Perturbation of Dual Group Generalized Inverse and Group Inverse

Abstract

Symmetry plays a crucial role in the study of dual matrices and dual matrix group inverses. This paper is mainly divided into two parts. We present the definition of the spectral norm of a dual real matrix $\widehat{A}$, (which is usually represented in the form $\widehat{A}=A+\epsilon A_0$, A and $A_0$ are, respectively, the standard part and the infinitesimal part of $\widehat{A}$) and two matrix decompositions over dual rings. The group inverse has been extensively investigated and widely applied in the solution of singular linear systems and computations of various aspects of Markov chains. The forms of the dual group generalized inverse (DGGI for short) are given by using two matrix decompositions. The relationships among the range, the null space, and the DGGI of dual real matrices are also discussed under symmetric conditions. We use the above-mentioned facts to provide the symmetric expression of the perturbed dual real matrix and apply the dual spectral norm to discuss the perturbation of the DGGI. In the real field, we present the symmetric expression of the group inverse after the matrix perturbation under the rank condition. We also estimate the error between the group inverse and the DGGI with respect to the P-norm. Especially, we find that the error is the infinitesimal quantity of the square of a real number, which is small enough and not equal to 0.

1. Introduction

Throughout this paper, the following standard notations will be adopted. The symbols ${\mathbb{R}}^{m\times n}$ will denote the set of $m\times n$ real matrices. For a given matrix $A\in {\mathbb{R}}^{m\times n}$, let $A^{\top }\in {\mathbb{R}}^{n\times m}$ be the transpose of A; rank$\left(A\right)$ for the rank of A; $I_n$ for the n-by-n identity matrix; $\mathcal{R}\left(A\right)$ and $\mathcal{N}\left(A\right)$ denote the range and the null space of A, and $O(·)$ for the infinitesimal. The symbols ${∥A∥}_2$ and ${∥A∥}_F$ are the spectral norm and the Frobenius norm of A, respectively. The induced norm

$${∥A∥}_P={\parallel P^{-1}AP\parallel }_F$$

is a matrix norm, which is called the P-norm of A, where P is an invertible matrix. More details of the P-norm can be referred in [1,2]. The symbol Ind$\left(A\right)$ stands for the smallest positive integer k satisfying rank$\left(A^{k+1}\right)$ = rank$\left(A^k\right)$, where $A\in {\mathbb{R}}^{n\times n}$. A dual number $\widehat{a}$ has the following form

$$\begin{array}{c}\hfill \widehat{a}=a+\epsilon a_0,\end{array}$$

where a and $a_0$ are real numbers; $\epsilon$ is the hypercomplex unit basis; $\epsilon$ satisfies

$$\begin{array}{c}\hfill \epsilon \ne 0,0\epsilon =\epsilon 0=0,1\epsilon =\epsilon 1=\epsilon ,{\epsilon }^2=0.\end{array}$$

If $a\ne 0$, we say that $\widehat{a}$ is appreciable; otherwise, we say that $\widehat{a}$ is infinitesimal.

An m-by-n dual real matrix, $\widehat{A}$ is usually represented in the form $\widehat{A}=A+\epsilon A_0$, where $A,A_0\in {\mathbb{R}}^{m\times n}$ are, respectively, the standard part and the infinitesimal part of $\widehat{A}$. When the standard part $A\ne 0$, we say that $\widehat{A}$ is appreciable; otherwise, we say that $\widehat{A}$ is infinitesimal. For the sake of convenience, we denote ${\mathbb{D}\mathbb{R}}^{m\times n}$ as the set of $m\times n$ dual real matrices. An n-by-n dual real matrix $\widehat{U}$, which satisfies ${\widehat{U}}^{\top }\widehat{U}=\widehat{U}{\widehat{U}}^{\top }=I_n$, is called T-$orthogonal$. We say that $\widehat{U}\in {\mathbb{D}\mathbb{R}}^{n\times s}\left(s\le n\right)$ is partially T-$orthogonal$, if $\widehat{U}$ satisfies ${\widehat{U}}^{\top }\widehat{U}=I_s$. More details of the above matrices can be found in [3,4].

Dual real matrices and dual quaternion matrices have been hot topics. They have attracted the attention of many scholars owing to their significant applicability to various areas of science and engineering, for instance, in kinematic analysis and synthesis of machines and mechanisms, times-series analysis, traveling wave identification, and brain dynamics [5,6,7,8,9,10,11,12,13,14].

The increasing applicability of dual real matrices in various fields of science and engineering also promotes the research on its application related to linear algebra [15,16]. Gutin [17] generalizes the singular value decomposition (SVD for short) of dual real matrices and proves that the dual T-SVD exists universally for all dual real matrices. Qi et al. [3,18,19,20,21,22] propose the singular value decomposition (DQ-SVD for short) and the Jordan form for the dual quaternion matrices. They point out that the singular values of the matrix are all non-negative dual numbers. Qi et al. [3,23] introduce the rank and appreciable rank of a general dual quaternion matrix $\widehat{A_Q}=A+\epsilon A_0$, and denote them by Rank $\left(\widehat{A_Q}\right)$ and ARank $\left(\widehat{A_Q}\right)$ (see Lemma 2.6), respectively. The rank is the number of positive singular values of $\widehat{A_Q}$, and the appreciable rank is the number of positive singular values of the standard part A. Furthermore, the appreciable rank of a dual quaternion matrix is equal to the rank of the standard part of the dual quaternion matrix. When discussing dual real matrices, we continue to use the above two symbols. The number of positive singular values of a dual real matrix is called its rank, and the number of appreciable positive singular values is called its appreciable rank. For example, let

$$\begin{array}{c}\hfill \widehat{A}=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]+\epsilon \left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right],\end{array}$$

(1)

then, singular values of the dual real matrix $\widehat{A}$ are 1 and $\epsilon$, thus

$$\begin{array}{c}\hfill \mathrm{Rank}\left(\widehat{A}\right)=2,\mathrm{ARank}\left(\widehat{A}\right)=1.\end{array}$$

(2)

Angeles [24] developed the application of the dual generalized inverse in kinematics synthesis. Pennestrì, Udwadia, Falco, Valentini and Stefanelli [15,25,26,27] discussed some theoretical issues, and used different procedures and concise formulas to compute the dual generalized inverse. Udwadia et al. [28] introduced the dual Moore–Penrose generalized inverse (DMPGI for short) and prove its existence. The DMPGI of $\widehat{A}$ is a unique dual real matrix $\widehat{X}$, which satisfies

$$\begin{array}{c}\hfill \widehat{A}\widehat{X}\widehat{A}=\widehat{A},\widehat{X}\widehat{A}\widehat{X}=\widehat{X},{\left(\widehat{A}\widehat{X}\right)}^{\top }=\widehat{A}\widehat{X},{\left(\widehat{X}\widehat{A}\right)}^{\top }=\widehat{X}\widehat{A},\end{array}$$

and the unique matrix $\widehat{X}$ is denoted by $\widehat{X}={\widehat{A}}^{\mathrm{DMPGI}}$. Cui et al. [29] established perturbations of Moore–Penrose inverse and dual Moore–Penrose generalized inverse.

Zhong and Zhang [30] introduced the dual group generalized inverse (DGGI for short). The DGGI of $\widehat{A}$ is a unique dual real matrix $\widehat{X}$ that satisfies

$$\begin{array}{c}\hfill \widehat{A}\widehat{X}\widehat{A}=\widehat{A},\widehat{X}\widehat{A}\widehat{X}=\widehat{X},\widehat{A}\widehat{X}=\widehat{X}\widehat{A},\end{array}$$

(3)

the unique matrix is denoted as $\widehat{X}={\widehat{A}}^{\mathrm{DGGI}}$. The group inverse has been extensively investigated and widely applied in the solution of singular linear systems and computations of various aspects of Markov chains [31,32,33,34,35,36,37,38,39].

When the infinitesimal part $A_0$ of $\widehat{A}$ is 0, $\widehat{A}=A$ is a real matrix, and obviously the unique matrix X is also a real matrix. Zhong and Zhang define the range and the null space of a dual real matrix and give equivalent conditions for the existence of its DGGI. They define the range and the null space of a dual real matrix $\widehat{A}=A+\epsilon A_0\in {\mathbb{D}\mathbb{R}}^{n\times n}$ as follows [30].

$$\begin{array}{cc}\hfill \mathcal{R}\left(\widehat{A}\right)& =\{\widehat{y}\in {\mathbb{D}\mathbb{R}}^n:\widehat{y}=\widehat{A}\widehat{x},\widehat{x}=x+\epsilon x_0\in {\mathbb{D}\mathbb{R}}^n\}\hfill \\ \hfill & =\{Ax+\epsilon (A{x}_0+A_{0}x):x,x_0\in {\mathbb{R}}^n\},\hfill \\ \hfill \mathcal{N}\left(\widehat{A}\right)& =\{\widehat{x}=x+\epsilon x_0\in {\mathbb{D}\mathbb{R}}^n:\widehat{A}\widehat{x}=0\}\hfill \\ \hfill & =\{x+\epsilon x_0:Ax=0,A{x}_0+A_{0}x=0,x,x_0\in {\mathbb{R}}^n\}\hfill \end{array}$$

Specifically, when the DGGI exists, they indicate that $\mathcal{R}\left(\widehat{A}\right)\cap \mathcal{N}\left(\widehat{A}\right)=\left\{0\right\}$.

The above conclusion also provides theoretical support for the perturbation of the dual group generalized inverse. Moreover, Zhong and Zhang apply relevant conclusions of the DGGI to solve the linear dual equations and obtain a solution of consistent dual matrix equations and a dual minimum P-norm least-squares solution of inconsistent dual matrix equations. Recently, Wang and Jiang [40] extended the sharp order to dual ring by using dual group generalized inverse. Wang et al. [41] investigated the QLY linear least-squares problem, giving its compact formula. More details of dual generalized inverses and their applications can be found in [42,43,44].

It is well known that perturbation theory plays an important role in the matrix theory. The excellent monograph [45] systematically discusses the theory, methods, and new developments of matrix perturbation analysis. Many scholars study the perturbation of generalized inverses. Wei and Deng [46] explore the perturbation of the generalized Drazin invertible matrices. Some explicit generalized Drazin inverse expressions for the perturbations are obtained under certain constraints of the perturbing matrices.

Wei [47,48,49] investigates the additive perturbation of group inverse with its oblique projection and provides explicit expressions and perturbation bounds for the group inverse of the perturbation matrix $B=A+E\in {\mathbb{R}}^{n\times n}$ in both special and general cases, respectively. By using perturbation results, we can consider two types of perturbation.

The first type is to consider a special perturbation bound for the DGGI of the dual real matrix $\widehat{B}=\widehat{A}+\widehat{E}$, where $\widehat{A}$ and $\widehat{E}$ are n-by-n dual real matrices; $\widehat{E}$ is considered as the perturbed part of $\widehat{A}$. However, the perturbation in this case must be discussed on the assumption that DGGI exists.

For $\widehat{A}=A+ϵA_0$, if $ϵ$ is viewed as an infinitesimal quantity and $ϵA_0$ is considered as the perturbed part of A, we can obtain the group inverse of $\widehat{A}$. Therefore, the second type can not only be analyzed as described above, but can also be explored via the perturbation of the group inverse in the general case. Even when the DGGI of $\widehat{A}$ exists, we can reveal the difference between the group inverse and the DGGI.

The organization of this paper is as follows. In Section 2, we briefly outline the Jordan canonical form of real matrices and some conclusions of matrix perturbations, and summarize the singular value decomposition of dual real matrices. In Section 3, we present the definition of dual spectral norm and two types of matrix decompositions over dual rings, and use two decompositions to present the general form of the DGGI. Under some special conditions, the relationships among the range, the null space, and the DGGI of dual real matrices are discussed, and we utilize the above results to discuss the perturbation of $\widehat{B}=\widehat{A}+\widehat{E}$ on dual rings.

In addition, the upper bound of ${\displaystyle \frac{{∥{\widehat{B}}^{\mathrm{DGGI}}-{\widehat{A}}^{\mathrm{DGGI}}∥}_2}{{∥{\widehat{A}}^{\mathrm{DGGI}}∥}_2}}$ is estimated by using the dual spectral norm. In Section 4, we consider the group inverse of $\widehat{A}=A+ϵA_0$ under some rank conditions and reveal the difference between the group inverse and the DGGI. At the same time, we provide some numerical examples to illustrate the above results. We also analyze the error between the group inverse and the DGGI by using the P-norm.

2. Preliminary

In this section, we provide some preliminary results that would be used in the following sections.

Lemma 1

([50]). Let $A\in {\mathbb{R}}^{n\times n}$, $\mathrm{Ind}\left(A\right)=1$ and $\mathrm{rank}\left(A\right)=r$. Then,

$$\begin{array}{c}\hfill A=P\left[\begin{array}{cc}D& 0\\ 0& 0\end{array}\right]P^{-1},A^{\#}=P\left[\begin{array}{cc}{D}^{-1}& 0\\ 0& 0\end{array}\right]P^{-1},\end{array}$$

(4)

where $P\in {\mathbb{R}}^{n\times n}$ is an invertible matrix such that $P^{-1}AP$ is the Jordan canonical form of A, and $D\in {\mathbb{R}}^{r\times r}$ is nonsingular.

The symmetric expression of the perturbation for the group inverse has been developed as follows.

Lemma 2

([45]). Let $B=A+E\in {\mathbb{R}}^{n\times n}$ such that $\mathrm{Ind}\left(B\right)=\mathrm{Ind}\left(A\right)=1$ and $\mathrm{rank}\left(B\right)=\mathrm{rank}\left(A\right)$. If $I_n+A^{\#}E$ is invertible, then

$$\begin{array}{cc}\hfill B^{\#}=& {\left(I_n+Y\right)}^{-1}{\left(I_n+A^{\#}E\right)}^{-1}{A}^{\#}{\left(I_n+Y\right)}^{-1}\hfill \\ \hfill & +\left(I_n-A{A}^{\#}\right){\left(I_n+E{A}^{\#}\right)}^{-1}E{A}^{\#}{\left(I_n+Y\right)}^{-1}{\left(I_n+A^{\#}E\right)}^{-1}{A}^{\#}{\left(I_n+Y\right)}^{-1}\hfill \\ \hfill & +{\left(I_n+Y\right)}^{-1}{A}^{\#}{\left(I_n+E{A}^{\#}\right)}^{-1}{\left(I_n+Y\right)}^{-1}{A}^{\#}E{\left(I_n+A^{\#}E\right)}^{-1}\left(I_n-A{A}^{\#}\right)\hfill \\ \hfill & +\left(I_n-A{A}^{\#}\right){\left(I_n+E{A}^{\#}\right)}^{-1}E{A}^{\#}{\left(I_n+Y\right)}^{-1}{A}^{\#}{\left(I_n+E{A}^{\#}\right)}^{-1}\hfill \\ \hfill & \times {\left(I_n+Y\right)}^{-1}{A}^{\#}E{\left(I_n+A^{\#}E\right)}^{-1}\left(I_n-A{A}^{\#}\right),\hfill \end{array}$$

(5)

where $Y=A^{\#}E{\left(I_n+A^{\#}E\right)}^{-1}\left(I_n-A{A}^{\#}\right){\left(I_n+E{A}^{\#}\right)}^{-1}E{A}^{\#}$.

The special case of the perturbation can be found in the following lemma.

Lemma 3

([37,51]). Let $B=A+E\in {\mathbb{R}}^{n\times n}$ with $\mathrm{Ind}\left(A\right)=1$.

(1)

If $I_n+E{A}^{\#}$ is invertible and $E=EA{A}^{\#}$, then,

$$\begin{array}{c}\hfill B^{\#}=A^{\#}-A^{\#}E{A}^{\#}{\left(I_n+E{A}^{\#}\right)}^{-1}-\left(I_n-A{A}^{\#}\right){\left(I_n+E{A}^{\#}\right)}^{-1}{A}^{\#}{\left(I_n+E{A}^{\#}\right)}^{-1}.\end{array}$$

(2)

If $I_n+A^{\#}E$ is invertible and $E=A{A}^{\#}E$, then,

$$\begin{array}{c}\hfill B^{\#}=A^{\#}-{\left(I_n+A^{\#}E\right)}^{-1}{A}^{\#}E{A}^{\#}-{\left(I_n+A^{\#}E\right)}^{-1}{A}^{\#}{\left(I_n+A^{\#}E\right)}^{-1}\left(I_n-A{A}^{\#}\right).\end{array}$$

(3)

If $I_n+A^{\#}E$ is invertible and $A{A}^{\#}E=E=EA{A}^{\#}$, then, $B^{\#}={\left(I_n+A^{\#}E\right)}^{-1}{A}^{\#}=A^{\#}{\left(I_n+E{A}^{\#}\right)}^{-1}$.

Now we consider the dual group generalized inverse.

Lemma 4

([41]). Let $\widehat{A}=A+\epsilon A_0\in {\mathbb{D}\mathbb{R}}^{n\times n}$ and $\mathrm{rank}$$\left(A\right)=r$, then the dual index of $\widehat{A}$ is equal to 1, which is equivalent to $\mathrm{Ind}\left(A\right)=1$, and $\mathrm{rank}$$\left(A\right)$ = $\mathrm{rank}$$\left(\left[\begin{array}{cc}A& A_0\left(I_n-A{A}^{\#}\right)\end{array}\right]\right)$.

Lemma 5

([30]). Let $\widehat{A}=A+\epsilon A_0$ be a dual real matrix with A, $A_0\in {\mathbb{R}}^{n\times n}$ and $\mathrm{Ind}\left(A\right)=1$. Then, the following conditions are equivalent:

(1)

The DGGI of $\widehat{A}$ exists;

(2)

$\left(I_n-A{A}^{\#}\right)A_0\left(I_n-A{A}^{\#}\right)=0$;

(3)

The DMPGI of $\widehat{A}$ exists.

