A Mirsky-Type Unitarily Invariant Norm Inequality for Dual Quaternion Matrices and Its Applications

Abstract

In this paper, we present a Mirsky-type unitarily invariant norm inequality for dual quaternion matrices, which can be regarded as a singular value perturbation theorem for dual quaternion matrices. By using this unitarily invariant norm inequality, we obtain some other unitarily invariant norm inequalities for dual quaternion matrices, including the low-rank approximation problem, eigenvalue perturbation theorem and polar decomposition. Some inequalities for the difference between singular values and eigenvalues of two dual quaternion matrices under the Frobenius norm are extended.

1. Introduction

In 1873, Clifford [1] introduced the definition of a dual quaternion, which is a combination of two quaternions algebraically combined via a new symbol $ϵ$ that verifies $ϵ\ne 0$ and ${ϵ}^2=0$, i.e., a dual quaternion has the form $q_1+ϵq_2$, where $q_1$ and $q_2$ are both ordinary quaternions. Nowadays, the algebra of dual quaternions has developed into a thriving subalgebra of Clifford algebras, also known as geometric algebras [2].

Dual quaternions have been found to have direct applications in many areas. The problem of studying the displacement of rigid bodies is one of the most important issues in various research domains like robotics [3,4], kinematics [5,6,7] and astrodynamics [8,9,10]. The general displacement of a rigid body can be represented in terms of points or in terms of lines. When investigating line-based methods, dual quaternions stand out as a powerful, concise and elegant tool to represent and operate on rigid-body movement, since dual quaternions are a better representation of rigid body displacements than those treating rotation and translation components independently. Based upon these, dual quaternions were used by many researchers to describe the kinematics of rigid bodies and mechanisms. Yang and Freudenstein introduced the use of dual quaternions for the analysis of spatial mechanisms [11], and Yang also applied them to a serial mechanism [12]. In recent years, dual quaternions have been used in the kinematic analysis and synthesis of mechanisms, and the number of articles applying dual quaternions has increased rapidly. For example, Yacob and Semere used dual quaternions to compensate for variations in the machining process [13]. Kinematics calibration using dual quaternions has been accomplished both in serial robots [14,15] and parallel robots [16]. A controller for a formation consisting of a ground vehicle being escorted by an aerial one was provided by using dual quaternions in formation control [17]. Daniilidis introduced the dual quaternion approach for estimating hand-eye calibration in naturally singular configurations [18]. We refer the readers to reference [19] for a more detailed introduction into the applications of dual quaternions.

A dual quaternion matrix is a matrix in which all of the entries are dual quaternions. Applications of dual quaternion algebra require theoretical and computational background. Some theoretical and computational findings of dual quaternion matrices have been successfully used in many areas. In 2011, Wang [20] proposed the study of dual quaternion matrices in his research on formation control in 3D space. Additionally, in an unpublished manuscript, Wang, Yu and Zheng [21] studied the application of dual quaternion matrices in the multiple rigid-bodies rendezvous problem and proposed three dual quaternion matrices. Recently, many articles were dedicated to establishing some theoretical and computational results for dual quaternion matrices, including matrix decompositions, eigenvalues, singular values and norms. Qi and Luo [22] presented a spectral decomposition for a dual quaternion Hermitian matrix and also the singular value decomposition for a general dual quaternion matrix. Ling, Qi and Yan [23] gave a minimax principle for eigenvalues of dual quaternion Hermitian matrices. Ding et al. [24] proposed a practical method for computing the singular value decomposition of dual quaternion matrices. Ding, Li and Wei [25] presented an eigenvalue decomposition algorithm for dual quaternion Hermitian matrices. Cui and Qi [26] proposed a power method for computing the dominant eigenvalue of a dual quaternion Hermitian matrix and applied it to the simultaneous location and mapping problem. Ling, Pan and Qi [27] introduced a new metric function for dual quaternion matrices and proposed two implementable proximal point algorithms for finding approximate solutions of dual quaternion overdetermined equations. Ling, He and Qi [28] studied some basic properties of dual quaternion matrices, including the polar decomposition theorem, the minimax principle and Weyl’s type monotonicity inequality for singular values, spectral norm and the Pythagoras theorem, and the best low-rank approximations for dual quaternion matrices were also presented. Ling et al. [29] established a von Neumann-type trace inequality and a Hoffman–Wielandt-type inequality for general dual quaternion matrices. Investigations of the solutions and applications of the dual quaternion matrix equation $AXB=C$ can be found in [30,31].

As an important tool in defining metric functions of dual quaternion matrices and dual quaternion vectors, the unitarily invariant norm plays a pivotal role in least-squares problems of dual quaternion equations, low-rank approximation problems, and so on. Additionally, it is frequently used in analyzing rigid-body motion in kinematics, such as the minimum-norm displacement. The study of unitarily invariant norms of dual quaternion matrices has also recently received attention. Cheng and Hu [32] investigated unitarily invariant norms of dual quaternion matrices and introduced the symmetric gauge function on dual quaternions. The unitarily invariant property of norms of dual quaternion matrices was characterized by utilizing the symmetric gauge function. They also introduced the definition of some vital unitarily invariant norms of dual quaternion matrices such as Schatten p-norm and Fan k-norm. Ling, He and Qi [28] discussed the low-rank approximation of a given dual quaternion matrix. Two theorems were presented to characterize the relationship between the approximation degree of the best low-rank approximations under the Frobenius norm and the spectral norm and the best low-rank approximation of dual quaternion matrices in a given subspace. In [29], the authors addressed the concept of the spectral norm of dual quaternion matrices and showed that the spectral norm of a dual quaternion matrix A is exactly the largest singular value of A. Furthermore, Hoffman–Wielandt-type inequalities for two dual quaternion matrices were presented. Zhu, Wang and Kou [33] discussed the norm of the general solution to matrix equation $AXB=C$ over the dual quaternion algebra. Expressions for both the least-norm solution and the least norm of the general solution in two different cases were derived.

Unitarily invariant norms and singular values are two closely related topics. For example, the spectral norm of a dual quaternion matrix is equal to its largest singular value [29]. The singular value decomposition given in [22] is fundamental in dual quaternion matrix research. It is also of importance to consider the effects of errors on the singular value decomposition of dual quaternion matrices, especially the errors initially present in the dual quaternion matrix, as this kind of error can be large compared to a rounding error. One way to evaluate the difference between the singular values of the perturbed dual quaternion matrix and its original is to give a perturbation bound. To this end, in this paper, given two dual quaternion matrices A, B, with ${\sigma }_j$ and ${\mu }_j$, respectively, being singular values of A and B, we shall give an upper bound for the difference between ${\sigma }_j$ and ${\mu }_j$ in terms of a unitarily invariant norm of $A-B$, which is analogous to the Mirsky inequality for complex matrices [34]. The main result can be stated as: for two dual quaternion matrices A, B of appropriate sizes,

$$\begin{array}{c}\hfill \parallel \mathrm{diag}\left(\sigma \left(A\right)\right)-\mathrm{diag}\left(\sigma \left(B\right)\right)\parallel \le \parallel A-B\parallel \end{array}$$

(1)

holds for any unitarily invariant norm $\parallel ·\parallel$, where $\mathrm{diag}\left(\sigma \left(A\right)\right)$ is the diagonal matrix whose main diagonal entries are the singular values of A with nonincreasing order. The inequality (1) establishes one uniform bound for all differences $|{\sigma }_j\left(A\right)-{\sigma }_j\left(B\right)|$ regardless of the magnitudes of the singular values. It also conveys to us some messages about the relationship between dual quaternion matrices and their singular values. For example, if we take the norm to be the spectral norm in (1), then it tells us that the maximum of $|{\sigma }_j\left(A\right)-{\sigma }_j\left(B\right)|$ is bounded by the largest singular value of $A-B$. Furthermore, the validation of the inequality in (1) not only generalizes the result in ([29], Theorem 5.1), but also settles an unsolved problem in [29], which is the aim of this paper. Additionally, we give some applications of the Mirsky-type inequality (1), including the low-rank approximation, and some other unitarily invariant norm inequalities for dual quaternion matrices. Compared with the results in [29], we show that the norm inequality in (1) holds for any unitarily invariant norm and any dual quaternion matrices A and B, while it was shown in [29] that the inequality in (1) holds for Frobenius norm and the case when $A-B$ is appreciable, and the case when A and B are both infinitesimal. The inequality in (1) also gives an upper bound for the difference between the singular values of two dual quaternion matrices A and B in terms of a unitarily invariant norm of $A-B$.

The rest of this paper is organized as follows: In Section 2, we introduce some basic knowledge of dual numbers, dual quaternions and dual quaternion matrices, and some lemmas concerning eigenvalues, singular values and norms of dual quaternion matrices. In Section 2, we first establish some equalities and inequalities for the spectral norm and trace norm of dual quaternion matrices. By using these equalities and inequalities, we give the main result of this paper in Theorem 1. Moreover, by taking the unitarily invariant norm in Theorem 1 to be the Frobenius norm, we obtain some corresponding results in [28,29] as corollaries. Finally, we make some conclusions and remarks in Section 4.

