Left and Right Inverse Eigenpairs Problem for 𝜅-Hermitian Matrices

Abstract

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 𝐶^𝑛×𝑚 be the set of all 𝑛×𝑚 complex matrices, 𝑈𝐶^𝑛×𝑛, 𝐻𝐶^𝑛×𝑛, 𝑆𝐻𝐶^𝑛×𝑛 denote the set of all 𝑛×𝑛 unitary matrices, hermitian matrices, skew-hermitian matrices, respectively. Let 𝐴, 𝐴^𝐻, and 𝐴⁺ be the conjugate, conjugate transpose, and the Moore-Penrose generalized inverse of 𝐴, respectively. For 𝐴,𝐵∈𝐶^𝑛×𝑚, ⟨𝐴,𝐵⟩=re(tr(𝐵^𝐻𝐴)), where re(tr(𝐵^𝐻𝐴)) denotes the real part of tr(𝐵^𝐻𝐴), the inner product of matrices 𝐴 and 𝐵. The induced matrix norm is called Frobenius norm. That is, ‖𝐴‖=⟨𝐴,𝐴⟩^1/2=(tr(𝐴^𝐻𝐴))^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,𝑥_𝑖), 𝑖=1,…,ℎ; (𝜇_𝑗,𝑦_𝑗), 𝑗=1,…,𝑙, a special matrix set 𝑆, finding a matrix 𝐴∈𝑆 such that 𝐴𝑥_𝑖=𝜆_𝑖𝑥_𝑖𝑦,𝑖=1,…,ℎ,_𝑇𝑗𝐴=𝜇_𝑗𝑦^𝑇_𝑗,𝑗=1,…,𝑙. (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 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=𝐾^𝐻=𝐾, 𝐾²=𝐼_𝑛, a matrix 𝐴∈𝐶^𝑛×𝑛 is said to be 𝜅-hermitian matrices (skew 𝜅-hermitian matrices) if and only if 𝑎_𝑖𝑗=𝑎_{𝑘(𝑗)𝑘(𝑖)}(𝑎_𝑖𝑗=−𝑎_{𝑘(𝑗)𝑘(𝑖)}), 𝑖,𝑗=1,…,𝑛. We denote the set of 𝜅-hermitian matrices (skew 𝜅-hermitian matrices) by 𝐾𝐻𝐶^𝑛×𝑛(𝑆𝐾𝐻𝐶^𝑛×𝑛).

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].

From Definition 1, it is also easy to prove the following conclusions. (1)𝐴∈𝐾𝐻𝐶^𝑛×𝑛 if and only if 𝐴=𝐾𝐴^𝐻𝐾. (2)𝐴∈𝑆𝐾𝐻𝐶^𝑛×𝑛 if and only if 𝐴=−𝐾𝐴^𝐻𝐾. (3) If 𝐾 is a fixed permutation matrix, then 𝐾𝐻𝐶^𝑛×𝑛 and 𝑆𝐾𝐻𝐶^𝑛×𝑛 are the closed linear subspaces of 𝐶^𝑛×𝑛 and satisfy 𝐶^𝑛×𝑛=𝐾𝐻𝐶^𝑛×𝑛𝑆𝐾𝐻𝐶^𝑛×𝑛. (2)The notation 𝑉₁⊕𝑉₂ stands for the orthogonal direct sum of linear subspace 𝑉₁ and 𝑉₂. (4)𝐴∈𝐾𝐻𝐶^𝑛×𝑛 if and only if there is a matrix 𝐴∈𝐻𝐶^𝑛×𝑛 such that 𝐴=𝐾𝐴. (5)𝐴∈𝑆𝐾𝐻𝐶^𝑛×𝑛 if and only if there is a matrix 𝐴∈𝑆𝐻𝐶^𝑛×𝑛 such that 𝐴=𝐾𝐴.