Furthermore, if the DGGI of $\widehat{A}$ exists, then

$$\begin{array}{c}\hfill {\widehat{A}}^{\mathrm{DGGI}}=A^{\#}-\epsilon \left[A^{\#}{A}_0{A}^{\#}-{\left(A^{\#}\right)}^2{A}_0\left(I_n-A{A}^{\#}\right)-\left(I_n-A{A}^{\#}\right)A_0{\left(A^{\#}\right)}^2\right].\end{array}$$

(6)

Lemma 6

([17] Dual T-SVD). Let $\widehat{A}=A+\epsilon A_0\in {\mathbb{D}\mathbb{R}}^{n\times n}$ with $\mathrm{Rank}\left(\widehat{A}\right)=t$. Then, there exist T-orthogonal matrices $\widehat{U}$, $\widehat{V}\in {\mathbb{D}\mathbb{R}}^{n\times n}$ such that

$$\begin{array}{c}\hfill \widehat{A}=\widehat{U}\left[\begin{array}{cc}{\widehat{\mathsf{\Sigma }}}_t& 0\\ 0& 0\end{array}\right]{\widehat{V}}^{\top },\end{array}$$

(7)

where ${\widehat{\mathsf{\Sigma }}}_t\in {\mathbb{D}\mathbb{R}}^{t\times t}$ is a diagonal matrix, having the unique form ${\widehat{\mathsf{\Sigma }}}_t=$ diag $\left({\sigma }_1,\dots ,{\sigma }_r,\dots ,{\sigma }_t\right)$, $r\le t\le n$, ${\sigma }_1\ge {\sigma }_2\ge \dots \ge {\sigma }_r$ are positive appreciable dual numbers, and ${\sigma }_{r+1}\ge {\sigma }_{r+2}\ge \dots \ge {\sigma }_t$ are positive infinitesimal dual numbers.

Remark 1.

From (7), we know that

$$\begin{array}{c}\hfill \mathrm{Rank}\left(\widehat{A}\right)=\mathrm{Rank}\left({\widehat{\mathsf{\Sigma }}}_t\right)=t,\mathrm{ARank}\left(\widehat{A}\right)=\mathrm{ARank}\left({\widehat{\mathsf{\Sigma }}}_t\right)=r,\end{array}$$

(8)

and main diagonal entries of ${\widehat{\mathsf{\Sigma }}}_t$ are positive appreciable dual numbers if and only if

$$\begin{array}{c}\hfill \mathrm{Rank}\left({\widehat{\mathsf{\Sigma }}}_t\right)=\mathrm{ARank}\left({\widehat{\mathsf{\Sigma }}}_t\right)=r=t.\end{array}$$

(9)

Then, we obtain $\widehat{A}\in {\mathbb{D}\mathbb{R}}^{n\times n}$ is invertible if and only if ${\widehat{\mathsf{\Sigma }}}_t$ is an n-by-n invertible matrix if and only if the main diagonal of ${\widehat{\mathsf{\Sigma }}}_t$ has exactly n positive appreciable dual numbers. From (8) and (9), we can verify that $\widehat{A}$ is invertible if and only if

$$\begin{array}{c}\hfill n=\mathrm{Rank}\left(\widehat{A}\right)=\mathrm{Rank}\left({\widehat{\mathsf{\Sigma }}}_t\right)=\mathrm{ARank}\left({\widehat{\mathsf{\Sigma }}}_t\right)=\mathrm{ARank}\left(\widehat{A}\right).\end{array}$$

For any dual real matrix, its dual T-SVD always exists (see [17]). Therefore, for an arbitrary matrix $\widehat{A}\in {\mathbb{D}\mathbb{R}}^{n\times n}$, $\widehat{A}$ is invertible if and only if $\mathrm{Rank}\left(\widehat{A}\right)=\mathrm{ARank}\left(\widehat{A}\right)=n$.

3. Symmetric Perturbation of $\widehat{\mathit{E}}$ to $\widehat{\mathit{A}}$ in Dual Ring

In this section, we first introduce the definition of the spectral norm of a dual real matrix, and then provide a characterization such that ${\widehat{B}}^{\mathrm{DGGI}}$ has the forms as in (32a) and (32b). Finally, we establish the perturbation bounds for the DGGI from the symmetric point of view,

$$\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}=\widehat{E}=\widehat{E}\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}.$$

In [23,52,53], the definition of absolute value of dual real numbers is proposed. Afterwards, the norms of dual quaternion vectors and dual quaternion matrices are given. Next, we define the spectral norms of dual real vectors and dual real matrices.

For $\widehat{x}=x+\epsilon x_0\in {\mathbb{D}\mathbb{R}}^n$, where x, $x_0\in {\mathbb{R}}^n$ are the standard part and the infinitesimal part of $\widehat{x}$, respectively. Suppose that $\widehat{x}={\left({\widehat{x}}_1,{\widehat{x}}_2,\dots ,{\widehat{x}}_n\right)}^{\top }$, ${\widehat{x}}_i=x_i+\epsilon {\left(x_0\right)}_i$ for $i=1,2,\cdots ,n$. For the spectral norm of $\widehat{x}$, if not all of ${\widehat{x}}_i$ are infinitesimal, we define

$$\begin{array}{c}\hfill {∥\widehat{x}∥}_2=\sqrt{\sum _{i=1}^n{\left|{\widehat{x}}_i\right|}^2}.\end{array}$$

(10)

If all ${\widehat{x}}_i$ are infinitesimal, then

$$\begin{array}{c}\hfill {∥\widehat{x}∥}_2=\sqrt{\sum _{i=1}^n{\left|{\left(x_0\right)}_i\right|}^2}\epsilon .\end{array}$$

(11)

We see that the spectral norm ${∥\widehat{x}∥}_2$, defined by (10) and (11), satisfies the three properties of norms. Next, we further study the property of this norm.

Theorem 1.

Suppose that $\widehat{U}\in {\mathbb{D}\mathbb{R}}^{m\times n}$ is partially T-orthogonal, and $\widehat{x}\in {\mathbb{D}\mathbb{R}}^n$. Then,

$$\begin{array}{c}\hfill {∥\widehat{U}\widehat{x}∥}_2={∥\widehat{x}∥}_2.\end{array}$$

(12)

Proof. 

Let $\widehat{x}=x+\epsilon x_0$ be appreciable. It follows from (10) that

$$\begin{array}{c}\hfill {∥\widehat{x}∥}_2^2=\sum _{i=1}^n{\left|{\widehat{x}}_i\right|}^2={\widehat{x}}^{\top }\widehat{x}.\end{array}$$

Moreover, suppose that $\widehat{U}=U+\epsilon U_0$. By a direct calculation, we have $U^{\top }U=I_n$. Then, the standard part $Ux$ of $\widehat{U}\widehat{x}$ is not equal to 0; that is, $\widehat{U}\widehat{x}$ is also appreciable. Accordingly, applying (10) again, we obtain ${∥\widehat{U}\widehat{x}∥}_2^2={\left(\widehat{U}\widehat{x}\right)}^{\top }\widehat{U}\widehat{x}={\widehat{x}}^{\top }\widehat{x}$. Thus, ${∥\widehat{U}\widehat{x}∥}_2={∥\widehat{x}∥}_2$ holds in this case.

If $\widehat{x}$ is infinitesimal, i.e., $\widehat{x}=\epsilon x_0$, then $\widehat{U}\widehat{x}=\epsilon U{x}_0$, which means that $\widehat{U}\widehat{x}$ is also infinitesimal. Furthermore, ${∥\widehat{U}\widehat{x}∥}_2={∥U{x}_0∥}_2={∥x_0∥}_2$ in terms of $U^{\top }U=I_n$. Therefore, we still have ${∥\widehat{U}\widehat{x}∥}_2={∥\widehat{x}∥}_2$ in this case. □

Definition 1

([52]). For a given $\widehat{A}=A+\epsilon A_0\in {\mathbb{D}\mathbb{R}}^{n\times n}$, the spectral norm of a dual real matrix can be defined as

$$\begin{array}{c}\hfill {∥\widehat{A}∥}_2=\underset{{\displaystyle \genfrac{}{}{0pt}{}{\widehat{x}\in {\mathbb{D}\mathbb{R}}^n}{{∥\widehat{x}∥}_2=1}}}{max}{∥\widehat{A}\widehat{x}∥}_2.\end{array}$$

(13)

As ${∥\widehat{A}∥}_2$ is induced by the ${\ell }_2$-norm on dual vector spaces, and hence the norm ${∥\widehat{A}∥}_2$ is a matrix norm. Moreover, it follows from (13) that

$$\begin{array}{c}\hfill {∥\widehat{A}\widehat{x}∥}_2\le {∥\widehat{A}∥}_2{∥\widehat{x}∥}_2\end{array}$$

(14)

for any appreciable $\widehat{x}\in {\mathbb{D}\mathbb{R}}^n$.

Next, we discuss some properties on the spectral norm of dual real matrices.

Theorem 2.

Let $\widehat{A}=A+\epsilon A_0\in {\mathbb{D}\mathbb{R}}^{n\times n}$ with $\mathrm{Rank}\left(\widehat{A}\right)=t$. It holds that ${∥\widehat{A}∥}_2={\sigma }_1\left(\widehat{A}\right)$, where ${\sigma }_1\left(\widehat{A}\right)$ is the largest singular value of $\widehat{A}$.

Proof. 

Let $\widehat{A}=\widehat{U}\widehat{\mathsf{\Sigma }}{\widehat{V}}^{\top }$ be a dual T-SVD of $\widehat{A}$, in which $\widehat{U},\widehat{V}\in {\mathbb{D}\mathbb{R}}^{n\times n}$ are T-orthogonal, $\widehat{\mathsf{\Sigma }}=\mathrm{diag}\left({\sigma }_1\left(\widehat{A}\right),\dots ,{\sigma }_n\left(\widehat{A}\right)\right)$ with ${\sigma }_1\left(\widehat{A}\right)\ge \dots \ge {\sigma }_n\left(\widehat{A}\right)\ge 0$.

Applying Theorem 1 and (13), we have

$$\begin{array}{cc}\hfill {∥\widehat{A}∥}_2& =\underset{{\displaystyle \genfrac{}{}{0pt}{}{\widehat{x}\in {\mathbb{D}\mathbb{R}}^n}{{∥\widehat{x}∥}_2=1}}}{max}{∥\widehat{\mathsf{\Sigma }}{\widehat{V}}^{\top }\widehat{x}∥}_2=\underset{{\displaystyle \genfrac{}{}{0pt}{}{\widehat{y}\in {\mathbb{D}\mathbb{R}}^n}{{∥\widehat{V}\widehat{y}∥}_2=1}}}{max}{∥\widehat{\mathsf{\Sigma }}\widehat{y}∥}_2\hfill \\ \hfill & =\underset{{\displaystyle \genfrac{}{}{0pt}{}{\widehat{y}\in {\mathbb{D}\mathbb{R}}^n}{{∥\widehat{y}∥}_2=1}}}{max}\sqrt{\sum _{i=1}^n{\sigma }_i^2\left(\widehat{A}\right){\left|{\widehat{y}}_i\right|}^2}\hfill \\ \hfill & \le \underset{{\displaystyle \genfrac{}{}{0pt}{}{\widehat{y}\in {\mathbb{D}\mathbb{R}}^n}{{∥\widehat{y}∥}_2=1}}}{max}{\sigma }_1\left(\widehat{A}\right)\sqrt{\sum _{i=1}^n{\left|{\widehat{y}}_i\right|}^2}={\sigma }_1\left(\widehat{A}\right).\hfill \end{array}$$

However, ${∥\widehat{A}\widehat{y}∥}_2={\sigma }_1\left(\widehat{A}\right)$ for $\widehat{y}=e_n$, where $e_n$ is an identity vector of order n. It follows (14) that ${∥\widehat{A}∥}_2\ge {\sigma }_1\left(\widehat{A}\right)$. Hence, we conclude that ${∥\widehat{A}∥}_2={\sigma }_1\left(\widehat{A}\right)$ and complete the proof. □

Theorem 3.

Let $\widehat{A}=A+\epsilon A_0\in {\mathbb{D}\mathbb{R}}^{p\times p}$. For any partially T-orthogonal matrices $\widehat{U},\widehat{V}\in {\mathbb{D}\mathbb{R}}^{n\times p}\left(p\le n\right)$, we have

$$\begin{array}{c}\hfill {∥\widehat{U}\widehat{A}{\widehat{V}}^{\top }∥}_2={∥\widehat{A}∥}_2.\end{array}$$

Proof. 

It follows from ${\widehat{U}}^{\top }\widehat{U}=I_p$ that ${\left(\widehat{U}\widehat{A}\right)}^{\top }\widehat{U}\widehat{A}={\widehat{A}}^{\top }\widehat{A}$. By applying Theorem 2, we know that ${∥\widehat{U}\widehat{A}∥}_2$ is exactly the square root of the largest eigenvalue of ${\left(\widehat{U}\widehat{A}\right)}^{\top }\widehat{U}\widehat{A}$. Consequently, we obtain ${∥\widehat{U}\widehat{A}∥}_2={∥\widehat{A}∥}_2$. Similarly, we can prove that ${∥\widehat{A}{\widehat{V}}^{\top }∥}_2={∥\widehat{A}∥}_2$.

To sum up, the T-orthogonal invariance ${∥\widehat{U}\widehat{A}{\widehat{V}}^{\top }∥}_2={∥\widehat{A}∥}_2$ holds. □

If $\widehat{B}=B+\epsilon B_0\in {\mathbb{D}\mathbb{R}}^{n\times n}$ with ${∥\widehat{B}∥}_2<1$, then all the eigenvalues of $\widehat{B}$ are less than unity in absolute value. $I_n+\widehat{B}$ is nonsingular, and ${∥{\left(I_n+\widehat{B}\right)}^{-1}∥}_2⩽{\left(1-{∥\widehat{B}∥}_2\right)}^{-1}$.

Theorem 4.

For a given matrix $\widehat{A}\in {\mathbb{D}\mathbb{R}}^{n\times n}$ with $\mathrm{Rank}\left(\widehat{A}\right)=t$, there exists a T-$orthogonal$ matrix $\widehat{U}\in {\mathbb{D}\mathbb{R}}^{n\times n}$ such that

$$\begin{array}{c}\hfill \widehat{A}=\widehat{U}\left[\begin{array}{cc}\widehat{T}& \widehat{S}\\ 0& 0\end{array}\right]{\widehat{U}}^{\top },\end{array}$$

(15)

in which $\widehat{T}={\widehat{\mathsf{\Sigma }}}_t\widehat{K}$, $\widehat{S}={\widehat{\mathsf{\Sigma }}}_t\widehat{L}$, ${\widehat{\mathsf{\Sigma }}}_t$ is as shown in Lemma 6. $\widehat{K}\in {\mathbb{D}\mathbb{R}}^{t\times t}$ and $\widehat{L}\in {\mathbb{D}\mathbb{R}}^{t\times (n-t)}$ satisfy $\widehat{K}{\widehat{K}}^{\top }+\widehat{L}{\widehat{L}}^{\top }=I_t$.

Proof. 

Let $\widehat{A}=\widehat{U}\left[\begin{array}{cc}{\widehat{\mathsf{\Sigma }}}_t& 0\\ 0& 0\end{array}\right]{\widehat{V}}^{\top }$ be the dual T-SVD of $\widehat{A}$, where $\widehat{U}$ and $\widehat{V}$ are T-$orthogonal$, and ${\widehat{\mathsf{\Sigma }}}_t$ is as shown in Lemma 6. Then,

$$\begin{array}{c}\hfill \widehat{A}=\widehat{U}\left[\begin{array}{cc}{\widehat{\mathsf{\Sigma }}}_t& 0\\ 0& 0\end{array}\right]{\widehat{V}}^{\top }\widehat{U}{\widehat{U}}^{\top }=\widehat{U}\left[\begin{array}{cc}{\widehat{\mathsf{\Sigma }}}_t& 0\\ 0& 0\end{array}\right]{\widehat{W}}^{\top }{\widehat{U}}^{\top }=\widehat{U}\left[\begin{array}{cc}{\widehat{\mathsf{\Sigma }}}_t\widehat{K}& {\widehat{\mathsf{\Sigma }}}_t\widehat{L}\\ 0& 0\end{array}\right]{\widehat{U}}^{\top },\end{array}$$

where $\widehat{W}={\widehat{V}}^{\top }\widehat{U}=\left[\begin{array}{cc}\widehat{K}& \widehat{L}\\ \widehat{M}& \widehat{N}\end{array}\right]$ is T-$orthogonal$. Hence, (15) is clear. Furthermore, we have

$$\begin{array}{c}\hfill {\widehat{V}}^{\top }\widehat{U}{\left({\widehat{V}}^{\top }\widehat{U}\right)}^{\top }=\left[\begin{array}{cc}\widehat{K}{\widehat{K}}^{\top }+\widehat{L}{\widehat{L}}^{\top }& \widehat{K}{\widehat{M}}^{\top }+\widehat{L}{\widehat{N}}^{\top }\\ \widehat{M}{\widehat{K}}^{\top }+\widehat{N}{\widehat{L}}^{\top }& \widehat{M}{\widehat{M}}^{\top }+\widehat{N}{\widehat{N}}^{\top }\end{array}\right]=I_n,\end{array}$$

that is, $\widehat{K}{\widehat{K}}^{\top }+\widehat{L}{\widehat{L}}^{\top }=I_t$. □

Theorem 5.

Let $\widehat{A}=A+\epsilon A_0\in {\mathbb{D}\mathbb{R}}^{n\times n}$ with $\mathrm{Ind}\left(\widehat{A}\right)$ = 1, $\mathrm{Rank}\left(\widehat{A}\right)=t$. Suppose that the form of $\widehat{A}$ is as in (15). Then $\widehat{T}$ is invertible. Furthermore, ${\widehat{A}}^{\mathrm{DGGI}}$ has the following decomposition

$$\begin{array}{c}\hfill {\widehat{A}}^{\mathrm{DGGI}}=\widehat{U}\left[\begin{array}{cc}{\widehat{T}}^{-1}& {\widehat{T}}^{-2}\widehat{S}\\ 0& 0\end{array}\right]{\widehat{U}}^{\top }.\end{array}$$

(16)

Proof. 