2. Preliminaries

2.1. Dual Numbers

A dual number has the form $a=a_{\mathrm{st}}+a_{\mathrm{in}}ϵ$, where real numbers $a_{\mathrm{st}}$ and $a_{\mathrm{in}}$ are, respectively, called the standard part and the infinitesimal part of a, and $ϵ$ is the infinitesimal unit satisfying $ϵ\ne 0$ and ${ϵ}^2=0$. If $a_{\mathrm{st}}\ne 0$, we say that a is appreciable; otherwise, we say that a is infinitesimal. Denote by $\widehat{\mathbb{R}}$ the set of dual numbers. In [35], a total order for dual numbers was introduced. Given two dual numbers $a=a_{\mathrm{st}}+a_{\mathrm{in}}ϵ$, $b=b_{\mathrm{st}}+b_{\mathrm{in}}ϵ\in \widehat{\mathbb{R}}$. If $a_{\mathrm{st}}>b_{\mathrm{st}}$, or $a_{\mathrm{st}}=b_{\mathrm{st}}$ and $a_{\mathrm{in}}>b_{\mathrm{in}}$, then we have $a>b$. If $a_{\mathrm{st}}=b_{\mathrm{st}}$ and $a_{\mathrm{in}}=b_{\mathrm{in}}$, then $a=b$. The total order provides us a way to compare two dual numbers, and thus it is a foundation for us to establish unitarily invariant norm inequalities for dual quaternion matrices. If $a>0$, then we say that a is a positive dual number, and if $a\ge 0$, we say that a is a non-negative dual number. Denote the set of non-negative dual numbers by ${\widehat{\mathbb{R}}}_{+}$, and the set of positive dual numbers is denoted by ${\widehat{\mathbb{R}}}_{++}$. If $a=a_{\mathrm{st}}+a_{\mathrm{in}}ϵ\in \widehat{\mathbb{R}}$ is positive and appreciable, then the square root of a is defined as

$$\begin{array}{c}\hfill \sqrt{a}=\sqrt{a_{\mathrm{st}}}+{\displaystyle \frac{a_{\mathrm{in}}}{2\sqrt{a_{\mathrm{st}}}}}ϵ.\end{array}$$

For $a=a_{\mathrm{st}}+a_{\mathrm{in}}ϵ$, $b=b_{\mathrm{st}}+b_{\mathrm{in}}ϵ\in \widehat{\mathbb{R}}$, we denote $a+b=a_{\mathrm{st}}+b_{\mathrm{st}}+(a_{\mathrm{in}}+b_{\mathrm{in}})ϵ$, $ab=a_{\mathrm{st}}{b}_{\mathrm{st}}+(a_{\mathrm{st}}{b}_{\mathrm{in}}+a_{\mathrm{in}}{b}_{\mathrm{st}})ϵ$. The absolute value of $a\in \widehat{\mathbb{R}}$ is defined by

$$\begin{array}{c}\hfill \left|a\right|=\left\{\begin{array}{cc}|a_{\mathrm{st}}|+{\displaystyle \frac{a_{\mathrm{st}}}{|a_{\mathrm{st}}|}}{a}_{\mathrm{in}}ϵ,\hfill & \mathrm{if}{a}_{\mathrm{st}}\ne 0,\hfill \\ |a_{\mathrm{in}}|ϵ,\hfill & \mathrm{otherwise}.\hfill \end{array}\right.\end{array}$$

Proposition 1

([28,35]). Let $p,q\in \widehat{\mathbb{R}}$. Then, we have the following conclusions.

(i)

$\left|p\right|=p$ if $p\ge 0$; $\left|p\right|>p$ otherwise.

(ii)

$\left|p\right|=\sqrt{p^2}$ if p is appreciable.

(iii)

If $p,q\in {\widehat{\mathbb{R}}}_{++}$ and are both appreciable, then $\sqrt{pq}=\sqrt{p}\sqrt{q}$.

(iv)

If $p,q\in {\widehat{\mathbb{R}}}_{++}$ are appreciable, then $p-q\in {\widehat{\mathbb{R}}}_{+}$ implies $\sqrt{p}-\sqrt{q}\in {\widehat{\mathbb{R}}}_{+}$.

(v)

For any $p,q\in {\widehat{\mathbb{R}}}_{+}$ satisfying $p-q\in {\widehat{\mathbb{R}}}_{+}$, it holds that $p^k-q^k\in {\widehat{\mathbb{R}}}_{+}$ for any positive integer k.

(vi)

$\left|pq\right|=\left|p\right|\left|q\right|$.

(vii)

$|p+q|\le \left|p\right|+\left|q\right|$.

2.2. Dual Quaternions and Dual Quaternion Matrices

Denote by $\mathbb{Q}$ the set of quaternions. For given $u=u_0+u_1\mathbf{i}+u_2\mathbf{j}+u_3\mathbf{k}\in \mathbb{Q}$, the conjugate of u is $\overline{u}=u_0-u_1\mathbf{i}-u_2\mathbf{j}-u_3\mathbf{k}$. The norm of $u\in \mathbb{Q}$ is defined as $\left|u\right|=\sqrt{\overline{u}u}=\sqrt{u_0^2+u_1^2+u_2^2+u_3^2}$. A dual quaternion has the form $q=q_{\mathrm{st}}+q_{\mathrm{in}}ϵ$, where $q_{\mathrm{st}}$, $q_{\mathrm{in}}\in \mathbb{Q}$ are the standard part and the infinitesimal part of q, respectively. We refer to $\widehat{\mathbb{Q}}$ as the set of dual quaternions. Similar to dual numbers, if $q_{\mathrm{st}}\ne 0$, we say that q is appreciable; otherwise, we say that q is infinitesimal. The conjugate of $q=q_{\mathrm{st}}+q_{\mathrm{in}}ϵ$ is $\overline{q}={\overline{q}}_{\mathrm{st}}+{\overline{q}}_{\mathrm{in}}ϵ$. The magnitude of $q\in \widehat{\mathbb{Q}}$ is given by

$$\begin{array}{c}\hfill \left|q\right|=\left\{\begin{array}{cc}|q_{\mathrm{st}}|+{\displaystyle \frac{q_{\mathrm{st}}{\overline{q}}_{\mathrm{in}}+q_{\mathrm{in}}{\overline{q}}_{\mathrm{st}}}{2|q_{\mathrm{st}}|}}ϵ,\hfill & \mathrm{if}{q}_{\mathrm{st}}\ne 0,\hfill \\ |q_{\mathrm{in}}|ϵ,\hfill & \mathrm{otherwise}.\hfill \end{array}\right.\end{array}$$

A dual quaternion matrix is a matrix in which all entries are dual quaternions. Denote by ${\widehat{\mathbb{Q}}}^{m\times n}$ the set of all $m\times n$ dual quaternion matrices. Especially, ${\widehat{\mathbb{Q}}}^n={\widehat{\mathbb{Q}}}^{n\times 1}$. Then $A\in {\widehat{\mathbb{Q}}}^{m\times n}$ can be written as $A=A_{\mathrm{st}}+A_{\mathrm{in}}ϵ$, where $A_{\mathrm{st}}$ and $A_{\mathrm{in}}$ are, respectively, the standard part and the infinitesimal part of A. If $A_{\mathrm{st}}\ne 0$, then A is said to be appreciable; otherwise, A is said to be infinitesimal. The transpose of $A=\left(a_{ij}\right)\in {\widehat{\mathbb{Q}}}^{m\times n}$ is denoted as $A^{\mathrm{T}}=\left(a_{ji}\right)$. The conjugate of A is denoted as $\overline{A}=\left({\overline{a}}_{ij}\right)$. The conjugate transpose of $A\in {\widehat{\mathbb{Q}}}^{m\times n}$ is $A^{*}={\overline{A}}^{\mathrm{T}}$. It is obvious that $A^{\mathrm{T}}=A_{\mathrm{st}}^{\mathrm{T}}+A_{\mathrm{in}}^{\mathrm{T}}ϵ$, and $A^{*}=A_{\mathrm{st}}^{*}+A_{\mathrm{in}}^{*}ϵ$. A square dual quaternion matrix $A\in {\widehat{\mathbb{Q}}}^{n\times n}$ is called Hermitian if $A^{*}=A$, and unitary if $A^{*}A=I_n$, where $I_n$ is the identity matrix of order n. For two dual quaternion matrices $A\in {\widehat{\mathbb{Q}}}^{m\times n}$, $B\in {\widehat{\mathbb{Q}}}^{n\times m}$, we have ${\left(AB\right)}^{*}=B^{*}{A}^{*}$. In general, ${\left(AB\right)}^{\mathrm{T}}\ne B^{\mathrm{T}}{A}^{\mathrm{T}}$ and $\overline{AB}\ne \overline{A}\overline{B}$.

For given $u={(u_1,u_2,\dots ,u_n)}^{\mathrm{T}}$, $v={(v_1,v_2,\dots ,v_n)}^{\mathrm{T}}\in {\widehat{\mathbb{Q}}}^n$, we denote by $\langle u,v\rangle$ the dual quaternion-valued inner product of u and v, i.e., $\langle u,v\rangle ={\sum }_{i=1}^n{\overline{v}}_i{u}_i$. The Frobenius norm of $A=A_{\mathrm{st}}+A_{\mathrm{in}}ϵ\in {\widehat{\mathbb{Q}}}^{m\times n}$ is given by

$$\begin{array}{c}\hfill {\parallel A\parallel }_F=\left\{\begin{array}{cc}\sqrt{{\displaystyle \sum _{i=1}^m}{\displaystyle \sum _{j=1}^n}{\left|a_{ij}\right|}^2},\hfill & \mathrm{if}{A}_{\mathrm{st}}\ne 0,\hfill \\ \parallel A_{\mathrm{in}}{\parallel }_Fϵ,\hfill & \mathrm{otherwise}.\hfill \end{array}\right.\end{array}$$

For a dual quaternion matrix $A\in {\widehat{\mathbb{Q}}}^{n\times n}$, if there exists a $\lambda \in \widehat{\mathbb{Q}}$ and an appreciable vector $x\in {\widehat{\mathbb{Q}}}^n$ such that $Ax=x\lambda$, then we say that $\lambda$ is a right eigenvalue of A, with x being a corresponding right eigenvector. If $\lambda$ is a dual number, then we have $Ax=\lambda x$, i.e., $\lambda$ is also a left eigenvalue of A. In this case, $\lambda$ is simply called an eigenvalue of A, and x is a corresponding eigenvector. In [22], it was shown that an $n\times n$ dual quaternion Hermitian matrix has exactly n eigenvalues. For a dual quaternion Hermitian matrix $A\in {\widehat{\mathbb{Q}}}^{n\times n}$, we always denote the eigenvalues of A in decreasing order by ${\lambda }_1\left(A\right)\ge {\lambda }_2\left(A\right)\ge \dots \ge {\lambda }_n\left(A\right)$. The trace function on a square dual quaternion matrix $A\in {\widehat{\mathbb{Q}}}^{n\times n}$ is $\mathrm{tr}\left(A\right)={\sum }_{i=1}^n{a}_{ii}$. If $A\in {\widehat{\mathbb{Q}}}^{n\times n}$ is Hermitian, then the trace of A is also equal to ${\sum }_{i=1}^n{\lambda }_i\left(A\right)$, i.e., is equal to the sum of all eigenvalues of A.

