Solving the Dual Generalized Commutative Quaternion Matrix Equation AXB = C

Abstract

Dual generalized commutative quaternions have broad application prospects in many fields. Additionally, the matrix equation $AXB=C$ has important applications in mathematics and engineering, especially in control systems, economics, computer science, and other disciplines. However, research on the matrix equation $AXB=C$ over the dual generalized commutative quaternions remains relatively insufficient. In this paper, we derive the necessary and sufficient conditions for the solvability of the dual generalized commutative quaternion matrix equation $AXB=C$. Furthermore, we provide the general solution expression for this matrix equation, when it is solvable. Finally, a numerical algorithm and an example are provided to confirm the reliability of the main conclusions.

1. Introduction

In 1843, the renowned mathematician William Rowan Hamilton [1] introduced the concept of quaternions, making a groundbreaking discovery. The set of quaternions is typically denoted as

$$\mathbb{H}=\{q=q_0+q_{1}i+q_{2}j+q_{3}k:i^2=j^2=k^2=-1,ijk=-1,q_0,q_1,q_2,q_3\in \mathbb{R}\}.$$

It is easy to verify that $\mathbb{H}$ is a four-dimensional non-commutative division ring over the real number field $\mathbb{R}$. Quaternions exhibit a diverse array of applications, spanning not only the realm of mathematics but also playing a pivotal role in mechanics, quantum physics, signal processing, and numerous other disciplines [2,3,4]. Nevertheless, the non-commutativity of quaternion multiplication poses significant challenges in academic research. Corrado Segre [5] expanded on the definition of quaternions, overcoming the difficulty that quaternions are not commutative and proposing the concept of commutative quaternions of the following form.

$$\mathbb{S}=\{a=a_0+a_{1}i+a_{2}j+a_{3}k:a_0,a_1,a_2,a_3\in \mathbb{R}\},$$

here $i,j,k$ satisfying the following conditions.

$$i^2=k^2=-1,j^2=1,ij=ji=k,jk=kj=i,ki=ik=-j.$$

Presently, commutative quaternions have garnered widespread application in the domains of digital signal and image processing [6,7,8,9]. Tian et al. [10] expanded upon the notion of commutative quaternions, introducing the concept of generalized commutative quaternions, which is defined as

$${\mathbb{S}}_g=\{a=a_0+a_{1}i+a_{2}j+a_{3}k:a_0,a_1,a_2,a_3\in \mathbb{R}\},$$

where $i,j$ and k satisfy the following fundamental properties.

$$i^2=\alpha ,j^2=\beta ,k^2=ijk=\alpha \beta ,ij=ji=k,jk=kj=\beta i,ki=ik=\alpha j,\alpha ,\beta \in \mathbb{R}\setminus \left\{0\right\}.$$

Specifically, when $\alpha =-1,\beta =1$, the generalized commutative quaternion algebra ${\mathbb{S}}_g$ reduces to the commutative quaternion algebra $\mathbb{S}$. Similarly, the set of generalized commutative quaternion matrices can be written as

$${\mathbb{S}}_g^{m\times n}=\{X=X_0+X_{1}i+X_{2}j+X_{3}k:X_0,X_1,X_2,X_3\in {\mathbb{R}}^{m\times n}\},$$

here $i,j$ and k satisfy the same conditions as above.

In 1873, Clifford [11] introduced the concept of dual numbers as an extension of real numbers. This is a development that has garnered significant attention due to their pivotal role in kinematic synthesis, robotics, and numerous other disciplines [12,13]. As an extension of quaternions, dual quaternions have found extensive applications in theoretical kinematics, 3D computer graphics, robotics, and numerous other domains [14,15,16]. Similarly, we can further extend the generalized commutative quaternion concept by incorporating dual numbers, thus arriving at the concept of the dual generalized commutative quaternion. For the definition of dual numbers and dual generalized commutative quaternions, please refer to Section 2.

As everyone knows, the classical complex matrix equation

$$AXB=C$$

(1)

has a wide range of applications in the field of control systems, which has attracted the attention of many scholars. Roger Penrose [17] provided the general solution and solvable conditions for the matrix Equation $\left(1\right)$. In 2003, Liao and Bai [18] conducted in-depth research on the least squares solution of matrix Equation $\left(1\right)$ based on their work, especially with regard to symmetric positive semidefinite matrices. Subsequently, the centrosymmetric solution to the matrix Equation $\left(1\right)$ was formulated by Peng [19]. Building upon this foundation, Deng et al. [20] delved into the generalized formulation of the Hermitian solution pertaining to the same matrix equation. Moreover, many scholars have studied various particular solutions of matrix equation (1), as detailed in references [21,22,23,24,25]. Additionally, corresponding numerical algorithms for these specific solutions have been proposed, as discussed in references [26,27,28,29,30]. Later, the research on matrix Equation $\left(1\right)$ was gradually extended to the quaternion algebras. Ivan Kyrchei [31] first derived explicit determinantal representation formulas for the solution of the quaternionic matrix Equation (1). Subsequently, Ivan Kyrchei [32] formulated determinantal representation formulas for the quaternion two-sided generalized Sylvester matrix equation $A_1{X}_1{B}_1+A_2{X}_2{B}_2=C.$ Recently, Chen et al. [33] shifted the perspective to the dual quaternions, providing the necessary and sufficient conditions for the equation to be solvable. In addition, Si and Wang [34] studied the solvability and general solution over dual split quaternions. At present, the research results are still being further expanded.

Up until now, there is a paucity of research information pertaining to the Equation $\left(1\right)$ in the context of dual generalized commutative quaternions. Therefore, this article aims to establish the necessary and sufficient conditions for the solvability of the Equation $\left(1\right)$ over dual generalized commutative quaternions; furthermore, we present the expression of the general solution of the Equation $\left(1\right)$ when it is consistent.

