The $\mathfrak{m}$-CCE Inverse in Minkowski Space and Its Applications

Abstract

In this paper, we introduce a new generalized inverse called $\mathfrak{m}$-CCE inverse which presents a generalization of the CCE inverse in Minkowski space by using the $\mathfrak{m}$-core-EP decomposition and the Minkowski inverse. We first show the existence and the uniqueness of the generalized inverse. Then, a number of basic properties and diverse characterizations are derived for the $\mathfrak{m}$-CCE inverse as well as its limit and integral expressions. Additionally, applications of the $\mathfrak{m}$-CCE inverse are given in solving a system of linear equations. Applying the generalized inverse, we introduce a binary relation based on the $\mathfrak{m}$-CCE inverse.

1. Introduction

Generalized inverses are studied in many branches of mathematical field: matrix theory, operator theory, $C^{*}$-algebras, rings, etc. Additionally, generalized inverses turn out to be a powerful tool in many applications such as Markov chains, differential equations, difference equations, electrical networks, optimal control, and so on.

Penrose [1] earliest defined the Moore–Penrose inverse using four matrix equations in 1955. Then, a wave of research on generalized inverses has begun. In 1958, Drazin [2] introduced the Drazin inverse with spectral properties in associative rings and semigroups. The Drazin inverse of a group matrix is also known as its group inverse [3]. Using the Moore–Penrose inverse and the Drazin inverse, Malik and Thome [4] defined the DMP inverse in 2014. The core inverse was firstly proposed by Baksalary and Trenkler [5] in 2010 as an alternative to the group inverse. Manjunatha Prasad and Mohana [6] then extended the core inverse in 2014 by introducing the core-EP inverse. In 2016, Wang [7] introduced the core-EP decomposition, which plays a crucial role in the investigation of generalized inverses. In 2018, Mehdipour introduced the CMP inverse by applying the Moore–Penrose inverse and the Drazin inverse. In 2020, Zuo [8] defined the CCE inverse based on the core-EP decomposition.

In studying polarized hight, Renardy [9] investigated the singular value decomposition in Minkowski space in order to quickly verify that a Mueller matrix can map the forward light cone into itself. Subsequently, the Minkowski inverse in Minkowski space was established by Meenakshi [10], who also gave a condition for a Mueller matrix to have a singular value decomposition in Minkowski space according to its Minkowski inverse. Recently, Wang and Liu defined the $\mathfrak{m}$-core inverse [11], the $\mathfrak{m}$-core-EP inverse [12], the $\mathfrak{m}$-WG inverse [13] and the $\mathfrak{m}$-WG^° inverse [14] in Minkowski space, which can be regarded as extensions of the core inverse [5], the core-EP inverse [6], the weak group inverse [15] and the m-weak group inverse [16], respectively. In 2023, Zuo [17] promoted the DMP inverse to Minkowski space; we call it the $\mathfrak{m}$-DMP inverse afterward.

Inspired by the study of the CCE inverse and the generalized inverses in Minkowski space, the intention of this paper is to introduce a new generalized inverse in Minkowski space, called the $\mathfrak{m}$-CCE inverse, and discuss its properties, characterizations, representations and applications.

2. Preliminaries

In this paper, we denote the Minkowski space by $\mathcal{M}$, which is an n-dimensional complex vector space with the metric matrix $G=$ diag $(1,-I_{n-1})$, where $I_{n-1}$ denotes the identity matrix of order $n-1$. It is easy to check that $G^{*}=G$ and $G^2=I_n$. We define that $A^0=I_n$ for $A\in {\mathbb{R}}^{n\times n}$(including for $A=0_n$), where ${\mathbb{R}}^{n\times n}$ represents the set of all real $n\times n$ matrices.

Let ${\mathbb{C}}_n$, ${\mathbb{C}}_{m\times n}$ be the sets of all complex n-dimensional vectors and complex $m\times n$ matrices. * means the conjugate transpose of a matrix. The symbols $A^{*}$, $R\left(A\right)$, $N\left(A\right)$ and $\mathrm{rank}\left(A\right)$ denote the conjugate transpose, range, null space, and rank, respectively, of $A\in {\mathbb{C}}_{m\times n}$. Let $A\left\{2\right\}=\{X\in {\mathbb{C}}_{n\times m}|XAX=X\}$. If $X\in A\left\{2\right\}$, we call X an outer inverse of A, which is denoted by $A^{\left(2\right)}$. If an outer inverse X of A satisfies $\mathcal{R}\left(X\right)=\mathcal{T}$ and $\mathcal{N}\left(X\right)=\mathit{S}$, then X is called the outer inverse with the prescribed range $\mathcal{T}$ and null space $\mathit{S}$, which is denoted by $A_{\mathcal{T},\mathit{S}}^{\left(2\right)}$. The Minkowski adjoint of matrix A is denoted by $A^{\sim }$, and it is defined as $A^{\sim }=G_2{A}^{*}{G}_1$, where $G_1$ and $G_2$ are Minkowski metric matrices with orders m and n, respectively. For a $n\times n$ matrix A, the index of A is the smallest non-negative integer k such that $\mathrm{rank}\left(A^{k+1}\right)=\mathrm{rank}\left(A^k\right)$, which is denoted as $\mathrm{Ind}\left(A\right)$. Notice that $\mathrm{Ind}\left(A\right)=0$ if and only if A is invertible. For subspaces $\mathcal{T}$ and $\mathit{S}$ satisfying their direct sum as ${\mathbb{C}}_{n\times 1}$, i.e., $\mathit{S}\oplus \mathcal{T}={\mathbb{C}}_{n\times 1}$, the projector onto $\mathcal{T}$ along $\mathit{S}$ is indicated by $P_{\mathcal{T},\mathit{S}}$.

The Moore–Penrose inverse denoted by $A^{†}=X\in {\mathbb{C}}_{n\times m}$ of $A\in {\mathbb{C}}_{m\times n}$ is the unique matrix satisfying the following matrix equations [1]

$$\left(1\right)AXA=A,\left(2\right)XAX=X,\left(3\right){\left(AX\right)}^{*}=AX,\left(4\right){\left(XA\right)}^{*}=XA.$$

The Drazin inverse denoted by $A^D=X\in {\mathbb{C}}_{n\times n}$ of $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$ is the unique matrix satisfying the following matrix equations [2]

$$\left(1\right)A^{k+1}X=A^k,\left(2\right)XAX=X,\left(3\right)AX=XA.$$

When $k\le 1$, $A^D$ is called the group inverse of A denoted by $A^{\#}$ [3].

The DMP inverse denoted by $A^{D,†}=X\in {\mathbb{C}}_{n\times n}$ of $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$ is the unique matrix satisfying the following matrix equations [4]

$$\left(1\right)XAX=X,\left(2\right)XA=A^{D}A,\left(3\right)A^{k}X=A^k{A}^{†},$$

notice that $A^{D,†}=A^{D}A{A}^{†}$ and $A^{†,D}=A^{†}A{A}^D$ represents the dual DMP (or MPD) inverse of A.

The core-EP inverse denoted by $A^{†◯}=X\in {\mathbb{C}}_{n\times n}$ of $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$ is the unique matrix satisfying the following matrix equations [6].

$$\left(1\right)XAX=X,\left(2\right)X{A}^{k+1}=A^k,\left(3\right){\left(AX\right)}^{*}=AX,\left(4\right)R\left(X\right)\subseteq R\left(A^k\right),$$

notice that $A^{†◯}=A^D{A}^k{\left(A^k\right)}^{†}$.

The CMP inverse denoted by $A^{C,†}=X\in {\mathbb{C}}_{n\times n}$ of $A\in {\mathbb{C}}_{n\times n}$ is the unique matrix satisfying the following matrix equations [18]

$$\left(1\right)XAX=X,\left(2\right)AXA=A{A}^{D}A,\left(3\right)AX=A{A}^{D}A{A}^{†},\left(4\right)XA=A^{†}A{A}^{D}A,$$

notice that $A^{C,†}=A^{†}A{A}^{D}AA{A}^{†}$.

The CCE inverse denoted by $A^{C, †◯}=X\in {\mathbb{C}}_{n\times n}$ of $A\in {\mathbb{C}}_{n\times n}$ is the unique matrix satisfying the following matrix equations [8]

$$\left(1\right)XAX=X,\left(2\right)XA=A^{†}A{A}^{†◯}A,\left(3\right)AX=A{A}^{†◯}A{A}^{†},$$

notice that $A^{C, †◯}=A^{†}A{A}^{†◯}A{A}^{†}$. Evidently, the core-EP inverse underpins the CCE inverse by supplying essential properties and acting as its building block, whereas the CCE inverse constitutes an extension of the core-EP inverse. From Theorem 3.9 of [8], we obtain that $A^{C, †◯}=A^{†◯}$ if and only if $A^k{A}^{†,D}=A^{†,D}{A}^k$, where $k=\mathrm{Ind}\left(A\right)$.

The Minkowski inverse denoted by $A^{\mathfrak{m}}=X\in {\mathbb{C}}_{n\times m}$ of $A\in {\mathbb{C}}_{m\times n}$ with $\mathrm{Ind}\left(A\right)=k$ and $\mathrm{rank}\left(A^{\sim }A{A}^{\sim }\right)=\mathrm{rank}\left(A\right)$ in Minkowski space is the unique matrix satisfying the following matrix equations [10]

$$\left(1\right)AXA=A,\left(2\right)XAX=X,\left(3\right){\left(AX\right)}^{\sim }=AX,\left(4\right){\left(XA\right)}^{\sim }=XA.$$

The $\mathfrak{m}$-DMP inverse denoted by $A^{D,\mathfrak{m}}=X\in {\mathbb{C}}_{n\times n}$ of $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$ and $\mathrm{rank}\left(A^{\sim }A{A}^{\sim }\right)=\mathrm{rank}\left(A\right)$ in Minkowski space is the unique matrix satisfying the following matrix equations [17]

$$\left(1\right)XAX=X,\left(2\right)XA=A^{D}A,\left(3\right)A^{k}X=A^k{A}^{\mathfrak{m}},$$

notice that $A^{D,\mathfrak{m}}=A^{D}A{A}^{\mathfrak{m}}$ and $A^{\mathfrak{m},D}=A^{\mathfrak{m}}A{A}^D$ represents the dual $\mathfrak{m}$-DMP (or $\mathfrak{m}$-MPD) inverse of A.

The $\mathfrak{m}$-core inverse denoted by $A^{m◯}=X\in {\mathbb{C}}_{n\times n}$ of $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=1$ and $\mathrm{rank}\left(A\right)=\mathrm{rank}\left(A^{\sim }A\right)$ in Minkowski space is the unique matrix satisfying the following matrix equations [11]

$$\left(1\right)AXA=A,\left(2\right)A{X}^2=X,\left(3\right){\left(AX\right)}^{\sim }=AX.$$

The $\mathfrak{m}$-core-EP inverse denoted by $A^{E◯}=X\in {\mathbb{C}}_{n\times n}$ of $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$ and $\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left({\left(A^k\right)}^{\sim }{A}^k\right)$ in Minkowski space is the unique matrix satisfying the following matrix equations [12]

$$\left(1\right)XAX=X,\left(2\right)X{A}^{k+1}=A^k,\left(3\right){\left(AX\right)}^{\sim }=AX,\left(4\right)R\left(X\right)\subseteq R\left(A^k\right),$$

notice that $A^{E◯}=A^k{A}^D{\left(A^k\right)}^{m◯}$.

Lemma 1 

([17]). Let $A\in {\mathbb{C}}_{m\times n}$. Then, the following conditions are equivalent:

 (1) 

$A^{\mathfrak{m}}$ exists;

 (2) 

$\mathrm{Rank}\left(A^{\sim }A\right)=\mathrm{rank}\left(A{A}^{\sim }\right)=\mathrm{rank}\left(A\right)$;

 (3) 

$\mathrm{Rank}\left(A^{\sim }A{A}^{\sim }\right)=\mathrm{rank}\left(A\right)$.

