i \\ 4 & - 3 i & 0 \end{matrix}], \\ C_{11} = [\begin{matrix} - 1 - i & 2 & 5 \\ - 1 & 2 + i & 3 + i \\ 4 + i & 2 i & - 3 i \end{matrix}], C_{21} = [\begin{matrix} 2 i & - 3 & 0 \\ 3 + i & 0 & 2 \\ 1 + 2 i & 2 + i & 1 + 2 i \end{matrix}], \\ A_{12} = [\begin{matrix} 1 + 2 i & - i & 1 \\ 4 & 2 i & i \\ 3 - i & 3 + i & 1 + i \end{matrix}], A_{22} = [\begin{matrix} i & 1 - i & 1 + i \\ - 3 i & 3 & 3 i \\ 2 - i & 1 + i & 2 \end{matrix}], \\ B_{11} = [\begin{matrix} - 1 + i & 2 + i \\ 3 & - 2 i \\ 1 + i & 2 \end{matrix}], B_{21} = [\begin{matrix} 2 i & 4 \\ 5 i & 1 + 3 i \\ 2 - i & 0 \end{matrix}], \\ B_{12} = [\begin{matrix} 1 + 3 i & 2 - i \\ 1 + i & 1 + 2 i \\ 4 & 2 i \end{matrix}], B_{22} = [\begin{matrix} - 2 & 1 \\ 1 + i & 2 - 2 i \\ 0 & i \end{matrix}], \\ D_{11} = [\begin{matrix} 1 + i & 0 \\ 2 i & - 3 i \\ 3 + i & - 1 - i \end{matrix}], D_{21} = [\begin{matrix} 1 + 2 i & - i \\ 3 - i & 2 + i \\ 0 & 1 + i \end{matrix}], \\ E_{1} = [\begin{matrix} 27 + 17 i & 6 - 11 i \\ - 35 + 102 i & 24 + 10 i \\ 51 + 73 i & 6 + 26 i \end{matrix}], E_{2} = [\begin{matrix} 27 - 5 i & - 17 - 15 i \\ 9 + 21 i & 44 - 20 i \\ - 10 - 14 i & - 4 \end{matrix}] \end{matrix}

taking

X_{1} (1) = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}], X_{2} (1) = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] a n d J = [\begin{matrix} 0 & 0 & - 1 \\ 0 & - i & 0 \\ 1 & 0 & 0 \end{matrix}]

We apply Algorithm 1 to compute $X_{1}, X_{2}$ .

After iterating 23 steps, we obtain

X_{1} (23) = [\begin{matrix} - 1 & i & 2 \\ 0 & 0 & 1 \\ 2 & 0 & 1 \end{matrix}], X_{2} (23) = [\begin{matrix} 1 + i & 1 + 2 i & 0 \\ 1 - 2 i & 0 & 2 + i \\ 0 & 2 - i & - 1 + i \end{matrix}]

which satisfies the system of matrix equation (16).

The obtained results are presented in Figure 1, where

\begin{array}{l} r_{k} = ‖ R_{i} (k) ‖ (Residual) \\ δ_{k} = \frac{| | X_{j} (k) - X_{j}^{*} | |}{‖ X_{j}^{*} ‖} (Relative error) \end{array}

From Figure 1, it is clear that the error $δ_{k}$ is becoming smaller and approaches zero as iteration number k increases. This indicates that the proposed algorithm is effective and convergent.

Figure 1.

The residual and the relative error versus k (iteration number).

Example 2: Consider the generalized coupled Sylvester – conjugate matrix equations

\begin{matrix} A_{11} X_{1} B_{11} + C_{11} \bar{X_{1}} D_{11} + C_{12} \bar{X_{2}} D_{12} = E_{1} \\ C_{21} \bar{X_{1}} D_{21} + A_{22} X_{2} B_{22} + C_{22} \bar{X_{2}} D_{22} = E_{2} \end{matrix}}

(17)