We recall some theorems concerning the eigenvalues of dual quaternion Hermitian matrices in [23].

Lemma 1

([23]). Let $A=A_{\mathrm{st}}+A_{\mathrm{in}}ϵ\in {\widehat{\mathbb{Q}}}^{m\times m}$ be Hermitian. Then, there exists a unitary matrix $U\in {\widehat{\mathbb{Q}}}^{m\times m}$ and a diagonal matrix $\mathrm{\Sigma }\in {\widehat{\mathbb{R}}}^{m\times m}$ such that $A=U\mathrm{\Sigma }{U}^{*}$, where

$$\begin{array}{c}\hfill \mathrm{\Sigma }:=\mathrm{diag}\left({\lambda }_1+{\lambda }_{1,1}ϵ,\dots ,{\lambda }_1+{\lambda }_{1,k_1}ϵ,{\lambda }_2+{\lambda }_{2,1}ϵ,\dots ,{\lambda }_r+{\lambda }_{r,k_r}ϵ\right),\end{array}$$

where ${\lambda }_1>{\lambda }_2>\dots >{\lambda }_r$ are real numbers, ${\lambda }_i$ is a $k_i$-multiple eigenvalue of $A_{\mathrm{st}}$, ${\lambda }_{i,1}\ge {\lambda }_{i,2}\ge \dots \ge {\lambda }_{i,k_i}$ are also real numbers. Counting possible multiplicities of ${\lambda }_{i,j}$, the form Σ is unique.

The 2-norm of a dual quaternion vector $x=x_{\mathrm{st}}+x_{\mathrm{in}}ϵ\in {\widehat{\mathbb{Q}}}^n$ is defined as [35]

$$\begin{array}{c}\hfill {\parallel x\parallel }_2=\left\{\begin{array}{cc}\sqrt{{\displaystyle \sum _{i=1}^n}{\left|x_i\right|}^2},& \mathrm{if}{x}_{\mathrm{st}}\ne 0,\\ \parallel x_{\mathrm{in}}{\parallel }_2ϵ,& \mathrm{otherwise}.\end{array}\right.\end{array}$$

(2)

Lemma 2

([23]). Let $A\in {\widehat{\mathbb{Q}}}^{m\times m}$ be Hermitian, and ${\lambda }_1\ge {\lambda }_2\ge \dots \ge {\lambda }_m$ be eigenvalues of A. Then we have

$$\begin{array}{c}\hfill {\lambda }_1={\max \{\parallel x\parallel }_2^{-2}\left(x^{*}Ax\right)|x\in {\widehat{\mathbb{Q}}}^m\setminus \left[\mathbf{0}\right]\}\end{array}$$

and

$$\begin{array}{c}\hfill {\lambda }_m={\min \{\parallel x\parallel }_2^{-2}\left(x^{*}Ax\right)|x\in {\widehat{\mathbb{Q}}}^m\setminus \left[\mathbf{0}\right]\},\end{array}$$

where $\left[\mathbf{0}\right]$ is the set of all infinitesimal vectors in ${\widehat{\mathbb{Q}}}^m$.

Lemma 3

([23]). Let $A\in {\widehat{\mathbb{Q}}}^{m\times m}$ be Hermitian, and ${\lambda }_1\ge {\lambda }_2\ge \dots \ge {\lambda }_m$ be eigenvalues of A. Then for $k=2,3,\dots ,m$, we have

$$\begin{array}{c}\hfill {\lambda }_k=\underset{B\in {\widehat{\mathbb{Q}}}^{m\times (k-1)}}{min}\underset{x\in \mathcal{N}\left(B^{*}\right)\setminus \left[\mathbf{0}\right]}{max}{\parallel x\parallel }_2^{-2}\left(x^{*}Ax\right),\end{array}$$

and it attains ${\lambda }_k$ when $B=[u_1,u_2,\dots ,u_{k-1}]$; and for $k=1,2,\dots ,m-1$, we have

$$\begin{array}{c}\hfill {\lambda }_{m-k}=\underset{C\in {\widehat{\mathbb{Q}}}^{m\times k}}{max}\underset{x\in \mathcal{N}\left(C^{*}\right)\setminus \left[\mathbf{0}\right]}{min}{\parallel x\parallel }_2^{-2}\left(x^{*}Ax\right),\end{array}$$

and it attains ${\lambda }_{m-k}$ when $C=[u_{m-k+1},u_{m-k+2},\dots ,u_m]$, where $u_i$ is the ith column of the unitary matrix U in Lemma 1. Here, for given $W\in {\widehat{\mathbb{Q}}}^{p\times q}$, $N\left(W\right):=\{z\in {\widehat{\mathbb{Q}}}^q|Wz=0\}$.

The following singular value decomposition (SVD) of dual quaternion matrices can be found in [22].

Lemma 4

([22]). For a given $A\in {\widehat{\mathbb{Q}}}^{m\times n}$, there exists a dual quaternion unitary matrix $U\in {\widehat{\mathbb{Q}}}^{m\times m}$ and a dual quaternion unitary matrix $V\in {\widehat{\mathbb{Q}}}^{n\times n}$, such that

$$\begin{array}{c}\hfill A=U{\left[\begin{array}{cc}{\mathrm{\Sigma }}_t& O\\ O& O\end{array}\right]}_{m\times n}{V}^{*},\end{array}$$

where ${\mathrm{\Sigma }}_t\in {\widehat{\mathbb{R}}}^{t\times t}$ is a diagonal matrix, taking the form ${\mathrm{\Sigma }}_t=\mathrm{diag}({\sigma }_1,\dots ,{\sigma }_r,\dots ,{\sigma }_t)$, $r\le t\le p:=\min \{m,n\}$, ${\sigma }_1\ge {\sigma }_2\ge \dots \ge {\sigma }_r$ are positive appreciable dual numbers, and ${\sigma }_{r+1}\ge {\sigma }_{r+2}\ge \dots \ge {\sigma }_t$ are positive infinitesimal dual numbers. Counting possible multiplicities of the diagonal entries, the form ${\mathrm{\Sigma }}_t$ is unique.

The non-negative dual numbers ${\sigma }_1,\dots ,{\sigma }_r,\dots ,{\sigma }_t$ and possibly ${\sigma }_{t+1}=\dots ={\sigma }_p=0$ (if $t<p$) in Lemma 4 are called the singular values of A, where r and t are, respectively, the appreciable rank and rank of A, denoted by ARank(A) and Rank(A). We call the vector $\sigma \left(A\right):={({\sigma }_1,{\sigma }_2,\dots ,{\sigma }_p)}^{\mathrm{T}}$ with ${\sigma }_1\ge {\sigma }_2\ge \dots \ge {\sigma }_p$, the singular value vector of A.

A norm $\parallel ·\parallel$ on ${\widehat{\mathbb{Q}}}^{m\times n}$ is called unitarily invariant if $\parallel UA{V}^{*}\parallel =\parallel A\parallel$ for every $A\in {\widehat{\mathbb{Q}}}^{m\times n}$, and every unitary matrix $U\in {\widehat{\mathbb{Q}}}^{m\times m}$, $V\in {\widehat{\mathbb{Q}}}^{n\times n}$. Noting that any unitarily invariant norm can be represented in terms of a norm of $\sigma \left(A\right)$ ([32], Theorem 4.1). Hence, a unitarily invariant norm on ${\widehat{\mathbb{Q}}}^{m\times n}$ is a dual number-valued function: ${\widehat{\mathbb{Q}}}^{m\times n}\to {\widehat{\mathbb{R}}}_{+}$. In [32], two important unitarily invariant norms were introduced, i.e., the Schatten p-norm and the Fan k-norm.

For $1\le p<\infty$, the Schatten p-norm of $A\in {\widehat{\mathbb{Q}}}^{m\times n}$ is given by

$$\begin{array}{c}\hfill {\parallel A\parallel }_p=\left\{\begin{array}{cc}{\left({\displaystyle \sum _{j=1}^{\min \{m,n\}}}{s_j\left(A\right)}^p\right)}^{{\displaystyle \frac{1}{p}}},& \mathrm{if}{A}_{\mathrm{st}}\ne 0,\\ {\left({\displaystyle \sum _{j=1}^{\min \{m,n\}}}{s_j\left(A_{\mathrm{in}}\right)}^p\right)}^{{\displaystyle \frac{1}{p}}}ϵ,& \mathrm{otherwise}.\end{array}\right.\end{array}$$

In particular, the Schatten 2-norm of a dual quaternion matrix A is the Frobenius norm of A.

The class of Fan-k norms is defined as

$$\begin{array}{c}\hfill {\parallel A\parallel }_{\left(k\right)}={\displaystyle \sum _{j=1}^k}{\sigma }_j\left(A\right),1\le j\le p,\end{array}$$

where $p=\min \{m,n\}$. In particular, ${\parallel A\parallel }_{\left(1\right)}:={\parallel A\parallel }_2$ is the spectral norm of A, ${\parallel A\parallel }_{\left(p\right)}:={\parallel A\parallel }_1$ is the trace norm of A.

The main importance of Fan-k norm lies in the following:

Lemma 5

([32]). Let $A,B\in {\widehat{\mathbb{Q}}}^{m\times n}$. If ${\parallel A\parallel }_{\left(k\right)}\le {\parallel B\parallel }_{\left(k\right)}$ for each $1\le k\le \min \{m,n\}$, then $\parallel A\parallel \le \parallel B\parallel$ holds for any unitarily invariant norm $\parallel ·\parallel$ on ${\widehat{\mathbb{Q}}}^{m\times n}$.

Lemma 6

([23]). (Cauchy–Schwarz inequality on ${\widehat{\mathbb{Q}}}^m$) For any $u,v\in {\widehat{\mathbb{Q}}}^m$, it holds that

$$\begin{array}{c}\hfill {\parallel u\parallel }_2{\parallel v\parallel }_2-\left|\langle u,v\rangle \right|\in {\widehat{\mathbb{R}}}_{+},\end{array}$$

i.e., $\left|\langle u,v\rangle \right|\le {\parallel u\parallel }_2{\parallel v\parallel }_2$.