Lemma 2 

([10]). Let $A\in {\mathbb{C}}_{m\times n}$ with $\mathrm{rank}\left(A^{\sim }A{A}^{\sim }\right)=\mathrm{rank}\left(A\right)$. Then,

 (1) 

$R\left(A^{\mathfrak{m}}\right)=R\left(A^{\sim }\right)$ and $N\left(A^{\mathfrak{m}}\right)=N\left(A^{\sim }\right)$;

 (2) 

$A{A}^{\mathfrak{m}}=P_{R\left(A\right),N\left(A^{\sim }\right)}$;

 (3) 

$A^{\mathfrak{m}}A=P_{R\left(A^{\sim }\right),N\left(A\right)}$.

Lemma 3 

([19]). Let $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$, $\mathrm{rank}\left(A^{\sim }A{A}^{\sim }\right)=\mathrm{rank}\left(A\right)$. Then,

 (1) 

$R\left(A^D\right)=R\left(A^k\right)$ and $N\left(A^D\right)=N\left(A^k\right)$;

 (2) 

$A{A}^D=A^{D}A=P_{R\left(A^k\right),N\left(A^k\right)}$.

Lemma 4 

([20]). Let $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$ and $\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left({\left(A^k\right)}^{\sim }{A}^k\right)$. Then,

 (1) 

$R\left(A^{E◯}\right)=R\left(A^k\right)$ and $N\left(A^{E◯}\right)=N\left({\left(A^k\right)}^{\sim }\right)$;

 (2) 

$A{A}^{E◯}=P_{R\left(A^k\right),N\left({\left(A^k\right)}^{\sim }\right)}$.

Lemma 5 

([21]). If X is an outer inverse of A, then

$$R\left(XA\right)=R\left(X\right)andN\left(AX\right)=N\left(X\right).$$

Lemma 6 

([19]). Let $A\in {\mathbb{C}}_{n\times n}$. Let L and M be complementary subspaces of ${\mathbb{C}}_n$, i.e., $L\oplus M={\mathbb{C}}_n$.Then,

 (1) 

$P_{L,M}A=A\iff R\left(A\right)\subseteq L$;

 (2) 

$A=A{P}_{L,M}\iff M\subseteq N\left(A\right)$.

Lemma 7 

([22]). Every matrix $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{rank}\left(A\right)=r>0$ has a Hartwig–Spindelböck decomposition:

$$A=U\left(\begin{array}{cc}\mathrm{\Sigma }K& \mathrm{\Sigma }L\\ 0& 0\end{array}\right)U^{*},$$

(1)

where $U\in {\mathbb{C}}_{n\times n}$ is a unitary matrix, $\mathrm{\Sigma }=$ diag $({\sigma }_1{I}_{k_1},{\sigma }_2{I}_{k_2},\dots ,{\sigma }_t{I}_{k_t})$ is a diagonal matrix, the elements on the diagonal ${\sigma }_1>{\sigma }_2>\cdots >{\sigma }_t>0$ being the singular values of the matrix A, $k_1+k_2+\cdots +k_t=r=\mathrm{rank}\left(A\right)$, and $K\in {\mathbb{C}}_{r\times r}$ and $L\in {\mathbb{C}}_{r\times (n-r)}$ satisfy $K{K}^{*}+L{L}^{*}=I_r$.

Lemma 8 

([17]). Let $A\in {\mathbb{C}}_{n\times n}$ be given by (1), let $\mathrm{\Delta }=\left(\begin{array}{cc}K& L\end{array}\right)U^{*}GU\left(\begin{array}{c}{K}^{*}\\ L^{*}\end{array}\right)$, and let the partition of the Minkowski metric matrix $G\in {\mathbb{C}}_{n\times n}$ be

$$G=U\left(\begin{array}{cc}{G}_1& G_2\\ G_2^{*}& G_4\end{array}\right)U^{*},$$

where $G_1\in {\mathbb{C}}_{r\times r}$, $G_2\in {\mathbb{C}}_{r\times (n-r)}$ and $G_4\in {\mathbb{C}}_{(n-r)\times (n-r)}$. If Δ and $G_1$ are nonsingular, then

$$A^{\mathfrak{m}}=GU\left(\begin{array}{cc}{K}^{*}{\left(G_1\mathrm{\Sigma }\mathrm{\Delta }\right)}^{-1}& 0\\ L^{*}{\left(G_1\mathrm{\Sigma }\mathrm{\Delta }\right)}^{-1}& 0\end{array}\right)U^{*}G.$$

The $\mathfrak{m}$-core-EP inverse of A is given as [14]

$$A^{E◯}=U\left(\begin{array}{cc}{\left(\mathrm{\Sigma }K\right)}^{E◯}& 0\\ 0& 0\end{array}\right)U^{*}.$$

3. The $\mathfrak{m}$-CCE Inverse in Minkowski Space

This section defines a new generalized inverse named $\mathfrak{m}$-CCE inverse in Minkowski space. In order to do this, we recall the $\mathfrak{m}$-core-EP decomposition firstly. Then, we consider a system of equations.

Lemma 9 

([12]). Let $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$ and $\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left({\left(A^k\right)}^{\sim }{A}^k\right)=r$. Then, matrix A has an $\mathfrak{m}$-core-EP decomposition $A=A_1+A_2$, and it has the following matrix form

$$A=U\left(\begin{array}{cc}T& S\\ 0& N\end{array}\right)U^{*},$$

where $\mathrm{Ind}\left(A_1\right)⩽1$, $\mathrm{rank}\left(A_1\right)=\mathrm{rank}\left(A_1^{\sim }{A}_1\right)$, $A_2^k=0$, $A_1^{\sim }{A}_2=A_2{A}_1=0$ and $U\in {\mathbb{C}}_{n\times n}$ is unitary. Furthermore,

$$A_1=U\left(\begin{array}{cc}T& S+G_1^{-1}{G}_{2}N\\ 0& 0\end{array}\right)U^{*},A_2=U\left(\begin{array}{cc}0& -G_1^{-1}{G}_{2}N\\ 0& N\end{array}\right)U^{*},$$

where $G_1$ and $G_2$ hold the form as Lemma 8, $T\in {\mathbb{C}}_{r\times r}$ is nonsingular, $S\in {\mathbb{C}}_{r\times (n-r)}$, and $N\in {\mathbb{C}}_{(n-r)\times (n-r)}$ is nilpotent such that $N^k=0$. In addition, the $\mathfrak{m}$-core-EP decomposition of matrix A is unique and $A_1=A{A}^{E◯}A$, $A_2=A-A{A}^{E◯}A$.

Let $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$, $\mathrm{rank}\left(A^{\sim }A{A}^{\sim }\right)=\mathrm{rank}\left(A\right)$ and $\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left({\left(A^k\right)}^{\sim }{A}^k\right)$ having the decomposition as Lemma 9. Let $X\in {\mathbb{C}}_{n\times n}$. Consider the following system of equations.

$$XAX=X,AX=A_1{A}^{\mathfrak{m}},XA=A^{\mathfrak{m}}{A}_1.$$

(2)

Theorem 1. 

The system (2) has a unique solution $X=A^{\mathfrak{m}}{A}_1{A}^{\mathfrak{m}}$.

Proof. 

It is easy to check that $X=A^{\mathfrak{m}}{A}_1{A}^{\mathfrak{m}}$ satisfies the three equations in system (2). For uniqueness, we suppose that X satisfies (2). Then,

$$X=XAX=X{A}_1{A}^{\mathfrak{m}}=XA{A}^{E◯}A{A}^{\mathfrak{m}}=A^{\mathfrak{m}}{A}_1{A}^{E◯}A{A}^{\mathfrak{m}}=A^{\mathfrak{m}}A{A}^{E◯}A{A}^{E◯}A{A}^{\mathfrak{m}}=A^{\mathfrak{m}}A{A}^{E◯}A{A}^{\mathfrak{m}}=A^{\mathfrak{m}}{A}_1{A}^{\mathfrak{m}}.$$

Hence, $X=A^{\mathfrak{m}}{A}_1{A}^{\mathfrak{m}}$ is the unique solution of (2). □

Definition 1. 

Let $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$, $\mathrm{rank}\left(A^{\sim }A{A}^{\sim }\right)=\mathrm{rank}\left(A\right)$ and $\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left({\left(A^k\right)}^{\sim }{A}^k\right)$. The $\mathfrak{m}$-CCE inverse of A in Minkowski space, which is denoted by $A^{C,E◯}$, is defined as

$$A^{C,E◯}=A^{\mathfrak{m}}{A}_1{A}^{\mathfrak{m}}=A^{\mathfrak{m}}A{A}^{E◯}A{A}^{\mathfrak{m}}.$$

Example 1. 

Let

$$A=\left(\begin{array}{ccc}0& 4& -1\\ -1& 3& -1\\ -2& -2& 0\end{array}\right).$$

Then, $k=\mathrm{Ind}\left(A\right)=2$, $\mathrm{rank}\left(A\right)=\mathrm{rank}\left(A^{\sim }A{A}^{\sim }\right)=2$ and $\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left({\left(A^k\right)}^{\sim }{A}^k\right)=1$. Also,

$$A^{†}=\left(\begin{array}{ccc}-{\displaystyle \frac{1}{54}}& -{\displaystyle \frac{11}{54}}& -{\displaystyle \frac{10}{27}}\\ {\displaystyle \frac{7}{54}}& {\displaystyle \frac{5}{54}}& -{\displaystyle \frac{2}{27}}\\ -{\displaystyle \frac{1}{27}}& -{\displaystyle \frac{2}{27}}& -{\displaystyle \frac{2}{27}}\end{array}\right),A^D=\left(\begin{array}{ccc}-{\displaystyle \frac{2}{27}}& {\displaystyle \frac{14}{27}}& -{\displaystyle \frac{4}{27}}\\ -{\displaystyle \frac{1}{27}}& {\displaystyle \frac{7}{27}}& -{\displaystyle \frac{2}{27}}\\ {\displaystyle \frac{2}{27}}& -{\displaystyle \frac{14}{27}}& {\displaystyle \frac{4}{27}}\end{array}\right),A^{D,†}=\left(\begin{array}{ccc}{\displaystyle \frac{2}{9}}& {\displaystyle \frac{2}{9}}& 0\\ {\displaystyle \frac{1}{9}}& {\displaystyle \frac{1}{9}}& 0\\ -{\displaystyle \frac{2}{9}}& -{\displaystyle \frac{2}{9}}& 0\end{array}\right),$$

$$A^{†◯}=\left(\begin{array}{ccc}{\displaystyle \frac{4}{27}}& {\displaystyle \frac{2}{27}}& -{\displaystyle \frac{4}{27}}\\ {\displaystyle \frac{2}{27}}& {\displaystyle \frac{1}{27}}& -{\displaystyle \frac{2}{27}}\\ -{\displaystyle \frac{4}{27}}& -{\displaystyle \frac{2}{27}}& {\displaystyle \frac{4}{27}}\end{array}\right),A^{C,†}=\left(\begin{array}{ccc}{\displaystyle \frac{1}{6}}& {\displaystyle \frac{1}{6}}& 0\\ {\displaystyle \frac{1}{6}}& {\displaystyle \frac{1}{6}}& 0\\ 0& 0& 0\end{array}\right),A^{C, †◯}=\left(\begin{array}{ccc}{\displaystyle \frac{1}{9}}& {\displaystyle \frac{1}{18}}& -{\displaystyle \frac{1}{9}}\\ {\displaystyle \frac{1}{9}}& {\displaystyle \frac{1}{18}}& -{\displaystyle \frac{1}{9}}\\ 0& 0& 0\end{array}\right),$$

