Matrix Equation’s Reflexive and Anti-Reflexive Solutions over Quaternions

Abstract

We consider when the quaternion matrix equation $AXB+CXD=E$ has a reflexive (or anti-reflexive) solution with respect to a given generalized reflection matrix. We adopt a real representation method to derive the solutions when it is solvable. Moreover, we obtain the explicit expressions of the least-squares reflexive (or anti-reflexive) solutions.

1. Introduction

Recall that an $n\times n$ matrix P over the complex number field $\mathbb{C}$ is said to be a generalized reflection matrix if P satisfies the following two conditions: $P^2=I_n$ and $P=P^{*}$ (the conjugate transpose of P). For a given such generalized reflection matrix P, two kinds of special subspaces in ${\mathbb{C}}^{n\times n}$ have been considered (see [1,2,3]).

$$\begin{array}{c}{\mathbb{C}}_r^{n\times n}\left(P\right)=\left\{A\in {\mathbb{C}}^{n\times n}\mid A=PAP\right\}\mathrm{and}{\mathbb{C}}_a^{n\times n}\left(P\right)=\left\{A\in {\mathbb{C}}^{n\times n}\mid A=-PAP\right\}.\end{array}$$

A matrix $A\in {\mathbb{C}}_r^{n\times n}\left(P\right)$ (or $A\in {\mathbb{C}}_a^{n\times n}\left(P\right)$) is called reflexive (or anti-reflexive) about a generalized reflection matrix P. It is well known that a reflexive (or anti-reflexive) matrix is centrosymmetric (or centroskew symmetric) when P is provided by $J_n$, which is an $n\times n$ backward identity matrix having the elements along the southwest-northeast diagonal and with the remaining elements being zeros. Many properties of a generalized reflexive matrix P and anti-reflexive matrix have been widely used in engineering and scientific research (for example, see [4]).

The reflexive and anti-reflexive solutions to the different kinds of complex matrix equations have been studied previously (see e.g., [4,5,6,7,8,9,10], etc.) For instance, in [8], the reflexive and anti-reflexive solutions of the matrix equation $AX=B$ were studied with respect to the generalized reflection matrix P. Moreover, the authors discussed the related problems of the nearness of the matrix. In [5,9], the authors dealt with the complex matrix equation $AXB=C$. The authors in [11,12,13] considered certain quaternion matrix equations. For instance, in [12], the authors derived the sufficient and necessary conditions for the existence and general expressions of reflexive (or anti-reflexive) solutions of the system $A_1{X}_1=C_1,A_2{X}_2=C_2,A{X}_{1}B+C{X}_{2}D=E$. As a more general form of reflexive and anti-reflexive solutions, the $(P,Q)$ reflexive (or anti-reflexive) solutions to matrix equation have been widely investigated (see e.g., [11,13,14,15,16,17,18,19]). In [13], the authors investigated the $(P,Q)$- maximal and minimal rank reflexive (or anti-reflexive) solutions to the system of quaternion matrix equations $AX=B,XC=D$ using the rank method. In [11], an iterative method was proposed to derive the $(P,Q)$-reflextive solution to the system of quaternion matrix equations

$$\sum _{l=1}^M(A_{ls}{X}_l{B}_{ls}+C_{ls}\widetilde{X_l}{D}_{ls})=F_s,s=1,2,\cdots ,N,$$

where $\widetilde{X_l}$ denotes the $\eta$-conjugate of $X_l$. The authors of [17] proposed an iterative algorithm to find the $(P,Q)$-reflexive solution to the system of quaternion matrix equations

$$\sum (A_{l}X{B}_l+C_s{X}^T{D}_s)=F.$$

For more related works, interested readers may refer to [20,21]. The real Hamilton quaternion $\mathbb{H}$ is a four-dimensional division algebra over the real number field $\mathbb{R}$, that is,

$$\mathbb{H}=\left\{a_1+a_{2}i+a_{3}j+a_{4}k\mid i^2=j^2=k^2=-1,ijk=-1,a_1,a_2,a_3,a_4\in \mathbb{R}\right\}.$$

The main obstacle in the study of quaternion matrices comes from the noncommutative multiplication of quaternions. One efficient method for overcoming this difficulty is using real representation methods. In this paper, we mainly study the least-squares reflexive (or anti-reflexive) solutions to the quaternion matrix equation

$$AXB+CXD=E$$

(1)

about the generalized reflection quaternion matrix P following the method of real representation.

The structure of this paper is as follows: in Section 2, we introduce a real representation and several properties that are used in this paper; in Section 3, we explore the solvability conditions and reflexive (or anti-reflexive) solutions to (1) and $AX=B$; next, the explicit expressions of the least-squares reflexive (or anti-reflexive) solutions to (1) are derived in Section 4; finally, a numerical example is provided in Section 5.

2. Real Representation

The real representations of a quaternion matrix can convert problems from the quaternion skew field $\mathbb{H}$ to the complex number field $\mathbb{C}$. The definition of a real representation adopted in this paper is as follows:

Definition 1. 

Let $X=X_0+X_{1}i+X_{2}j+X_{3}k\in {\mathbb{H}}^{m\times n},X_i\in {\mathbb{R}}^{m\times n},i=0,1,2,3$. The real representation is defined as

$${\mathbf{X}}^{\tau }=\left[\begin{array}{cccc}\hfill X_0& \hfill -X_1& -X_2& \hfill -X_3\\ \hfill X_1& \hfill X_0& -X_3& \hfill X_2\\ \hfill X_2& \hfill X_3& X_0& \hfill -X_1\\ \hfill X_3& \hfill -X_2& X_1& \hfill X_0\end{array}\right].$$

To further discuss its properties, we need to use the following special matrices:

$$\begin{array}{cc}\hfill Q_n& =\left[\begin{array}{cccc}0& -I_n& 0& 0\\ I_n& 0& 0& 0\\ 0& 0& 0& I_n\\ 0& 0& -I_n& 0\end{array}\right],\hfill \\ \hfill R_n& =\left[\begin{array}{cccc}0& 0& I_n& 0\\ 0& 0& 0& I_n\\ -I_n& 0& 0& 0\\ 0& -I_n& 0& 0\end{array}\right],\hfill \\ \hfill S_n& =\left[\begin{array}{cccc}0& 0& 0& -I_n\\ 0& 0& -I_n& 0\\ 0& -I_n& 0& 0\\ I_n& 0& 0& 0\end{array}\right].\hfill \end{array}$$

These special matrices have the following properties:

Lemma 1. 