For clarity, hereby introduce some symbols. Let ${\mathbb{R}}^{m\times n},{\mathbb{S}}_g^{m\times n}$ denote the set of all $m\times n$ matrices with real-valued entries and the set of all $m\times n$ matrices whose elements belong to the set ${\mathbb{S}}_g$, respectively. For $A\in {\mathbb{R}}^{m\times n}$, its transpose, conjugate transpose, and rank are represented by $A^{\mathrm{T}},A^{*}$ and $rank\left(A\right)$, respectively. Additionally, I and O refer to the identity matrix and zero matrix, respectively. The Kronecker product of two matrices $A=\left(a_{ij}\right)\in {\mathbb{R}}^{m\times n}$ and $B=\left(b_{ij}\right)\in {\mathbb{R}}^{s\times t}$ is defined as $A\otimes B=\left(a_{ij}B\right)\in {\mathbb{R}}^{ms\times nt}$. The vectorization operator is denoted by $\mathrm{Vec}\left(A\right)$, which is expressed as $\mathrm{Vec}\left(A\right)=(x_1^{\mathrm{T}},x_2^{\mathrm{T}},...,x_n^{\mathrm{T}})$, where $x_i(i=1,2,\cdots ,n)$ represents the i-th column vector of A. The Moore-Penrose inverse is denoted as $A^{†}$ and satisfies the following equations.

$$A{A}^{†}A=A,A^{†}A{A}^{†}=A^{†},{\left(A{A}^{†}\right)}^{*}=A{A}^{†},{\left(A^{†}A\right)}^{*}=A^{†}A.$$

Lastly, we denote $L_A=I-A^{†}A,R_A=I-A{A}^{†}$.

The structure of this article is outlined as follows. We devote Section 2 to revisiting several definitions, fundamental properties and lemmas that serve as the foundation for our subsequent analysis. In Section 3, we derive the necessary and sufficient conditions for the solvability of the matrix Equation (1) over the dual generalized commutative quaternions. Furthermore, we present the general expression for the solution when it is consistent. Finally, accompanying algorithms and numerical examples are offered in Section 4 to demonstrate the application and validity of our theoretical results.

2. Preliminaries

This section delves into a review of definitions about dual numbers, dual generalized commutative quaternions, and their attendant propositions. Furthermore, we also introduce some lemmas and properties that will be used in subsequent proofs.

Definition 1

([35]). Suppose that $x_0,x_1\in \mathbb{R},$ we say that x is a dual number if x is in the form

$$x=x_0+x_1ϵ,$$

where ϵ is an infinitesimal unit, satisfying ${ϵ}^2=0.$ Denote that $x_0$ and $x_1$ as the real part of x and the dual part of x, respectively.

The multiplication of infinitesimal unit $ϵ$ with real numbers, complex numbers, and quaternions exhibits commutativity. The collection of dual numbers is denoted as

$$\mathbb{D}=\{x=x_0+x_1ϵ|x_0,x_1\in \mathbb{R},{ϵ}^2=0\}.$$

For $x=x_0+x_1ϵ,y=y_0+y_1ϵ\in \mathbb{D},$ if $x_0=y_0$ and $x_1=y_1,$ then $x=y.$ The addition and multiplication operations of dual numbers satisfy the following properties.

$$\begin{array}{cc}\hfill & x+y=x_0+y_0+(x_1+y_1)ϵ,\hfill \\ \hfill & xy=x_0{y}_0+(x_0{y}_1+x_1{y}_0)ϵ.\hfill \end{array}$$

Definition 2.

If $z_0,z_1\in {\mathbb{S}}_g,$ then z is a dual generalized commutative quaternion if z is in the form

$$z=z_0+z_1ϵ$$

where ϵ is an infinitesimal unit, satisfying ${ϵ}^2=0.$

The set of dual generalized commutative quaternions is formally defined as

$${\mathbb{DS}}_g=\{z=z_0+z_1ϵ|{ϵ}^2=0,z_0,z_1\in {\mathbb{S}}_g\}.$$

The addition and multiplication rules for the dual generalized commutative quaternions are similar to the dual number.

Let $X_0,X_1\in {\mathbb{S}}_g^{m\times n}.$ X is called a dual generalized commutative quaternion matrix if it has the form $X=X_0+X_1ϵ.$ We use ${\mathbb{DS}}_g^{m\times n}$ to represent all dual generalized commutative quaternion matrices. For $X,Y\in {\mathbb{DS}}_g^{m\times n},$ we have $X=Y$, if $X_0=Y_0,X_1=Y_1$. Furthermore,

$$\begin{array}{cc}\hfill & X+Y=X_0+Y_0+(X_1+Y_1)ϵ,\hfill \\ \hfill & XY=X_0{Y}_0+(X_0{Y}_1+X_1{Y}_0)ϵ.\hfill \end{array}$$

Definition 3

([36]). For $X=X_0+X_{1}i+X_{2}j+X_{3}k\in {\mathbb{S}}_g^{m\times n}$, where $X_0,X_1,X_2,X_3\in {\mathbb{R}}^{m\times n},\alpha ,\beta \in \mathbb{R}\setminus \left\{0\right\}$, there are three kinds of real representations of X denoted by $X^{{\sigma }_{\eta }},\eta \in \{i,j,k\}$, where

$$X^{{\sigma }_i}=\left(\begin{array}{cccc}{X}_0& \alpha X_1& \beta X_2& \alpha \beta X_3\\ X_1& X_0& \beta X_3& \beta X_2\\ X_2& \alpha X_3& X_0& \alpha X_1\\ X_3& X_2& X_1& X_0\end{array}\right),$$

$$X^{{\sigma }_j}=V_m{X}^{{\sigma }_i}=\left(\begin{array}{cccc}-X_0& -\alpha X_1& -\beta X_2& -\alpha \beta X_3\\ X_1& X_0& \beta X_3& \beta X_2\\ X_2& \alpha X_3& X_0& \alpha X_1\\ -X_3& -X_2& -X_1& -X_0\end{array}\right),$$

$$X^{{\sigma }_k}=W_m{X}^{{\sigma }_i}=\left(\begin{array}{cccc}{X}_0& \alpha X_1& \beta X_2& \alpha \beta X_3\\ X_1& X_0& \beta X_3& \beta X_2\\ -X_2& -\alpha X_3& -X_0& -\alpha X_1\\ -X_3& -X_2& -X_1& -X_0\end{array}\right).$$

Here,

$$V_m=\left(\begin{array}{cccc}-I_m& 0& 0& 0\\ 0& I_m& 0& 0\\ 0& 0& I_m& 0\\ 0& 0& 0& -I_m\end{array}\right),$$

(2)

$$W_m=\left(\begin{array}{cccc}{I}_m& 0& 0& 0\\ 0& I_m& 0& 0\\ 0& 0& -I_m& 0\\ 0& 0& 0& -I_m\end{array}\right).$$

(3)

.

