Existence Criteria and Related Relation of the gMP Inverse of Matrices

Abstract

New characterizations of the gMP inverse are provided by the core part of the core-EP decomposition. We also answer the question as to whether X is the gMP inverse of A under the conditions of $\mathcal{R}\left(X\right)\subseteq \mathcal{R}\left(A^{*}{A}^k\right)$ or $\mathcal{N}\left({\left(A^k\right)}^{*}\right)\subseteq \mathcal{N}\left(X\right)$. We investigate the relationship between the core-EP inverse and the gMP inverse.Using the gMP inverse, the gMP relation is investigated in view of the core-EP decomposition.

1. Introduction

The set of all $m\times n$ complex matrices is denoted by ${\mathbb{C}}^{m\times n}$. The conjugate transpose of A is $A^{*}$. $\mathcal{R}\left(A\right)$ and $\mathcal{N}\left(A\right)$ denotes the column and row spaces, respectively. The index of A is denoted as ind(A), which is the smallest integer (k) such that $rank\left(A^k\right)=rank\left(A^{k+1}\right)$. The Moore–Penrose inverse of A is denoted by $X=A^{†}$ [1,2], and X satisfies $AXA=A,XAX=X,{\left(AX\right)}^{*}=AX$ and ${\left(XA\right)}^{*}=XA$. The Drazin inverse of A is denoted by $A^D$ [3], which is a matrix that satisfies $AXA=A,X{A}^{k+1}=A^k$ and $AX=XA$, where k is the index of A. The set of all square matrices with $ind\left(A\right)=k$ is denoted by ${\mathbb{C}}_k^{n\times n}$. The core-EP inverse of A was introduced in [4], and the symbol of this inverse is $A^{\overline{)†}}$; X is called the core-EP inverse of A if $XAX=X,\mathcal{R}\left(X\right)=\mathcal{R}\left(X^{*}\right)=\mathcal{R}\left(A^k\right)$, where $A\in {\mathbb{C}}_k^{n\times n}$ and $X\in {\mathbb{C}}^{n\times n}$. The MPCEP inverse of an operator was introduced by Chen, Mosić and Xu [5], and this concept was expanded on quaternion matrices by Kyrchei, Mosić and Stanimirović [6,7]. The MPCEP inverse of A is denoted by $A^{†,\overline{)†}}$, which is a matrix ($X\in {\mathbb{C}}^{n\times n}$) such that $XAX=X,AX=A{A}^{\overline{)†}}\mathrm{and}XA=A^{†}A{A}^{\overline{)†}}A$. Let X be the MPCEP inverse of A; the authors of [8] proved that $XAX=X$, $\mathcal{R}\left(X^{*}\right)=\mathcal{R}\left(A^k\right)$ and $\mathcal{R}\left(X\right)=\mathcal{R}\left(A^{†}{A}^k\right)$. A matrix (X) is called the generalized Moore–Penrose inverse (or gMP) of A [9] (Theorem 1 and Definition 1) if

$$XAX=X,AX=A{\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}\mathrm{and}XA={\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}A.$$

The symbol of the gMP inverse is $A^{\diamond }$. More properties of the gMP inverse can be seen in [10,11].

The $(B,C)$ inverse of A is denoted by $A^{\parallel (B,C)}$ [12,13,14], which is a matrix (Y) that satisfies $YAB=B,CAY=C,\mathcal{N}\left(C\right)\subseteq \mathcal{N}\left(Y\right)\mathrm{and}\mathcal{R}\left(Y\right)\subseteq \mathcal{R}\left(B\right)$, where $A,B,C,Y\in {\mathbb{C}}^{n\times n}$. The $(B,C)$ inverse of A is an outer inverse of A [13,15]. In [16], $A^{\parallel (B,C)}=B{\left(CAB\right)}^{-}C$, where ${\left(CAB\right)}^{-}\in CAB\left\{1\right\}$, and the symbol $CAB\left\{1\right\}$ is the set of all inner inverses of $CAB$. The gMP inverse for bounded linear operators was introduced by Stojanović and Mosić [9], which extends the Moore–Penrose inverse of the operator.

Let $A\in {\mathbb{C}}^{n\times n}$. The core-nilpotent decomposition can be seen in [17]; this matrix decomposition is unique. Let $A=A_1+A_2$ be the core-nilpotent decomposition of A; the core part of A is $A_1=A{A}^{D}A$ [18] and $A_2=A-A_1$ is the nilpotent part of A. The core-EP decomposition of $A\in {\mathbb{C}}^{n\times n}$ with $\mathrm{ind}\left(A\right)=k$ was introduced by Wang [19] (Theorem 2.1), which says that A can be written as $A={\widehat{A}}_1+{\widehat{A}}_2$, where ${\widehat{A}}_1$ is group-invertible, ${\widehat{A}}_2^k=0$, and ${\widehat{A}}_1^{*}{\widehat{A}}_2={\widehat{A}}_2{\widehat{A}}_1=0$. Let $A\in {\mathbb{C}}_k^{n\times n}$. The EP-nilpotent decomposition of A was introduced in [20] by Wang and Liu.

2. New Characterizations of the gMP Inverse by Using the Core Part of the Core-EP Decomposition

In order to investigate the gMP relation, we need the following lemmas. The core-EP decomposition is unique, and there exists a matrix ($U\in {\mathbb{C}}^{n\times n}$) such that

$${\widehat{A}}_1=U\left[\begin{array}{cc}T& S\\ 0& 0\end{array}\right]U^{*}\mathrm{and}{\widehat{A}}_2=U\left[\begin{array}{cc}0& 0\\ 0& N\end{array}\right]U^{*},$$

(1)

where $U^{*}U=E$, the $r\times r$ matrix (T) is invertible, the $(n-r)\times (n-r)$ matrix (N) is nilpotent and r is number of nonzero eigenvalues of A, that is, $r=rank\left(A^k\right)$ [19].

For the convenience of readers, we provide a simple proof of the following lemma.

Lemma 1.

Let $A\in {\mathbb{C}}_k^{n\times n}$ and A have the matrix decomposition of (1); then,

(1)

$E+T^{-1}S{\left(T^{-1}S\right)}^{*}$ is invertible;

(2)

$T{T}^{*}+S{S}^{*}$ is invertible;

(3)

$T+S{S}^{*}{\left(T^{-1}\right)}^{*}$ is invertible.

Proof. 

“(1)” It is obvious that $T^{-1}S{\left(T^{-1}S\right)}^{*}$ is Hermitian. Then, for any $\mathit{x}\in {\mathbb{C}}^n$, we have

$${\mathit{x}}^{*}\left(T^{-1}S{\left(T^{-1}S\right)}^{*}\right)\mathit{x}=\langle {\left(T^{-1}S\right)}^{*}\mathit{x},{\left(T^{-1}S\right)}^{*}\mathit{x}\rangle ⩾0,$$

which suggests that $T^{-1}S{\left(T^{-1}S\right)}^{*}$ is a positive semi-definite matrix. Then, we have $E+T^{-1}S{\left(T^{-1}S\right)}^{*}$, which is a positive definite matrix. Thus, $E+T^{-1}S{\left(T^{-1}S\right)}^{*}$ is invertible.

“(2)” It is easy to check that $T{T}^{*}+S{S}^{*}$ is a positive definite matrix; then, we can conclude that $T{T}^{*}+S{S}^{*}$ is invertible.

