On the solutions of the dual matrix equation $ A^\top XA = B $

Min Zeng; Yongxin Yuan; Min Zeng; Yongxin Yuan

doi:10.3934/mmc.2023018

Mathematical Modelling and Control

2023, Volume 3, Issue 3: 210-217. doi: 10.3934/mmc.2023018

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Research article

On the solutions of the dual matrix equation $A^\top XA = B$

Min Zeng ,
Yongxin Yuan ^,

School of Mathematics and Statistics, Hubei Normal University, Huangshi, 435002, China

Received: 04 November 2022 Revised: 19 February 2023 Accepted: 05 March 2023 Published: 01 September 2023

Let $\mathbb{D}^{m \times n} = \{A = A_{1}+\varepsilon A_{2}|A_{1}, A_{2}\in \mathbb{R}^{m \times n}\}$ be the set of all $m\times n$ real dual matrices. In this paper, the following problems are considered. Problem I: Given dual matrices $A = A_{1}+\varepsilon A_{2}\in \mathbb{D}^{m\times n}$ and $B = B_{1}+\varepsilon B_{2}\in \mathbb{D}^{n\times n}$ , find $X\in S$ such that the dual matrix equation $A^\top XA = B$ is satisfied, where $S = \{X\in \mathbb{D}^{m \times m}|CX = D, C, D\in \mathbb{D}^{p \times m}\}$ . Problem II: Given dual matrices $A = A_{1}+\varepsilon A_{2}\in \mathbb{D}^{m\times n}, B = B_{1}+\varepsilon B_{2}\in \mathbb{D}^{n\times n}$ and $\tilde{X} = \tilde{X}_{1}+\varepsilon \tilde{X}_{2}\in \mathbb{D}^{m\times m}$ , with $B_{i} = B^\top_{i}, i = 1, 2$ , find $\hat{X}\in T$ such that $\|\hat{X}-\tilde{X}\|_{{\rm D}} = \mathop{\min}\limits_{X\in T} \|X-\tilde{X}\|_{{\rm D}} = \mathop{\min}\limits_{X\in T}\sqrt{\Vert X_{1}-\tilde{X}_{1} \Vert^{2}+\Vert X_{2}-\tilde{X}_{2}\Vert^{2}}$ , where $T = \{X = X_{1}+\varepsilon X_{2}\in \mathbb{D}^{m \times m}|A^\top XA = B \ \ \mbox{s. t.} \ X_{i} = X^\top_{i}, i = 1, 2\}$ . We derive the solvability conditions and the representation of the general solution of Problem I using the Moore-Penrose inverse. Also, we deduce the solvability conditions and the explicit formula of $T$ and the unique approximation solution $\hat{X}$ of Problem II by applying the Moore-Penrose inverse and Kronecker product of matrices. Finally, we give a numerical example to show the correctness of our result.

Keywords:

Citation: Min Zeng, Yongxin Yuan. On the solutions of the dual matrix equation $A^\top XA = B$ [J]. Mathematical Modelling and Control, 2023, 3(3): 210-217. doi: 10.3934/mmc.2023018

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Abstract

1. Introduction

We will adopt the following terminology. ${\mathbb{R}}^{m \times n}$ denotes the set of all $m \times n$ real matrices. ${I_n}$ denotes the identity matrix of size $n$ . $A^\top, A^\dagger, {\rm tr}(A)$ and $\Vert A \Vert$ represent the transpose, the Moore-Penrose inverse, the trace and the Frobenius norm of the matrix $A$ , respectively. Given two matrices $A = [a_{ij}]\in {\mathbb{R}}^{m \times n}$ and $B\in {\mathbb{R}}^{p \times q}$ , the Kronecker product of $A$ and $B$ is defined by $A\otimes B = [a_{ij}B]\in {\mathbb{R}}^{mp \times nq}$ . Also, for a matrix $A = (a_1, a_2, \cdots, a_n)\in \mathbb{R}^{m \times n}, a_i\in \mathbb{R}^{m}, i = 1, 2, \cdots, n,$ the stretch function ${\rm vec}(A)$ is defined as ${\rm vec}(A) = (a_1^\top, a_2^\top, \cdots, a_n^\top)^\top$ . Further, the symbols $E_A$ and $F_A$ stand for two orthogonal projectors $E_A = I_m -AA^\dagger, \; F_A = I_n-A^\dagger A$ induced by $A\in {\mathbb{R}}^{m \times n}$ .

Many scholars considered the following matrix equation

$\begin{align} A^\top XA = B \end{align}$

(1.1)

in real and complex matrix spaces. For example, Dai and Lancaster ^[1] considered symmetric, positive semi-definite, and positive definite solutions of the matrix equation (1.1) with the help of the singular value decomposition. Peng et al. ^[2] provided the necessary and sufficient conditions and the expression of the symmetric ortho-symmetric solutions of the matrix equation (1.1) by applying the generalized singular value decomposition. Li ^[3] gave the necessary and sufficient conditions and the expressions for the D-symmetric solutions of the matrix equation (1.1) on a linear manifold using the generalized singular value decomposition.

In 1873, Clifford ^[4] introduced dual numbers. Subsequently, the dual algebra develops rapidly and has been widely applied to kinematic analysis ^[5], robotics ^[6], screw motion ^[7] and rigid body motion analysis ^[8,9]. The set of the dual numbers is usually denoted by

$\mathbb{D} = \{a = a_{1}+\varepsilon a_{2}|a_{1},a_{2}\in\mathbb{R}\}.$

The real unit $\varepsilon$ is subjected to the rules: $\varepsilon\neq0, \; 0\varepsilon = \varepsilon0 = 0, \; 1\varepsilon = \varepsilon1 = \varepsilon, \; \varepsilon^{2} = 0.$ For the operation rules about the dual numbers, the readers can see Ref. ^[5]. A matrix whose elements are dual numbers is called a dual matrix, namely, the set of all $m\times n$ real dual matrices is

$\mathbb{D}^{m \times n} = \{A = A_{1}+\varepsilon A_{2}|A_{1},A_{2}\in \mathbb{R}^{m \times n}\}.$

The operational rules for dual matrices are similar to those of dual numbers. Dual matrices have important applications in kinematic analysis ^[5,10], spatial kinematics ^[11,12] and robotics ^[6,13]. The solutions of linear dual equations are widely used in kinematic analysis and sensor calibration problems. For instance, Angeles ^[10] applied the dual algebra to compute the parameters of the serew of a rigid body between two finitely-separated positions and of the instant screw. Condurache and Burlacu ^[14] solved the $AX = XB$ sensor calibration problem by means of the orthogonal dual tensor method. Condurache and Ciureanu ^[15] explored the $AX = YB$ sensor calibration problem using dual algebra.

Furthermore, many authors considered the solutions of the dual matrix equation $Ax = b$ . Udwadia ^[16] considered this equation using the dual generalized inverses. Zhong and Zhang ^[17] introduced the dual group-inverse solution of $Ax = b$ . Pennestrì and Valentini ^[18] proposed to solve this dual equation by applying the QR-decomposition.