$$A^{\mathfrak{m}}=\left(\begin{array}{ccc}-{\displaystyle \frac{3}{16}}& -{\displaystyle \frac{1}{16}}& -{\displaystyle \frac{1}{2}}\\ {\displaystyle \frac{19}{16}}& -{\displaystyle \frac{15}{16}}& {\displaystyle \frac{1}{2}}\\ -{\displaystyle \frac{1}{4}}& {\displaystyle \frac{1}{4}}& 0\end{array}\right),A^{D,\mathfrak{m}}=\left(\begin{array}{ccc}2& -{\displaystyle \frac{14}{9}}& {\displaystyle \frac{8}{9}}\\ 1& -{\displaystyle \frac{7}{9}}& {\displaystyle \frac{4}{9}}\\ -2& {\displaystyle \frac{14}{9}}& -{\displaystyle \frac{8}{9}}\end{array}\right),A^{E◯}=\left(\begin{array}{ccc}-{\displaystyle \frac{4}{3}}& {\displaystyle \frac{2}{3}}& -{\displaystyle \frac{4}{3}}\\ -{\displaystyle \frac{2}{3}}& {\displaystyle \frac{1}{3}}& -{\displaystyle \frac{2}{3}}\\ {\displaystyle \frac{4}{3}}& -{\displaystyle \frac{2}{3}}& {\displaystyle \frac{4}{3}}\end{array}\right),$$

$$A^{C,E◯}=\left(\begin{array}{ccc}-{\displaystyle \frac{9}{8}}& {\displaystyle \frac{9}{16}}& -{\displaystyle \frac{9}{8}}\\ -{\displaystyle \frac{7}{8}}& {\displaystyle \frac{7}{16}}& -{\displaystyle \frac{7}{8}}\\ {\displaystyle \frac{1}{2}}& -{\displaystyle \frac{1}{4}}& {\displaystyle \frac{1}{2}}\end{array}\right).$$

This shows that $\mathfrak{m}$-CCE inverse is different from some known generalized inverse.

4. Characterizations and a Canonical Form of the $\mathfrak{m}$-CEE Inverse

More characterizations for the Moore–Penrose inverse, the Drazin inverse, the DMP inverse, the CCE inverse, the Minkowski inverse, the $\mathfrak{m}$-core-EP inverse and the $\mathfrak{m}$-core inverse can be found in [23,24,25,26], and we now present characterizations for the $\mathfrak{m}$-CCE inverse.

Theorem 2. 

Let $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$, $\mathrm{rank}\left(A^{\sim }A{A}^{\sim }\right)=\mathrm{rank}\left(A\right)$ and $\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left({\left(A^k\right)}^{\sim }{A}^k\right)$. Then,

 (1) 

$R\left(A^{C, E◯}\right)=R\left(A^{\mathfrak{m}}{A}^k\right)$;

 (2) 

$N\left(A^{C, E◯}\right)=N\left({\left(A^k\right)}^{\sim }\right)$;

 (3) 

$A^{C, E◯}=A_{R\left(A^{\mathfrak{m}}{A}^k\right),N\left({\left(A^k\right)}^{\sim }\right)}^{\left(2\right)}=A_{R\left(A^{\mathfrak{m}}{A}^k{\left(A^k\right)}^{\sim }\right),N\left(A^{\mathfrak{m}}{A}^k{\left(A^k\right)}^{\sim }\right)}^{\left(2\right)}$.

Proof. 

(1) From Lemmas 4 and 5, we know

$$\begin{array}{cc}\hfill R\left(A^{C, E◯}\right)& =R\left(A^{C, E◯}A\right)=R\left(A^{\mathfrak{m}}A{A}^{E◯}A{A}^{\mathfrak{m}}A\right)=R\left(A^{\mathfrak{m}}A{A}^{E◯}A\right)\hfill \\ \hfill & =A^{\mathfrak{m}}AR\left(A^{E◯}A\right)=A^{\mathfrak{m}}AR\left(A^k\right)=A^{\mathfrak{m}}R\left(A^{k+1}\right)=A^{\mathfrak{m}}R\left(A^k\right)=R\left(A^{\mathfrak{m}}{A}^k\right).\hfill \end{array}$$

(2) According to Lemma 2, we know that

$$N\left(A{A}^{\mathfrak{m}}\right)=N\left(A^{\sim }\right)\subseteq N\left({\left(A^k\right)}^{\sim }\right).$$

Using Lemma 6, we obtain that

$${\left(A^k\right)}^{\sim }A{A}^{\mathfrak{m}}={\left(A^k\right)}^{\sim },$$

which implies

$$N\left({\left(A^k\right)}^{\sim }A{A}^{\mathfrak{m}}\right)=N\left({\left(A^k\right)}^{\sim }\right).$$

From Lemma 5, we have that

$$N\left(A^{C, E◯}\right)=N\left(A{A}^{C, E◯}\right)=N\left(A{A}^{\mathfrak{m}}A{A}^{E◯}A{A}^{\mathfrak{m}}\right)=N\left(A{A}^{E◯}A{A}^{\mathfrak{m}}\right).$$

This implies

$$x\in N\left(A^{C, E◯}\right)\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}A{A}^{\mathfrak{m}}x\in N\left(A{A}^{E◯}\right)=N\left({\left(A^k\right)}^{\sim }\right).$$

Furthermore,

$$x\in N\left(A^{C, E◯}\right)\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}x\in N\left({\left(A^k\right)}^{\sim }A{A}^{\mathfrak{m}}\right)=N\left({\left(A^k\right)}^{\sim }\right).$$

Hence, we obtain $N\left(A^{C, E◯}\right)=N\left({\left(A^k\right)}^{\sim }\right)$.

(3)

On the one hand, according to item (1) and item (2), we obtain

$$A^{C, E◯}=A_{R\left(A^{\mathfrak{m}}{A}^k\right),N\left({\left(A^k\right)}^{\sim }\right)}^{\left(2\right)},$$

on the other hand, we can obtain $R\left(A^{\mathfrak{m}}{A}^k\right)=R\left(A^{\mathfrak{m}}{A}^k{\left(A^k\right)}^{\sim }\right)$ from

$$\begin{array}{cc}\hfill R\left(A^{\mathfrak{m}}{A}^k\right)& =R\left(A^{\mathfrak{m}}{A}^k{\left(A^k\right)}^{\mathfrak{m}}{A}^k\right)\subseteq R\left(A^{\mathfrak{m}}{A}^k{\left(A^k\right)}^{\mathfrak{m}}\right)=A^{\mathfrak{m}}{A}^{k}R\left({\left(A^k\right)}^{\mathfrak{m}}\right)\hfill \\ \hfill & =A^{\mathfrak{m}}{A}^{k}R\left({\left(A^k\right)}^{\sim }\right)=R\left(A^{\mathfrak{m}}{A}^k{\left(A^k\right)}^{\sim }\right)\subseteq R\left(A^{\mathfrak{m}}{A}^k\right).\hfill \end{array}$$

In addition, since G denoted as a Minkowski metric matrice is nonsingular, it follows that

$$\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left({\left(A^k\right)}^{*}\right)=\mathrm{rank}\left(G{\left(A^k\right)}^{*}G\right)=\mathrm{rank}\left({\left(A^k\right)}^{\sim }\right).$$

It is obvious that

$$\mathrm{rank}\left(A^{\mathfrak{m}}{A}^k\right)⩽\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left(A^{k+1}\right)=\mathrm{rank}\left(A{A}^k\right)=\mathrm{rank}\left(A{A}^{\mathfrak{m}}A{A}^k\right)⩽\mathrm{rank}\left(A^{\mathfrak{m}}{A}^{k+1}\right)⩽\mathrm{rank}\left(A^{\mathfrak{m}}{A}^k\right),$$

which implies

$$\mathrm{rank}\left(A^{\mathfrak{m}}{A}^k\right)=\mathrm{rank}\left(A^k\right).$$

As a consequence, we obtain

$$\mathrm{rank}\left(A^{\mathfrak{m}}{A}^k{\left(A^k\right)}^{\sim }\right)=\mathrm{rank}\left(A^{\mathfrak{m}}{A}^k\right)=\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left({\left(A^k\right)}^{\sim }\right).$$

Due to $N\left({\left(A^k\right)}^{\sim }\right)\subseteq N\left(A^{\mathfrak{m}}{A}^k{\left(A^k\right)}^{\sim }\right)$, we deduce that

$$N\left({\left(A^k\right)}^{\sim }\right)=N\left(A^{\mathfrak{m}}{A}^k{\left(A^k\right)}^{\sim }\right).$$

In conclusion, we obtain $A^{C,E◯}=A_{R\left(A^{\mathfrak{m}}{A}^k\right),N\left({\left(A^k\right)}^{\sim }\right)}^{\left(2\right)}=A_{R\left(A^{\mathfrak{m}}{A}^k{\left(A^k\right)}^{\sim }\right),N\left(A^{\mathfrak{m}}{A}^k{\left(A^k\right)}^{\sim }\right)}^{\left(2\right)}$. □

Theorem 3. 

Let $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$, $\mathrm{rank}\left(A^{\sim }A{A}^{\sim }\right)=\mathrm{rank}\left(A\right)$ and $\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left({\left(A^k\right)}^{\sim }{A}^k\right)$. Then,

 (1) 

$A{A}^{C,E◯}=P_{R\left(A^k\right),N\left({\left(A^k\right)}^{\sim }\right)}$;

 (2) 

$A^{C,E◯}A=P_{R\left(A^{\mathfrak{m}}{A}^k\right),N\left({\left(A^k\right)}^{\sim }A\right)}$.

Proof. 

(1) Using Theorem 2, we possess

$$R\left(A{A}^{C,E◯}\right)=AR\left(A^{C,E◯}\right)=AR\left(A^{\mathfrak{m}}{A}^k\right)=R\left(A{A}^{\mathfrak{m}}{A}^k\right)=R\left(A^k\right).$$

According to Theorem 2(2), we know

$$N\left(A{A}^{C,E◯}\right)=N\left({\left(A^k\right)}^{\sim }\right).$$

Hence, we know $A{A}^{C,E◯}=P_{R\left(A^k\right),N\left({\left(A^k\right)}^{\sim }\right)}$.

(2)

According to Theorem 2(1), we know

$$R\left(A^{C,E◯}A\right)=R\left(A^{\mathfrak{m}}{A}^k\right).$$

It is obvious that

$$N\left(A^{C,E◯}A\right)=N\left(A^{\mathfrak{m}}A{A}^{E◯}A\right)\subseteq N\left(A{A}^{\mathfrak{m}}A{A}^{E◯}A\right)=N\left(A{A}^{E◯}A\right)\subseteq N\left(A^{E◯}A{A}^{E◯}A\right)=N\left(A^{E◯}A\right)\subseteq N\left(A^{\mathfrak{m}}A{A}^{E◯}A\right),$$

which shows that

$$N\left(A^{C,E◯}A\right)=N\left(A^{\mathfrak{m}}A{A}^{E◯}A\right)=N\left(A^{E◯}A\right).$$

Hence, we can infer that

$$x\in N\left(A^{C,E◯}A\right)\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}Ax\in N\left(A^{E◯}\right)=N\left({\left(A^k\right)}^{\sim }\right);$$

furthermore,

$$x\in N\left(A^{C,E◯}A\right)\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}x\in N\left({\left(A^k\right)}^{\sim }A\right).$$

So, $N\left(A^{C,E◯}A\right)=N\left({\left(A^k\right)}^{\sim }A\right)$. In conclusion, $A^{C,E◯}A=P_{R\left(A^{\mathfrak{m}}{A}^k\right),N\left({\left(A^k\right)}^{\sim }A\right)}$. □

Theorem 4. 