We denote $\widehat{U}=U+\epsilon U_0$, ${\widehat{U}}^{\top }=U^{\top }+\epsilon {U_0}^{\top }$ and $\left[\begin{array}{cc}\widehat{T}& \widehat{S}\end{array}\right]=\left[\begin{array}{cc}T& S\end{array}\right]+\epsilon \left[\begin{array}{cc}{T}_0& S_0\end{array}\right]$, then we know that the standard part $A=U\left[\begin{array}{cc}T& S\\ 0& 0\end{array}\right]U^{\top }$ of $\widehat{A}$. Since $\mathrm{Ind}\left(\widehat{A}\right)=1$, by Lemma 4, we have $\mathrm{Ind}\left(A\right)=1$; that is, T is nonsingular. Then, $\widehat{T}$ is also nonsingular. It follows from (3) and (15) that ${\widehat{A}}^{\mathrm{DGGI}}$ has the form as in (16). □

Theorem 6.

Let $\widehat{A}=A+\epsilon A_0\in {\mathbb{D}\mathbb{R}}^{n\times n}$ with $\mathrm{Ind}\left(\widehat{A}\right)$ = 1, $\mathrm{Rank}\left(\widehat{A}\right)=t$. Then, there exist dual invertible matrices $\widehat{P}\in {\mathbb{D}\mathbb{R}}^{n\times n}$ and $\widehat{T}\in {\mathbb{D}\mathbb{R}}^{t\times t}$ such that

$$\begin{array}{c}\hfill \widehat{A}=\widehat{P}\left[\begin{array}{cc}\widehat{T}& 0\\ 0& 0\end{array}\right]{\widehat{P}}^{-1}.\end{array}$$

(17)

Furthermore,

$$\begin{array}{c}\hfill {\widehat{A}}^{\mathrm{DGGI}}=\widehat{P}\left[\begin{array}{cc}{\widehat{T}}^{-1}& 0\\ 0& 0\end{array}\right]{\widehat{P}}^{-1}.\end{array}$$

(18)

Proof. 

According to Theorems 4 and 5, we know that $\widehat{A}$ has the following decomposition

$$\begin{array}{c}\hfill \widehat{A}=\widehat{U}\left[\begin{array}{cc}\widehat{T}& \widehat{S}\\ 0& 0\end{array}\right]{\widehat{U}}^{\top },\end{array}$$

(19)

where $\widehat{U}$ is T-$orthogonal$ and $\widehat{T}$ is invertible. From the above equation, we deduce that

$$\begin{array}{c}\hfill \widehat{A}=\widehat{U}\left[\begin{array}{cc}{I}_t& -{\widehat{T}}^{-1}\widehat{S}\\ 0& I_{n-t}\end{array}\right]\left[\begin{array}{cc}\widehat{T}& 0\\ 0& 0\end{array}\right]\left[\begin{array}{cc}{I}_t& {\widehat{T}}^{-1}\widehat{S}\\ 0& I_{n-t}\end{array}\right]{\widehat{U}}^{\top }.\end{array}$$

(20)

Denote

$$\begin{array}{c}\hfill \widehat{P}=\widehat{U}\left[\begin{array}{cc}{I}_t& -{\widehat{T}}^{-1}\widehat{S}\\ 0& I_{n-t}\end{array}\right].\end{array}$$

Since dual real matrix $\widehat{U}$ is T-$orthogonal$, it is clear that ${\widehat{U}}^{\top }={\widehat{U}}^{-1}$ and dual real matrix $\widehat{P}$ is invertible. Then, $\widehat{A}$ has the form as in (17). Applying (3) and (17), it is easy to verify that the form of ${\widehat{A}}^{\mathrm{DGGI}}$ is as in (18). □

It is well known that there is a symmetric class of matrices over dual ring, whose dual index is equal to 1. Therefore, we further characterize dual T-SVD on the condition that dual index is equal to 1.

Theorem 7.

Let $\widehat{A}=A+\epsilon A_0\in {\mathbb{D}\mathbb{R}}^{n\times n}$ with $\mathrm{Ind}\left(\widehat{A}\right)$ = 1, $\mathrm{Rank}\left(\widehat{A}\right)=t$. Suppose that the form of $\widehat{A}$ is as in (15). Then, $\widehat{T}$ is invertible, and

$$\begin{array}{c}\hfill \mathrm{Rank}\left(\widehat{A}\right)=\mathrm{ARank}\left(\widehat{A}\right)=\mathrm{Rank}\left({\widehat{\mathsf{\Sigma }}}_t\right)=\mathrm{ARank}\left({\widehat{\mathsf{\Sigma }}}_t\right)=t.\end{array}$$

(21)

Proof. 

Since the form of $\widehat{A}$ is given in (7), we have

$$\begin{array}{c}\hfill \mathrm{Rank}\left(\widehat{A}\right)=\mathrm{Rank}\left({\widehat{\mathsf{\Sigma }}}_t\right)=t\mathrm{and}\mathrm{ARank}\left(\widehat{A}\right)=\mathrm{ARank}\left({\widehat{\mathsf{\Sigma }}}_t\right)=r.\end{array}$$

(22)

By $\mathrm{Ind}\left(\widehat{A}\right)$ = 1, it follows from Theorem 6 that $\widehat{A}$ can be represented in the form as in (17). Then,

$$\begin{array}{c}\hfill \mathrm{Rank}\left(\widehat{A}\right)=\mathrm{Rank}\left(\widehat{T}\right)\mathrm{and}\mathrm{ARank}\left(\widehat{A}\right)=\mathrm{ARank}\left(\widehat{T}\right).\end{array}$$

(23)

Since $\widehat{T}\in {\mathbb{D}}^{t\times t}$ is invertible, applying Remark 1, we obtain $\mathrm{Rank}\left(\widehat{T}\right)=\mathrm{ARank}\left(\widehat{T}\right)=t$. From (22) and (23), we know that $\mathrm{Rank}\left(\widehat{A}\right)=\mathrm{ARank}\left(\widehat{A}\right)=\mathrm{Rank}\left({\widehat{\mathsf{\Sigma }}}_t\right)=\mathrm{ARank}\left({\widehat{\mathsf{\Sigma }}}_t\right)=t$ and ${\widehat{\mathsf{\Sigma }}}_t$ is invertible. □

In the above theorem, when $\mathrm{Ind}\left(\widehat{A}\right)$ = 1 is replaced with the condition that the DGGI of $\widehat{A}$ exists, the theorem is also applicable. The reason is that $\mathrm{Ind}\left(\widehat{A}\right)$ = 1 is proved to be equivalent to the existence of the DGGI. In the following theorems, we characterize the range and the null space of dual real matrices by using some rank conditions. These results provide help for the subsequent perturbation analysis.

Theorem 8.

Let $\widehat{B}=\widehat{A}+\widehat{E}\in {\mathbb{D}\mathbb{R}}^{n\times n}$, $\mathrm{Rank}\left(\widehat{B}\right)=\mathrm{Rank}\left(\widehat{A}\right)=t$. Furthermore, let DGGIs of $\widehat{A}$ and $\widehat{B}$ exist. If $\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}=\widehat{E}$, then $\mathcal{R}\left(\widehat{A}\right)=\mathcal{R}\left(\widehat{B}\right)$.

Proof. 

Since $\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}=\widehat{E}$, we deduce that

$$\begin{array}{c}\hfill \widehat{B}=\widehat{A}+\widehat{E}=\widehat{A}+\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}=\widehat{A}\left(I_n+{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}\right).\end{array}$$

(24)

It follows from (24) that $\mathcal{R}\left(\widehat{B}\right)\subseteq \mathcal{R}\left(\widehat{A}\right)$; that is, there exists $\widehat{X}\in {\mathbb{D}\mathbb{R}}^{n\times n}$ such that $\widehat{B}=\widehat{A}\widehat{X}$.

Let the dual T-SVD of $\widehat{A}$ be as in (7), then

$$\begin{array}{c}\hfill \widehat{B}=\widehat{A}\widehat{X}=\widehat{U}\left[\begin{array}{cc}{\widehat{\mathsf{\Sigma }}}_t& 0\\ 0& 0\end{array}\right]{\widehat{V}}^{\top }\widehat{V}\left[\begin{array}{cc}{\widehat{X}}_1& {\widehat{X}}_2\\ {\widehat{X}}_3& {\widehat{X}}_4\end{array}\right]{\widehat{V}}^{\top }=\widehat{U}\left[\begin{array}{cc}{\widehat{T}}_1& {\widehat{T}}_2\\ 0& 0\end{array}\right]{\widehat{V}}^{\top },\end{array}$$

(25)

where ${\widehat{T}}_1={\widehat{\mathsf{\Sigma }}}_t{\widehat{X}}_1$, ${\widehat{T}}_2={\widehat{\mathsf{\Sigma }}}_t{\widehat{X}}_2$, ${\widehat{X}}_1\in {\mathbb{D}\mathbb{R}}^{t\times t}$ and ${\widehat{X}}_4\in {\mathbb{D}\mathbb{R}}^{(n-t)\times (n-t)}$. Since the DGGI of $\widehat{A}$ exists, applying Lemma 6 and Theorem 7, we know that ${\widehat{\mathsf{\Sigma }}}_t$ is invertible. From $\mathrm{Rank}\left(\widehat{B}\right)=\mathrm{Rank}\left(\widehat{A}\right)=t$, we have

$$\begin{array}{c}\hfill \mathrm{Rank}\left(\widehat{A}\right)=\mathrm{Rank}\left({\widehat{\mathsf{\Sigma }}}_t\right)=t=\mathrm{Rank}\left(\widehat{B}\right)=\mathrm{Rank}\left(\left[\begin{array}{cc}{\widehat{T}}_1& {\widehat{T}}_2\end{array}\right]\right).\end{array}$$

That is to say, the partitioned dual matrix $\left[\begin{array}{cc}{\widehat{T}}_1& {\widehat{T}}_2\end{array}\right]$ is of row full rank. Since the DGGI of $\widehat{B}$ exists, it follows from Lemma 5 and (25) that

$$\begin{array}{c}\hfill \left[\begin{array}{cc}{\widehat{T}}_1& {\widehat{T}}_2\\ 0& 0\end{array}\right]{\left[\begin{array}{cc}{\widehat{T}}_1& {\widehat{T}}_2\\ 0& 0\end{array}\right]}^{\mathrm{DMPGI}}=\left[\begin{array}{cc}{I}_t& 0\\ 0& 0\end{array}\right].\end{array}$$

Then, there exists a dual real matrix

$$\begin{array}{c}\hfill \widehat{Y}=\widehat{V}{\left[\begin{array}{cc}{\widehat{T}}_1& {\widehat{T}}_2\\ 0& 0\end{array}\right]}^{\mathrm{DMPGI}}\left[\begin{array}{cc}{\widehat{\mathsf{\Sigma }}}_t& 0\\ 0& 0\end{array}\right]{\widehat{V}}^{\top }\end{array}$$

such that $\widehat{B}\widehat{Y}=\widehat{A}$; that is, $\mathcal{R}\left(\widehat{A}\right)\subseteq \mathcal{R}\left(\widehat{B}\right)$. To sum up, we prove that $\mathcal{R}\left(\widehat{A}\right)=\mathcal{R}\left(\widehat{B}\right)$. □

Theorem 9.

Let $\widehat{B}=\widehat{A}+\widehat{E}\in {\mathbb{D}\mathbb{R}}^{n\times n}$, $\mathrm{Rank}\left(\widehat{B}\right)=\mathrm{Rank}\left(\widehat{A}\right)=t$. Furthermore, let DGGIs of $\widehat{A}$ and $\widehat{B}$ exist. If $\widehat{E}=\widehat{E}\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}$, then $\mathcal{N}\left(\widehat{A}\right)=\mathcal{N}\left(\widehat{B}\right)$.

Proof. 

Since $\widehat{E}=\widehat{E}\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}$, we obtain

$$\begin{array}{c}\hfill \widehat{B}=\widehat{A}+\widehat{E}=\widehat{A}+\widehat{E}\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}=\left(I_n+\widehat{E}{\widehat{A}}^{\mathrm{DGGI}}\right)\widehat{A}.\end{array}$$

(26)

From (26), we have $\mathcal{N}\left(\widehat{A}\right)\subseteq \mathcal{N}\left(\widehat{B}\right)$; that is, there exists $\widehat{Z}\in {\mathbb{D}\mathbb{R}}^{n\times n}$ such that $\widehat{B}=\widehat{Z}\widehat{A}$.

Let the dual T-SVD of $\widehat{A}$ be as in (7), then we have

$$\begin{array}{c}\hfill \widehat{B}=\widehat{Z}\widehat{A}=\widehat{U}\left[\begin{array}{cc}{\widehat{Z}}_1& {\widehat{Z}}_2\\ {\widehat{Z}}_3& {\widehat{Z}}_4\end{array}\right]{\widehat{U}}^{\top }\widehat{U}\left[\begin{array}{cc}{\widehat{\mathsf{\Sigma }}}_t& 0\\ 0& 0\end{array}\right]{\widehat{V}}^{\top }=\widehat{U}\left[\begin{array}{cc}{\widehat{M}}_1& 0\\ {\widehat{M}}_3& 0\end{array}\right]{\widehat{V}}^{\top },\end{array}$$

(27)

where ${\widehat{M}}_1={\widehat{Z}}_1{\widehat{\mathsf{\Sigma }}}_t$, ${\widehat{M}}_3={\widehat{Z}}_3{\widehat{\mathsf{\Sigma }}}_t$, ${\widehat{Z}}_1\in {\mathbb{D}\mathbb{R}}^{t\times t}$, and ${\widehat{Z}}_4\in {\mathbb{D}\mathbb{R}}^{(n-t)\times (n-t)}$. Since the DGGI of $\widehat{A}$ exists, applying Lemma 6 and Theorem 7, we know that ${\widehat{\mathsf{\Sigma }}}_t$ is invertible. From $\mathrm{Rank}\left(\widehat{B}\right)=\mathrm{Rank}\left(\widehat{A}\right)=t$, we have

$$\begin{array}{c}\hfill \mathrm{Rank}\left(\widehat{A}\right)=\mathrm{Rank}\left({\widehat{\mathsf{\Sigma }}}_t\right)=t=\mathrm{Rank}\left(\widehat{B}\right)=\mathrm{Rank}\left(\left[\begin{array}{c}{\widehat{M}}_1\\ {\widehat{M}}_3\end{array}\right]\right).\end{array}$$

That is, the partitioned dual matrix $\left[\begin{array}{c}{\widehat{M}}_1\\ {\widehat{M}}_3\end{array}\right]$ is of column full rank. Since the DGGI of $\widehat{B}$ exists, it follows from Lemma 5 and (27) that

$$\begin{array}{c}\hfill {\left[\begin{array}{cc}{\widehat{M}}_1& 0\\ {\widehat{M}}_3& 0\end{array}\right]}^{\mathrm{DMPGI}}\left[\begin{array}{cc}{\widehat{M}}_1& 0\\ {\widehat{M}}_3& 0\end{array}\right]=\left[\begin{array}{cc}{I}_t& 0\\ 0& 0\end{array}\right].\end{array}$$

Then, there exists a dual real matrix

$$\begin{array}{c}\hfill \widehat{\mathsf{\Phi }}=\widehat{U}\left[\begin{array}{cc}{\widehat{\mathsf{\Sigma }}}_t& 0\\ 0& 0\end{array}\right]{\left[\begin{array}{cc}{\widehat{M}}_1& 0\\ {\widehat{M}}_3& 0\end{array}\right]}^{\mathrm{DMPGI}}{\widehat{U}}^{\top }\end{array}$$

such that $\widehat{A}=\widehat{\mathsf{\Phi }}\widehat{B}$; that is, $\mathcal{N}\left(\widehat{B}\right)\subseteq \mathcal{N}\left(\widehat{A}\right)$. Therefore, $\mathcal{N}\left(\widehat{A}\right)=\mathcal{N}\left(\widehat{B}\right)$ holds. □

Next, we provide more concise statements. If ${\widehat{A}}^{\mathrm{DGGI}}$ exists, then $\mathcal{R}\left({\widehat{A}}^{\mathrm{DGGI}}\right)=\mathcal{R}\left(\widehat{A}\right)$ and $\mathcal{N}\left({\widehat{A}}^{\mathrm{DGGI}}\right)=\mathcal{N}\left(\widehat{A}\right)$ [28]. It is also obvious that $\mathrm{Rank}\left(\widehat{A}\right)=\mathrm{Rank}\left({\widehat{A}}^{\mathrm{DGGI}}\right)$. By applying Theorems 8 and 9, we derive the following Corollaries 1 and 2.

Corollary 1.

Let $\widehat{B}=\widehat{A}+\widehat{E}\in {\mathbb{D}\mathbb{R}}^{n\times n}$, $\mathrm{Rank}\left(\widehat{B}\right)=\mathrm{Rank}\left(\widehat{A}\right)=t$. Furthermore, let DGGIs of $\widehat{A}$ and $\widehat{B}$ exist.

(1) If $\mathcal{R}\left(\widehat{B}\right)\subseteq \mathcal{R}\left(\widehat{A}\right)$, then $\mathcal{R}\left(\widehat{A}\right)=\mathcal{R}\left(\widehat{B}\right)$.

(2) If $\mathcal{N}\left(\widehat{A}\right)\subseteq \mathcal{N}\left(\widehat{B}\right)$, then $\mathcal{N}\left(\widehat{A}\right)=\mathcal{N}\left(\widehat{B}\right)$.

Corollary 2.