“(3)” Since $T+S{S}^{*}{\left(T^{-1}\right)}^{*}=T{\left(T^{-1}T\right)}^{*}+S{S}^{*}{\left(T^{-1}\right)}^{*}=T{T}^{*}{\left(T^{-1}\right)}^{*}+S{S}^{*}{\left(T^{-1}\right)}^{*}=(T{T}^{*}+S{S}^{*}){\left(T^{-1}\right)}^{*}$, $T+S{S}^{*}{\left(T^{-1}\right)}^{*}$ is invertible by $T{T}^{*}+S{S}^{*}$, and T is invertible. □

The following lemma was proposed in [19] (Theorem 2.3).

Lemma 2.

Let $A\in {\mathbb{C}}_k^{n\times n}$ and A have the matrix decomposition of (1); then, ${\widehat{A}}_1=A^k{\left(A^k\right)}^{†}A$ and ${\widehat{A}}_2=A-A^k{\left(A^k\right)}^{†}A$.

Lemma 3

([19] Corollary 3.3). Let $A\in {\mathbb{C}}_k^{n\times n}$. Then, $A{A}^{\overline{)†}}=A^k{\left(A^k\right)}^{†}$.

Lemma 4

([10] Lemma 2.5). Let $A\in {\mathbb{C}}_k^{n\times n}$ and A have the matrix decomposition of (1). Then, $A^{\diamond }=U\left[\begin{array}{cc}{T}^{*}{(T{T}^{*}+S{S}^{*})}^{-1}& 0\\ S^{*}{(T{T}^{*}+S{S}^{*})}^{-1}& 0\end{array}\right]U^{*}.$

Theorem 1.

Let $A\in {\mathbb{C}}_k^{n\times n}$ and A have the matrix decomposition of (1). Then, A is gMP-invertible with $A^{\diamond }=X$ if and only if X satisfies

$$X=XA{A}^{\overline{)†}}\mathit{and}{\widehat{A}}_1={\widehat{A}}_1{\left(XA\right)}^{*}.$$

(2)

Proof. 

“⇒” It is trivial according to [9] (Theorem 1) and ${\widehat{A}}_1=U\left[\begin{array}{cc}T& S\\ 0& 0\end{array}\right]U^{*}$.

“⇐” Let $X=U\left[\begin{array}{cc}{X}_1& X_2\\ X_3& X_4\end{array}\right]U^{*}.$ According to $A{A}^{\overline{)†}}=U\left[\begin{array}{cc}{E}_t& 0\\ 0& 0\end{array}\right]U^{*}$ and $X=XA{A}^{\overline{)†}}$, we have $X=U\left[\begin{array}{cc}{X}_1& 0\\ X_3& 0\end{array}\right]U^{*}$, where $t=rank\left(T\right)=rank\left(A^k\right)$. The conditions ${\widehat{A}}_1={\widehat{A}}_1{\left(XA\right)}^{*}$ and ${\widehat{A}}_1=U\left[\begin{array}{cc}T& S\\ 0& 0\end{array}\right]U^{*}$ yield

$$\left\{\begin{array}{cc}\hfill T& =T{T}^{*}{X}_1^{*}+S{S}^{*}{X}_1^{*}\hfill \\ \hfill S& =T{T}^{*}{X}_3^{*}+S{S}^{*}{X}_3^{*}\hfill \end{array}\right.$$

(3)

The equality (3) yields

$$\left\{\begin{array}{cc}\hfill X_1& =T^{*}{(T{T}^{*}+S{S}^{*})}^{-1}\hfill \\ \hfill X_3& =S^{*}{(T{T}^{*}+S{S}^{*})}^{-1}\hfill \end{array}\right.$$

(4)

Thus, $X=U\left[\begin{array}{cc}{T}^{*}{(T{T}^{*}+S{S}^{*})}^{-1}& 0\\ S^{*}{(T{T}^{*}+S{S}^{*})}^{-1}& 0\end{array}\right]U^{*}$, which indicates that the proof is finished by Lemma 4. □

Theorem 2.

Let $A\in {\mathbb{C}}_k^{n\times n}$ and A have the matrix decomposition of (1). Then, A is gMP-invertible with $A^{\diamond }=X$ if and only if X satisfies $\mathcal{N}\left({\left(A^k\right)}^{*}\right)\subseteq \mathcal{N}\left(X\right)$ and $XA{A}^{*}{A}^k=A^{*}{A}^k$.

Proof. 

Suppose X is the gMP inverse of A; then, $X=XA{A}^{\overline{)†}}$ and ${\widehat{A}}_1={\widehat{A}}_1{\left(XA\right)}^{*}$ according to Theorem 1. According to Lemma 3, we have

$$\begin{array}{cc}\hfill X=XA{A}^{\overline{)†}}& \iff X^{*}={\left(XA{A}^{\overline{)†}}\right)}^{*}={\left(A{A}^{\overline{)†}}\right)}^{*}{X}^{*}={\left[A^k{\left(A^k\right)}^{†}\right]}^{*}{X}^{*}=A^k{\left(A^k\right)}^{†}{X}^{*}\hfill \\ \hfill & \iff \mathcal{R}\left(X^{*}\right)\subseteq \mathcal{R}\left(A^k\right)\iff \mathcal{N}\left({\left(A^k\right)}^{*}\right)\subseteq \mathcal{N}\left(X\right).\hfill \end{array}$$

The condition ${\widehat{A}}_1={\widehat{A}}_1{\left(XA\right)}^{*}$ holds if and only if $A{A}^{\overline{)†}}A=A{A}^{\overline{)†}}A{\left(XA\right)}^{*}$ according to Lemmas 2 and 3. Taking involution on $A{A}^{\overline{)†}}A=A{A}^{\overline{)†}}A{\left(XA\right)}^{*}$ yields $A^{*}A{A}^{\overline{)†}}=XA{A}^{*}A{A}^{\overline{)†}}$, which implies $A^{*}{A}^k{\left(A^k\right)}^{†}=XA{A}^{*}{A}^k{\left(A^k\right)}^{†}$. Post-multiplying by $A^k$ on $A^{*}{A}^k{\left(A^k\right)}^{†}=XA{A}^{*}{A}^k{\left(A^k\right)}^{†}$ yields $XA{A}^{*}{A}^k=A^{*}{A}^k$. □

Theorem 3.

Let $A\in {\mathbb{C}}_k^{n\times n}$ and A have the matrix decomposition of (1). Then, A is gMP-invertible with $A^{\diamond }=X$ if and only if X satisfies

$$A{A}^{\overline{)†}}={\widehat{A}}_{1}X\mathit{and}{X}^{*}=X^{*}{A}^{\overline{)†}}A.$$

(5)

Proof. 

“⇒” It is trivial according to [9] (Theorem 1) and ${\widehat{A}}_1=U\left[\begin{array}{cc}T& S\\ 0& 0\end{array}\right]U^{*}$.