Let $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$, $\mathrm{rank}\left(A^{\sim }A{A}^{\sim }\right)=\mathrm{rank}\left(A\right)$ and $\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left({\left(A^k\right)}^{\sim }{A}^k\right)$. Then,

 (1) 

$A^{C,E◯}$ is an outer inverse of A;

 (2) 

$A^{C,E◯}$ is a reflexive g-inverse of $A_1$;

 (3) 

$A_1{A}^{C,E◯}=A_1{A}^{\mathfrak{m}}$, $A^{C,E◯}{A}_1=A^{\mathfrak{m}}{A}_1$.

Proof. 

(1) This is evident.

(2)

Using the properties of $A^{C,E◯}$ and $A_1$, we can compute directly that

$$A_1{A}^{C,E◯}{A}_1=A{A}^{E◯}A{A}^{\mathfrak{m}}A{A}^{E◯}A{A}^{\mathfrak{m}}A{A}^{E◯}A=A{A}^{E◯}A=A_1,$$

$$A^{C,E◯}{A}_1{A}^{C,E◯}=A^{\mathfrak{m}}A{A}^{E◯}A{A}^{\mathfrak{m}}A{A}^{E◯}A{A}^{\mathfrak{m}}A{A}^{E◯}A{A}^{\mathfrak{m}}=A^{\mathfrak{m}}A{A}^{E◯}A{A}^{\mathfrak{m}}=A^{C,E◯}.$$

In conclusion, we observe that $A^{C,E◯}$ is a reflexive g-inverse of $A_1$.

(3)

It is obvious that

$$A_1{A}^{C,E◯}=A{A}^{E◯}A{A}^{\mathfrak{m}}A{A}^{E◯}A{A}^{\mathfrak{m}}=A{A}^{E◯}A{A}^{\mathfrak{m}}=A_1{A}^{\mathfrak{m}},$$

$$A^{C,E◯}{A}_1=A^{\mathfrak{m}}A{A}^{E◯}A{A}^{\mathfrak{m}}A{A}^{E◯}A=A^{\mathfrak{m}}A{A}^{E◯}A=A^{\mathfrak{m}}{A}_1.$$

Hence, we obtain the results. □

Theorem 5. 

Let $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$, $\mathrm{rank}\left(A^{\sim }A{A}^{\sim }\right)=\mathrm{rank}\left(A\right)$ and $\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left({\left(A^k\right)}^{\sim }{A}^k\right)$. Then, $A^{C,E◯}$ presents a unique solution of

$$AX=P_{R\left(A^k\right),N\left({\left(A^k\right)}^{\sim }\right)},R\left(X\right)\subseteq R\left(A^{\mathfrak{m}}A\right).$$

(3)

Proof. 

Because of Theorems 2 and 3, we observe that $A^{C,E◯}$ satisfies (3). To prove the uniqueness, we assume that $X_1$ and $X_2$ satisfy (3). Then,

$$A{X}_1=A{X}_2=P_{R\left(A^k\right),N\left({\left(A^k\right)}^{\sim }\right)},i.e.,A(X_1-X_2)=0,$$

which implies $R(X_1-X_2)\subseteq N\left(A\right)\subseteq N\left(A^{\mathfrak{m}}A\right)$. From

$$R\left(X_1\right)\subseteq R\left(A^{\mathfrak{m}}A\right),R\left(X_2\right)\subseteq R\left(A^{\mathfrak{m}}A\right),$$

we have that

$$R(X_1-X_2)\subseteq R\left(A^{\mathfrak{m}}A\right).$$

Above all, we obtain $R(X_1-X_2)\subseteq R\left(A^{\mathfrak{m}}A\right)\cap N\left(A^{\mathfrak{m}}A\right)=0$. So $X_1=X_2$. □

Theorem 6. 

Let $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$, $\mathrm{rank}\left(A^{\sim }A{A}^{\sim }\right)=\mathrm{rank}\left(A\right)$ and $\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left({\left(A^k\right)}^{\sim }{A}^k\right)$. Then, the following statements are equivalent:

 (1) 

$X=A^{C,E◯}$;

 (2) 

$N\left(X\right)=N\left({\left(A^k\right)}^{\sim }\right)$, $XA=A^{\mathfrak{m}}{A}_1$;

 (3) 

$N\left(X\right)=N\left({\left(A^k\right)}^{\sim }\right)$, $X{A}^k=A^{\mathfrak{m}}{A}^k$;

 (4) 

$N\left(X\right)=N\left({\left(A^k\right)}^{\sim }\right)$, $XA{A}^{E◯}=A^{\mathfrak{m}}A{A}^{E◯}$.

Proof. 

(1) ⇒ (2). It is easily obtained by (2) and Theorem 2.

(2) ⇒ (3).

Post-multiplying $XA=A^{\mathfrak{m}}{A}_1$ by $A^{k-1}$, we earn

$$X{A}^k=A^{\mathfrak{m}}{A}_1{A}^{k-1}=A^{\mathfrak{m}}A{A}^{E◯}A{A}^{k-1}=A^{\mathfrak{m}}A{A}^{E◯}{A}^k=A^{\mathfrak{m}}{A}^k,$$

where the last equality follows from Lemmas 4 and 6.

(3) ⇒ (4).

We know $A{A}^{E◯}$ is an idempotent from $A^{E◯}A{A}^{E◯}=A^{E◯}$. It follows that

$$XA{A}^{E◯}=X{A}^k{\left(A^{E◯}\right)}^k=A^{\mathfrak{m}}A{A}^{E◯}.$$

(4) ⇒ (1).

Since Theorem 3, we obtain

$$N\left(X\right)=N\left({\left(A^k\right)}^{\sim }\right)=N\left(A{A}^{C,E◯}\right)=N\left(A{A}^{E◯}A{A}^{\mathfrak{m}}\right),$$

using Lemma 6, we obtain

$$X=XA{A}^{E◯}A{A}^{\mathfrak{m}}.$$

Applying $XA{A}^{E◯}=A^{\mathfrak{m}}A{A}^{E◯}$ to $X=XA{A}^{E◯}A{A}^{\mathfrak{m}}$, we have

$$X=XA{A}^{E◯}A{A}^{\mathfrak{m}}=A^{\mathfrak{m}}A{A}^{E◯}A{A}^{\mathfrak{m}}=A^{C,E◯}.$$

This finishes the proof. □

Theorem 7. 

Let $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$, $\mathrm{rank}\left(A^{\sim }A{A}^{\sim }\right)=\mathrm{rank}\left(A\right)$ and $\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left({\left(A^k\right)}^{\sim }{A}^k\right)$. Then, the following statements are equivalent:

 (1) 

$X=A^{C,E◯}$;

 (2) 

$X{A}_{1}X=X$, $A_{1}X=A_1{A}^{\mathfrak{m}}$, $X{A}_1=A^{\mathfrak{m}}{A}_1$;

 (3) 

$A^{\mathfrak{m}}{A}_{1}X=X$, $AX=A_1{A}^{\mathfrak{m}}$, $XA=A^{\mathfrak{m}}{A}_1$;

 (4) 

$XAX=X$, $AX=A_1{A}^{\mathfrak{m}}$, $XA=A^{\mathfrak{m}}{A}_1$.

Proof. 

(1) ⇒ (2). It is obvious by Theorem 4.

(2) ⇒ (3).

Applying $A_{1}X=A_1{A}^{\mathfrak{m}}$ and $X{A}_1=A^{\mathfrak{m}}{A}_1$ to $X{A}_{1}X=X$, we gain

$$X=X{A}_{1}X=A^{\mathfrak{m}}{A}_{1}X=A^{\mathfrak{m}}{A}_1{A}^{\mathfrak{m}}=A^{C,E◯}.$$

According to Definition 1, we realize

$$AX=A_1{A}^{\mathfrak{m}}andXA=A^{\mathfrak{m}}{A}_1.$$

(3) ⇒ (4). Applying $XA=A^{\mathfrak{m}}{A}_1$ to $A^{\mathfrak{m}}{A}_{1}X=X$, we gain

$$X=A^{\mathfrak{m}}{A}_{1}X=XAX.$$

(4) ⇒ (1). It is easily obtained by Theorem 1. This completes the proof. □

Theorem 8. 

Let $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$, $\mathrm{rank}\left(A^{\sim }A{A}^{\sim }\right)=\mathrm{rank}\left(A\right)$ and $\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left({\left(A^k\right)}^{\sim }{A}^k\right)$. Then,

 (1) 

$A_1=A\iff \mathrm{Ind}\left(A\right)⩽1$;

 (2) 

$A^{C,E◯}=A^{\mathfrak{m}}\iff \mathrm{Ind}\left(A\right)⩽1$.

Proof. 

(1) The result is easily obtained by Lemma 9.

(2)

It is obvious that

$$A^{C,E◯}=A^{\mathfrak{m}}\iff A^{\mathfrak{m}}A{A}^{E◯}A{A}^{\mathfrak{m}}=A^{\mathfrak{m}}\iff A{A}^{E◯}A=A{A}^{\mathfrak{m}}A\iff A_1=A.$$

From item (1), we know $A^{C,E◯}=A^{\mathfrak{m}}\iff \mathrm{Ind}\left(A\right)⩽1$. □

Theorem 9. 

Let $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$, $\mathrm{rank}\left(A^{\sim }A{A}^{\sim }\right)=\mathrm{rank}\left(A\right)$ and $\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left({\left(A^k\right)}^{\sim }{A}^k\right)$. Then,

 (1) 

$A^{C,E◯}=A{A}^{D,\mathfrak{m}}\iff A^{\mathfrak{m}}{A}^k=A^k$;

 (2) 

$A_1=A^{C,E◯}{A}_1\iff A^{\mathfrak{m}}{A}^k=A^k$;

 (3) 

$A_1=A^{\mathfrak{m},D}{A}_1\iff A^{\mathfrak{m}}{A}^k=A^k$;

 (4) 

$A_1=A^{D,\mathfrak{m}}{A}_1\iff A^k=A^{k+1}$;

 (5) 

$A{A}^{C,E◯}=A{A}^{\mathfrak{m}}\iff A^{C,E◯}=A^{\mathfrak{m}}$;

 (6) 

$A^{C,E◯}A=A^{\mathfrak{m}}A\iff A^{C,E◯}=A^{\mathfrak{m}}$.

Proof. 

From Lemmas 3 and 4, we acquire

$$R\left(A{A}^D\right)=R\left(A{A}^{E◯}\right)=R\left(A^k\right).$$

Using Lemma 6, we attain

$$A{A}^{D}A{A}^{E◯}=A{A}^{E◯}.$$

It can be easily obtained that

$$R\left(A_1\right)=R\left(A{A}^{E◯}A\right)=AR\left(A^{E◯}A\right)=AR\left(A^{E◯}\right)=AR\left(A^k\right)=R\left(A^{k+1}\right)=R\left(A^k\right).$$

Hence, we secure the following proof.