Let $\widehat{B}=\widehat{A}+\widehat{E}\in {\mathbb{D}\mathbb{R}}^{n\times n}$, $\mathrm{Rank}\left(\widehat{B}\right)=\mathrm{Rank}\left(\widehat{A}\right)=t$. Furthermore, let DGGIs of $\widehat{A}$ and $\widehat{B}$ exist.

(1) If $\mathcal{R}\left({\widehat{B}}^{\mathrm{DGGI}}\right)\subseteq \mathcal{R}\left({\widehat{A}}^{\mathrm{DGGI}}\right)$, then $\mathcal{R}\left({\widehat{A}}^{\mathrm{DGGI}}\right)=\mathcal{R}\left({\widehat{B}}^{\mathrm{DGGI}}\right)$.

(2) If $\mathcal{N}\left({\widehat{A}}^{\mathrm{DGGI}}\right)\subseteq \mathcal{N}\left({\widehat{B}}^{\mathrm{DGGI}}\right)$, then $\mathcal{N}\left({\widehat{A}}^{\mathrm{DGGI}}\right)=\mathcal{N}\left({\widehat{B}}^{\mathrm{DGGI}}\right)$.

From Theorems 8 and 9, we know that if $\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}=\widehat{E}=\widehat{E}\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}$, then $\mathcal{R}\left(\widehat{A}\right)=\mathcal{R}\left(\widehat{B}\right)$ and $\mathcal{N}\left(\widehat{A}\right)=\mathcal{N}\left(\widehat{B}\right)$. Based on the above special matrix equations, we have the following theorem, which shows the relationship between $\widehat{B}{\widehat{B}}^{\mathrm{DGGI}}$ and $\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}$.

Theorem 10.

Let $\widehat{B}=\widehat{A}+\widehat{E}\in {\mathbb{D}\mathbb{R}}^{n\times n}$, $\mathrm{Rank}\left(\widehat{B}\right)=\mathrm{Rank}\left(\widehat{A}\right)=t$. Furthermore, let DGGIs of $\widehat{A}$ and $\widehat{B}$ exist. If $\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}=\widehat{E}=\widehat{E}\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}$, then $\widehat{B}{\widehat{B}}^{\mathrm{DGGI}}=\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}$.

Proof. 

Since $\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}=\widehat{E}$, it is obvious that $\mathcal{R}\left(\widehat{B}\right)\subseteq \mathcal{R}\left(\widehat{A}\right)$. Then, there exists $\widehat{X}\in {\mathbb{D}\mathbb{R}}^{n\times n}$ such that $\widehat{B}=\widehat{A}\widehat{X}$. Applying Theorem 6, we know that $\widehat{A}$ and ${\widehat{A}}^{\mathrm{DGGI}}$ have the forms as in (17) and (18), respectively. Correspondingly, we write

$$\begin{array}{c}\hfill \widehat{B}=\widehat{P}\left[\begin{array}{cc}{\widehat{B}}_1& {\widehat{B}}_2\\ {\widehat{B}}_3& {\widehat{B}}_4\end{array}\right]{\widehat{P}}^{-1},\widehat{X}=\widehat{P}\left[\begin{array}{cc}{\widehat{X}}_1& {\widehat{X}}_2\\ {\widehat{X}}_3& {\widehat{X}}_4\end{array}\right]{\widehat{P}}^{-1},\end{array}$$

(28)

where ${\widehat{B}}_1,{\widehat{X}}_1\in {\mathbb{D}\mathbb{R}}^{t\times t}$. Therefore,

$$\begin{array}{c}\hfill \widehat{P}\left[\begin{array}{cc}{\widehat{B}}_1& {\widehat{B}}_2\\ {\widehat{B}}_3& {\widehat{B}}_4\end{array}\right]{\widehat{P}}^{-1}=\widehat{P}\left[\begin{array}{cc}\widehat{T}& 0\\ 0& 0\end{array}\right]\left[\begin{array}{cc}{\widehat{X}}_1& {\widehat{X}}_2\\ {\widehat{X}}_3& {\widehat{X}}_4\end{array}\right]{\widehat{P}}^{-1}=\widehat{P}\left[\begin{array}{cc}\widehat{T}{\widehat{X}}_1& \widehat{T}{\widehat{X}}_2\\ 0& 0\end{array}\right]{\widehat{P}}^{-1},\end{array}$$

(29)

where $\widehat{T}\in {\mathbb{D}\mathbb{R}}^{t\times t}$ is invertible. It follows that ${\widehat{B}}_3=0$ and ${\widehat{B}}_4=0$. Therefore,

$$\begin{array}{c}\hfill \widehat{B}=\widehat{P}\left[\begin{array}{cc}{\widehat{B}}_1& {\widehat{B}}_2\\ 0& 0\end{array}\right]{\widehat{P}}^{-1}.\end{array}$$

(30)

By applying $\widehat{E}=\widehat{E}\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}$ and Theorem 9, we obtain $\mathcal{N}\left(\widehat{A}\right)=\mathcal{N}\left(\widehat{B}\right)$. Then, $\widehat{A}\widehat{x}=0\Rightarrow \widehat{B}\widehat{x}=0$. Since the DGGI of $\widehat{A}$ exists, the solution $\widehat{x}=\left(I_n-\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}\right)\widehat{y}$ of equation $\widehat{A}\widehat{x}=0$ is the solution of the equation $\widehat{B}\widehat{x}=0$, for any arbitrary dual vector $\widehat{y}$. Thus, we obtain that $\widehat{B}\left(I_n-\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}\right)=0$.

Moreover, from (17), (18), and (30), we obtain ${\widehat{B}}_2=0$. Therefore,

$$\begin{array}{c}\hfill \widehat{B}=\widehat{P}\left[\begin{array}{cc}{\widehat{B}}_1& 0\\ 0& 0\end{array}\right]{\widehat{P}}^{-1}.\end{array}$$

Since the DGGI of $\widehat{B}$ exists, according to Theorem 7 and $\mathrm{Rank}\left(\widehat{B}\right)=\mathrm{Rank}\left(\widehat{A}\right)$, we deduce that

$$\begin{array}{c}\hfill \mathrm{Rank}\left(\widehat{T}\right)=\mathrm{Rank}\left(\widehat{A}\right)=\mathrm{Rank}\left(\widehat{B}\right)=\mathrm{ARank}\left(\widehat{B}\right)=\mathrm{ARank}\left({\widehat{B}}_1\right)=\mathrm{Rank}\left({\widehat{B}}_1\right)=t;\end{array}$$

that is, ${\widehat{B}}_1$ is invertible. Then,

$$\begin{array}{cc}\hfill {\widehat{B}}^{\mathrm{DGGI}}& =\widehat{P}\left[\begin{array}{cc}{\widehat{B}}_1^{-1}& 0\\ 0& 0\end{array}\right]{\widehat{P}}^{-1}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \widehat{B}{\widehat{B}}^{\mathrm{DGGI}}& =\widehat{P}\left[\begin{array}{cc}{\widehat{B}}_1& 0\\ 0& 0\end{array}\right]\left[\begin{array}{cc}{\widehat{B}}_1^{-1}& 0\\ 0& 0\end{array}\right]{\widehat{P}}^{-1}=\widehat{P}\left[\begin{array}{cc}{I}_t& 0\\ 0& 0\end{array}\right]{\widehat{P}}^{-1}=\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}.\hfill \end{array}$$

□

Corollary 3.

Let $\widehat{B}=\widehat{A}+\widehat{E}\in {\mathbb{D}\mathbb{R}}^{n\times n}$, $\mathrm{Rank}\left(\widehat{B}\right)=\mathrm{Rank}\left(\widehat{A}\right)$. Furthermore, let DGGIs of $\widehat{A}$ and $\widehat{B}$ exist. If $\mathcal{R}\left(\widehat{A}\right)=\mathcal{R}\left(\widehat{B}\right)$ and $\mathcal{N}\left(\widehat{A}\right)=\mathcal{N}\left(\widehat{B}\right)$, then

$$\begin{array}{c}\hfill \widehat{B}{\widehat{B}}^{\mathrm{DGGI}}=\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}.\end{array}$$

(31)

As stated in the previous theorem, it is clear that $\mathcal{R}\left(\widehat{A}\right)=\mathcal{R}\left(\widehat{B}\right)\Rightarrow \mathcal{R}\left(\widehat{B}\right)\subseteq \mathcal{R}\left(\widehat{A}\right)$, i.e., there exists $\widehat{X}\in {\mathbb{D}\mathbb{R}}^{n\times n}$ such that $\widehat{B}=\widehat{A}\widehat{X}$. If $\mathcal{N}\left(\widehat{A}\right)=\mathcal{N}\left(\widehat{B}\right)$. Then, we obtain that $\mathcal{N}\left(\widehat{A}\right)\subseteq \mathcal{N}\left(\widehat{B}\right)$, namely, $\widehat{A}\widehat{x}=0\Rightarrow \widehat{B}\widehat{x}=0$. Hence, the argument is the same as Theorem 10.

From the derivation of the above conclusions, we know that the DGGI plays a crucial role. Therefore, when discussing the perturbation of dual real matrices, we also assume that it exists.

Theorem 11.

Let $\widehat{B}=\widehat{A}+\widehat{E}\in {\mathbb{D}\mathbb{R}}^{n\times n}$. Furthermore, let DGGIs of $\widehat{A}$ and $\widehat{B}$ exist. Then,

$$\begin{array}{cc}\hfill {\widehat{B}}^{\mathrm{DGGI}}& ={\left(I_n+{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}\right)}^{-1}{\widehat{A}}^{\mathrm{DGGI}}\hfill \end{array}$$

(32a)

$$\begin{array}{cc}\hfill & ={\widehat{A}}^{\mathrm{DGGI}}{\left(I_n+\widehat{E}{\widehat{A}}^{\mathrm{DGGI}}\right)}^{-1}\hfill \end{array}$$

(32b)

if and only if

$$\begin{array}{c}\hfill \mathrm{Rank}\left(\widehat{B}\right)=\mathrm{Rank}\left(\widehat{A}\right)and\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}=\widehat{E}=\widehat{E}\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}.\end{array}$$

(33)

Proof. 

Let $\widehat{A}=A+\epsilon A_0$, $\widehat{E}=E+\epsilon E_0$ and $\widehat{B}=B+\epsilon B_0\in {\mathbb{D}\mathbb{R}}^{n\times n}$.

“⇐” Suppose that (33) holds. We will show that $I_n+{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}$ is invertible.

It is easy to obtain that

$$\begin{array}{c}\hfill \widehat{B}=\widehat{A}+\widehat{E}=\widehat{A}\left(I_n+{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}\right)=\left(I_n+\widehat{E}{\widehat{A}}^{\mathrm{DGGI}}\right)\widehat{A}.\end{array}$$

From Theorems 8–10, we have

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\widehat{B}{\widehat{B}}^{\mathrm{DGGI}}=\widehat{A}{\widehat{A}}^{\mathrm{DGGI}},\\ \mathcal{R}\left(\widehat{A}\right)=\mathcal{R}\left(\widehat{B}\right),\\ \mathcal{N}\left(\widehat{A}\right)=\mathcal{N}\left(\widehat{B}\right).\end{array}\right.\end{array}$$

From the existence of ${\widehat{A}}^{\mathrm{DGGI}}$, we know that $\mathcal{R}\left(\widehat{A}\right)\cap \mathcal{N}\left(\widehat{A}\right)=\left\{0\right\}$. Furthermore, $\mathcal{R}\left(\widehat{B}\right)\cap \mathcal{N}\left(\widehat{B}\right)=\left\{0\right\}$.

If $I_n+{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}$ is singular, then there exists a nonzero dual vector $\widehat{x}$ such that

$$\begin{array}{c}\hfill \left(I_n+{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}\right)\widehat{x}=0,\end{array}$$

that is, $\widehat{x}=-{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}\widehat{x}\in \mathcal{R}\left({\widehat{A}}^{\mathrm{DGGI}}\right)=\mathcal{R}\left(\widehat{A}\right)$, and $\widehat{A}\left(I_n+{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}\right)\widehat{x}=0$⇒$\widehat{B}\widehat{x}=0$, $\widehat{x}\in \mathcal{N}\left(\widehat{B}\right)=\mathcal{N}\left(\widehat{A}\right)$. So, $\widehat{x}\in \mathcal{R}\left(\widehat{A}\right)\cap \mathcal{N}\left(\widehat{A}\right)=\left\{0\right\}$ is contradiction. We conclude that $I_n+{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}$ is nonsingular. Then,

$$\begin{array}{cc}\hfill {\widehat{B}}^{\mathrm{DGGI}}-{\widehat{A}}^{\mathrm{DGGI}}& ={\widehat{A}}^{\mathrm{DGGI}}\left(\widehat{B}-\widehat{A}\right){\widehat{B}}^{\mathrm{DGGI}}+{\widehat{B}}^{\mathrm{DGGI}}-{\widehat{A}}^{\mathrm{DGGI}}-{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}{\widehat{B}}^{\mathrm{DGGI}}\hfill \\ \hfill & =\left(I_n-{\widehat{A}}^{\mathrm{DGGI}}\widehat{A}\right){\widehat{B}}^{\mathrm{DGGI}}+{\widehat{A}}^{\mathrm{DGGI}}\left(\widehat{B}{\widehat{B}}^{\mathrm{DGGI}}-I_n\right)-{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}{\widehat{B}}^{\mathrm{DGGI}}\hfill \\ \hfill & =-{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}{\widehat{B}}^{\mathrm{DGGI}}.\hfill \end{array}$$

From the above equation, we obtain $\left(I_n+{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}\right){\widehat{B}}^{\mathrm{DGGI}}={\widehat{A}}^{\mathrm{DGGI}}$. Since $I_n+{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}$ is nonsingular,

$${\widehat{B}}^{\mathrm{DGGI}}={\left(I_n+{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}\right)}^{-1}{\widehat{A}}^{\mathrm{DGGI}}.$$

Similarly, Equation (32b) can be proved.

“⇒” Suppose that (32a) and (32b) hold. We deduce that

$$\begin{array}{cc}\hfill \mathrm{Rank}\left(\widehat{B}\right)& =\mathrm{Rank}\left({\widehat{B}}^{\mathrm{DGGI}}\right)=\mathrm{Rank}\left({\left(I_n+{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}\right)}^{-1}{\widehat{A}}^{\mathrm{DGGI}}\right)=\mathrm{Rank}\left({\widehat{A}}^{\mathrm{DGGI}}\right)\hfill \\ \hfill & =\mathrm{Rank}\left(\widehat{A}\right).\hfill \end{array}$$

(34)

Next, we will prove that $\mathcal{R}\left(\widehat{B}\right)=\mathcal{R}\left(\widehat{A}\right)$. Obviously, by ${\widehat{B}}^{\mathrm{DGGI}}={\widehat{A}}^{\mathrm{DGGI}}{\left(I_n+\widehat{E}{\widehat{A}}^{\mathrm{DGGI}}\right)}^{-1}$, we have

$$\begin{array}{c}\hfill \mathcal{R}\left({\widehat{B}}^{\mathrm{DGGI}}\right)\subseteq \mathcal{R}\left({\widehat{A}}^{\mathrm{DGGI}}\right).\end{array}$$

(35)

Since ${\widehat{A}}^{\mathrm{DGGI}}$ and ${\widehat{B}}^{\mathrm{DGGI}}$ exist, we know that DGGIs of ${\widehat{A}}^{\mathrm{DGGI}}$ and ${\widehat{B}}^{\mathrm{DGGI}}$ exist. Furthermore, ${\left({\widehat{B}}^{\mathrm{DGGI}}\right)}^{\mathrm{DGGI}}=\widehat{B}$ and ${\left({\widehat{A}}^{\mathrm{DGGI}}\right)}^{\mathrm{DGGI}}=\widehat{A}$. By applying (34), (35), and Corollary 2, we obtain

$$\begin{array}{c}\hfill \mathcal{R}\left(\widehat{B}\right)=\mathcal{R}\left({\widehat{B}}^{\mathrm{DGGI}}\right)=\mathcal{R}\left({\widehat{A}}^{\mathrm{DGGI}}\right)=\mathcal{R}\left(\widehat{A}\right).\end{array}$$

(36)

Furthermore, it follows from ${\widehat{B}}^{\mathrm{DGGI}}={\left(I_n+{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}\right)}^{-1}{\widehat{A}}^{\mathrm{DGGI}}$ that

$$\begin{array}{cc}\hfill \widehat{A}\widehat{x}=0\iff {\widehat{A}}^{\mathrm{DGGI}}\widehat{x}=0& \iff {\left(I_n+{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}\right)}^{-1}{\widehat{A}}^{\mathrm{DGGI}}\widehat{x}=0\iff {\widehat{B}}^{\mathrm{DGGI}}\widehat{x}=0\iff \widehat{B}\widehat{x}=0,\hfill \end{array}$$

that is,

$$\begin{array}{c}\hfill \mathcal{N}\left(\widehat{B}\right)=\mathcal{N}\left(\widehat{A}\right).\end{array}$$

(37)

According to (36), (37), and Corollary 3, we obtain $\widehat{B}{\widehat{B}}^{\mathrm{DGGI}}=\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}$.

By a direct calculation, we obtain

$$\begin{array}{c}\hfill \widehat{A}{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}=\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}\left(\widehat{B}-\widehat{A}\right)=\widehat{B}{\widehat{B}}^{\mathrm{DGGI}}\widehat{B}-\widehat{A}=\widehat{E}\end{array}$$

and

$$\begin{array}{c}\hfill \widehat{E}\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}=\left(\widehat{B}-\widehat{A}\right)\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}=\widehat{B}\widehat{B}{\widehat{B}}^{\mathrm{DGGI}}-\widehat{A}=\widehat{E}.\end{array}$$

Thus, Equation (33) holds. □

Corollary 4.

