An Efficient Algorithm for the Reflexive Solution of the Quaternion Matrix Equation 𝐴𝑋𝐵+𝐶𝑋^𝐻𝐷=𝐹

Abstract

We propose an iterative algorithm for solving the reflexive solution of the quaternion matrix equation 𝐴𝑋𝐵+𝐶𝑋^𝐻𝐷=𝐹. When the matrix equation is consistent over reflexive matrix 𝑋, a reflexive solution can be obtained within finite iteration steps in the absence of roundoff errors. By the proposed iterative algorithm, the least Frobenius norm reflexive solution of the matrix equation can be derived when an appropriate initial iterative matrix is chosen. Furthermore, the optimal approximate reflexive solution to a given reflexive matrix 𝑋₀ can be derived by finding the least Frobenius norm reflexive solution of a new corresponding quaternion matrix equation. Finally, two numerical examples are given to illustrate the efficiency of the proposed methods.

1. Introduction

Throughout the paper, the notations ℝ^𝑚×𝑛 and ℍ^𝑚×𝑛 represent the set of all 𝑚×𝑛 real matrices and the set of all 𝑚×𝑛 matrices over the quaternion algebra ℍ={𝑎₁+𝑎₂𝑖+𝑎₃𝑗+𝑎₄𝑘∣𝑖²=𝑗²=𝑘²=𝑖𝑗𝑘=−1,𝑎₁,𝑎₂,𝑎₃,𝑎₄∈ℝ}. We denote the identity matrix with the appropriate size by 𝐼. We denote the conjugate transpose, the transpose, the conjugate, the trace, the column space, the real part, the 𝑚𝑛×1 vector formed by the vertical concatenation of the respective columns of a matrix 𝐴 by 𝐴^𝐻,𝐴^𝑇,𝐴,tr(𝐴),𝑅(𝐴),Re(𝐴),andvec(𝐴), respectively. The Frobenius norm of 𝐴 is denoted by ‖𝐴‖, that is, ‖𝐴‖=tr(𝐴^𝐻𝐴). Moreover, 𝐴⊗𝐵 and 𝐴⊙𝐵 stand for the Kronecker matrix product and Hadmard matrix product of the matrices 𝐴 and 𝐵.

Let 𝑄∈ℍ^𝑛×𝑛 be a generalized reflection matrix, that is, 𝑄²=𝐼 and 𝑄^𝐻=𝑄. A matrix 𝐴 is called reflexive with respect to the generalized reflection matrix 𝑄, if 𝐴=𝑄𝐴𝑄. It is obvious that any matrix is reflexive with respect to 𝐼. Let ℝℍ^𝑛×𝑛(𝑄) denote the set of order 𝑛 reflexive matrices with respect to 𝑄. The reflexive matrices with respect to a generalized reflection matrix 𝑄 have been widely used in engineering and scientific computations [1, 2].

In the field of matrix algebra, quaternion matrix equations have received much attention. Wang et al. [3] gave necessary and sufficient conditions for the existence and the representations of P-symmetric and P-skew-symmetric solutions of quaternion matrix equations 𝐴_𝑎𝑋=𝐶_𝑎 and 𝐴_𝑏𝑋𝐵_𝑏=𝐶_𝑏. Yuan and Wang [4] derived the expressions of the least squares 𝜂-Hermitian solution with the least norm and the expressions of the least squares anti-𝜂-Hermitian solution with the least norm for the quaternion matrix equation 𝐴𝑋𝐵+𝐶𝑋𝐷=𝐸. Jiang and Wei [5] derived the explicit solution of the quaternion matrix equation 𝑋−𝐴𝑋𝐵=𝐶. Li and Wu [6] gave the expressions of symmetric and skew-antisymmetric solutions of the quaternion matrix equations 𝐴₁𝑋=𝐶₁ and 𝑋𝐵₃=𝐶₃. Feng and Cheng [7] gave a clear description of the solution set of the quaternion matrix equation 𝐴𝑋−𝑋𝐵=0.

The iterative method is a very important method to solve matrix equations. Peng [8] constructed a finite iteration method to solve the least squares symmetric solutions of linear matrix equation 𝐴𝑋𝐵=𝐶. Also Peng [9–11] presented several efficient iteration methods to solve the constrained least squares solutions of linear matrix equations 𝐴𝑋𝐵=𝐶 and 𝐴𝑋𝐵+𝐶𝑌𝐷=𝐸, by using Paige’s algorithm [12] as the frame method. Duan et al. [13–17] proposed iterative algorithms for the (Hermitian) positive definite solutions of some nonlinear matrix equations. Ding et al. proposed the hierarchical gradient-based iterative algorithms [18] and hierarchical least squares iterative algorithms [19] for solving general (coupled) matrix equations, based on the hierarchical identification principle [20]. Wang et al. [21] proposed an iterative method for the least squares minimum-norm symmetric solution of 𝐴X𝐵=𝐸. Dehghan and Hajarian constructed finite iterative algorithms to solve several linear matrix equations over (anti)reflexive [22–24], generalized centrosymmetric [25, 26], and generalized bisymmetric [27, 28] matrices. Recently, Wu et al. [29–31] proposed iterative algorithms for solving various complex matrix equations.

However, to the best of our knowledge, there has been little information on iterative methods for finding a solution of a quaternion matrix equation. Due to the noncommutative multiplication of quaternions, the study of quaternion matrix equations is more complex than that of real and complex equations. Motivated by the work mentioned above and keeping the interests and wide applications of quaternion matrices in view (e.g., [32–45]), we, in this paper, consider an iterative algorithm for the following two problems.

Problem 1. For given matrices 𝐴,𝐶∈ℍ^𝑚×𝑛,𝐵,𝐷∈ℍ^𝑛×𝑝,𝐹∈ℍ^𝑚×𝑝 and the generalized reflection matrix 𝑄, find 𝑋∈ℝℍ^𝑛×𝑛(𝑄), such that 𝐴𝑋𝐵+𝐶𝑋^𝐻𝐷=𝐹. (1)

Problem 2. When Problem 1 is consistent, let its solution set be denoted by 𝑆_𝐻. For a given reflexive matrix 𝑋₀∈ℝℍ^𝑛×𝑛(𝑄), find 𝑋̆∈ℝℍ^𝑛×𝑛(𝑄), such that 𝑋̆−𝑋₀=min_{𝑋∈𝑆𝐻}𝑋−𝑋₀. (2)

The remainder of this paper is organized as follows. In Section 2, we give some preliminaries. In Section 3, we introduce an iterative algorithm for solving Problem 1. Then we prove that the given algorithm can be used to obtain a reflexive solution for any initial matrix within finite steps in the absence of roundoff errors. Also we prove that the least Frobenius norm reflexive solution can be obtained by choosing a special kind of initial matrix. In addition, the optimal reflexive solution of Problem 2 by finding the least Frobenius norm reflexive solution of a new matrix equation is given. In Section 4, we give two numerical examples to illustrate our results. In Section 5, we give some conclusions to end this paper.

2. Preliminary

In this section, we provide some results which will play important roles in this paper. First, we give a real inner product for the space ℍ^𝑚×𝑛 over the real field ℝ.

Theorem 3. In the space ℍ^𝑚×𝑛 over the field ℝ, a real inner product can be defined as ⟨𝐴,𝐵⟩=Retr𝐵^𝐻𝐴 (3) for 𝐴,𝐵∈ℍ^𝑚×𝑛. This real inner product space is denoted as (ℍ^𝑚×𝑛,ℝ,⟨⋅,⋅⟩).