We observe that the solutions of the dual matrix equation seems to be rarely considered. Therefore, in this paper, we will consider two problems of the dual matrix equation (1.1), that is :

Problem I. Given dual matrices $A = A_{1}+\varepsilon A_{2}\in \mathbb{D}^{m\times n}$ and $B = B_{1}+\varepsilon B_{2}\in \mathbb{D}^{n\times n}$ , find $X\in S$ such that the dual matrix equation (1.1) is satisfied, where $S = \{X\in \mathbb{D}^{m \times m}|CX = D, C, D\in \mathbb{D}^{p \times m}\}$ .

Problem II. Given dual matrices $A = A_{1}+\varepsilon A_{2}\in \mathbb{D}^{m\times n}, B = B_{1}+\varepsilon B_{2}\in \mathbb{D}^{n\times n}$ and $\tilde{X} = \tilde{X}_{1}+\varepsilon \tilde{X}_{2}\in \mathbb{D}^{m\times m}$ , with $B_{i} = B^\top_{i}, i = 1, 2$ , find $\hat{X}\in T$ such that $\|\hat{X}-\tilde{X}\|_{{\rm D}} = \mathop{\min}\limits_{X\in T} \|X-\tilde{X}\|_{{\rm D}} = \mathop{\min}\limits_{X\in T}\sqrt{\Vert X_{1}-\tilde{X}_{1} \Vert^{2} +\Vert X_{2}-\tilde{X}_{2}\Vert^{2}}$ , where $T = \{X = X_{1}+\varepsilon X_{2}\in \mathbb{D}^{m \times m}|A^\top XA = B \ \ \mbox{s. t.} \ X_{i} = X^\top_{i}, i = 1, 2\}$ .

The outline of the rest of this paper is as follows. In Section 2, we introduce some lemmas. In Section 3, the solvability conditions and the representation of the general solution of Problem I are derived by applying the Moore-Penrose inverse. In Section 4, by utilizing the Moore-Penrose inverse and Kronecker product of matrices, we obtain the unique approximation solution $\hat{X}$ of Problem II. In Section 5, a numerical algorithm to solve Problem II and a numerical example are provided. Some concluding remarks are given in Section 6.

2. Preliminaries

First, we should point out that ${\|P\|_{\rm D}} = \sqrt{\Vert P_{1} \Vert^{2}+\Vert P_{2} \Vert^{2}}$ is indeed a matrix norm for the the dual matrix $P = P_{1}+\varepsilon P_{2}$ . In fact, for all $k\in {\mathbb{R}}$ and for all the $m$ -by- $p$ dual matrices $P = P_{1}+\varepsilon P_{2}$ and $Q = Q_{1}+\varepsilon Q_{2},$ where $P_{i}, \; Q_{i}\in \mathbb{R}^{m\times p}\; (i = 1, 2),$ we have ● ${\|P\|_{\rm D}} = \sqrt{\Vert P_{1} \Vert^{2}+\Vert P_{2} \Vert^{2}} \geq 0$ and ${\| P \|_{\rm D}} = 0 \Leftrightarrow P_1 = 0, P_2 = 0$ ;

● ${\|kP\|_{\rm D}} = \sqrt{\Vert kP_{1} \Vert^{2}+\Vert kP_{2} \Vert^{2}} = \sqrt{k^{2}(\Vert P_{1} \Vert^{2}+\Vert P_{2} \Vert^{2})} = |k|\sqrt{\Vert P_{1} \Vert^{2}+\Vert P_{2} \Vert^{2}} = {|k|\cdot\|P \|_{\rm D}}$ ;

● Since

$\begin{equation*} \begin{aligned} {\|P+Q\|_{\rm D}^{2}} = &\Vert P_{1}+P_{2} \Vert^{2}+\Vert Q_{1}+Q_{2}\Vert^{2}\\ \leq &(\Vert P_{1} \Vert+\Vert P_{2} \Vert)^{2}+(\Vert Q_{1} \Vert+\Vert Q_{2} \Vert)^{2} \\ = &\Vert P_{1} \Vert^{2}+\Vert P_{2} \Vert^{2}+\Vert Q_{1} \Vert^{2}+\Vert Q_{2} \Vert^{2}\\ &+2 (\Vert P_{1} \Vert\cdot \Vert P_{2} \Vert+\Vert Q_{1} \Vert\cdot \Vert Q_{2} \Vert),\\ {(\|P\|_{\rm D}+\|Q\|_{\rm D})^{2}} = &\Vert P_{1} \Vert^{2}+\Vert P_{2} \Vert^{2} +\Vert Q_{1} \Vert^{2}+\Vert Q_{2} \Vert^{2}\\ &+2\sqrt{\Vert P_{1} \Vert^{2}+ \Vert P_{2} \Vert^{2}}\cdot\sqrt{\Vert Q_{1} \Vert^{2}+ \Vert Q_{2} \Vert^{2}} , \end{aligned} \end{equation*}$

and

$\Vert P_{1} \Vert\cdot \Vert P_{2} \Vert +\Vert Q_{1} \Vert\cdot \Vert Q_{2} \Vert\leq \sqrt{\Vert P_{1} \Vert^{2}+ \Vert P_{2} \Vert^{2}}\cdot\sqrt{\Vert Q_{1} \Vert^{2}+ \Vert Q_{2} \Vert^{2}}.$

Thus, the inequality ${\|P+Q\|_{\rm D} \leq \|P\|_{\rm D}+\|Q\|_{\rm D}}$ follows.

Next, in order to solve Problems I and II, we introduce the following lemmas.

Lemma 2.1. ^[19] If $A\in \mathbb{R}^{m\times p}, B\in \mathbb{R}^{q\times n}$ and $D\in \mathbb{R}^{m\times n}$ . Then the matrix equation $AXB = D$ has a solution $X\in \mathbb{R}^{p\times q}$ if and only if $AA^\dagger DB^\dagger B = D$ . In this case, the general solution is $X = A^\dagger DB^\dagger +F_{A}V_{1}+V_{2}E_{B}$ , where $V_{1}, V_{2}$ are arbitrary matrices.

Lemma 2.2. ^[20] Let $A\in \mathbb{R}^{m\times n}, B\in \mathbb{R}^{p\times q}, C\in \mathbb{R}^{m\times r}, D\in \mathbb{R}^{s\times q}$ and $E\in \mathbb{R}^{m\times q}.$ Then the linear matrix equation $AXB+CYD = E$ is consistent if and only if

$E_{G}E_{A}E = 0,E_{A}EF_{D} = 0,E_{C}EF_{B} = 0,EF_{B}F_{H} = 0,$

where $G = E_{A}C, \; H = DF_{B}.$ In this case, the general solution is

$\begin{align*} Y = &G^\dagger E_{A}ED^\dagger+(F_{G}C^\dagger+F_{C}G^\dagger E_{A})EF_{B}H^\dagger+W\\ &-C^\dagger CF_{G}WHH^\dagger-G^\dagger GWDD^\dagger,\\ X = &A^\dagger(E-CYD)B^\dagger+Z-A^\dagger AZBB^\dagger, \end{align*}$

where $W, \; Z$ are arbitrary matrices.