(1)

According to the properties of $A^{D,\mathfrak{m}}$ and $A^{C,E◯}$, we know

$$\begin{array}{cc}\hfill A^{C,E◯}=A{A}^{D,\mathfrak{m}}& \iff A^{\mathfrak{m}}A{A}^{E◯}A{A}^{\mathfrak{m}}=A{A}^{D}A{A}^{\mathfrak{m}}\hfill \\ \hfill & \iff A^{\mathfrak{m}}A{A}^{E◯}A=A{A}^{D}A\hfill \\ \hfill & \iff A^{\mathfrak{m}}A{A}^{E◯}=A{A}^{D}A{A}^{E◯}\hfill \\ \hfill & \iff A^{\mathfrak{m}}A{A}^{E◯}=A{A}^{E◯}\hfill \\ \hfill & \iff (A^{\mathfrak{m}}-I)A{A}^{E◯}=0\hfill \\ \hfill & \iff R\left(A{A}^{E◯}\right)\subseteq N(A^{\mathfrak{m}}-I)\hfill \\ \hfill & \iff R\left(A^k\right)\subseteq N(A^{\mathfrak{m}}-I)\hfill \\ \hfill & \iff A^{\mathfrak{m}}{A}^k=A^k.\hfill \end{array}$$

(2) According to the properties of $A_1$ and $A^{C,E◯}$, we have

$$\begin{array}{cc}\hfill A_1=A^{C,E◯}{A}_1& \iff A_1=A^{\mathfrak{m}}A{A}^{E◯}A{A}^{\mathfrak{m}}A{A}^{E◯}A\hfill \\ \hfill & \iff A_1=A^{\mathfrak{m}}A{A}^{E◯}A\hfill \\ \hfill & \iff A_1=A^{\mathfrak{m}}{A}_1\hfill \\ \hfill & \iff (I-A^{\mathfrak{m}})A_1=0\hfill \\ \hfill & \iff R\left(A_1\right)\subseteq N(I-A^{\mathfrak{m}})\hfill \\ \hfill & \iff R\left(A^k\right)\subseteq N(I-A^{\mathfrak{m}})\hfill \\ \hfill & \iff A^{\mathfrak{m}}{A}^k=A^k.\hfill \end{array}$$

(3) According to the properties of $A_1$ and $A^{\mathfrak{m},D}$, we obtain

$$\begin{array}{cc}\hfill A_1=A^{\mathfrak{m},D}{A}_1& \iff A_1=A^{\mathfrak{m}}A{A}^{D}A{A}^{E◯}A\hfill \\ \hfill & \iff A_1=A^{\mathfrak{m}}A{A}^{E◯}A\hfill \\ \hfill & \iff A_1=A^{\mathfrak{m}}{A}_1\hfill \\ \hfill & \iff A^{\mathfrak{m}}{A}^k=A^k.\hfill \end{array}$$

(4) According to the properties of $A_1$ and $A^{D,\mathfrak{m}}$, we know

$$\begin{array}{cc}\hfill A_1=A^{D,\mathfrak{m}}{A}_1& \iff A_1=A^{D}A{A}^{\mathfrak{m}}A{A}^{E◯}A\hfill \\ \hfill & \iff A_1=A^{D}A{A}^{E◯}A\hfill \\ \hfill & \iff A_1=A^D{A}_1\hfill \\ \hfill & \iff (I-A^D)A_1=0\hfill \\ \hfill & \iff R\left(A_1\right)\subseteq N(I-A^D)\hfill \\ \hfill & \iff R\left(A^k\right)\subseteq N(I-A^D)\hfill \\ \hfill & \iff (I-A^D)A^k=0\hfill \\ \hfill & \iff (I-A^D)A^{k+1}=0\hfill \\ \hfill & \iff A^{k+1}=A^D{A}^{k+1}=A^k.\hfill \end{array}$$

(5) Pre-multiplying $A{A}^{C,E◯}=A{A}^{\mathfrak{m}}$ by $A^{\mathfrak{m}}$, we earn

$$A^{\mathfrak{m}}A{A}^{C,E◯}=A^{\mathfrak{m}}A{A}^{\mathfrak{m}}=A^{\mathfrak{m}}.$$

It follows that $A^{C,E◯}=A^{\mathfrak{m}}$ from $A^{\mathfrak{m}}A{A}^{C,E◯}=A^{C,E◯}$. The converse is obvious.

(6)

Because $A^{C,E◯}A=A^{\mathfrak{m}}A$ and the fact that $A^{C,E◯}A{A}^{\mathfrak{m}}=A^{C,E◯}$, we realize

$$A^{C,E◯}=A^{C,E◯}A{A}^{\mathfrak{m}}=A^{\mathfrak{m}}A{A}^{\mathfrak{m}}=A^{\mathfrak{m}}.$$

The converse is obvious. □

Theorem 10. 

Let $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$ be given by (1) with $\mathrm{rank}\left(A\right)=\mathrm{rank}\left(A^{\sim }A{A}^{\sim }\right)$ and $\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left({\left(A^k\right)}^{\sim }{A}^k\right)$. Then,

$$A^{C,E◯}=GU\left(\begin{array}{cc}{K}^{*}{\mathrm{\Delta }}^{-1}K{\left(\mathrm{\Sigma }K\right)}^{E◯}{G}_1^{-1}& 0\\ L^{*}{\mathrm{\Delta }}^{-1}K{\left(\mathrm{\Sigma }K\right)}^{E◯}{G}_1^{-1}& 0\end{array}\right)U^{*}G,$$

where Δ and $G_1$ hold the same forms as Lemma 8.

Proof. 

Using (1) and Lemma 8, we own

$$\begin{array}{cc}\hfill A^{\mathfrak{m}}A{A}^{E◯}& =GU\left(\begin{array}{cc}{K}^{*}{\left(G_1\mathrm{\Sigma }\mathrm{\Delta }\right)}^{-1}& 0\\ L^{*}{\left(G_1\mathrm{\Sigma }\mathrm{\Delta }\right)}^{-1}& 0\end{array}\right)U^{*}GU\left(\begin{array}{cc}\mathrm{\Sigma }K& \mathrm{\Sigma }L\\ 0& 0\end{array}\right)\left(\begin{array}{cc}{\left(\mathrm{\Sigma }K\right)}^{E◯}& 0\\ 0& 0\end{array}\right)U^{*}\hfill \\ \hfill & =GU\left(\begin{array}{cc}{K}^{*}{\mathrm{\Delta }}^{-1}{\mathrm{\Sigma }}^{-1}{G}_1^{-1}& 0\\ L^{*}{\mathrm{\Delta }}^{-1}{\mathrm{\Sigma }}^{-1}{G}_1^{-1}& 0\end{array}\right)\left(\begin{array}{cc}{G}_1& G_2\\ G_2^{*}& G_4\end{array}\right)\left(\begin{array}{cc}\mathrm{\Sigma }K{\left(\mathrm{\Sigma }K\right)}^{E◯}& 0\\ 0& 0\end{array}\right)U^{*}\hfill \\ \hfill & =GU\left(\begin{array}{cc}{K}^{*}{\mathrm{\Delta }}^{-1}{\mathrm{\Sigma }}^{-1}& K^{*}{\mathrm{\Delta }}^{-1}{\mathrm{\Sigma }}^{-1}{G}_1^{-1}{G}_2\\ L^{*}{\mathrm{\Delta }}^{-1}{\mathrm{\Sigma }}^{-1}& L^{*}{\mathrm{\Delta }}^{-1}{\mathrm{\Sigma }}^{-1}{G}_1^{-1}{G}_2\end{array}\right)\left(\begin{array}{cc}\mathrm{\Sigma }K{\left(\mathrm{\Sigma }K\right)}^{E◯}& 0\\ 0& 0\end{array}\right)U^{*}\hfill \\ \hfill & =GU\left(\begin{array}{cc}{K}^{*}{\mathrm{\Delta }}^{-1}K{\left(\mathrm{\Sigma }K\right)}^{E◯}& 0\\ L^{*}{\mathrm{\Delta }}^{-1}K{\left(\mathrm{\Sigma }K\right)}^{E◯}& 0\end{array}\right)U^{*},\hfill \end{array}$$

$$\begin{array}{cc}\hfill A{A}^{\mathfrak{m}}& =U\left(\begin{array}{cc}\mathrm{\Sigma }K& \mathrm{\Sigma }L\\ 0& 0\end{array}\right)U^{*}GU\left(\begin{array}{cc}{K}^{*}{\left(G_1\mathrm{\Sigma }\mathrm{\Delta }\right)}^{-1}& 0\\ L^{*}{\left(G_1\mathrm{\Sigma }\mathrm{\Delta }\right)}^{-1}& 0\end{array}\right)U^{*}G\hfill \\ \hfill & =U\left(\begin{array}{c}\mathrm{\Sigma }\\ 0\end{array}\right)\left(\begin{array}{cc}K& L\end{array}\right)U^{*}GU\left(\begin{array}{c}{K}^{*}\\ L^{*}\end{array}\right)\left(\begin{array}{cc}{\left(G_1\mathrm{\Sigma }\mathrm{\Delta }\right)}^{-1}& 0\end{array}\right)U^{*}G\hfill \\ \hfill & =U\left(\begin{array}{c}\mathrm{\Sigma }\\ 0\end{array}\right)\mathrm{\Delta }\left(\begin{array}{cc}{\mathrm{\Delta }}^{-1}{\mathrm{\Sigma }}^{-1}{G}_1^{-1}& 0\end{array}\right)U^{*}G\hfill \\ \hfill & =U\left(\begin{array}{c}\mathrm{\Sigma }\\ 0\end{array}\right)\left(\begin{array}{cc}{\mathrm{\Sigma }}^{-1}{G}_1^{-1}& 0\end{array}\right)U^{*}G\hfill \\ \hfill & =U\left(\begin{array}{cc}{G}_1^{-1}& 0\\ 0& 0\end{array}\right)U^{*}G,\hfill \end{array}$$

$$\begin{array}{cc}\hfill and{A}^{C,E◯}& =A^{\mathfrak{m}}A{A}^{E◯}A{A}^{\mathfrak{m}}\hfill \\ \hfill & =GU\left(\begin{array}{cc}{K}^{*}{\mathrm{\Delta }}^{-1}K{\left(\mathrm{\Sigma }K\right)}^{E◯}& 0\\ L^{*}{\mathrm{\Delta }}^{-1}K{\left(\mathrm{\Sigma }K\right)}^{E◯}& 0\end{array}\right)\left(\begin{array}{cc}{G}_1^{-1}& 0\\ 0& 0\end{array}\right)U^{*}G\hfill \\ \hfill & =GU\left(\begin{array}{cc}{K}^{*}{\mathrm{\Delta }}^{-1}K{\left(\mathrm{\Sigma }K\right)}^{E◯}{G}_1^{-1}& 0\\ L^{*}{\mathrm{\Delta }}^{-1}K{\left(\mathrm{\Sigma }K\right)}^{E◯}{G}_1^{-1}& 0\end{array}\right)U^{*}G.\hfill \end{array}$$

This finishes the proof. □

5. Representations

In this section, we present some representations for $\mathfrak{m}$-CCE inverse.

Lemma 10 

([27]). Let $A\in {\mathbb{C}}_{m\times n}$, $X\in {\mathbb{C}}_{n\times p}$ and $Y\in {\mathbb{C}}_{p\times m}$. If $A_{R\left(XY\right),N\left(XY\right)}^{\left(2\right)}$ exists, then

$$A_{R\left(XY\right),N\left(XY\right)}^{\left(2\right)}=\underset{\lambda \to 0}{lim}X{(\lambda I_p+YAX)}^{-1}Y.$$

Lemma 11 

([28]). Let $A\in {\mathbb{C}}_{m\times n}$. Then,

$$A^{\mathfrak{m}}={\int }_0^{\infty }exp(-A^{\sim }At)A^{\sim }\mathrm{d}t,$$

$$A^{\mathfrak{m}}=\underset{t\to 0}{lim}{\left(t{I}_n+A^{\sim }A\right)}^{-1}{A}^{\sim }.$$

Lemma 12 

([19]). Let $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$. If $A=B_1{C}_1$ is a full-rank decomposition and $C_i{B}_i=B_{i+1}{C}_{i+1},(i=1,2,\dots ,k-1)$ are also full-rank decompositions. Then,

 (1) 

$C_k{B}_k$ is invertible;

 (2) 

$A^k=B_1\cdots B_k{C}_k\cdots C_1$;

 (3) 

$A^D=B_1\cdots B_k{\left(C_k{B}_k\right)}^{-k-1}{C}_k\cdots C_1$.

Lemma 13 

([29]). Let $A=B_1{C}_1\in C_{m\times n}$ be a rank factorization of rank r and $\mathrm{rank}\left(A{A}^{\sim }\right)=\mathrm{rank}\left(A^{\sim }A\right)=\mathrm{rank}\left(A\right)=r$. Then,

$$A^{\mathfrak{m}}=C_1^{\sim }{\left(C_1{C}_1^{\sim }\right)}^{-1}{\left(B_1^{\sim }{B}_1\right)}^{-1}{B}_1^{\sim }.$$

Theorem 11. 