Proof. (1) For 𝐴∈ℍ^𝑚×𝑛, let 𝐴=𝐴₁+𝐴₂𝑖+𝐴₃𝑗+𝐴₄𝑘, then ⟨𝐴,𝐴⟩=Retr𝐴^𝐻𝐴=tr𝐴^𝑇₁𝐴₁+𝐴^𝑇₂𝐴₂+𝐴^𝑇₃𝐴₃+𝐴^𝑇₄𝐴₄. (4)
It is obvious that ⟨𝐴,𝐴⟩>0 and ⟨𝐴,𝐴⟩=0⇔𝐴=0.
(2) For 𝐴,𝐵∈ℍ^𝑚×𝑛, let 𝐴=𝐴₁+𝐴₂𝑖+𝐴₃𝑗+𝐴₄𝑘 and 𝐵=𝐵₁+𝐵₂𝑖+𝐵₃𝑗+𝐵₄𝑘, then we have ⟨𝐴,𝐵⟩=Retr𝐵^𝐻𝐴=tr𝐵^𝑇₁𝐴₁+𝐵^𝑇₂𝐴₂+𝐵^𝑇₃𝐴₃+𝐵^𝑇₄𝐴₄=tr𝐴^𝑇₁𝐵₁+𝐴^𝑇₂𝐵₂+𝐴^𝑇₃𝐵₃+𝐴^𝑇₄𝐵₄=Retr𝐴^𝐻𝐵=⟨𝐵,𝐴⟩. (5)
(3) For 𝐴,𝐵,𝐶∈ℍ^𝑚×𝑛⟨𝐴+𝐵,𝐶⟩=Retr𝐶^𝐻(𝐶𝐴+𝐵)=Retr^𝐻𝐴+𝐶^𝐻𝐵=Retr𝐶^𝐻𝐴+Retr𝐶^𝐻𝐵=⟨𝐴,𝐶⟩+⟨𝐵,𝐶⟩. (6)
(4) For 𝐴,𝐵∈ℍ^𝑚×𝑛 and 𝑎∈ℝ, ⟨𝑎𝐴,𝐵⟩=Retr𝐵^𝐻(𝑎𝐴)=Retr𝑎𝐵^𝐻𝐴=𝑎Retr𝐵^𝐻𝐴=𝑎⟨𝐴,𝐵⟩. (7)
All the above arguments reveal that the space ℍ^𝑚×𝑛 over field ℝ with the inner product defined in (3) is an inner product space.

Let ‖⋅‖_Δ represent the matrix norm induced by the inner product ⟨⋅,⋅⟩. For an arbitrary quaternion matrix 𝐴∈ℍ^𝑚×𝑛, it is obvious that the following equalities hold: ‖𝐴‖_Δ=⟨𝐴,𝐴⟩=Retr𝐴^𝐻𝐴=tr𝐴^𝐻𝐴=‖𝐴‖, (8) which reveals that the induced matrix norm is exactly the Frobenius norm. For convenience, we still use ‖⋅‖ to denote the induced matrix norm.

Let 𝐸_𝑖𝑗 denote the 𝑚×𝑛 quaternion matrix whose (𝑖,𝑗) entry is 1, and the other elements are zeros. In inner product space (ℍ^𝑚×𝑛,ℝ,⟨⋅,⋅⟩), it is easy to verify that 𝐸_𝑖𝑗,𝐸_𝑖𝑗𝑖, 𝐸_𝑖𝑗𝑗,𝐸_𝑖𝑗𝑘,𝑖=1,2,…,𝑚, 𝑗=1,2,…,𝑛, is an orthonormal basis, which reveals that the dimension of the inner product space (ℍ^𝑚×𝑛,ℝ,⟨⋅,⋅⟩) is 4𝑚𝑛.

Next, we introduce a real representation of a quaternion matrix.

For an arbitrary quaternion matrix 𝑀=𝑀₁+𝑀₂𝑖+𝑀₃𝑗+𝑀₄𝑘, a map 𝜙(⋅), from ℍ^𝑚×𝑛 to ℝ^4𝑚×4𝑛, can be defined as 𝜙(𝑀)=𝑀₁−𝑀₂−𝑀₃−𝑀₄𝑀₂𝑀₁−𝑀₄𝑀₃𝑀₃𝑀₄𝑀₁−𝑀₂𝑀₄−𝑀₃𝑀₂𝑀₁. (9)

Lemma 4 (see [41]). Let 𝑀 and 𝑁 be two arbitrary quaternion matrices with appropriate size. The map 𝜙(⋅) defined by (9) satisfies the following properties.(a)𝑀=𝑁⇔𝜙(𝑀)=𝜙(𝑁). (b)𝜙(𝑀+𝑁)=𝜙(𝑀)+𝜙(𝑁),𝜙(𝑀𝑁)=𝜙(𝑀)𝜙(𝑁),𝜙(𝑘𝑀)=𝑘𝜙(𝑀),𝑘∈ℝ.(c)𝜙(𝑀^𝐻)=𝜙^𝑇(𝑀). (d)𝜙(𝑀)=𝑇_𝑚⁻¹𝜙(𝑀)𝑇_𝑛=𝑅_𝑚⁻¹𝜙(𝑀)𝑅_𝑛=𝑆_𝑚⁻¹𝜙(𝑀)𝑆_𝑛, where 𝑇_𝑡=0−𝐼_𝑡𝐼00_𝑡000000𝐼_𝑡00−𝐼_𝑡0,𝑅_𝑡=0−𝐼_2𝑡𝐼_2𝑡0,𝑆_𝑡=000−𝐼_𝑡00𝐼_𝑡00−𝐼_𝑡𝐼00_𝑡000,𝑡=𝑚,𝑛. (10)

By (9), it is easy to verify that 𝜙(𝑀)=2‖𝑀‖. (11)

Finally, we introduce the commutation matrix.

A commutation matrix 𝑃(𝑚,𝑛) is a 𝑚𝑛×𝑚𝑛 matrix which has the following explicit form: 𝑃(𝑚,𝑛)=^𝑚𝑛_𝑖=1𝑗=1𝐸_𝑖𝑗⊗𝐸^𝑇_𝑖𝑗=𝐸^𝑇_{𝑖𝑗𝑖=1,…,𝑚𝑗=1,…,𝑛}. (12) Moreover, 𝑃(𝑚,𝑛) is a permutation matrix and 𝑃(𝑚,𝑛)=𝑃^𝑇(𝑛,𝑚)=𝑃⁻¹(𝑛,𝑚). We have the following lemmas on the commutation matrix.

Lemma 5 (see [46]). Let 𝐴 be a 𝑚×𝑛 matrix. There is a commutation matrix 𝑃(𝑚,𝑛) such that vec𝐴^𝑇=𝑃(𝑚,𝑛)vec(𝐴). (13)

Lemma 6 (see [46]). Let 𝐴 be a 𝑚×𝑛 matrix and 𝐵 a 𝑝×𝑞 matrix. There exist two commutation matrices 𝑃(𝑚,𝑝) and 𝑃(𝑛,𝑞) such that 𝐵⊗𝐴=𝑃^𝑇𝑚,𝑝(𝐴⊗𝐵)𝑃𝑛,𝑞. (14)

3. Main Results

3.1. The Solution of Problem 1

In this subsection, we will construct an algorithm for solving Problem 1. Then some lemmas will be given to analyse the properties of the proposed algorithm. Using these lemmas, we prove that the proposed algorithm is convergent.

Algorithm 7 (Iterative algorithm for Problem 1). (1)Choose an initial matrix 𝑋(1)∈ℝℍ^𝑛×𝑛(𝑄).(2)Calculate 𝑅(1)=𝐹−𝐴𝑋(1)𝐵−𝐶𝑋^𝐻(1)𝑃(1)=1𝐷;2𝐴^𝐻𝑅(1)𝐵^𝐻+𝐷𝑅^𝐻(1)𝐶+𝑄𝐴^𝐻𝑅(1)𝐵^𝐻𝑄+𝑄𝐷𝑅^𝐻(1)𝐶𝑄;𝑘∶=1. (15)(3)If 𝑅(𝑘)=0, then stop and 𝑋(𝑘) is the solution of Problem 1; else if 𝑅(𝑘)≠0 and 𝑃(𝑘)=0, then stop and Problem 1 is not consistent; else 𝑘∶=𝑘+1.(4)Calculate 𝑋(𝑘)()+‖𝑅(=𝑋𝑘−1𝑘−1)‖²‖𝑃(𝑘−1)‖²𝑃();𝑅(𝑘)()−‖𝑅(𝑘−1=𝑅𝑘−1𝑘−1)‖²‖𝑃(𝑘−1)‖²×𝐴𝑃(𝑘−1)𝐵+𝐶𝑃^𝐻()𝐷;𝑃(𝑘)=1𝑘−12𝐴^𝐻𝑅(𝑘)𝐵^𝐻+𝐷𝑅^𝐻(𝑘)𝐶+𝑄𝐴^𝐻𝑅(𝑘)𝐵^𝐻𝑄+𝑄𝐷𝑅^𝐻(𝑘)+‖𝑅(𝑘𝐶𝑄)‖²‖𝑅(𝑘−1)‖²𝑃(𝑘−1). (16)(5)Go to Step (3).