where

\begin{matrix} A_{11} = [\begin{matrix} 1 & - i & 1 & - 3 i & i \\ 2 - i & 2 & i & 1 + i & 1 + 2 i \\ 2 i & 1 & 1 & 2 i & i \\ i & 0 & 2 i & 1 - i & - 2 \end{matrix}], B_{11} = [\begin{matrix} 1 & 2 i & i \\ 2 & 0 & 0 \\ 1 & 2 - i & i \\ i & - i & - 3 i \\ 1 + i & 1 - i & i \end{matrix}], C_{11} = [\begin{matrix} - i & 5 & - 1 & i & 0 \\ - 3 i & 0 & i & 1 + i & 1 \\ 2 - i & 1 + 2 i & 2 - i & 1 & i \\ 2 & 1 + i & - 3 i & - 2 i & 2 \end{matrix}], \\ D_{11} = [\begin{matrix} 1 & - 1 & 1 \\ 2 - i & - 3 i & i \\ 0 & i & 2 \\ 3 & 0 & 2 i \\ 2 & 1 & 1 - i \end{matrix}], C_{12} = [\begin{matrix} - 1 - i & 1 + 2 i & 0 & i & - i \\ 2 & 3 - i & 2 i & 2 & 0 \\ 2 + i & - 3 i & 1 & 0 & 2 - i \\ 0 & 0 & 2 - i & i & 2 i \end{matrix}], D_{12} = [\begin{matrix} 2 i & - 1 - i & 3 \\ 3 i & - 1 & 2 - i \\ - 2 & 4 + i & 1 + i \\ 0 & 2 & 2 i \\ 1 + i & 2 + i & i \end{matrix}], \\ C_{21} = [\begin{matrix} i & 1 & i & - i & 2 i \\ 2 & 1 & 0 & 2 & - 1 \\ 1 & 1 - i & - 5 i & 1 & 0 \\ - 1 & 2 - i & - i & 0 & - i \end{matrix}], D_{21} = [\begin{matrix} 1 + i & 2 i & - 1 + i \\ 3 & - 2 & 2 + i \\ 2 & 5 & 2 \\ - i & 3 + i & 0 \\ 1 + 3 i & 0 & - 2 i \end{matrix}], A_{22} = [\begin{matrix} i & 2 & 2 i & 1 - i & 3 \\ 0 & 2 - i & 0 & 2 & 0 \\ 2 & 2 & i & 0 & i \\ 1 & i & 1 & i & 2 - i \end{matrix}], \\ B_{22} = [\begin{matrix} 1 - i & 2 i & 1 - i \\ i & 3 i & 2 + i \\ 1 & 1 & - 2 \\ 2 i & 0 & - 3 i \\ 0 & - i & 0 \end{matrix}], C_{22} = [\begin{matrix} 2 - i & 1 & - i & - 1 & 2 i \\ 1 & 2 & 1 & 3 & 4 \\ 1 & i & 0 & 1 + i & 0 \\ i & 0 & - i & - 2 i & 2 - i \end{matrix}], D_{22} = [\begin{matrix} i & - i & 3 \\ 1 - i & 1 & i \\ 2 & - 2 & - i \\ 3 - i & 1 + i & 2 - i \\ 2 i & 0 & 0 \end{matrix}], \\ E_{1} = [\begin{matrix} - 5 - 3 i & 82 - i & 9 - 35 i \\ 43 + 40 i & 93 + 5 i & 48 + 31 i \\ 28 + 50 i & 20 + 29 i & 55 \\ 47 - 11 i & 17 + 26 i & 51 - 12 i \end{matrix}], E_{2} = [\begin{matrix} - 10 + 9 i & - 28 + 63 i & - 3 i \\ 35 - 9 i & - 8 - 3 i & 49 - 32 i \\ 55 + 27 i & 39 - 4 i & 7 - i \\ - 21 - 8 i & - 46 - 35 i & - 14 - 12 i \end{matrix}] \end{matrix}

Taking

\begin{array}{l} X_{1} (1) = [\begin{matrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{matrix}], X_{2} (1) = [\begin{matrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{matrix}] \\ a n d J = [\begin{matrix} 0 & 0 & 0 & 0 & - 1 \\ 0 & i & 0 & 0 & 0 \\ 0 & 0 & - i & 0 & 0 \\ 0 & 0 & 0 & - i & 0 \\ 1 & 0 & 0 & 0 & 0 \end{matrix}] \end{array}

We apply Algorithm 1 to compute $X_{1}, X_{2}$ .

After iterating 76 steps, we obtain

\begin{array}{l} X_{1} (76) = [\begin{matrix} 1 & 0 & 2 - i & i & 1 \\ 2 & - 2 i & i & - 2 & 1 + i \\ 1 - i & i & 1 & - i & - i \\ 1 & 4 & 1 + i & 1 & 1 - i \\ i & - 3 i & 1 & 3 & 2 + i \end{matrix}], \\ X_{2} (76) = [\begin{matrix} 1 & - i & - i & 1 - i & i \\ - i & 1 + i & 1 + i & 1 & 1 \\ i & 1 & 2 + i & i & 0 \\ 2 & 2 i & 0 & 2 & 0 \\ 0 & i & 1 & - 1 + 2 i & - 2 i \end{matrix}] \end{array}

which satisfies the system of matrix equation (17).

The obtained results are presented in Figure 2, where

r_{k} = ‖ R_{i} (k) ‖ (Residual)

From Figure 2, it is clear that the error is becoming smaller and approaches zero as iteration number k increases. This indicates that the proposed algorithm is convergent.

Figure 2.

The residual versus k (iteration number).

Conclusions

An iterative algorithm for solving the generalized coupled Sylvester – conjugate matrix equation (1) over Hamiltonian matrices is presented. We have proven that the iterative algorithms always converge to the solution for any arbitrary initial generalized Hamiltonian matrices $X_{j} (1), j = 1, 2, \dots, q$ . We stated and proved some lemmas and theorems where the solutions are obtained. The obtained results show that the methods are very neat and efficient. The proposed method is illustrated by numerical examples. The examples we tested using MATLAB to verify our theoretical results.’

Footnotes

Acknowledgements

The authors would like to express their heartfelt thanks to the editor and anonymous referees for their useful comments.

Declaration of conflicting interest

The authors declare that there is no conflict of interest.

Funding

This research received no specific grant from any funding agency in the public, commercial, or not-for-profit sectors.

ORCID iD

Ahmed ME Bayoumi

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Finite iterative Hamiltonian solutions of the generalized coupled Sylvester – conjugate matrix equations

Abstract

Keywords

Introduction

Preliminaries

Main results

Numerical example

Conclusions

Footnotes

Acknowledgements

Declaration of conflicting interest

Funding

ORCID iD

References