Characterizations of the Image-Kernel (p, q)-Inverses

Abstract

Necessary and sufficient conditions for the existence of the image-kernel (p, q)-inverses in a ring with involution are investigated, and some new expressions for these inverses are given. Several new properties of the image-kernel (p, q)-inverses are presented too.

1 Introduction

Throughout this paper, let ${\mathcal {R}}$ be a ring with the unit 1. The sets of all idempotents ($p^2=p$) and invertible elements of ${\mathcal {R}}$ will be denoted by ${\mathcal {R}}^\bullet$ and ${\mathcal {R}}^{-1}$, respectively.

Let $a\in {\mathcal {R}}$ and let $p,q\in {\mathcal {R}}^\bullet$. An element $b\in {\mathcal {R}}$ is the image-kernel (p, q)-inverse of a if

$$\begin{aligned} bab=b,\quad ba{\mathcal {R}}=p{\mathcal {R}}\quad \mathrm{and}\quad (1-ab){\mathcal {R}}=q{\mathcal {R}}. \end{aligned}$$

If the image-kernel (p, q)-inverse b exists, it is unique [6] and denoted by $a^{\times }_{p,q}$. We recall the reader that the image-kernel (p, q)-inverse of Kantún-Montiel [6] coincides with the (p, q, l)-outer generalized inverse of Cao and Xue [1]. For more details about the image-kernel (p, q)-inverses, see [9, 11, 12].

The special case of the image-kernel (p, q)-inverse is the (p, q)-outer generalized inverse. Precisely, for $p,q\in {\mathcal {R}}^\bullet$, the (p, q)-outer generalized inverse of a is a unique element $b\in {\mathcal {R}}$ (in the case when it exists) satisfying

$$\begin{aligned} bab=b, \quad ba=p\quad \mathrm{and}\quad 1-ab=q. \end{aligned}$$

In this case, we write $b=a^{(2)}_{p,q}$ [4]. Outer generalized inverses can be found in [3, 5, 8, 13].

If the image-kernel (p, q)-inverse b of a satisfies the equations $a=aba$, then b is a reflexive image-kernel (p, q)-inverse of a which will be denoted by $a^{(1,\times )}_{p,q}$. Evidently, $a^{(1,\times )}_{p,q}$ is unique if it exists.

An equivalent condition for the existence of the image-kernel (p, q)-inverse without explicit reference to ideals is given in the following lemma.

Lemma 1.1

[11, Theorem 2.1] Let $p,q\in {\mathcal {R}}^\bullet$ and let $a\in {\mathcal {R}}$. Then, the following statements are equivalent:

• (i) $a^{\times }_{p,q}$ exists,

• (ii) there exists some $b\in {\mathcal {R}}$ such that

  $$\begin{aligned} b=pb,\quad bap=p, \quad bq=0,\quad 1-q=(1-q)ab. \end{aligned}$$

Observe that the element b in the part (ii) of Lemma 1.1 satisfies $b=a^{\times }_{p,q}$.

An involution $a\mapsto a^*$ in a ring ${\mathcal {R}}$ is an anti-isomorphism of degree 2, that is,

$$\begin{aligned} (a^*)^*=a, \quad (a+b)^*=a^*+b^*, \quad (ab)^*=b^*a^*, \end{aligned}$$

for any $a, b\in {\mathcal {R}}$. An element $a\in {\mathcal {R}}$ is self-adjoint if $a^*=a$. An element $q\in {\mathcal {R}}$ is a projection if it is self-adjoint idempotent ($q^*=q=q^2$).

Let $p\in {\mathcal {R}}^\bullet$. The range projection of p is a projection $p^\bot$ such that $p^\bot p=p$ and $pp^\bot =p^\bot$ [7]. In a ring ${\mathcal {R}}$ with involution, $p^\bot =p$ for any projection p.

Lemma 1.2

([7, Theorem 2.1]) Let p be an idempotent in a ring ${\mathcal {R}}$ with involution. Then, the following conditions are equivalent:

• (i) $p+p^*-1$ is invertible in ${\mathcal {R}}$.

• (ii) $p^\bot$ and $(p^*)^\bot$ exist.

The range projections are unique, given by the formulae

$$\begin{aligned} p^\bot = p(p+p^*-1)^{-1},\qquad (p^*)^\bot =(p+p^*-1)^{-1}p. \end{aligned}$$

If a ring ${\mathcal {R}}$ with involution has the GN-property ($1 + x^*x\in {\mathcal {R}}^{-1}$ for all $x\in {\mathcal {R}}$), then every idempotent has a unique range projection.

Lemma 1.3

([2, Lemma 2.2]) Let $p\in {\mathcal {R}}^\bullet$ such that $p^\bot$ exists. If $f_p=1+p-p^\bot$, then $f_p\in {\mathcal {R}}^{-1}$ and $f_p^{-1}=f^*_{1-p^*}$.

For $p_1, p_2\in {\mathcal {R}}^\bullet$ such that $p_1p_2=p_2p_1=0$, the equality $p_1 + p_2=1$ is called a decomposition of the identity of the ring ${\mathcal {R}}$.