Lemma 8. Assume that the sequences {𝑅(𝑖)} and {𝑃(𝑖)} are generated by Algorithm 7, then ⟨𝑅(𝑖),𝑅(𝑗)⟩=0 and ⟨𝑃(𝑖),𝑃(𝑗)⟩=0  for 𝑖,𝑗=1,2,…,𝑖≠𝑗.

Proof. Since ⟨𝑅(𝑖),𝑅(𝑗)⟩=⟨𝑅(𝑗),𝑅(𝑖)⟩ and ⟨𝑃(𝑖),𝑃(𝑗)⟩=⟨𝑃(𝑗),𝑃(𝑖)⟩ for 𝑖,𝑗=1,2,…, we only need to prove that ⟨𝑅(𝑖),𝑅(𝑗)⟩=0 and ⟨𝑃(𝑖),𝑃(𝑗)⟩=0 for 1≤𝑖<𝑗.
Now we prove this conclusion by induction.
Step 1. We show that ⟨𝑅(𝑖)(⟨𝑃(𝑖)(,𝑅𝑖+1)⟩=0,,𝑃𝑖+1)⟩=0for𝑖=1,2,…. (17) We also prove (17) by induction. When 𝑖=1, we have ⟨𝑅(1)(2𝑅,𝑅)⟩=Retr^𝐻(2)𝑅(1𝑅(1)−‖𝑅(1)=Retr)‖²‖𝑃(1)‖²𝐴𝑃(1)𝐵+𝐶𝑃^𝐻(1)𝐷^𝐻(1=‖𝑅(1×𝑅))‖²−‖𝑅(1)‖²‖𝑃(1)‖²×Retr𝑃^𝐻(1)𝐴^𝐻𝑅(1)𝐵^𝐻+𝐷𝑅^𝐻(1)𝐶=‖𝑅(1)‖²−‖𝑅(1)‖²‖𝑃(1)‖²×Retr𝑃^𝐻(1)𝐴^𝐻𝑅(1)𝐵^𝐻+𝐷𝑅^𝐻(1)𝐶+𝑄𝐴^𝐻𝑅(1)×𝐵^𝐻𝑄+𝑄𝐷𝑅^𝐻(1)𝐶𝑄×(2)⁻¹=‖𝑅(1)‖²−‖𝑅(1)‖²‖𝑃(1)‖²‖𝑃(1)‖²=0. (18) Also we can write ⟨𝑃(1)(2𝑃,𝑃)⟩=Retr^𝐻(2)𝑃(11)=Retr2𝐴^𝐻𝑅(2)𝐵^𝐻+𝐷𝑅^𝐻(2)𝐶+𝑄𝐴^𝐻𝑅(2)𝐵^𝐻𝑄+𝑄𝐷𝑅^𝐻(2)+‖𝑅(2𝐶𝑄)‖²‖𝑅(1)‖²𝑃(1)^𝐻𝑃(1=‖𝑅(2))‖²‖𝑅(1)‖²‖𝑃(1)‖²+Retr𝑃^𝐻(1)×𝐴^𝐻𝑅(2)𝐵^𝐻+𝐷𝑅^𝐻(2)𝐶+𝑄𝐴^𝐻𝑅(2)𝐵^𝐻𝑄+𝑄𝐷𝑅^𝐻(2)𝐶𝑄×(2)⁻¹=‖𝑅(2)‖²‖𝑅(1)‖²‖𝑃(1)‖²+Retr𝑃^𝐻(1)𝐴^𝐻𝑅(2)𝐵^𝐻+𝐷𝑅^𝐻(2)𝐶=‖𝑅(2)‖²‖𝑅(1)‖²‖𝑃(1)‖²+Retr𝑅^𝐻(2)𝐴𝑃(1)𝐵+𝐶𝑃^𝐻(1)𝐷=‖𝑅(2)‖²‖𝑅(1)‖²‖𝑃(1)‖²+‖𝑃(1)‖²‖𝑅(1)‖²Retr𝑅^𝐻(2𝑅(1)(2=‖𝑅(2)(−𝑅)))‖²‖𝑅(1)‖²‖𝑃(1)‖²−‖𝑃(1)‖²‖𝑅(1)‖²‖𝑅(2)‖²=0. (19) Now assume that conclusion (17) holds for 1≤𝑖≤𝑠−1, then ⟨𝑅(𝑠)(𝑅,𝑅𝑠+1)⟩=Retr^𝐻()𝑅(𝑠𝑅(𝑠)−‖𝑅(𝑠𝑠+1)=Retr)‖²‖𝑃(𝑠)‖²𝐴𝑃(𝑠)𝐵+𝐶𝑃^𝐻(𝑠)𝐷^𝐻(𝑠=‖𝑅(×𝑅)s)‖²−‖𝑅(𝑠)‖²‖𝑃(𝑠)‖²×Retr𝑃^𝐻(𝑠)𝐴^𝐻𝑅(𝑠)𝐵^𝐻+𝐷𝑅^𝐻(𝑠)𝐶=‖𝑅(𝑠)‖²−‖𝑅(𝑠)‖²‖𝑃(𝑠)‖²×Retr𝑃^𝐻(𝑠)×𝐴^𝐻𝑅(𝑠)𝐵^𝐻+𝐷𝑅^𝐻(𝑠)𝐶+𝑄𝐴^𝐻𝑅(𝑠)𝐵^𝐻𝑄+𝑄𝐷𝑅^𝐻(𝑠)𝐶𝑄×(2)⁻¹=‖𝑅(𝑠)‖²−‖𝑅(𝑠)‖²‖𝑃(𝑠)‖²×Retr𝑃^𝐻(𝑠)𝑃(𝑠)−‖𝑅(𝑠)‖²‖𝑅(𝑠−1)‖²𝑃(𝑠−1)=0. (20) And it can also be obtained that ⟨𝑃(𝑠)(𝑃,𝑃𝑠+1)⟩=Retr^𝐻()𝑃(𝑠𝐴𝑠+1)=Retr^𝐻𝑅(𝑠+1)𝐵^𝐻+𝐷𝑅^𝐻(𝑠+1)𝐶+𝑄𝐴^𝐻𝑅(𝑠+1)𝐵^𝐻𝑄+𝑄𝐷𝑅^𝐻(𝑠+1)𝐶𝑄×(2)⁻¹+‖𝑅(𝑠+1)‖²‖𝑅(𝑠)‖²𝑃(𝑠)^𝐻𝑃(𝑠=‖𝑅()𝑠+1)‖²‖𝑅(𝑠)‖²‖𝑃(𝑠)‖²+Retr𝑃^𝐻(𝑠)𝐴^𝐻𝑅(𝑠+1)𝐵^𝐻+𝐷𝑅^𝐻()𝐶=‖𝑅(𝑠+1𝑠+1)‖²‖𝑅(𝑠)‖²‖𝑃(𝑠)‖²+Retr𝑅^𝐻(𝑠+1)𝐴𝑃(𝑠)𝐵+𝐶𝑃^𝐻(𝑠)𝐷=‖𝑅(𝑠+1)‖²‖𝑅(𝑠)‖²‖𝑃(𝑠)‖²+‖𝑃(𝑠)‖²‖𝑅(𝑠)‖²Retr𝑅^𝐻(𝑠+1)(𝑅(𝑠)−𝑅(𝑠+1))=0. (21) Therefore, the conclusion (17) holds for 𝑖=1,2,….
Step 2. Assume that ⟨𝑅(𝑖),𝑅(𝑖+𝑟)⟩=0 and ⟨𝑃(𝑖),𝑃(𝑖+𝑟)⟩=0 for 𝑖≥1 and 𝑟≥1. We will show that ⟨𝑅(𝑖),𝑅(𝑖+𝑟+1)⟩=0,⟨𝑃(𝑖),𝑃(𝑖+𝑟+1)⟩=0. (22) We prove the conclusion (22) in two substeps.
