# Dual Quaternion Matrix Equation AXB = C with Applications

^{1}

^{2}

^{*}

## Abstract

**:**

## 1. Introduction

**0**are defined by the rank of a given quaternion matrix A, the conjugate transpose of A, identity matrix, and zero matrix with appropriate sizes, respectively. The Moore–Penrose inverse of $A\in {\mathbb{H}}^{l\times k}$ is denoted as ${A}^{\u2020}$, which is defined as the solution of $AYA=A,\phantom{\rule{4pt}{0ex}}YAY=Y,\phantom{\rule{4pt}{0ex}}{\left(AY\right)}^{*}=AY$, and ${\left(YA\right)}^{*}=YA.$ Moreover, let ${L}_{A}=I-{A}^{\u2020}A$ and ${R}_{A}=I-A{A}^{\u2020}$ represent two projectors along A.

## 2. Preliminaries

**Definition**

**1**

**Definition**

**2**

**Definition**

**3**

**Definition**

**4**

**Proposition**

**1**

**Definition**

**5**

**Definition**

**6**

**Proposition**

**2.**

- (1)
- ${\left({Z}^{\varphi}\right)}^{\u2020}={\left({Z}^{\u2020}\right)}^{\varphi}$,
- (2)
- ${\left({L}_{Z}\right)}^{\varphi}={R}_{{Z}^{\varphi}}$,
- (3)
- ${\left({R}_{Z}\right)}^{\varphi}={L}_{{Z}^{\varphi}}$.

**Proof.**

**Definition**

**7.**

**Proposition**

**3**

- (1)
- ${(\alpha X+\beta Y)}^{\varphi}={X}^{\varphi}\varphi \left(\alpha \right)+{Y}^{\varphi}\varphi \left(\beta \right),$
- (2)
- ${(X\alpha +Y\beta )}^{\varphi}=\varphi \left(\alpha \right){X}^{\varphi}+\varphi \left(\beta \right){Y}^{\varphi},$
- (3)
- ${\left(XY\right)}^{\varphi}={Y}_{\varphi}{X}_{\varphi},$
- (4)
- ${\left({X}^{\varphi}\right)}^{\varphi}=X$.

**Proposition**

**4.**

- (1)
- ${(X+Y)}^{\varphi}={X}^{\varphi}+{Y}^{\varphi}$,
- (2)
- ${\left(XY\right)}^{\varphi}={Y}^{\varphi}{X}^{\varphi}$,
- (3)
- ${\left({X}^{\varphi}\right)}^{\varphi}=X$.

**Proof.**

**Lemma**

**1**

**Lemma**

**2**

- (1)
- The quaternion matrix equation$${A}_{1}{X}_{1}{B}_{1}+{A}_{1}{X}_{2}{B}_{2}+{A}_{2}{X}_{3}{B}_{2}=C$$
- (2)
- ${R}_{M}A=0,\phantom{\rule{4pt}{0ex}}{R}_{{A}_{1}}C{L}_{{B}_{2}}=0,\phantom{\rule{4pt}{0ex}}{C}_{1}{L}_{B}=0.$
- (3)
- $$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& r\left(\begin{array}{ccc}{A}_{1}& {A}_{2}& C\end{array}\right)=r\left(\begin{array}{cc}{A}_{1}& {A}_{2}\end{array}\right),\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& r\left(\begin{array}{cc}{B}_{2}& 0\\ C& {A}_{1}\end{array}\right)=r\left(\begin{array}{c}{B}_{2}\end{array}\right)+r\left(\begin{array}{c}{A}_{1}\end{array}\right),\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& r\left(\begin{array}{c}{C}_{1}\\ {B}_{1}\\ {B}_{2}\end{array}\right)=r\left(\begin{array}{c}{B}_{1}\\ {B}_{2}\end{array}\right).\hfill \end{array}$$

**Lemma**

**3.**

- (1)
- $r\left(\begin{array}{c}{R}_{A}B\end{array}\right)=r\left(\begin{array}{cc}A& B\end{array}\right)-r\left(A\right).$
- (2)
- $r\left(\begin{array}{c}C{L}_{B}\end{array}\right)=r\left(\begin{array}{c}B\\ C\end{array}\right)-r\left(B\right).$
- (3)
- $r\left(\begin{array}{c}{R}_{A}B{L}_{C}\end{array}\right)=r\left(\begin{array}{cc}B& A\\ C& 0\end{array}\right)-r\left(A\right)-r\left(C\right).$
- (4)
- $r\left(\begin{array}{c}{R}_{A}B\\ C\end{array}\right)=r\left(\begin{array}{cc}B& A\\ C& 0\end{array}\right)-r\left(A\right).$
- (5)
- $r\left(\begin{array}{cc}A{L}_{B}& C\end{array}\right)=r\left(\begin{array}{cc}A& C\\ B& 0\end{array}\right)-r\left(B\right).$
- (6)
- $r\left(\begin{array}{cc}A& B{L}_{D}\\ {R}_{E}C& 0\end{array}\right)=r\left(\begin{array}{ccc}A& B& 0\\ C& 0& E\\ 0& D& 0\end{array}\right)-r\left(D\right)-r\left(E\right).$