The following lemmas are useful in the subsequent proof process.

Lemma 1

([36,37]). For $X,Y\in {\mathbb{S}}_g^{m\times n},Z\in {\mathbb{S}}_g^{n\times s},\lambda \in \mathbb{R},\alpha \in \mathbb{R}\setminus \left\{0\right\}$ and $\beta \in \mathbb{R}\setminus \left\{0\right\}$, set

$$G_n=\left(\begin{array}{cccc}{I}_n& 0& 0& 0\\ 0& -\alpha I_n& 0& 0\\ 0& 0& \beta I_n& 0\\ 0& 0& 0& -\alpha \beta I_n\end{array}\right),$$

$$R_n=\left(\begin{array}{cccc}0& \alpha I_n& 0& 0\\ I_n& 0& 0& 0\\ 0& 0& 0& \alpha I_n\\ 0& 0& I_n& 0\end{array}\right),$$

$$S_n=\left(\begin{array}{cccc}0& 0& \beta I_n& 0\\ 0& 0& 0& \beta I_n\\ I_n& 0& 0& 0\\ 0& I_n& 0& 0\end{array}\right),$$

$$T_n=\left(\begin{array}{cccc}0& 0& 0& \alpha \beta I_n\\ 0& 0& \beta I_n& 0\\ 0& \alpha I_n& 0& 0\\ I_n& 0& 0& 0\end{array}\right).$$

Then the following statements are satisfied.

$\left(1\right)X=Y\iff X^{{\sigma }_{\eta }}=Y^{{\sigma }_{\eta }},\eta \in \{i,j,k\},$

$\left(2\right){(X+Y)}^{{\sigma }_{\eta }}=X^{{\sigma }_{\eta }}+Y^{{\sigma }_{\eta }},\eta \in \{i,j,k\},$

$\left(3\right){\left(\lambda X\right)}^{{\sigma }_{\eta }}={\left(X\lambda \right)}^{{\sigma }_{\eta }}=\lambda X^{{\sigma }_{\eta }},\eta \in \{i,j,k\},$

$\left(4\right){\left(XZ\right)}^{{\sigma }_i}=X^{{\sigma }_i}{Z}^{{\sigma }_i},{\left(XZ\right)}^{{\sigma }_j}=X^{{\sigma }_j}{V}_n{Z}^{{\sigma }_j},{\left(XZ\right)}^{{\sigma }_k}=X^{{\sigma }_k}{W}_n{Z}^{{\sigma }_k},V_n$ and $W_n$ are in the form of $\left(2\right)$ and $\left(3\right)$, respectively,

$\begin{array}{cc}\hfill \left(5\right)& \left(a\right)R_m^{-1}{X}^{{\sigma }_i}{R}_n=X^{{\sigma }_i},S_m^{-1}{X}^{{\sigma }_i}{S}_n=X^{{\sigma }_i},T_m^{-1}{X}^{{\sigma }_i}{T}_n=X^{{\sigma }_i},\hfill \\ & \left(b\right)R_m^{-1}{X}^{{\sigma }_j}{R}_n=-X^{{\sigma }_j},S_m^{-1}{X}^{{\sigma }_j}{S}_n=-X^{{\sigma }_j},T_m^{-1}{X}^{{\sigma }_j}{T}_n=X^{{\sigma }_j},\hfill \\ & \left(c\right)R_m^{-1}{X}^{{\sigma }_k}{R}_n=X^{{\sigma }_k},S_m^{-1}{X}^{{\sigma }_k}{S}_n=-X^{{\sigma }_k},T_m^{-1}{X}^{{\sigma }_k}{T}_n=-X^{{\sigma }_k},\hfill \end{array}$

$\left(6\right){\left(X^{i*}\right)}^{{\sigma }_i}=G_n^{-1}{\left(X^{{\sigma }_i}\right)}^{\mathrm{T}}{G}_m,{\left(X^{j*}\right)}^{{\sigma }_j}=G_n^{-1}{\left(X^{{\sigma }_j}\right)}^{\mathrm{T}}{G}_m,{\left(X^{k*}\right)}^{{\sigma }_k}=G_n^{-1}{\left(X^{{\sigma }_k}\right)}^{\mathrm{T}}{G}_m,$

$\begin{array}{cc}\hfill \left(7\right)& \left(a\right)X={\displaystyle \frac{1}{4}}\left(\begin{array}{cccc}{I}_m& I_{m}i& I_{m}j& I_{m}k\end{array}\right)X^{{\sigma }_i}\left(\begin{array}{c}{I}_n\\ {\displaystyle \frac{1}{\alpha }}{I}_{n}i\\ {\displaystyle \frac{1}{\beta }}{I}_{n}j\\ {\displaystyle \frac{1}{\alpha \beta }}{I}_{n}k\end{array}\right),\hfill \\ & \left(b\right)X={\displaystyle \frac{1}{4}}\left(\begin{array}{cccc}-I_m& I_{m}i& I_{m}j& -I_{m}k\end{array}\right)X^{{\sigma }_j}\left(\begin{array}{c}{I}_n\\ {\displaystyle \frac{1}{\alpha }}{I}_{n}i\\ {\displaystyle \frac{1}{\beta }}{I}_{n}j\\ {\displaystyle \frac{1}{\alpha \beta }}{I}_{n}k\end{array}\right),\hfill \\ & \left(c\right)X={\displaystyle \frac{1}{4}}\left(\begin{array}{cccc}{I}_m& I_{m}i& -I_{m}j& -I_{m}k\end{array}\right)X^{{\sigma }_k}\left(\begin{array}{c}{I}_n\\ {\displaystyle \frac{1}{\alpha }}{I}_{n}i\\ {\displaystyle \frac{1}{\beta }}{I}_{n}j\\ {\displaystyle \frac{1}{\alpha \beta }}{I}_{n}k\end{array}\right),\hfill \end{array}$

$\begin{array}{cc}\hfill \left(8\right)& \left(a\right)R_n{V}_n=-V_n{R}_n,R_n{G}_n^{-1}=-G_n^{-1}{R}_n^{\mathrm{T}},R_n{W}_n=W_n{R}_n,R_n^{-1}={\displaystyle \frac{1}{\alpha }}{R}_n,\hfill \\ & \left(b\right)S_n{V}_n=-V_n{S}_n,S_n{G}_n^{-1}=G_n^{-1}{S}_n^{\mathrm{T}},S_n{W}_n=-W_n{S}_n,S_n^{-1}={\displaystyle \frac{1}{\beta }}{S}_n,\hfill \\ & \left(c\right)T_n{V}_n=V_n{T}_n,T_n{G}_n^{-1}=-G_n^{-1}{T}_n^{\mathrm{T}},T_n{W}_n=-W_n{T}_n,T_n^{-1}={\displaystyle \frac{1}{\alpha \beta }}{T}_n.\hfill \end{array}$

Lemma 2

([38]). Let $A\in {\mathbb{R}}^{m\times s},B\in {\mathbb{R}}^{r\times n},C\in {\mathbb{R}}^{m\times n}.$ Then the following statements are equivalent:

$\left(1\right)$ The matrix equation $AXB=C$ is consistent.