“⇐” Let $X=U\left[\begin{array}{cc}{X}_1& X_2\\ X_3& X_4\end{array}\right]U^{*}.$ According to $A{A}^{\overline{)†}}=U\left[\begin{array}{cc}{E}_t& 0\\ 0& 0\end{array}\right]U^{*}$ and $A{A}^{\overline{)†}}={\widehat{A}}_{1}X$, we have

$$U\left[\begin{array}{cc}{E}_t& 0\\ 0& 0\end{array}\right]U^{*}=U\left[\begin{array}{cc}T& S\\ 0& 0\end{array}\right]\left[\begin{array}{cc}{X}_1& X_2\\ X_3& X_4\end{array}\right]U^{*}=U\left[\begin{array}{cc}T{X}_1+S{X}_3& T{X}_2+S{X}_4\\ 0& 0\end{array}\right]U^{*},$$

which yields

$$\left\{\begin{array}{cc}\hfill E_t& =T{X}_1+S{X}_3\hfill \\ \hfill 0& =T{X}_2+S{X}_4\hfill \end{array}\right.$$

(6)

According to $A^{\overline{)†}}A=U\left[\begin{array}{cc}{E}_t& T^{-1}S\\ 0& 0\end{array}\right]U^{*}$ and $X^{*}=X^{*}{A}^{\overline{)†}}A$, we have

$$U\left[\begin{array}{cc}{X}_1^{*}& X_3^{*}\\ X_2^{*}& X_4^{*}\end{array}\right]U^{*}=U\left[\begin{array}{cc}{X}_1^{*}& X_3^{*}\\ X_2^{*}& X_4^{*}\end{array}\right]\left[\begin{array}{cc}{E}_t& T^{-1}S\\ 0& 0\end{array}\right]U^{*}=U\left[\begin{array}{cc}{X}_1^{*}& X_1^{*}{T}^{-1}S\\ X_2^{*}& X_2^{*}{T}^{-1}S\end{array}\right]U^{*},$$

which yields

$$\left\{\begin{array}{cc}\hfill X_3^{*}& =X_1^{*}{T}^{-1}S\hfill \\ \hfill X_4^{*}& =X_2^{*}{T}^{-1}S\hfill \end{array}\right.$$

(7)

According to Lemma 1 and equalities (6) and (7), we have

$$\left\{\begin{array}{cc}\hfill X_1& =T^{*}{(T{T}^{*}+S{S}^{*})}^{-1}\hfill \\ \hfill X_2& =0\hfill \\ \hfill X_3& =S^{*}{(T{T}^{*}+S{S}^{*})}^{-1}\hfill \\ \hfill X_4& =0\hfill \end{array}\right.$$

(8)

Thus, $X=U\left[\begin{array}{cc}{T}^{*}{(T{T}^{*}+S{S}^{*})}^{-1}& 0\\ S^{*}{(T{T}^{*}+S{S}^{*})}^{-1}& 0\end{array}\right]U^{*}$ according to (8), which indicates that the proof is finished by Lemma 4. □

Theorem 4.

Let $A\in {\mathbb{C}}_k^{n\times n}$, and A have the matrix decomposition of (1). Then, $X\in {\mathbb{C}}^{n\times n}$ is the gMP inverse of A if and only if $\mathcal{R}\left(X\right)\subseteq \mathcal{R}\left(A^{*}{A}^k\right)$ and ${\left(A^k\right)}^{*}AX={\left(A^k\right)}^{*}$.

Proof. 

Suppose X is the right gMP inverse of A; then, $A{A}^{\overline{)†}}={\widehat{A}}_{1}X$ and $X^{*}=X^{*}{A}^{\overline{)†}}A$ according to Theorem 3. According to Lemmas 2 and 3, we have

$$\begin{array}{cc}\hfill A{A}^{\overline{)†}}={\widehat{A}}_{1}X& \iff A{A}^{\overline{)†}}={\left(AX\right)}^{*}A{A}^{\overline{)†}}\hfill \\ \hfill & \iff A^k{\left(A^k\right)}^{†}={\left(AX\right)}^{*}{A}^k{\left(A^k\right)}^{†}\hfill \\ \hfill & \iff A^k={\left(AX\right)}^{*}{A}^k\hfill \\ \hfill & \iff {\left(A^k\right)}^{*}AX={\left(A^k\right)}^{*}.\hfill \end{array}$$

Then, $\mathcal{R}\left(X\right)\subseteq \mathcal{R}\left({\left(A^{\overline{)†}}A\right)}^{*}\right)$ by ${\left(A^{\overline{)†}}A\right)}^{*}$ is idempotent according to [9]. Thus, $\mathcal{R}\left(X\right)\subseteq \mathcal{R}\left(A^{*}{A}^k\right)$ holds according to $\mathcal{R}\left(A^{*}{A}^k\right)=\mathcal{R}\left({\left(A^{\overline{)†}}A\right)}^{*}\right)$. □

3. New Characterizations of the gMP Inverse by Using Equations and Subspaces

Motivated by Theorem 2 in [9], in the following theorem, we show that the condition $\mathcal{R}\left(X\right)=\mathcal{R}\left(A^{*}{A}^k\right)$ in [10] (Theorem 3.1(b)–(e)) can be relaxed as the condition $\mathcal{R}\left(X\right)\subseteq \mathcal{R}\left(A^{*}{A}^k\right)$.

Theorem 5.

Let $A\in {\mathbb{C}}_k^{n\times n}$ and $X\in {\mathbb{C}}^{n\times n}$. Then, the following statements are equivalent:

(1)

X is the gMP inverse of A;

(2)

$\mathcal{R}\left(X\right)\subseteq \mathcal{R}\left(A^{*}{A}^k\right)$, $AX=A{\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}$;

(3)

$\mathcal{R}\left(X\right)\subseteq \mathcal{R}\left(A^{*}{A}^k\right)$, $A^{\overline{)†}}AX=A^{\overline{)†}}$;

(4)

$\mathcal{R}\left(X\right)\subseteq \mathcal{R}\left(A^{*}{A}^k\right)$, ${\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}AX={\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}$;

(5)

$\mathcal{R}\left(X\right)\subseteq \mathcal{R}\left(A^{*}{A}^k\right)$, ${\left(A^{\overline{)†}}A\right)}^{*}{A}^{\overline{)†}}AX={\left(A^{\overline{)†}}A\right)}^{*}{A}^{\overline{)†}}$.

Proof. 

“$\left(1\right)\Rightarrow$(2)–(4)” It is obvious according to [10] (Theorem 3.1).

“$\left(2\right)\Rightarrow \left(1\right)$” We have ${\left(A^{\overline{)†}}\right)}^{*}=A^{k}U$ for some $U\in {\mathbb{C}}^{n\times n}$ according to $\mathcal{R}\left({\left(A^{\overline{)†}}\right)}^{*}\right)=\mathcal{R}\left(A^k\right)$; then,

$${\left(A^{\overline{)†}}A\right)}^{*}=A^{*}{A}^{k}U\mathrm{implies}\mathcal{R}\left({\left(A^{\overline{)†}}A\right)}^{*}\right)\subseteq \mathcal{R}\left(A^{*}{A}^k\right).$$

(9)

According to Lemma 3, we have

$$A^{*}{A}^k=A^{*}{A}^k{\left(A^k\right)}^{†}{A}^k=A^{*}{\left(A^{\overline{)†}}\right)}^{*}{A}^{*}{A}^k,\mathrm{which}\mathrm{implies}\mathcal{R}\left(A^{*}{A}^k\right)\subseteq \mathcal{R}\left({\left(A^{\overline{)†}}A\right)}^{*}\right).$$

(10)

According to $\mathcal{R}\left({\left(A^{\overline{)†}}A\right)}^{*}\right)=\mathcal{R}\left({\left(A^{\overline{)†}}A\right)}^{†}\right)$, equalities (9) and (10) yield

$$\mathcal{R}\left(A^{*}{A}^k\right)=\mathcal{R}\left({\left(A^{\overline{)†}}A\right)}^{†}\right).$$

(11)

Assuming $\mathcal{R}\left(X\right)\subseteq \mathcal{R}\left(A^{*}{A}^k\right)$ and equality (11) yield

$$\mathcal{R}\left(X\right)\subseteq \mathcal{R}\left(A^{*}{A}^k\right)=\mathcal{R}\left({\left(A^{\overline{)†}}A\right)}^{†}\right).$$

(12)

Equality (12) implies $X={\left(A^{\overline{)†}}A\right)}^{†}V$ for some $V\in {\mathbb{C}}^{n\times n}$. Then,

$$X={\left(A^{\overline{)†}}A\right)}^{†}V={\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}A{\left(A^{\overline{)†}}A\right)}^{†}V={\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}AX.$$

(13)

By assuming $AX=A{\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}$, equality (13) and Theorem 2 in [9], we have X, which is the gMP inverse of A.