Let $\widehat{B}=\widehat{A}+\widehat{E}\in {\mathbb{D}\mathbb{R}}^{n\times n}$ with $\mathrm{Ind}\left(\widehat{A}\right)=1$. Furthermore, let DGGIs of $\widehat{A}$ and $\widehat{B}$ exist. Suppose $\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}=\widehat{E}=\widehat{E}\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}$ and ${∥{\widehat{A}}^{\mathrm{DGGI}}∥}_2{∥\widehat{E}∥}_2<1$ hold, then

$$\begin{array}{c}\hfill {\displaystyle \frac{{∥{\widehat{B}}^{\mathrm{DGGI}}-{\widehat{A}}^{\mathrm{DGGI}}∥}_2}{{∥{\widehat{A}}^{\mathrm{DGGI}}∥}_2}}⩽{\displaystyle \frac{{∥{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}∥}_2}{1-{∥{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}∥}_2}}and\widehat{B}{\widehat{B}}^{\mathrm{DGGI}}=\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}.\end{array}$$

Proof. 

According to Theorem 11, we know that $\widehat{B}{\widehat{B}}^{\mathrm{DGGI}}=\widehat{A}{\widehat{A}}^{\mathrm{DGGI}}$, and we can estimate that

$$\begin{array}{cc}\hfill {\displaystyle \frac{{∥{\widehat{B}}^{\mathrm{DGGI}}-{\widehat{A}}^{\mathrm{DGGI}}∥}_2}{{∥{\widehat{A}}^{\mathrm{DGGI}}∥}_2}}& ={\displaystyle \frac{{∥{(I_n+{\widehat{A}}^{\mathrm{DGGI}}\widehat{E})}^{-1}{\widehat{A}}^{\mathrm{DGGI}}-{\widehat{A}}^{\mathrm{DGGI}}∥}_2}{{∥{\widehat{A}}^{\mathrm{DGGI}}∥}_2}}\hfill \\ \hfill & \le {\displaystyle \frac{{∥{(I_n+{\widehat{A}}^{\mathrm{DGGI}}\widehat{E})}^{-1}-I_n∥}_2{∥{\widehat{A}}^{\mathrm{DGGI}}∥}_2}{{∥{\widehat{A}}^{\mathrm{DGGI}}∥}_2}}\hfill \\ \hfill & ={∥{(I_n+{\widehat{A}}^{\mathrm{DGGI}}\widehat{E})}^{-1}\left[I_n-(I_n+{\widehat{A}}^{\mathrm{DGGI}}\widehat{E})\right]∥}_2\hfill \\ \hfill & \le {∥{(I_n+{\widehat{A}}^{\mathrm{DGGI}}\widehat{E})}^{-1}∥}_2{∥{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}∥}_2\hfill \\ \hfill & \le {\displaystyle \frac{{∥{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}∥}_2}{1-{∥{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}∥}_2}}.\hfill \end{array}$$

□

Corollary 5.

Under the same assumption of Corollary 4, then

$$\begin{array}{c}\hfill {\displaystyle \frac{{∥{\widehat{A}}^{\mathrm{DGGI}}∥}_2}{1+{∥{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}∥}_2}}⩽{∥{\widehat{A}}^{\mathrm{DGGI}}∥}_2⩽{\displaystyle \frac{{∥{\widehat{A}}^{\mathrm{DGGI}}∥}_2}{1-{∥{\widehat{A}}^{\mathrm{DGGI}}\widehat{E}∥}_2}}.\end{array}$$

4. General Perturbation of $\mathit{\epsilon }{\mathit{A}}_{\mathbf{0}}$ to $\mathit{A}$ in Real Field

By studying the symmetries of dual matrices within dual matrix group inverse, we can uncover deeper connections between geometric transformations and algebraic structures. In this section, let $\widehat{A}=A+ϵA_0\in {\mathbb{R}}^{n\times n}$ with $\mathrm{Ind}\left(A\right)=1$, where $ϵA_0$ is regarded as a perturbation of the real matrix A, $ϵ$ is a small positive real number and ${ϵ}^2>0$. We characterize the group inverse of $\widehat{A}$ under the condition of equal rank. Next, we consider the error between (6) and the group inverse of $\widehat{A}$, in which $ϵ$ is a small positive real number. At the same time, we illustrate these conclusions with examples.

Theorem 12.

Let $\widehat{A}=A+ϵA_0\in {\mathbb{R}}^{n\times n}$, $\mathrm{Ind}\left(\widehat{A}\right)=\mathrm{Ind}\left(A\right)=1$ and $\mathrm{Rank}\left(\widehat{A}\right)=\mathrm{rank}\left(A\right)=r$. If $I_n+ϵA^{\#}{A}_0$ is nonsingular, then

$$\begin{array}{cc}\hfill {\widehat{A}}^{\#}=& {\left(I_n+{ϵ}^{2}Y\right)}^{-1}{\left(I_n+ϵA^{\#}{A}_0\right)}^{-1}{A}^{\#}{\left(I_n+{ϵ}^{2}Y\right)}^{-1}\hfill \\ \hfill & +ϵ\left(I_n-A{A}^{\#}\right){\left(I_n+ϵA_0{A}^{\#}\right)}^{-1}{A}_0{A}^{\#}{\left(I_n+{ϵ}^{2}Y\right)}^{-1}{\left(I_n+ϵA^{\#}{A}_0\right)}^{-1}{A}^{\#}{\left(I_n+{ϵ}^{2}Y\right)}^{-1}\hfill \\ \hfill & +ϵ{\left(I_n+{ϵ}^{2}Y\right)}^{-1}{A}^{\#}{\left(I_n+ϵA_0{A}^{\#}\right)}^{-1}{\left(I_n+{ϵ}^{2}Y\right)}^{-1}{A}^{\#}{A}_0{\left(I_n+ϵA^{\#}{A}_0\right)}^{-1}\left(I_n-A{A}^{\#}\right)\hfill \\ \hfill & +{ϵ}^2\left(I_n-A{A}^{\#}\right){\left(I_n+ϵA_0{A}^{\#}\right)}^{-1}{A}_0{A}^{\#}{\left(I_n+{ϵ}^{2}Y\right)}^{-1}{A}^{\#}{\left(I_n+ϵA_0{A}^{\#}\right)}^{-1}\hfill \\ \hfill & \times {\left(I_n+{ϵ}^{2}Y\right)}^{-1}{A}^{\#}{A}_0{\left(I_n+ϵA^{\#}{A}_0\right)}^{-1}\left(I_n-A{A}^{\#}\right),\hfill \end{array}$$

(38)

where $Y=A^{\#}{A}_0{\left(I_n+ϵA^{\#}{A}_0\right)}^{-1}\left(I_n-A{A}^{\#}\right){\left(I_n+ϵA_0{A}^{\#}\right)}^{-1}{A}_0{A}^{\#}$. Furthermore,

$$\begin{array}{c}\hfill {\widehat{A}}^{\#}=P\left[\begin{array}{cc}T& ϵT{D}^{-1}{A}_2-{ϵ}^{2}T{\Delta }_1\\ ϵA_3{D}^{-1}T-{ϵ}^2{\Delta }_{2}T& {ϵ}^4{\Delta }_{2}T{\Delta }_1-{ϵ}^3{\Delta }_{2}T{D}^{-1}{A}_2-{ϵ}^3{A}_3{D}^{-1}T{\Delta }_1+{ϵ}^2{A}_3{D}^{-1}T{D}^{-1}{A}_2\end{array}\right]P^{-1}.\end{array}$$

(39)

where

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\Delta }_1=D^{-1}{A}_1{\left(D+ϵA_1\right)}^{-1}{A}_2\hfill \\ {\Delta }_2=A_3{\left(D+ϵA_1\right)}^{-1}{A}_1{D}^{-1}\hfill \\ M={ϵ}^2{\Delta }_1{\Delta }_2-ϵD^{-1}{A}_2{\Delta }_2-ϵ{\Delta }_1{A}_3{D}^{-1}+D^{-1}{A}_2{A}_3{D}^{-1}\hfill \\ T={\left(I_r+{ϵ}^{2}M\right)}^{-1}{\left(D+ϵA_1\right)}^{-1}{\left(I_r+{ϵ}^{2}M\right)}^{-1},\hfill \end{array}\right.\end{array}$$

(40)

in which D is as shown in Lemma 1, $A_1$, $A_2$, and $A_3$ are the appropriate block matrices for the decomposition of $A_0$, ${\Delta }_1\in {\mathbb{R}}^{r\times (n-r)}$, ${\Delta }_2\in {\mathbb{R}}^{(n-r)\times r}$, and $M,T\in {\mathbb{R}}^{r\times r}$.

Proof. 

According to Lemma 2, we can obtain Equation (38). Let A be represented in the form as in (4). Denote $A_0=P\left[\begin{array}{cc}{A}_1& A_2\\ A_3& A_4\end{array}\right]P^{-1}$, where $A_1\in {\mathbb{R}}^{r\times r}$. Since $I_n+ϵA^{\#}{A}_0$ is invertible, $D+ϵA_1$ is nonsingular. Then,

$$\begin{array}{cc}\hfill Y=& P\left[\begin{array}{cc}{D}^{-1}& 0\\ 0& 0\end{array}\right]\left[\begin{array}{cc}{A}_1& A_2\\ A_3& A_4\end{array}\right]{\left[\begin{array}{cc}{I}_r+ϵD^{-1}{A}_1& ϵD^{-1}{A}_2\\ 0& I_{n-r}\end{array}\right]}^{-1}\left[\begin{array}{cc}0& 0\\ 0& I_{n-r}\end{array}\right]\hfill \\ \hfill & \times {\left[\begin{array}{cc}{I}_r+ϵA_1{D}^{-1}& 0\\ ϵA_3{D}^{-1}& I_{n-r}\end{array}\right]}^{-1}\left[\begin{array}{cc}{A}_1& A_2\\ A_3& A_4\end{array}\right]\left[\begin{array}{cc}{D}^{-1}& 0\\ 0& 0\end{array}\right]P^{-1}\hfill \\ \hfill =& P\left[\begin{array}{cc}{ϵ}^2{\Delta }_1{\Delta }_2-ϵD^{-1}{A}_2{\Delta }_2-ϵ{\Delta }_1{A}_3{D}^{-1}+D^{-1}{A}_2{A}_3{D}^{-1}& 0\\ 0& 0\end{array}\right]P^{-1},\hfill \end{array}$$

(41)

where ${\Delta }_1=D^{-1}{A}_1{\left(D+ϵA_1\right)}^{-1}{A}_2$, ${\Delta }_2=A_3{\left(D+ϵA_1\right)}^{-1}{A}_1{D}^{-1}$. Let $M={ϵ}^2{\Delta }_1{\Delta }_2-ϵD^{-1}{A}_2{\Delta }_2-ϵ{\Delta }_1{A}_3{D}^{-1}+D^{-1}{A}_2{A}_3{D}^{-1}$. Since $I_n$ is nonsingular and ${ϵ}^{2}Y$ is the infinitesimal quantity, $I_n+{ϵ}^{2}Y$ is nonsingular. Then,

$$\begin{array}{c}\hfill {\left(I_n+{ϵ}^{2}Y\right)}^{-1}=P\left[\begin{array}{cc}{\left(I_r+{ϵ}^{2}M\right)}^{-1}& 0\\ 0& I_{n-r}\end{array}\right]P^{-1}.\end{array}$$

(42)

Denote $T={\left(I_r+{ϵ}^{2}M\right)}^{-1}{\left(D+ϵA_1\right)}^{-1}{\left(I_r+{ϵ}^{2}M\right)}^{-1}$. By a direct calculation, we obtain

$$\begin{array}{cc}\hfill & {\left(I_n+{ϵ}^{2}Y\right)}^{-1}{\left(I_n+ϵA^{\#}{A}_0\right)}^{-1}{A}^{\#}{\left(I_n+{ϵ}^{2}Y\right)}^{-1}\hfill \\ \hfill =& P\left[\begin{array}{cc}{\left(I_r+{ϵ}^{2}M\right)}^{-1}& 0\\ 0& 0\end{array}\right]{\left[\begin{array}{cc}{I}_r+ϵD^{-1}{A}_1& ϵD^{-1}{A}_2\\ 0& I_{n-r}\end{array}\right]}^{-1}\left[\begin{array}{cc}{D}^{-1}& 0\\ 0& 0\end{array}\right]\left[\begin{array}{cc}{\left(I_r+{ϵ}^{2}M\right)}^{-1}& 0\\ 0& 0\end{array}\right]P^{-1}\hfill \\ \hfill =& P\left[\begin{array}{cc}T& 0\\ 0& 0\end{array}\right]P^{-1},\hfill \end{array}$$

(43)

$$\begin{array}{cc}\hfill & \left(I_n-A{A}^{\#}\right){\left(I_n+ϵA_0{A}^{\#}\right)}^{-1}{A}_0{A}^{\#}{\left(I_n+{ϵ}^{2}Y\right)}^{-1}{\left(I_n+ϵA^{\#}{A}_0\right)}^{-1}{A}^{\#}{\left(I_n+{ϵ}^{2}Y\right)}^{-1}\hfill \\ \hfill =& P\left[\begin{array}{cc}0& 0\\ 0& I_{n-r}\end{array}\right]{\left[\begin{array}{cc}{I}_r+ϵA_1{D}^{-1}& 0\\ ϵA_3{D}^{-1}& I_{n-r}\end{array}\right]}^{-1}\left[\begin{array}{cc}{A}_1{D}^{-1}& 0\\ A_3{D}^{-1}& 0\end{array}\right]\left[\begin{array}{cc}{\left(I_r+{ϵ}^{2}M\right)}^{-1}& 0\\ 0& 0\end{array}\right]\hfill \\ \hfill & \times {\left[\begin{array}{cc}{I}_r+ϵD^{-1}{A}_1& ϵD^{-1}{A}_2\\ 0& I_{n-r}\end{array}\right]}^{-1}\left[\begin{array}{cc}{D}^{-1}& 0\\ 0& 0\end{array}\right]\left[\begin{array}{cc}{\left(I_r+{ϵ}^{2}M\right)}^{-1}& 0\\ 0& 0\end{array}\right]P^{-1}\hfill \\ \hfill =& P\left[\begin{array}{cc}0& 0\\ A_3{D}^{-1}T-ϵ{\Delta }_{2}T& 0\end{array}\right]P^{-1},\hfill \end{array}$$

(44)

$$\begin{array}{cc}\hfill & {\left(I_n+{ϵ}^{2}Y\right)}^{-1}{A}^{\#}{\left(I_n+ϵA_0{A}^{\#}\right)}^{-1}{\left(I_n+{ϵ}^{2}Y\right)}^{-1}{A}^{\#}{A}_0{\left(I_n+ϵA^{\#}{A}_0\right)}^{-1}\left(I_n-A{A}^{\#}\right)\hfill \\ \hfill =& P\left[\begin{array}{cc}{\left(I_r+{ϵ}^{2}M\right)}^{-1}& 0\\ 0& 0\end{array}\right]\left[\begin{array}{cc}{D}^{-1}& 0\\ 0& 0\end{array}\right]{\left[\begin{array}{cc}{I}_r+ϵA_1{D}^{-1}& 0\\ ϵA_3{D}^{-1}& I_{n-r}\end{array}\right]}^{-1}\left[\begin{array}{cc}{\left(I_r+{ϵ}^{2}M\right)}^{-1}& 0\\ 0& 0\end{array}\right]\hfill \\ \hfill & \times \left[\begin{array}{cc}{D}^{-1}{A}_1& D^{-1}{A}_2\\ 0& 0\end{array}\right]{\left[\begin{array}{cc}{I}_r+ϵD^{-1}{A}_1& ϵD^{-1}{A}_2\\ 0& I_{n-r}\end{array}\right]}^{-1}\left[\begin{array}{cc}0& 0\\ 0& I_{n-r}\end{array}\right]P^{-1}\hfill \\ \hfill =& P\left[\begin{array}{cc}0& T{D}^{-1}{A}_2-ϵT{\Delta }_1\\ 0& 0\end{array}\right]P^{-1},\hfill \end{array}$$

(45)

and

$$\begin{array}{cc}\hfill & \left(I_n-A{A}^{\#}\right){\left(I_n+ϵA_0{A}^{\#}\right)}^{-1}{A}_0{A}^{\#}{\left(I_n+{ϵ}^{2}Y\right)}^{-1}{A}^{\#}{\left(I_n+ϵA_0{A}^{\#}\right)}^{-1}{\left(I_n+{ϵ}^{2}Y\right)}^{-1}\hfill \\ \hfill & \times A^{\#}{A}_0{\left(I_n+ϵA^{\#}{A}_0\right)}^{-1}\left(I_n-A{A}^{\#}\right)\hfill \\ \hfill =& P\left[\begin{array}{cc}0& 0\\ 0& I_{n-r}\end{array}\right]{\left[\begin{array}{cc}{I}_r+ϵA_1{D}^{-1}& 0\\ ϵA_3{D}^{-1}& I_{n-r}\end{array}\right]}^{-1}\left[\begin{array}{cc}{A}_1{D}^{-1}& 0\\ A_3{D}^{-1}& 0\end{array}\right]\left[\begin{array}{cc}{\left(I_r+{ϵ}^{2}M\right)}^{-1}& 0\\ 0& 0\end{array}\right]\hfill \\ \hfill & \times \left[\begin{array}{cc}{D}^{-1}& 0\\ 0& 0\end{array}\right]{\left[\begin{array}{cc}{I}_r+ϵA_1{D}^{-1}& 0\\ ϵA_3{D}^{-1}& I_{n-r}\end{array}\right]}^{-1}\left[\begin{array}{cc}{\left(I_r+{ϵ}^{2}M\right)}^{-1}& 0\\ 0& 0\end{array}\right]\left[\begin{array}{cc}{D}^{-1}{A}_1& D^{-1}{A}_2\\ 0& 0\end{array}\right]\hfill \\ \hfill & \times {\left[\begin{array}{cc}{I}_r+ϵD^{-1}{A}_1& ϵD^{-1}{A}_2\\ 0& I_{n-r}\end{array}\right]}^{-1}\left[\begin{array}{cc}0& 0\\ 0& I_{n-r}\end{array}\right]P^{-1}\hfill \\ \hfill =& P\left[\begin{array}{cc}0& 0\\ 0& {ϵ}^2{\Delta }_{2}T{\Delta }_1-ϵ{\Delta }_{2}T{D}^{-1}{A}_2-ϵA_3{D}^{-1}T{\Delta }_1+A_3{D}^{-1}T{D}^{-1}{A}_2\end{array}\right]P^{-1}.\hfill \end{array}$$

(46)

Substituting (43)–(46) into (38), we obtain (39). □

Example 1.