Substep 2.1. In this substep, we show that ⟨𝑅(1),𝑅(𝑟+2)⟩=0,⟨𝑃(1),𝑃(𝑟+2)⟩=0. (23) It follows from Algorithm 7 that ⟨𝑅(1)(𝑅,𝑅𝑟+2)⟩=Retr^𝐻()𝑅(1𝑅()−‖𝑅(𝑟+2)=Retr𝑟+1𝑟+1)‖²‖𝑃(𝑟+1)‖²×𝐴𝑃(𝑟+1)𝐵+𝐶𝑃^𝐻(𝑟+1)𝐷^𝐻𝑅(1𝑅)=Retr^𝐻()𝑅(1−‖𝑅(𝑟+1)𝑟+1)‖²‖𝑃(𝑟+1)‖²×Retr𝑃^𝐻(𝑟+1)𝐴^𝐻𝑅(1)𝐵^𝐻+𝐷𝑅^𝐻(1)𝐶=Retr𝑅^𝐻()𝑅(1−‖𝑅(𝑟+1)𝑟+1)‖²‖𝑃(𝑟+1)‖²×Retr𝑃^𝐻(𝑟+1)𝐴^𝐻𝑅(1)𝐵^𝐻+𝐷𝑅^𝐻(1)𝐶+𝑄𝐴^𝐻𝑅(1)𝐵^𝐻𝑄+𝑄𝐷𝑅^𝐻(1)𝐶𝑄×(2)⁻¹=Retr𝑅^𝐻()𝑅(1−‖𝑅(𝑟+1)𝑟+1)‖²‖𝑃(𝑟+1)‖²Retr𝑃^𝐻(𝑟+1)𝑃(1)=0. (24) Also we can write ⟨𝑃(1)(𝑃,𝑃𝑟+2)⟩=Retr^𝐻()𝑃(1𝐴𝑟+2)=Retr^𝐻𝑅(𝑟+2)𝐵^𝐻+𝐷𝑅^𝐻(𝑟+2)𝐶+𝑄𝐴^𝐻𝑅(𝑟+2)𝐵^𝐻𝑄+𝑄𝐷𝑅^𝐻()×(2)𝑟+2𝐶𝑄₋₁+‖𝑅(𝑟+2)‖²‖𝑅(𝑟+1)‖²𝑃(𝑟+1)^𝐻𝑃(1𝐴)=Retr^𝐻𝑅(𝑟+2)𝐵^𝐻+𝐷𝑅^𝐻(𝑟+2)𝐶^𝐻𝑃(1+‖𝑅()𝑟+2)‖²‖𝑅(𝑟+1)‖²Retr𝑃^𝐻()𝑃(1𝑅𝑟+1)=Retr^𝐻(𝑟+2)𝐴𝑃(1)𝐵+𝐶𝑃^𝐻(1)𝐷+‖𝑅(𝑟+2)‖²‖𝑅(𝑟+1)‖²Retr𝑃^𝐻()𝑃(1=‖𝑃(1𝑟+1))‖²‖𝑅(1)‖²Retr𝑅^𝐻(𝑅(1)(2+‖𝑅(𝑟+2)(−𝑅))𝑟+2)‖²‖𝑅(𝑟+1)‖²Retr𝑃^𝐻(𝑟+1)𝑃(1)=0. (25)Substep 2.2. In this substep, we prove the conclusion (22) in Step 2. It follows from Algorithm 7 that ⟨𝑅(𝑖)(𝑅,𝑅𝑖+𝑟+1)⟩=Retr^𝐻()𝑅(𝑖𝑅()−‖𝑅(𝑖+𝑟+1)=Retr𝑖+𝑟𝑖+𝑟)‖²‖𝑃(𝑖+𝑟)‖²×𝐴𝑃(𝑖+𝑟)𝐵+𝐶𝑃^𝐻(𝑖+𝑟)𝐷^𝐻𝑅(𝑖𝑅)=Retr^𝐻()𝑅(𝑖−‖𝑅(𝑖+𝑟)𝑖+𝑟)‖²‖𝑃(𝑖+𝑟)‖²×Retr𝑃^𝐻(𝑖+𝑟)𝐴^𝐻𝑅(𝑖)𝐵^𝐻+𝐷𝑅^𝐻(𝑖)𝐶‖𝑅(=−𝑖+𝑟)‖²‖𝑃(𝑖+𝑟)‖²×Retr𝑃^𝐻(𝑖+𝑟)𝐴^𝐻𝑅(𝑖)𝐵^𝐻+𝐷𝑅^𝐻(𝑖)𝐶+𝑄𝐴^𝐻𝑅(𝑖)𝐵^𝐻𝑄+𝑄𝐷𝑅^𝐻(𝑖)𝐶𝑄×(2)⁻¹‖𝑅(=−𝑖+𝑟)‖²‖𝑃(𝑖+𝑟)‖²×Retr𝑃^𝐻()𝑃(𝑖)−‖𝑅(𝑖𝑖+𝑟)‖²‖𝑅(𝑖−1)‖²𝑃()=‖𝑅(𝑖−1𝑖+𝑟)‖²‖𝑅(𝑖)‖²‖𝑃(𝑖+𝑟)‖²‖𝑅(𝑖−1)‖²Retr𝑃^𝐻()𝑃(,⟨𝑃(𝑖)(𝑃𝑖+𝑟𝑖−1),𝑃𝑖+𝑟+1)⟩=Retr^𝐻()𝑃(𝑖1𝑖+𝑟+1)=Retr2𝐴^𝐻𝑅(𝑖+𝑟+1)𝐵^𝐻+𝐷𝑅^𝐻(𝑖+𝑟+1)𝐶+𝑄𝐴^𝐻𝑅(𝑖+𝑟+1)𝐵^𝐻𝑄+𝑄𝐷𝑅^𝐻()+‖‖𝑖+𝑟+1𝐶𝑄𝑅(𝑖+𝑟+1)²‖𝑅(𝑖+𝑟)‖²𝑃(𝑖+𝑟)^𝐻𝑃(𝑖𝐴)=Retr^𝐻𝑅(𝑖+𝑟+1)𝐵^𝐻+𝐷𝑅^𝐻(𝑖+𝑟+1)𝐶^𝐻𝑃(𝑖𝑅)=Retr^𝐻(𝑖+𝑟+1)𝐴𝑃(𝑖)B+𝐶𝑃^𝐻(𝑖)𝐷=‖𝑃(𝑖)‖²‖𝑅(𝑖)‖²Retr𝑅^𝐻(𝑅(𝑖)(=‖𝑃(𝑖𝑖+𝑟+1)(−𝑅𝑖+1)))‖²‖𝑅(𝑖)‖²Retr𝑅^𝐻()𝑅(𝑖−‖𝑃(𝑖𝑖+𝑟+1))‖²‖𝑅(𝑖)‖²Retr𝑅^𝐻()𝑅(=‖𝑃(𝑖𝑖+𝑟+1𝑖+1))‖²‖𝑅(𝑖+𝑟)‖²‖𝑅(𝑖)‖²‖𝑅(𝑖)‖²‖𝑃(𝑖+𝑟)‖²‖𝑅(𝑖−1)‖²×Retr𝑃^𝐻(𝑖+𝑟)𝑃(𝑖−1). (26) Repeating the above process (26), we can obtain ⟨𝑅(𝑖),𝑅(𝑖+𝑟+1)⟩=⋯=𝛼Retr𝑃^𝐻()𝑃(1;⟨𝑃(𝑖)(𝑃𝑟+2),𝑃𝑖+𝑟+1)⟩=⋯=𝛽Retr^𝐻(𝑟+2)𝑃(1). (27) Combining these two relations with (24) and (25), it implies that (22) holds. So, by the principle of induction, we know that Lemma 8 holds.