Lemma 7

([28]). For any A, $B\in {\widehat{\mathbb{Q}}}^{m\times n}$, we have

$$\begin{array}{c}\hfill {\sigma }_{i+j-1}(A+B)\le {\sigma }_i\left(A\right)+{\sigma }_j\left(B\right),{\sigma }_{i+j-1}\left(A^{*}B\right)\le {\sigma }_i\left(A\right){\sigma }_j\left(B\right),\end{array}$$

for any $1\le i,j\le \min \{m,n\}$ and $i+j\le \min \{m,n\}+1$.

3. A Mirsky-Type Norm Inequality for Dual Quaternion Matrices

In this section, we are concerned with a Mirsky-type norm inequality for dual quaternion matrices, which can be regarded as a perturbation theorem of singular values.

In order to establish the main result, we need the following lemmas.

Lemma 8.

Let $A=\mathrm{diag}(d_1,d_2,\dots ,d_n)\in {\widehat{\mathbb{R}}}^{n\times n}$. Then ${\parallel A\parallel }_2=\underset{1\le j\le n}{max}\left|d_j\right|$.

Proof. 

Without loss of generality, let $\left|d_k\right|=\underset{1\le j\le n}{max}\left|d_j\right|$, where k is a positive integer satisfying $1\le k\le n$. We consider the following two cases.

Case 1. A is appreciable. In this case, $d_k$ is also appreciable. It follows from [29] that

$$\begin{array}{c}\hfill {\parallel A\parallel }_2=\underset{x\in {\widehat{\mathbb{Q}}}^n,{\parallel x\parallel }_2=1}{max}{\parallel Ax\parallel }_2.\end{array}$$

(3)

If $Ax$ is infinitesimal for some $x\in {\widehat{\mathbb{Q}}}^n$ satisfying ${\parallel x\parallel }_2=1$, then by (2), ${\parallel Ax\parallel }_2$ is also infinitesimal, and thus ${\parallel Ax\parallel }_2\le \left|d_k\right|$. If $Ax$ is appreciable for $x\in {\widehat{\mathbb{Q}}}^n$ satisfying ${\parallel x\parallel }_2=1$, then it can be observed from (2) and items (ii)–(iii), (v)–(vi) in Proposition 1 that

$$\begin{array}{ccc}\hfill {\parallel Ax\parallel }_2& =& \sqrt{|d_1{x}_1{|}^2+|d_2{x}_2{|}^2+\dots +{\left|d_n{x}_n\right|}^2}\hfill \\ & \le & \sqrt{|d_k{|}^2|x_1{|}^2+|d_k{|}^2|x_2{|}^2+\dots +|d_k{|}^2{\left|x_n\right|}^2}\hfill \\ & =& \sqrt{|d_k{|}^2}\sqrt{|x_1{|}^2+|x_2{|}^2+\dots +{\left|x_n\right|}^2}\hfill \\ & =& |d_k|.\hfill \end{array}$$

Consequently, ${\parallel Ax\parallel }_2\le \left|d_k\right|$ for any $x\in {\widehat{\mathbb{Q}}}^n$ satisfying ${\parallel x\parallel }_2=1$, i.e.,

$$\underset{x\in {\widehat{\mathbb{Q}}}^n,{\parallel x\parallel }_2=1}{max}{\parallel Ax\parallel }_2\le \left|d_k\right|.$$

On the other hand, taking $x=e_k$ in (3), where $e_k$ is the kth column of $I_n$, then $A{e}_k$ is appreciable and ${\parallel Ax\parallel }_2=\sqrt{|d_k{|}^2}=\left|d_k\right|$. Therefore, ${\parallel A\parallel }_2=\underset{x\in {\widehat{\mathbb{Q}}}^n,{\parallel x\parallel }_2=1}{max}{\parallel Ax\parallel }_2=|d_k|=\underset{1\le j\le n}{max}\left|d_j\right|$.

Case 2. A is infinitesimal. In this case, $A=A_{\mathrm{in}}ϵ$ for some real diagonal matrix $A_{\mathrm{in}}$. Therefore, by ([36], Theorem 5.10), ${\parallel A\parallel }_2=\parallel A_{\mathrm{in}}{\parallel }_2ϵ=\underset{1\le j\le n}{max}\left|{\left(d_j\right)}_{\mathrm{in}}\right|ϵ=\underset{1\le j\le n}{max}\left|d_j\right|$. □

Corollary 1.

Let $A\in {\widehat{\mathbb{Q}}}^{n\times n}$ be Hermitian, with ${\lambda }_1$, ${\lambda }_2$, …, ${\lambda }_n$ being eigenvalues of A. Then ${\parallel A\parallel }_2=\underset{1\le j\le n}{max}\left|{\lambda }_j\right|$.

Proof. 

Let $A\in {\widehat{\mathbb{Q}}}^{n\times n}$ be Hermitian, and $A=U\mathrm{\Sigma }{U}^{*}$ be the spectral decomposition of A as in Lemma 1, where $U\in {\widehat{\mathbb{Q}}}^{n\times n}$ is unitary, and $\mathrm{\Sigma }=\mathrm{diag}({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n)$. Then, by Lemma 8, ${\parallel A\parallel }_2=\parallel U\mathrm{\Sigma }{U}^{*}{\parallel }_2={\parallel \mathrm{\Sigma }\parallel }_2=\underset{1\le j\le n}{max}\left|{\lambda }_j\right|$. □

For the given $x={(x_1,x_2,\dots ,x_n)}^{\mathrm{T}}\in {\widehat{\mathbb{Q}}}^n$, denote by $\mathrm{diag}\left(x\right)$ the diagonal matrix whose diagonal entries are $x_1$, $x_2$, …, $x_n$.

Lemma 9.

Let A, $B\in {\widehat{\mathbb{Q}}}^{m\times n}$. Then

$$\begin{array}{c}\hfill \parallel \mathrm{diag}\left(\sigma \left(A\right)\right)-\mathrm{diag}\left(\sigma \left(B\right)\right){\parallel }_2\le {\parallel A-B\parallel }_2.\end{array}$$

(4)

Proof. 

By Lemma 8, $\parallel \mathrm{diag}\left(\sigma \left(A\right)\right)-\mathrm{diag}\left(\sigma \left(B\right)\right){\parallel }_2=\underset{1\le i\le p}{max}|{\sigma }_i\left(A\right)-{\sigma }_i\left(B\right)|$, where $p=\min \{m,n\}$. Hence, in order to prove the inequality in (4), we need only to show that

$$\begin{array}{c}\hfill \underset{1\le i\le n}{max}|{\sigma }_i\left(A\right)-{\sigma }_i\left(B\right)|\le {\sigma }_1(A-B).\end{array}$$

(5)

Let $j=1$. Then the first inequality in Lemma 7 becomes ${\sigma }_i(A+B)\le {\sigma }_i\left(A\right)+{\sigma }_1\left(B\right)$, or equivalently,

$$\begin{array}{c}\hfill {\sigma }_i\left(A\right)-{\sigma }_i\left(B\right)\le {\sigma }_1(A-B).\end{array}$$

(6)

Exchanging the roles of A and B in (6) we get

$$\begin{array}{c}\hfill {\sigma }_i\left(B\right)-{\sigma }_i\left(A\right)\le {\sigma }_1(A-B).\end{array}$$

(7)

Case 1: $A-B$ is appreciable. In this case, ${(A-B)}_{\mathrm{st}}\ne 0$. By the singular value decomposition of dual quaternion matrices in Lemma 4, we know that the standard parts of the singular values of a dual quaternion matrix are exactly the singular values of the standard part of that dual quaternion matrix. Hence, $A-B$ has a positive appreciable singular value, and ${\sigma }_1(A-B)$ is also appreciable. We consider the following two subcases: (i) $\sigma \left(A\right)-\sigma \left(B\right)$ is infinitesimal. In this subcase, ${\sigma }_i\left(A\right)-{\sigma }_i\left(B\right)$ is infinitesimal for all $1\le i\le p$. Then, ${\sigma }_1(A-B)$ being appreciable implies that the inequality in (5) holds. (ii) $\sigma \left(A\right)-\sigma \left(B\right)$ is appreciable. In this subcase, $\underset{1\le i\le n}{max}|{\sigma }_i\left(A\right)-{\sigma }_i\left(B\right)|$ is also appreciable. Let $k\in {\mathbb{Z}}^{+}$ be such that $|{\sigma }_k\left(A\right)-{\sigma }_k\left(B\right)|=\underset{1\le i\le n}{max}|{\sigma }_i\left(A\right)-{\sigma }_i\left(B\right)|$. Assume that ${\sigma }_k\left(A\right)-{\sigma }_k\left(B\right)=p_1+ϵq_1$. Then $|{\sigma }_k\left(A\right)-{\sigma }_k\left(B\right)|=|p_1|+\mathrm{sgn}\left(p_1\right)q_1ϵ$. Let ${\sigma }_1(A-B)=p_2+ϵq_2$. By (6) and (7), $|p_1|\le p_2$. If $|p_1|<p_2$, then the inequality in (5) holds. If $|p_1|=p_2$, then $p_2=p_1$ or $p_2=-p_1$. If $p_2=p_1$, then $|{\sigma }_k\left(A\right)-{\sigma }_k\left(B\right)|=p_1+q_1ϵ$. In this case, by (6) and (7), $q_1\le q_2$. Hence, in this case, the inequality in (5) also holds. The case $p_2=-p_1$ can be proved similarly.