Proof. (1) From Definition 1, if 𝐴=(𝑎_𝑖𝑗)∈𝐾𝐻𝐶^𝑛×𝑛, then 𝑎_𝑖𝑗=𝑎_{𝑘(𝑗)𝑘(𝑖)}, this implies 𝐴=𝐾𝐴^𝐻𝐾, for 𝐾𝐴^𝐻𝐾=(𝑎_{𝑘(𝑗)𝑘(𝑖)}).
(2) With the same method, we can prove (2). So, the proof is omitted.
(3) (a) For any 𝐴∈𝐶^𝑛×𝑛, there exist 𝐴₁∈𝐾𝐻𝐶^𝑛×𝑛, 𝐴₂∈𝑆𝐾𝐻𝐶^𝑛×𝑛 such that 𝐴=𝐴₁+𝐴₂, (3) where 𝐴₁=(1/2)(A+𝐾𝐴^𝐻𝐾), 𝐴₂=(1/2)(𝐴−𝐾𝐴^𝐻𝐾).  (b) If there exist another 𝐴₁∈𝐾𝐻𝐶^𝑛×𝑛, 𝐴₂∈𝑆𝐾𝐻𝐶^𝑛×𝑛 such that 𝐴=𝐴₁+𝐴₂, (4) (3)-(4) yields 𝐴₁−𝐴₁=−𝐴₂−𝐴₂. (5) Multiplying (5) on the left and on the right by 𝐾, respectively, and according to (1) and (2), we obtain 𝐴₁−𝐴₁=𝐴₂−𝐴₂. (6) Combining (5) and (6) gives 𝐴₁=𝐴₁, 𝐴₂=𝐴₂.  (c) For any 𝐴₁∈𝐾𝐻𝐶^𝑛×𝑛, 𝐴₂∈𝑆𝐾𝐻𝐶^𝑛×𝑛, we have 𝐴₁,𝐴₂=retr𝐴^𝐻₂𝐴₁=retr𝐾𝐴^𝐻₂𝐾𝐾𝐴₁𝐾=retr−𝐴^𝐻₂𝐴₁=−𝐴₁,𝐴₂. (7) This implies ⟨𝐴₁,𝐴₂⟩=0. Combining (a), (b), and (c) gives (3).
(4) Let 𝐴=𝐾𝐴, if 𝐴∈𝐾𝐻𝐶^𝑛×𝑛, then 𝐴^𝐻=𝐴∈𝐻𝐶^𝑛×𝑛. If 𝐴^𝐻=𝐴∈𝐻𝐶^𝑛×n, then 𝐴=𝐾𝐴 and 𝐾𝐴^𝐻𝐾=𝐾𝐴^𝐻𝐾𝐾=𝐾𝐴=𝐴∈𝐾𝐻𝐶^𝑛×𝑛.
(5) With the same method, we can prove (5). So, the proof is omitted.

In this paper, we suppose that 𝐾 is a fixed permutation matrix and assume (𝜆_𝑖,𝑥_𝑖), 𝑖=1,…,ℎ, be right eigenpairs of 𝐴; (𝜇_𝑗,𝑦_𝑗), 𝑗=1,…,𝑙, be left eigenpairs of 𝐴. If we let 𝑋=(𝑥₁,…,𝑥_ℎ)∈𝐶^𝑛×ℎ, Λ=diag(𝜆₁,…,𝜆_ℎ)∈𝐶^ℎ×ℎ; 𝑌=(𝑦₁,…,𝑦_𝑙)∈𝐶^𝑛×𝑙, Γ=diag(𝜇₁,…,𝜇_𝑙)∈𝐶^𝑙×𝑙, then the problems studied in this paper can be described as follows.

Problem 2. Giving 𝑋∈𝐶^𝑛×ℎ, Λ=diag(𝜆₁,…,𝜆_ℎ)∈𝐶^ℎ×ℎ; 𝑌∈𝐶^𝑛×𝑙, Γ=diag(𝜇₁,…,𝜇_𝑙)∈𝐶^𝑙×𝑙, find 𝐴∈𝐾𝐻𝐶^𝑛×𝑛 such that 𝑌𝐴𝑋=𝑋Λ,_𝑇𝐴=Γ𝑌^𝑇. (8)

Problem 3. Giving 𝐵∈𝐶^𝑛×𝑛, find 𝐴∈𝑆_𝐸 such that 𝐵−𝐴=min_{∀𝐴∈𝑆𝐸}‖𝐵−𝐴‖, (9) where 𝑆_𝐸 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 𝐾𝐻𝐶^𝑛×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 𝐾𝐻𝐶^𝑛×𝑛