## 3. The Solution of Matrix Equation (1)

**Theorem**

**1.**

- (1)
- Dual quaternion matrix Equation (1) is consistent.
- (2)
- $$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& {R}_{{A}_{0}}{C}_{0}=0,\phantom{\rule{4pt}{0ex}}{C}_{0}{L}_{{B}_{0}}=0,\hfill \end{array}$$$$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& {R}_{M}N=0,\phantom{\rule{4pt}{0ex}}{R}_{{A}_{0}}{C}_{2}{L}_{{B}_{0}}=0,\phantom{\rule{4pt}{0ex}}F{L}_{E}=0.\hfill \end{array}$$
- (3)
- $$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& r\left(\begin{array}{cc}{A}_{0}& {C}_{0}\end{array}\right)=r\left(\begin{array}{c}{A}_{0}\end{array}\right),r\left(\begin{array}{c}{B}_{0}\\ {C}_{0}\end{array}\right)=r\left(\begin{array}{c}{B}_{0}\end{array}\right),\hfill \end{array}$$$$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& r\left(\begin{array}{ccc}{A}_{1}& {A}_{0}& {C}_{1}\\ {A}_{0}& 0& {C}_{0}\end{array}\right)=r\left(\begin{array}{cc}{A}_{1}& {A}_{0}\\ {A}_{0}& 0\end{array}\right),\hfill \end{array}$$$$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& r\left(\begin{array}{cc}{C}_{1}& {A}_{0}\\ {B}_{0}& 0\end{array}\right)=r\left(\begin{array}{c}{A}_{0}\end{array}\right)+r\left(\begin{array}{c}{B}_{0}\end{array}\right),\hfill \end{array}$$$$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& r\left(\begin{array}{cc}{B}_{1}& {B}_{0}\\ {B}_{0}& 0\\ {C}_{1}& {C}_{0}\end{array}\right)=r\left(\begin{array}{cc}{B}_{1}& {B}_{0}\\ {B}_{0}& 0\end{array}\right).\hfill \end{array}$$