Case 2: $A-B$ is infinitesimal. By ([22], Proposition 6.2), the appreciable rank of $A-B$ is equal to the rank of ${(A-B)}_{\mathrm{st}}$. Therefore, if $A-B$ is infinitesimal, then the appreciable rank of $A-B$ is zero, which implies that all the singular values of $A-B$ are infinitesimal, ${\sigma }_1(A-B)$ is also infinitesimal. Moreover, $A-B$ being infinitesimal implies that $A_{\mathrm{st}}=B_{\mathrm{st}}$. On the other hand, by the singular value decomposition of dual quaternion matrices in Lemma 4, again, we know that the standard parts of the singular values of A, B are, respectively, the singular values of $A_{\mathrm{st}}$ and $B_{\mathrm{st}}$. Hence,

$$\begin{array}{ccc}\hfill \sigma \left(A\right)-\sigma \left(B\right)& =& \left({\left[\sigma \left(A\right)\right]}_{\mathrm{st}}-{\left[\sigma \left(B\right)\right]}_{\mathrm{st}}\right)+\left({\left[\sigma \left(A\right)\right]}_{\mathrm{in}}-{\left[\sigma \left(B\right)\right]}_{\mathrm{in}}\right)\hfill \\ & =& \left[\sigma \left(A_{\mathrm{st}}\right)-\sigma \left(A_{\mathrm{st}}\right)\right]+\left({\left[\sigma \left(A\right)\right]}_{\mathrm{in}}-{\left[\sigma \left(B\right)\right]}_{\mathrm{in}}\right)\hfill \\ & =& {\left[\sigma \left(A\right)\right]}_{\mathrm{in}}-{\left[\sigma \left(B\right)\right]}_{\mathrm{in}}\hfill \end{array}$$

is infinitesimal, and $\underset{1\le i\le n}{max}|{\sigma }_i\left(A\right)-{\sigma }_i\left(B\right)|$ is also infinitesimal. Suppose $\underset{1\le i\le n}{max}|{\sigma }_i\left(A\right)-{\sigma }_i\left(B\right)|=|p|ϵ$ and ${\sigma }_1(A-B)=qϵ$. Then it follows from (6) and (7) that $\left|p\right|\le q$. Therefore, $\underset{1\le i\le n}{max}|{\sigma }_i\left(A\right)-{\sigma }_i\left(B\right)|\le {\sigma }_1(A-B)$. □

Corollary 2.

Let $A\in {\widehat{\mathbb{Q}}}^{m\times n}$, $p=\min \{m,n\}$. Then, for any $1\le k\le p$,

$$\begin{array}{c}\hfill {\sigma }_k\left(A\right)={\min \{\parallel A-X\parallel }_2\mid \mathrm{Rank}\left(X\right)\le k-1,X\in {\widehat{\mathbb{Q}}}^{m\times n}\}.\end{array}$$

Proof. 

Let k be a fixed integer that satisfies $1\le k\le p$. For any $X\in {\widehat{\mathbb{Q}}}^{m\times n}$ with $\mathrm{Rank}\left(X\right)\le k-1$, according to Lemmas 8 and 9, we have

$$\begin{array}{ccc}\hfill {\parallel A-X\parallel }_2& \ge & \parallel \mathrm{diag}\left(\sigma \left(A\right)\right)-\mathrm{diag}\left(\sigma \left(X\right)\right){\parallel }_2\hfill \\ & =& \parallel \mathrm{diag}\left({\sigma }_1\left(A\right)-{\sigma }_1\left(X\right),\dots ,{\sigma }_{k-1}\left(A\right)-{\sigma }_{k-1}\left(X\right),{\sigma }_k\left(A\right),\dots ,{\sigma }_p\left(A\right)\right){\parallel }_2\hfill \\ & \ge & {\sigma }_k\left(A\right).\hfill \end{array}$$

On the other hand, let $t=\mathrm{Rank}\left(A\right)$ and $A=U\mathrm{diag}({\sigma }_1\left(A\right),\dots ,{\sigma }_t\left(A\right),0,\dots ,0)V^{*}$ be the singular value decomposition of A. Denote $X_0=U\mathrm{diag}({\sigma }_1\left(A\right),\dots ,{\sigma }_{k-1}\left(A\right),0,\dots ,0)V^{*}$, then $X_0\in {\widehat{\mathbb{Q}}}^{m\times n}$ and $\mathrm{Rank}\left(X_0\right)\le k-1$. By Lemma 8, we have

$$\begin{array}{c}\hfill \parallel A-X_0{\parallel }_2={\parallel U\mathrm{diag}(0,\dots ,0,{\sigma }_k\left(A\right),\dots ,{\sigma }_t\left(A\right),0,\dots ,0)V^{*}\parallel }_2={\sigma }_k\left(A\right).\end{array}$$

Therefore, for any $k=1,2,\dots ,p$,

$$\begin{array}{c}\hfill {\sigma }_k\left(A\right)={\min \{\parallel A-X\parallel }_2\mid \mathrm{Rank}\left(X\right)\le k-1,X\in {\widehat{\mathbb{Q}}}^{m\times n}\}.\end{array}$$

□

Lemma 10

([35]). Let $A\in {\widehat{\mathbb{Q}}}^{m\times n}$. We have

$$\begin{array}{c}\hfill \underset{X\in {\mathbb{U}\widehat{\mathbb{Q}}}^{m\times n}}{max}\mathrm{tr}(A^{*}X+X^{*}A)=2{\parallel A\parallel }_1,\end{array}$$

where $\mathbb{U}{\widehat{\mathbb{Q}}}^{m\times n}=\{X\in {\widehat{\mathbb{Q}}}^{m\times n}\mid X{X}^{*}=I_m\}$.

Lemma 11.

For a Hermitian matrix $A\in {\widehat{\mathbb{Q}}}^{n\times n}$. It holds that

$$\begin{array}{c}\hfill {\parallel A\parallel }_1=\sum _{i=1}^n\left|{\lambda }_i\left(A\right)\right|.\end{array}$$

Proof. 

By [23], for any $A\in {\widehat{\mathbb{Q}}}^{n\times n}$, ${\lambda }_i\left({\displaystyle \frac{A+A^{*}}{2}}\right)\le {\sigma }_i\left(A\right)$, $1\le i\le n$. In particular, if A is Hermitian, then ${\lambda }_i\left(A\right)\le {\sigma }_i\left(A\right)$ for any $1\le i\le n$. Replacing A by $-A$ in the inequality ${\lambda }_i\left(A\right)\le {\sigma }_i\left(A\right)$ we obtain the inequality $-{\lambda }_i\left(A\right)\le {\sigma }_i\left(A\right)$. Combining these two inequalities we conclude that $|{\lambda }_i\left(A\right)|\le {\sigma }_i\left(A\right)$ for any Hermitian matrix A and any $1\le i\le n$. Hence,

$${\parallel A\parallel }_1=\sum _{i=1}^n{\sigma }_i\left(A\right)\ge \sum _{i=1}^n\left|{\lambda }_i\left(A\right)\right|.$$

Conversely, let $A=U^{*}\mathrm{diag}({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n)U$ be the spectral decomposition of A as in Lemma 1. Then, for any unitary matrix $X\in {\widehat{\mathbb{Q}}}^{n\times n}$, it follows from Proposition 1 (vii) and Lemma 6 that

$$\begin{array}{ccc}\hfill \mathrm{tr}(A^{*}X+X^{*}A)& =& \mathrm{tr}\left[U^{*}\mathrm{diag}({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n)UX+X^{*}{U}^{*}\mathrm{diag}({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n)U\right]\hfill \\ & =& \mathrm{tr}\left[U^{*}\mathrm{diag}({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n)W+W^{*}\mathrm{diag}({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n)U\right]\hfill \\ & =& \sum _{i=1}^n\sum _{j=1}^n{\lambda }_i{\overline{U}}_{ij}{W}_{ij}+\sum _{i=1}^n\sum _{j=1}^n{\lambda }_i{\overline{W}}_{ij}{U}_{ij}\hfill \\ & =& \sum _{i=1}^n{\lambda }_i\left(\langle W_i^{\mathrm{T}},U_i^{\mathrm{T}}\rangle +\langle U_i^{\mathrm{T}},W_i^{\mathrm{T}}\rangle \right)\hfill \\ & \le & \left|\sum _{i=1}^n{\lambda }_i\left(\langle W_i^{\mathrm{T}},U_i^{\mathrm{T}}\rangle +\langle U_i^{\mathrm{T}},W_i^{\mathrm{T}}\rangle \right)\right|\hfill \\ & \le & \sum _{i=1}^n\left|{\lambda }_i\right|\left(|\langle W_i^{\mathrm{T}},U_i^{\mathrm{T}}\rangle |+|\langle U_i^{\mathrm{T}},W_i^{\mathrm{T}}\rangle |\right)\hfill \\ & \le & \sum _{i=1}^n|{\lambda }_i\left|\right(\parallel W_i^{\mathrm{T}}{\parallel }_2\parallel U_i^{\mathrm{T}}{\parallel }_2+\parallel U_i^T{\parallel }_2\parallel W_i^T{\parallel }_2)\hfill \\ & =& 2\sum _{i=1}^n\left|{\lambda }_i\right|,\hfill \end{array}$$

where $W=UX$ is unitary, $W_i^{\mathrm{T}}$ and $V_i^{\mathrm{T}}$ are, respectively, the ith row of W and V, $1\le i\le n$.

Now, it can be seen from Lemma 10 that

$$\begin{array}{c}\hfill {2\parallel A\parallel }_1=\underset{X\in {\mathbb{U}\widehat{\mathbb{Q}}}^{n\times n}}{max}\mathrm{tr}(A^{*}X+X^{*}A)\le 2\sum _{i=1}^n\left|{\lambda }_i\right|,\end{array}$$

i.e., ${\parallel A\parallel }_1\le {\sum }_{i=1}^n\left|{\lambda }_i\right|$. Therefore, ${\parallel A\parallel }_1={\sum }_{i=1}^n\left|{\lambda }_i\right|$. □

Let $A=U\mathrm{diag}({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n)U^{*}$ be the spectral decomposition of a Hermitian matrix $A\in {\widehat{\mathbb{Q}}}^{n\times n}$, where ${\lambda }_i$’s ($1\le i\le n$) are eigenvalues of A. Suppose ${\lambda }_1\ge \dots \ge {\lambda }_k\ge 0\ge {\lambda }_{k+1}\ge \dots \ge {\lambda }_n$. Then A can be written as

$$\begin{array}{ccc}\hfill A& =& U\mathrm{diag}({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_k,0,\dots ,0)U^{*}-U\mathrm{diag}(0,\dots ,0,-{\lambda }_{k+1},\dots ,-{\lambda }_n)U^{*}\hfill \\ & :=& A_{+}-A_{-}.\hfill \end{array}$$

The decomposition $A=A_{+}-A_{-}$ is called the Jordan decomposition of A. Clearly, $A_{+}$ and $A_{-}$ are positive semidefinite.