Lemma 2.3. ^[21] If $A\in \mathbb{R}^{m\times n}, D\in \mathbb{R}^{m\times m}.$ Then the matrix equation $AXA^\top = D$ has a symmetric solution if and only if $D = D^\top, E_{A}D = 0,$ in this case, the general symmetric solution is $X = A^\dagger D(A^\dagger)^\top+F_{A}V+V^\top F_{A},$ where $V$ is an arbitrary matrix.

Lemma 2.4. ^[22] Suppose that $A, B$ are two real matrices, and $X$ is an unknown variable matrix. Then

$\begin{align*} &\frac{\partial {\rm tr}(BX)}{\partial X} = B^\top, \; \frac{\partial {\rm tr}\left(X^\top B^\top\right)}{\partial X} = B^\top,\\ &\frac{\partial {\rm tr}(AXBX)}{\partial X} = (BXA+AXB)^\top,\\ &\frac{\partial {\rm tr}\left(AX^\top BX^\top\right)}{\partial X} = BX^\top A+AX^\top B,\\ &\frac{\partial {\rm tr}\left(AXBX^\top\right)}{\partial X} = AXB+A^\top XB^\top. \end{align*}$

Lemma 2.5. ^[23] Let $A\in \mathbb{R}^{m\times n}, B\in \mathbb{R}^{n\times l}, C\in \mathbb{R}^{l\times s}$ . Then

${\rm vec}(ABC) = \left(C^\top \otimes A\right){\rm vec}(B).$

Lemma 2.6. ^[24] Let $V\in \mathbb{R}^{m\times n}$ , then ${\rm vec}(V^\top) = T_{mn}{\rm vec}(V)$ , where

$T_{mn} = \left[ \begin{array}{cccc} J^\top_{11} & J^\top_{12} & \cdots & J^\top_{1n} \\ J^\top_{21} & J^\top_{22} & \cdots & J^\top_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ J^\top_{m1} & J^\top_{m2} & \cdots & J^\top_{mn} \\ \end{array} \right]\in \mathbb{R}^{mn \times mn}$

with $J_{ij}, i = 1, \cdots, m, j = 1, \cdots, n$ is an $m\times n$ matrix with the element at position $(i, j)$ is $1$ and the others are $0$ , $T_{mn}$ can be uniquely determined by $m$ and $n$ .

3. Solving Problem I

Theorem 3.1. Given dual matrices $A = A_{1}+\varepsilon A_{2}\in \mathbb{D}^{m\times n}, \; B = B_{1}+\varepsilon B_{2}\in \mathbb{D}^{n\times n}, \; C = C_{1}+\varepsilon C_{2}\in \mathbb{D}^{p \times m}$ and $D = D_{1}+\varepsilon D_{2}\in \mathbb{D}^{p \times m}, i = 1, 2$ , if write

$\begin{align*} G_{1}& = E_{C_{1}}C_{2}F_{C_{1}},\; G_{2} = A^\top_{1}F_{C_{1}}F_{G_{1}},\\ G_{3}& = A^\top_{2}F_{C_{1}}F_{G_{1}}F_{G_{2}}-A^\top_{1}C^\dagger_{1}C_{2}F_{C_{1}}F_{G_{1}}F_{G_{2}},\\ J_{1}& = C^\dagger_{1}D_{1}+F_{C_{1}}G^\dagger_{1}E_{C_{1}}(D_{2}-C_{2}C^\dagger_{1}D_{1}),\\ J_{2}& = (C^\dagger_{1}-C^\dagger_{1}C_{2}F_{C_{1}}G^\dagger_{1}E_{C_{1}})(D_{2} -C_{2}C^\dagger_{1}D_{1}),\\ J_{3}& = J_{1}+F_{C_{1}}F_{G_{1}}G^\dagger_{2}(B_{1}-A^\top_{1}J_{1}A_{1})A^\dagger_{1},\\ J_{4}& = J_{2}-C^\dagger_{1}C_{2}F_{C_{1}}F_{G_{1}}G^\dagger_{2}(B_{1}-A^\top_{1}J_{1}A_{1})A^\dagger_{1},\\ J_{5}& = B_{2}-A^\top_{2}J_{3}A_{1}-A^\top_{1}J_{4}A_{1}-A^\top_{1}J_{3}A_{2},\\ M& = [G_{3}\ \ A^\top_{1}F_{C_{1}}], \; M^\dagger = \left[ \begin{array}{c} M_{1} \\ M_{2} \\ \end{array} \right],\\ N& = E_{A_{1}}A_{2},\; K = E_{M}G_{2},\; H = NF_{A_{1}},\\ J_{6}& = J_{3}+F_{C_{1}}F_{G_{1}}F_{G_{2}}M_{1}J_{5}A^\dagger_{1}\\ &-F_{C_{1}}F_{G_{1}}F_{G_{2}}M_{1}G_{2}K^\dagger E_{M}J_{5}N^\dagger NA^\dagger_{1}\\ &-F_{C_{1}}F_{G_{1}}F_{G_{2}}M_{1}G_{2}F_{K}G^\dagger_{2}J_{5}F_{A_{1}}H^\dagger NA^\dagger_{1}\\ &+F_{C_{1}}F_{G_{1}}K^\dagger E_{M}J_{5}N^\dagger E_{A_{1}}\\ &+F_{C_{1}}F_{G_{1}}(F_{K}G^\dagger_{2} +F_{G_{2}}K^\dagger E_{M})J_{5}F_{A_{1}}H^\dagger E_{A_{1}},\\ J_{7}& = J_{4}-C^\dagger_{1}C_{2}F_{C_{1}}F_{G_{1}}F_{G_{2}}M_{1}J_{5}A^\dagger_{1}\\ &+C^\dagger_{1}C_{2}F_{C_{1}}F_{G_{1}}F_{G_{2}}M_{1}G_{2}K^\dagger E_{M}J_{5}N^\dagger NA^\dagger_{1}\\ &+C^\dagger_{1}C_{2}F_{C_{1}}F_{G_{1}}F_{G_{2}}M_{1}G_{2}F_{K}G^\dagger_{2} J_{5}F_{A_{1}}H^\dagger NA^\dagger_{1}\\ &-C^\dagger_{1}C_{2}F_{C_{1}}F_{G_{1}}K^\dagger E_{M}J_{5}N^\dagger E_{A_{1}}\\ &-C^\dagger_{1}C_{2}F_{C_{1}}F_{G_{1}}(F_{K}G^\dagger_{2}+F_{G_{2}}K^\dagger E_{M})J_{5}F_{A_{1}}H^\dagger E_{A_{1}}\\ &+F_{C_{1}}M_{2}J_{5}A^\dagger_{1} -F_{C_{1}}M_{2}G_{2}K^\dagger E_{M}J_{5}N^\dagger N A^\dagger_{1}\\ &-F_{C_{1}}M_{2}G_{2}F_{K}G^\dagger_{2}J_{5}F_{A_{1}}H^\dagger N A^\dagger_{1}. \end{align*}$