Let $p_1 + p_2=1$ and $q_1 + q_2=1$ be two decompositions of the identity of a ring ${\mathcal {R}}$. We can represent any $x\in {\mathcal {R}}$ as a matrix

$$\begin{aligned} x=\left[ \begin{array}{cc} x_{11}&{}x_{12}\\ x_{21}&{}x_{22} \end{array}\right] _{p_1\times q_1}, \end{aligned}$$

where $x_{ij}=p_ixq_j\in p_i{\mathcal {R}}q_j$. If any $y\in {\mathcal {R}}$ is written in a matrix form $y=[y_{ij}]_{q,p}$, $y_{ij}=q_iyp_j\in q_i{\mathcal {R}}p_j$, $i,j=\overline{1,2}$, we can use usual matrix rules in order to add and multiply x and y, because $p_1p_2=p_2p_1=0$ and $q_1q_2=q_2q_1=0$. In a ring with involution, notice that

$$\begin{aligned} x^*=\left[ \begin{array}{cc} x^*_{11}&{}x^*_{21}\\ x^*_{12}&{}x^*_{22} \end{array}\right] _{q^*_1\times p^*_1}. \end{aligned}$$

An element $a\in {\mathcal {R}}$ is group invertible if there exists a unique element $a^\#\in {\mathcal {R}}$ such that

$$\begin{aligned} aa^\# a=a,\quad a^\# aa^\#=a^\#,\quad aa^\#=a^\#a. \end{aligned}$$

We use ${\mathcal {R}}^\#$ to denote the set of all group invertible elements of ${\mathcal {R}}$. If $a\in {\mathcal {R}}^\#$, then $a^\pi =1-aa^\#$ is the spectral idempotent of a.

By elementary computations, we can check the following result which was proved for Banach algebra elements in [10].

Lemma 1.4

• (i) Let $x=\left[ \begin{array}{cc} a&{}b\\ 0&{}s \end{array}\right] _{p\times p}\in {\mathcal {R}}$. If $a\in (p{\mathcal {R}}p)^\#$, $s\in ((1-p){\mathcal {R}}(1-p))^\#$, and $a^\pi bs^\pi =0$, then

  $$\begin{aligned} x^\#=\left[ \begin{array}{cc} a^\#&{}(a^\#)^2bs^\pi -a^\#bs^\#+a^\pi b(s^\#)^{2}\\ 0&{}s^\# \end{array}\right] _{p\times p}. \end{aligned}$$

• (ii) Let $x=\left[ \begin{array}{cc} a&{}0\\ c&{}s \end{array}\right] _{p\times p}\in {\mathcal {R}}$. If $a\in (p{\mathcal {R}}p)^\#$, $s\in ((1-p){\mathcal {R}}(1-p))^\#$ and $s^\pi ca^\pi =0$, then

  $$\begin{aligned} x^\#=\left[ \begin{array}{cc} a^\#&{}0\\ s^\pi c(a^\#)^2-s^\#ca^\#+(s^\#)^{2}ca^\pi &{}s^\# \end{array}\right] _{p\times p}. \end{aligned}$$

Several explicit expressions for the (P, Q)-outer generalized inverses of bounded linear operators in Hilbert spaces are given in [2].

In this paper, equivalent conditions for the existence of the image-kernel (p, q)-inverses and new expressions for them in a ring with involution are presented, which generalize some results from [2]. Also, some properties of the image-kernel (p, q)-inverses are investigated.

2 Characterizations of the Image-Kernel (p, q)-Inverses

In the first theorem, we present necessary and sufficient conditions for the existence of the image-kernel (p, q)-inverse $a^{\times }_{p,q}$ and the reflexive image-kernel (p, q)-inverse $a^{(1,\times )}_{p,q}$ in a ring with involution, giving new representation for these inverse.

Theorem 2.1

Let $p,q\in {\mathcal {R}}^\bullet$ such that $p^\bot$ and $q^\bot$ exist and let $f_p=1+p-p^\bot$ and $f_q=1+q-q^\bot$. For $a\in {\mathcal {R}}$, then $a^{\times }_{p,q}$ exists if and only if

$$\begin{aligned} a=f_q^{-1} \left[ \begin{array}{cc} a_{1}&{}a_{2}\\ a_{3}&{}a_{4} \end{array}\right] _{(1-q^\bot )\times p^\bot }f_p \end{aligned}$$

(1)

and $(a_{1})^{(2)}_{p^\bot ,q^\bot }$ exists.

In this case,

$$\begin{aligned} a^{\times }_{p,q}=f_p^{-1} \left[ \begin{array}{cc}(a_{1})^{(2)}_{p^\bot ,q^\bot }&{}0\\ 0&{}0\end{array}\right] _{p^\bot \times (1-q^\bot )}f_q. \end{aligned}$$

(2)

Furthermore, $a^{(1,\times )}_{p,q}$ exists if and only if $a_4=a_3(a_{1})^{(2)}_{p^\bot ,q^\bot }a_2$ in (1). Then, $a^{(1,\times )}_{p,q}$ is represented as in (2).