$Q_n,R_n,S_n$ are all orthogonal matrices. Moreover, for any $A\in {\mathbb{H}}^{n\times n},$

$$A^{\tau }{Q}_n=Q_n{A}^{\tau },A^{\tau }{R}_n=R_n{A}^{\tau },A^{\tau }{S}_n=S_n{A}^{\tau }.$$

Next, we summarize the properties of the above real representation, shown below. We denote $A^T$ as the transpose of A.

Proposition 1. 

Let $A,B\in {\mathbb{H}}^{m\times n},C\in {\mathbb{H}}^{m\times s},a\in \mathbb{R}$. Then, the real representation has the following properties:

(a)

${(A+B)}^{\tau }=A^{\tau }+B^{\tau },{\left(aA\right)}^{\tau }=a{A}^{\tau }$;

(b)

${\left(AC\right)}^{\tau }=A^{\tau }{C}^{\tau }$;

(c)

$Q_m{}^T{A}^{\tau }{Q}_n=A^{\tau },R_m{}^T{A}^{\tau }{R}_n=A^{\tau },S_m{}^T{A}^{\tau }{S}_n=A^{\tau }$;

(d)

${\left(A^{*}\right)}^{\tau }={\left(A^{\tau }\right)}^T$;

(e)

For a generalized reflection matrix $P\in {\mathbb{H}}^{n\times n}$, we can preserve $P^{\tau }\in {\mathbb{C}}^{4n\times 4n}$ as a generalized reflection matrix, that is,

$${\left(P^{\tau }\right)}^{*}=P^{\tau },{\left(P^{\tau }\right)}^2=I_{4n}.$$

Proof. 

As Properties (a)–(d) are well known, we only need to prove Property (e). Applying (b) to $P^2=I_n$ and $P^{*}=P$ results in ${\left(P^{\tau }\right)}^2=I_{4n},{\left(P^{*}\right)}^{\tau }=P^{\tau }$. Note that $P^{\tau }$ is a real matrix; thus, combining with (d) yields

$${\left(P^{*}\right)}^{\tau }={\left(P^{\tau }\right)}^T={\left(P^{\tau }\right)}^{*}=P^{\tau }.$$

□

Proposition 2 provides a method to generate a quaternion matrix X from its real representation matrix $X^{\tau }$.

Proposition 2. 

Let $X\in {\mathbb{H}}^{m\times n}$; then,

$$X=\frac{1}{4}\left[I_{n}i{I}_{n}j{I}_{n}k{I}_n\right]X^{\tau }\left[\begin{array}{c}{I}_n\\ -i{I}_n\\ -j{I}_n\\ -k{I}_n\end{array}\right].$$

3. Existence and Expression of Solutions

In Section 3, we mainly consider the consistency of the quaternion Equation (1) using the real representation matrix introduced in Section 2. As an application, we explore the general solution form of $AX=B$ when it is consistent.

Theorem 1. 

Let $A,C\in {\mathbb{H}}^{m\times n},B,D\in {\mathbb{H}}^{n\times p},E\in {\mathbb{H}}^{m\times p}$ and $P\in {\mathbb{H}}^{n\times n}$ be a generalized reflection matrix. Then,

(a)

The real matrix equation

$$A^{\tau }Y{B}^{\tau }+C^{\tau }Y{D}^{\tau }=E^{\tau }$$

(2)

is consistent if and only if the quaternion matrix Equation (1) is consistent.

(b)

If $Y=\pm P^{\tau }Y{P}^{\tau }\in {\mathbb{R}}^{4n\times 4n}$ is the solution to (2), then the quaternion matrix Equation (1) has a solution $X=\pm PXP\in {\mathbb{H}}^{n\times n}$, which is provided by

$$X=\frac{1}{16}\left[I_{n}i{I}_{n}j{I}_{n}k{I}_n\right]\left(Y+Q_{n}Y{Q}_n{}^T+R_{n}Y{R}_n{}^T+S_{n}Y{S}_n{}^T\right)\left[\begin{array}{c}{I}_n\\ -i{I}_n\\ -j{I}_n\\ -k{I}_n\end{array}\right].$$

(3)

Proof. 

First, we show that any reflexive (or anti-reflexive) solution $\mathrm{Y}$ of the real matrix equation

$$A^{\tau }Y{B}^{\tau }+C^{\tau }Y{D}^{\tau }=E^{\tau }$$

with respect to $P^{\tau }$ can generate a reflexive (or anti-reflexive) solution X of the quaternion matrix Equation (1) with respect to P. That is, if (2) is consistent, then (1) is consistent as well.

Assume that (2) has a reflexive (or anti-reflexive) solution Y; then, Y satisfies (2) and

$$\pm P^{\tau }Y{P}^{\tau }=Y.$$

(4)

Applying (c) in Proposition 1 to (2), we obtain

$$\begin{array}{c}\left(Q_m^T{A}^{\tau }{Q}_n\right)Y\left(Q_n^T{B}^{\tau }{Q}_p\right)+\left(Q_m^T{C}^{\tau }{Q}_n\right)Y\left(Q_n^T{D}^{\tau }{Q}_p\right)=\left(Q_m^T{E}^{\tau }{Q}_p\right),\end{array}$$

(5)

$$\left(R_m^T{A}^{\tau }{R}_n\right)Y\left(R_n^T{B}^{\tau }{R}_p\right)+\left(R_m^T{C}^{\tau }{R}_n\right)Y\left(R_n^T{D}^{\tau }{R}_p\right)=\left(R_m^T{E}^{\tau }{R}_p\right),$$

$$\begin{array}{c}\left(S_m^T{A}^{\tau }{S}_n\right)Y\left(S_n^T{B}^{\tau }{S}_p\right)+\left(S_m^T{C}^{\tau }{S}_n\right)Y\left(S_n^T{D}^{\tau }{S}_p\right)=\left(S_m^T{E}^{\tau }{S}_p\right).\end{array}$$