Lemma 4. Denoting 𝑀=𝐾𝐸𝐾𝐺𝐸, and 𝐸∈𝐻𝐶^𝑛×𝑛, one has the following conclusions. (1)If 𝐺∈𝐾𝐻𝐶^𝑛×𝑛, then 𝑀∈𝐾𝐻𝐶^𝑛×𝑛. (2)If 𝐺∈𝑆𝐾𝐻𝐶^𝑛×𝑛, then 𝑀∈𝑆𝐾𝐻𝐶^𝑛×𝑛. (3)If 𝐺=𝐺₁+𝐺₂, where 𝐺₁∈𝐾𝐻𝐶^𝑛×𝑛, 𝐺₂∈𝑆𝐾𝐻𝐶^𝑛×𝑛, then 𝑀∈𝐾𝐻𝐶^𝑛×𝑛 if and only if 𝐾𝐸𝐾𝐺₂𝐸=0. In addition, one has 𝑀=𝐾𝐸𝐾𝐺₁𝐸.

Proof. (1) 𝐾𝑀^𝐻𝐾=𝐾𝐸𝐺^𝐻𝐾𝐸𝐾𝐾=𝐾𝐸(𝐾𝐺𝐾)𝐾𝐸=𝐾𝐸𝐾𝐺𝐸=𝑀.
Hence, we have 𝑀∈𝐾𝐻𝐶^𝑛×𝑛.
(2) 𝐾𝑀^𝐻𝐾=𝐾𝐸𝐺^𝐻𝐾𝐸𝐾𝐾=𝐾𝐸(−𝐾𝐺𝐾)𝐾𝐸=−𝐾𝐸𝐾𝐺𝐸=−𝑀.
Hence, we have 𝑀∈𝑆𝐾𝐻𝐶^𝑛×𝑛.
(3) 𝑀=𝐾𝐸𝐾(𝐺₁+𝐺₂)𝐸=𝐾𝐸𝐾𝐺₁𝐸+𝐾𝐸𝐾𝐺₂𝐸, we have 𝐾𝐸𝐾𝐺₁𝐸∈𝐾𝐻𝐶^𝑛×𝑛, 𝐾𝐸𝐾𝐺₂𝐸∈𝑆𝐾𝐻𝐶^𝑛×𝑛 from (1) and (2). If 𝑀∈𝐾𝐻𝐶^𝑛×𝑛, then 𝑀−𝐾𝐸𝐾𝐺₁𝐸∈𝐾𝐻𝐶^𝑛×𝑛, while 𝑀−𝐾𝐸𝐾𝐺₁𝐸=𝐾𝐸𝐾𝐺₂𝐸∈𝑆𝐾𝐻𝐶^𝑛×𝑛. Therefore from the conclusion (3) of Definition 1, we have 𝐾𝐸𝐾𝐺₂𝐸=0, that is, 𝑀=𝐾𝐸𝐾𝐺₁𝐸. On the contrary, if 𝐾𝐸𝐾𝐺₂𝐸=0, it is clear that 𝑀=𝐾𝐸𝐾𝐺₁𝐸∈𝐾𝐻𝐶^𝑛×𝑛. The proof is completed.

Lemma 5. Let 𝐴∈𝐾𝐻𝐶^𝑛×𝑛, if (𝜆,𝑥) is a right eigenpair of 𝐴, then (𝜆,𝐾𝑥) is a left eigenpair of 𝐴.

Proof. If (𝜆,𝑥) is a right eigenpair of 𝐴, then we have 𝐴𝑥=𝜆𝑥. (10) From the conclusion (1) of Definition 1, it follows that 𝐾𝐴^𝐻𝐾𝑥=𝜆𝑥. (11) This implies (𝐾𝑥)^𝑇𝐴=𝜆(𝐾𝑥)^𝑇. (12) So (𝜆,𝐾𝑥) is a left eigenpair of 𝐴.