Proof

$\Rightarrow$: We have the following representations of p and q:

$$\begin{aligned} p=\left[ \begin{array}{cc} p^\bot &{}p_1\\ 0&{}0\end{array}\right] _{p^\bot \times p^\bot },\quad q=\left[ \begin{array}{cc}0&{}0\\ q_1&{}q^\bot \end{array}\right] _{(1-q^\bot )\times (1-q^\bot )}. \end{aligned}$$

(3)

Assume that

$$\begin{aligned} a=\left[ \begin{array}{cc}a_{1}&{}a_{2}+a_1p_1\\ a_{3}-q_1a_1&{}a_{4}+a_3p_1-q_1a_2-q_1a_1p_1 \end{array}\right] _{(1-q^\bot )\times p^\bot }, \end{aligned}$$

for $a_1\in (1-q^\bot ){\mathcal {R}}p^\bot$, $a_2\in (1-q^\bot ){\mathcal {R}}(1-p^\bot )$, $a_3\in q^\bot {\mathcal {R}}p^\bot$, $a_4\in q^\bot {\mathcal {R}}(1-p^\bot )$, and

$$\begin{aligned} a^{\times }_{p,q}=b= \left[ \begin{array}{cc}b_{1}&{}b_{2}\\ b_{3}&{}b_{4} \end{array}\right] _{p^\bot \times (1-q^\bot )}. \end{aligned}$$

Since $b=pb$ and

$$\begin{aligned} b=b(1-q)=b \left[ \begin{array}{cc}1-q^\bot &{}0\\ -q_1&{}0 \end{array}\right] _{(1-q^\bot )\times (1-q^\bot )}, \end{aligned}$$

we obtain $b_2=0$, $b_3=0$, and $b_4=0$. The equality $bap=p$ is equivalent to $b_1a_1p^\bot =p^\bot$ and $b_1a_1p_1=p_1$, that is, $b_1a_1=p^\bot$. Similarly, $(1-q)=(1-q)ab$ if and only if $1-q^\bot =a_1b_1$. From $b=bab$, we get $b_1=b_1a_1b_1$. Thus, $b_1=(a_{1})^{(2)}_{p^\bot ,q^\bot }$. Also, observe that

$$\begin{aligned} f_q^{-1} \left[ \begin{array}{cc} a_{1}&{}a_{2}\\ a_{3}&{}a_{4} \end{array}\right] _{(1-q^\bot )\times p^\bot }f_p= & {} \left[ \begin{array}{cc} 1-q^\bot &{}0\\ -q_1&{}q^\bot \end{array}\right] _{(1-q^\bot )\times (1-q^\bot )} \left[ \begin{array}{cc} a_{1}&{}a_{2}\\ a_{3}&{}a_{4}\end{array}\right] _{(1-q^\bot )\times p^\bot }\\&\times \left[ \begin{array}{cc}p^\bot &{}p_1\\ 0&{}1-p^\bot \end{array}\right] _{p^\bot \times p^\bot }\\= & {} \left[ \begin{array}{cc} a_{1}&{}a_{2}+a_1p_1\\ a_{3}-q_1a_1&{}a_{4}+a_3p_1-q_1a_2-q_1a_1p_1 \end{array}\right] _{(1-q^\bot )\times p^\bot }\\= & {} a \end{aligned}$$

and

$$\begin{aligned} f_p^{-1}\left[ \begin{array}{cc} (a_{1})^{(2)}_{p^\bot ,q^\bot }&{}0\\ 0&{}0\end{array}\right] _{p^\bot \times (1-q^\bot )}f_q= & {} \left[ \begin{array}{cc}p^\bot &{}-p_1\\ 0&{}1-p^\bot \end{array}\right] _{p^\bot \times p^\bot }\left[ \begin{array}{cc}(a_{1})^{(2)}_{p^\bot ,q^\bot }&{}0\\ 0&{}0\end{array}\right] _{p^\bot \times (1-q^\bot )}\\&\times \left[ \begin{array}{cc}1-q^\bot &{}0\\ q_1&{}q^\bot \end{array}\right] _{(1-q^\bot )\times (1-q^\bot )}\\ {}= & {} \left[ \begin{array}{cc}(a_{1})^{(2)}_{p^\bot ,q^\bot }&{}0\\ 0&{}0\end{array}\right] _{p^\bot \times (1-q^\bot )}=b=a^{\times }_{p,q}. \end{aligned}$$

Using (1) and (2), if $a^{(1,\times )}_{p,q}$ exists, then $a=aa^{\times }_{p,q}a$ gives $a_4=a_3(a_{1})^{(2)}_{p^\bot ,q^\bot }a_2$ and $a^{(1,\times )}_{p,q}=a^{\times }_{p,q}$.