Premultiplying $Q_m$ and postmultiplying ${Q_p}^T$ to both sides of (5), we obtain

$$\begin{array}{c}{A}^{\tau }\left(Q_{n}Y{Q}_n{}^T\right)B^{\tau }+C^{\tau }\left(Q_{n}Y{Q}_n{}^T\right)D^{\tau }=E^{\tau }.\hfill \end{array}$$

In a similar way, we can obtain

$$A^{\tau }\left(R_{n}Y{R}_n{}^T\right)B^{\tau }+C^{\tau }\left(R_{n}Y{R}_n{}^T\right)D^{\tau }=E^{\tau }.$$

$$\begin{array}{c}{A}^{\tau }\left(S_{n}Y{S}_n{}^T\right)B^{\tau }+C^{\tau }\left(S_{n}Y{S}_n{}^T\right)D^{\tau }=E^{\tau }.\hfill \end{array}$$

Moreover, using Lemma 1 and (4) yields

$$Q_{n}Y{Q_n}^T=\pm P^{\tau }\left(Q_{n}Y{Q_n}^T\right)P^{\tau },$$

$$R_{n}Y{R_n}^T=\pm P^{\tau }\left(R_{n}Y{R_n}^T\right)P^{\tau },$$

$$S_{n}Y{S_n}^T=\pm P^{\tau }\left(S_{n}Y{S_n}^T\right)P^{\tau }.$$

By virtue of Property (e) of Proposition 1, $P^{\tau }$ is a generalized reflection matrix. Thus, the above equations imply that $Q_{n}Y{Q}_n{}^T$, $R_{n}Y{R}_n{}^T$ and $S_{n}Y{S}_n{}^T$ are reflexive (or anti-reflexive) solutions of (2) with respect to $P^{\tau }$, as is

$$\mathcal{Y}=\frac{1}{4}\left(Y+Q_{n}Y{Q}_n{}^T+R_{n}Y{R}_n{}^T+S_{n}Y{S}_n{}^T\right).$$

(6)

Assume that $Y={\left(Y_{ij}\right)}_{4\times 4}$ has the form

$$Y=\left[\begin{array}{cccc}{Y}_{11}\hfill & Y_{12}\hfill & Y_{13}\hfill & Y_{14}\hfill \\ Y_{21}\hfill & Y_{22}\hfill & Y_{23}\hfill & Y_{24}\hfill \\ Y_{31}\hfill & Y_{32}\hfill & Y_{33}\hfill & Y_{34}\hfill \\ Y_{41}\hfill & Y_{42}\hfill & Y_{43}\hfill & Y_{44}\hfill \end{array}\right]$$

Then, by substituting it in (6) we can obtain

$$\mathcal{Y}=\left[\begin{array}{cccc}\hfill Z_0& \hfill -Z_1& \hfill -Z_2& \hfill -Z_3\\ \hfill Z_1& \hfill Z_0& \hfill -Z_3& \hfill Z_2\\ \hfill Z_2& \hfill Z_3& \hfill Z_0& \hfill -Z_1\\ \hfill Z_3& \hfill -Z_2& \hfill Z_1& \hfill Z_0\end{array}\right],$$

where

$$\begin{array}{cc}\hfill Z_0& =\frac{1}{4}\left(Y_{11}+Y_{22}+Y_{33}+Y_{44}\right),\hfill \\ \hfill Z_1& =\frac{1}{4}\left(Y_{21}-Y_{12}+Y_{43}-Y_{34}\right),\hfill \\ \hfill Z_2& =\frac{1}{4}\left(Y_{31}-Y_{42}-Y_{13}+Y_{24}\right),\hfill \\ \hfill Z_3& =\frac{1}{4}\left(Y_{41}+Y_{32}-Y_{23}-Y_{14}\right).\hfill \end{array}$$

Now, we construct a new reflexive (anti-reflexive) quaternion matrix $\mathrm{X}$ using $Z_i$:

$$X=Z_0+Z_{1}i+Z_{2}j+Z_{3}k=\frac{1}{4}\left[I_{n}i{I}_{n}j{I}_{n}k{I}_n\right]\mathcal{Y}\left[\begin{array}{c}{I}_n\\ -i{I}_n\\ -j{I}_n\\ -k{I}_n\end{array}\right].$$

It is easy to verify that $X^{\tau }=\mathcal{Y}$. Note that $\mathcal{Y}$ is a solution of (2). By virtue of Property (b) of Proposition 1, we obtain

$$\begin{array}{c}{A}^{\tau }\mathcal{Y}{B}^{\tau }+C^{\tau }\mathcal{Y}{D}^{\tau }=A^{\tau }{X}^{\tau }{B}^{\tau }+C^{\tau }{X}^{\tau }{D}^{\tau }={(AXB+CXD)}^{\tau }=E^{\tau },\end{array}$$

$$\begin{array}{c}\pm P^{\tau }\mathcal{Y}{P}^{\tau }=\pm P^{\tau }{X}^{\tau }{P}^{\tau }=\pm {\left(PXP\right)}^{\tau }=X^{\tau },\end{array}$$

and X is a reflexive (anti-reflexive) solution to (1) with respect to P.

In order to prove that when (1) is consistent, (2) is consistent as well, we assume that $X_0$ is a reflexive (or anti-reflexive) solution to $AXB+CXD=E$ with respect to a generalized reflection matrix $P\in {\mathbb{H}}^{n\times n}$, i.e., $\pm P{X}_{0}P=X_0$.

Applying Property (b) of Proposition 1 yields

$$A^{\tau }{\left(X_0\right)}^{\tau }{B}^{\tau }+C^{\tau }{\left(X_0\right)}^{\tau }{D}^{\tau }=E^{\tau }$$

and

$$\pm P^{\tau }{X}_0^{\tau }{P}^{\tau }=X_0^{\tau }.$$

Observe that $P^{\tau }$ is a generalized reflection matrix. Thus, $X_0^{\tau }$ is a reflexive (or anti-reflexive) solution to (2) with regard to $P^{\tau }$. Finally, it is shown that (1) is consistent if and only if (2) is consistent; if (2) is consistent, then the solution in (3) is our required result. □

Next, we apply Theorem 1 to find the consistency conditions and solutions to the matrix equation

$$AX=B.$$

(7)

We start with a few useful lemmas (see, e.g., [8]). Throughout this paper, $R^{†}$ denotes the Moore–Penrose inverse of a complex matrix R.