Lemma 9. Assume that Problem 1 is consistent, and let 𝑋∈ℝℍ^𝑛×𝑛(𝑄) be its solution. Then, for any initial matrix 𝑋(1)∈ℝℍ^𝑛×𝑛(𝑄), the sequences {𝑅(𝑖)}, {𝑃(𝑖)}, and {𝑋(𝑖)} generated by Algorithm 7 satisfy 𝑃(𝑖),𝑋−𝑋(𝑖)=‖𝑅(𝑖)‖²,𝑖=1,2,…. (28)

Proof. We also prove this conclusion by induction.
When 𝑖=1, it follows from Algorithm 7 that 𝑃(1),(1(1)𝑋−𝑋)=Retr𝑋−𝑋^𝐻𝑃(1(1)=Retr𝑋−𝑋)^𝐻×𝐴^𝐻𝑅(1)𝐵^𝐻+𝐷𝑅^𝐻(1)𝐶+𝑄𝐴^𝐻𝑅(1)𝐵^𝐻𝑄+𝑄𝐷𝑅^𝐻(1)𝐶𝑄×(2)⁻¹=Retr𝑋−𝑋(1)^𝐻𝐴^𝐻𝑅(1)𝐵^𝐻+𝐷𝑅^𝐻(1)𝐶=Retr𝑅^𝐻(1)𝐴𝑋−𝑋(1)𝐵+𝐶𝑋−𝑋(1)^𝐻𝐷=Retr𝑅^𝐻(1)𝑅(1=‖𝑅(1))‖². (29) This implies that (28) holds for 𝑖=1.
Now it is assumed that (28) holds for 𝑖=𝑠, that is 𝑃(𝑠),𝑋−𝑋(𝑠)=‖𝑅(𝑠)‖². (30) Then, when 𝑖=𝑠+1𝑃(),(()𝑠+1𝑋−𝑋𝑠+1)=Retr𝑋−𝑋𝑠+1^𝐻𝑃(()𝑠+1)=Retr𝑋−𝑋𝑠+1^𝐻×12𝐴^𝐻𝑅(𝑠+1)𝐵^𝐻+𝐷𝑅^𝐻(𝑠+1)𝐶+𝑄𝐴^𝐻𝑅(𝑠+1)𝐵^𝐻𝑄+𝑄𝐷𝑅^𝐻()+‖‖𝑠+1𝐶𝑄𝑅(𝑠+1)²‖𝑅(𝑠)‖²𝑃(𝑠)=Retr𝑋−𝑋(𝑠+1)^𝐻×𝐴^𝐻𝑅(𝑠+1)𝐵^𝐻+𝐷𝑅^𝐻()𝐶+‖𝑅(𝑠+1𝑠+1)‖²‖𝑅(𝑠)‖²Retr𝑋−𝑋(𝑠+1)^𝐻𝑃(𝑠𝑅)=Retr^𝐻()×𝐴()𝐵()𝑠+1𝑋−𝑋𝑠+1+𝐶𝑋−𝑋𝑠+1^𝐻𝐷+‖𝑅(𝑠+1)‖²‖𝑅(𝑠)‖²Retr𝑋−𝑋(𝑠)^𝐻𝑃(𝑠𝑋()(𝑠)−Retr(𝑠+1−𝑋))^𝐻𝑃(𝑠=‖𝑅()𝑠+1)‖²+‖𝑅(𝑠+1)‖²‖𝑅(𝑠)‖²×‖𝑅(𝑠)‖²−‖𝑅(𝑠)‖²‖𝑃(𝑠)‖²Retr𝑃^𝐻(𝑠)𝑃(𝑠=‖𝑅()𝑠+1)‖². (31) Therefore, Lemma 9 holds by the principle of induction.

From the above two lemmas, we have the following conclusions.

Remark 10. If there exists a positive number 𝑖 such that 𝑅(𝑖)≠0 and 𝑃(𝑖)=0, then we can get from Lemma 9 that Problem 1 is not consistent. Hence, the solvability of Problem 1 can be determined by Algorithm 7 automatically in the absence of roundoff errors.

Theorem 11. Suppose that Problem 1 is consistent. Then for any initial matrix 𝑋(1)∈ℝℍ^𝑛×𝑛(𝑄), a solution of Problem 1 can be obtained within finite iteration steps in the absence of roundoff errors.

Proof. In Section 2, it is known that the inner product space (ℍ^𝑚×𝑝,𝑅,⟨⋅,⋅⟩) is 4𝑚𝑝-dimensional. According to Lemma 9, if 𝑅(𝑖)≠0,  𝑖=1,2,…,4𝑚𝑝, then we have 𝑃(𝑖)≠0,𝑖=1,2,…,4𝑚𝑝. Hence 𝑅(4𝑚𝑝+1) and 𝑃(4𝑚𝑝+1) can be computed. From Lemma 8, it is not difficult to get 𝑅(𝑖),𝑅𝑗=0for𝑖,𝑗=1,2,…,4𝑚𝑝,𝑖≠𝑗. (32) Then 𝑅(1),𝑅(2),…,𝑅(4𝑚𝑝) is an orthogonal basis of the inner product space (ℍ^𝑚×𝑝,𝑅,⟨⋅,⋅⟩). In addition, we can get from Lemma 8 that 𝑅(𝑖),𝑅4𝑚𝑝+1=0for𝑖=1,2,…,4𝑚𝑝. (33) It follows that 𝑅(4𝑚𝑝+1)=0, which implies that 𝑋(4𝑚𝑝+1) is a solution of Problem 1.

3.2. The Solution of Problem 2

In this subsection, firstly we introduce some lemmas. Then, we will prove that the least Frobenius norm reflexive solution of (1) can be derived by choosing a suitable initial iterative matrix. Finally, we solve Problem 2 by finding the least Frobenius norm reflexive solution of a new-constructed quaternion matrix equation.

Lemma 12 (see [47]). Assume that the consistent system of linear equations 𝑀𝑦=𝑏 has a solution 𝑦₀∈𝑅(𝑀^𝑇) then 𝑦₀ is the unique least Frobenius norm solution of the system of linear equations.

Lemma 13. Problem 1 is consistent if and only if the system of quaternion matrix equations 𝐴𝑋𝐵+𝐶𝑋^𝐻𝐷=𝐹,𝐴𝑄𝑋𝑄𝐵+𝐶𝑄𝑋^𝐻𝑄𝐷=𝐹 (34) is consistent. Furthermore, if the solution sets of Problem 1 and (34) are denoted by 𝑆_𝐻 and 𝑆¹_𝐻, respectively, then, we have 𝑆_𝐻⊆𝑆¹_𝐻.

Proof. First, we assume that Problem 1 has a solution 𝑋. By 𝐴𝑋𝐵+𝐶𝑋^𝐻𝐷=𝐹 and 𝑄𝑋𝑄=𝑋, we can obtain 𝐴𝑋𝐵+𝐶𝑋^𝐻𝐷=𝐹 and 𝐴𝑄𝑋𝑄𝐵+𝐶𝑄𝑋^𝐻𝑄𝐷=𝐹, which implies that 𝑋 is a solution of quaternion matrix equations (34), and 𝑆_𝐻⊆𝑆¹_𝐻.
Conversely, suppose (34) is consistent. Let 𝑋 be a solution of (34). Set 𝑋_𝑎=(𝑋+𝑄𝑋𝑄)/2. It is obvious that 𝑋_𝑎∈ℝℍ^𝑛×𝑛(𝑄). Now we can write 𝐴𝑋_𝑎𝐵+𝐶𝑋^𝐻_𝑎𝐷=12𝐴(𝑋+𝑄𝑋𝑄)𝐵+𝐶(𝑋+𝑄𝑋𝑄)^𝐻𝐷=12𝐴𝑋𝐵+𝐶𝑋^𝐻𝐷+𝐴𝑄𝑋𝑄𝐵+𝐶𝑄𝑋^𝐻=1𝑄𝐷2[𝐹+𝐹]=𝐹. (35) Hence 𝑋_𝑎 is a solution of Problem 1. The proof is completed.

Lemma 14. The system of quaternion matrix equations (34) is consistent if and only if the system of real matrix equations 𝜙(𝐴)𝑋_𝑖𝑗4×4𝜙(𝐵)+𝜙(𝐶)𝑋_𝑖𝑗^𝑇_4×4𝜙(𝐷)(𝐹),𝜙(𝐴)𝜙(𝑄𝑋=𝜙)_𝑖𝑗4×4𝜙(𝑄)𝜙(𝐵)+𝜙(𝐶)𝜙(𝑄)𝑋_𝑖𝑗^𝑇_4×4𝜙(𝑄)𝜙(𝐷)=𝜙(𝐹) (36) is consistent, where 𝑋_𝑖𝑗∈ℍ^𝑛×𝑛,𝑖,𝑗=1,2,3,4, are submatrices of the unknown matrix. Furthermore, if the solution sets of (34) and (36) are denoted by 𝑆¹_𝐻 and 𝑆²_𝑅, respectively, then, we have 𝜙(𝑆¹_𝐻)⊆𝑆²_𝑅.