Then Problem I is solvable if and only if

$\begin{align} &E_{C_{1}}D_{1} = 0,\; E_{G_{1}}E_{C_{1}}(D_{2}-C_{2}C^\dagger_{1}D_{1}) = 0, \end{align}$

(3.1)

$\begin{align} &G_{2}G^\dagger_{2}B_{1}A^\dagger_{1}A_{1}+E_{G_{2}}A^\top_{1}J_{1}A_{1} = B_{1}, \end{align}$

(3.2)

$\begin{align} &E_{K}E_{M}J_{5} = 0,\; E_{M}J_{5}F_{N} = 0,\; E_{G_{2}}J_{5}F_{A_{1}} = 0,\; J_{5}F_{A_{1}}F_{H} = 0. \end{align}$

(3.3)

In this case, the general solution of Problem I can be expressed as $X = X_{1}+\varepsilon X_{2}$ , where

$\begin{align} X_{1} = &J_{6}-F_{C_{1}}F_{G_{1}}F_{G_{2}}M_{1}G_{2}F_{K}W_{6}E_{H}NA^\dagger_{1} +F_{C_{1}}F_{G_{1}}F_{G_{2}}W_{71} \\ &-F_{C_{1}}F_{G_{1}}F_{G_{2}}M_{1}G_{3}W_{71}A_{1}A^\dagger_{1} -F_{C_{1}}F_{G_{1}}K^\dagger KW_{6}NN^\dagger E_{A_{1}}\\ &-F_{C_{1}}F_{G_{1}}F_{G_{2}}M_{1}A^\top_{1}F_{C_{1}}W_{72}A_{1}A^\dagger_{1}\\ &+F_{C_{1}}F_{G_{1}}W_{6}E_{A_{1}} -F_{C_{1}}F_{G_{1}}G^\dagger_{2}G_{2}F_{K}W_{6}HH^\dagger E_{A_{1}}, \end{align}$

(3.4)

$\begin{align} X_{2} = &J_{7}+C^\dagger_{1}C_{2}F_{C_{1}}F_{G_{1}}F_{G_{2}}M_{1}G_{2}F_{K}W_{6}E_{H}NA^\dagger_{1}\\ &-C^\dagger_{1}C_{2}F_{C_{1}}F_{G_{1}}F_{G_{2}}W_{71}+F_{C_{1}}W_{72}\\ &+C^\dagger_{1}C_{2}F_{C_{1}}F_{G_{1}}F_{G_{2}}M_{1}G_{3}W_{71}A_{1}A^\dagger_{1}\\ &+C^\dagger_{1}C_{2}F_{C_{1}}F_{G_{1}}F_{G_{2}}M_{1}A^\top_{1}F_{C_{1}}W_{72}A_{1}A^\dagger_{1}\\ &-F_{C_{1}}M_{2}G_{3}W_{71}A_{1}A^\dagger_{1} -F_{C_{1}}M_{2}A^\top_{1}F_{C_{1}}W_{72}A_{1}A^\dagger_{1}\\ &+C^\dagger_{1}C_{2}F_{C_{1}}F_{G_{1}}G^\dagger_{2}G_{2}F_{K}W_{6}HH^\dagger E_{A_{1}} \\ &+C^\dagger_{1}C_{2}F_{C_{1}}F_{G_{1}}K^\dagger KW_{6}NN^\dagger E_{A_{1}} \\ &-F_{C_{1}}M_{2}G_{2}F_{K}W_{6}E_{H}NA^\dagger_{1} -C^\dagger_{1}C_{2}F_{C_{1}}F_{G_{1}}W_{6}E_{A_{1}}, \end{align}$

(3.5)

and $W_{6}, \; W_{71}, \; W_{72}$ are arbitrary matrices.

Proof. By separating the dual matrix equations $CX = D$ and (1.1) into the real part and the dual part leads to the following four equations:

$\begin{align} C_{1}X_{1} = &D_{1}, \end{align}$

(3.6)

$\begin{align} C_{2}X_{1}+C_{1}X_{2} = &D_{2}, \end{align}$

(3.7)

$\begin{align} A^\top_{1}X_{1}A_{1} = &B_{1}, \end{align}$

(3.8)

$\begin{align} A^\top_{1}X_{2}A_{1}+A^\top_{2}X_{1}A_{1}+A^\top_{1}X_{1}A_{2} = &B_{2}. \end{align}$

(3.9)

By using Lemma 2.1, Eq. (3.6) is solvable if and only if the first condition of (3.1) is satisfied, and the general solution is

$\begin{align} X_{1} = C^\dagger_{1}D_{1}+F_{C_{1}}W_{1}, \end{align}$

(3.10)

where $W_{1}$ is an arbitrary matrix. Plugging (3.10) into (3.7), we have

$\begin{align} C_{1}X_{2} = D_{2}-C_{2}C^\dagger_{1}D_{1}-C_{2}F_{C_{1}}W_{1}. \end{align}$

(3.11)

By Lemma 2.1, Eq. (3.11) with respect to $X_{2}$ is solvable if and only if

$\begin{align} G_{1}W_{1} = E_{C_{1}}(D_{2}-C_{2}C^\dagger_{1}D_{1}), \end{align}$

(3.12)

In this case, the general solution is

$\begin{align} X_{2} = C^\dagger_{1}(D_{2}-C_{2}C^\dagger_{1}D_{1}) -C^\dagger_{1}C_{2}F_{C_{1}}W_{1}+F_{C_{1}}W_{2}, \end{align}$

(3.13)

where $W_{2}$ is an arbitrary matrix. By applying Lemma 2.1, Eq. (3.12) is solvable if and only if the second condition of (3.1) is satisfied, and the general solution is

$\begin{align} W_{1} = G^\dagger_{1}E_{C_{1}}(D_{2}-C_{2}C^\dagger_{1}D_{1})+F_{G_{1}}W_{3}, \end{align}$

(3.14)

where $W_{3}$ is an arbitrary matrix. Substituting (3.14) into (3.10) and (3.13) yields

$\begin{align} X_{1}& = J_{1}+F_{C_{1}}F_{G_{1}}W_{3}, \end{align}$

(3.15)

$\begin{align} X_{2}& = J_{2}-C^\dagger_{1}C_{2}F_{C_{1}}F_{G_{1}}W_{3}+F_{C_{1}}W_{2}. \end{align}$

(3.16)

Inserting (3.15) into (3.8) yields

$\begin{align} G_{2}W_{3}A_{1} = B_{1}-A^\top_{1}J_{1}A_{1}. \end{align}$

(3.17)

Using Lemma 2.1 again, Eq. (3.17) with respect to $W_{3}$ is solvable if and only if (3.2) is satisfied, the general solution is

$\begin{align} W_{3} = G^\dagger_{2}(B_{1}-A^\top_{1}J_{1}A_{1})A^\dagger_{1}+F_{G_{2}}W_{4}+W_{5}E_{A_{1}}, \end{align}$

(3.18)

where $W_{4}$ and $W_{5}$ are arbitrary matrices. Plugging (3.18) into (3.15) and (3, 16) leads to

$\begin{align} X_{1} = &J_{3}+F_{C_{1}}F_{G_{1}}F_{G_{2}}W_{4}+F_{C_{1}}F_{G_{1}}W_{5}E_{A_{1}}, \end{align}$