Lemma 2. 

Given $A,B\in {\mathbb{C}}^{m\times n}$ and a generalized reflection matrix $P\in {\mathbb{C}}^{n\times n}$, we assume that P can be expressed as

$$P=U\left[\begin{array}{cc}{I}_r& 0\\ 0& -I_{n-r}\end{array}\right]U^{*}.$$

(8)

Moreover, $AU$ and $BU$ have the following partition from

$$AU=\left[A_1,A_2\right],A_1\in {\mathbb{C}}^{m\times r},A_2\in {\mathbb{C}}^{m\times (n-r)},$$

(9)

$$BU=\left[B_1,B_2\right],B_1\in {\mathbb{C}}^{m\times r},B_2\in {\mathbb{C}}^{m\times (n-r)}.$$

(10)

Then, $AX=B$ has a solution $X\in {\mathbb{C}}_r^{n\times n}\left(P\right)$ if and only if

$$A_1{A}_1^{†}{B}_1=B_1,A_2{A}_2^{†}{B}_2=B_2.$$

In this case it has the general solution

$$X=U\left[\begin{array}{cc}{A}_1^{†}{B}_1+\left(I_r-A_1^{†}{A}_1\right)G_1& 0\\ 0& A_2^{†}{B}_2+\left(I_{n-r}-A_2^{†}{A}_2\right)G_2\end{array}\right]U^{*},$$

where $G_1\in {\mathbb{C}}^{r\times r},G_2\in {\mathbb{C}}^{(n-r)\times (n-r)}$are arbitrary matrices.

Lemma 3. 

Given $A,B\in {\mathbb{C}}^{m\times n}$ and a generalized reflection matrix P of size n, we assume that $P\in {\mathbb{C}}^{n\times n}$ can be expressed as (8) and that $AU$ and $BU$ have the following partition from (9) and (10); then, $AX=B$ has a solution $X\in {\mathbb{C}}_a^{n\times n}\left(P\right)$ if and only if

$$A_1{A}_1^{†}{B}_2=B_2,A_2{A}_2^{†}{B}_1=B_1.$$

In this case, it has the general solution

$$X=U\left[\begin{array}{cc}0& A_1^{†}{B}_2+\left(I_r-A_1^{†}{A}_1\right)G_1\\ A_2^{†}{B}_1+\left(I_{n-r}-A_2^{†}{A}_2\right)G_2& 0\end{array}\right]U^{*},$$

where $G_1\in {\mathbb{C}}^{r\times (n-r)},G_2\in {\mathbb{C}}^{(n-r)\times r}$are arbitrary matrices.

For a generalized reflection matrix $P\in {\mathbb{H}}^{n\times n}$, from Properties (b) and (d) of Proposition 1 it can be shown that $P^{\tau }\in {\mathbb{C}}^{4n\times 4n}$ can be preserved as a generalized reflection matrix, that is,

$${\left(P^{\tau }\right)}^{*}=P^{\tau },{\left(P^{\tau }\right)}^2=I_{4n}.$$

Thus, below we can assume that $P^{\tau }$ can be decomposed as

$$P^{\tau }=U\left[\begin{array}{cc}{I}_{4r}& 0\\ 0& -I_{4n-4r}\end{array}\right]U^{*},$$

then decompose $A^{\tau }$ and $B^{\tau }$ as follows:

$$A^{\tau }U=\left[\begin{array}{cc}{\mathcal{A}}_1\hfill & {\mathcal{A}}_2\hfill \end{array}\right],{\mathcal{A}}_1\in {\mathbb{C}}^{4m\times 4r},{\mathcal{A}}_2\in {\mathbb{C}}^{4m\times (4n-4r)},$$

$$B^{\tau }U=\left[\begin{array}{cc}{\mathcal{B}}_1\hfill & {\mathcal{B}}_2\hfill \end{array}\right],{\mathcal{B}}_1\in {\mathbb{B}}^{4m\times 4r},{\mathcal{B}}_2\in {\mathbb{B}}^{4m\times (4n-4r)}.$$

Combining Theorem 1 with Lemma 2 (or Lemma 3), we can derive the following two corollaries.

Corollary 1. 

Let $A,B\in {\mathbb{H}}^{m\times n}$ and $P\in {\mathbb{H}}^{n\times n}$ be a generalized reflection matrix. If

$${\mathcal{A}}_1^{\tau }{\left({{\mathcal{A}}_1}^{\tau }\right)}^{†}{{\mathcal{B}}_1}^{\tau }={{\mathcal{B}}_1}^{\tau },{\mathcal{A}}_2^{\tau }{\left({\mathcal{A}}_2^{\tau }\right)}^{†}{\mathcal{B}}_2^{\tau }={\mathcal{B}}_2^{\tau }$$

hold, then the quaternion matrix Equation (7) has a solution $X=PXP\in {\mathbb{H}}^{n\times n}$, which is provided by

$$X=\frac{1}{16}\left[I_{n}i{I}_{n}j{I}_{n}k{I}_n\right]\left(Y+Q_{n}Y{Q}_n{}^T+R_{n}Y{R}_n{}^T+S_{n}Y{S}_n{}^T\right)\left[\begin{array}{c}{I}_n\\ -i{I}_n\\ -j{I}_n\\ -k{I}_n\end{array}\right],$$

(11)

where

$$Y=U\left[\begin{array}{cc}{\left({{\mathcal{A}}_1}^{\tau }\right)}^{†}{{\mathcal{B}}_1}^{\tau }+(I_{4r}-{\left({{\mathcal{A}}_1}^{\tau }\right)}^{†}{\mathcal{A}}_1^{\tau })G_1& 0\\ 0& {\left({{\mathcal{A}}_2}^{\tau }\right)}^{†}{{\mathcal{B}}_2}^{\tau }+(I_{4n-4r}-{\left({{\mathcal{A}}_2}^{\tau }\right)}^{†}{\mathcal{A}}_2^{\tau })G_2\end{array}\right]U^{*}$$

(12)

and $G_1\in {\mathbb{H}}^{4r\times 4r},G_2\in {\mathbb{H}}^{(4n-4r)\times (4n-4r)}$are arbitrary matrices.