$\left(2\right)$ $A{A}^{†}C{B}^{†}B=C.$

$\left(3\right)$ $A{A}^{†}C=C,C{B}^{†}B=C.$

In that case, the general solution of the matrix equation $AXB=C$ can be expressed as

$$X=A^{†}C{B}^{†}+L_{A}V+U{R}_B,$$

where $U,V$ are arbitrary real matrices with appropriate order.

Lemma 3

([17]). Suppose that $A\in {\mathbb{R}}^{m\times n},b\in {\mathbb{R}}^m$, then the matrix equation

$$Ax=b$$

(4)

is consistent if and only if

$$A{A}^{†}b=b.$$

In that case, the general solution to the matrix Equation $\left(4\right)$ can be expressed as $x=A^{†}b+L_{A}u$, where u is an arbitrary real vector with appropriate size. Moreover, the matrix Equation $\left(4\right)$ has a unique solution $x=A^{†}b$ if $rank\left(A\right)=n$.

Lemma 4

([39]). For $A\in {\mathbb{C}}^{m\times n},B\in {\mathbb{C}}^{n\times s},$ and $C\in {\mathbb{C}}^{s\times t},$ then we obtain

$$\mathrm{Vec}\left(ABC\right)=\left(C^{\mathrm{T}}\otimes A\right)\mathrm{Vec}\left(B\right).$$

3. Solving the Matrix Equation $\left(1\right)$ over ${\mathbb{DS}}_g$

In this section, we consider solving matrix Equation $\left(1\right)$ in combination with the previous lemmas.

Theorem 1.

Let $A=A_0+A_1ϵ\in {\mathbb{DS}}_g^{m\times n},B=B_0+B_1ϵ\in {\mathbb{DS}}_g^{n\times s},C=C_0+C_1ϵ\in {\mathbb{DS}}_g^{m\times s}$. Set

$$\begin{array}{cc}\hfill & A_{0\eta }=\left\{\begin{array}{c}{A}_0^{{\sigma }_i},\eta =i\hfill \\ A_0^{{\sigma }_j}{V}_n,\eta =j\hfill \\ A_0^{{\sigma }_k}{W}_n,\eta =k\hfill \end{array}\right.,A_{1\eta }=\left\{\begin{array}{c}{A}_1^{{\sigma }_i},\eta =i\hfill \\ A_1^{{\sigma }_j}{V}_n,\eta =j\hfill \\ A_1^{{\sigma }_k}{W}_n,\eta =k\hfill \end{array}\right.,\hfill \\ \hfill & B_{0\eta }=\left\{\begin{array}{c}{B}_0^{{\sigma }_i},\eta =i\hfill \\ B_0^{{\sigma }_j}{V}_s,\eta =j\hfill \\ B_0^{{\sigma }_k}{W}_s,\eta =k\hfill \end{array}\right.,B_{1\eta }=\left\{\begin{array}{c}{B}_1^{{\sigma }_i},\eta =i\hfill \\ B_1^{{\sigma }_j}{V}_s,\eta =j\hfill \\ B_1^{{\sigma }_k}{W}_s,\eta =k\hfill \end{array}\right.,\hfill \end{array}$$

$$C_{0\eta }=C_0^{{\sigma }_{\eta }},C_{1\eta }=C_1^{{\sigma }_{\eta }},$$

$$C_{00}=A_{0\eta }{A}_{0\eta }^{†}{C}_{0\eta }{B}_{0\eta }^{†}{B}_{1\eta }+A_{1\eta }{A}_{0\eta }^{†}{C}_{0\eta }{B}_{0\eta }^{†}{B}_{0\eta },$$

$$A_{00}=A_{1\eta }{L}_{A_{0\eta }},B_{00}=R_{B_{0\eta }}{B}_{1\eta },$$

$$L=B_{0\eta }^{\mathrm{T}}\otimes A_{00},M=B_{00}^{\mathrm{T}}\otimes A_{0\eta },N=B_{0\eta }^{\mathrm{T}}\otimes A_{0\eta },$$

$$a=\mathrm{Vec}\left(V\right),b=\mathrm{Vec}\left(U\right),c=\mathrm{Vec}\left(X_{1\eta }\right),\sigma =\left(\begin{array}{c}a\\ b\\ c\end{array}\right),$$

$$Q=\left(\begin{array}{ccc}L& M& N\end{array}\right),d=\mathrm{Vec}\left(C_{1\eta }-C_{00}\right),$$

$V_n$ and $V_s$ are in the form of $\left(2\right)$, while $W_n$ and $W_s$ are in the form of $\left(3\right)$. Then, the following statements are equivalent:

$\left(1\right)$ The dual generalized commutative quaternion matrix Equation $\left(1\right)$ is consistent.

$\left(2\right)$ The system of matrix equations

$$\left\{\begin{array}{c}{A}_{0\eta }{X}_{0\eta }{B}_{0\eta }=C_{0\eta },\hfill \\ A_{0\eta }{X}_{0\eta }{B}_{1\eta }+A_{0\eta }{X}_{1\eta }{B}_{0\eta }+A_{1\eta }{X}_{0\eta }{B}_{0\eta }=C_{1\eta }.\hfill \end{array}\right.$$

is consistent.