“$\left(3\right)\Rightarrow \left(1\right)$” From the proof of $\left(2\right)\Rightarrow \left(1\right)$, we have $X={\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}AX$, so X is the gMP inverse of A based on the assumption of $A^{\overline{)†}}AX=A^{\overline{)†}}$ and Theorem 2 in [9].

“$\left(4\right)\Rightarrow \left(1\right)$” From the proof of $\left(2\right)\Rightarrow \left(1\right)$, we have $X={\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}AX$; then by assuming ${\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}AX={\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}$, we have

$$X={\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}AX={\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}};$$

thus, X is the gMP inverse of A according to [9] (Theorem 1).

“$\left(4\right)\iff \left(5\right)$” is trivial according to $\mathcal{N}\left({\left(A^{\overline{)†}}A\right)}^{*}\right)=\mathcal{N}\left({\left(A^{\overline{)†}}A\right)}^{†}\right)$. □

Motivated by Theorem 2 in [9], in the following theorem, we show that the condition of $\mathcal{N}\left(X\right)=\mathcal{N}\left({\left(A^k\right)}^{*}\right)$ in Theorem 3.2(b)–(e) of [10] can be relaxed as the condition $\mathcal{N}\left({\left(A^k\right)}^{*}\right)\subseteq \mathcal{N}\left(X\right)$.

Theorem 6.

Let $A\in {\mathbb{C}}_k^{n\times n}$ and $X\in {\mathbb{C}}^{n\times n}$. Then, the following are equivalent:

(1)

X is the gMP inverse of A;

(2)

$\mathcal{N}\left({\left(A^k\right)}^{*}\right)\subseteq \mathcal{N}\left(X\right)$, $XA={\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}A$;

(3)

$\mathcal{N}\left({\left(A^k\right)}^{*}\right)\subseteq \mathcal{N}\left(X\right)$, $XA{A}^{\overline{)†}}={\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}$;

(4)

$\mathcal{N}\left({\left(A^k\right)}^{*}\right)\subseteq \mathcal{N}\left(X\right)$, $XA{\left(A^{\overline{)†}}A\right)}^{*}={\left(A^{\overline{)†}}A\right)}^{*}$;

(5)

$\mathcal{N}\left({\left(A^k\right)}^{*}\right)\subseteq \mathcal{N}\left(X\right)$, $XA{\left(A^{\overline{)†}}A\right)}^{†}={\left(A^{\overline{)†}}A\right)}^{†}$.

Proof. 

“$\left(1\right)\Rightarrow \left(2\right)$” It is obvious according to [10] (Theorem 3.2(b)).

“$\left(2\right)\Rightarrow \left(3\right)$” Post-multiplying by $A^{\overline{)†}}$ on $XA={\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}A$ yields $XA{A}^{\overline{)†}}={\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}$, and $A^{\overline{)†}}$ is an outer inverse of A.

“$\left(3\right)\Rightarrow \left(4\right)$” According to Lemma 3, we have $A{A}^{\overline{)†}}=A^k{\left(A^k\right)}^{†}={\left({\left(A^k\right)}^{†}\right)}^{*}{\left(A^k\right)}^{*}$ and ${\left(A^k\right)}^{*}={\left(A^k{\left(A^k\right)}^{†}{A}^k\right)}^{*}={\left(A^k\right)}^{*}A{A}^{\overline{)†}}$, which implies that $\mathcal{N}\left(A{A}^{\overline{)†}}\right)=\mathcal{N}\left({\left(A^k\right)}^{*}\right)$, so $\mathcal{N}\left(A{A}^{\overline{)†}}\right)\subseteq \mathcal{N}\left(X\right)$ according to $\mathcal{N}\left({\left(A^k\right)}^{*}\right)\subseteq \mathcal{N}\left(X\right)$. Since $A{A}^{\overline{)†}}(E-A{A}^{\overline{)†}})=O$, $X(E-A{A}^{\overline{)†}})=O$ according to $\mathcal{N}\left(A{A}^{\overline{)†}}\right)\subseteq \mathcal{N}\left(X\right)$, which implies that $X=XA{A}^{\overline{)†}}$. Thus,

$$\begin{array}{cc}\hfill XA{\left(A^{\overline{)†}}A\right)}^{*}& =\left(XA{A}^{\overline{)†}}\right)A{\left(A^{\overline{)†}}A\right)}^{*}=\left[{\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}\right]A{\left(A^{\overline{)†}}A\right)}^{*}\hfill \\ \hfill & =\left[{\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}A\right]{\left(A^{\overline{)†}}A\right)}^{*}={\left[{\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}A\right]}^{*}{\left(A^{\overline{)†}}A\right)}^{*}\hfill \\ \hfill & ={\left[A^{\overline{)†}}A{\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}A\right]}^{*}\hfill \\ \hfill & ={\left(A^{\overline{)†}}A\right)}^{*}.\hfill \end{array}$$

“$\left(4\right)\Rightarrow \left(5\right)$”

$$\begin{array}{cc}\hfill XA{\left(A^{\overline{)†}}A\right)}^{†}& =XA{\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}A{\left(A^{\overline{)†}}A\right)}^{†}=XA{\left[{\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}A\right]}^{*}{\left(A^{\overline{)†}}A\right)}^{†}\hfill \\ \hfill & =XA{\left(A^{\overline{)†}}A\right)}^{*}{\left[{\left(A^{\overline{)†}}A\right)}^{†}\right]}^{*}{\left(A^{\overline{)†}}A\right)}^{†}={\left(A^{\overline{)†}}A\right)}^{*}{\left[{\left(A^{\overline{)†}}A\right)}^{†}\right]}^{*}{\left(A^{\overline{)†}}A\right)}^{†}\hfill \\ \hfill & ={\left[{\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}A\right]}^{*}{\left(A^{\overline{)†}}A\right)}^{†}={\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}A{\left(A^{\overline{)†}}A\right)}^{†}={\left(A^{\overline{)†}}A\right)}^{†}.\hfill \end{array}$$

“$\left(5\right)\Rightarrow \left(1\right)$” According to the proof of “$\left(3\right)\Rightarrow \left(4\right)$” $X=XA{A}^{\overline{)†}}$,

$$\begin{array}{cc}\hfill X=XA{A}^{\overline{)†}}& =XA\left(A^{\overline{)†}}A{A}^{\overline{)†}}\right)=XA\left(A^{\overline{)†}}A\right)A^{\overline{)†}}=XA\left[A^{\overline{)†}}A{\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}A\right]A^{\overline{)†}}\hfill \\ \hfill & =\left(XA{A}^{\overline{)†}}\right)A{\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}=XA{\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}\hfill \\ \hfill & ={\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}},\hfill \end{array}$$

The proof is finished by Theorem 1 of [9]. □

The condition $\mathcal{R}\left(A^{*}{A}^k\right)=\mathcal{R}\left(X\right)$ in [10] (Theorem 3.3(b)) can be relaxed as the condition $\mathcal{R}\left(A^{*}{A}^k\right)\subseteq \mathcal{R}\left(X\right)$.

Theorem 7.

Let $A\in {\mathbb{C}}_k^{n\times n}$ and $X\in {\mathbb{C}}^{n\times n}$. Then, X is the gMP inverse of A if and only if $XAX=X$, $\mathcal{R}\left(A^{*}{A}^k\right)\subseteq \mathcal{R}\left(X\right)$ and $\mathcal{N}\left({\left(A^k\right)}^{*}\right)=\mathcal{N}\left(X\right)$.

Proof. 

“⇒” is obvious according to [10] (Theorem 3.3(b)).