$\Leftarrow$: Suppose that a is represented by (1). We can easily verify that the element on the right side of (2) is the image-kernel (p, q)-inverse of a. $\square$

We now characterize the existence of the image-kernel (p, q)-inverses $c^{\times }_{1-q,1-p}$ and $c^{(1,\times )}_{1-q,1-p}$.

Theorem 2.2

Let $p,q\in {\mathcal {R}}^\bullet$ such that $p^\bot$ and $q^\bot$ exist and let $f_p=1+p-p^\bot$ and $f_q=1+q-q^\bot$. For $c\in {\mathcal {R}}$, then $c^{\times }_{1-q,1-p}$ exists if and only if

$$\begin{aligned} c=f_p^{-1}\left[ \begin{array}{cc}c_{1}&{}c_{2}\\ c_{3}&{}c_{4}\end{array}\right] _{p^\bot \times (1-q^\bot )}f_q \end{aligned}$$

(4)

and $(c_{1})^{(2)}_{1-q^\bot ,1-p^\bot }$ exists.

In this case,

$$\begin{aligned} c^{\times }_{1-q,1-p}=f_q^{-1} \left[ \begin{array}{cc}(c_{1})^{(2)}_{1-q^\bot ,1-p^\bot }&{}0\\ 0&{}0\end{array}\right] _{(1-q^\bot )\times p^\bot }f_p. \end{aligned}$$

(5)

Furthermore, $c^{(1,\times )}_{1-q,1-p}$ exists if and only if $c_4=c_3(c_{1})^{(2)}_{1-q^\bot ,1-p^\bot }c_2$ in (4). Then, $c^{(1,\times )}_{1-q,1-p}$ is represented as in (5).

Proof

Suppose that p and q are represented as in (3). Let

$$\begin{aligned} c=\left[ \begin{array}{cc}c_{1}+c_2q_1-p_1c_3-p_1c_4q_1&{}c_{2}-p_1c_4\\ c_{3}+c_4q_1&{}c_{4}\end{array}\right] _{p^\bot \times (1-q^\bot )}, \end{aligned}$$

where $c_1\in p^\bot {\mathcal {R}}(1-q^\bot )$, $c_2\in p^\bot {\mathcal {R}}q^\bot$, $c_3\in (1-p^\bot ){\mathcal {R}}(1-q^\bot )$, $c_4\in (1-p^\bot ){\mathcal {R}}q^\bot$, and

$$\begin{aligned} c^{\times }_{1-q,1-p}=b^{\prime }=\left[ \begin{array}{cc}b^{\prime }_{1}&{}b^{\prime }_{2}\\ b^{\prime }_{3}&{}b^{\prime }_{4}\end{array}\right] _{(1-q^\bot )\times p^\bot }. \end{aligned}$$

Using the equalities

$$\begin{aligned} b^{\prime }=(1-q)b^{\prime }=\left[ \begin{array}{cc}1-q^\bot &{}0\\ -q_1&{}0\end{array}\right] _{(1-q^\bot )\times (1-q^\bot )}b^{\prime } \end{aligned}$$

and

$$\begin{aligned} 0=b^{\prime }(1-p)=b^{\prime } \left[ \begin{array}{cc}0&{}-p_1\\ 0&{}1-p^\bot \end{array}\right] _{p^\bot \times p^\bot }, \end{aligned}$$

we deduce that $b^{\prime }_2=b^{\prime }_1p_1$, $b^{\prime }_3=-q_1b^{\prime }_1$, and $b^{\prime }_4=-q_1b^{\prime }_1p_1$. So,

$$\begin{aligned} b^{\prime }=\left[ \begin{array}{cc}b^{\prime }_{1}&{}b^{\prime }_1p_1\\ -q_1b^{\prime }_1&{}-q_1b^{\prime }_1p_1\end{array}\right] _{(1-q^\bot )\times p^\bot }. \end{aligned}$$

Notice that $b^{\prime }c(1-q)=(1-q)$ holds if and only if $b^{\prime }_1c_1=1-q^\bot$. Also, $p=pcb'$ is equivalent to $c_1b^{\prime }_1=p^\bot$. Because $b^{\prime }cb^{\prime }=b^{\prime }$ yields $b^{\prime }_1c_1b^{\prime }_1=b^{\prime }_1$, we conclude that $b^{\prime }_1=(c_1)^{(2)}_{1-q^\bot , 1-p^\bot }$. Therefore,

$$\begin{aligned} f_p^{-1}\left[ \begin{array}{cc}c_{1}&{}c_{2}\\ c_{3}&{}c_{4}\end{array}\right] _{p^\bot \times (1\!-\!q^\bot )}f_q\!=\!\left[ \begin{array}{cc}c_{1}\!+\!c_2q_1\!-\!p_1c_3-p_1c_4q_1&{}c_{2}\!-\!p_1c_4\\ c_{3}\!+\!c_4q_1&{}c_{4}\end{array}\right] _{p^\bot \times (1-q^\bot )}\!=\!c \end{aligned}$$

and

$$\begin{aligned} f_q^{-1} \left[ \begin{array}{cc}(c_{1})^{(2)}_{1-q^\bot ,1-p^\bot }&{}0\\ 0&{}0\end{array}\right] _{(1-q^\bot )\times p^\bot }f_p= & {} \left[ \begin{array}{cc}b^{\prime }_{1}&{}b^{\prime }_1p_1\\ -q_1b^{\prime }_1&{}-q_1b^{\prime }_1p_1\end{array}\right] _{(1-q^\bot )\times p^\bot }\\= & {} b'=c^{\times }_{1-q,1-p}. \end{aligned}$$

Observe that $cc^{(1,\times )}_{1-q,1-p}c=c$ is equivalent to $c_4=c_3(c_{1})^{(2)}_{1-q^\bot ,1-p^\bot }c_2$. $\square$

As Theorems 2.1 and 2.2, we verify the following result.