Let

$$\widehat{A}=A+ϵA_0=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& -1& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]+ϵ\left[\begin{array}{cccc}1& 0& 1& 0\\ 0& 2& -2& 0\\ 0& 1& -1& 0\\ 0& 0& 0& 0\end{array}\right].$$

According to Theorem 12, we can obtain

$$\begin{array}{c}\hfill Y=A^{\#}{A}_0{\left(I_4+ϵA^{\#}{A}_0\right)}^{-1}\left(I_4-A{A}^{\#}\right){\left(I_4+ϵA_0{A}^{\#}\right)}^{-1}{A}_0{A}^{\#}=\left[\begin{array}{cccc}0& {\displaystyle \frac{1}{{\left(1+ϵ\right)}^2}}& {\displaystyle \frac{-1}{{\left(1+ϵ\right)}^2}}& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right].\end{array}$$

Furthermore,

$$\left\{\begin{array}{c}{\left(I_4+{ϵ}^{2}Y\right)}^{-1}={\left[\begin{array}{cccc}1& {\displaystyle \frac{{ϵ}^2}{{\left(1+ϵ\right)}^2}}& {\displaystyle \frac{-{ϵ}^2}{{\left(1+ϵ\right)}^2}}& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]}^{-1}=\left[\begin{array}{cccc}1& {\displaystyle \frac{-{ϵ}^2}{{\left(1+ϵ\right)}^2}}& {\displaystyle \frac{{ϵ}^2}{{\left(1+ϵ\right)}^2}}& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right],\hfill \\ {\left(I_4+{ϵ}^{2}Y\right)}^{-1}{\left(I_4+ϵA^{\#}{A}_0\right)}^{-1}{A}^{\#}{\left(I_4+{ϵ}^{2}Y\right)}^{-1}=\left[\begin{array}{cccc}{\displaystyle \frac{1}{1+ϵ}}& {\displaystyle \frac{-2{ϵ}^2}{{\left(1+ϵ\right)}^3}}& {\displaystyle \frac{2{ϵ}^2}{{\left(1+ϵ\right)}^3}}& 0\\ 0& {\displaystyle \frac{1}{1+ϵ}}& {\displaystyle \frac{-1}{1+ϵ}}& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right],\hfill \\ \left(I_4-A{A}^{\#}\right){\left(I_4+ϵA_0{A}^{\#}\right)}^{-1}{A}_0{A}^{\#}{\left(I_4+{ϵ}^{2}Y\right)}^{-1}{\left(I_4+ϵA^{\#}{A}_0\right)}^{-1}{A}^{\#}{\left(I_4+{ϵ}^{2}Y\right)}^{-1}\hfill \\ =\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& {\displaystyle \frac{1}{{\left(1+ϵ\right)}^2}}& {\displaystyle \frac{-1}{{\left(1+ϵ\right)}^2}}& 0\\ 0& {\displaystyle \frac{1}{{\left(1+ϵ\right)}^2}}& {\displaystyle \frac{-1}{{\left(1+ϵ\right)}^2}}& 0\\ 0& 0& 0& 0\end{array}\right],\hfill \\ {\left(I_4+{ϵ}^{2}Y\right)}^{-1}{A}^{\#}{\left(I_4+ϵA_0{A}^{\#}\right)}^{-1}{\left(I_4+{ϵ}^{2}Y\right)}^{-1}{A}^{\#}{A}_0{\left(I_4+ϵA^{\#}{A}_0\right)}^{-1}\left(I_4-A{A}^{\#}\right)\hfill \\ =\left[\begin{array}{cccc}0& 0& {\displaystyle \frac{1}{{\left(1+ϵ\right)}^2}}& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right],\hfill \end{array}\right.$$

and

$$\begin{array}{cc}\hfill & \left(I_4-A{A}^{\#}\right){\left(I_4+ϵA_0{A}^{\#}\right)}^{-1}{A}_0{A}^{\#}{\left(I_4+{ϵ}^{2}Y\right)}^{-1}{A}^{\#}{\left(I_4+ϵA_0{A}^{\#}\right)}^{-1}{\left(I_4+{ϵ}^{2}Y\right)}^{-1}{A}^{\#}{A}_0\hfill \\ \hfill & \times {\left(I_4+ϵA^{\#}{A}_0\right)}^{-1}\left(I_4-A{A}^{\#}\right)=0.\hfill \end{array}$$

Then,

$$\begin{array}{c}\hfill {\widehat{A}}^{\#}=\left[\begin{array}{cccc}{\displaystyle \frac{1}{1+ϵ}}& {\displaystyle \frac{-2{ϵ}^2}{{\left(1+ϵ\right)}^3}}& {\displaystyle \frac{3{ϵ}^2+ϵ}{{\left(1+ϵ\right)}^3}}& 0\\ 0& {\displaystyle \frac{1+2ϵ}{{\left(1+ϵ\right)}^2}}& {\displaystyle \frac{-1-2ϵ}{{\left(1+ϵ\right)}^2}}& 0\\ 0& {\displaystyle \frac{ϵ}{{\left(1+ϵ\right)}^2}}& {\displaystyle \frac{-ϵ}{{\left(1+ϵ\right)}^2}}& 0\\ 0& 0& 0& 0\end{array}\right].\end{array}$$

If we consider ${ϵ}^2$ to be 0, then applying (38), we can simplify as [28]

$$\begin{array}{c}\hfill {\widehat{A}}^{\#}=A^{\#}-ϵ\left[A^{\#}{A}_0{A}^{\#}-{\left(A^{\#}\right)}^2{A}_0\left(I_n-A{A}^{\#}\right)-\left(I_n-A{A}^{\#}\right)A_0{\left(A^{\#}\right)}^2\right].\end{array}$$

In this case, ${\widehat{A}}^{\mathrm{DGGI}}$ and ${\widehat{A}}^{\#}$ are consistent in form.

Next, we reveal the difference between the ${\widehat{A}}^{\#}$ and ${\widehat{A}}^{\mathrm{DGGI}}$, where ${\widehat{A}}^{\mathrm{DGGI}}$ is considered a notation and its form is given in (6), $ϵ$ is a small positive real number, and ${ϵ}^2>0$.

Theorem 13.

Let $\widehat{A}=A+ϵA_0\in {\mathbb{R}}^{n\times n}$ such that $\mathrm{Ind}\left(\widehat{A}\right)=\mathrm{Ind}\left(A\right)=1$, the DGGI of $\widehat{A}$ exist, $\mathrm{Rank}\left(\widehat{A}\right)$ $=\mathrm{rank}\left(\mathit{A}\right)=\mathrm{r}$, and $I_n+ϵA^{\#}{A}_0$ be nonsingular. Then,

$$\begin{array}{c}\hfill {\widehat{A}}^{\#}-{\widehat{A}}^{\mathrm{DGGI}}=P\left[\begin{array}{cc}{N}_1& N_2\\ N_3& N_4\end{array}\right]P^{-1},\end{array}$$

(47)

where

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{N}_1={\left(I_r+{ϵ}^{2}M\right)}^{-1}{\left(D+ϵA_1\right)}^{-1}{\left(I_r+{ϵ}^{2}M\right)}^{-1}-D^{-1}+ϵD^{-1}{A}_1{D}^{-1},\hfill \\ N_2=ϵ{\left(I_r+{ϵ}^{2}M\right)}^{-1}{\left(D+ϵA_1\right)}^{-1}{\left(I_r+{ϵ}^{2}M\right)}^{-1}{D}^{-1}{A}_2\hfill \\ -{ϵ}^2{\left(I_r+{ϵ}^{2}M\right)}^{-1}{\left(D+ϵA_1\right)}^{-1}{\left(I_r+{ϵ}^{2}M\right)}^{-1}{D}^{-1}{A}_1{\left(D+ϵA_1\right)}^{-1}{A}_2\hfill \\ -ϵD^{-2}{A}_2,\hfill \\ N_3=ϵA_3{D}^{-1}{\left(I_r+{ϵ}^{2}M\right)}^{-1}{\left(D+ϵA_1\right)}^{-1}{\left(I_r+{ϵ}^{2}M\right)}^{-1}\hfill \\ -{ϵ}^2{A}_3{\left(D+ϵA_1\right)}^{-1}{A}_1{D}^{-1}{\left(I_r+{ϵ}^{2}M\right)}^{-1}{\left(D+ϵA_1\right)}^{-1}{\left(I_r+{ϵ}^{2}M\right)}^{-1}\hfill \\ -ϵA_3{D}^{-2},\hfill \\ N_4={ϵ}^2{A}_3{D}^{-1}{\left(I_r+{ϵ}^{2}M\right)}^{-1}{\left(D+ϵA_1\right)}^{-1}{\left(I_r+{ϵ}^{2}M\right)}^{-1}{A}_2\hfill \\ -{ϵ}^3{A}_3{\left(D+ϵA_1\right)}^{-1}{A}_1{D}^{-1}{\left(I_r+{ϵ}^{2}M\right)}^{-1}{\left(D+ϵA_1\right)}^{-1}{\left(I_r+{ϵ}^{2}M\right)}^{-1}{D}^{-1}{A}_2\hfill \\ -{ϵ}^3{A}_3{D}^{-1}{\left(I_r+{ϵ}^{2}M\right)}^{-1}{\left(D+ϵA_1\right)}^{-1}{\left(I_r+{ϵ}^{2}M\right)}^{-1}{D}^{-1}{A}_1{\left(D+ϵA_1\right)}^{-1}{A}_2\hfill \\ +{ϵ}^4{A}_3{\left(D+ϵA_1\right)}^{-1}{A}_1{D}^{-1}{\left(I_r+{ϵ}^{2}M\right)}^{-1}{\left(D+ϵA_1\right)}^{-1}\hfill \\ \times {\left(I_r+{ϵ}^{2}M\right)}^{-1}{D}^{-1}{A}_1{\left(D+ϵA_1\right)}^{-1}{A}_2,\hfill \end{array}\right.\end{array}$$

(48)

in which D, $A_1$, $A_2$, $A_3$, and M are as shown in Theorem 12, $N_1\in {\mathbb{R}}^{r\times r}$, $N_2\in {\mathbb{R}}^{r\times (n-r)}$, $N_3\in {\mathbb{R}}^{(n-r)\times r}$, and $N_4\in {\mathbb{R}}^{(n-r)\times (n-r)}$. Furthermore,

$$\begin{array}{c}\hfill {\parallel {\widehat{A}}^{\#}-{\widehat{A}}^{\mathrm{DGGI}}\parallel }_P=\sqrt{{∥N_1∥}_F^2+{∥N_2∥}_F^2+{∥N_3∥}_F^2+{∥N_4∥}_F^2}\in O\left({ϵ}^2\right).\end{array}$$

(49)

Proof. 

By applying Theorem 12 and Lemma 5, we obtain

$$\begin{array}{cc}\hfill & {\widehat{A}}^{\#}-{\widehat{A}}^{\mathrm{DGGI}}\hfill \\ \hfill =& P\left[\begin{array}{cc}T& ϵT{D}^{-1}{A}_2-{ϵ}^{2}T{\Delta }_1\\ ϵA_3{D}^{-1}T-{ϵ}^2{\Delta }_{2}T& {ϵ}^4{\Delta }_{2}T{\Delta }_1-{ϵ}^3{\Delta }_{2}T{D}^{-1}{A}_2-{ϵ}^3{A}_3{D}^{-1}T{\Delta }_1+{ϵ}^2{A}_3{D}^{-1}T{D}^{-1}{A}_2\end{array}\right]P^{-1}\hfill \\ \hfill & -P\left[\begin{array}{cc}{D}^{-1}-ϵD^{-1}{A}_1{D}^{-1}& ϵD^{-2}{A}_2\\ ϵA_3{D}^{-2}& 0\end{array}\right]P^{-1}\hfill \\ \hfill =& P\left[\begin{array}{cc}T-D^{-1}+ϵD^{-1}{A}_1{D}^{-1}& ϵT{D}^{-1}{A}_2-{ϵ}^{2}T{\Delta }_1-ϵD^{-2}{A}_2\\ ϵA_3{D}^{-1}T-{ϵ}^2{\Delta }_{2}T-ϵA_3{D}^{-2}& {ϵ}^2{A}_3{D}^{-1}T{D}^{-1}{A}_2-{ϵ}^3{A}_3{D}^{-1}T{\Delta }_1-{ϵ}^3{\Delta }_{2}T{D}^{-1}{A}_2+{ϵ}^4{\Delta }_{2}T{\Delta }_1\end{array}\right]P^{-1},\hfill \end{array}$$

where T, ${\Delta }_1$, and ${\Delta }_2$ are as shown in Theorem 12.

Subtracting $T={\left(I_r+{ϵ}^{2}M\right)}^{-1}{\left(D+ϵA_1\right)}^{-1}{\left(I_r+{ϵ}^{2}M\right)}^{-1}$ into the above formula, it can be reduced to (47), in which M has the form as in (40). We obtain

$${\parallel {\widehat{A}}^{\#}-{\widehat{A}}^{\mathrm{DGGI}}\parallel }_P=\sqrt{{∥N_1∥}_F^2+{∥N_2∥}_F^2+{∥N_3∥}_F^2+{∥N_4∥}_F^2}.$$

Moreover,

$$\begin{array}{cc}\hfill {\left(I_r+{ϵ}^{2}M\right)}^{-1}& =I_r-{ϵ}^{2}M+{ϵ}^4{M}^2-\cdots +O\left({ϵ}^{2n}{M}^n\right),\hfill \\ \hfill {\left(I_r+ϵD^{-1}{A}_1\right)}^{-1}& =I_r-ϵD^{-1}{A}_1+{ϵ}^2{\left(D^{-1}{A}_1\right)}^2-\cdots +O\left({ϵ}^n{\left(D^{-1}{A}_1\right)}^n\right).\hfill \end{array}$$