(3.19)

$\begin{align} X_{2} = &J_{4}-C^\dagger_{1}C_{2}F_{C_{1}}F_{G_{1}}F_{G_{2}}W_{4} -C^\dagger_{1}C_{2}F_{C_{1}}F_{G_{1}}W_{5}E_{A_{1}}+F_{C_{1}}W_{2}. \end{align}$

(3.20)

Then, by substituting (3.19) and (3.20) into (3.9), we can get

$\begin{align} MLA_{1}+G_{2}W_{5}N = J_{5}, \end{align}$

(3.21)

where $L = \left[ \begin{array}{c} W_{4} \\ W_{2} \\ \end{array} \right]$ , by Lemma 2.2, Eq. (3.21) with respects to $L$ and $W_{5}$ is solvable if and only if the conditions of (3.3) holds, and the general solution is

$\begin{align} W_{5} = &K^\dagger E_{M}J_{5}N^\dagger+(F_{K}G^\dagger_{2}+F_{G_{2}}K^\dagger E_{M})J_{5}F_{A_{1}}H^\dagger+W_{6}\\ &-G^\dagger_{2}G_{2}F_{K}W_{6}HH^\dagger -K^\dagger KW_{6}NN^\dagger, \end{align}$

(3.22)

$\begin{align} L = &M^\dagger (J_{5}-G_{2}W_{5}N)A^\dagger_{1}+W_{7}-M^\dagger MW_{7}A_{1}A^\dagger_{1}, \end{align}$

(3.23)

where $W_{6}$ and $W_{7}$ are arbitrary matrices. Then

$\begin{align} W_{4} = &M_{1}(J_{5}-G_{2}W_{5}N)A^\dagger_{1}+W_{71} -M_{1}(G_{3}W_{71} \\ &+A^\top_{1}F_{C_{1}}W_{72})A_{1}A^\dagger_{1}, \end{align}$

(3.24)

$\begin{align} W_{2} = &M_{2}(J_{5}-G_{2}W_{5}N)A^\dagger_{1}+W_{72} -M_{2}(G_{3}W_{71} \\ &+A^\top_{1}F_{C_{1}}W_{72})A_{1}A^\dagger_{1}, \end{align}$

(3.25)

where $W_{7} = \left[ \begin{array}{c} W_{71} \\ W_{72} \\ \end{array} \right]$ with $W_{71}\in \mathbb{R}^{m\times n}$ . Inserting (3.22), (3.24) and (3.25) into (3.19) and (3.20), we can easily obtain the expressions (3.4) and (3.5). □

4. Solving Problem II

Theorem 4.1. Given dual matrices $A = A_{1}+\varepsilon A_{2}\in \mathbb{D}^{m\times n}, B = B_{1}+\varepsilon B_{2}\in \mathbb{D}^{n\times n}$ and $\tilde{X} = \tilde{X}_{1}+\varepsilon \tilde{X}_{2}\in \mathbb{D}^{m\times m}$ with $B_{i} = B^\top_{i}, i = 1, 2$ , if write

$\begin{align*} P = &B_{2}-A^\top_{2}(A^\top_{1})^\dagger B_{1}A^\dagger_{1}A_{1}-A^\dagger_{1}A_{1}B_{1}A^\dagger_{1}A_{2},\\ Q = &F_{A_{1}}A^\top_{2}E_{A_{1}},\Theta = F_{Q}E_{A_{1}}A_{2}A^\dagger_{1},\\ V_{1} = &(A^\top_{1})^\dagger B_{1}A^\dagger_{1}+E_{A_{1}}Q^\dagger F_{A_{1}}PA^\dagger_{1} +(A^\dagger_{1})^\top P^\top F_{A_{1}}(Q^\dagger )^\top E_{A_{1}},\\ V_{2} = &(A^\top_{1})^\dagger PA^\dagger_{1}-(A^\dagger_{1})^\top A^\top_{2}E_{A_{1}}Q^\dagger F_{A_{1}}PA^\dagger_{1}\\ &-(A^\dagger_{1})^\top P^\top F_{A_{1}}(Q^\dagger )^\top E_{A_{1}}A_{2}A^\dagger_{1},\\ R_{1} = &\frac{1}{2}E_{A_{1}}\left((\tilde{X}_{2}+\tilde{X}_{2}^\top)-(V^\top_{2}+V_{2})\right),\\ R_{2} = &\frac{1}{2}F_{Q}E_{A_{1}}\left((\tilde{X}_{1}+\tilde{X}_{1}^\top)-(V^\top_{1}+V_{1})\right)\\ &-\frac{1}{2}\Theta\left((\tilde{X}_{2}+\tilde{X}_{2}^\top)-(V^\top_{2}+V_{2})\right)A_{1}A^\dagger_{1},\\ R_{3} = &\frac{1}{2}E_{A_{1}}\left((\tilde{X}_{1}+\tilde{X}_{1}^\top)-(V^\top_{1}+V_{1})\right)E_{A_{1}}. \end{align*}$

Then dual matrix equation $(1.1)$ has a symmetric solution if and only if

$\begin{align} F_{A_{1}}B_{1} = 0,QQ^\dagger F_{A_{1}}PA^\dagger_{1}A_{1} = F_{A_{1}}P. \end{align}$

(4.1)

and the general symmetric solution set of dual matrix equation $(1.1)$ can be expressed as

$\begin{align} T = \{X = X_{1}+\varepsilon X_{2}\in \mathbb{D}^{m \times m}|A^\top XA = B,X_{i} = X^\top_{i} ,i = 1,2\}, \end{align}$

(4.2)

where

$\begin{align} X_{1}& = V_{1}+E_{A_{1}}(F_{Q}U_{3}+U_{4}E_{A_{1}})+(F_{Q}U_{3}+U_{4}E_{A_{1}})^\top E_{A_{1}}, \end{align}$

(4.3)

$\begin{align} X_{2}& = V_{2}-\Theta^\top U_{3}A_{1}A^\dagger_{1} -A_{1}A^\dagger_{1}U^\top_{3}\Theta+E_{A_{1}}U_{2}+U^\top_{2}E_{A_{1}}, \end{align}$

(4.4)

with $U_{2}, \; U_{3}, \; U_{4}$ are arbitrary matrices. In this case, Problem II has the unique solution $\hat{X}$ , and $\hat{X}$ admits the following representation:

$\begin{align} \hat{X} = \hat{X}_{1}+\varepsilon \hat{X}_{2}, \end{align}$

(4.5)

where

$\begin{align} \hat{X}_{1}& = V_{1}+E_{A_{1}}(F_{Q}U_{3}+U_{4}E_{A_{1}})+(F_{Q}U_{3}+U_{4}E_{A_{1}})^\top E_{A_{1}}, \end{align}$

(4.6)

$\begin{align} \hat{X}_{2}& = V_{2}-\Theta^\top U_{3}A_{1}A^\dagger_{1} -A_{1}A^\dagger_{1}U^\top_{3}\Theta+E_{A_{1}}U_{2}+U^\top_{2}E_{A_{1}}, \end{align}$

(4.7)

and $U_{2}, U_{3}$ and $U_{4}$ are determined by solving the unique solution of the equation

$\begin{align} \Delta \left[ \begin{array}{c} {\rm vec}(U_{2}) \\ {\rm vec}(U_{3})\\ {\rm vec}(U_{4}) \\ \end{array} \right] = R, \end{align}$

(4.8)

with $\Delta$ and $R$ being defined as in $(4.25)$ .