$\left(3\right)$ The following equations

$$\begin{array}{c}{A}_{0\eta }{A}_{0\eta }^{†}{C}_{0\eta }{B}_{0\eta }^{†}{B}_{0\eta }=C_{0\eta },\hfill \\ A_{0\eta }{A}_{0\eta }^{†}{C}_{0\eta }=C_{0\eta },C_{0\eta }{B}_{0\eta }^{†}{B}_{0\eta }=C_{0\eta },\hfill \\ Q{Q}^{†}d=d,\hfill \end{array}$$

hold. In this case, the general solution to the dual generalized commutative quaternion matrix Equation $\left(1\right)$ can be formulated as

$$X=X_0+X_1ϵ,$$

(5)

$\left(a\right)$ when $\eta =i$,

$$X_0={\displaystyle \frac{1}{16}}\left(\begin{array}{cccc}{I}_n& I_{n}i& I_{n}j& I_{n}k\end{array}\right)\left(X_{0i}+R_n^{-1}{X}_{0i}{R}_n+S_n^{-1}{X}_{0i}{S}_n+T_n^{-1}{X}_{0i}{T}_n\right)\left(\begin{array}{c}{I}_n\\ {\displaystyle \frac{1}{\alpha }}{I}_{n}i\\ {\displaystyle \frac{1}{\beta }}{I}_{n}j\\ {\displaystyle \frac{1}{\alpha \beta }}{I}_{n}k\end{array}\right),$$

$$X_1={\displaystyle \frac{1}{16}}\left(\begin{array}{cccc}{I}_n& I_{n}i& I_{n}j& I_{n}k\end{array}\right)\left(X_{1i}+R_n^{-1}{X}_{1i}{R}_n+S_n^{-1}{X}_{1i}{S}_n+T_n^{-1}{X}_{1i}{T}_n\right)\left(\begin{array}{c}{I}_n\\ {\displaystyle \frac{1}{\alpha }}{I}_{n}i\\ {\displaystyle \frac{1}{\beta }}{I}_{n}j\\ {\displaystyle \frac{1}{\alpha \beta }}{I}_{n}k\end{array}\right),$$

$\left(b\right)$ when $\eta =j$,

$$X_0={\displaystyle \frac{1}{16}}\left(\begin{array}{cccc}-I_n& I_{n}i& I_{n}j& -I_{n}k\end{array}\right)\left(X_{0j}-R_n^{-1}{X}_{0j}{R}_n-S_n^{-1}{X}_{0j}{S}_n+T_n^{-1}{X}_{0j}{T}_n\right)\left(\begin{array}{c}{I}_n\\ {\displaystyle \frac{1}{\alpha }}{I}_{n}i\\ {\displaystyle \frac{1}{\beta }}{I}_{n}j\\ {\displaystyle \frac{1}{\alpha \beta }}{I}_{n}k\end{array}\right),$$

$$X_1={\displaystyle \frac{1}{16}}\left(\begin{array}{cccc}-I_n& I_{n}i& I_{n}j& -I_{n}k\end{array}\right)\left(X_{1j}-R_n^{-1}{X}_{1j}{R}_n-S_n^{-1}{X}_{1j}{S}_n+T_n^{-1}{X}_{1j}{T}_n\right)\left(\begin{array}{c}{I}_n\\ {\displaystyle \frac{1}{\alpha }}{I}_{n}i\\ {\displaystyle \frac{1}{\beta }}{I}_{n}j\\ {\displaystyle \frac{1}{\alpha \beta }}{I}_{n}k\end{array}\right),$$

$\left(c\right)$ when $\eta =k$,

$$X_0={\displaystyle \frac{1}{16}}\left(\begin{array}{cccc}{I}_n& I_{n}i& -I_{n}j& -I_{n}k\end{array}\right)\left(X_{0k}+R_n^{-1}{X}_{0k}{R}_n-S_n^{-1}{X}_{0k}{S}_n-T_n^{-1}{X}_{0k}{T}_n\right)\left(\begin{array}{c}{I}_n\\ {\displaystyle \frac{1}{\alpha }}{I}_{n}i\\ {\displaystyle \frac{1}{\beta }}{I}_{n}j\\ {\displaystyle \frac{1}{\alpha \beta }}{I}_{n}k\end{array}\right),$$

$$X_1={\displaystyle \frac{1}{16}}\left(\begin{array}{cccc}{I}_n& I_{n}i& -I_{n}j& -I_{n}k\end{array}\right)\left(X_{1k}+R_n^{-1}{X}_{1k}{R}_n-S_n^{-1}{X}_{1k}{S}_n-T_n^{-1}{X}_{1k}{T}_n\right)\left(\begin{array}{c}{I}_n\\ {\displaystyle \frac{1}{\alpha }}{I}_{n}i\\ {\displaystyle \frac{1}{\beta }}{I}_{n}j\\ {\displaystyle \frac{1}{\alpha \beta }}{I}_{n}k\end{array}\right).$$

In above $\left(a\right),\left(b\right)$ and $\left(c\right)$,

$$X_{0\eta }=A_{0\eta }^{†}{C}_{0\eta }{B}_{0\eta }^{†}+L_{A_{0\eta }}V+U{R}_{B_{0\eta }},$$

$$\sigma =\left(\begin{array}{c}a\\ b\\ c\end{array}\right)=\left(\begin{array}{c}\mathrm{Vec}\left(V\right)\\ \mathrm{Vec}\left(U\right)\\ \mathrm{Vec}\left(X_{1\eta }\right)\end{array}\right)=Q^{†}d+L_{Q}u,u\in {\mathbb{R}}^{48n^2},$$

where $U,V,X_{1\eta },u$ are arbitrary matrices over $\mathbb{R}$ with appropriate sizes.

Proof. 

Firstly, we demonstrate $\left(1\right)\iff \left(2\right).$ We prove the case of $\eta =i$ in detail, and the other two cases can be proved similarly.