Proof. Suppose that (34) has a solution 𝑋=𝑋₁+𝑋₂𝑖+𝑋₃𝑗+𝑋₄𝑘. (37) Applying (𝑎), (𝑏), and (𝑐) in Lemma 4 to (34) yields 𝜙(𝐴)𝜙(𝑋)𝜙(𝐵)+𝜙(𝐶)𝜙^𝑇(𝑋)𝜙(𝐷)(𝐹),𝜙(𝐴)𝜙(𝑄)𝜙(𝑋)𝜙(𝑄)𝜙(𝐵)(𝐶)𝜙(𝑄)𝜙=𝜙+𝜙^𝑇(𝑋)𝜙(𝑄)𝜙(𝐷)=𝜙(𝐹), (38) which implies that 𝜙(𝑋) is a solution of (36) and 𝜙(𝑆¹_𝐻)⊆𝑆²_𝑅. Conversely, suppose that (36) has a solution 𝑋̆=𝑋_𝑖𝑗4×4. (39) By (𝑑) in Lemma 4, we have that 𝑇_𝑚⁻¹𝜙(𝐴)𝑇_𝑛𝑋̆𝑇_𝑛⁻¹𝜙(𝐵)𝑇_𝑝+𝑇_𝑚⁻¹𝜙(𝐶)𝑇_𝑛𝑋̆^𝑇𝑇_𝑛⁻¹𝜙(𝐷)𝑇_𝑝=𝑇_𝑚⁻¹𝜙(𝐹)𝑇_𝑝,𝑅_𝑚⁻¹𝜙(𝐴)𝑅_𝑛𝑋̆𝑅_𝑛⁻¹𝜙(𝐵)𝑅_𝑝+𝑅_𝑚⁻¹𝜙(𝐶)𝑅_𝑛𝑋̆^𝑇𝑅_𝑛⁻¹𝜙(𝐷)𝑅_𝑝=𝑅_𝑚⁻¹𝜙(𝐹)𝑅_𝑝,𝑆_𝑚⁻¹𝜙(𝐴)𝑆_𝑛𝑋̆𝑆_𝑛⁻¹𝜙(𝐵)𝑆_𝑝+S_𝑚⁻¹𝜙(𝐶)𝑆_𝑛𝑋̆^𝑇𝑆_𝑛⁻¹𝜙(𝐷)𝑆_𝑝=𝑆_𝑚⁻¹𝜙(𝐹)𝑆_𝑝,𝑇_𝑚⁻¹𝜙(𝐴)𝜙(𝑄)𝑇_𝑛𝑋̆𝑇_𝑛⁻¹𝜙(𝑄)𝜙(𝐵)𝑇_𝑝+𝑇_𝑚⁻¹𝜙(𝐶)𝜙(𝑄)𝑇_𝑛𝑋̆^𝑇𝑇_𝑛⁻¹𝜙(𝑄)𝜙(𝐷)𝑇_𝑝=𝑇_𝑚⁻¹𝜙(𝐹)𝑇_𝑝,𝑅_𝑚⁻¹𝜙(𝐴)𝜙(𝑄)𝑅_𝑛𝑋̆𝑅_𝑛⁻¹𝜙(𝑄)𝜙(𝐵)𝑅_𝑝+𝑅_𝑚⁻¹𝜙(𝐶)𝜙(𝑄)𝑅_𝑛𝑋̆^𝑇𝑅_𝑛⁻¹𝜙(𝑄)𝜙(𝐷)𝑅_𝑝=𝑅_𝑚⁻¹𝜙(𝐹)𝑅_𝑝,𝑆_𝑚⁻¹𝜙(𝐴)𝜙(𝑄)𝑆_𝑛𝑋̆𝑆_𝑛⁻¹𝜙(𝑄)𝜙(𝐵)𝑆_𝑝+𝑆_𝑚⁻¹𝜙(𝐶)𝜙(𝑄)𝑆_𝑛𝑋̆^𝑇𝑆_𝑛⁻¹𝜙(𝑄)𝜙(𝐷)𝑆_𝑝=𝑆_𝑚⁻¹𝜙(𝐹)𝑆_𝑝. (40) Hence 𝜙(𝐴)𝑇_𝑛𝑋̆𝑇_𝑛⁻¹𝜙(𝐵)+𝜙(𝐶)𝑇_𝑛𝑋̆𝑇_𝑛^−1𝑇𝜙(𝐷)(𝐹),𝜙(𝐴)𝑅=𝜙_𝑛𝑋̆𝑅_𝑛⁻¹𝜙(𝐵)+𝜙(𝐶)R_𝑛𝑋̆𝑅_𝑛^−1𝑇𝜙(𝐷)(𝐹),𝜙(𝐴)𝑆=𝜙_𝑛𝑋̆𝑆_𝑛⁻¹𝜙(𝐵)+𝜙(𝐶)𝑆_𝑛𝑋̆𝑆_𝑛^−1𝑇𝜙(𝐷)(𝐹),𝜙(𝐴)𝜙(𝑄)𝑇=𝜙_𝑛𝑋̆𝑇_𝑛⁻¹𝜙(𝑄)𝜙(𝐵)+𝜙(𝐶)𝜙(𝑄)𝑇_𝑛𝑋̆𝑇_𝑛^−1𝑇𝜙(𝑄)𝜙(𝐷)(𝐹),𝜙(𝐴)𝜙(𝑄)𝑅=𝜙_𝑛𝑋̆𝑅_𝑛⁻¹𝜙(𝑄)𝜙(𝐵)+𝜙(𝐶)𝜙(𝑄)𝑅_𝑛𝑋̆𝑅_𝑛^−1𝑇𝜙(𝑄)𝜙(𝐷)(𝐹),𝜙(𝐴)𝜙(𝑄)𝑆=𝜙_𝑛𝑋̆𝑆_𝑛⁻¹𝜙(𝑄)𝜙(𝐵)+𝜙(𝐶)𝜙(𝑄)𝑆_𝑛𝑋̆𝑆_𝑛^−1𝑇𝜙(𝑄)𝜙(𝐷)=𝜙(𝐹), (41) which implies that 𝑇_𝑛𝑋̆𝑇_𝑛⁻¹,𝑅_𝑛𝑋̆𝑅_𝑛⁻¹, and 𝑆_𝑛𝑋̆𝑆_𝑛⁻¹ are also solutions of (36). Thus, 14𝑋̆+𝑇_𝑛𝑋̆𝑇_𝑛⁻¹+𝑅_𝑛𝑋̆𝑅_𝑛⁻¹+𝑆_𝑛𝑋̆𝑆_𝑛⁻¹ (42) is also a solution of (36), where 𝑋̆+𝑇_𝑛𝑋̆𝑇_𝑛⁻¹+𝑅_𝑛𝑋̆𝑅_𝑛⁻¹+𝑆_𝑛𝑋̆𝑆_𝑛⁻¹=𝑋_𝑖𝑗4×4𝑋,𝑖,𝑗=1,2,3,4,₁₁=𝑋₁₁+𝑋₂₂+𝑋₃₃+𝑋₄₄,𝑋₁₂=𝑋₁₂−𝑋₂₁+𝑋₃₄−𝑋₄₃,𝑋₁₃=𝑋₁₃−𝑋₂₄−𝑋₃₁+𝑋₄₂,𝑋₁₄=𝑋₁₄+𝑋₂₃−𝑋₃₂−𝑋₄₁,𝑋₂₁=−𝑋₁₂+𝑋₂₁−𝑋₃₄+𝑋₄₃,𝑋₂₂=𝑋₁₁+𝑋₂₂+𝑋₃₃+𝑋₄₄,𝑋₂₃=𝑋₁₄+𝑋₂₃−𝑋₃₂−𝑋₄₁,𝑋₂₄=−𝑋₁₃+𝑋₂₄+𝑋₃₁−𝑋₄₂,𝑋₃₁=−𝑋₁₃+𝑋₂₄+𝑋₃₁−𝑋₄₂,𝑋₃₂=−𝑋₁₄−𝑋₂₃+𝑋₃₂+𝑋₄₁,𝑋₃₃=𝑋₁₁+𝑋₂₂+𝑋₃₃+𝑋₄₄,𝑋₃₄=𝑋₁₂−𝑋₂₁+𝑋₃₄−𝑋₄₃,𝑋₄₁=−𝑋₁₄−𝑋₂₃+𝑋₃₂+𝑋₄₁,𝑋₄₂=𝑋₁₃−𝑋₂₄−𝑋₃₁+𝑋₄₂,𝑋₄₃=−𝑋₁₂+𝑋₂₁−𝑋₃₄+𝑋₄₃,𝑋₄₄=𝑋₁₁+𝑋₂₂+𝑋₃₃+𝑋₄₄. (43) Let 1𝑋=4𝑋₁₁+𝑋₂₂+𝑋₃₃+𝑋₄₄+14−𝑋₁₂+𝑋₂₁−𝑋₃₄+𝑋₄₃𝑖+14−𝑋₁₃+𝑋₂₄+𝑋₃₁−𝑋₄₂𝑗+14−𝑋₁₄−𝑋₂₃+𝑋₃₂+𝑋₄₁𝑘. (44) Then it is not difficult to verify that 𝜙𝑋=14𝑋̆+𝑇_𝑛𝑋̆𝑇_𝑛⁻¹+𝑅_𝑛𝑋̆𝑅_𝑛⁻¹+𝑆_𝑛𝑋̆𝑆_𝑛⁻¹. (45) We have that 𝑋 is a solution of (34) by (𝑎) in Lemma 4. The proof is completed.