The following lemma can convert a singular value problem for general dual quaternion matrices to an eigenvalue problem for dual quaternion Hermitian matrices. The proof is analogous to that in ([37], Theorem 7.3.3). For the completeness of presentation, we give a proof below.

Lemma 12.

If the singular values of $A\in {\widehat{\mathbb{Q}}}^{m\times n}$ are ${\sigma }_1$, ${\sigma }_2$, …, ${\sigma }_p$, $p=\min \{m,n\}$, then the eigenvalues of the dual quaternion Hermitian matrix

$$\begin{array}{c}\hfill \left[\begin{array}{cc}0& A^{*}\\ A& 0\end{array}\right]\end{array}$$

are ${\sigma }_1$, …, ${\sigma }_p$, $\underset{|n-m|}{\underbrace{0,\dots ,0}}$, $-{\sigma }_p$, …, $-{\sigma }_1$.

Proof. 

Let $A=U{\mathrm{\Sigma }}_{m\times n}{V}^{*}$ be the singular value decomposition of A, where $U\in {\widehat{\mathbb{Q}}}^{m\times m}$ and $V\in {\widehat{\mathbb{Q}}}^{n\times n}$ are unitary. Suppose $m\ge n$. Then we can rewrite $\mathrm{\Sigma }$ as $\mathrm{\Sigma }=\left[\begin{array}{c}{\mathrm{\Sigma }}_{n\times n}\\ 0\end{array}\right]$. Partiting the unitary matrix U as $U=\left[\begin{array}{cc}{U}_1& U_2\end{array}\right]$, where $U_1\in {\widehat{\mathbb{Q}}}^{m\times n}$. Denote $\tilde{U}={\displaystyle \frac{1}{\sqrt{2}}}{U}_1$ and $\tilde{V}={\displaystyle \frac{1}{\sqrt{2}}}V$. We now construct a matrix $Q\in {\widehat{\mathbb{Q}}}^{(m+n)\times (m+n)}$ as follows:

$$\begin{array}{c}\hfill Q=\left[\begin{array}{ccc}\tilde{U}& -\tilde{U}& U_2\\ \tilde{V}& \tilde{V}& 0\end{array}\right].\end{array}$$

Then, it is not difficult to check that Q is unitary. Moreover, a direct calculation shows that

$$\begin{array}{c}\hfill \left[\begin{array}{cc}0& A^{*}\\ A& 0\end{array}\right]=Q\left[\begin{array}{ccc}{\mathrm{\Sigma }}_m& 0& 0\\ 0& -{\mathrm{\Sigma }}_m& 0\\ 0& 0& 0_{n-m}\end{array}\right]Q^{*}.\end{array}$$

The case $m\le n$ can be similarly proved by applying the above process to $A^{*}$. □

Lemma 13.

Let A, $B\in {\widehat{\mathbb{Q}}}^{m\times n}$. Then

$$\begin{array}{c}\hfill \parallel \mathrm{diag}\left(\sigma \left(A\right)\right)-\mathrm{diag}\left(\sigma \left(B\right)\right){\parallel }_1\le {\parallel A-B\parallel }_1.\end{array}$$

Proof. 

Let $G,H\in {\widehat{\mathbb{Q}}}^{s\times s}$ be Hermitian. We first show that

$$\begin{array}{c}\hfill \sum _{j=1}^s|{\lambda }_j\left(G\right)-{\lambda }_j\left(H\right)|\le {\parallel G-H\parallel }_1.\end{array}$$

(8)

By Lemma 11 and the Jordan decomposition of $G-H$,

$$\begin{array}{c}\hfill {\parallel G-H\parallel }_1=\sum _{i=1}^s\left|{\lambda }_i(G-H)\right|=\mathrm{tr}{(G-H)}_{+}+\mathrm{tr}{(G-H)}_{-}.\end{array}$$

Let $C=G+{(G-H)}_{-}=H+{(G-H)}_{+}$. According to Lemma 3, for $2\le j\le s$, we have

$$\begin{array}{ccc}\hfill {\lambda }_j\left(C\right)& =& \underset{B\in {\widehat{\mathbb{Q}}}^{m\times (j-1)}}{min}\underset{x\in N\left(B^{*}\right)\setminus \left[\mathbf{0}\right]}{max}{\parallel x\parallel }^{-2}\left(x^{*}Cx\right)\hfill \\ & =& \underset{B\in {\widehat{\mathbb{Q}}}^{m\times (j-1)}}{min}\underset{x\in N\left(B^{*}\right)\setminus \left[\mathbf{0}\right]}{max}{\parallel x\parallel }^{-2}\left[x^{*}Gx+x^{*}{(G-H)}_{-}x\right]\hfill \\ & \ge & \underset{B\in {\widehat{\mathbb{Q}}}^{m\times (j-1)}}{min}\underset{x\in N\left(B^{*}\right)\setminus \left[\mathbf{0}\right]}{max}{\parallel x\parallel }^{-2}\left(x^{*}Gx\right)\hfill \\ & =& {\lambda }_j\left(G\right).\hfill \end{array}$$

Similarly, by Lemma 2, we have

$$\begin{array}{ccc}\hfill {\lambda }_1\left(C\right)& =& \underset{x\in {\widehat{\mathbb{Q}}}^m\setminus \left[\mathbf{0}\right]}{max}{\parallel x\parallel }^{-2}\left(x^{*}Cx\right)\hfill \\ & =& \underset{x\in {\widehat{\mathbb{Q}}}^m\setminus \left[\mathbf{0}\right]}{max}{\parallel x\parallel }^{-2}[x^{*}Gx+x^{*}{(G-H)}_{-}x]\hfill \\ & \ge & \underset{x\in {\widehat{\mathbb{Q}}}^m\setminus \left[\mathbf{0}\right]}{max}{\parallel x\parallel }^{-2}\left(x^{*}Gx\right)\hfill \\ & =& {\lambda }_1\left(G\right).\hfill \end{array}$$

Hence, ${\lambda }_j\left(C\right)\ge {\lambda }_j\left(G\right)$ for any $1\le j\le s$.

In a similar way, we conclude that ${\lambda }_j\left(C\right)\ge {\lambda }_j\left(H\right)$ for any $1\le j\le s$.

By Proposition 1 (i), we know that

$$\begin{array}{c}\hfill |{\lambda }_j\left(G\right)-{\lambda }_j\left(H\right)|=\left\{\begin{array}{cc}{\lambda }_j\left(G\right)-{\lambda }_j\left(H\right),\hfill & \mathrm{if}{\lambda }_j\left(G\right)-{\lambda }_j\left(H\right)\in {\widehat{\mathbb{R}}}_{+},\hfill \\ {\lambda }_j\left(H\right)-{\lambda }_j\left(G\right),\hfill & \mathrm{otherwise},\hfill \end{array}\right.\end{array}$$

Hence, $|{\lambda }_j\left(G\right)-{\lambda }_j\left(H\right)|+{\lambda }_j\left(G\right)+{\lambda }_j\left(H\right)=2{\lambda }_j\left(G\right)$ or $2{\lambda }_j\left(H\right)$. In either case, we have

$$\begin{array}{c}\hfill |{\lambda }_j\left(G\right)-{\lambda }_j\left(H\right)|+{\lambda }_j\left(G\right)+{\lambda }_j\left(H\right)\le 2{\lambda }_j\left(C\right)={\lambda }_j\left(2C\right),\end{array}$$

i.e.,

$$\begin{array}{c}\hfill |{\lambda }_j\left(G\right)-{\lambda }_j\left(H\right)|\le {\lambda }_j\left(2C\right)-{\lambda }_j\left(G\right)-{\lambda }_j\left(H\right).\end{array}$$

(9)

By the inequality in (9), we know that

$$\begin{array}{ccc}\hfill \sum _{j=1}^s|{\lambda }_j\left(G\right)-{\lambda }_j\left(H\right)|& \le & \sum _{j=1}^s[{\lambda }_j\left(2C\right)-{\lambda }_j\left(G\right)-{\lambda }_j\left(H\right)]\hfill \\ & =& \mathrm{tr}\left(2C\right)-\mathrm{tr}\left(G\right)-\mathrm{tr}\left(H\right)\hfill \\ & =& \mathrm{tr}(2C-G-H)\hfill \\ & =& \mathrm{tr}\left[\right(C-G)+(C-H\left)\right]\hfill \\ & =& \mathrm{tr}[{(G-H)}_{-}+{(G-H)}_{+}]\hfill \\ & =& \mathrm{tr}\left[{(G-H)}_{-}\right]+\mathrm{tr}\left[{(G-H)}_{+}\right]\hfill \\ & =& {\parallel G-H\parallel }_1.\hfill \end{array}$$

Let

$$\begin{array}{c}\hfill \mathcal{A}=\left[\begin{array}{cc}0& A^{*}\\ A& 0\end{array}\right],\mathcal{B}=\left[\begin{array}{cc}0& B^{*}\\ B& 0\end{array}\right].\end{array}$$

Then $\mathcal{A},\mathcal{B}\in {\widehat{\mathbb{Q}}}^{(m+n)\times (m+n)}$ are Hermitian. Replacing G and H, respectively, by $\mathcal{A}$ and $\mathcal{B}$ in (8), we get

$$\begin{array}{c}\hfill \sum _{j=1}^{m+n}|{\lambda }_j\left(\mathcal{A}\right)-{\lambda }_j\left(\mathcal{B}\right)|\le {\parallel \mathcal{A}-\mathcal{B}\parallel }_1.\end{array}$$

(10)

It can be seen from Lemmas 11 and 12 and (10) that $2{\sum }_{i=1}^p|{\sigma }_i\left(A\right)-{\sigma }_i\left(B\right)|\le {\sum }_{i=1}^{p}2{\sigma }_i(A-B)$, i.e., $\parallel \mathrm{diag}\left(\sigma \left(A\right)\right)-\mathrm{diag}\left(\sigma \left(B\right)\right){\parallel }_1\le {\parallel A-B\parallel }_1$. □

We are now in a position to prove the Mirsky-type norm inequality for dual quaternion matrices.

Theorem 1.