Assuming $X\in {\mathbb{DS}}_g^{n\times n}$ is the solution of the dual generalized commutative quaternion matrix Equation $\left(1\right)$, then X can be expressed as

$$X=X_0+X_1ϵ.$$

Substituting $\left(5\right)$ into the matrix Equation $\left(1\right)$, we obtain

$$\begin{array}{cc}\left(1\right)& \iff \left(A_0+A_1ϵ\right)\left(X_0+X_1ϵ\right)\left(B_0+B_1ϵ\right)=C_0+C_1ϵ,\hfill \\ & \iff A_0{X}_0{B}_0+\left(A_0{X}_0{B}_1+A_0{X}_1{B}_0+A_1{X}_0{B}_0\right)ϵ=C_0+C_1ϵ,\hfill \end{array}$$

i.e.,

$$\left\{\begin{array}{c}{A}_0{X}_0{B}_0=C_0,\hfill \\ A_0{X}_0{B}_1+A_0{X}_1{B}_0+A_1{X}_0{B}_0=C_1.\hfill \end{array}\right.$$

(6)

By applying the formula $\left(2\right)$ and $\left(4\right)$ of Lemma 1 to the system $\left(6\right)$, we obtain system

$$\left\{\begin{array}{c}{A}_{0i}{X}_{0i}{B}_{0i}=C_{0i},\hfill \\ A_{0i}{X}_{0i}{B}_{1i}+A_{0i}{X}_{1i}{B}_{0i}+A_{1i}{X}_{0i}{B}_{0i}=C_{1i}.\hfill \end{array}\right.$$

(7)

At this time, $(X_{0i},X_{1i})$ is the solution of the system $\left(7\right)$.

Conversely, if the system $\left(7\right)$ has a solution $(X_{0i},X_{1i})$, which can be expressed as

$$X_{0i}=\left(\begin{array}{cccc}{a}_{11}& a_{12}& a_{13}& a_{14}\\ a_{21}& a_{22}& a_{23}& a_{24}\\ a_{31}& a_{32}& a_{33}& a_{34}\\ a_{41}& a_{42}& a_{43}& a_{44}\end{array}\right),a_{uv}\in {\mathbb{R}}^{n\times n},u,v=1,2,3,4,$$

$$X_{1i}=\left(\begin{array}{cccc}{b}_{11}& b_{12}& b_{13}& b_{14}\\ b_{21}& b_{22}& b_{23}& b_{24}\\ b_{31}& b_{32}& b_{33}& b_{34}\\ b_{41}& b_{42}& b_{43}& b_{44}\end{array}\right),b_{uv}\in {\mathbb{R}}^{n\times n},u,v=1,2,3,4.$$

Using the formula $\left(5\right)$ of Lemma 1, we derive

$$\left\{\begin{array}{c}{A}_{0i}{R}_n^{-1}{X}_{0i}{R}_n{B}_{0i}=C_{0i},\hfill \\ A_{0i}{R}_n^{-1}{X}_{0i}{R}_n{B}_{1i}+A_{0i}{R}_n^{-1}{X}_{1i}{R}_n{B}_{0i}+A_{1i}{R}_n^{-1}{X}_{0i}{R}_n{B}_{0i}=C_{1i}.\hfill \end{array}\right.$$

Therefore, $(R_n^{-1}{X}_{0i}{R}_n,R_n^{-1}{X}_{1i}{R}_n)$ is also a solution to the system $\left(7\right)$. Similarly, $(S_n^{-1}{X}_{0i}{S}_n,S_n^{-1}{X}_{1i}{S}_n)$ and $(T_n^{-1}{X}_{0i}{T}_n,T_n^{-1}{X}_{1i}{T}_n)$ are also solutions to the system $\left(7\right)$.

Let

$$\begin{array}{c}{Z}_1={\displaystyle \frac{1}{4}}(X_{0i}+R_n^{-1}{X}_{0i}{R}_n+S_n^{-1}{X}_{0i}{S}_n+T_n^{-1}{X}_{0i}{T}_n),\hfill \\ Z_2={\displaystyle \frac{1}{4}}(X_{1i}+R_n^{-1}{X}_{1i}{R}_n+S_n^{-1}{X}_{1i}{S}_n+T_n^{-1}{X}_{1i}{T}_n).\hfill \end{array}$$

Then $(Z_1,Z_2)$ is also a solution to the system $\left(7\right)$. That is

$$\left\{\begin{array}{c}{A}_{0i}{Z}_1{B}_{0i}=C_{0i},\hfill \\ A_{0i}{Z}_1{B}_{1i}+A_{0i}{Z}_2{B}_{0i}+A_{1i}{Z}_1{B}_{0i}=C_{1i}.\hfill \end{array}\right.$$

By direct computation, we obtain

$$Z_1=\left(\begin{array}{cccc}{c}_0& \alpha c_1& \beta c_2& \alpha \beta c_3\\ c_1& c_0& \beta c_3& \beta c_2\\ c_2& \alpha c_3& c_0& \alpha c_1\\ c_3& c_2& c_1& c_0\end{array}\right),$$

where

$$\begin{array}{c}{c}_0={\displaystyle \frac{1}{4}}(a_{11}+a_{22}+a_{33}+a_{44}),c_1={\displaystyle \frac{1}{4}}\left({\displaystyle \frac{1}{\alpha }}{a}_{12}+{\displaystyle \frac{1}{\alpha }}{a}_{34}+a_{21}+a_{43}\right),\hfill \\ c_2={\displaystyle \frac{1}{4}}\left({\displaystyle \frac{1}{\beta }}{a}_{13}+{\displaystyle \frac{1}{\beta }}{a}_{24}+a_{31}+a_{42}\right),c_3={\displaystyle \frac{1}{4}}\left({\displaystyle \frac{1}{\alpha \beta }}{a}_{14}+{\displaystyle \frac{1}{\beta }}{a}_{23}+{\displaystyle \frac{1}{\alpha }}{a}_{32}+a_{41}\right),\hfill \end{array}$$

and

$$Z_2=\left(\begin{array}{cccc}{d}_0& \alpha d_1& \beta d_2& \alpha \beta d_3\\ d_1& d_0& \beta d_3& \beta d_2\\ d_2& \alpha d_3& d_0& \alpha d_1\\ d_3& d_2& d_1& d_0\end{array}\right).$$

Here

$$\begin{array}{c}{d}_0={\displaystyle \frac{1}{4}}(b_{11}+b_{22}+b_{33}+b_{44}),d_1={\displaystyle \frac{1}{4}}\left({\displaystyle \frac{1}{\alpha }}{b}_{12}+{\displaystyle \frac{1}{\alpha }}{b}_{34}+b_{21}+b_{43}\right),\hfill \\ d_2={\displaystyle \frac{1}{4}}\left({\displaystyle \frac{1}{\beta }}{b}_{13}+{\displaystyle \frac{1}{\beta }}{b}_{24}+b_{31}+b_{42}\right),d_3={\displaystyle \frac{1}{4}}\left({\displaystyle \frac{1}{\alpha \beta }}{b}_{14}+{\displaystyle \frac{1}{\beta }}{b}_{23}+{\displaystyle \frac{1}{\alpha }}{b}_{32}+b_{41}\right).\hfill \end{array}$$