Lemma 15. There exists a permutation matrix P(4n,4n) such that (36) is equivalent to ×vec𝑋_𝑖𝑗4×4=^{𝜙𝑇(𝐵)(𝐴)+𝜙⊗𝜙𝑇(𝐷)(𝐶𝑃()𝜙⊗𝜙)4𝑛,4𝑛𝑇(𝐵)𝜙(𝑄)(𝐴)𝜙(𝑄)+𝜙⊗𝜙𝑇(𝐷)𝜙(𝑄)(𝐶)𝜙(𝑄𝑃()⊗𝜙)4𝑛,4𝑛}𝜙(𝐹𝜙(𝐹.vec)vec) (46)

Lemma 15 is easily proven through Lemma 5. So we omit it here.

Theorem 16. When Problem 1 is consistent, let its solution set be denoted by 𝑆_𝐻. If 𝑋^∘∈𝑆_𝐻, and 𝑋^∘ can be expressed as 𝑋^∘=𝐴^𝐻𝐺𝐵^𝐻+𝐷𝐺^𝐻𝐶+𝑄𝐴^𝐻𝐺𝐵^𝐻𝑄+𝑄𝐷𝐺^𝐻𝐶𝑄,𝐺∈ℍ^𝑚×𝑝, (47) then, 𝑋^∘ is the least Frobenius norm solution of Problem 1.

Proof. By (𝑎),  (𝑏), and (𝑐) in Lemmas 4, 5, and 6, we have that vec𝜙𝑋^∘=vec𝜙^𝑇(𝐴)𝜙(𝐺)𝜙^𝑇(𝐵)+𝜙(𝐷)𝜙^𝑇(𝐺)𝜙(𝐶)+𝜙(𝑄)𝜙^𝑇(𝐴)𝜙(𝐺)𝜙^𝑇(𝐵)𝜙(𝑄)+𝜙(𝑄)𝜙(𝐷)𝜙^𝑇(𝐺)𝜙(𝐶)𝜙(𝑄)=𝜙(𝐵)⊗𝜙^𝑇(𝐴)+𝜙^𝑇(𝐶)(𝐷)𝑃,𝜙(𝑄)𝜙(𝐵)(𝑄)𝜙⊗𝜙4𝑚,4𝑝⊗𝜙^𝑇(𝐴)+𝜙(𝑄)𝜙^𝑇(𝐶)(𝑄)𝜙(𝐷)𝑃×𝜙(𝐺)𝜙(𝐺)=⊗𝜙4𝑚,4𝑝vecvec_{𝜙𝑇(𝐵)⊗𝜙(𝐴)+(𝜙𝑇𝜙(𝐷)⊗𝜙(𝐶))𝑃(4𝑛,4𝑛)𝑇(𝐵)𝜙(𝑄)⊗𝜙(𝐴)𝜙(𝑄)+(𝜙𝑇(𝐷)𝜙(𝑄)⊗𝜙(𝐶)𝜙(𝑄))𝑃(4𝑛,4𝑛)𝑇}×𝜙(𝐺)𝜙(𝐺)𝜙vecvec∈𝑅^𝑇(𝐵)(𝐴)𝜙⊗𝜙_𝑇(𝐵)𝜙(𝑄)(𝐴)𝜙(𝑄)+𝜙⊗𝜙_𝑇(𝐷)(𝐶𝑃()+𝜙⊗𝜙)4𝑛,4𝑛_𝑇(𝐷)𝜙(𝑄)⊗𝜙(𝐶)𝜙(𝑄)𝑃(4𝑛,4𝑛)^𝑇. (48) By Lemma 12, 𝜙(𝑋^∘) is the least Frobenius norm solution of matrix equations (46).
Noting (11), we derive from Lemmas 13, 14, and 15 that 𝑋^∘ is the least Frobenius norm solution of Problem 1.

From Algorithm 7, it is obvious that, if we consider 𝑋(1)=𝐴^𝐻𝐺𝐵^𝐻+𝐷𝐺^𝐻𝐶+𝑄𝐴^𝐻𝐺𝐵^𝐻𝑄+𝑄𝐷𝐺^𝐻𝐶𝑄,𝐺∈ℍ^𝑚×𝑝, (49) then all 𝑋(𝑘) generated by Algorithm 7 can be expressed as 𝑋(𝑘)=𝐴^𝐻𝐺_𝑘𝐵^𝐻+𝐷𝐺^𝐻_𝑘𝐶+𝑄𝐴^𝐻𝐺_𝑘𝐵^𝐻𝑄+𝑄𝐷𝐺^𝐻_𝑘𝐶𝑄,𝐺_𝑘∈ℍ^𝑚×𝑝. (50)

Using the above conclusion and considering Theorem 16, we propose the following theorem.

Theorem 17. Suppose that Problem 1 is consistent. Let the initial iteration matrix be 𝑋(1)=𝐴^𝐻𝐺𝐵^H+𝐷𝐺^𝐻𝐶+𝑄𝐴^𝐻𝐺𝐵^𝐻𝑄+𝑄𝐷𝐺^𝐻𝐶𝑄, (51) where 𝐺 is an arbitrary quaternion matrix, or especially, 𝑋(1)=0, then the solution 𝑋^∗, generated by Algorithm 7, is the least Frobenius norm solution of Problem 1.

Now we study Problem 2. When Problem 1 is consistent, the solution set of Problem 1 denoted by 𝑆_𝐻 is not empty. Then, For a given reflexive matrix 𝑋₀∈ℝℍ^𝑛×𝑛(𝑄), 𝐴𝑋𝐵+𝐶𝑋^𝐻𝐷=𝐹⟺𝐴𝑋−𝑋₀𝐵+𝐶𝑋−𝑋₀^𝐻𝐷=𝐹−𝐴𝑋₀𝐵−𝐶𝑋^𝐻₀𝐷. (52) Let 𝑋̇=𝑋−𝑋₀ and 𝐹̇=𝐹−𝐴𝑋₀𝐵−𝐶𝑋^𝐻₀𝐷, then Problem 2 is equivalent to finding the least Frobenius norm reflexive solution of the quaternion matrix equation 𝐴𝑋̇𝐵+𝐶𝑋̇^𝐻𝐷=𝐹̇. (53) By using Algorithm 7, let the initial iteration matrix 𝑋̇(1)=𝐴^𝐻𝐺𝐵^𝐻+𝐷𝐺^𝐻𝐶+𝑄𝐴^𝐻𝐺𝐵^𝐻𝑄+𝑄𝐷𝐺^𝐻𝐶Q, where 𝐺 is an arbitrary quaternion matrix in ℍ^𝑚×𝑝, or especially, 𝑋̇(1)=0, we can obtain the least Frobenius norm reflexive solution 𝑋̇^∗ of (53). Then we can obtain the solution of Problem 2, which is 𝑋̆=𝑋̇^∗+𝑋₀. (54)

4. Examples

In this section, we give two examples to illustrate the efficiency of the theoretical results.