Proof. In the first step, we need to find the general symmetric solution of the dual matrix equation (1.1).

The dual matrix equation (1.1) is equivalent to Equations (3.8)–(3.9). Using Lemma 2.3, Eq. (3.8) has a symmetric solution if and only if the first condition of (4.1) is satisfied, and the general symmetric solution is

$\begin{align} X_{1} = (A^\top_{1})^\dagger B_{1}A^\dagger_{1}+E_{A_{1}}U_{1}+U^\top_{1}E_{A_{1}}, \end{align}$

(4.9)

where $U_{1}$ is an arbitrary matrix. Inserting (4.9) into (3.9) yields

$\begin{align} A^\top_{1}X_{2}A_{1} = P-A^\top_{2}F_{A_{1}}U^\top_{1}A_{1}-A^\top_{1}U_{1}F_{A_{1}}A_{2}. \end{align}$

(4.10)

Using Lemma 2.3 again, Eq. (4.10) has a symmetric solution if and only if

$\begin{align} QU_{1}A_{1} = F_{A_{1}}P, \end{align}$

(4.11)

the general symmetric solution is

$\begin{align} X_{2} = &(A^\top_{1})^\dagger PA^\dagger_{1}-(A^\top_{1})^\dagger A^\top_{2}E_{A_{1}}U_{1}A_{1}A^\dagger_{1}-A_{1}A^\dagger_{1}U^\top_{1}E_{A_{1}}A_{2}A^\dagger_{1}\\ &+E_{A_{1}}U^\top_{2}+U_{2}E_{A_{1}}, \end{align}$

(4.12)

where $U_{2}$ is an arbitrary matrix. By Lemma 2.1, Eq. (4.11) with unknown matrix $U_{1}$ has a solution if and only if the second condition of (4.1) is satisfied, the general solution is

$\begin{align} U_{1} = Q^\dagger F_{A_{1}}PA^\dagger_{1}+F_{Q}U_{3}+U_{4}E_{A_{1}}, \end{align}$

(4.13)

where $U_{3}, \; U_{4}$ are arbitrary matrices. By substituting (4.13) into (4.9) and (4.12), we can get (4.3) and (4.4).

In the second step, we need to solve the minimization problem. For the given dual matrix $\tilde{X}\in \mathbb{D}^{m\times m}$ and any matrix $X \in T$ in (4.2), we have

$\begin{align*} &f(U_{2},U_{3},U_{4})\\ = &{\|X-\tilde{X}\|^{2}_{\rm D}}\\ = &\|X_{1}-\tilde{X}_{1}\|^{2}+\|X_{2}-\tilde{X}_{2}\|^{2}\\ = &\|V_{1}+E_{A_{1}}(F_{Q}U_{3}+U_{4}E_{A_{1}})+(F_{Q}U_{3}+U_{4}E_{A_{1}})^\top E_{A_{1}} -\tilde{X}_{1}\|\\ &+\|V_{2}-\Theta^\top U_{3}A_{1}A^\dagger_{1} -A_{1}A^\dagger_{1}U^\top_{3}\Theta+E_{A_{1}}U_{2}+U^\top_{2}E_{A_{1}} -\tilde{X}_{2}\|\\ = &{\rm tr}(V^\top_{1}V_{1}+U^\top_{3}F_{Q}E_{A_{1}}F_{Q}U_{3} +E_{A_{1}}U^\top_{4}E_{A_{1}}U_{4}E_{A_{1}}\\ &+E_{A_{1}}F_{Q}U_{3}U^\top_{3}F_{Q}E_{A_{1}} +E_{A_{1}}U_{4}E_{A_{1}}U^\top_{4}E_{A_{1}}+\tilde{X}^\top_{1}\tilde{X}_{1} +V^\top_{2}V_{2}\\ &+A_{1}A^\dagger_{1}U^\top_{3}\Theta\Theta^\top U_{3}A_{1}A^\dagger_{1} +\Theta^\top U_{3}A_{1}A^\dagger_{1}U^\top_{3}\Theta +U^\top_{2}E_{A_{1}}U_{2}\\ &+E_{A_{1}}U_{2}U^\top_{2}E_{A_{1}}+\tilde{X}^\top_{2}\tilde{X}_{2} +2V^\top_{1}E_{A_{1}}F_{Q}U_{3}+2V^\top_{1}E_{A_{1}}U_{4}E_{A_{1}}\\ &+2V^\top_{1}U^\top_{3}F_{Q}E_{A_{1}}+2V^\top_{1}E_{A_{1}}U^\top_{4}E_{A_{1}} -2V^\top_{1}\tilde{X}_{1}\\ &+2U^\top_{3}F_{Q}E_{A_{1}}U_{4}E_{A_{1}} +2U^\top_{3}F_{Q}E_{A_{1}}U^\top_{3}F_{Q}E_{A_{1}}\\ &+2U^\top_{3}F_{Q}E_{A_{1}}U^\top_{4}E_{A_{1}} -2U^\top_{3}F_{Q}E_{A_{1}}\tilde{X}_{1} +2U^\top_{2}E_{A_{1}}U^\top_{2}E_{A_{1}}\\ &+2E_{A_{1}}U^\top_{4}E_{A_{1}}U^\top_{3}F_{Q}E_{A_{1}} +2E_{A_{1}}U^\top_{4}E_{A_{1}}U^\top_{4}E_{A_{1}}\\ &-2E_{A_{1}}U^\top_{4}E_{A_{1}}\tilde{X}_{1}+2E_{A_{1}}F_{Q}U_{3}E_{A_{1}}U^\top_{4}E_{A_{1}} -2E_{A_{1}}F_{Q}U_{3}\tilde{X}_{1}\\ &-2E_{A_{1}}U_{4}E_{A_{1}}\tilde{X}_{1}-2V^\top_{2}\Theta^\top U_{3}A_{1}A^\dagger_{1} -2V^\top_{2}A_{1}A^\dagger_{1}U^\top_{3}\Theta\\ &+2V^\top_{2}E_{A_{1}}U_{2} +2V^\top_{2}U^\top_{2}E_{A_{1}} -2U^\top_{2}E_{A_{1}}\tilde{X}_{2}\\ &+2U^\top_{3}\Theta U^\top_{3}\Theta -2V^\top_{2}\tilde{X}_{2}-2 E_{A_{1}}U_{2}\tilde{X}_{2} +2\Theta^\top U_{3}A_{1}A^\dagger_{1}\tilde{X}_{2}\\ &+2A_{1}A^\dagger_{1}U^\top_{3}\Theta\tilde{X}_{2} ). \end{align*}$