Construct $X_0,X_1\in {\mathbb{S}}_g^{n\times n}$, and they satisfy

$$X_0^{{\sigma }_i}=X_{0i}=Z_1,X_1^{{\sigma }_i}=X_{1i}=Z_2,$$

i.e., $(X_{0i},X_{1i})$ is the solution of the system $\left(7\right)$. In addition, applying the formula $\left(2\right)$ and $\left(4\right)$ of Lemma 1 to the system $\left(7\right)$, we get the system $\left(6\right)$. Therefore,

$$X=X_0+X_1ϵ,$$

is the solution of the matrix Equation $\left(1\right)$. In this case, we employ the item $\left(7\right)$ in Lemma 1 to express the specific forms of X. It can be written as

$$\begin{array}{cc}\hfill X_0=& c_0+c_{1}i+c_{2}j+c_{3}k\hfill \\ \hfill =& {\displaystyle \frac{1}{4}}\left(\begin{array}{cccc}{I}_n& I_{n}i& I_{n}j& I_{n}k\end{array}\right)Z_1\left(\begin{array}{c}{I}_n\\ {\displaystyle \frac{1}{\alpha }}{I}_{n}i\\ {\displaystyle \frac{1}{\beta }}{I}_{n}j\\ {\displaystyle \frac{1}{\alpha \beta }}{I}_{n}k\end{array}\right)\hfill \\ \hfill =& {\displaystyle \frac{1}{16}}\left(\begin{array}{cccc}{I}_n& I_{n}i& I_{n}j& I_{n}k\end{array}\right)(X_{0i}+R_n^{-1}{X}_{0i}{R}_n+S_n^{-1}{X}_{0i}{S}_n+T_n^{-1}{X}_{0i}{T}_n)\left(\begin{array}{c}{I}_n\\ {\displaystyle \frac{1}{\alpha }}{I}_{n}i\\ {\displaystyle \frac{1}{\beta }}{I}_{n}j\\ {\displaystyle \frac{1}{\alpha \beta }}{I}_{n}k\end{array}\right),\hfill \end{array}$$

$$\begin{array}{cc}\hfill X_1=& d_0+d_{1}i+d_{2}j+d_{3}k\hfill \\ \hfill =& {\displaystyle \frac{1}{4}}\left(\begin{array}{cccc}{I}_n& I_{n}i& I_{n}j& I_{n}k\end{array}\right)Z_2\left(\begin{array}{c}{I}_n\\ {\displaystyle \frac{1}{\alpha }}{I}_{n}i\\ {\displaystyle \frac{1}{\beta }}{I}_{n}j\\ {\displaystyle \frac{1}{\alpha \beta }}{I}_{n}k\end{array}\right)\hfill \\ \hfill =& {\displaystyle \frac{1}{16}}\left(\begin{array}{cccc}{I}_n& I_{n}i& I_{n}j& I_{n}k\end{array}\right)(X_{1i}+R_n^{-1}{X}_{1i}{R}_n+S_n^{-1}{X}_{1i}{S}_n+T_n^{-1}{X}_{1i}{T}_n)\left(\begin{array}{c}{I}_n\\ {\displaystyle \frac{1}{\alpha }}{I}_{n}i\\ {\displaystyle \frac{1}{\beta }}{I}_{n}j\\ {\displaystyle \frac{1}{\alpha \beta }}{I}_{n}k\end{array}\right).\hfill \end{array}$$

Next, we divide the system $\left(7\right)$ into two parts,

$$A_{0i}{X}_{0i}{B}_{0i}=C_{0i},$$

(8)

and

$$A_{0i}{X}_{0i}{B}_{1i}+A_{0i}{X}_{1i}{B}_{0i}+A_{1i}{X}_{0i}{B}_{0i}=C_{1i}.$$

(9)

Applying the Lemma 2 to the matrix Equation $\left(8\right)$ yields

$$X_{0i}=A_{0i}^{†}{C}_{0i}{B}_{0i}^{†}+L_{A_{0i}}V+U{R}_{B_{0i}},$$

(10)

where $U,V\in {\mathbb{R}}^{4n\times 4n}$ are arbitrary matrices. Substitute (10) into the $\left(9\right)$ to obtain

$$A_{00}V{B}_{0i}+A_{0i}U{B}_{00}+A_{0i}{X}_{1i}{B}_{0i}=C_{1i}-C_{00}.$$

(11)

Using the Vec operator on both sides of Equation (11) and Lemma 4, we derive

$$\left(B_{0i}^{\mathrm{T}}\otimes A_{00}\right)\mathrm{Vec}\left(V\right)+\left(B_{00}^{\mathrm{T}}\otimes A_{0i}\right)\mathrm{Vec}\left(U\right)+\left(B_{0i}^{\mathrm{T}}\otimes A_{0i}\right)\mathrm{Vec}\left(X_{1i}\right)=\mathrm{Vec}\left(C_{1i}-C_{00}\right).$$

(12)

Therefore, $\left(12\right)$ can be expressed as

$$La+Mb+Nc=d.$$

That is

$$Q\sigma =d.$$

(13)

Thus, $\left(13\right)$ is solvable if and only if

$$Q{Q}^{†}d=d.$$

At this time, the general solution of matrix Equation (13) is given by

$$\sigma =Q^{†}d+L_{Q}u,$$

where $u\in {\mathbb{R}}^{48n^2}$ is an arbitrary vector.

Finally, utilizing Lemmas 2 and 3, we establish the equivalence between $\left(2\right)\iff \left(3\right)$ easily. □

4. Numerical Example

To further elaborate on the key findings of this article, we now present a numerical algorithm and an example.

Example 1.