Theorem 2.3

Let $p,q\in {\mathcal {R}}^\bullet$ such that $p^\bot$ and $q^\bot$ exist and let $f_p=1+p-p^\bot$ and $f_q=1+q-q^\bot$. For $a\in {\mathcal {R}}$, then

• (i) $a^{\times }_{p,1-q}$ exists if and only if a is represented as in (1) and $(a_{3})^{(2)}_{p^\bot ,1-q^\bot }$ exists.

  In this case,

  $$\begin{aligned} a^{\times }_{p,1-q}= & {} f_p^{-1} \left[ \begin{array}{cc}0&{}(a_{3})^{(2)}_{p^\bot ,1-q^\bot }\\ 0&{}0\end{array}\right] _{p^\bot \times (1-q^\bot )}f_q\\= & {} \left[ \begin{array}{cc}(a_{3})^{(2)}_{p^\bot ,1-q^\bot }q_1&{}(a_{3})^{(2)}_{p^\bot ,1-q^\bot }\\ 0&{}0\end{array}\right] _{p^\bot \times (1-q^\bot )}, \end{aligned}$$

  where $q_1=q^\bot q(1-q^\bot )=q-q^\bot$.

• (ii) $a^{\times }_{1-p,q}$ exists if and only if a is represented as in (1) and $(a_{2})^{(2)}_{1-p^\bot ,q^\bot }$ exists.

  In this case,

  $$\begin{aligned}a^{\times }_{1-p,q}= & {} f_p^{-1} \left[ \begin{array}{cc}0&{}0\\ (a_{2})^{(2)}_{1-p^\bot ,q^\bot }&{}0\end{array}\right] _{p^\bot \times (1-q^\bot )}f_q\\ {}= & {} \left[ \begin{array}{cc}-p_1(a_{2})^{(2)}_{1-p^\bot ,q^\bot }&{}0\\ (a_{2})^{(2)}_{1-p^\bot ,q^\bot }&{}0\end{array}\right] _{p^\bot \times (1-q^\bot )}, \end{aligned}$$

  where $p_1=p^\bot p(1-p^\bot )=p-p^\bot$.

• (iii) $a^{\times }_{1-p,1-q}$ exists if and only if a is represented as in (1) and $(a_{4})^{(2)}_{1-p^\bot ,1-q^\bot }$ exists.

  In this case,

  $$\begin{aligned} a^{\times }_{1-p,1-q}= & {} f_p^{-1} \left[ \begin{array}{cc}0&{}0\\ 0&{}(a_{4})^{(2)}_{1-p^\bot ,1-q^\bot }\end{array}\right] _{p^\bot \times (1-q^\bot )}f_q\\ {}= & {} \left[ \begin{array}{cc}-p_1(a_{4})^{(2)}_{1-p^\bot ,1-q^\bot }q_1&{}-p_1(a_{4})^{(2)}_{1-p^\bot , 1-q^\bot }\\ (a_{4})^{(2)}_{1-p^\bot ,1-q^\bot }q_1&{}(a_{4})^{(2)}_{1-p^\bot ,1-q^\bot } \end{array}\right] _{p^\bot \times (1-q^\bot )}, \end{aligned}$$

  where $p_1$ and $q_1$ are defined as in parts (i) and (ii).

  Furthermore, $a^{(1,\times )}_{1-p,1-q}$ exists if and only if $a_1=a_2(a_{4})^{(2)}_{1-p^\bot ,1-q^\bot }a_3$ in (1). Then, $a^{(1,\times )}_{1-p,1-q}=a^{\times }_{1-p,1-q}$.

Some interesting properties of the image-kernel (p, q)-inverse are studied in the next theorem.

Theorem 2.4

Let $a\in {\mathcal {R}}$ and $p,q\in {\mathcal {R}}^\bullet$. Then,

• (i) $a^{\times }_{p,q}$ exists if and only if $(a^*)^{\times }_{1-q^*,1-p^*}$ exists. In addition, $(a^{\times }_{p,q})^*=(a^*)^{\times }_{1-q^*,1-p^*}$.

• (ii) If $a^{\times }_{p,q}$ exists, then $(a^{\times }_{p,q})^2=a^{\times }_{p,q}$ if and only if $a^{\times }_{p,q}p=p$.