“⇐” By assuming $\mathcal{R}\left(A^{*}{A}^k\right)\subseteq \mathcal{R}\left(X\right)$, we have $A^{*}{A}^k=XU$ for some $U\in {\mathbb{C}}^{n\times n}$. Then, $A^{*}{A}^k=XU=XAXU=XA{A}^{*}{A}^k$; thus, X is the gMP inverse of A according to $\mathcal{N}\left({\left(A^k\right)}^{*}\right)=\mathcal{N}\left(X\right)$ and [10] (Theorem 3.2(f)). □

The condition $\mathcal{N}\left({\left(A^k\right)}^{*}\right)=\mathcal{N}\left(X\right)$ in [10] (Theorem 3.3(b)) can be relaxed as the condition $\mathcal{N}\left({\left(A^k\right)}^{*}\right)\supseteq \mathcal{N}\left(X\right)$.

Theorem 8.

Let $A\in {\mathbb{C}}_k^{n\times n}$ and $X\in {\mathbb{C}}^{n\times n}$. Then, X is the gMP inverse of A if and only if $XAX=X$, $\mathcal{R}\left(A^{*}{A}^k\right)=\mathcal{R}\left(X\right)$ and $\mathcal{N}\left({\left(A^k\right)}^{*}\right)\supseteq \mathcal{N}\left(X\right)$.

Proof. 

“⇒” is obvious according to [10] (Theorem 3.3(b)).

“⇐” By assuming $XAX=X$, we have $X(E-AX)=O$. By assuming $X(E-AX)=O$ and the assumption condition $\mathcal{N}\left({\left(A^k\right)}^{*}\right)\supseteq \mathcal{N}\left(X\right)$, we have ${\left(A^k\right)}^{*}(E-AX)=O$, that is, ${\left(A^k\right)}^{*}AX={\left(A^k\right)}^{*}$; thus, X is the gMP inverse of A according to $\mathcal{R}\left(A^{*}{A}^k\right)=\mathcal{R}\left(X\right)$ and [10] (Theorem 3.1(f)). □

The relationship between the core-EP inverse and the gMP inverse is investigated in the following theorem. Note that Stojanović and Mosić proposed a condition such the core-EP inverse coincides with the gMP inverse, that is, $A^{\overline{)†}}=A^{\diamond }$ if and only if $\mathcal{R}\left(A^D\right)\subseteq \mathcal{R}\left({\left(A^{\overline{)†}}A\right)}^{*}\right)$. According to Theorem 5, we have $\mathcal{R}\left(A^{*}{A}^k\right)=\mathcal{R}\left({\left(A^{\overline{)†}}A\right)}^{*}\right)=\mathcal{R}\left({\left(A^{\overline{)†}}A\right)}^{†}\right)$. One can also prove that $\mathcal{R}\left(A^D\right)=\mathcal{R}\left(A^k\right)=\mathcal{R}\left(A_1\right)=\mathcal{R}\left({\widehat{A}}_1\right)=\mathcal{R}\left({\tilde{A}}_1\right)$; thus, we have the following:

Proposition 1.

Let $A\in {\mathbb{C}}_k^{n\times n}$ and $X\in {\mathbb{C}}^{n\times n}$. Then, the core-EP inverse coincides with the gMP inverse if and only if $\mathcal{R}\left(A^k\right)\subseteq \mathcal{R}\left(A^{*}{A}^k\right)$. Moreover, the condition $\mathcal{R}\left(A^k\right)$ can be replaced by $\mathcal{R}\left(A_1\right)$, $\mathcal{R}\left({\widehat{A}}_1\right)$ or $\mathcal{R}\left({\tilde{A}}_1\right)$.

According to the definition of the core-EP inverse and [10] (Theorem 3.3(b)), we have the following table. Note that the relationship between the core-EP inverse and the gMP inverse can be determined by the following

Table 1. The gMP inverse ($A^{\diamond }$), the MPCEP inverse ($A^{†,\overline{)†}}$) and the core-EP inverse ($A^{\overline{)†}}$).

Three Generalized Inverses	Same Conditions	Different Coditions
$A^{\diamond }$	$XAX=X,\mathcal{R}\left(X^{*}\right)=\mathcal{R}\left(A^k\right)$	$\mathcal{R}\left(X\right)=\mathcal{R}\left(A^{*}{A}^k\right)$
$A^{†,\overline{)†}}$	$XAX=X,\mathcal{R}\left(X^{*}\right)=\mathcal{R}\left(A^k\right)$	$\mathcal{R}\left(X\right)=\mathcal{R}\left(A^{†}{A}^k\right)$
$A^{\overline{)†}}$	$XAX=X,\mathcal{R}\left(X^{*}\right)=\mathcal{R}\left(A^k\right)$	$\mathcal{R}\left(X\right)=\mathcal{R}\left(A^k\right)$

. We can also determine the the relationship between the MPCEP inverse and the gMP inverse according to [8].

For the convenience of readers, in the following proposition, some properties of the gMP inverse are collected.

Proposition 2.

Let $A\in {\mathbb{C}}_k^{n\times n}$, A have the matrix decomposition of (1) and $X\in {\mathbb{C}}^{n\times n}$ be the gMP inverse of A. Then,

(1)

$A^{\overline{)†}}=A^{\overline{)†}}AX$ and $X=XA{A}^{\overline{)†}}$;

(2)

${\widehat{A}}_{1}X=A{A}^{\overline{)†}}$ and $X{\widehat{A}}_1=XA={\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}A$;

(3)

$AXA\in A^{\overline{)†}}\{1,2\}$;

(4)

$\mathcal{R}\left(X\right)=\mathcal{R}\left(A^{*}{A}^k\right)=\mathcal{R}\left(X{\widehat{A}}_1\right)=\mathcal{R}\left({\left(A^{\overline{)†}}A\right)}^{†}\right)=\mathcal{R}\left({\left(A^{\overline{)†}}A\right)}^{*}\right)$;

(5)

$\mathcal{N}\left(X\right)=\mathcal{N}\left({\left(A^k\right)}^{*}\right)=\mathcal{N}\left({\widehat{A}}_{1}X\right)=\mathcal{N}\left(A{A}^{\overline{)†}}\right)$.

Proof. 

“(1)” is obvious according to [9] (Theorem 2(iv),(vii)).

“(2)” Pre-multiplying by A on $A^{\overline{)†}}=A^{\overline{)†}}AX$, we have $A{A}^{\overline{)†}}=A{A}^{\overline{)†}}AX$, which yields ${\widehat{A}}_{1}X=A{A}^{\overline{)†}}$ according to Lemma 2. Post-multiplying by A on $X=XA{A}^{\overline{)†}}$, we have $XA=XA{A}^{\overline{)†}}A$, which yields $X{\widehat{A}}_1=XA={\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}A$ according to [9] (Theorem 1).

“(3)” We have $X={\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}$ according to [9] (Theorem 1); then,

$$\begin{array}{ccc}& & A^{\overline{)†}}\left(AXA\right)A^{\overline{)†}}=A^{\overline{)†}}\left(A{\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}A\right)A^{\overline{)†}}=\left[A^{\overline{)†}}A{\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}A\right]A^{\overline{)†}}=A^{\overline{)†}};\hfill \\ & & \left(AXA\right)A^{\overline{)†}}\left(AXA\right)=\left(A{\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}A\right)A^{\overline{)†}}\left(A{\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}A\right)=AXA.\hfill \end{array}$$

“(4)” We have $X{\widehat{A}}_1={\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}A$ according to (2), so we have $\mathcal{R}\left(X{\widehat{A}}_1\right)=\mathcal{R}\left({\left(A^{\overline{)†}}A\right)}^{*}\right)$. Thus, the proof is finished by Theorem 3.3 of [10] and the proof of $\left(2\right)\Rightarrow \left(1\right)$ in Theorem 5.