Let $\alpha =-1,\beta =-1.$ Take $\eta =i$ as an example, the other two cases are similar.

$$\begin{array}{c}A=A_0+A_1ϵ=\left(\begin{array}{cc}1-i-2j& -2-2k\end{array}\right)+\left(\begin{array}{cc}-i+k& j+k\end{array}\right)ϵ,\hfill \\ B=B_0+B_1ϵ=\left(\begin{array}{cc}1+2j+k& i+2k\\ i+2k& 2+i+2j+k\end{array}\right)+\left(\begin{array}{cc}1+2i+j+k& 1+j+2k\\ 2i+j& 1+i+j+3k\end{array}\right)ϵ,\hfill \\ C=C_0+C_1ϵ=\left(\begin{array}{cc}1+2i+j+k& 1+4j+3k\end{array}\right)+\left(\begin{array}{cc}1+i+j+2k& i+2j+2k\end{array}\right)ϵ.\hfill \end{array}$$

Utilizing MATLAB8.6 and Algorithm, we obtain

$$\begin{array}{c}{A}_{0i}{A}_{0i}^{†}{C}_{0i}{B}_{0i}^{†}{B}_{0i}-C_{0i}=8.8668\times {10}^{-15},\hfill \\ A_{0i}{A}_{0i}^{†}{C}_{0i}-C_{0i}=2.9286\times {10}^{-15},\hfill \\ C_{0i}{B}_{0i}^{†}{B}_{0i}-C_{0i}=8.5821\times {10}^{-15},\hfill \\ Q{Q}^{†}d-d=1.4093\times {10}^{-14}.\hfill \end{array}$$

In this case, the general solution of the matrix Equation $\left(1\right)$ can be formulated as

$$X=X_0+X_1ϵ,$$

where

$$\begin{array}{c}{X}_0={\displaystyle \frac{1}{16}}\left(\begin{array}{cccc}{I}_2& I_{2}i& I_{2}j& I_{2}k\end{array}\right)(X_{0i}+R_2^{-1}{X}_{0i}{R}_2+S_2^{-1}{X}_{0i}{S}_2+T_2^{-1}{X}_{0i}{T}_2)\left(\begin{array}{c}{I}_2\\ {\displaystyle \frac{1}{\alpha }}{I}_{2}i\\ {\displaystyle \frac{1}{\beta }}{I}_{2}j\\ {\displaystyle \frac{1}{\alpha \beta }}{I}_{2}k\end{array}\right),\hfill \\ X_1={\displaystyle \frac{1}{16}}\left(\begin{array}{cccc}{I}_2& I_{2}i& I_{2}j& I_{2}k\end{array}\right)(X_{1i}+R_2^{-1}{X}_{1i}{R}_2+S_2^{-1}{X}_{1i}{S}_2+T_2^{-1}{X}_{1i}{T}_2)\left(\begin{array}{c}{I}_2\\ {\displaystyle \frac{1}{\alpha }}{I}_{2}i\\ {\displaystyle \frac{1}{\beta }}{I}_{2}j\\ {\displaystyle \frac{1}{\alpha \beta }}{I}_{2}k\end{array}\right),\hfill \end{array}$$

$$\begin{array}{cc}\hfill X_{0i}=& \left(\begin{array}{cccccccc}0.12573& 0.043649& -0.17884& -0.022075& -0.17501& -0.13434& -0.17171& 0.0257\\ 0.049808& -0.18161& -0.042146& -0.042912& 0.042146& 0.042912& 0.049808& -0.18161\\ 0.17884& 0.022075& 0.12573& 0.043649& 0.17171& -0.0257& -0.17501& -0.13434\\ 0.042146& 0.042912& 0.049808& -0.18161& -0.049808& 0.18161& 0.042146& 0.042912\\ 0.17501& 0.13434& 0.17171& -0.0257& 0.12573& 0.043649& -0.17884& -0.022075\\ -0.042146& -0.042912& -0.049808& 0.18161& 0.049808& -0.18161& -0.042146& -0.042912\\ -0.17171& 0.0257& 0.17501& 0.13434& 0.17884& 0.022075& 0.12573& 0.043649\\ 0.049808& -0.18161& -0.042146& -0.042912& 0.042146& 0.042912& 0.049808& -0.18161\end{array}\right)\hfill \\ & +L_{A_{0i}}V+U{R}_{B_{0i}},\hfill \end{array}$$

$$\mathrm{Vec}\left(X_{1i}\right)=\left(\begin{array}{c}{X}_{1i1}\\ X_{1i2}\\ X_{1i3}\\ X_{1i4}\end{array}\right),$$

$$X_{1i1}=\left(\begin{array}{c}-0.13404\\ -0.029119\\ -0.090283\\ -0.11942\\ -0.13544\\ 0.11942\\ 0.20831\\ -0.029119\\ -0.14232\\ 0.021957\\ 0.058871\\ 0.099797\\ 0.097791\\ -0.099797\\ 0.081445\\ 0.021957\end{array}\right),X_{1i2}=\left(\begin{array}{c}0.090283\\ 0.11942\\ -0.13404\\ -0.029119\\ -0.20831\\ 0.029119\\ -0.13544\\ 0.11942\\ -0.058871\\ -0.099797\\ -0.14232\\ 0.021957\\ -0.081445\\ -0.021957\\ 0.097791\\ -0.099797\end{array}\right),X_{1i3}=\left(\begin{array}{c}0.13544\\ -0.11942\\ -0.20831\\ 0.029119\\ -0.13404\\ -0.029119\\ -0.090283\\ -0.11942\\ -0.097791\\ 0.099797\\ -0.081445\\ -0.021957\\ -0.14232\\ 0.021957\\ 0.058871\\ 0.099797\end{array}\right),X_{1i4}=\left(\begin{array}{c}0.20831\\ -0.029119\\ 0.13544\\ -0.11942\\ 0.090283\\ 0.11942\\ -0.13404\\ -0.029119\\ 0.081445\\ 0.021957\\ -0.097791\\ 0.099797\\ -0.058871\\ -0.099797\\ -0.14232\\ 0.021957\end{array}\right).$$

Here $U,V$ are arbitrary matrices over $\mathbb{R}$ with appropriate sizes.

5. Conclusions

In this article, we conducted an in-depth study of the solution to the dual generalized quaternion matrix Equation (1), and derived a general solution expression when it is consistent. This research not only provides a new perspective and method for understanding and solving this special type of matrix equation, but also contributes new knowledge to the field of dual generalized commutative quaternion algebra. To show our research results more intuitively, we provide a specific numerical example. Through this example, we can clearly see how to apply the existence conditions and general solution expressions to solve the dual generalized quaternion matrix equation.

We will study more complex matrix equations and tensor equations over the dual generalized commutative quaternions in the future, which may have broader applications in physics and engineering fields.