Example 18. Consider the quaternion matrix equation 𝐴𝑋𝐵+𝐶𝑋^𝐻𝐷=𝐹, (55) with ,,,,.𝐴=1+𝑖+𝑗+𝑘2+2𝑗+2𝑘𝑖+3𝑗+3𝑘2+𝑖+4𝑗+4𝑘3+2𝑖+2𝑗+𝑘3+3𝑖+𝑗1+7𝑖+3𝑗+𝑘6−7𝑖+2𝑗−4𝑘𝐵=−2+𝑖+3𝑗+𝑘3+3𝑖+4𝑗+𝑘5−4𝑖−6𝑗−5𝑘−1−5𝑖+4𝑗+3𝑘−1+2𝑖+𝑗+𝑘2+2𝑗+6𝑘3−2𝑖+𝑗+𝑘4+𝑖−2𝑗−4𝑘𝐶=1+𝑖+𝑗+𝑘2+2𝑖+2𝑗𝑘2+2𝑖+2𝑗+𝑘−1+𝑖+3𝑗+3𝑘−2+3𝑖+4𝑗+2𝑘6−2𝑖+6𝑗+4𝑘3+𝑖+𝑗+5𝑘𝐷=−1+4𝑗+2𝑘1+2𝑖+3𝑗+3𝑘7+6𝑖+5𝑗+6𝑘3+7𝑖−𝑗+9𝑘4+𝑖+6𝑗+𝑘7+2𝑖+9𝑗+𝑘1+𝑖+3𝑗−3𝑘1+3𝑖+2𝑗+2𝑘𝐹=1+3𝑖+2𝑗+𝑘2−2𝑖+3𝑗+3𝑘−1+4𝑖+2𝑗+𝑘−2+2𝑖−1𝑗 (56) We apply Algorithm 7 to find the reflexive solution with respect to the generalized reflection matrix. 𝑄=.0.2800.96𝑘00−10.00−0.96𝑘0−0.280000−1 (57) For the initial matrix 𝑋(1)=𝑄, we obtain a solution, that is 𝑋^∗()=,=𝑋200.27257−0.23500𝑖+0.034789𝑗+0.054677𝑘0.085841−0.064349𝑖+0.10387𝑗−0.26871𝑘0.028151−0.013246𝑖−0.091097𝑗+0.073137𝑘−0.46775+0.048363𝑖−0.015746𝑗+0.21642𝑘0.043349+0.079178𝑖+0.085167𝑗−0.84221𝑘0.35828−0.13849𝑖−0.085799𝑗+0.11446𝑘0.13454−0.028417𝑖−0.063010𝑗−0.10574𝑘0.10491−0.039417𝑖+0.18066𝑗−0.066963𝑘−0.043349+0.079178𝑖+0.085167𝑗+0.84221𝑘−0.014337−0.14868𝑖−0.011146𝑗−0.00036397𝑘0.097516+0.12146𝑖−0.017662𝑗−0.037534𝑘−0.064483+0.070885𝑖−0.089331𝑗+0.074969𝑘−0.21872+0.18532𝑖+0.011399𝑗+0.029390𝑘0.00048529+0.014861𝑖−0.19824𝑗−0.019117𝑘−0.14099+0.084013𝑖−0.037890𝑗−0.17938𝑘−0.63928−0.067488𝑖+0.042030𝑗+0.10106𝑘 (58) with corresponding residual ‖𝑅(20)‖=2.2775×10⁻¹¹. The convergence curve for the Frobenius norm of the residuals 𝑅(𝑘) is given in Figure 1, where 𝑟(𝑘)=‖𝑅(𝑘)‖.

Figure 1 

The convergence curve for the Frobenius norm of the residuals from Example 18.

Example 19. In this example, we choose the matrices 𝐴,𝐵,𝐶,𝐷,𝐹, and 𝑄 as same as in Example 18. Let 𝑋₀=^{1−0.36𝑖−0.75𝑘0−0.5625𝑖−0.48𝑘0.36𝑖1.280.48𝑗−0.96𝑗}_{01−0.48𝑗10.64−0.75𝑗0.48𝑘0.96𝑗0.640.72}∈ℝℍ^𝑛×𝑛(𝑄). (59) In order to find the optimal approximation reflexive solution to the given matrix 𝑋₀, let 𝑋̇=𝑋−𝑋₀ and 𝐹̇=𝐹−𝐴𝑋₀𝐵−𝐶𝑋^𝐻₀𝐷. Now we can obtain the least Frobenius norm reflexive solution 𝑋̇^∗ of the quaternion matrix equation 𝐴𝑋̇𝐵+𝐶𝑋̇^𝐻𝐷=𝐹̇, by choosing the initial matrix 𝑋̇(1)=0, that is 𝑋̇^∗=𝑋̇()=,20−0.47933+0.021111𝑖+0.11703𝑗−0.066934𝑘−0.52020−0.19377𝑖+0.17420𝑗−0.59403𝑘−0.31132−0.11705𝑖−0.33980𝑗+0.11991𝑘−0.86390−0.21715𝑖−0.037089𝑗+0.40609𝑘−0.24134−0.025941𝑖−0.31930𝑗−0.26961𝑘−0.44552+0.13065𝑖+0.14533𝑗+0.39015𝑘−0.11693−0.27043𝑖+0.22718𝑗−0.0000012004𝑘−0.44790−0.31260𝑖−1.0271𝑗+0.27275𝑘0.24134−0.025941𝑖−0.31930𝑗+0.26961𝑘0.089010−0.67960𝑖+0.43044𝑗−0.045772𝑘−0.089933−0.25485𝑖+0.087788𝑗−0.23349𝑘−0.17004−0.37655𝑖+0.80481𝑗+0.37636𝑘−0.63660+0.16515𝑖−0.13216𝑗+0.073845𝑘−0.034329+0.32283𝑖+0.50970𝑗−0.066757𝑘0.00000090029+0.17038𝑖+0.20282𝑗−0.087697𝑘−0.44735+0.21846𝑖−0.21469𝑗−0.15347𝑘 (60)
with corresponding residual ‖𝑅̇(20)‖=4.3455×10⁻¹¹. The convergence curve for the Frobenius norm of the residuals 𝑅̇(𝑘) is given in Figure 2, where 𝑟(𝑘)=‖𝑅̇(𝑘)‖.
Therefore, the optimal approximation reflexive solution to the given matrix 𝑋₀ is 𝑋̆=𝑋̇^∗+𝑋₀=.0.52067+0.021111𝑖+0.11703𝑗−0.066934𝑘−0.52020−0.55377𝑖+0.17420𝑗−1.3440𝑘−0.31132+0.24295𝑖−0.33980𝑗+0.11991𝑘0.41610−0.21715𝑖−0.037089𝑗+0.40609𝑘−0.24134−0.025941𝑖−0.31930𝑗−0.26961𝑘0.55448+0.13065𝑖−0.33467𝑗+0.39015𝑘−0.11693−0.27043𝑖+0.22718𝑗+0.48000𝑘−0.44790−0.31260𝑖−0.067117𝑗+0.27275𝑘0.24134−0.025941𝑖−0.31930𝑗+0.26961𝑘0.089010−1.2421𝑖+0.43044𝑗−0.52577𝑘−0.089933−0.25485𝑖+0.56779𝑗−0.23349𝑘−0.17004−0.37655𝑖−0.15519𝑗+0.37636𝑘0.36340+0.16515𝑖−0.13216𝑗+0.073845𝑘0.60567+0.32283𝑖−0.24030𝑗−0.066757𝑘0.64000+0.17038𝑖+0.20282𝑗−0.087697𝑘0.27265+0.21846𝑖−0.21469𝑗−0.15347𝑘 (61)
The results show that Algorithm 7 is quite efficient.

Figure 2 

The convergence curve for the Frobenius norm of the residuals from Example 19.

5. Conclusions

In this paper, an algorithm has been presented for solving the reflexive solution of the quaternion matrix equation 𝐴𝑋𝐵+𝐶𝑋^𝐻𝐷=𝐹. By this algorithm, the solvability of the problem can be determined automatically. Also, when the quaternion matrix equation 𝐴𝑋𝐵+𝐶𝑋^𝐻𝐷=𝐹 is consistent over reflexive matrix 𝑋, for any reflexive initial iterative matrix, a reflexive solution can be obtained within finite iteration steps in the absence of roundoff errors. It has been proven that by choosing a suitable initial iterative matrix, we can derive the least Frobenius norm reflexive solution of the quaternion matrix equation 𝐴𝑋𝐵+𝐶𝑋^𝐻𝐷=𝐹 through Algorithm 7. Furthermore, by using Algorithm 7, we solved Problem 2. Finally, two numerical examples were given to show the efficiency of the presented algorithm.

Acknowledgments

This research was supported by Grants from Innovation Program of Shanghai Municipal Education Commission (13ZZ080), the National Natural Science Foundation of China (11171205), the Natural Science Foundation of Shanghai (11ZR1412500), the Discipline Project at the corresponding level of Shanghai (A. 13-0101-12-005), and Shanghai Leading Academic Discipline Project (J50101).