From Lemma 5, without loss of the generality, we may assume that Problem 2 is as follows. 𝑋∈𝐶^𝑛×ℎ,Λ=diag𝜆₁,…,𝜆_ℎ∈𝐶^ℎ×ℎ,𝑌=𝐾𝑋∈𝐶^𝑛×ℎ,Γ=Λ∈𝐶^ℎ×ℎ. (13)

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

Lemma 6. If 𝑋, Λ, 𝑌, Γ are given by (13), then Problem 2 is equivalent to the following problem. If 𝑋, Λ, 𝑌, Γ are given by (13), find 𝐾𝐴∈𝐻𝐶^𝑛×𝑛 such that 𝐾𝐴𝑋=𝐾𝑋Λ. (14)

Lemma 7 (see [11]). If giving 𝑋∈𝐶^𝑛×ℎ, 𝐵∈𝐶^𝑛×ℎ, then matrix equation 𝐴𝑋=𝐵 has solution 𝐴∈𝐻𝐶^𝑛×𝑛 if and only if 𝐵=𝐵𝑋⁺𝑋,𝐵^𝐻𝑋=𝑋^𝐻𝐵. (15) Moreover, the general solution 𝐴 can be expressed as 𝐴=𝐵𝑋⁺+(𝐵𝑋⁺)^𝐻𝐼_𝑛−𝑋𝑋⁺+𝐼_𝑛−𝑋𝑋⁺𝐺𝐼_𝑛−𝑋𝑋⁺,∀𝐺∈𝐻𝐶^𝑛×𝑛. (16)

Theorem 8. If 𝑋, Λ, 𝑌, Γ are given by (13), then Problem 2 has a solution in 𝐾𝐻𝐶^𝑛×𝑛 if and only if 𝑋^𝐻𝐾𝑋Λ=Λ𝑋^𝐻𝐾𝑋,𝑋Λ=𝑋Λ𝑋⁺𝑋. (17) Moreover, the general solution can be expressed as 𝐴=𝐴₀+𝐾𝐸𝐾𝐺𝐸,∀𝐺∈𝐾𝐻𝐶^𝑛×𝑛, (18) where 𝐴₀=𝑋Λ𝑋⁺+𝐾(𝑋Λ𝑋⁺)^𝐻𝐾𝐸,𝐸=𝐼_𝑛−𝑋𝑋⁺. (19)