Corollary 2. 

Let $A,B\in {\mathbb{H}}^{m\times n}$, and $P\in {\mathbb{H}}^{n\times n}$ be a generalized reflection matrix. If the equations

$${\mathcal{A}}_1^{\tau }{\left({{\mathcal{A}}_1}^{\tau }\right)}^{†}{{\mathcal{B}}_2}^{\tau }={\mathcal{B}}_2^{\tau },{\mathcal{A}}_2^{\tau }{\left({\mathcal{A}}_2^{\tau }\right)}^{†}{\mathcal{B}}_1^{\tau }={\mathcal{B}}_1^{\tau }$$

hold, then the quaternion matrix Equation (7) has a solution $X=-PXP\in {\mathbb{H}}^{n\times n}$, which is provided by

$$X=\frac{1}{16}\left[I_{n}i{I}_{n}j{I}_{n}k{I}_n\right]\left(Y+Q_{n}Y{Q}_n^T+R_{n}Y{R}_n{}^T+S_{n}Y{S}_n{}^T\right)\left[\begin{array}{c}{I}_n\\ -i{I}_n\\ -j{I}_n\\ -k{I}_n\end{array}\right],$$

(13)

where

$$Y=U\left[\begin{array}{cc}0& {\left({{\mathcal{A}}_1}^{\tau }\right)}^{†}{{\mathcal{B}}_2}^{\tau }+(I_{4r}-{\left({{\mathcal{A}}_1}^{\tau }\right)}^{†}{\mathcal{A}}_1^{\tau })G_1\\ {\left({{\mathcal{A}}_2}^{\tau }\right)}^{†}{{\mathcal{B}}_1}^{\tau }+(I_{4n-4r}-{\left({{\mathcal{A}}_2}^{\tau }\right)}^{†}{\mathcal{A}}_2^{\tau })G_2& 0\end{array}\right]U^{*}$$

and $G_1\in {\mathbb{C}}^{4r\times (4n-4r)},G_2\in {\mathbb{C}}^{(4n-4r)\times 4r}$are arbitrary matrices.

4. Least-Squares Reflexive (or Anti-Reflexive) Solution

In this section, we derive the least-squares reflexive (or anti-reflexive) solutions to the quaternion matrix Equation (1). Let $M=M_1+M_{2}i+M_{3}j+M_{4}k\in {\mathbb{H}}^{p\times m}$, where $M_i\in {\mathbb{R}}^{p\times m}$. We define

$$M_c^{\tau }=\left[\begin{array}{c}{M}_1\\ M_2\\ M_3\\ M_4\end{array}\right].$$

(14)

It is easy to check that for any $N\in {\mathbb{H}}^{m\times q}$,

$${\left(MN\right)}_c^{\tau }=M^{\tau }{N_c}^{\tau }.$$

(15)

Now, using this real representation (14), we can define the Frobenius norm of the quaternion matrix M as

$${\parallel M\left|\right|}_F\equiv \frac{1}{2}\parallel M^{\tau }{\left|\right|}_F\equiv {\parallel {M_c}^{\tau }\parallel }_F.$$

(16)

Let $A,C\in {\mathbb{H}}^{m\times n},B,D\in {\mathbb{H}}^{n\times p},E\in {\mathbb{H}}^{m\times p}$ and $P\in {\mathbb{H}}^{n\times n}$ be a generalized reflection matrix. Next, we find the least-squares reflexive (or anti-reflexive) solutions $X=\pm PXP\in {\mathbb{H}}^{n\times n}$ such that

$$\begin{array}{c}\hfill {\parallel AXB+CXD-E\parallel }_F=\underset{X_0\in {\mathbb{H}}^{n\times n},X_0=\pm P{X}_{0}P}{min}{\parallel A{X}_{0}B+C{X}_{0}D-E\parallel }_F,\end{array}$$

The lemma concerning the least-squares solutions is as follows:

Lemma 4 

([22]). The solutions of the least-squares problem of a complex matrix equation $AX=B$ are

$$X=A^{†}B+(I-A^{†}A)Z,$$

in which Z is arbitrary.

Next, we introduce two important decompositions for a quaternion reflexive (or anti-reflexive) solution with respect to P.

Lemma 5 

([12]). Let $P\in {\mathbb{H}}^{n\times n}$ be nontrivial involuntary matrices with the decompositions

$$P=T\left[\begin{array}{cc}{I}_r& 0\\ 0& -I_{n-r}\end{array}\right]T^{-1}.$$

Then:

(a)

$X\in {\mathbb{H}}^{n\times n}$ is reflexive if and only if X can be expressed as

$$X=T\left[\begin{array}{cc}{X}_1& 0\\ 0& X_2\end{array}\right]T^{-1},$$

where $X_1\in {\mathbb{H}}^{r\times r},X_2\in {\mathbb{H}}^{(n-r)\times (n-r)}.$

(b)

$Y\in {\mathbb{H}}^{n\times n}$ is anti-reflexive if and only if Y can be expressed as

$$Y=T\left[\begin{array}{cc}0& Y_1\\ Y_2& 0\end{array}\right]T^{-1},$$

where $Y_1\in {\mathbb{H}}^{r\times (n-r)},Y_2\in {\mathbb{H}}^{(n-r)\times r}$.

Next, we present the main result of this section. According to the decomposition of a reflexive (or anti-reflexive) matrix X in Lemma 5, we can partition the following matrices:

$$AT=[T_1,T_2],T^{-1}B=\left[\begin{array}{c}{S}_1\\ S_2\end{array}\right],$$

(17)

$$CT=[E_1,E_2],T^{-1}D=\left[\begin{array}{c}{F}_1\\ F_2\end{array}\right],$$

(18)

where $T_1,E_1\in {\mathbb{H}}^{m\times r}$, $T_2,E_2\in {\mathbb{H}}^{m\times (n-r)}$, $S_1,F_1\in {\mathbb{H}}^{r\times p}$, $S_2,F_2\in {\mathbb{H}}^{(n-r)\times p}$.

Theorem 2. 