Therefore, $f(U_{2}, U_{3}, U_{4})$ is minimized if and only if $\frac{{\partial f(U_{2}, U_{3}, U_{4})}}{{\partial U_{2}}} = 0, \frac{{\partial f(U_{2}, U_{3}, U_{4})}}{{\partial U_{3}}} = 0, \frac{{\partial f(U_{2}, U_{3}, U_{4})}}{{\partial U_{4}}} = 0,$ which implies that

$\begin{align} &E_{A_{1}}U_{2}+E_{A_{1}}U^\top_{2}E_{A_{1}} = R_{1}, \end{align}$

(4.14)

$\begin{align} &F_{Q}E_{A_{1}}F_{Q}U_{3}+F_{Q}E_{A_{1}}U^\top_{3}F_{Q}E_{A_{1}}+\Theta\Theta^\top U_{3}A_{1}A^\dagger_{1}+\Theta U^\top_{3}\Theta\\ &+F_{Q}E_{A_{1}}U_{4}E_{A_{1}}+F_{Q}E_{A_{1}}U^\top_{4}E_{A_{1}} = R_{2}, \end{align}$

(4.15)

$\begin{align} &E_{A_{1}}F_{Q}U_{3}E_{A_{1}}+E_{A_{1}}U^\top_{3}F_{Q}E_{A_{1}}+E_{A_{1}}U_{4}E_{A_{1}}+E_{A_{1}}U^\top_{4}E_{A_{1}} = R_{3}. \end{align}$

(4.16)

By applying the Kronecker product and stretching function, (4.14)–(4.16) can be equivalently written as

$\begin{align} &\Delta_{11}{\rm vec}(U_{2}) = {\rm vec}(R_{1}), \end{align}$

(4.17)

$\begin{align} &\Delta_{22}{\rm vec}(U_{3})+\Delta_{23}{\rm vec}(U_{4}) = {\rm vec}(R_{2}), \end{align}$

(4.18)

$\begin{align} &\Delta_{32}{\rm vec}(U_{3})+\Delta_{33}{\rm vec}(U_{4}) = {\rm vec}(R_{3}), \end{align}$

(4.19)

where

$\begin{align} &\Delta_{11} = I_{m}\otimes E_{A_{1}}+T_{m^{2}}(E_{A_{1}}\otimes E_{A_{1}}), \end{align}$

(4.20)

$\begin{align} &\Delta_{22} = I_{m}\otimes {(F_{Q}E_{A_{1}}F_{Q})}+{(A_{1}A^\dagger_{1})}\otimes{(\Theta\Theta^\top)} +T_{m^{2}}\left(\Theta\otimes \Theta^\top \right.\\ +&\left.{(F_{Q}E_{A_{1}})}\otimes {(E_{A_{1}}F_{Q})}\right), \end{align}$

(4.21)

$\begin{align} &\Delta_{23} = E_{A_{1}}\otimes {(F_{Q}E_{A_{1}})}+T_{m^{2}}({(F_{Q}E_{A_{1}})}\otimes E_{A_{1}}), \end{align}$

(4.22)

$\begin{align} &\Delta_{32} = E_{A_{1}}\otimes {(E_{A_{1}}F_{Q})}+T_{m^{2}}\left(E_{A_{1}}\otimes {(E_{A_{1}}F_{Q})}\right), \end{align}$

(4.23)

$\begin{align} &\Delta_{33} = E_{A_{1}}\otimes E_{A_{1}}+T_{m^{2}}(E_{A_{1}}\otimes E_{A_{1}}), \end{align}$

(4.24)

with $T_{m^{2}}$ is the $m^{2}\times m^{2}$ commutation matrix which is defined by Lemma 2.6. Let

$\begin{align} \Delta = \left[ \begin{array}{ccc} \Delta_{11} & 0 & 0 \\ 0 & \Delta_{22} & \Delta_{23} \\ 0 & \Delta_{32} & \Delta_{33} \\ \end{array} \right], \; R = \left[ \begin{array}{c} R_{1}\\ R_{2} \\ R_{3} \\ \end{array} \right]. \end{align}$

(4.25)

Then, (4.17)–(4.19) can be expressed as the equation of (4.8). The proof is complete. □

5. Numerical algorithm and numerical example

Based on Theorem 4.1, we can formulate the following algorithm to solve Problem II.

Algorithm 1

1) Input matrices $A_{i}, B_{i}$ and $\tilde{X}_{i}, i = 1, 2.$

2) Calculate $P, Q, \Theta, V_{1}, V_{2}, R_{1}, R_{2}$ and $R_{3}$ according to Theorem 4.1.

3) If the conditions (4.1) are satisfied, then continue, otherwise, Problem II has no solution, and stop.

4) Compute the matrices $\Delta_{11}, \Delta_{22}, \Delta_{23}, \Delta_{32}$ and $\Delta_{33}$ by (4.20)–(4.24).

5) Compute the matrices $\Delta$ and $R$ by (4.25).

6) Solving Eq. (4.8), we obtain $U_{2} = {\rm reshape(vec}(U_{2})), \; U_{3} = {\rm reshape(vec}(U_{3}))$ and $U_{4} = {\rm reshape(vec}(U_{4}))$ .

7) Compute the matrices $\hat{X}_{1}, \hat{X}_{2}$ on the basis of (4.6)–(4.7).

8) Calculate the unique approximation solution $\hat{X} = \hat{X}_{1}+\varepsilon \hat{X}_{2}$ .

Example 5.1. Let $m = 6, n = 6$ , and the matrices $A_{1}, A_{2}, B_{1}, \; B_{2}, \tilde{X}_{1}, \tilde{X}_{2}$ be given by