“(5)” The equality $\mathcal{N}\left({\widehat{A}}_{1}X\right)=\mathcal{N}\left(A{A}^{\overline{)†}}\right)$ is trivial according to the condition ${\widehat{A}}_{1}X=A{A}^{\overline{)†}}$ in (2). The equality $\mathcal{N}\left(X\right)=\mathcal{N}\left({\left(A^k\right)}^{*}\right)$ holds according to [10] (Theorem 3.3). According to Lemma 3, we have $A{A}^{\overline{)†}}={\left[{\left(A^k\right)}^{†}\right]}^{*}{\left(A^k\right)}^{*}$ and ${\left(A^k\right)}^{*}={\left(A^k\right)}^{*}{A}^k{\left(A^k\right)}^{†}={\left(A^k\right)}^{*}A{A}^{\overline{)†}}$, which yields $\mathcal{N}\left({\left(A^k\right)}^{*}\right)=\mathcal{N}\left(A{A}^{\overline{)†}}\right)$. □

Remark 1.

The condition $\mathcal{R}\left(X\right)=\mathcal{R}\left({\left(A^{\overline{)†}}A\right)}^{†}\right)$ in Proposition 2 is very significant for investigating the gMP inverse of a complex matrix. Since the condition $\mathcal{R}\left(X\right)=\mathcal{R}\left(A^{*}{A}^k\right)$ yields $X=A^{*}{A}^k{\left(A^{*}{A}^k\right)}^{†}X$ and $\mathcal{R}\left(X\right)=\mathcal{R}\left({\left(A^{\overline{)†}}A\right)}^{†}\right)$ yields $X={\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}AX$, one can see that $A^{\overline{)†}}A$ is an idempotent matrix [9].

Let $A\in {\mathbb{C}}_k^{n\times n}$ and $X\in {\mathbb{C}}^{n\times n}$ be the gMP inverse of A. According to Proposition 2, $AXA$ is a $\{1,2\}$ inverse of A. It is easy to check that $AXA{A}^{\overline{)†}}=A{\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}$ and $A^{\overline{)†}}AXA=A^{\overline{)†}}A$. In the following example, we show that $AXA$ is not a $\{3,4\}$ inverse of A.

Example 1.

Let $A=\left[\begin{array}{ccc}2& 0& 0\\ 1& 0& -1\\ 1& 0& 0\end{array}\right]$ in ${\mathbb{C}}^{3\times 3}$. Then, $A^{\overline{)†}}=\left[\begin{array}{ccc}\frac{8}{21}& \frac{2}{21}& \frac{4}{21}\\ \frac{2}{21}& \frac{1}{42}& \frac{1}{21}\\ \frac{4}{21}& \frac{1}{21}& \frac{2}{21}\end{array}\right]$ and ${\left(A^{\overline{)†}}A\right)}^{†}=\left[\begin{array}{ccc}\frac{44}{61}& \frac{11}{61}& \frac{22}{61}\\ 0& 0& 0\\ -\frac{4}{61}& -\frac{1}{61}& -\frac{2}{61}\end{array}\right]$. Therefore, $AXA{A}^{\overline{)†}}=A{\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}=\left[\begin{array}{ccc}\frac{44}{61}& \frac{11}{61}& \frac{22}{61}\\ \frac{24}{61}& \frac{6}{61}& \frac{12}{61}\\ \frac{22}{61}& \frac{11}{122}& \frac{11}{61}\end{array}\right]$ and $A^{\overline{)†}}AXA=A^{\overline{)†}}A=\left[\begin{array}{ccc}\frac{22}{21}& 0& -\frac{2}{21}\\ \frac{11}{42}& 0& -\frac{1}{42}\\ \frac{11}{21}& 0& -\frac{1}{21}\end{array}\right].$ Thus, $\left[\begin{array}{ccc}\frac{44}{61}& \frac{11}{61}& \frac{22}{61}\\ \frac{24}{61}& \frac{6}{61}& \frac{12}{61}\\ \frac{22}{61}& \frac{11}{122}& \frac{11}{61}\end{array}\right]$ and $\left[\begin{array}{ccc}\frac{22}{21}& 0& -\frac{2}{21}\\ \frac{11}{42}& 0& -\frac{1}{42}\\ \frac{11}{21}& 0& -\frac{1}{21}\end{array}\right]$ are not Hermitian matrices, so $AXA$ is not a $\{3,4\}$ inverse of A.

For maximal classes, the gMP inverse was investigated by Stojanović and Mosić in [9] (Theorem 6). In the following theorem, we show that ${\left(UA\right)}^{†}U$ is the inverse along ${\left(UA\right)}^{*}$ and U.

Theorem 9.

Let $A,U\in {\mathbb{C}}^{n\times n}$. Then, ${\left(UA\right)}^{†}U$ is the inverse along ${\left(UA\right)}^{*}$ and U.

Proof. 

Since ${\left(UA\right)}^{†}={\left(UA\right)}^{*}{\left[UA{\left(UA\right)}^{*}\right]}^{†}$,

$${\left(UA\right)}^{†}U={\left(UA\right)}^{*}{\left[UA{\left(UA\right)}^{*}\right]}^{†}U=B{\left(CAB\right)}^{†}C,$$

where $B={\left(UA\right)}^{*}$ and $C=U$. □

Remark 2.

Let $A,M,N\in {\mathbb{C}}^{n\times n}$. Then, ${\left(MA\right)}^{†}N$ is the inverse along ${\left(MA\right)}^{*}$ and N. The proof of a such fact is similar to the proof Theorem 9.

4. The gMP Relation

Several necessary and sufficient conditions for the binary relation based on the gMP inverse are obtained.

Definition 1.

Let $A,B\in {\mathbb{C}}^{n\times n}$. Then, A is below B under the gMP relation if

$$A^{\diamond }A=A^{\diamond }B\mathit{and}A{A}^{\diamond }=B{A}^{\diamond },$$

(14)

where $A^{\diamond }$ is the gMP inverse of A. If A is below B under the gMP relation, then the symbol $A\stackrel{\diamond }{\le }B$ denotes this relation.

Example 2.

The gMP relation is not a partial ordering. Let $A=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right]\in {\mathbb{C}}^{3\times 3}$ and $B=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right]\in {\mathbb{C}}^{3\times 3}$. Then, $ind\left(A\right)=2$ and $ind\left(B\right)=1$. It is not difficult to check that $A^{\diamond }=B^{\diamond }=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right]$. Moreover, the binary relation $\stackrel{\diamond }{\le }$ is not anti-symmetric according to $A\ne B$ in view of $A\stackrel{\diamond }{\le }B$ and $B\stackrel{\diamond }{\le }A$.

The core-EP inverse can be expressed by the core-EP decomposition as the following lemma.

Lemma 5

([19] Theorem 3.2). Let $A\in {\mathbb{C}}_k^{n\times n}$ if $A={\widehat{A}}_1+{\widehat{A}}_2$ is the core-EP decomposition of A and ${\widehat{A}}_1,{\widehat{A}}_2$, as in (1). Then, $A^{\overline{)†}}=U\left[\begin{array}{cc}{T}^{-1}& 0\\ 0& 0\end{array}\right]U^{*}$.