Proof

(i) By Theorem 1.1, $a^{\times }_{p,q}$ exists if and only if there exists some $b\in {\mathcal {R}}$ such that

$$\begin{aligned} b=pb,\quad bap=p, \quad bq=0,\quad 1-q=(1-q)ab, \end{aligned}$$

which is equivalent to the existence of some $b\in {\mathcal {R}}$ such that

$$\begin{aligned} b^*(1-p^*)=0,\quad p^*a^*b^*=p^*, \quad (1-q^*)b^*=b^*,\quad 1-q^*=b^*a^*(1-q^*), \end{aligned}$$

that is, the existence of $(a^*)^{\times }_{1-q^*,1-p^*}$.

(ii) If $(a^{\times }_{p,q})^2=a^{\times }_{p,q}$, we get

$$\begin{aligned} p=a^{\times }_{p,q}ap=(a^{\times }_{p,q})^2ap=a^{\times }_{p,q}(a^{\times }_{p,q}ap) =a^{\times }_{p,q}p. \end{aligned}$$

Conversely, the equality $a^{\times }_{p,q}p=p$ gives

$$\begin{aligned} a^{\times }_{p,q}=pa^{\times }_{p,q}=a^{\times }_{p,q}pa^{\times }_{p,q} =(a^{\times }_{p,q})^2. \end{aligned}$$

$\square$

In the following result, we consider the image-kernel $(p,1-p)$-inverse.

Theorem 2.5

Let $a\in {\mathcal {R}}$, $p\in {\mathcal {R}}^\bullet$ such that $p^\bot$ exists and let $f_p=1+p-p^\bot$. If $a^{\times }_{p,1-p}$ exists, then

• (i) $a^{\times }_{p,1-p}a=aa^{\times }_{p,1-p}$ if and only if

  $$\begin{aligned} a=f_p^{-1}\left[ \begin{array}{cc} a_{1}&{}0\\ 0&{}a_{4}\end{array}\right] _{p^\bot \times p^\bot }f_p. \end{aligned}$$

  (6)

• (ii) $a^{\times }_{p,1-p}$ is self-adjoint if and only if $p=p^*$ and $p^\bot ap^\bot$ is self-adjoint.

Proof

Assume that p is represented as in (3),

$$\begin{aligned} a=\left[ \begin{array}{cc}a_{1}-p_1a_3&{}a_1p_1+a_{2}-p_1a_3p_1-p_1a_4\\ a_{3}&{}a_3p_1-a_{4}\end{array}\right] _{p^\bot \times p^\bot }, \end{aligned}$$

(7)

for $a_1\in p^\bot {\mathcal {R}}p^\bot$, $a_2\in p^\bot {\mathcal {R}}(1-p^\bot )$, $a_3\in (1-p^\bot ){\mathcal {R}}p^\bot$, $a_4\in (1-p^\bot ){\mathcal {R}}(1-p^\bot )$, and

$$\begin{aligned} a^{\times }_{p,1-p}=b=\left[ \begin{array}{cc}b_{1}&{}b_{2}\\ b_{3}&{}b_{4}\end{array}\right] _{p^\bot \times p^\bot }. \end{aligned}$$

Using $b=pb$, we obtain that $b_3=0$ and $b_4=0$. By the equality $b(1-p)=0$, we deduce that $b_2=b_1p_1$. Since

$$\begin{aligned} p=bap=\left[ \begin{array}{cc}b_{1}a_1&{}b_{1}a_1p_1\\ 0&{}0\end{array}\right] _{p^\bot \times p^\bot } \end{aligned}$$

and

$$\begin{aligned} p=pab=\left[ \begin{array}{cc}a_1b_{1}&{}a_1b_{1}p_1\\ 0&{}0\end{array}\right] _{p^\bot \times p^\bot }, \end{aligned}$$

then $b_{1}a_1=p^\bot$ and $a_1b_{1}=p^\bot$. From $bab=b$, it follows that $b_1a_1b_1=b_1$. Thus, $b_1=(a_{1})^{(2)}_{p^\bot ,1-p^\bot }$,

$$\begin{aligned} f_p^{-1}\left[ \begin{array}{cc}(a_{1})^{(2)}_{p^\bot ,1-p^\bot }&{}0\\ 0&{}0\end{array}\right] _{p^\bot \times p^\bot }f_p= & {} \left[ \begin{array}{cc} (a_{1})^{(2)}_{p^\bot ,1-p^\bot }&{}(a_{1})^{(2)}_{p^\bot ,1-p^\bot }p_1\\ 0&{}0\end{array}\right] _{p^\bot \times p^\bot }\\= & {} b=a^{\times }_{p,1-p} \end{aligned}$$

and

$$\begin{aligned} f_p^{-1}\left[ \begin{array}{cc}a_{1}&{}a_{2}\\ a_{3}&{}a_{4}\end{array}\right] _{p^\bot \times p^\bot }f_p= \left[ \begin{array}{cc}a_{1}-p_1a_3&{}a_1p_1+a_{2}-p_1a_3p_1-p_1a_4\\ a_{3}&{}a_3p_1-a_{4}\end{array}\right] _{p^\bot \times p^\bot }=a. \end{aligned}$$