Proof. Necessity: If there is a matrix 𝐴∈𝐾𝐻𝐶^𝑛×𝑛 such that (𝐴𝑋=𝑋Λ, 𝑌^𝑇𝐴=Γ𝑌^𝑇), then from Lemma 6, there exists a matrix 𝐾𝐴∈𝐻𝐶^𝑛×𝑛 such that 𝐾𝐴𝑋=𝐾𝑋Λ, and according to Lemma 7, we have 𝐾𝑋Λ=𝐾𝑋Λ𝑋⁺𝑋,(𝐾𝑋Λ)^𝐻𝑋=𝑋^𝐻(𝐾𝑋Λ). (20) It is easy to see that (20) is equivalent to (17).
Sufficiency: If (17) holds, then (20) holds. Hence, matrix equation 𝐾𝐴𝑋=𝐾𝑋Λ has solution 𝐾𝐴∈𝐻𝐶^𝑛×𝑛. Moreover, the general solution can be expressed as follows: 𝐾𝐴=𝐾𝑋Λ𝑋⁺+(𝐾𝑋Λ𝑋⁺)^𝐻𝐼_𝑛−𝑋𝑋⁺+𝐼_𝑛−𝑋𝑋⁺𝐺𝐼_𝑛−𝑋𝑋⁺,∀𝐺∈𝐻𝐶^𝑛×𝑛. (21) Let 𝐴₀=𝑋Λ𝑋⁺+𝐾(𝑋Λ𝑋⁺)^𝐻𝐾𝐸,𝐸=𝐼_𝑛−𝑋𝑋⁺. (22) This implies 𝐴=𝐴₀+𝐾𝐸𝐺𝐸. Combining the definition of 𝐾, 𝐸 and the first equation of (17), we have 𝐾𝐴^𝐻₀𝐾=𝐾(𝑋Λ𝑋⁺)^𝐻𝐾+𝑋Λ𝑋⁺−𝐾(𝑋𝑋⁺)^𝐻𝐾𝑋Λ𝑋⁺=𝑋Λ𝑋⁺+𝐾𝑋Λ𝑋^+𝐻𝐾−𝐾𝑋Λ𝑋^+𝑇𝐾𝑋𝑋⁺=𝐴₀. (23) Hence, 𝐴₀∈𝐾𝐻𝐶^𝑛×𝑛. Combining the definition of 𝐾, 𝐸, (13) and (17), we have 𝐴₀𝑋=𝑋Λ𝑋⁺𝑋+𝐾𝑋Λ𝑋^+𝐻𝐾𝐼_𝑛−𝑋𝑋⁺𝑌𝑋=𝑋Λ,_𝑇𝐴₀=𝑋^𝐻𝐾𝑋Λ𝑋⁺+𝑋^𝐻𝐾𝐾(𝑋Λ𝑋⁺)^𝐻=𝐾𝐸Λ𝑋^𝐻𝐾𝑋𝑋⁺+(𝐾𝑋Λ𝑋⁺𝑋)^𝐻𝐼_𝑛−𝑋𝑋⁺=Λ𝑋^𝐻𝐾𝑋𝑋⁺+Λ𝑋^𝐻𝐾𝐼_𝑛−𝑋𝑋⁺=Λ𝑋^𝐻𝐾=Γ𝑌^𝑇. (24) Therefore, 𝐴₀ is a special solution of Problem 2. Combining the conclusion (4) of Definition 1, Lemma 4, and 𝐸=𝐼_𝑛−𝑋𝑋⁺∈𝐻𝐶^𝑛×𝑛, it is easy to prove that 𝐴=𝐴₀+𝐾𝐸𝐾𝐺𝐸∈𝐾𝐻𝐶^𝑛×𝑛 if and only if 𝐺∈𝐾𝐻𝐶^𝑛×𝑛. 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 𝑆_𝐸 of Problem 2 is a nonempty closed convex set if Problem 2 has a solution in 𝐾𝐻𝐶^𝑛×𝑛. We claim that for any given 𝐵∈𝑅^𝑛×𝑛, there exists a unique optimal approximation for Problem 3.

Theorem 9. Giving 𝐵∈𝐶^𝑛×𝑛, if the conditions of 𝑋, 𝑌, Λ, Γ are the same as those in Theorem 8, then Problem 3 has a unique solution 𝐴∈𝑆_𝐸. Moreover, 𝐴 can be expressed as 𝐴=𝐴₀+𝐾𝐸𝐾𝐵₁𝐸, (25) where 𝐴₀, 𝐸 are given by (19) and 𝐵₁=(1/2)(𝐵+𝐾𝐵^𝐻𝐾).