Lemma 6

([21] Lemma 2). If $M\in {\mathbb{C}}^{m\times n}$ is partitioned as $M=\left[\begin{array}{cc}A& O\\ B& C\end{array}\right]$, then $M^{†}=\left[\begin{array}{cc}{K}^{†}{A}^{*}& K^{†}{B}^{*}\\ O& C^{†}\end{array}\right]$ if and only if $B^{*}C=0$, where $K=A^{*}A+B^{*}B.$

It is well known that for a complex matrix ($A\in {\mathbb{C}}^{m\times n}$), we have ${\left(A^{*}\right)}^{†}={\left(A^{†}\right)}^{*}.$ Thus, $A^{†}={\left[{\left(A^{*}\right)}^{†}\right]}^{*}$.

Lemma 7.

Let $N\in {\mathbb{C}}^{m\times n}$ be partitioned as $N=\left[\begin{array}{cc}D& G\\ O& F\end{array}\right]$. Then, $N^{†}=\left[\begin{array}{cc}{D}^{*}{L}^{†}& O\\ G^{*}{L}^{†}& F^{†}\end{array}\right]$ if and only if $G{F}^{*}=0$, where $L=D{D}^{*}+G{G}^{*}.$

Theorem 10.

Let $A\in {\mathbb{C}}_k^{n\times n}$ and A have the matrix decomposition of (1). Then, $A\stackrel{\diamond }{\le }B$ if and only if B can be written as

$$B=U\left[\begin{array}{cc}T& S\\ (N-B_4){\left(T^{-1}S\right)}^{*}& B_4\end{array}\right]U^{*},\mathit{for}\mathit{some}{B}_4\in {\mathbb{C}}^{(n-r)\times (n-r)}.$$

(15)

Proof. 

According to [9] (Theorem 1), we have $A\stackrel{\diamond }{\le }B$ if and only if

$$\left\{\begin{array}{cc}\hfill {\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}A& ={\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}B\hfill \\ \hfill A{\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}& =B{\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}\hfill \end{array}\right.$$

(16)

We have

$$A^{\overline{)†}}A=U\left[\begin{array}{cc}{T}^{-1}& O\\ O& O\end{array}\right]\left[\begin{array}{cc}T& S\\ O& N\end{array}\right]U^{*}=U\left[\begin{array}{cc}E& T^{-1}S\\ O& O\end{array}\right]U^{*}\triangleq U\left[\begin{array}{cc}E& Z\\ O& O\end{array}\right]U^{*}$$

(17)

according to Lemma 5, where $Z=T^{-1}S$. According to Lemmas 1, 7 and (17), we have

$${\left(A^{\overline{)†}}A\right)}^{†}=U\left[\begin{array}{cc}{E}^{*}{(E{E}^{*}+Z{Z}^{*})}^{-1}& O\\ Z^{*}{(E{E}^{*}+Z{Z}^{*})}^{-1}& O\end{array}\right]U^{*}\triangleq U\left[\begin{array}{cc}{V}^{-1}& O\\ Z^{*}{V}^{-1}& O\end{array}\right]U^{*},$$

(18)

where $V=E+Z{Z}^{*}$. According to (17) and (18), we have

$$\begin{array}{cc}\hfill {\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}A& =U\left[\begin{array}{cc}{V}^{-1}& O\\ Z^{*}{V}^{-1}& O\end{array}\right]\left[\begin{array}{cc}E& Z\\ O& O\end{array}\right]U^{*}\hfill \\ \hfill & =U\left[\begin{array}{cc}{V}^{-1}& V^{-1}Z\\ Z^{*}{V}^{-1}& Z^{*}{V}^{-1}Z\end{array}\right]U^{*}.\hfill \end{array}$$

(19)

Let $B=U\left[\begin{array}{cc}{B}_1& B_2\\ B_3& B_4\end{array}\right]U^{*}$. According to Lemma 5 and (18), we have

$$\begin{array}{cc}\hfill {\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}B& =U\left[\begin{array}{cc}{V}^{-1}& 0\\ Z^{*}{V}^{-1}& 0\end{array}\right]\left[\begin{array}{cc}{T}^{-1}& O\\ O& O\end{array}\right]\left[\begin{array}{cc}{B}_1& B_2\\ B_3& B_4\end{array}\right]U^{*}\hfill \\ \hfill & =U\left[\begin{array}{cc}{V}^{-1}{T}^{-1}& O\\ Z^{*}{V}^{-1}{T}^{-1}& O\end{array}\right]\left[\begin{array}{cc}{B}_1& B_2\\ B_3& B_4\end{array}\right]U^{*}\hfill \\ \hfill & =U\left[\begin{array}{cc}{V}^{-1}{T}^{-1}{B}_1& V^{-1}{T}^{-1}{B}_2\\ Z^{*}{V}^{-1}{T}^{-1}{B}_1& Z^{*}{V}^{-1}{T}^{-1}{B}_2\end{array}\right]U^{*}.\hfill \end{array}$$

(20)

according to (16), (19) and (20), we have

$$\left\{\begin{array}{cc}\hfill V^{-1}& =V^{-1}{T}^{-1}{B}_1\hfill \\ \hfill V^{-1}Z& =V^{-1}{T}^{-1}{B}_2\hfill \\ \hfill Z^{*}{V}^{-1}& =Z^{*}{V}^{-1}{T}^{-1}{B}_1\hfill \\ \hfill Z^{*}{V}^{-1}Z& =Z^{*}{V}^{-1}{T}^{-1}{B}_2,\hfill \end{array}\right.$$

(21)

which is equivalent to

$$\left\{\begin{array}{cc}\hfill V^{-1}& =V^{-1}{T}^{-1}{B}_1\hfill \\ \hfill V^{-1}Z& =V^{-1}{T}^{-1}{B}_2\hfill \end{array}\right.$$

(22)

Note that $Z=T^{-1}S$, so (22) is equivalent to

$$\left\{\begin{array}{cc}\hfill V^{-1}& =V^{-1}{T}^{-1}{B}_1\hfill \\ \hfill V^{-1}{T}^{-1}S& =V^{-1}{T}^{-1}{B}_2\hfill \end{array}\right.$$

(23)

As V and T are invertible, (23) is equivalent to

$$\left\{\begin{array}{cc}\hfill E& =T^{-1}{B}_1\hfill \\ \hfill S& =B_2\hfill \end{array},\right.$$

(24)

that is,

$$\left\{\begin{array}{cc}\hfill T& =B_1\hfill \\ \hfill S& =B_2\hfill \end{array}\right.$$

(25)

According to (25), matrix B can be written as

$$B=U\left[\begin{array}{cc}T& S\\ B_3& B_4\end{array}\right]U^{*}.$$

(26)

According Lemma 5, (18) and (26), we have

$$\begin{array}{cc}\hfill A{\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}& =U\left[\begin{array}{cc}T& S\\ O& N\end{array}\right]\left[\begin{array}{cc}{V}^{-1}& 0\\ Z^{*}{V}^{-1}& 0\end{array}\right]\left[\begin{array}{cc}{T}^{-1}& O\\ O& O\end{array}\right]U^{*}\hfill \\ \hfill & =U\left[\begin{array}{cc}T& S\\ O& N\end{array}\right]\left[\begin{array}{cc}{V}^{-1}{T}^{-1}& O\\ Z^{*}{V}^{-1}{T}^{-1}& O\end{array}\right]U^{*}\hfill \\ \hfill & =U\left[\begin{array}{cc}T{V}^{-1}{T}^{-1}+S{Z}^{*}{V}^{-1}{T}^{-1}& O\\ N{Z}^{*}{V}^{-1}{T}^{-1}& O\end{array}\right]U^{*}.\hfill \end{array}$$