Let $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=1$ and $\mathrm{rank}\left(A\right)=\mathrm{rank}\left(A^{\sim }A\right)$. If A has a full-rank decomposition as Lemma 12, then

$$A^{m◯}=B_1{\left(C_1{B}_1\right)}^{-1}{\left(B_1^{\sim }{B}_1\right)}^{-1}{B}_1^{\sim }.$$

Proof. 

Let $X=B_1{\left(C_1{B}_1\right)}^{-1}{\left(B_1^{\sim }{B}_1\right)}^{-1}{B}_1^{\sim }$. We must verify that X satisfies $AXA=A$, $A{X}^2=X$ and

${\left(AX\right)}^{\sim }=AX$. Then,

$$AXA=B_1{C}_1{B}_1{\left(C_1{B}_1\right)}^{-1}{\left(B_1^{\sim }{B}_1\right)}^{-1}{B}_1^{\sim }{B}_1{C}_1=B_1{C}_1=A;$$

$$A{X}^2=B_1{C}_1{B}_1{\left(C_1{B}_1\right)}^{-1}{\left(B_1^{\sim }{B}_1\right)}^{-1}{B}_1^{\sim }{B}_1{\left(C_1{B}_1\right)}^{-1}{\left(B_1^{\sim }{B}_1\right)}^{-1}{B}_1^{\sim }=B_1{\left(C_1{B}_1\right)}^{-1}{\left(B_1^{\sim }{B}_1\right)}^{-1}{B}_1^{\sim }=X;$$

$$AX=B_1{C}_1{B}_1{\left(C_1{B}_1\right)}^{-1}{\left(B_1^{\sim }{B}_1\right)}^{-1}{B}_1^{\sim }=B_1{\left(B_1^{\sim }{B}_1\right)}^{-1}{B}_1^{\sim },$$

$${\left(AX\right)}^{\sim }={\left(B_1{\left(B_1^{\sim }{B}_1\right)}^{-1}{B}_1^{\sim }\right)}^{\sim }=B_1{\left({\left(B_1^{\sim }{B}_1\right)}^{-1}\right)}^{\sim }{B}_1^{\sim }=B_1{\left({\left(B_1^{\sim }{B}_1\right)}^{\sim }\right)}^{-1}{B}_1^{\sim }=B_1{\left(B_1^{\sim }{B}_1\right)}^{-1}{B}_1^{\sim }=AX.$$

Thus, we obtain the result. □

Theorem 12. 

Let $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$ and $\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left({\left(A^k\right)}^{\sim }{A}^k\right)$. If A has a full-rank decomposition as Lemma 12, then

$$A^{E◯}=BCB{\left(C_k{B}_k\right)}^{-k-1}{\left(B^{\sim }B\right)}^{-1}{B}^{\sim },$$

where $B=B_1\cdots B_k$, $C=C_k\cdots C_1$.

Proof. 

Let $B=B_1\cdots B_k$, $C=C_k\cdots C_1$. Then, $A^k=BC$ and $A^D=B{\left(C_k{B}_k\right)}^{-k-1}C$. According to Theorem 11, we know

$${\left(A^k\right)}^{m◯}={\left(BC\right)}^{m◯}=B{\left(CB\right)}^{-1}{\left(B^{\sim }B\right)}^{-1}{B}^{\sim }.$$

It follows from $A^{E◯}=A^k{A}^D{\left(A^k\right)}^{m◯}$ that

$$A^{E◯}=BCB{\left(C_k{B}_k\right)}^{-k-1}CB{\left(CB\right)}^{-1}{\left(B^{\sim }B\right)}^{-1}{B}^{\sim }=BCB{\left(C_k{B}_k\right)}^{-k-1}{\left(B^{\sim }B\right)}^{-1}{B}^{\sim }.$$

This finishes the proof. □

Theorem 13. 

Let $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$, $\mathrm{rank}\left(A\right)=\mathrm{rank}\left(A^{\sim }A{A}^{\sim }\right)$ and $\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left({\left(A^k\right)}^{\sim }{A}^k\right)$. If A has a full-rank decomposition as Lemma 12, then

$$A^{C,E◯}=C_1^{\sim }{\left(C_1{C}_1^{\sim }\right)}^{-1}{C}_{1}BCB{\left(C_k{B}_k\right)}^{-k-1}{\left(B^{\sim }B\right)}^{-1}{B}^{\sim },$$

where $B=B_1\cdots B_k$, $C=C_k\cdots C_1$.

Proof. 

Using Lemma 13, Theorem 12 and the fact that $A^{C,E◯}=A^{\mathfrak{m}}A{A}^{E◯}A{A}^{\mathfrak{m}}$, we have

$$\begin{array}{cc}\hfill A^{C,E◯}& =C_1^{\sim }{\left(C_1{C}_1^{\sim }\right)}^{-1}{\left(B_1^{\sim }{B}_1\right)}^{-1}{B}_1^{\sim }{B}_1{C}_{1}BCB{\left(C_k{B}_k\right)}^{-k-1}{\left(B^{\sim }B\right)}^{-1}{B}^{\sim }{B}_1{C}_1{C}_1^{\sim }{\left(C_1{C}_1^{\sim }\right)}^{-1}{\left(B_1^{\sim }{B}_1\right)}^{-1}{B}_1^{\sim }\hfill \\ \hfill & =C_1^{\sim }{\left(C_1{C}_1^{\sim }\right)}^{-1}{C}_{1}BCB{\left(C_k{B}_k\right)}^{-k-1}{\left(B^{\sim }B\right)}^{-1}{B}^{\sim }{B}_1{\left(B_1^{\sim }{B}_1\right)}^{-1}{B}_1^{\sim }\hfill \\ \hfill & =C_1^{\sim }{\left(C_1{C}_1^{\sim }\right)}^{-1}{C}_{1}BCB{\left(C_k{B}_k\right)}^{-k-1}{\left(B^{\sim }B\right)}^{-1}{B}_k^{\sim }\cdots B_2^{\sim }{B}_1^{\sim }\hfill \\ \hfill & =C_1^{\sim }{\left(C_1{C}_1^{\sim }\right)}^{-1}{C}_{1}BCB{\left(C_k{B}_k\right)}^{-k-1}{\left(B^{\sim }B\right)}^{-1}{B}^{\sim }.\hfill \end{array}$$

This finishes the proof. □

Example 2. 

Let

$$A=\left(\begin{array}{ccc}-2& 14& -4\\ -1& 7& -2\\ 2& -14& 4\end{array}\right)$$

Then, $\mathrm{Ind}\left(A\right)=1$ and $\mathrm{rank}\left(A\right)=\mathrm{rank}\left(A^{\sim }A\right)=1$. In addition,

$$B_1=\left(\begin{array}{c}-2\\ -1\\ 2\end{array}\right),C_1=\left(\begin{array}{ccc}1& -7& 2\end{array}\right),A^{\mathfrak{m}}=\left(\begin{array}{ccc}-{\displaystyle \frac{4}{9}}& {\displaystyle \frac{2}{9}}& -{\displaystyle \frac{4}{9}}\\ -{\displaystyle \frac{2}{9}}& {\displaystyle \frac{1}{9}}& -{\displaystyle \frac{2}{9}}\\ {\displaystyle \frac{4}{9}}& -{\displaystyle \frac{2}{9}}& {\displaystyle \frac{4}{9}}\end{array}\right),$$

$$B_1{\left(C_1{B}_1\right)}^{-1}{\left(B_1^{\sim }{B}_1\right)}^{-1}{B}_1^{\sim }=\left(\begin{array}{ccc}-{\displaystyle \frac{4}{9}}& {\displaystyle \frac{2}{9}}& -{\displaystyle \frac{4}{9}}\\ -{\displaystyle \frac{2}{9}}& {\displaystyle \frac{1}{9}}& -{\displaystyle \frac{2}{9}}\\ {\displaystyle \frac{4}{9}}& -{\displaystyle \frac{2}{9}}& {\displaystyle \frac{4}{9}}\end{array}\right)=A^{m◯}.$$

Example 3. 

Let us test the matrix A given in Example 1. Then, $k=\mathrm{Ind}\left(A\right)=2$, and

$$B_1=\left(\begin{array}{cc}0& 4\\ -1& 3\\ -2& -2\end{array}\right),C_1=\left(\begin{array}{ccc}1& 0& {\displaystyle \frac{1}{4}}\\ 0& 1& -{\displaystyle \frac{1}{4}}\end{array}\right),B_2=\left(\begin{array}{c}-{\displaystyle \frac{1}{2}}\\ -{\displaystyle \frac{1}{2}}\end{array}\right),C_2=\left(\begin{array}{cc}1& -7\end{array}\right),$$

$$B=B_1{B}_2=\left(\begin{array}{c}-2\\ -1\\ 2\end{array}\right),C=C_2{C}_1=\left(\begin{array}{ccc}1& -7& 2\end{array}\right),$$

$$BCB{\left(C_k{B}_k\right)}^{-k-1}{\left(B^{\sim }B\right)}^{-1}{B}^{\sim }=\left(\begin{array}{ccc}-{\displaystyle \frac{4}{3}}& {\displaystyle \frac{2}{3}}& -{\displaystyle \frac{4}{3}}\\ -{\displaystyle \frac{2}{3}}& {\displaystyle \frac{1}{3}}& -{\displaystyle \frac{2}{3}}\\ {\displaystyle \frac{4}{3}}& -{\displaystyle \frac{2}{3}}& {\displaystyle \frac{4}{3}}\end{array}\right)=A^{E◯},$$

$$C_1^{\sim }{\left(C_1{C}_1^{\sim }\right)}^{-1}{C}_{1}BCB{\left(C_k{B}_k\right)}^{-k-1}{\left(B^{\sim }B\right)}^{-1}{B}^{\sim }=\left(\begin{array}{ccc}-{\displaystyle \frac{9}{8}}& {\displaystyle \frac{9}{16}}& -{\displaystyle \frac{9}{8}}\\ -{\displaystyle \frac{7}{8}}& {\displaystyle \frac{7}{16}}& -{\displaystyle \frac{7}{8}}\\ {\displaystyle \frac{1}{2}}& -{\displaystyle \frac{1}{4}}& {\displaystyle \frac{1}{2}}\end{array}\right)=A^{C,E◯},$$

where $A^{E◯}$ and $A^{C,E◯}$ have been shown in Example 1.

Theorem 14. 

Let $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$, $\mathrm{rank}\left(A\right)=\mathrm{rank}\left(A^{\sim }A{A}^{\sim }\right)$ and $\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left({\left(A^k\right)}^{\sim }{A}^k\right)$. Let A be the full-rank decomposition as Lemma 12. Then,

$$A^{C,E◯}={\int }_0^{\infty }exp(-C_1^{\sim }{C}_{1}t)C_1^{\sim }\mathrm{d}tM,$$

where $B=B_1\cdots B_k$, $C=C_k\cdots C_1$, $M=C_{1}BCB{\left(C_k{B}_k\right)}^{-k-1}{\left(B^{\sim }B\right)}^{-1}{B}^{\sim }$.

Proof. 