$\begin{aligned} & A_1 = \left[\begin{array}{llllll} 1.5712 & 0.5686 & 1.5930 & 1.2840 & 0.9838 & 0.6992 \\ 0.8688 & 0.5996 & 0.9163 & 0.7342 & 0.7349 & 0.6207 \\ 1.0319 & 0.1179 & 1.0144 & 0.8216 & 0.4750 & 0.2494 \\ 1.6420 & 0.6331 & 1.6696 & 1.3452 & 1.0542 & 0.7626 \\ 0.9667 & 0.4320 & 0.9903 & 0.7970 & 0.6603 & 0.4976 \\ 0.8451 & 0.0925 & 0.8303 & 0.6726 & 0.3863 & 0.2009 \end{array}\right], \\ & A_2 = \left[\begin{array}{llllll} 0.2920 & 0.3395 & 0.4177 & 0.1280 & 0.4607 & 0.1206 \\ 0.4317 & 0.9516 & 0.9831 & 0.9991 & 0.9816 & 0.5895 \\ 0.0155 & 0.9203 & 0.3015 & 0.1711 & 0.1564 & 0.2262 \\ 0.9841 & 0.0527 & 0.7011 & 0.0326 & 0.8555 & 0.3846 \\ 0.1672 & 0.7379 & 0.6663 & 0.5612 & 0.6448 & 0.5830 \\ 0.1062 & 0.2691 & 0.5391 & 0.8819 & 0.3763 & 0.2518 \end{array}\right], \end{aligned} \\ \begin{aligned} B_1 & = \left[\begin{array}{rrrrrr} 53.5564 & 18.5678 & 54.1964 & 43.6984 & 32.9892 & 23.1646 \\ 18.5678 & 6.5279 & 18.8010 & 15.1578 & 11.4979 & 8.1054 \\ 54.1964 & 18.8010 & 54.8454 & 44.2215 & 33.3910 & 23.4507 \\ 43.6984 & 15.1578 & 44.2215 & 35.6556 & 26.9221 & 18.9071 \\ 32.9892 & 11.4979 & 33.3910 & 26.9221 & 20.3610 & 14.3185 \\ 23.1646 & 8.1054 & 23.4507 & 18.9071 & 14.3185 & 10.0803 \end{array}\right], \\ B_2 & = \left[\begin{array}{rrrrrr} 82.7260 & 46.3330 & 94.2048 & 73.4880 & 66.5812 & 44.0999 \\ 46.3330 & 22.3895 & 50.6798 & 40.1518 & 34.1939 & 23.0895 \\ 94.2048 & 50.6798 & 105.9817 & 82.9702 & 73.9555 & 49.2920 \\ 73.4880 & 40.1518 & 82.9702 & 64.8924 & 58.2046 & 38.7950 \\ 66.5812 & 34.1939 & 73.9555 & 58.2046 & 50.8615 & 34.1143 \\ 44.0999 & 23.0895 & 49.2920 & 38.7950 & 34.1143 & 22.8048 \end{array}\right], \end{aligned} \\ \tilde{X}_1 = \left[\begin{array}{llllll} 0.4243 & 0.4735 & 0.7655 & 0.6476 & 0.4501 & 0.2564 \\ 0.4609 & 0.1527 & 0.1887 & 0.6790 & 0.4587 & 0.6135 \\ 0.7702 & 0.3411 & 0.2875 & 0.6358 & 0.6619 & 0.5822 \\ 0.3225 & 0.6074 & 0.0911 & 0.9452 & 0.7703 & 0.5407 \\ 0.7847 & 0.1917 & 0.5762 & 0.2089 & 0.3502 & 0.8699 \\ 0.4714 & 0.7384 & 0.6834 & 0.7093 & 0.6620 & 0.2648 \end{array}\right], \\ \tilde{X}_2 = \left[\begin{array}{llllll} 0.6074 & 0.0911 & 0.9452 & 0.7703 & 0.5407 & 0.5447 \\ 0.1917 & 0.5762 & 0.2089 & 0.3502 & 0.8699 & 0.6473 \\ 0.7384 & 0.6834 & 0.7093 & 0.6620 & 0.2648 & 0.5439 \\ 0.2428 & 0.5466 & 0.2362 & 0.4162 & 0.3181 & 0.7210 \\ 0.9174 & 0.4257 & 0.1194 & 0.8419 & 0.1192 & 0.5225 \\ 0.2691 & 0.6444 & 0.6073 & 0.8329 & 0.9398 & 0.9937 \end{array}\right] .$

It is easy to verity that the conditions (4.1) hold:

$\begin{align*} &\|F_{A_{1}}B_{1}\| = 2.2693\times 10^{-14},\\ &\|QQ^\dagger F_{A_{1}}PA^\dagger_{1}A_{1}-F_{A_{1}}P\| = 2.6989\times 10^{-14}. \end{align*}$

By using Algorithm 1, we can obtain the unique approximation solution $\hat{X} = \hat{X}_{1}+\varepsilon \hat{X}_{2}$ of Problem II as follows:

$\hat{X}_1 = \left[\begin{array}{llllll} 1.4250 & 1.2127 & 1.4706 & 1.2908 & 1.2738 & 0.7912 \\ 1.2127 & 1.1359 & 0.3465 & 0.8545 & 0.8276 & 0.8239 \\ 1.4706 & 0.3465 & 1.1821 & 1.3277 & 1.0543 & 1.1221 \\ 1.2908 & 0.8545 & 1.3277 & 1.4092 & 0.8873 & 1.2891 \\ 1.2738 & 0.8276 & 1.0543 & 0.8873 & 0.7528 & 1.0798 \\ 0.7912 & 0.8239 & 1.1221 & 1.2891 & 1.0798 & 0.4615 \end{array}\right], \\ \hat{X}_2 = \left[\begin{array}{llllll} 1.7453 & 0.5624 & 1.3277 & 1.7978 & 1.0231 & 0.9802 \\ 0.5624 & 0.4391 & 0.5530 & 0.6539 & 0.6188 & 0.3080 \\ 1.3277 & 0.5530 & 1.1629 & 1.0739 & 0.9511 & 1.1491 \\ 1.7978 & 0.6539 & 1.0739 & 2.1033 & 0.5943 & 1.2673 \\ 1.0231 & 0.6188 & 0.9511 & 0.5943 & 0.7730 & 0.6016 \\ 0.9802 & 0.3080 & 1.1491 & 1.2673 & 0.6016 & 0.5225 \end{array}\right].$

The absolute errors are estimated by

$\begin{align*} &\|A^\top_{1}X_{1}A_{1}-B_{1}\| = 2.9543\times 10^{-12},\\ &\|A^\top_{1}X_{2}A_{1}+A^\top_{2}X_{1}A_{1}+A^\top_{1}X_{1}A_{2}-B_{2}\| = 1.2922\times 10^{-12}, \end{align*}$

which implies that $\hat{X}$ is the unique approximation solution of Problem II.

6. Conclusions

Solving dual matrix equations is often required in kinematic analysis and sensor calibration. In this paper, the solvability conditions and explicit solutions of Problem I are obtained using the Moore-Penrose inverse (see Theorem 3.1). Further, by applying the Moore-Penrose inverse and Kronecker product of matrices, we obtain the unique approximation solution of Problem II (see Theorem 4.1).

Conflict of interest

All authors declare that they have no conflicts of interest.

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Mathematical Modelling and Control

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Mathematical Modelling and Control

On the solutions of the dual matrix equation $A^\top XA = B$

Related Papers:

Abstract

1. Introduction

2. Preliminaries

3. Solving Problem I

4. Solving Problem II

5. Numerical algorithm and numerical example

6. Conclusions

Conflict of interest

References

Reader Comments

通讯作者: 陈斌, bchen63@163.com

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Catalog

Abstract

1. Introduction

2. Preliminaries

3. Solving Problem I

4. Solving Problem II

5. Numerical algorithm and numerical example

6. Conclusions

Conflict of interest

References

Mathematical Modelling and Control

On the solutions of the dual matrix equation A⊤XA=B A^\top XA = B

Related Papers:

Abstract

1. Introduction

2. Preliminaries

3. Solving Problem I

4. Solving Problem II

5. Numerical algorithm and numerical example

6. Conclusions

Conflict of interest

References

Reader Comments

通讯作者: 陈斌, bchen63@163.com

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Catalog

Abstract

1. Introduction

2. Preliminaries

3. Solving Problem I

4. Solving Problem II

5. Numerical algorithm and numerical example

6. Conclusions

Conflict of interest

References

On the solutions of the dual matrix equation $A^\top XA = B$