Let A, $B\in {\widehat{\mathbb{Q}}}^{m\times n}$. Then

$$\begin{array}{c}\hfill \parallel \mathrm{diag}\left(\sigma \left(A\right)\right)-\mathrm{diag}\left(\sigma \left(B\right)\right)\parallel \le \parallel A-B\parallel \end{array}$$

(11)

holds for any unitarily invariant norm $\parallel ·\parallel$.

Proof. 

By Lemma 5, it suffices to prove that the inequality (11) holds for Fan k-norms, $1\le k\le p=\min \{m,n\}$.

Let $A-B=U\mathrm{diag}({\sigma }_1,{\sigma }_2,\dots ,{\sigma }_p)V^{*}$ be the singular value decomposition of $A-B$, where $U\in {\widehat{\mathbb{Q}}}^{m\times m}$ and $V\in {\widehat{\mathbb{Q}}}^{n\times n}$ are dual quaternion unitary matrices, ${\sigma }_1\ge {\sigma }_2\ge \dots \ge {\sigma }_p$ are singular values of $A-B$. Then, for any fixed k satisfying $1\le k\le p$,

$$\begin{array}{ccc}\hfill A-B& =& U\mathrm{diag}({\sigma }_1-{\sigma }_k,{\sigma }_2-{\sigma }_k,\dots ,{\sigma }_k-{\sigma }_k,0,\dots ,0)V^{*}\hfill \\ & +& U\mathrm{diag}({\sigma }_k,{\sigma }_k,\dots ,{\sigma }_k,{\sigma }_{k+1},\dots ,{\sigma }_p)V^{*}\hfill \\ & :=& X+Y.\hfill \end{array}$$

Hence, ${\parallel A-B\parallel }_{\left(k\right)}={\sum }_{i=1}^k{\sigma }_i={\sum }_{i=1}^k({\sigma }_i-{\sigma }_k)+k{\sigma }_k={\parallel X\parallel }_1+k{\parallel Y\parallel }_2$.

It can be seen from the equality

$$\begin{array}{ccc}\hfill \mathrm{diag}\left(\sigma \left(A\right)\right)-\mathrm{diag}\left(\sigma \left(B\right)\right)& =& \left[\mathrm{diag}\left(\sigma (X+B)\right)-\mathrm{diag}\left(\sigma \left(B\right)\right)\right]\hfill \\ & +& \left[\mathrm{diag}\left(\sigma \left(A\right)\right)-\mathrm{diag}\left(\sigma (X+B)\right)\right]\hfill \end{array}$$

and Lemmas 9 and 13 that

$$\begin{array}{ccc}\hfill \parallel \mathrm{diag}\left(\sigma \left(A\right)\right)-\mathrm{diag}\left(\sigma \left(B\right)\right){\parallel }_{\left(k\right)}& \le & \parallel \mathrm{diag}\left(\sigma (X+B)\right)-\mathrm{diag}\left(\sigma \left(B\right)\right){\parallel }_{\left(k\right)}\hfill \\ & +& \parallel \mathrm{diag}\left(\sigma \left(A\right)\right)-\mathrm{diag}\left(\sigma (X+B)\right){\parallel }_{\left(k\right)}\hfill \\ & \le & \parallel \mathrm{diag}\left(\sigma (X+B)\right)-\mathrm{diag}\left(\sigma \left(B\right)\right){\parallel }_1\hfill \\ & +& k\parallel \mathrm{diag}\left(\sigma \left(A\right)\right)-\mathrm{diag}\left(\sigma (X+B)\right){\parallel }_2\hfill \\ & \le & {\parallel X+B-B\parallel }_1+k{\parallel A-X-B\parallel }_2\hfill \\ & =& {\parallel X\parallel }_1+k{\parallel Y\parallel }_2\hfill \\ & =& {\parallel A-B\parallel }_{\left(k\right)},\hfill \end{array}$$

which completes the proof. □

Corollary 3.

Let A, $B\in {\widehat{\mathbb{Q}}}^{m\times n}$. Then

$$\begin{array}{c}\hfill \parallel \mathrm{diag}\left(\sigma \left(A\right)\right)-\mathrm{diag}\left(\sigma \left(B\right)\right){\parallel }_F\le {\parallel A-B\parallel }_F.\end{array}$$

(12)

Remark 1.

Notice that $\parallel \mathrm{diag}\left(\sigma \left(A\right)\right)-\mathrm{diag}\left(\sigma \left(B\right)\right){\parallel }_F={\parallel \sigma \left(A\right)-\sigma \left(B\right)\parallel }_2$ for any A, $B\in {\widehat{\mathbb{Q}}}^{m\times n}$. Hence, the inequality in (12) can be rewritten as ${\parallel \sigma \left(A\right)-\sigma \left(B\right)\parallel }_2\le {\parallel A-B\parallel }_F$, which was shown to be true for the case $A-B$ and is appreciable in ([29], Theorem 5.1) and the case A and B are both infinitesimal. However, if both A and B are appreciable, but $A-B$ is infinitesimal, whether the inequality (12) holds is unknown. We have shown in Corollary 3 that ${\parallel \sigma \left(A\right)-\sigma \left(B\right)\parallel }_2\le {\parallel A-B\parallel }_F$ holds for any A, $B\in {\widehat{\mathbb{Q}}}^{m\times n}$. Hence, the result in ([29], Theorem 5.1) is a special case of Corollary 3. It is also of interest to consider the converse of Theorem 1. i.e., if $\parallel \mathrm{diag}\left(\sigma \left(A\right)\right)-\mathrm{diag}\left(\sigma \left(B\right)\right)\parallel \le \delta$ for some constant δ, does it imply anything about $\parallel A-B\parallel$?

The inequality in Theorem 1 is very useful in proving some unitarily invariant norm inequalities that involve the difference between two dual quaternion matrices. We will list some applications of Theorem 1 below. One of important applications of Theorem 1 is the low-rank approximation problem, which has applications in many fields, such as image compression and signal processing.

Corollary 4.

Suppose that $A\in {\widehat{\mathbb{Q}}}^{m\times n}$ has an SVD as follows:

$$\begin{array}{c}\hfill A=U{\left[\begin{array}{cc}{\mathrm{\Sigma }}_t& 0\\ 0& 0\end{array}\right]}_{m\times n}{V}^{*},\end{array}$$

where $U\in {\widehat{\mathbb{Q}}}^{m\times m}$ and $V\in {\widehat{\mathbb{Q}}}^{n\times n}$ are dual quaternion unitary matrices, and ${\mathrm{\Sigma }}_t\in {\widehat{\mathbb{R}}}^{t\times t}$ is a diagonal matrix, taking the form ${\mathrm{\Sigma }}_t=\mathrm{diag}({\sigma }_1,\dots ,{\sigma }_r,\dots ,{\sigma }_t)$ with $r\le t\le p:=\min \{m,n\}$, ${\sigma }_1\ge {\sigma }_2\ge \dots \ge {\sigma }_r$ are positive appreciable dual numbers, and ${\sigma }_{r+1}\ge {\sigma }_{r+2}\ge \dots \ge {\sigma }_t$ are positive infinitesimal dual numbers. Let $A_k=U\left[\begin{array}{cc}{\mathrm{\Sigma }}_k& 0\\ 0& 0\end{array}\right]V^{*}$ be with $k\le t$, where ${\mathrm{\Sigma }}_k=\mathrm{diag}({\sigma }_1,{\sigma }_2,\dots ,{\sigma }_k)$. Then, for any $B\in {\widehat{\mathbb{Q}}}^{m\times n}$ with rank at most k,

$$\begin{array}{c}\hfill \parallel A-A_k\parallel \le \parallel A-B\parallel \end{array}$$

(13)

holds for any unitarily invariant norm.

Proof. 

Let k be an integer satisfying $1\le k\le t$, and let $B\in {\widehat{\mathbb{Q}}}^{m\times n}$ be with $\mathrm{Rank}\left(B\right)\le k$. Suppose $\sigma \left(B\right)={({\mu }_1,{\mu }_2,\dots ,{\mu }_k,0,\dots ,0)}^{\mathrm{T}}\in {\widehat{\mathbb{R}}}^p$, where ${\mu }_1\ge {\mu }_2\ge \dots \ge {\mu }_k\ge 0$. For any unitarily invariant norm $\parallel ·\parallel$, by Theorem 1,

$$\begin{array}{c}\hfill \parallel A-B\parallel \ge \parallel \mathrm{diag}\left(\sigma \left(A\right)\right)-\mathrm{diag}\left(\sigma \left(B\right)\right)\parallel =\parallel {\mathrm{\Lambda }}_1\oplus {\mathrm{\Lambda }}_2\parallel ,\end{array}$$

(14)

where ${\mathrm{\Lambda }}_1=\mathrm{diag}({\sigma }_1-{\mu }_1,\dots ,{\sigma }_k-{\mu }_k)$, ${\mathrm{\Lambda }}_2=\mathrm{diag}({\sigma }_{k+1},\dots ,{\sigma }_t,0,\dots ,0)$, ${\mathrm{\Lambda }}_1\oplus {\mathrm{\Lambda }}_2=\mathrm{diag}({\mathrm{\Lambda }}_1,{\mathrm{\Lambda }}_2)$ is the direct sum of ${\mathrm{\Lambda }}_1$ and ${\mathrm{\Lambda }}_2$.

Noting that $\sigma ({\mathrm{\Lambda }}_1\oplus {\mathrm{\Lambda }}_2)=\sigma \left({\mathrm{\Lambda }}_1\right)\cup \sigma \left({\mathrm{\Lambda }}_2\right)$. Then ${\sigma }_j({\mathrm{\Lambda }}_1\oplus {\mathrm{\Lambda }}_2)\ge {\sigma }_j(O\oplus {\mathrm{\Lambda }}_2)$ for any $1\le j\le t$, where O is a zero matrix of appropriate size. Therefore, by Lemma 5, we have

$$\begin{array}{c}\hfill \parallel {\mathrm{\Lambda }}_1\oplus {\mathrm{\Lambda }}_2\parallel \ge \parallel O\oplus {\mathrm{\Lambda }}_2\parallel =\parallel U(O\oplus {\mathrm{\Lambda }}_2)V^{*}\parallel =\parallel A-A_k\parallel .\end{array}$$

(15)

Now, it can be seen from (14) and (15) that $\parallel A-A_k\parallel \le \parallel A-B\parallel$ holds for any unitarily invariant norm. □

It was shown in ([28], Theorem 7.1) that the inequality in (13) holds for two special unitarily invariant norms, i.e., the spectrum norm ${\parallel ·\parallel }_2$ and the Frobenius norm ${\parallel ·\parallel }_F$. Therefore, the results in ([28], Theorem 7.1) are special cases of Corollary 4.