Let $A,C\in {\mathbb{H}}^{m\times n},B,D\in {\mathbb{H}}^{n\times p},E\in {\mathbb{H}}^{m\times p}$ and $P\in {\mathbb{H}}^{n\times n}$ be a generalized reflection matrix. Then, the general expression of the least-squares reflexive solution to the matrix Equation (1) is provided by

$$X=T\left[\begin{array}{cc}{X}_1& 0\\ 0& X_2\end{array}\right]T^{-1},$$

where T is provided in Lemma 5,

$$\mathrm{V}\mathrm{e}\mathrm{c}\left(X_1\right)=\left[\begin{array}{cccccccc}{I}_{r^2}\hfill & 0_1\hfill & i{I}_{r^2}\hfill & 0_1\hfill & j{I}_{r^2}\hfill & 0_1\hfill & k{I}_{r^2}\hfill & 0_1\hfill \end{array}\right]y,$$

$$\mathrm{V}\mathrm{e}\mathrm{c}\left(X_2\right)=\left[\begin{array}{cccccccc}{0}_2\hfill & I_{{(n-r)}^2}\hfill & 0_2\hfill & i{I}_{{(n-r)}^2}\hfill & 0_2\hfill & j{I}_{{(n-r)}^2}\hfill & 0_2\hfill & k{I}_{{(n-r)}^2}\hfill \end{array}\right]y,$$

$$\begin{gathered} \begin{array}{cc}& y=P^{†}b+(I-P^{†}P)z,\end{array} \\ P={[({S_1}^T\otimes T_1+{F_1}^T\otimes E_1),({S_2}^T\otimes T_2+{F_2}^T\otimes E_2)]}^{\tau },b={\left(\mathrm{V}\mathrm{e}\mathrm{c}\left(B\right)\right)}_c^{\tau }, \\ {0}_1 \mathit{is} \mathit{the} \mathit{zero} \mathit{matrix} \mathit{with} \mathit{size} r^2\times {(n-r)}^2, \\ {0}_2 \mathit{is} \mathit{the} \mathit{zero} \mathit{matrix} \mathit{with} \mathit{size} {(n-r)}^2\times r^2, \\ z\in {\mathbb{C}}^{{4r}^2+{4(n-r)}^2} \mathit{is} \mathit{arbitrary}. \end{gathered}$$

Proof. 

By virtue of (a) in Lemma 5, we have

$$\begin{array}{cc}\hfill {\parallel AXB+CXD-E\parallel }_F& =\parallel AT\left[\begin{array}{cc}{X}_1& 0\\ 0& X_2\end{array}\right]T^{-1}B+CT\left[\begin{array}{cc}{X}_1& 0\\ 0& X_2\end{array}\right]T^{-1}{D-E\parallel }_F\hfill \\ \hfill & =\parallel T_1{X}_1{S}_1+T_2{X}_2{S}_2+E_1{X}_1{F}_1+E_2{X}_2{F}_2{-E\parallel }_F\hfill \end{array}$$

Then, (1) has a least-squares reflexive solution X if and only if

$$T_1{X}_1{S}_1+T_2{X}_2{S}_2+E_1{X}_1{F}_1+E_2{X}_2{F}_2=E$$

has a least-squares solution pair $\{X_1,X_2\}$. Using the vector operation with the above equation results in

$$({S_1}^T\otimes T_1+{F_1}^T\otimes E_1)\mathrm{V}\mathrm{e}\mathrm{c}\left(X_1\right)+({S_2}^T\otimes T_2+{F_2}^T\otimes E_2)\mathrm{V}\mathrm{e}\mathrm{c}\left(X_2\right)=\mathrm{V}\mathrm{e}\mathrm{c}\left(E\right)$$

i.e.,

$$[({S_1}^T\otimes T_1+{F_1}^T\otimes E_1),({S_2}^T\otimes T_2+{F_2}^T\otimes E_2)]\left[\begin{array}{c}\mathrm{V}\mathrm{e}\mathrm{c}\left(X_1\right)\\ \mathrm{V}\mathrm{e}\mathrm{c}\left(X_2\right)\end{array}\right]=\mathrm{V}\mathrm{e}\mathrm{c}\left(E\right).$$

(19)

Taking the real representations on both sides of (19),

$${[({S_1}^T\otimes T_1+{F_1}^T\otimes E_1),({S_2}^T\otimes T_2+{F_2}^T\otimes E_2)]}^{\tau }{\left[\begin{array}{c}\mathrm{V}\mathrm{e}\mathrm{c}\left(X_1\right)\\ \mathrm{V}\mathrm{e}\mathrm{c}\left(X_2\right)\end{array}\right]}^{\tau }={\mathrm{V}\mathrm{e}\mathrm{c}\left(E\right)}^{\tau },$$

It follows from (15) and (16) that

$$\begin{array}{cc}\hfill & \parallel {[({S_1}^T\otimes T_1+{F_1}^T\otimes E_1),({S_2}^T\otimes T_2+{F_2}^T\otimes E_2)]}^{\tau }{\left(\left[\begin{array}{c}\mathrm{V}\mathrm{e}\mathrm{c}\left(X_1\right)\\ \mathrm{V}\mathrm{e}\mathrm{c}\left(X_2\right)\end{array}\right]\right)}^{\tau }-{\left(\mathrm{V}\mathrm{e}\mathrm{c}\right(E\left)\right)}^{\tau }{\parallel }_F\hfill \\ \hfill =& 2\parallel {[({S_1}^T\otimes T_1+{F_1}^T\otimes E_1),({S_2}^T\otimes T_2+{F_2}^T\otimes E_2)]}^{\tau }{\left(\left[\begin{array}{c}\mathrm{V}\mathrm{e}\mathrm{c}\left(X_1\right)\\ \mathrm{V}\mathrm{e}\mathrm{c}\left(X_2\right)\end{array}\right]\right)}_c^{\tau }-{\left(\mathrm{V}\mathrm{e}\mathrm{c}\left(E\right)\right)}_c^{\tau }{\parallel }_F,\hfill \end{array}$$