Proof. Denoting 𝐸₁=𝐼_𝑛−𝐸, it is easy to prove that matrices 𝐸 and 𝐸₁ are orthogonal projection matrices satisfying 𝐸𝐸₁=0. It is clear that matrices 𝐾𝐸𝐾 and 𝐾𝐸₁𝐾 are also orthogonal projection matrices satisfying (𝐾𝐸𝐾)(𝐾𝐸₁𝐾)=0. According to the conclusion (3) of Definition 1, for any 𝐵∈𝐶^𝑛×𝑛, there exists unique 𝐵₁∈𝐾𝐻𝐶^𝑛×𝑛,𝐵₂∈𝑆𝐾𝐻𝐶^𝑛×𝑛 (26) such that 𝐵=𝐵₁+𝐵₂,𝐵₁,𝐵₂=0, (27) where 𝐵₁=12𝐵+𝐾𝐵^𝐻𝐾,𝐵₂=12𝐵−𝐾𝐵^𝐻𝐾. (28) Combining Theorem 8, for any 𝐴∈𝑆_𝐸, we have ‖𝐵−𝐴‖²=𝐵−𝐴₀−𝐾𝐸𝐾𝐺𝐸²=𝐵₁+𝐵₂−𝐴₀−𝐾𝐸𝐾𝐺𝐸²=𝐵₁−𝐴₀−𝐾𝐸𝐾𝐺𝐸²+𝐵₂²=𝐵₁−𝐴₀𝐸+𝐸₁−𝐾𝐸𝐾𝐺𝐸²+𝐵₂²=𝐵₁−𝐴₀𝐸−𝐾𝐸𝐾𝐺𝐸²+𝐵₁−𝐴₀𝐸₁²+𝐵₂²=𝐾𝐸+𝐸₁𝐾𝐵₁−𝐴₀𝐸−𝐾𝐸𝐾𝐺𝐸²+𝐵₁−𝐴₀𝐸₁²+𝐵₂²=𝐾𝐸𝐾𝐵₁−𝐴₀𝐸−𝐾𝐸𝐾𝐺𝐸²+𝐾𝐸₁𝐾𝐵₁−𝐴₀𝐸²+(𝐵₁−𝐴₀)𝐸₁²+𝐵₂². (29) It is easy to prove that 𝐾𝐸𝐾𝐴₀𝐸=0 according to the definitions of 𝐴₀, 𝐸. So we have ‖𝐵−𝐴‖²=𝐾𝐸𝐾𝐵₁𝐸−𝐾𝐸𝐾𝐺𝐸²+𝐾𝐸₁𝐾𝐵₁−𝐴₀𝐸²+(𝐵₁−𝐴₀)𝐸₁²+𝐵₂². (30) Obviously, min_{𝐴∈𝑆𝐸}‖𝐵−𝐴‖ is equivalent to min_{𝐺∈𝐾𝐻𝐶𝑛×𝑛}𝐾𝐸𝐾𝐵₁𝐸−𝐾𝐸𝐾𝐺𝐸. (31) Since 𝐸𝐸₁=0,(𝐾𝐸𝐾)(𝐾𝐸₁𝐾)=0, it is clear that 𝐺=𝐵₁+𝐾𝐸₁𝐾𝐺𝐸₁, for any 𝐺∈𝐾𝐻𝐶^𝑛×𝑛, is a solution of (31). Substituting this result to (18), we can obtain (25).

Algorithm 10. (1) Input 𝑋, Λ, 𝑌, Γ according to (13). (2) Compute 𝑋^𝐻𝐾𝑋Λ, Λ𝑋^𝐻𝐾𝑋, 𝑋Λ𝑋⁺𝑋, 𝑋Λ, if (17) holds, then continue; otherwise stop. (3) Compute 𝐴₀ according to (19), and compute 𝐵₁ according to (28). (4) According to (25) calculate 𝐴.