**Proof.**

- Now, we turn to prove that $\left(10\right)\iff \left(12\right)-\left(14\right)$. Let ${X}_{0}={A}_{0}^{\u2020}{C}_{0}{B}_{0}^{\u2020}.$ Then, it is easy to verify that ${X}_{0}$ is a particular solution to the matrix equation ${A}_{0}{X}_{0}{B}_{0}={C}_{0}.$ By Lemma 3 and block elementary operations, we obtain$$\begin{array}{cc}\hfill {R}_{M}N=0& \iff r\left({R}_{M}N\right)=0\iff r\left(\begin{array}{cc}{R}_{{A}_{0}}{A}_{2}& {R}_{{A}_{0}}{C}_{2}\end{array}\right)=r\left({R}_{{A}_{0}}{A}_{2}\right),\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \iff r\left(\begin{array}{ccc}{A}_{0}& {A}_{2}& {C}_{2}\end{array}\right)=r\left(\begin{array}{cc}{A}_{0}& {A}_{2}\end{array}\right),\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \iff r\left(\begin{array}{ccc}{A}_{0}& {A}_{1}{L}_{{A}_{0}}& {C}_{2}\end{array}\right)=r\left(\begin{array}{cc}{A}_{0}& {A}_{1}{L}_{{A}_{0}}\end{array}\right),\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \iff r\left(\begin{array}{ccc}{A}_{1}& {A}_{0}& {C}_{2}\\ {A}_{0}& 0& 0\end{array}\right)=r\left(\begin{array}{cc}{A}_{1}& {A}_{0}\\ {A}_{0}& 0\end{array}\right),\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \iff r\left(\begin{array}{ccc}{A}_{1}& {A}_{0}& {C}_{1}-{A}_{0}{A}_{0}^{\u2020}{C}_{0}{B}_{0}^{\u2020}{B}_{1}-{A}_{1}{A}_{0}^{\u2020}{C}_{0}{B}_{0}^{\u2020}{B}_{0}\\ {A}_{0}& 0& 0\end{array}\right)=r\left(\begin{array}{cc}{A}_{1}& {A}_{0}\\ {A}_{0}& 0\end{array}\right),\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \iff r\left(\begin{array}{ccc}{A}_{1}& {A}_{0}& {C}_{1}\\ {A}_{0}& 0& {A}_{0}{A}_{0}^{\u2020}{C}_{0}{B}_{0}^{\u2020}{B}_{0}\end{array}\right)=r\left(\begin{array}{cc}{A}_{1}& {A}_{0}\\ {A}_{0}& 0\end{array}\right),\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \iff r\left(\begin{array}{ccc}{A}_{1}& {A}_{0}& {C}_{1}\\ {A}_{0}& 0& {C}_{0}\end{array}\right)=r\left(\begin{array}{cc}{A}_{1}& {A}_{0}\\ {A}_{0}& 0\end{array}\right),\hfill \end{array}$$$$\begin{array}{cc}\hfill {R}_{{A}_{0}}{C}_{2}{L}_{{B}_{0}}=0& \iff r\left({R}_{{A}_{0}}{C}_{2}{L}_{{B}_{0}}\right)=0\iff r\left(\begin{array}{cc}{C}_{2}& {A}_{0}\\ {B}_{0}& 0\end{array}\right)=r\left({A}_{0}\right)+r\left({B}_{0}\right),\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \iff r\left(\begin{array}{cc}{C}_{1}-{C}_{11}-{C}_{22}& {A}_{0}\\ {B}_{0}& 0\end{array}\right)=r\left({A}_{0}\right)+r\left({B}_{0}\right),\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \iff r\left(\begin{array}{cc}{C}_{1}-{A}_{0}{A}_{0}^{\u2020}{C}_{0}{B}_{0}^{\u2020}{B}_{1}-{A}_{1}{A}_{0}^{\u2020}{C}_{0}{B}_{0}^{\u2020}{B}_{0}& {A}_{0}\\ {B}_{0}& 0\end{array}\right)=r\left({A}_{0}\right)+r\left({B}_{0}\right),\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \iff r\left(\begin{array}{cc}{C}_{1}& {A}_{0}\\ {B}_{0}& 0\end{array}\right)=r\left({A}_{0}\right)+r\left({B}_{0}\right),\hfill \\ \hfill F{L}_{E}=0& \iff r\left(F{L}_{E}\right)=0\iff r\left(\begin{array}{c}E\\ F\end{array}\right)=r\left(E\right)\iff r\left(\begin{array}{c}{B}_{2}{L}_{{B}_{0}}\\ {C}_{2}{L}_{{B}_{0}}\end{array}\right)=r\left({B}_{2}{L}_{{B}_{0}}\right),\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \iff r\left(\begin{array}{c}{B}_{0}\\ {B}_{2}\\ {C}_{2}\end{array}\right)=r\left(\begin{array}{c}{B}_{0}\\ {B}_{2}\end{array}\right)\iff r\left(\begin{array}{c}{B}_{0}\\ {R}_{{B}_{0}}{B}_{1}\\ {C}_{2}\end{array}\right)=r\left(\begin{array}{c}{B}_{0}\\ {R}_{{B}_{0}}{B}_{1}\end{array}\right),\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \iff r\left(\begin{array}{cc}{B}_{1}& {B}_{0}\\ {B}_{0}& 0\\ {C}_{2}& 0\end{array}\right)=r\left(\begin{array}{cc}{B}_{1}& {B}_{0}\\ {B}_{0}& 0\end{array}\right),\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \iff r\left(\begin{array}{cc}{B}_{1}& {B}_{0}\\ {B}_{0}& 0\\ {C}_{1}-{A}_{0}{A}_{0}^{\u2020}{C}_{0}{B}_{0}^{\u2020}{B}_{1}-{A}_{1}{A}_{0}^{\u2020}{C}_{0}{B}_{0}^{\u2020}{B}_{0}& 0\end{array}\right)=r\left(\begin{array}{cc}{B}_{1}& {B}_{0}\\ {B}_{0}& 0\end{array}\right),\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \iff r\left(\begin{array}{cc}{B}_{1}& {B}_{0}\\ {B}_{0}& 0\\ {C}_{1}& {A}_{0}{A}_{0}^{\u2020}{C}_{0}{B}_{0}^{\u2020}{B}_{0}\end{array}\right)=r\left(\begin{array}{cc}{B}_{1}& {B}_{0}\\ {B}_{0}& 0\end{array}\right),\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \iff r\left(\begin{array}{cc}{B}_{1}& {B}_{0}\\ {B}_{0}& 0\\ {C}_{1}& {C}_{0}\end{array}\right)=r\left(\begin{array}{cc}{B}_{1}& {B}_{0}\\ {B}_{0}& 0\end{array}\right).\hfill \end{array}$$