By substituting the above two formulas into $N_1$, we can achieve

$$\begin{array}{cc}\hfill N_1=& {\left(I_r+{ϵ}^{2}M\right)}^{-1}{\left(D+ϵA_1\right)}^{-1}{\left(I_r+{ϵ}^{2}M\right)}^{-1}-D^{-1}+ϵD^{-1}{A}_1{D}^{-1}\hfill \\ \hfill =& {\left(I_r+{ϵ}^{2}M\right)}^{-1}{\left(I_r+ϵD^{-1}{A}_1\right)}^{-1}{D}^{-1}{\left(I_r+{ϵ}^{2}M\right)}^{-1}-D^{-1}+ϵD^{-1}{A}_1{D}^{-1}\hfill \\ \hfill =& D^{-1}-ϵD^{-1}{A}_1{D}^{-1}+{ϵ}^2{\left(D^{-1}{A}_1\right)}^2{D}^{-1}-\cdots +O\left({ϵ}^n{\left(D^{-1}{A}_1\right)}^n{D}^{-1}\right)\hfill \\ \hfill & -{ϵ}^{2}M{D}^{-1}+{ϵ}^{3}M{D}^{-1}{A}_1{D}^{-1}-{ϵ}^{4}M{\left(D^{-1}{A}_1\right)}^2{D}^{-1}+\cdots -O\left({ϵ}^{n+2}M{\left(D^{-1}{A}_1\right)}^n{D}^{-1}\right)\hfill \\ \hfill & +{ϵ}^4{M}^2{D}^{-1}-{ϵ}^5{M}^2{D}^{-1}{A}_1{D}^{-1}+{ϵ}^6{M}^2{\left(D^{-1}{A}_1\right)}^2{D}^{-1}-\cdots +O\left({ϵ}^{n+4}{M}^2{\left(D^{-1}{A}_1\right)}^n{D}^{-1}\right)\hfill \\ \hfill & -\cdots +O\left({ϵ}^{2n}{M}^n{D}^{-1}\right)-O\left({ϵ}^{2n+1}{M}^n{D}^{-1}{A}_1{D}^{-1}\right)+O\left({ϵ}^{2n+2}{M}^n{\left(D^{-1}{A}_1\right)}^2{D}^{-1}\right)\hfill \\ \hfill & -\cdots +O\left({ϵ}^{3n}{M}^n{\left(D^{-1}{A}_1\right)}^n{D}^{-1}\right)\hfill \\ \hfill & -{ϵ}^2{D}^{-1}M+{ϵ}^3{D}^{-1}{A}_1{D}^{-1}M-{ϵ}^4{\left(D^{-1}{A}_1\right)}^2{D}^{-1}M+\cdots -O\left({ϵ}^{n+2}{\left(D^{-1}{A}_1\right)}^n{D}^{-1}M\right)\hfill \\ \hfill & +{ϵ}^{4}M{D}^{-1}M-{ϵ}^{5}M{D}^{-1}{A}_1{D}^{-1}M+{ϵ}^{6}M{\left(D^{-1}{A}_1\right)}^2{D}^{-1}M-\cdots \hfill \\ \hfill & +O\left({ϵ}^{n+4}M{\left(D^{-1}{A}_1\right)}^n{D}^{-1}M\right)\hfill \\ \hfill & -{ϵ}^6{M}^2{D}^{-1}M+{ϵ}^7{M}^2{D}^{-1}{A}_1{D}^{-1}M-{ϵ}^8{M}^2{\left(D^{-1}{A}_1\right)}^2{D}^{-1}M+\cdots -O\left({ϵ}^{n+6}{M}^2{\left(D^{-1}{A}_1\right)}^n{D}^{-1}M\right)\hfill \\ \hfill & +\cdots -O\left({ϵ}^{2n+2}{M}^n{D}^{-1}M\right)+O\left({ϵ}^{2n+3}{M}^n{D}^{-1}{A}_1{D}^{-1}M\right)-O\left({ϵ}^{2n+4}{M}^n{\left(D^{-1}{A}_1\right)}^2{D}^{-1}M\right)\hfill \\ \hfill & +\cdots -O\left({ϵ}^{3n+2}{M}^n{\left(D^{-1}{A}_1\right)}^n{D}^{-1}M\right)\hfill \\ \hfill & +{ϵ}^4{D}^{-1}{M}^2-{ϵ}^5{D}^{-1}{A}_1{D}^{-1}{M}^2+{ϵ}^6{\left(D^{-1}{A}_1\right)}^2{D}^{-1}{M}^2-\cdots +O\left({ϵ}^{n+4}{\left(D^{-1}{A}_1\right)}^n{D}^{-1}{M}^2\right)\hfill \\ \hfill & -{ϵ}^{6}M{D}^{-1}{M}^2+{ϵ}^{7}M{D}^{-1}{A}_1{D}^{-1}{M}^2-{ϵ}^{8}M{\left(D^{-1}{A}_1\right)}^2{D}^{-1}{M}^2+\cdots -O\left({ϵ}^{n+6}M{\left(D^{-1}{A}_1\right)}^n{D}^{-1}{M}^2\right)\hfill \\ \hfill & +{ϵ}^8{M}^2{D}^{-1}{M}^2-{ϵ}^9{M}^2{D}^{-1}{A}_1{D}^{-1}{M}^2+{ϵ}^{10}{M}^2{\left(D^{-1}{A}_1\right)}^2{D}^{-1}{M}^2-\cdots +O\left({ϵ}^{n+8}{M}^2{\left(D^{-1}{A}_1\right)}^n{D}^{-1}{M}^2\right)\hfill \\ \hfill & -\cdots +O\left({ϵ}^{2n+4}{M}^n{D}^{-1}{M}^2\right)-O\left({ϵ}^{2n+5}{M}^n{D}^{-1}{A}_1{D}^{-1}{M}^2\right)+O\left({ϵ}^{2n+6}{M}^n{\left(D^{-1}{A}_1\right)}^2{D}^{-1}{M}^2\right)\hfill \\ \hfill & -\cdots +O\left({ϵ}^{3n+4}{M}^n{\left(D^{-1}{A}_1\right)}^n{D}^{-1}{M}^2\right)\hfill \\ \hfill & -\cdots +O\left({ϵ}^{2n}{D}^{-1}{M}^n\right)-O\left({ϵ}^{2n+1}{D}^{-1}{A}_1{D}^{-1}{M}^n\right)+O\left({ϵ}^{2n+2}{\left(D^{-1}{A}_1\right)}^2{D}^{-1}{M}^n\right)\hfill \\ \hfill & -\cdots +O\left({ϵ}^{3n}{\left(D^{-1}{A}_1\right)}^n{D}^{-1}{M}^n\right)\hfill \\ \hfill & -O\left({ϵ}^{2n+2}M{D}^{-1}{M}^n\right)+O\left({ϵ}^{2n+3}M{D}^{-1}{A}_1{D}^{-1}{M}^n\right)-O\left({ϵ}^{2n+4}M{\left(D^{-1}{A}_1\right)}^2{D}^{-1}{M}^n\right)\hfill \\ \hfill & +\cdots -O\left({ϵ}^{3n+2}M{\left(D^{-1}{A}_1\right)}^n{D}^{-1}{M}^n\right)\hfill \\ \hfill & +O\left({ϵ}^{2n+4}{M}^2{D}^{-1}{M}^n\right)-O\left({ϵ}^{2n+5}{M}^2{D}^{-1}{A}_1{D}^{-1}{M}^n\right)+O\left({ϵ}^{2n+6}{M}^2{\left(D^{-1}{A}_1\right)}^2{D}^{-1}{M}^n\right)\hfill \\ \hfill & -\cdots +O\left({ϵ}^{3n+4}{M}^2{\left(D^{-1}{A}_1\right)}^n{D}^{-1}{M}^n\right)\hfill \\ \hfill & -\cdots +O\left({ϵ}^{4n}{M}^n{D}^{-1}{M}^n\right)-O\left({ϵ}^{4n+1}{M}^n{D}^{-1}{A}_1{D}^{-1}{M}^n\right)+O\left({ϵ}^{4n+2}{M}^n{\left(D^{-1}{A}_1\right)}^2{D}^{-1}{M}^n\right)\hfill \\ \hfill & -\cdots +O\left({ϵ}^{5n}{M}^n{\left(D^{-1}{A}_1\right)}^n{D}^{-1}{M}^n\right)-D^{-1}+ϵD^{-1}{A}_1{D}^{-1}.\hfill \end{array}$$

Then, ${∥N_1∥}_F^2={∥T-D^{-1}+ϵD^{-1}{A}_1{D}^{-1}∥}_F^2\in O\left({ϵ}^4\right)$. Similarly, we have

$${∥N_2∥}_F\in O\left({ϵ}^2\right),{∥N_3∥}_F\in O\left({ϵ}^2\right),{∥N_4∥}_F\in O\left({ϵ}^2\right).$$

Thus, (49) holds. □

Example 2.

Let

$$\begin{array}{c}\hfill \widehat{A}=A+ϵA_0=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& -1& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]+ϵ\left[\begin{array}{cccc}1& 0& 1& 0\\ 0& 2& -2& 0\\ 0& 1& -1& 0\\ 0& 0& 0& 0\end{array}\right],\end{array}$$

where the invertible matrix $P=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 1& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$ such that the Jordan canonical form of A is

$$\begin{array}{c}\hfill P^{-1}AP=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right].\end{array}$$

(50)

According to Theorem 12, we can calculate

$$\begin{array}{cc}\hfill {\widehat{A}}^{\#}-{\widehat{A}}^{\mathrm{DGGI}}& =\left[\begin{array}{cccc}{\displaystyle \frac{1}{1+ϵ}}& {\displaystyle \frac{-2{ϵ}^2}{{\left(1+ϵ\right)}^3}}& {\displaystyle \frac{3{ϵ}^2+ϵ}{{\left(1+ϵ\right)}^3}}& 0\\ 0& {\displaystyle \frac{1+2ϵ}{{\left(1+ϵ\right)}^2}}& {\displaystyle \frac{-1-2ϵ}{{\left(1+ϵ\right)}^2}}& 0\\ 0& {\displaystyle \frac{ϵ}{{\left(1+ϵ\right)}^2}}& {\displaystyle \frac{-ϵ}{{\left(1+ϵ\right)}^2}}& 0\\ 0& 0& 0& 0\end{array}\right]-\left[\begin{array}{cccc}1-ϵ& 0& ϵ& 0\\ 0& 1& -1& 0\\ 0& ϵ& -ϵ& 0\\ 0& 0& 0& 0\end{array}\right]\hfill \\ \hfill & =\left[\begin{array}{cccc}{\displaystyle \frac{{ϵ}^2}{1+ϵ}}& {\displaystyle \frac{-2{ϵ}^2}{{\left(1+ϵ\right)}^3}}& {\displaystyle \frac{-{ϵ}^3\left(3+ϵ\right)}{{\left(1+ϵ\right)}^3}}& 0\\ 0& {\displaystyle \frac{-{ϵ}^2}{{\left(1+ϵ\right)}^2}}& {\displaystyle \frac{{ϵ}^2}{{\left(1+ϵ\right)}^2}}& 0\\ 0& {\displaystyle \frac{-{ϵ}^2\left(2+ϵ\right)}{{\left(1+ϵ\right)}^2}}& {\displaystyle \frac{{ϵ}^2\left(2+ϵ\right)}{{\left(1+ϵ\right)}^2}}& 0\\ 0& 0& 0& 0\end{array}\right].\hfill \end{array}$$

Furthermore,

$$\begin{array}{cc}\hfill {\parallel {\widehat{A}}^{\#}-{\widehat{A}}^{\mathrm{DGGI}}\parallel }_P& ={∥\left[\begin{array}{cccc}{\displaystyle \frac{{ϵ}^2}{1+ϵ}}& {\displaystyle \frac{-2{ϵ}^2}{{\left(1+ϵ\right)}^3}}& {\displaystyle \frac{-{ϵ}^2\left(2+ϵ\right)}{{\left(1+ϵ\right)}^2}}& 0\\ 0& {\displaystyle \frac{{ϵ}^2}{1+ϵ}}& 0& 0\\ 0& {\displaystyle \frac{-{ϵ}^2\left(2+ϵ\right)}{{\left(1+ϵ\right)}^2}}& 0& 0\\ 0& 0& 0& 0\end{array}\right]∥}_F\hfill \\ \hfill & ={\displaystyle \frac{\sqrt{2}{ϵ}^2\sqrt{{\left(1+ϵ\right)}^4+{\left(1+ϵ\right)}^2{\left(2+ϵ\right)}^2+2}}{{\left(1+ϵ\right)}^3}}\in O\left({ϵ}^2\right).\hfill \end{array}$$

According to Theorem 13, we obtain

$$\begin{array}{cc}\hfill & {\Delta }_1=\left[\begin{array}{cc}{\displaystyle \frac{1}{1+ϵ}}& 0\\ 0& 0\end{array}\right],{\Delta }_2=\left[\begin{array}{cc}0& {\displaystyle \frac{1}{1+ϵ}}\\ 0& 0\end{array}\right],M=\left[\begin{array}{cc}0& {\displaystyle \frac{1}{{(1+ϵ)}^2}}\\ 0& 0\end{array}\right],{\left(I+{ϵ}^{2}M\right)}^{-1}=\left[\begin{array}{cc}1& {\displaystyle \frac{-{ϵ}^2}{{(1+ϵ)}^2}}\\ 0& 1\end{array}\right]\hfill \\ \hfill & N_1=\left[\begin{array}{cc}{\displaystyle \frac{{ϵ}^2}{1+ϵ}}& {\displaystyle \frac{-2{ϵ}^2}{{\left(1+ϵ\right)}^3}}\\ 0& {\displaystyle \frac{{ϵ}^2}{1+ϵ}}\end{array}\right],N_2=\left[\begin{array}{cc}{\displaystyle \frac{-{ϵ}^2\left(2+ϵ\right)}{{\left(1+ϵ\right)}^2}}& 0\\ 0& 0\end{array}\right],N_3=\left[\begin{array}{cc}0& {\displaystyle \frac{-{ϵ}^2\left(2+ϵ\right)}{{\left(1+ϵ\right)}^2}}\\ 0& 0\end{array}\right],N_4=0.\hfill \end{array}$$

Then, ${∥N_1∥}_F=\sqrt{{\displaystyle \frac{2{ϵ}^4}{{\left(1+ϵ\right)}^2}}+{\displaystyle \frac{4{ϵ}^4}{{\left(1+ϵ\right)}^6}}}$, ${∥N_2∥}_F=\sqrt{{\displaystyle \frac{{ϵ}^4{\left(2+ϵ\right)}^2}{{\left(1+ϵ\right)}^4}}}$ and ${∥N_3∥}_F=\sqrt{{\displaystyle \frac{{ϵ}^4{\left(2+ϵ\right)}^2}{{\left(1+ϵ\right)}^4}}}$. Therefore,

$$\begin{array}{cc}\hfill & \sqrt{{∥N_1∥}_F^2+{∥N_2∥}_F^2+{∥N_3∥}_F^2+{∥N_4∥}_F^2}={\displaystyle \frac{\sqrt{2}{ϵ}^2\sqrt{{\left(1+ϵ\right)}^4+{\left(1+ϵ\right)}^2{\left(2+ϵ\right)}^2+2}}{{\left(1+ϵ\right)}^3}}\in O\left({ϵ}^2\right).\hfill \end{array}$$

5. Symmetric Expressions for the Perturbation of Group Inverse under Range/Null Space Conditions

In this section, we give symmetric expressions for the group inverse of $\widehat{A}$ under the range or the null space conditions. We find that the error between ${\widehat{A}}^{\#}$ and ${\widehat{A}}^{\mathrm{DGGI}}$ is an infinitesimal quantity of ${ϵ}^2$ by using the matrix P-norm, where matrix P is nonsingular. Meanwhile, we give some examples to illustrate these results.

Based on Lemma 3 and Theorem 12, we have the following conclusions.

Corollary 6.

Let $\widehat{A}=A+ϵA_0\in {\mathbb{R}}^{n\times n}$ with $\mathrm{Ind}\left(A\right)=1$.

(1) If $I_n+ϵA_0{A}^{\#}$ is nonsingular and $A_0=A_{0}A{A}^{\#}$, then

$$\begin{array}{cc}\hfill {\widehat{A}}^{\#}& =A^{\#}-ϵA^{\#}{A}_0{A}^{\#}{\left(I_n+ϵA_0{A}^{\#}\right)}^{-1}-\left(I_n-A{A}^{\#}\right){\left(I_n+ϵA_0{A}^{\#}\right)}^{-1}{A}^{\#}{\left(I_n+ϵA_0{A}^{\#}\right)}^{-1}\hfill \\ \hfill & =P\left[\begin{array}{cc}{D}^{-1}& 0\\ 0& 0\end{array}\right]P^{-1}-P\left[\begin{array}{cc}ϵD^{-1}{A}_1{\left(D+ϵA_1\right)}^{-1}& 0\\ 0& 0\end{array}\right]P^{-1}-P\left[\begin{array}{cc}0& 0\\ -ϵA_3{\left(D+ϵA_1\right)}^{-2}& 0\end{array}\right]P^{-1}\hfill \\ \hfill & =P\left[\begin{array}{cc}{D}^{-1}-ϵD^{-1}{A}_1{\left(D+ϵA_1\right)}^{-1}& 0\\ ϵA_3{\left(D+ϵA_1\right)}^{-2}& 0\end{array}\right]P^{-1},\hfill \end{array}$$

where D, $A_1$, and $A_3$ are as shown in Theorem 12.

(2) If $I_n+ϵA^{\#}{A}_0$ is nonsingular and $A_0=A{A}^{\#}{A}_0$, then

$$\begin{array}{cc}\hfill {\widehat{A}}^{\#}& =A^{\#}-ϵ{\left(I_n+ϵA^{\#}{A}_0\right)}^{-1}{A}^{\#}{A}_0{A}^{\#}-{\left(I_n+ϵA^{\#}{A}_0\right)}^{-1}{A}^{\#}{\left(I_n+ϵA^{\#}{A}_0\right)}^{-1}\left(I_n-A{A}^{\#}\right)\hfill \\ \hfill & =P\left[\begin{array}{cc}{D}^{-1}& 0\\ 0& 0\end{array}\right]P^{-1}-P\left[\begin{array}{cc}ϵ{\left(D+ϵA_1\right)}^{-1}{A}_1{D}^{-1}& 0\\ 0& 0\end{array}\right]P^{-1}-P\left[\begin{array}{cc}0& -ϵ{\left(D+ϵA_1\right)}^{-2}{A}_2\\ 0& 0\end{array}\right]P^{-1}\hfill \\ \hfill & =P\left[\begin{array}{cc}{D}^{-1}-ϵ{\left(D+ϵA_1\right)}^{-1}{A}_1{D}^{-1}& ϵ{\left(D+ϵA_1\right)}^{-2}{A}_2\\ 0& 0\end{array}\right]P^{-1},\hfill \end{array}$$

where D, $A_1$, and $A_2$ are as shown in Theorem 12.

(3) If $\mathrm{Rank}\left(\widehat{A}\right)=\mathrm{Rank}\left(A\right)$ and $A{A}^{\#}{A}_0=A_0=A_{0}A{A}^{\#}$, then

$$\begin{array}{cc}\hfill {\widehat{A}}^{\#}& ={\left(I_n+ϵA^{\#}{A}_0\right)}^{-1}{A}^{\#}=A^{\#}{\left(I_n+ϵA_0{A}^{\#}\right)}^{-1}=P\left[\begin{array}{cc}{\left(D+ϵA_1\right)}^{-1}& 0\\ 0& 0\end{array}\right]P^{-1},\hfill \end{array}$$

where D and $A_1$ are as shown in Theorem 12.

Corollary 7.

Let $\widehat{A}=A+ϵA_0\in {\mathbb{R}}^{n\times n}$ with $\mathrm{Ind}\left(A\right)=1$, and the DGGI of $\widehat{A}$ exist.

(1) If $I_n+ϵA_0{A}^{\#}$ is nonsingular and $A_0=A_{0}A{A}^{\#}$, then

$$\begin{array}{c}\hfill {\widehat{A}}^{\#}-{\widehat{A}}^{\mathrm{DGGI}}=P\left[\begin{array}{cc}{N}_1& N_2\\ N_3& 0\end{array}\right]P^{-1},\end{array}$$

(51)

where

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{N}_1=ϵD^{-1}{A}_1{D}^{-1}-ϵD^{-1}{A}_1{\left(D+ϵA_1\right)}^{-1}\hfill \\ N_2=-ϵD^{-2}{A}_2\hfill \\ N_3=ϵA_3{\left(D+ϵA_1\right)}^{-2}-ϵA_3{D}^{-2},\hfill \end{array}\right.\end{array}$$

(52)

in which D, $A_1$, $A_2$, and $A_3$ are as shown in Theorem 12, $N_1\in {\mathbb{R}}^{r\times r}$, $N_2\in {\mathbb{R}}^{r\times (n-r)}$ and $N_3\in {\mathbb{R}}^{(n-r)\times r}$.