Example 11 (𝑛=8, ℎ=𝑙=4).  ,,,𝑋=0.5661−0.2014−0.1422𝑖0.1446+0.2138𝑖0.524−0.2627+0.1875𝑖0.5336−0.2110−0.4370𝑖−0.0897+0.3467𝑖−0.4132+0.2409𝑖0.0226−0.0271𝑖−0.1095+0.2115𝑖−0.3531−0.0642𝑖−0.0306+0.2109𝑖−0.3887−0.0425𝑖0.2531+0.2542𝑖0.0094+0.2991𝑖0.0842−0.1778𝑖−0.0004−0.3733𝑖0.3228−0.1113𝑖0.1669+0.1952𝑖0.0139−0.3757𝑖−0.2363+0.3856𝑖0.2583+0.0721𝑖0.1841−0.2202𝑖0.0460+0.3276𝑖−0.1114+0.0654𝑖−0.0521−0.2556𝑖−0.2351+0.3002𝑖0.0085−0.1079𝑖0.0974+0.3610𝑖0.5060−0.2901−0.0268𝑖Λ=−0.3967−0.4050𝑖0000−0.3967+0.4050𝑖00000.00010000−0.0001𝑖𝐾=0000000100000010000001000001000000001000001000000100000010000000𝑌=𝐾𝑋,Γ=Λ (32)
𝐵=
From the first column to the fourth column−0.5218+0.0406𝑖0.2267−0.0560𝑖−0.1202+0.0820𝑖−0.0072−0.3362𝑖0.3909−0.3288𝑖0.2823−0.2064𝑖−0.0438−0.0403𝑖0.2707+0.0547𝑖0.2162−0.1144𝑖−0.4307+0.2474𝑖−0.0010−0.0412𝑖0.2164−0.1314𝑖−0.1872−0.0599𝑖−0.0061+0.4698𝑖0.3605−0.0247𝑖0.4251+0.1869𝑖−0.1227−0.0194𝑖0.2477−0.0606𝑖0.3918+0.6340𝑖0.1226+0.0636𝑖−0.0893+0.4335𝑖0.0662+0.0199𝑖−0.0177−0.1412𝑖0.4047+0.2288𝑖0.1040−0.2015𝑖0.1840+0.2276𝑖0.2681−0.3526𝑖−0.5252+0.1022𝑖0.1808+0.2669𝑖0.2264+0.3860𝑖−0.1791+0.1976𝑖−0.0961−0.0117𝑖 (33)From the fifth column to the eighth column.−0.2638−0.4952𝑖−0.0863−0.1664𝑖0.2687+0.1958𝑖−0.2544−0.1099𝑖−0.2741−0.1656𝑖−0.0227+0.2684𝑖0.1846+0.2456𝑖−0.0298+0.5163𝑖−0.1495−0.3205𝑖0.1391+0.2434𝑖0.1942−0.5211𝑖−0.3052−0.1468𝑖−0.2554+0.2690𝑖−0.4222−0.1080𝑖0.2232+0.0774𝑖0.0965−0.0421𝑖0.3856−0.0619𝑖0.1217−0.0270𝑖0.1106−0.3090𝑖−0.1122+0.2379𝑖−0.1130+0.0766𝑖0.7102−0.0901𝑖0.1017+0.1397𝑖−0.0445+0.0038𝑖0.1216+0.0076𝑖0.2343−0.1772𝑖0.5242−0.0089𝑖−0.0613+0.0258𝑖−0.0750−0.3581𝑖0.0125+0.0964𝑖0.0779−0.1074𝑖0.6735−0.0266𝑖 (34)It is easy to see that matrices 𝑋, Λ, 𝑌, Γ satisfy (17). Hence, there exists the unique solution for Problem 3. Using the software “MATLAB”, we obtain the unique solution 𝐴 of Problem 3.
From the first column to the fourth column−0.1983+0.0491𝑖0.1648+0.0032𝑖0.0002+0.1065𝑖0.1308+0.2690𝑖0.1071−0.2992𝑖0.2106+0.2381𝑖−0.0533−0.3856𝑖−0.0946+0.0488𝑖0.1935−0.1724𝑖−0.0855−0.0370𝑖0.0200−0.0665𝑖−0.2155−0.0636𝑖0.0085−0.2373𝑖−0.0843−0.1920𝑖0.0136+0.0382𝑖−0.0328−0.0000𝑖−0.0529+0.1703𝑖0.1948−0.0719𝑖0.1266+0.1752𝑖0.0232−0.2351𝑖0.0855+0.1065𝑖0.0325−0.2068𝑖0.2624+0.0000𝑖0.0136−0.0382𝑖0.1283−0.1463𝑖−0.0467+0.0000𝑖0.0325+0.2067𝑖−0.0843+0.1920𝑖0.2498+0.0000𝑖0.1283+0.1463𝑖0.0855−0.1065𝑖0.0086+0.2373𝑖 (35)From the fifth column to the eighth column.0.2399−0.1019𝑖0.1928−0.1488𝑖−0.3480−0.2574𝑖0.1017−0.0000𝑖−0.1955+0.0644𝑖0.2925+0.1872𝑖0.3869+0.0000𝑖−0.3481+0.2574𝑖0.0074+0.0339𝑖−0.3132−0.0000𝑖0.2926−0.1872𝑖0.1928+0.1488𝑖0.0232+0.2351𝑖−0.2154+0.0636𝑖−0.0946−0.0489𝑖0.1309−0.2691𝑖−0.0545−0.0000𝑖0.0074−0.0339𝑖−0.1955−0.0643𝑖0.2399+0.1019𝑖0.1266−0.1752𝑖0.0200+0.0665𝑖−0.0533+0.3857𝑖0.0002−0.1065𝑖0.1949+0.0719𝑖−0.0855+0.0370𝑖0.2106−0.2381𝑖0.1648−0.0032𝑖−0.0529−0.1703𝑖0.1935+0.1724𝑖0.1071+0.2992𝑖−0.1983−0.0491𝑖 (36)

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.