Thus, we only need to find the least-squares solution of

$$Py=b,$$

where

$$P={[({S_1}^T\otimes T_1+{F_1}^T\otimes E_1),({S_2}^T\otimes T_2+{F_2}^T\otimes E_2)]}^{\tau },$$

$$y={\left(\left[\begin{array}{c}\mathrm{V}\mathrm{e}\mathrm{c}\left(X_1\right)\\ \mathrm{V}\mathrm{e}\mathrm{c}\left(X_2\right)\end{array}\right]\right)}_c^{\tau },b={({\mathrm{V}\mathrm{e}\mathrm{c}\left(E\right))}_c}^{\tau }.$$

It follows from Lemma 4 that $y=P^{†}b+(I-P^{†}P)z$ is the least-squares solution. Clearly,

$$\mathrm{V}\mathrm{e}\mathrm{c}\left(X_1\right)=\left[\begin{array}{cccccccc}{I}_{r^2}\hfill & 0_1\hfill & i{I}_{r^2}\hfill & 0_1\hfill & j{I}_{r^2}\hfill & 0_1\hfill & k{I}_{r^2}\hfill & 0_1\hfill \end{array}\right]y,$$

$$\mathrm{V}\mathrm{e}\mathrm{c}\left(X_2\right)=\left[\begin{array}{cccccccc}{0}_2\hfill & I_{{(n-r)}^2}\hfill & 0_2\hfill & i{I}_{{(n-r)}^2}\hfill & 0_2\hfill & j{I}_{{(n-r)}^2}\hfill & 0_2\hfill & k{I}_{{(n-r)}^2}\hfill \end{array}\right]y,$$

$0_1$ is the zero matrix with the size of $r^2\times {(n-r)}^2$ and $0_2$ is the zero matrix with the size of ${(n-r)}^2\times r^2$. Then,

$$X=T\left[\begin{array}{cc}{X}_1& 0\\ 0& X_2\end{array}\right]T^{-1}$$

is our required solution. □

Theorem 3. 

Let $A,C\in {\mathbb{H}}^{m\times n},B,D\in {\mathbb{H}}^{n\times p},E\in {\mathbb{H}}^{m\times p}$ and $P\in {\mathbb{H}}^{n\times n}$ be a generalized reflection matrix. Then, the general expression of the least-squares anti-reflexive solution to the matrix Equation (1) is provided by

$$X=T\left[\begin{array}{cc}0& Y_1\\ Y_2& 0\end{array}\right]T^{-1},$$

where T is provided in Lemma 5 and $T_1,T_2,S_1,S_2,E_1,E_2,F_1,F_2$ by (17) and (18).

$$Vec\left(Y_1\right)=\left[\begin{array}{cccccccc}{I}_{r(n-r)}\hfill & 0_3\hfill & i{I}_{r(n-r)}\hfill & 0_3\hfill & j{I}_{r(n-r)}\hfill & 0_3\hfill & k{I}_{r(n-r)}\hfill & 0_3\hfill \end{array}\right]y,$$

$$Vec\left(Y_2\right)=\left[\begin{array}{cccccccc}{0}_3\hfill & I_{r(n-r)}\hfill & 0_3\hfill & i{I}_{r(n-r)}\hfill & 0_3\hfill & j{I}_{r(n-r)}\hfill & 0_3\hfill & k{I}_{r(n-r)}\hfill \end{array}\right]y,$$

$$\begin{gathered} \begin{array}{cc}& y=P^{†}b+(I-P^{†}P)z,\end{array} \\ P={[({S_2}^T\otimes T_1+{F_2}^T\otimes E_1),({S_1}^T\otimes T_2+{F_1}^T\otimes E_2)]}^{\tau },b={\left(\mathrm{V}\mathrm{e}\mathrm{c}\left(E\right)\right)}_c^{\tau }, \\ {0}_3 \mathit{is} \mathit{the} \mathit{zero} \mathit{matrix} \mathit{with} \mathit{the} \mathit{size} \mathit{of} r(n-r)\times r(n-r), \\ z\in {\mathbb{C}}^{8r(n-r)} \mathit{is} \mathit{arbitrary}. \end{gathered}$$

5. Numerical Example

In Section 5, we provide a numerical example to explain our result (quoted to two decimal places).

Example 1. 

Let $P\in {\mathbb{H}}^{4\times 4}$,

$$P=\left[\begin{array}{cccc}0.28& 0& 0.96k& 0\\ 0& 0.28& 0& -0.96j\\ -0.96k& 0& -0.28& 0\\ 0& 0.96j& 0& -0.28\end{array}\right],$$

where P satisfies $P^{*}=P$ and $P^2=I$. For the quaternion matrix equation

$$AX=B,$$

where

$$A=[1,0,0,-1]+[0,2,4,0]i+[3,5,0,7]j+[0,4,0,0]k,$$

$$\begin{array}{cc}B=& [-0.06,-1.32,-0.86,-2.80]+[2.70,-0.72,-5.12,-0.48]i\\ & +[4.84,2.10,0.10,-1.76]j+[1.48,0.36,2.08,-0.96]k.\end{array}$$

Next we, find the reflexive solution with respect to $P.$

First, we convert the matrix equation over $\mathbb{H}$ into a matrix equation over $\mathbb{R}$ by real representation:

$$A^{\tau }Y=B^{\tau },$$

(20)

where $Y\in {\mathbb{C}}^{16\times 16}$ is an unknown matrix with

$$A^{\tau }=\left[\begin{array}{cccccccccccccccc}1& 0& 0& -1& 0& -2& -4& 0& -3& -5& 0& -7& 0& -4& 0& 0\\ 0& 2& 4& 0& 1& 0& 0& -1& 0& -4& 0& 0& 3& 5& 0& 7\\ 3& 5& 0& 7& 0& 4& 0& 0& 1& 0& 0& -1& 0& -2& -4& 0\\ 0& 4& 0& 0& -3& -5& 0& -7& 0& 2& 4& 0& 1& 0& 0& -1\end{array}\right]\in {\mathbb{C}}^{4\times 16},$$

$$\begin{array}{c}{B}^{\tau }=\left[\begin{array}{ccccccc}-0.06& -1.32& -0.86& -2.80& -2.70& -0.72& 5.12\\ 2.70& -0.72& -5.12& -0.48& -0.06& -1.32& -0.86\\ 4.84& 2.10& 0.10& -1.76& 1.48& 0.36& 2.08\\ 1.48& 0.36& 2.08& -0.96& -4.84& -2.10& -0.10\end{array}\right.\hfill \\ \left.\begin{array}{ccccccccc}0.48& -4.84& -2.10& -0.10& 1.76& -1.48& -0.36& -2.08& 0.96\\ -2.80& -1.48& -0.36& -2.08& 0.96& 4.84& 2.10& 0.10& -1.76\\ -0.96& -0.06& -1.32& -0.86& -2.80& -2.70& 0.72& 5.12& 0.48\\ 1.76& 2.70& -0.72& -5.12& -0.48& -0.06& -1.32& -0.86& -2.80\end{array}\right]\in {\mathbb{C}}^{4\times 16}.\hfill \end{array}$$