(27)

$$\begin{array}{cc}\hfill B{\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}& =U\left[\begin{array}{cc}T& S\\ B_3& B_4\end{array}\right]\left[\begin{array}{cc}{V}^{-1}& 0\\ Z^{*}{V}^{-1}& 0\end{array}\right]\left[\begin{array}{cc}{T}^{-1}& O\\ O& O\end{array}\right]U^{*}\hfill \\ \hfill & =U\left[\begin{array}{cc}T& S\\ B_3& B_4\end{array}\right]\left[\begin{array}{cc}{V}^{-1}{T}^{-1}& O\\ Z^{*}{V}^{-1}{T}^{-1}& O\end{array}\right]U^{*}\hfill \\ \hfill & =U\left[\begin{array}{cc}T{V}^{-1}{T}^{-1}+S{Z}^{*}{V}^{-1}{T}^{-1}& O\\ B_3{V}^{-1}{T}^{-1}+B_4{Z}^{*}{V}^{-1}{T}^{-1}& O\end{array}\right]U^{*}.\hfill \end{array}$$

(28)

According to (16), (27) and (28), we have

$$N{Z}^{*}{V}^{-1}{T}^{-1}=B_3{V}^{-1}{T}^{-1}+B_4{Z}^{*}{V}^{-1}{T}^{-1}.$$

(29)

As V and T are invertible, (29) is equivalent to

$$N{Z}^{*}=B_3+B_4{Z}^{*}.$$

(30)

Equality (30) yields $B_3=(N-B_4){\left(T^{-1}S\right)}^{*}$. □

For the convenience of the beginner, we provide a simple proof of the following theorem; the remaining proof is similar to the proof of Theorem 10.

Theorem 11.

Let $A\in {\mathbb{C}}_k^{n\times n}$, $B\in {\mathbb{C}}^{n\times n}$ and A have the matrix decomposition of (1). Then, $A\stackrel{\diamond }{\le }B$ if and only if B can be written as

$$B-A=U\left[\begin{array}{cc}0& 0\\ -X_4{\left(T^{-1}S\right)}^{*}& X_4\end{array}\right]U^{*},\mathit{for}\mathit{some}{X}_4\in {\mathbb{C}}^{(n-r)\times (n-r)}.$$

(31)

Proof. 

According to [9] (Theorem 1), we have $A\stackrel{\diamond }{\le }B$ if and only if

$$\left\{\begin{array}{cc}\hfill {\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}A& ={\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}B\hfill \\ \hfill A{\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}& =B{\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}},\hfill \end{array}\right.$$

(32)

which is equivalent to

$$\left\{\begin{array}{cc}\hfill {\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}(B-A)& =O\hfill \\ \hfill (B-A){\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}& =O\hfill \end{array}\right.$$

(33)

According to the proof of Theorem 10, we have

$${\left(A^{\overline{)†}}A\right)}^{†}=U\left[\begin{array}{cc}{E}^{*}{(E{E}^{*}+Z{Z}^{*})}^{-1}& O\\ Z^{*}{(E{E}^{*}+Z{Z}^{*})}^{-1}& O\end{array}\right]U^{*}\triangleq U\left[\begin{array}{cc}{V}^{-1}& O\\ Z^{*}{V}^{-1}& O\end{array}\right]U^{*},$$

(34)

where $V=E+Z{Z}^{*}$. Let $B-A=U\left[\begin{array}{cc}{X}_1& X_2\\ X_3& X_4\end{array}\right]U^{*}$. According to Lemma 5, (33) and (34), we have

$$\begin{array}{cc}\hfill {\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}(B-A)& =U\left[\begin{array}{cc}{V}^{-1}& 0\\ Z^{*}{V}^{-1}& 0\end{array}\right]\left[\begin{array}{cc}{T}^{-1}& O\\ O& O\end{array}\right]\left[\begin{array}{cc}{X}_1& X_2\\ X_3& X_4\end{array}\right]U^{*}\hfill \\ \hfill & =U\left[\begin{array}{cc}{V}^{-1}{T}^{-1}& O\\ Z^{*}{V}^{-1}{T}^{-1}& O\end{array}\right]\left[\begin{array}{cc}{X}_1& X_2\\ X_3& X_4\end{array}\right]U^{*}\hfill \\ \hfill & =U\left[\begin{array}{cc}{V}^{-1}{T}^{-1}{X}_1& V^{-1}{T}^{-1}{X}_2\\ Z^{*}{V}^{-1}{T}^{-1}{X}_1& Z^{*}{V}^{-1}{T}^{-1}{X}_2\end{array}\right]U^{*}\hfill \\ \hfill & =U\left[\begin{array}{cc}O& O\\ O& O\end{array}\right]U^{*}.\hfill \end{array}$$

(35)

$$\begin{array}{cc}\hfill (B-A){\left(A^{\overline{)†}}A\right)}^{†}{A}^{\overline{)†}}& =U\left[\begin{array}{cc}{X}_1& X_2\\ X_3& X_4\end{array}\right]\left[\begin{array}{cc}{V}^{-1}& 0\\ Z^{*}{V}^{-1}& 0\end{array}\right]\left[\begin{array}{cc}{T}^{-1}& O\\ O& O\end{array}\right]U^{*}\hfill \\ \hfill & =U\left[\begin{array}{cc}{X}_1& X_2\\ X_3& X_4\end{array}\right]\left[\begin{array}{cc}{V}^{-1}{T}^{-1}& O\\ Z^{*}{V}^{-1}{T}^{-1}& O\end{array}\right]U^{*}\hfill \\ \hfill & =U\left[\begin{array}{cc}{X}_1{V}^{-1}{T}^{-1}+X_2{Z}^{*}{V}^{-1}{T}^{-1}& O\\ X_3{V}^{-1}{T}^{-1}+X_4{Z}^{*}{V}^{-1}{T}^{-1}& O\end{array}\right]U^{*}\hfill \\ \hfill & =U\left[\begin{array}{cc}O& O\\ O& O\end{array}\right]U^{*}.\hfill \end{array}$$

(36)

Equality (35) implies

$$\left\{\begin{array}{cc}\hfill V^{-1}{T}^{-1}{X}_1& =O\hfill \\ \hfill V^{-1}{T}^{-1}{X}_2& =O\hfill \\ \hfill Z^{*}{V}^{-1}{T}^{-1}{X}_1& =O\hfill \\ \hfill Z^{*}{V}^{-1}{T}^{-1}{X}_2& =O\hfill \end{array}\right.$$

(37)

Equality (36) implies

$$\left\{\begin{array}{cc}\hfill X_1{V}^{-1}{T}^{-1}+X_2{Z}^{*}{V}^{-1}{T}^{-1}& =O\hfill \\ \hfill X_3{V}^{-1}{T}^{-1}+X_4{Z}^{*}{V}^{-1}{T}^{-1}& =O\hfill \end{array}\right.$$

(38)

As V and T are invertible, equality (37) implies $X_1=O$ and $X_2=O$. Equality (38) implies $X_3=-X_4{\left(T^{-1}S\right)}^{*}$. □

5. Conclusions

A one-sided gMP inverse for matrices was introduced in this paper. Some conditions are proposed such that a left (or right) gMP-invertible matrix is gMP-invertible. The necessary and sufficient conditions of $\mathcal{R}\left(X\right)\subseteq \mathcal{R}\left(A^{*}{A}^k\right)$ or $\mathcal{N}\left({\left(A^k\right)}^{*}\right)\subseteq \mathcal{N}\left(X\right)$ are proposed such that X is gMP-invertible. The relationship between the core-EP inverse and the gMP inverse is also proposed. The following future perspectives for research are proposed:

Part 1. The reverse order law of the gMP inverse;

Part 2. The rank properties of the gMP inverse, such as $rank(A{A}^{\diamond }-A^{\diamond }A)$;

Part 3. The weighted gMP inverse of matrices;

Part 4. Centralizer applications of the gMP inverse in rings.