(i) Since

$$\begin{aligned} a^{\times }_{p,1-p}a= & {} f_p^{-1}\left[ \begin{array}{cc} (a_{1})^{(2)}_{p^\bot ,1-p^\bot }a_1&{}(a_{1})^{(2)}_{p^\bot ,1-p^\bot }a_2\\ 0&{}0\end{array}\right] _{p^\bot \times p^\bot }f_p\\= & {} f_p^{-1}\left[ \begin{array}{cc}p^\bot &{}(a_{1})^{(2)}_{p^\bot ,1-p^\bot }a_2\\ 0&{}0\end{array}\right] _{p^\bot \times p^\bot }f_p \end{aligned}$$

and

$$\begin{aligned} aa^{\times }_{p,1-p}=f_p^{-1} \left[ \begin{array}{cc}a_1(a_{1})^{(2)}_{p^\bot ,1-p^\bot }&{}0\\ a_3(a_{1})^{(2)}_{p^\bot ,1-p^\bot }&{}0\end{array}\right] _{p^\bot \times p^\bot }f_p =f_p^{-1}\left[ \begin{array}{cc}p^\bot &{}0\\ a_3(a_{1})^{(2)}_{p^\bot ,1-p^\bot }&{}0 \end{array}\right] _{p^\bot \times p^\bot }f_p, \end{aligned}$$

notice that $a^{\times }_{p,1-p}a=aa^{\times }_{p,1-p}$ if and only if $a_{1}(a_{1})^{(2)}_{p^\bot ,1-p^\bot }a_2=0$ and $a_3(a_{1})^{(2)}_{p^\bot ,1-p^\bot }a_{1}=0$ which is equivalent to $a_2=0$ and $a_3=0$.

(ii) Using $a^{\times }_{p,1-p}=\left[ \begin{array}{cc}(a_{1})^{(2)}_{p^\bot ,1-p^\bot } &{}(a_{1})^{(2)}_{p^\bot ,1-p^\bot }p_1\\ 0&{}0\end{array}\right] _{p^\bot \times p^\bot }$, we observe that $(a^{\times }_{p,1-p})^*=a^{\times }_{p,1-p}$ is equivalent to $((a_{1})^{(2)}_{p^\bot ,1-p^\bot })^*=(a_{1})^{(2)}_{p^\bot ,1-p^\bot }$ and $p_1=0$. Also, $p=p^*$ if and only if $p_1=0$. When $p_1=0$ in (7), we have

$$\begin{aligned} a=\left[ \begin{array}{cc}a_{1}&{}a_{2}\\ a_{3}&{}a_{4}\end{array}\right] _{p^\bot \times p^\bot }. \end{aligned}$$

If $p^\bot ap^\bot =a_1=a_1^*$, then $(a_{1})^{(2)}_{p^\bot ,1-p^\bot }=(a_{1}^*)^{(2)}_{p^\bot ,1-p^\bot } =((a_{1})^{(2)}_{p^\bot ,1-p^\bot })^*$. Because $a_1=p^\bot a_1= a_1(a_{1})^{(2)}_{p^\bot ,1-p^\bot }a_1$, from $(a_{1})^{(2)}_{p^\bot ,1-p^\bot }=((a_{1})^{(2)}_{p^\bot ,1-p^\bot })^* =(a_{1}^*)^{(2)}_{p^\bot ,1-p^\bot }$,

$$\begin{aligned} a_1=[(a_{1})^{(2)}_{p^\bot ,1-p^\bot }]^{(2)}_{p^\bot ,1-p^\bot } =[(a_{1}^*)^{(2)}_{p^\bot ,1-p^\bot }]^{(2)}_{p^\bot ,1-p^\bot }=a_{1}^*. \end{aligned}$$

$\square$

New representations for $a^{\times }_{p,q}$ and $c^{\times }_{1-q,1-p}$ involving the corresponding group inverse are given now.

Theorem 2.6

Let $p,q\in {\mathcal {R}}^\bullet$ such that $p^\bot$ and $q^\bot$ exist and let $a,c\in {\mathcal {R}}$ such that $a^{\times }_{p,q}$ and $c^{\times }_{1-q,1-p}$ exist. Then,

• (i) $pc(1-q)a, pc(1-q)ap\in {\mathcal {R}}^\#$,

  $$\begin{aligned}&a^{\times }_{p,q}=[pc(1-q)a]^\#pc(1-q)=[pc(1-q)ap]^\#pc(1-q),\\&c^{\times }_{1-q,1-p}=(1-q)a[pc(1-q)ap]^\#; \end{aligned}$$

• (ii) $apc(1-q), (1-q)apc(1-q)\in {\mathcal {R}}^\#$,

  $$\begin{aligned}&a^{\times }_{p,q}=pc(1-q)[apc(1-q)]^\#=pc[(1-q)apc(1-q)]^\#,\\&c^{\times }_{1-q,1-p}=[(1-q)apc(1-q)]^\#(1-q)ap. \end{aligned}$$