Using MATLAB, we have

$${\mathcal{A}}_1^{\tau }{\left({{\mathcal{A}}_1}^{\tau }\right)}^{†}{{\mathcal{B}}_1}^{\tau }={\mathcal{B}}_1^{\tau },{\mathcal{A}}_2^{\tau }{\left({\mathcal{A}}_2^{\tau }\right)}^{†}{\mathcal{B}}_2^{\tau }={\mathcal{B}}_2^{\tau }.$$

Thus, the real matrix Equation (20) is solvable. By computing (12), we have a reflexive solution Y with regard to $P^{\tau }$:

$$\begin{array}{c}Y=\left[\begin{array}{cccccccc}0.68& -0.18& -0.24& 0.00& 0.00& 0.00& -0.24& 0.00\\ 0.18& 0.00& 0.00& -0.24& 0.14& 0.00& 0.48& 0.00\\ 0.24& 0.00& 0.82& 0.00& -0.24& 0.00& 0.00& -0.32\\ 0.48& 0.24& 0.14& 0.00& 0.00& 0.00& -0.32& 0.00\\ 0.00& 0.00& 0.24& 0.00& 0.68& -0.18& -0.24& 0.00\\ -0.14& 0.00& -0.48& 0.00& 0.18& 0.00& 0.00& -0.24\\ 0.24& 0.00& 0.00& 0.32& 0.24& 0.00& 0.82& 0.00\\ 0.00& 0.00& 0.32& 0.00& 0.48& 0.24& 0.14& 0.00\\ -0.18& 0.00& 0.00& -0.24& -0.18& 0.00& -0.24& 0.00\\ 0.36& 0.18& 0.48& 0.00& 0.00& 0.00& -0.24& 0.00\\ 0.00& 0.00& 0.32& 0.00& 0.24& -0.24& -0.32& 0.00\\ -0.24& 0.00& 0.00& 0.32& 0.48& 0.00& -0.64& 0.00\\ -0.18& 0.00& -0.24& 0.00& 0.18& 0.00& 0.00& 0.24\\ 0.00& 0.00& -0.24& 0.00& -0.36& -0.18& -0.48& 0.00\\ 0.24& -0.24& -0.32& 0.00& 0.00& 0.00& -0.32& 0.00\\ 0.48& 0.00& -0.64& 0.00& 0.24& 0.00& 0.00& -0.32\end{array}\right.\hfill \\ \left.\begin{array}{cccccccc}0.18& 0.00& 0.00& 0.24& 0.18& 0.00& 0.24& 0.00\\ -0.36& -0.18& -0.48& 0.00& 0.00& 0.00& 0.24& 0.00\\ 0.00& 0.00& -0.32& 0.00& -0.24& 0.24& 0.32& 0.00\\ 0.24& 0.00& 0.00& -0.32& -0.48& 0.00& 0.64& 0.00\\ 0.18& 0.00& 0.24& 0.00& -0.18& 0.00& 0.00& -0.24\\ 0.00& 0.00& 0.24& 0.00& 0.36& 0.18& 0.48& 0.00\\ -0.24& 0.24& 0.32& 0.00& 0.00& 0.00& 0.32& 0.00\\ -0.48& 0.00& 0.64& 0.00& -0.24& 0.00& 0.00& 0.32\\ 0.68& -0.18& -0.24& 0.00& 0.00& 0.00& -0.24& 0.00\\ 0.18& 0.00& 0.00& -0.24& 0.14& 0.00& 0.48& 0.00\\ 0.24& 0.00& 0.82& 0.00& -0.24& 0.00& 0.00& -0.32\\ 0.48& 0.24& 0.14& 0.00& 0.00& 0.00& 0.00& 0.00\\ 0.00& 0.00& 0.24& 0.00& 0.68& -0.18& -0.24& 0.00\\ -0.14& 0.00& -0.48& 0.00& 0.00& 0.00& 0.00& -0.24\\ 0.24& 0.00& 0.00& 0.32& 0.24& 0.00& 0.82& 0.00\\ 0.00& 0.00& 0.32& 0.00& 0.48& 0.24& 0.14& 0.00\end{array}\right].\hfill \end{array}$$

We can verify $Y=P^{\tau }Y{P}^{\tau }$, $A^{\tau }Y=B^{\tau }$. Now, per Corollary 1, we can construct our required X from Y by (11):

$$\begin{array}{cc}\hfill X=& \left[\begin{array}{cccc}0.68& -0.18& -0.24& 0.00\\ 0.18& 0.00& 0.00& -0.24\\ 0.24& 0.00& 0.82& 0.00\\ 0.48& 0.24& 0.14& 0.00\end{array}\right]+\left[\begin{array}{cccc}0.00& 0.00& 0.24& 0.00\\ -0.14& 0.00& -0.48& 0.00\\ 0.24& 0.00& 0.00& 0.32\\ 0.00& 0.00& 0.32& 0.00\end{array}\right]i\hfill \\ \hfill +& \left[\begin{array}{cccc}-0.18& 0.00& 0.00& -0.24\\ 0.36& 0.18& 0.48& 0.00\\ 0.00& 0.00& 0.32& 0.00\\ -0.24& 0.00& 0.00& 0.32\end{array}\right]j+\left[\begin{array}{cccc}-0.18& 0.00& -0.24& 0.00\\ 0.00& 0.00& -0.24& 0.00\\ 0.24& -0.24& -0.32& 0.00\\ 0.48& 0.00& -0.64& 0.00\end{array}\right]k.\hfill \end{array}$$

By direct computation, we find that X satisfies $X=PXP,AX=B.$