It is clear that

$$\mathrm{rank}\left(C_1{C}_1^{\sim }\right)⩽\mathrm{rank}\left(C_1\right)=\mathrm{rank}\left(A\right)=\mathrm{rank}\left(A^{\sim }A{A}^{\sim }\right)=\mathrm{rank}\left(C_1^{\sim }{B}_1^{\sim }{B}_1{C}_1{C}_1^{\sim }{B}_1^{\sim }\right)⩽\mathrm{rank}\left(C_1{C}_1^{\sim }\right),$$

which implies

$$\mathrm{rank}\left(C_1\right)=\mathrm{rank}\left(C_1{C}_1^{\sim }\right).$$

In addition, since $C_1^{\sim }$ is of full column rank, it follows that

$$\mathrm{rank}\left(C_1\right)=\mathrm{rank}\left(C_1{C}_1^{\sim }\right)=\mathrm{rank}\left(C_1^{\sim }{C}_1{C}_1^{\sim }\right).$$

This shows $C_1^{\mathfrak{m}}$ exists by Lemma 1. Also, From Lemma 1 and Theorem 2, we have

$$R\left(A^{C,E◯}\right)=R\left(A^{\mathfrak{m}}{A}^k\right)\subseteq R\left(A^{\mathfrak{m}}\right)=R\left(A^{\sim }\right)=R\left(C_1^{\sim }{B}_1^{\sim }\right)\subseteq R\left(C_1^{\sim }\right)=R\left(C_1^{\mathfrak{m}}\right)=R\left(C_1^{\mathfrak{m}}{C}_1\right).$$

According to Lemma 6, we obtain

$$A^{C,E◯}=C_1^{\mathfrak{m}}{C}_1{A}^{C,E◯}.$$

Using Lemma 11 and Theorem 13, we attain

$$\begin{array}{cc}\hfill A^{C,E◯}& ={\int }_0^{\infty }exp(-C_1^{\sim }{C}_{1}t)C_1^{\sim }\mathrm{d}t{C}_1{C}_1^{\sim }{\left(C_1{C}_1^{\sim }\right)}^{-1}{C}_{1}BCB{\left(C_k{B}_k\right)}^{-k-1}{\left(B^{\sim }B\right)}^{-1}{B}^{\sim }\hfill \\ \hfill & ={\int }_0^{\infty }exp(-C_1^{\sim }{C}_{1}t)C_1^{\sim }\mathrm{d}t{C}_{1}BCB{\left(C_k{B}_k\right)}^{-k-1}{\left(B^{\sim }B\right)}^{-1}{B}^{\sim }\hfill \\ \hfill & ={\int }_0^{\infty }exp(-C_1^{\sim }{C}_{1}t)C_1^{\sim }\mathrm{d}tM,\hfill \end{array}$$

where $M=C_{1}BCB{\left(C_k{B}_k\right)}^{-k-1}{\left(B^{\sim }B\right)}^{-1}{B}^{\sim }$. This finishes the proof. □

Theorem 15. 

Let $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$, $\mathrm{rank}\left(A^{\sim }A{A}^{\sim }\right)=\mathrm{rank}\left(A\right)$ and $\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left({\left(A^k\right)}^{\sim }{A}^k\right)$. Then,

 (1) 

$A^{C,E◯}={lim}_{\lambda \to 0}{\left(\lambda I_n+A^{\sim }A\right)}^{-1}{A}^{\sim }{A}^k{\left(\lambda I_n+{\left(A^k\right)}^{\sim }{A}^k\right)}^{-1}{\left(A^k\right)}^{\sim }$;

 (2) 

$A^{C,E◯}={lim}_{\lambda \to 0}{\left(\lambda I_n+A^{\sim }A\right)}^{-1}{A}^{\sim }{\left(\lambda I_n+A^k{\left(A^k\right)}^{\sim }A{\left(\lambda I_n+A^{\sim }A\right)}^{-1}{A}^{\sim }\right)}^{-1}{A}^k{\left(A^k\right)}^{\sim }$;

 (3) 

$A^{C,E◯}={lim}_{\lambda \to 0}{\left(\lambda I_n+A^{\sim }A\right)}^{-1}{A}^{\sim }{A}^k{\left(A^k\right)}^{\sim }{\left(\lambda I_n+A^k{\left(A^k\right)}^{\sim }\right)}^{-1}$;

 (4) 

$A^{C,E◯}={lim}_{\lambda \to 0}{\left(\lambda I_n+{\left(\lambda I_n+A^{\sim }A\right)}^{-1}{A}^{\sim }{A}^k{\left(A^k\right)}^{\sim }A\right)}^{-1}{\left(\lambda I_n+A^{\sim }A\right)}^{-1}{A}^{\sim }{A}^k{\left(A^k\right)}^{\sim }$.

Proof. 

We have $A^{C,E◯}=A_{R\left(A^{\mathfrak{m}}{A}^k{\left(A^k\right)}^{\sim }\right),N\left(A^{\mathfrak{m}}{A}^k{\left(A^k\right)}^{\sim }\right)}^{\left(2\right)}$ in Theorem 2 already. Combining with Lemmas 10 and 11(2), we have the following proof.

(1)

Let $X=A^{\mathfrak{m}}{A}^k$, $Y={\left(A^k\right)}^{\sim }$. Then,

$$\begin{array}{cc}\hfill A^{C,E◯}& =\underset{\lambda \to 0}{lim}{A}^{\mathfrak{m}}{A}^k{\left(\lambda I_n+{\left(A^k\right)}^{\sim }A{A}^{\mathfrak{m}}{A}^k\right)}^{-1}{\left(A^k\right)}^{\sim }\hfill \\ \hfill & =\underset{\lambda \to 0}{lim}{A}^{\mathfrak{m}}{A}^k{\left(\lambda I_n+{\left(A^k\right)}^{\sim }{A}^k\right)}^{-1}{\left(A^k\right)}^{\sim }\hfill \\ \hfill & =\underset{\lambda \to 0}{lim}{\left(\lambda I_n+A^{\sim }A\right)}^{-1}{A}^{\sim }{A}^k{\left(\lambda I_n+{\left(A^k\right)}^{\sim }{A}^k\right)}^{-1}{\left(A^k\right)}^{\sim }.\hfill \end{array}$$

(2) Let $X=A^{\mathfrak{m}}$, $Y=A^k{\left(A^k\right)}^{\sim }$. Then, we can obtain the result through a similar way.

(3)

Let $X=A^{\mathfrak{m}}{A}^k{\left(A^k\right)}^{\sim }$, $Y=I_n$. Then, we can obtain the result.

(4)

Let $X=I_n$, $Y=A^{\mathfrak{m}}{A}^k{\left(A^k\right)}^{\sim }$. Then, we can obtain the result. □

Example 4. 

Consider the matrix A given in Example 1. Then, $k=\mathrm{Ind}\left(A\right)=2$,

$$\begin{array}{cc}\hfill B& ={\left(\lambda I_n+A^{\sim }A\right)}^{-1}{A}^{\sim }{A}^k{\left(\lambda I_n+{\left(A^k\right)}^{\sim }{A}^k\right)}^{-1}{\left(A^k\right)}^{\sim }\hfill \\ \hfill & ={\left(\lambda I_n+A^{\sim }A\right)}^{-1}{A}^{\sim }{A}^k{\left(A^k\right)}^{\sim }{\left(\lambda I_n+A^k{\left(A^k\right)}^{\sim }\right)}^{-1}\hfill \\ \hfill & =\left(\begin{array}{ccc}{\displaystyle \frac{312(\lambda -3)}{{(\lambda -4)}^2(\lambda +52)}}& -{\displaystyle \frac{156(\lambda -3)}{{(\lambda -4)}^2(\lambda +52)}}& {\displaystyle \frac{312(\lambda -3)}{{(\lambda -4)}^2(\lambda +52)}}\\ {\displaystyle \frac{\left(104\right(\lambda -7\left)\right)}{{(\lambda -4)}^2(\lambda +52)}}& -{\displaystyle \frac{52(\lambda -7)}{{(\lambda -4)}^2(\lambda +52)}}& {\displaystyle \frac{104(\lambda -7)}{{(\lambda -4)}^2(\lambda +52)}}\\ -{\displaystyle \frac{104}{(\lambda -4)(\lambda +52)}}& {\displaystyle \frac{52}{(\lambda -4)(\lambda +52)}}& -{\displaystyle \frac{104}{(\lambda -4)(\lambda +52)}}\end{array}\right),\hfill \end{array}$$

$$\begin{array}{cc}\hfill C& ={\left(\lambda I_n+A^{\sim }A\right)}^{-1}{A}^{\sim }{\left(\lambda I_n+A^k{\left(A^k\right)}^{\sim }A{\left(\lambda I_n+A^{\sim }A\right)}^{-1}{A}^{\sim }\right)}^{-1}{A}^k{\left(A^k\right)}^{\sim }\hfill \\ \hfill & ={\left(\lambda I_n+{\left(\lambda I_n+A^{\sim }A\right)}^{-1}{A}^{\sim }{A}^k{\left(A^k\right)}^{\sim }A\right)}^{-1}{\left(\lambda I_n+A^{\sim }A\right)}^{-1}{A}^{\sim }{A}^k{\left(A^k\right)}^{\sim }\hfill \\ \hfill & =\left(\begin{array}{ccc}-{\displaystyle \frac{312(\lambda -3)}{-{\lambda }^3+8{\lambda }^2+348\lambda -832}}& {\displaystyle \frac{156(\lambda -3)}{-{\lambda }^3+8{\lambda }^2+348\lambda -832}}& -{\displaystyle \frac{312(\lambda -3)}{-{\lambda }^3+8{\lambda }^2+348\lambda -832}}\\ -{\displaystyle \frac{104(\lambda -7)}{-{\lambda }^3+8{\lambda }^2+348\lambda -832}}& {\displaystyle \frac{52(\lambda -7)}{-{\lambda }^3+8{\lambda }^2+348\lambda -832}}& -{\displaystyle \frac{104(\lambda -7)}{-{\lambda }^3+8{\lambda }^2+348\lambda -832}}\\ {\displaystyle \frac{104(\lambda -4)}{-{\lambda }^3+8{\lambda }^2+348\lambda -832}}& -{\displaystyle \frac{52(\lambda -4)}{-{\lambda }^3+8{\lambda }^2+348\lambda -832}}& {\displaystyle \frac{104(\lambda -4)}{-{\lambda }^3+8{\lambda }^2+348\lambda -832}}\end{array}\right).\hfill \end{array}$$

It is easy to check that ${lim}_{\lambda \to 0}B={lim}_{\lambda \to 0}C=A^{C,E◯}$, where $A^{C,E◯}$ has been shown in Example 1.

6. Applications

Theorem 16. 

Let $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$, $\mathrm{rank}\left(A^{\sim }A{A}^{\sim }\right)=\mathrm{rank}\left(A\right)$ and $\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left({\left(A^k\right)}^{\sim }{A}^k\right)$. Then, the general solution to the linear system

$$Ax=A{A}^{C,E◯}b,b\in {\mathbb{C}}_n$$

(4)

is $A^{C,E◯}b+(I-A^{\mathfrak{m}}A)y$, where $y\in {\mathbb{C}}_n$ is arbitrary.

Proof. 

On the one hand,

$$A(A^{C,E◯}b+(I-A^{\mathfrak{m}}A)y)=A{A}^{C,E◯}b+(A-A{A}^{\mathfrak{m}}A)y=A{A}^{C,E◯}b.$$

This implies $A^{C,E◯}b+(I-A^{\mathfrak{m}}A)y$ is a solution of (4). On the other hand, if (4) holds for some x, then

$$A^{\mathfrak{m}}Ax=A^{\mathfrak{m}}A{A}^{C,E◯}b=A^{\mathfrak{m}}A{A}^{\mathfrak{m}}A{A}^{E◯}A{A}^{\mathfrak{m}}b=A^{\mathfrak{m}}A{A}^{E◯}A{A}^{\mathfrak{m}}b=A^{C,E◯}b,$$

which yields

$$x=A^{C,E◯}b+x-A^{C,E◯}b=A^{C,E◯}b+x-A^{\mathfrak{m}}Ax=A^{C,E◯}b+(I-A^{\mathfrak{m}}A)x.$$

Thus, the solution x has the form $A^{C,E◯}b+(I-A^{\mathfrak{m}}A)y$, where $y\in {\mathbb{C}}_n$ is arbitrary. □

Theorem 17. 