Furthermore,

$$\begin{array}{c}\hfill {\parallel {\widehat{A}}^{\#}-{\widehat{A}}^{\mathrm{DGGI}}\parallel }_P=\sqrt{{∥N_1∥}_F^2+{∥N_2∥}_F^2+{∥N_3∥}_F^2}\in O\left({ϵ}^2\right).\end{array}$$

(53)

(2) If $I_n+ϵA^{\#}{A}_0$ is nonsingular and $A_0=A{A}^{\#}{A}_0$, then

$$\begin{array}{c}\hfill {\widehat{A}}^{\#}-{\widehat{A}}^{\mathrm{DGGI}}=P\left[\begin{array}{cc}{N}_1& N_2\\ N_3& 0\end{array}\right]P^{-1},\end{array}$$

(54)

where

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{N}_1=ϵD^{-1}{A}_1{D}^{-1}-ϵ{\left(D+ϵA_1\right)}^{-1}{A}_1{D}^{-1}\hfill \\ N_2=ϵ{\left(D+ϵA_1\right)}^{-2}{A}_2-ϵD^{-2}{A}_2\hfill \\ N_3=-ϵA_3{D}^{-2},\hfill \end{array}\right.\end{array}$$

(55)

in which D, $A_1$, $A_2$, and $A_3$ are as shown in Theorem 12, $N_1\in {\mathbb{R}}^{r\times r}$, $N_2\in {\mathbb{R}}^{r\times (n-r)}$ and $N_3\in {\mathbb{R}}^{(n-r)\times r}$.

Furthermore,

$$\begin{array}{c}\hfill {\parallel {\widehat{A}}^{\#}-{\widehat{A}}^{\mathrm{DGGI}}\parallel }_P=\sqrt{{∥N_1∥}_F^2+{∥N_2∥}_F^2+{∥N_3∥}_F^2}\in O\left({ϵ}^2\right).\end{array}$$

(56)

(3) If $\mathrm{Rank}\left(\widehat{A}\right)=\mathrm{rank}\left(A\right)$ and $A{A}^{\#}{A}_0=A_0=A_{0}A{A}^{\#}$, then

$$\begin{array}{c}\hfill {\widehat{A}}^{\#}-{\widehat{A}}^{\mathrm{DGGI}}=P\left[\begin{array}{cc}{N}_1& N_2\\ N_3& 0\end{array}\right]P^{-1},\end{array}$$

(57)

where

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{N}_1={\left(D+ϵA_1\right)}^{-1}-D^{-1}+ϵD^{-1}{A}_1{D}^{-1}\hfill \\ N_2=-ϵD^{-2}{A}_2\hfill \\ N_3=-ϵA_3{D}^{-2},\hfill \end{array}\right.\end{array}$$

(58)

in which D, $A_1$, $A_2$, and $A_3$ are as shown in Theorem 12, $N_1\in {\mathbb{R}}^{r\times r}$, $N_2\in {\mathbb{R}}^{r\times (n-r)}$, and $N_3\in {\mathbb{R}}^{(n-r)\times r}$.

Furthermore,

$$\begin{array}{c}\hfill {\parallel {\widehat{A}}^{\#}-{\widehat{A}}^{\mathrm{DGGI}}\parallel }_P=\sqrt{{∥N_1∥}_F^2+{∥N_2∥}_F^2+{∥N_3∥}_F^2}\in O\left({ϵ}^2\right).\end{array}$$

(59)

Example 3.

Let

$$\begin{array}{c}\hfill \widehat{A}=A+ϵA_0=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& -1& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]+ϵ\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 2& -2& 0\\ 0& 1& -1& 0\\ 0& 0& 0& 0\end{array}\right],\end{array}$$

where the Jordan canonical form of A is shown in (50).

According to $\left(1\right)$ of Corollaries 6 and 7, we can verify

$$\begin{array}{c}\hfill {\left(I_4+ϵA_0{A}^{\#}\right)}^{-1}={\left[\begin{array}{cccc}1+ϵ& 0& 0& 0\\ 0& 1+2ϵ& -2ϵ& 0\\ 0& ϵ& 1-ϵ& 0\\ 0& 0& 0& 1\end{array}\right]}^{-1}=\left[\begin{array}{cccc}{\displaystyle \frac{1}{1+ϵ}}& 0& 0& 0\\ 0& {\displaystyle \frac{1-ϵ}{1+ϵ}}& {\displaystyle \frac{2ϵ}{1+ϵ}}& 0\\ 0& {\displaystyle \frac{-ϵ}{1+ϵ}}& {\displaystyle \frac{1+2ϵ}{1+ϵ}}& 0\\ 0& 0& 0& 1\end{array}\right],\end{array}$$

then

$$\begin{array}{cc}\hfill {\widehat{A}}^{\#}& =A^{\#}-ϵA^{\#}{A}_0{A}^{\#}{\left(I_4+ϵA_0{A}^{\#}\right)}^{-1}-\left(I_4-A{A}^{\#}\right){\left(I_4+ϵA_0{A}^{\#}\right)}^{-1}{A}^{\#}{\left(I_4+ϵA_0{A}^{\#}\right)}^{-1}\hfill \\ \hfill & =\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& -1& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]-\left[\begin{array}{cccc}{\displaystyle \frac{ϵ}{1+ϵ}}& 0& 0& 0\\ 0& {\displaystyle \frac{ϵ}{1+ϵ}}& {\displaystyle \frac{-ϵ}{1+ϵ}}& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]-\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& {\displaystyle \frac{-ϵ}{{(1+ϵ)}^2}}& {\displaystyle \frac{ϵ}{{(1+ϵ)}^2}}& 0\\ 0& {\displaystyle \frac{-ϵ}{{(1+ϵ)}^2}}& {\displaystyle \frac{ϵ}{{(1+ϵ)}^2}}& 0\\ 0& 0& 0& 0\end{array}\right]\hfill \\ \hfill & =\left[\begin{array}{cccc}{\displaystyle \frac{1}{1+ϵ}}& 0& 0& 0\\ 0& {\displaystyle \frac{1+2ϵ}{{(1+ϵ)}^2}}& {\displaystyle \frac{-1-2ϵ}{{(1+ϵ)}^2}}& 0\\ 0& {\displaystyle \frac{ϵ}{{(1+ϵ)}^2}}& {\displaystyle \frac{-ϵ}{{(1+ϵ)}^2}}& 0\\ 0& 0& 0& 0\end{array}\right]\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\widehat{A}}^{\#}-{\widehat{A}}^{\mathrm{DGGI}}& =\left[\begin{array}{cccc}{\displaystyle \frac{1}{1+ϵ}}& 0& 0& 0\\ 0& {\displaystyle \frac{1+2ϵ}{{(1+ϵ)}^2}}& {\displaystyle \frac{-1-2ϵ}{{(1+ϵ)}^2}}& 0\\ 0& {\displaystyle \frac{ϵ}{{(1+ϵ)}^2}}& {\displaystyle \frac{-ϵ}{{(1+ϵ)}^2}}& 0\\ 0& 0& 0& 0\end{array}\right]-\left[\begin{array}{cccc}1-ϵ& 0& 0& 0\\ 0& 1& -1& 0\\ 0& ϵ& -ϵ& 0\\ 0& 0& 0& 0\end{array}\right]\hfill \\ \hfill & =\left[\begin{array}{cccc}{\displaystyle \frac{{ϵ}^2}{1+ϵ}}& 0& 0& 0\\ 0& {\displaystyle \frac{-{ϵ}^2}{{\left(1+ϵ\right)}^2}}& {\displaystyle \frac{{ϵ}^2}{{\left(1+ϵ\right)}^2}}& 0\\ 0& {\displaystyle \frac{-{ϵ}^3-2{ϵ}^2}{{\left(1+ϵ\right)}^2}}& {\displaystyle \frac{{ϵ}^3+2{ϵ}^2}{{\left(1+ϵ\right)}^2}}& 0\\ 0& 0& 0& 0\end{array}\right].\hfill \end{array}$$

Therefore,

$$\begin{array}{cc}\hfill {\parallel {\widehat{A}}^{\#}-{\widehat{A}}^{\mathrm{DGGI}}\parallel }_P& ={∥\left[\begin{array}{cccc}{\displaystyle \frac{{ϵ}^2}{1+ϵ}}& 0& 0& 0\\ 0& {\displaystyle \frac{{ϵ}^2}{1+ϵ}}& 0& 0\\ 0& {\displaystyle \frac{-{ϵ}^3-2{ϵ}^2}{{(1+ϵ)}^2}}& 0& 0\\ 0& 0& 0& 0\end{array}\right]∥}_F={\displaystyle \frac{{ϵ}^2\sqrt{2{(1+ϵ)}^2+{(2+ϵ)}^2}}{{(1+ϵ)}^2}},\hfill \end{array}$$

and we can obtain $N_1=\left[\begin{array}{cc}{\displaystyle \frac{{ϵ}^2}{1+ϵ}}& 0\\ 0& {\displaystyle \frac{{ϵ}^2}{1+ϵ}}\end{array}\right],N_2=0,N_3=\left[\begin{array}{cc}0& {\displaystyle \frac{-{ϵ}^3-2{ϵ}^2}{{(1+ϵ)}^2}}\\ 0& 0\end{array}\right]$. Then,

$$\begin{array}{c}\hfill \sqrt{{∥N_1∥}_F^2+{∥N_2∥}_F^2+{∥N_3∥}_F^2}={\displaystyle \frac{{ϵ}^2\sqrt{2{(1+ϵ)}^2+{(2+ϵ)}^2}}{{(1+ϵ)}^2}}\in O\left({ϵ}^2\right).\end{array}$$

Example 4.

Let

$$\begin{array}{c}\hfill \widehat{A}=A+ϵA_0=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& -1& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]+ϵ\left[\begin{array}{cccc}1& 0& 1& 0\\ 0& 1& -1& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right],\end{array}$$

where the Jordan canonical form of A is shown in (50).

According to $\left(2\right)$ of Corollaries 6 and 7, we can achieve

$$\begin{array}{c}\hfill {\left(I_4+ϵA^{\#}{A}_0\right)}^{-1}={\left[\begin{array}{cccc}1+ϵ& 0& ϵ& 0\\ 0& 1+ϵ& -ϵ& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]}^{-1}=\left[\begin{array}{cccc}{\displaystyle \frac{1}{1+ϵ}}& 0& {\displaystyle \frac{-ϵ}{1+ϵ}}& 0\\ 0& {\displaystyle \frac{1}{1+ϵ}}& {\displaystyle \frac{ϵ}{1+ϵ}}& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right],\end{array}$$

then

$$\begin{array}{cc}\hfill {\widehat{A}}^{\#}& =A^{\#}-ϵ{\left(I_4+ϵA^{\#}{A}_0\right)}^{-1}{A}^{\#}{A}_0{A}^{\#}-{\left(I_4+ϵA^{\#}{A}_0\right)}^{-1}{A}^{\#}{\left(I_4+ϵA^{\#}{A}_0\right)}^{-1}\left(I_4-A{A}^{\#}\right)\hfill \\ \hfill & =\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& -1& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]-\left[\begin{array}{cccc}{\displaystyle \frac{ϵ}{1+ϵ}}& 0& 0& 0\\ 0& {\displaystyle \frac{ϵ}{1+ϵ}}& {\displaystyle \frac{-ϵ}{1+ϵ}}& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]-\left[\begin{array}{cccc}0& 0& {\displaystyle \frac{-ϵ}{{(1+ϵ)}^2}}& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]\hfill \\ \hfill & =\left[\begin{array}{cccc}{\displaystyle \frac{1}{1+ϵ}}& 0& {\displaystyle \frac{ϵ}{{(1+ϵ)}^2}}& 0\\ 0& {\displaystyle \frac{1}{1+ϵ}}& {\displaystyle \frac{-1}{1+ϵ}}& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\widehat{A}}^{\#}-{\widehat{A}}^{\mathrm{DGGI}}& =\left[\begin{array}{cccc}{\displaystyle \frac{1}{1+ϵ}}& 0& {\displaystyle \frac{ϵ}{{(1+ϵ)}^2}}& 0\\ 0& {\displaystyle \frac{1}{1+ϵ}}& {\displaystyle \frac{-1}{1+ϵ}}& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]-\left[\begin{array}{cccc}1-ϵ& 0& ϵ& 0\\ 0& 1-ϵ& -1+ϵ& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]\hfill \\ \hfill & =\left[\begin{array}{cccc}{\displaystyle \frac{{ϵ}^2}{1+ϵ}}& 0& {\displaystyle \frac{-{ϵ}^3-2{ϵ}^2}{{\left(1+ϵ\right)}^2}}& 0\\ 0& {\displaystyle \frac{{ϵ}^2}{1+ϵ}}& {\displaystyle \frac{-{ϵ}^2}{1+ϵ}}& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right].\hfill \end{array}$$

Therefore,

$$\begin{array}{cc}\hfill {\parallel {\widehat{A}}^{\#}-{\widehat{A}}^{\mathrm{DGGI}}\parallel }_P& ={∥\left[\begin{array}{cccc}{\displaystyle \frac{{ϵ}^2}{1+ϵ}}& 0& {\displaystyle \frac{-{ϵ}^3-2{ϵ}^2}{{\left(1+ϵ\right)}^2}}& 0\\ 0& {\displaystyle \frac{{ϵ}^2}{1+ϵ}}& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]∥}_F={\displaystyle \frac{{ϵ}^2\sqrt{2{(1+ϵ)}^2+{(2+ϵ)}^2}}{{(1+ϵ)}^2}},\hfill \end{array}$$

and we can obtain $N_1=\left[\begin{array}{cc}{\displaystyle \frac{{ϵ}^2}{1+ϵ}}& 0\\ 0& {\displaystyle \frac{{ϵ}^2}{1+ϵ}}\end{array}\right],N_2=\left[\begin{array}{cc}{\displaystyle \frac{-{ϵ}^3-2{ϵ}^2}{{(1+ϵ)}^2}}& 0\\ 0& 0\end{array}\right],N_3=0$. Then,

$$\begin{array}{c}\hfill \sqrt{{∥N_1∥}_F^2+{∥N_2∥}_F^2+{∥N_3∥}_F^2}={\displaystyle \frac{{ϵ}^2\sqrt{2{(1+ϵ)}^2+{(2+ϵ)}^2}}{{(1+ϵ)}^2}}\in O\left({ϵ}^2\right).\end{array}$$

Example 5.

Let

$$\begin{array}{c}\hfill \widehat{A}=A+ϵA_0=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& -1& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]+ϵ\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& -1& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right],\end{array}$$

where the Jordan canonical form of A is shown in (50).

According to $\left(3\right)$ of Corollaries 6 and 7, we can have

$$\begin{array}{cc}\hfill {\left(I_4+ϵA^{\#}{A}_0\right)}^{-1}& ={\left[\begin{array}{cccc}1+ϵ& 0& 0& 0\\ 0& 1+ϵ& -ϵ& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]}^{-1}=\left[\begin{array}{cccc}{\displaystyle \frac{1}{1+ϵ}}& 0& 0& 0\\ 0& {\displaystyle \frac{1}{1+ϵ}}& {\displaystyle \frac{ϵ}{1+ϵ}}& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]\hfill \\ \hfill & ={\left(I_4+ϵA_0{A}^{\#}\right)}^{-1},\hfill \end{array}$$

then,

$$\begin{array}{cc}\hfill {\widehat{A}}^{\#}& ={\left(I_4+ϵA^{\#}{A}_0\right)}^{-1}{A}^{\#}=A^{\#}{\left(I_4+ϵA_0{A}^{\#}\right)}^{-1}=\left[\begin{array}{cccc}{\displaystyle \frac{1}{1+ϵ}}& 0& 0& 0\\ 0& {\displaystyle \frac{1}{1+ϵ}}& {\displaystyle \frac{-1}{1+ϵ}}& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right],\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\widehat{A}}^{\#}-{\widehat{A}}^{\mathrm{DGGI}}& =\left[\begin{array}{cccc}{\displaystyle \frac{1}{1+ϵ}}& 0& 0& 0\\ 0& {\displaystyle \frac{1}{1+ϵ}}& {\displaystyle \frac{-1}{1+ϵ}}& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]-\left[\begin{array}{cccc}1-ϵ& 0& 0& 0\\ 0& 1-ϵ& -1+ϵ& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]\hfill \\ \hfill & =\left[\begin{array}{cccc}{\displaystyle \frac{{ϵ}^2}{1+ϵ}}& 0& 0& 0\\ 0& {\displaystyle \frac{{ϵ}^2}{1+ϵ}}& {\displaystyle \frac{-{ϵ}^2}{1+ϵ}}& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right].\hfill \end{array}$$

Therefore,

$$\begin{array}{cc}\hfill {\parallel {\widehat{A}}^{\#}-{\widehat{A}}^{\mathrm{DGGI}}\parallel }_P& ={∥\left[\begin{array}{cccc}{\displaystyle \frac{{ϵ}^2}{1+ϵ}}& 0& 0& 0\\ 0& {\displaystyle \frac{{ϵ}^2}{1+ϵ}}& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]∥}_F={\displaystyle \frac{\sqrt{2}{ϵ}^2}{1+ϵ}},N_1=\left[\begin{array}{cc}{\displaystyle \frac{{ϵ}^2}{1+ϵ}}& 0\\ 0& {\displaystyle \frac{{ϵ}^2}{1+ϵ}}\end{array}\right],N_2=0,N_3=0.\hfill \end{array}$$

Then,

$$\begin{array}{c}\hfill \sqrt{{∥N_1∥}_F^2+{∥N_2∥}_F^2+{∥N_3∥}_F^2}={\displaystyle \frac{\sqrt{2}{ϵ}^2}{1+ϵ}}\in O\left({ϵ}^2\right).\end{array}$$

6. Concluding Remarks

Symmetry plays a crucial role in the study of dual matrices and dual matrix group inverses. We present the perturbation bounds of dual group generalized inverse and group inverse. It is not difficult for us to extend the matrix case to the tensor [54,55] with the help of the tensor Jordan canonical form, tensor singular value decomposition, tensor eigenvalues, tensor Schur decompostion, and tensor UTV decomposition.