**Corollary**

**1**

- (1)
- Dual quaternion matrix equation $AX=C$ is consistent.
- (2)
- $${R}_{{A}_{0}}{C}_{0}=0,\phantom{\rule{4pt}{0ex}}{R}_{M}N=0.$$
- (3)
- $$r\left(\begin{array}{cc}{A}_{0}& {C}_{0}\end{array}\right)=r\left(\begin{array}{c}{A}_{0}\end{array}\right),\phantom{\rule{4pt}{0ex}}r\left(\begin{array}{ccc}{A}_{1}& {A}_{0}& {C}_{1}\\ {A}_{0}& 0& {C}_{0}\end{array}\right)=r\left(\begin{array}{cc}{A}_{1}& {A}_{0}\\ {A}_{0}& 0\end{array}\right).$$

**Corollary**

**2**

- (1)
- Dual quaternion matrix equation $XB=C$ is consistent.
- (2)
- $${C}_{0}{L}_{{B}_{0}}=0,\phantom{\rule{4pt}{0ex}}F{L}_{E}=0.$$
- (3)
- $$r\left(\begin{array}{c}{B}_{0}\\ {C}_{0}\end{array}\right)=r\left(\begin{array}{c}{B}_{0}\end{array}\right),\phantom{\rule{4pt}{0ex}}r\left(\begin{array}{cc}{B}_{1}& {B}_{0}\\ {B}_{0}& 0\\ {C}_{1}& {C}_{0}\end{array}\right)=r\left(\begin{array}{cc}{B}_{1}& {B}_{0}\\ {B}_{0}& 0\end{array}\right).$$

**Remark**

**1.**

## 4. Applications

**Theorem**

**2.**

- (1)
- Dual quaternion matrix Equation (3) is consistent.
- (2)
- The following equalities are satisfied:$$\begin{array}{c}\hfill {R}_{{A}_{0}}{C}_{0}=0,\phantom{\rule{4pt}{0ex}}{R}_{M}N=0,\phantom{\rule{4pt}{0ex}}{R}_{{A}_{0}}{C}_{2}{L}_{{A}_{0}^{\varphi}}=0.\end{array}$$
- (3)
- The following rank equalities hold:$$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& r\left(\begin{array}{cc}{A}_{0}& {C}_{0}\end{array}\right)=r\left(\begin{array}{c}{A}_{0}\end{array}\right),\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& r\left(\begin{array}{ccc}{A}_{1}& {A}_{0}& {C}_{1}\\ {A}_{0}& 0& {C}_{0}\end{array}\right)=r\left(\begin{array}{cc}{A}_{1}& {A}_{0}\\ {A}_{0}& 0\end{array}\right),\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& r\left(\begin{array}{cc}{C}_{1}& {A}_{0}\\ {A}_{0}^{\varphi}& 0\end{array}\right)=r\left(\begin{array}{c}{A}_{0}\end{array}\right)+r\left(\begin{array}{c}{A}_{0}^{\varphi}\end{array}\right)=2r\left({A}_{0}\right).\hfill \end{array}$$

**Proof.**

## 5. Numerical Example

**Example**

**1.**

## 6. Conclusions

## Author Contributions

## Funding

## Data Availability Statement

## Conflicts of Interest

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**MDPI and ACS Style**

Chen, Y.; Wang, Q.-W.; Xie, L.-M.
Dual Quaternion Matrix Equation *AXB* = *C* with Applications. *Symmetry* **2024**, *16*, 287.
https://doi.org/10.3390/sym16030287

**AMA Style**

Chen Y, Wang Q-W, Xie L-M.
Dual Quaternion Matrix Equation *AXB* = *C* with Applications. *Symmetry*. 2024; 16(3):287.
https://doi.org/10.3390/sym16030287

**Chicago/Turabian Style**

Chen, Yan, Qing-Wen Wang, and Lv-Ming Xie.
2024. "Dual Quaternion Matrix Equation *AXB* = *C* with Applications" *Symmetry* 16, no. 3: 287.
https://doi.org/10.3390/sym16030287