Let $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$, $\mathrm{rank}\left(A^{\sim }A{A}^{\sim }\right)=\mathrm{rank}\left(A\right)$ and $\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left({\left(A^k\right)}^{\sim }{A}^k\right)$. Suppose $X\in {\mathbb{C}}_{n\times m}$ and $D\in {\mathbb{C}}_{n\times m}$. If $R\left(D\right)\subseteq R\left(A^k\right)$, then the restricted matrix equation

$$AX=D,R\left(X\right)\subseteq R\left(A^{\mathfrak{m}}{A}^k\right)$$

(5)

has the unique solution $X=A^{C,E◯}D$.

Proof. 

Since $R\left(D\right)\subseteq R\left(A^k\right)=R\left(A{A}^{C,E◯}\right)$, it follows that $A{A}^{C,E◯}D=D$, which implies $A^{C,E◯}D$ is a solution of $AX=D$. And it is obvious that

$$R\left(X\right)=R\left(A^{C,E◯}D\right)\subseteq R\left(A^{C,E◯}\right)=R\left(A^{\mathfrak{m}}{A}^k\right).$$

So $A^{C,E◯}D$ is a solution of (5). To prove the uniqueness, we assume Y is another solution of (5). Then, we acquire $R\left(Y\right)\subseteq R\left(A^{\mathfrak{m}}{A}^k\right)=R\left(A^{C,E◯}A\right)$. This implies

$$X=A^{C,E◯}D=A^{C,E◯}AY=Y.$$

This completes the proof. □

Theorem 18. 

Let $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$, $\mathrm{rank}\left(A^{\sim }A{A}^{\sim }\right)=\mathrm{rank}\left(A\right)$ and $\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left({\left(A^k\right)}^{\sim }{A}^k\right)$. Let $B\in {\mathbb{C}}_{n\times r}$ and $C^{*}\in {\mathbb{C}}_{n\times r}$ be of full column rank such that $R\left(B\right)=N\left({\left(A^k\right)}^{\sim }\right)$ and $N\left(C\right)=R\left(A^{\mathfrak{m}}{A}^k\right)$. Then, the bordered matrix

$$L=\left(\begin{array}{cc}A& B\\ C& 0\end{array}\right)$$

is nonsingular, and its inverse is given by

$$L^{-1}=\left(\begin{array}{cc}{A}^{C,E◯}& (I_n-A^{C,E◯}A)C^{†}\\ B^{†}(I_n-A{A}^{C,E◯})& B^{†}(A{A}^{C,E◯}A-A)C^{†}\end{array}\right).$$

Proof. 

From $R\left(A^{C,E◯}\right)=R\left(A^{\mathfrak{m}}{A}^k\right)=N\left(C\right)$, we realize $C{A}^{C,E◯}=0$. Since

$$R(I_n-A{A}^{C,E◯})=N\left(A{A}^{C,E◯}\right)=N\left({\left(A^k\right)}^{\sim }\right)=R\left(B\right)=R\left(B{B}^{†}\right),$$

it follows that $B{B}^{†}(I_n-A{A}^{C,E◯})=I_n-A{A}^{C,E◯}$.

Let

$$T=\left(\begin{array}{cc}{A}^{C,E◯}& (I_n-A^{C,E◯}A)C^{†}\\ B^{†}(I_n-A{A}^{C,E◯})& B^{†}(A{A}^{C,E◯}A-A)C^{†}\end{array}\right),$$

then

$$\begin{array}{cc}\hfill LT& =\left(\begin{array}{cc}A{A}^{C,E◯}+B{B}^{†}(I_n-A{A}^{C,E◯})& A(I_n-A^{C,E◯}A)C^{†}+B{B}^{†}(A{A}^{C,E◯}A-A)C^{†}\\ C{A}^{C,E◯}& C(I_n-A^{C,E◯}A)C^{†}\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}A{A}^{C,E◯}+I_n-A{A}^{C,E◯}& A(I_n-A^{C,E◯}A)C^{†}-(I_n-A{A}^{C,E◯})A{C}^{†}\\ 0& C{C}^{†}\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}{I}_n& 0\\ 0& I_r\end{array}\right)\hfill \end{array}$$

Hence, $T=L^{-1}$. □

Theorem 19. 

Let $A\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$, $\mathrm{rank}\left(A^{\sim }A{A}^{\sim }\right)=\mathrm{rank}\left(A\right)$ and $\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left({\left(A^k\right)}^{\sim }{A}^k\right)$. Let $B\in {\mathbb{C}}_{n\times r}$ and $C^{*}\in {\mathbb{C}}_{n\times r}$ be as Theorem 18, $X\in {\mathbb{C}}_{n\times m}$ and $D\in {\mathbb{C}}_{n\times m}$ be as Theorem 17. Then, the unique solution of (5) is given by $X=\left[x_{ij}\right]$, where

$$x_{ij}={\displaystyle \frac{det\left(\begin{array}{cc}A(i\to d_j)& B\\ C(i\to 0)& 0\end{array}\right)}{det\left(\begin{array}{cc}A& B\\ C& 0\end{array}\right)}},i=1,2,\dots ,n,j=1,2,\dots ,m,$$

(6)

where $d_j$ denotes the j-th column of D and $A(i\to d_j)$ and $C(i\to 0)$ mean to substitute the i-th column of A and C by $d_j$ and 0, respectively.

Proof. 

Since X is the solution of (5), we obtain that $R\left(X\right)\subseteq R\left(A^{\mathfrak{m}}{A}^k\right)=N\left(C\right)$, which implies $CX=0$. Then, (5) can be written as

$$\left(\begin{array}{cc}A& B\\ C& 0\end{array}\right)\left(\begin{array}{cc}X& 0\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}AX& 0\\ CX& 0\end{array}\right)=\left(\begin{array}{cc}D& 0\\ 0& 0\end{array}\right).$$

According to Theorem 18, we obtain that

$$\begin{array}{cc}\hfill \left(\begin{array}{cc}X& 0\\ 0& 0\end{array}\right)& ={\left(\begin{array}{cc}A& B\\ C& 0\end{array}\right)}^{-1}\left(\begin{array}{cc}D& 0\\ 0& 0\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}{A}^{C,E◯}& (I_n-A^{C,E◯}A)C^{†}\\ B^{†}(I_n-A{A}^{C,E◯})& B^{†}(A{A}^{C,E◯}A-A)C^{†}\end{array}\right)\left(\begin{array}{cc}D& 0\\ 0& 0\end{array}\right).\hfill \end{array}$$

Hence, $X=A^{C,E◯}D$ and (6) follows from the classical Cramer rule. □

7. A Binary Relation Based on the $\mathfrak{m}$-CCE Inverse

In this section, we study the binary relation for the $\mathfrak{m}$-CCE inverse.

Definition 2. 

Let $A,B\in {\mathbb{C}}_{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$, $\mathrm{rank}\left(A\right)=\mathrm{rank}\left(A^{\sim }A{A}^{\sim }\right)$ and $\mathrm{rank}\left(A^k\right)=\mathrm{rank}\left({\left(A^k\right)}^{\sim }{A}^k\right)$. We consider that A is below B under the relation $\stackrel{C,E◯}{\le }$ if

$$A^{C,E◯}A=A^{C,E◯}BandA{A}^{C,E◯}=B{A}^{C,E◯}.$$

The relation is denoted by A $\stackrel{C,E◯}{\le }$ B.

It is easy to know that the binary relation $\stackrel{C,E◯}{\le }$ is reflexive. According to the following examples, we observe it is neither anti-symmetric nor transitive.

Example 5. 

Consider a matrix $A\in {\mathbb{C}}_{3\times 3}$ as follows. Let B be the transpose of the matrix A.

$$A=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right),B=\left(\begin{array}{ccc}0& 0& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right).$$

It is easy to check that

$$A^{C,E◯}=B^{C,E◯}=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right).$$

By computing, we have

$$A^{C,E◯}A=A^{C,E◯}B=A{A}^{C,E◯}=B{A}^{C,E◯}=B^{C,E◯}B=B^{C,E◯}A=B{B}^{C,E◯}=A{B}^{C,E◯}=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right),$$

i.e., $A\stackrel{C,E◯}{\le }B$ and $B\stackrel{C,E◯}{\le }A$. But $A\ne B$. This means the relation $\stackrel{C,E◯}{\le }$ is not anti-symmetric.

Example 6. 

Consider the matrices

$$A=\left(\begin{array}{ccc}1& 2& 3\\ 0& 0& 1\\ 0& 0& 0\end{array}\right),B=\left(\begin{array}{ccc}1& 2& 3\\ 0& 0& 0\\ 0& 0& 0\end{array}\right),C=\left(\begin{array}{ccc}1& 2& 3\\ 5& 1& 1\\ 0& 0& 0\end{array}\right).$$

Then, we have

$$A^{C,E◯}=\left(\begin{array}{ccc}-{\displaystyle \frac{1}{3}}& 0& 0\\ {\displaystyle \frac{2}{3}}& 0& 0\\ 0& 0& 0\end{array}\right),B^{C,E◯}=\left(\begin{array}{ccc}-{\displaystyle \frac{1}{12}}& 0& 0\\ {\displaystyle \frac{1}{6}}& 0& 0\\ {\displaystyle \frac{1}{4}}& 0& 0\end{array}\right).$$

What is more, we have

$$A^{C,E◯}A=A^{C,E◯}B=\left(\begin{array}{ccc}-{\displaystyle \frac{1}{3}}& -{\displaystyle \frac{2}{3}}& -1\\ {\displaystyle \frac{2}{3}}& {\displaystyle \frac{4}{3}}& 2\\ 0& 0& 0\end{array}\right),$$

$$A{A}^{C,E◯}=B{A}^{C,E◯}=B{B}^{C,E◯}=C{B}^{C,E◯}=\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right),$$

$$B^{C,E◯}B=B^{C,E◯}C=\left(\begin{array}{ccc}-{\displaystyle \frac{1}{12}}& -{\displaystyle \frac{1}{6}}& -{\displaystyle \frac{1}{4}}\\ {\displaystyle \frac{1}{6}}& {\displaystyle \frac{1}{3}}& {\displaystyle \frac{1}{2}}\\ {\displaystyle \frac{1}{4}}& {\displaystyle \frac{1}{2}}& {\displaystyle \frac{3}{4}}\end{array}\right),$$

$$C{A}^{C,E◯}=\left(\begin{array}{ccc}1& 0& 0\\ -1& 0& 0\\ 0& 0& 0\end{array}\right),$$

i.e., $A\stackrel{C,E◯}{\le }B$, $B\stackrel{C,E◯}{\le }C$, But $A{A}^{C,E◯}\ne C{A}^{C,E◯}$, which means we cannot obtain $A\stackrel{C,E◯}{\le }C$. Thus, the relation $\stackrel{C,E◯}{\le }$ is not transitive. Above all, we know $\stackrel{C,E◯}{\le }$ is not a matrix partial order.

8. Conclusions

This paper introduces the $\mathfrak{m}$-CCE inverse, which is an extension of the CCE inverse in Minkowski space. Some of its properties, characterizations, representations, and applications in solving a system of linear equations are also shown in this paper. In addition, we prove that the $\mathfrak{m}$-CCE ordering is not a matrix partial ordering. Because of the wide range of research fields and application backgrounds of generalized inverses, we convince that further explorations of the $\mathfrak{m}$-CCE inverse will receive more attention and interest from researchers. Some possibilities for further research are given as follows:

• Further properties, characterizations and representations of the $\mathfrak{m}$-CCE inverse.
• We can further generalize the $\mathfrak{m}$-CCE inverse to tensors.
• The perturbation analysis and iterative methods for the $\mathfrak{m}$-CCE inverse will be two topics worth studying.
• It is equally interesting to discuss the $\mathfrak{m}$-CCE inverse for rectangular matrices.