Let $A\in {\widehat{\mathbb{Q}}}^{n\times n}$ be Hermitian with eigenvalues ${\lambda }_1\ge {\lambda }_2\ge \dots \ge {\lambda }_n$. Denote $\lambda \left(A\right)={({\lambda }_1,{\lambda }_2,\dots {\lambda }_n)}^{\mathrm{T}}$ as the eigenvalue vector of A. The following is a perturbation theorem for eigenvalues of dual quaternion Hermitian matrices.

Theorem 2.

Let $A,B\in {\widehat{\mathbb{Q}}}^{n\times n}$ be Hermitian. Then

$$\begin{array}{c}\hfill \parallel \mathrm{diag}\left(\lambda \left(A\right)\right)-\mathrm{diag}\left(\lambda \left(B\right)\right)\parallel \le \parallel A-B\parallel \end{array}$$

(16)

holds for any unitarily invariant norm.

Proof. 

Suppose that $A,B\in {\widehat{\mathbb{Q}}}^{n\times n}$ are Hermitian. Let d be a positive dual number that satisfies $d\ge \underset{1\le i\le n}{max}\left\{\right|{\lambda }_i\left(A\right)|,|{\lambda }_i\left(B\right)\left|\right\}$. Then, it is not difficult to see that the dual quaternion matrices $A+dI$ and $B+dI$ are both positive semidefinite. Moreover, the eigenvalues of a dual quaternion positive semidefinite matrix coincides with its singular values. Hence, by Theorem 1, for any unitarily invariant norm $\parallel ·\parallel$, we have

$$\begin{array}{ccc}\hfill \parallel \mathrm{diag}\left(\lambda \left(A\right)\right)-\mathrm{diag}\left(\lambda \left(B\right)\right)\parallel & =& \parallel \left[\mathrm{diag}\left(\lambda (A+dI)\right)-dI\right]-\left[\mathrm{diag}\left(\lambda (B+dI)\right)-dI\right]\parallel \hfill \\ & =& \parallel \mathrm{diag}\left(\sigma (A+dI)\right)-\mathrm{diag}\left(\sigma (B+dI)\right)\parallel \hfill \\ & \le & \parallel (A+dI)-(B+dI)\parallel \hfill \\ & =& \parallel A-B\parallel .\hfill \end{array}$$

□

Corollary 5

([29]). Let $A,B\in {\widehat{\mathbb{Q}}}^{n\times n}$. If both A and B are Hermitian matrices, then we have

$$\begin{array}{c}\hfill \parallel \mathrm{diag}\left(\lambda \left(A\right)\right)-\mathrm{diag}\left(\lambda \left(B\right)\right){\parallel }_F={\parallel \lambda \left(A\right)-\lambda \left(B\right)\parallel }_2\le {\parallel A-B\parallel }_F.\end{array}$$

Corollary 6.

Let $A,B\in {\widehat{\mathbb{Q}}}^{n\times n}$. If both A and B are Hermitian matrices, then we have

$$\begin{array}{c}\hfill \sum _{i=1}^n|{\lambda }_i\left(A\right)-{\lambda }_i\left(B\right)|\le \sum _{i=1}^n{\sigma }_i(A-B).\end{array}$$

Proof. 

It follows directly from Theorem 2 (taking the norm to be the trace norm ${\parallel ·\parallel }_1$) and Lemma 11. □

Corollary 7.

Let $A,B\in {\widehat{\mathbb{Q}}}^{n\times n}$ be Hermitian. Then

$$\begin{array}{c}\hfill \underset{1\le j\le n}{max}|{\lambda }_j\left(A\right)-{\lambda }_j\left(B\right)|\le \underset{1\le j\le n}{max}{\sigma }_j(A-B).\end{array}$$

Proof. 

It follows directly from Theorem 2 (taking the norm to be the spectral norm ${\parallel ·\parallel }_2$) and Lemma 8. □

For $A\in {\widehat{\mathbb{Q}}}^{n\times n}$, let $A=YD{Z}^{*}$ be the singular value decomposition of A, where $Y,Z\in {\widehat{\mathbb{Q}}}^{n\times n}$ are unitary, $D=\mathrm{diag}\left(\sigma \left(A\right)\right)$. Then A can be represented as

$$\begin{array}{c}\hfill A=YD{Z}^{*}=\left(Y{Z}^{*}\right)ZD{Z}^{*}:=UP,\end{array}$$

(17)

where $U=Y{Z}^{*}$ is unitary, $P=ZD{Z}^{*}$ is positive semidefinite. Similarly, A can also be represented as $A=YD{Z}^{*}=\left(YD{Y}^{*}\right)\left(Y{Z}^{*}\right):=QV$, where $V=Y{Z}^{*}$ is unitary, and $Q=YD{Y}^{*}$ is positive semidefinite. Analogous to complex matrices, $A=UP$ or $A=QV$ is called a polar decomposition of A.

Theorem 3.

Let $A=YD{Z}^{*}$ be the singular value decomposition of A, and let $A=UP$ be a polar decomposition of $A\in {\widehat{\mathbb{Q}}}^{n\times n}$ as in (17), where U is unitary and P is positive semidefinite. Then, for any unitarily invariant norm $\parallel ·\parallel$,

$$\begin{array}{c}\hfill \parallel A-U\parallel \le \parallel A-W\parallel \le \parallel A+U\parallel ,\end{array}$$

where $W\in {\widehat{\mathbb{Q}}}^{n\times n}$ is an arbitrary unitary matrix.

Proof. 

For any unitary matrix $W\in {\widehat{\mathbb{Q}}}^{n\times n}$, by [22], the identity matrix $I_n$ is a perfect Hermitian matrix, and then ${\sigma }_j\left(W\right)=1$ for all $1\le j\le n$. Therefore, by Theorem 1, for any unitarily invariant norm $\parallel ·\parallel$, we have

$$\begin{array}{c}\hfill \parallel A-W\parallel \ge \parallel \mathrm{diag}\left(\sigma \left(A\right)\right)-I_n\parallel .\end{array}$$

(18)

Let $A=UP$ be a polar decomposition of A as in (17), where $U=Y{Z}^{*}$ is unitary and $P=ZD{Z}^{*}$ is positive semidefinite. Therefore, for any unitarily invariant norm $\parallel ·\parallel$, we have

$$\begin{array}{c}\hfill \parallel A-U\parallel =\parallel UP-U\parallel =\parallel P-I_n\parallel .\end{array}$$

(19)

Moreover,

$$\begin{array}{c}\hfill \parallel P-I_n\parallel =\parallel ZD{Z}^{*}-Z{Z}^{*}\parallel =\parallel D-I_n\parallel =\parallel \mathrm{diag}\left(\sigma \left(A\right)\right)-I_n\parallel .\end{array}$$

(20)

Now, it can be seen from (18)–(20) that $\parallel A-U\parallel \le \parallel A-W\parallel$ holds for any unitarily invariant norm $\parallel ·\parallel$, and any unitary matrix $W\in {\widehat{\mathbb{Q}}}^{n\times n}$.

On the other hand, by Lemma 7,

$$\begin{array}{c}\hfill {\sigma }_{i+j-1}(A-W)\le {\sigma }_i\left(A\right)+{\sigma }_j(-W)={\sigma }_i\left(A\right)+1\end{array}$$

(21)

holds for any $1\le i,j\le n$ and $i+j\le n+1$.

Taking $j=1$ in (21), we get

$$\begin{array}{c}\hfill {\sigma }_i(A-W)\le {\sigma }_i\left(A\right)+1={\sigma }_i(D+I).\end{array}$$

(22)

Hence, by Lemma 5 and (22), for any unitarily invariant norm, it holds that

$$\begin{array}{c}\hfill \parallel A-W\parallel \le \parallel D+I\parallel .\end{array}$$

(23)

Furthermore,

$$\begin{array}{c}\hfill \parallel A+U\parallel =\parallel UP+U\parallel =\parallel P+I_n\parallel =\parallel ZD{Z}^{*}+Z{Z}^{*}\parallel =\parallel D+I_n\parallel .\end{array}$$

(24)

Therefore, the inequality $\parallel A-W\parallel \le \parallel A+U\parallel$ follows directly from (23) and (24). □

4. Conclusions

In this paper, we give a Mirsky-type unitarily invariant norm inequality for dual quaternion matrices. We establish the main result by firstly showing that the inequality holds for two special unitarily invariant norms, i.e., the spectral norm ${\parallel ·\parallel }_2$ and the trace norm ${\parallel ·\parallel }_1$. Furthermore, by applying the main result to the Frobenius norm ${\parallel ·\parallel }_F$, some results in [28,29] are extended. Our results may enrich the basic theory, especially the unitarily invariant norm theory for dual quaternion matrices. As we mentioned earlier, the applications of dual quaternion algebra require a theoretical background, we hope our results will be helpful in the problem of studying the displacement of rigid bodies, and also in other areas.

On the other hand, the inequality in (1) can be regarded as a perturbation theorem for singular values of dual quaternion matrices. We obtain an upper bound for the difference between singular values of two dual quaternion matrices A and B in terms of a unitarily invariant norm of $A-B$. The upper bound given in (1) works best in some circumstances. As we know, there are many results concerning the perturbation bound for the difference between singular values of complex matrices. However, as a new area of applied mathematics, there are few results for the perturbation problem of singular values of dual quaternion matrices. For theoretical and practice purposes, some further problems should be explored. For example, given two dual quaternion matrices $A_1$ and $A_2$, with $A_1=U_1{\mathrm{\Sigma }}_1{V}_1$ and $A_2=U_2{\mathrm{\Sigma }}_2{V}_2$, respectively, being their singular value decompositions, how do we compare $U_1$ and $U_2$, or $V_1$ and $V_2$? It is also worth studying the bounds for the relative differences between singular values of two dual quaternion matrices.