Proof

(i) Set $f_p=1+p-p^\bot$ and $f_q=1+q-q^\bot$. Applying Theorems 2.1 and 2.2,

$$\begin{aligned} pc(1-q)a= & {} f_p^{-1}\left[ \begin{array}{cc}p^\bot &{}0\\ 0&{}0\end{array}\right] _{p^\bot \times p^\bot }f_pf_p^{-1}\left[ \begin{array}{cc}c_{1}&{}c_{2}\\ c_{3}&{}c_{4}\end{array}\right] _{p^\bot \times (1-q^\bot )}f_q\\&\times f_q^{-1}\left[ \begin{array}{cc}1-q^\bot &{}0\\ 0&{}0\end{array}\right] _{(1-q^\bot ) \times (1-q^\bot )}f_qf_q^{-1}\left[ \begin{array}{cc}a_{1}&{}a_{2}\\ a_{3}&{}a_{4}\end{array}\right] _{(1-q^\bot )\times p^\bot }f_p\\= & {} f_p^{-1}\left[ \begin{array}{cc}c_{1}a_{1}&{}c_{1}a_{2}\\ 0&{}0\end{array}\right] _{p^\bot \times p^\bot }f_p. \end{aligned}$$

We prove that $c_{1}a_{1}$ is group invertible and $(c_{1}a_{1})^\#=(a_{1})^{(2)}_{p^\bot ,q^\bot }(c_{1})^{(2)}_{1-q^\bot ,1-p^\bot }$ by

$$\begin{aligned}&c_{1}a_{1}(a_{1})^{(2)}_{p^\bot ,q^\bot }(c_{1})^{(2)}_{1-q^\bot ,1-p^\bot } \!=\!c_{1}(1-q^\bot )(c_{1})^{(2)}_{1-q^\bot ,1-p^\bot } \!=\!c_{1}(c_{1})^{(2)}_{1-q^\bot ,1-p^\bot }\!=\!p^\bot ,\\&(a_{1})^{(2)}_{p^\bot ,q^\bot }(c_{1})^{(2)}_{1-q^\bot ,1-p^\bot }c_{1}a_{1}= (a_{1})^{(2)}_{p^\bot ,q^\bot }(1-q^\bot )a_{1}=(a_{1})^{(2)}_{p^\bot ,q^\bot }a_{1} =p^\bot ,\\&c_{1}a_{1}(a_{1})^{(2)}_{p^\bot ,q^\bot }(c_{1})^{(2)}_{1-q^\bot ,1-p^\bot }c_{1}a_{1} =p^\bot c_{1}a_{1}=c_{1}a_{1},\\&(a_{1})^{(2)}_{p^\bot ,q^\bot }(c_{1})^{(2)}_{1-q^\bot ,1-p^\bot }c_{1}a_{1} (a_{1})^{(2)}_{p^\bot ,q^\bot }(c_{1})^{(2)}_{1-q^\bot ,1-p^\bot } =p^\bot (a_{1})^{(2)}_{p^\bot ,q^\bot }(c_{1})^{(2)}_{1-q^\bot ,1-p^\bot }\\&\quad =(a_{1})^{(2)}_{p^\bot ,q^\bot }(c_{1})^{(2)}_{1-q^\bot ,1-p^\bot }. \end{aligned}$$

Since $(c_{1}a_{1})^\pi c_{1}a_2 =(p^\bot -c_{1}a_{1}(a_{1})^{(2)}_{p^\bot ,q^\bot }(c_{1})^{(2)}_{1-q^\bot ,1-p^\bot }) c_{1}a_2=0$, by Lemma 1.4,

$$\begin{aligned}&\left[ \begin{array}{cc}c_{1}a_{1}&{}c_{1}a_{2}\\ 0&{}0\end{array}\right] _{p^\bot \times p^\bot }^\#\\&\quad = \left[ \begin{array}{cc}(a_{1})^{(2)}_{p^\bot ,q^\bot }(c_{1})^{(2)}_{1-q^\bot ,1-p^\bot }&{} ((a_{1})^{(2)}_{p^\bot ,q^\bot }(c_{1})^{(2)}_{1-q^\bot ,1-p^\bot }))^2c_{1}a_{2}\\ 0&{}0\end{array}\right] _{p^\bot \times p^\bot }. \end{aligned}$$

Therefore, $pc(1-q)a\in {\mathcal {R}}^\#$ and

$$\begin{aligned}&[pc(1-q)a]^\#\\&\quad =f_p^{-1}\left[ \begin{array}{cc}(a_{1})^{(2)}_{p^\bot ,q^\bot } (c_{1})^{(2)}_{1-q^\bot ,1-p^\bot }&{} ((a_{1})^{(2)}_{p^\bot ,q^\bot }(c_{1})^{(2)}_{1-q^\bot ,1-p^\bot }))^2c_{1}a_{2}\\ 0&{}0\end{array}\right] _{p^\bot \times p^\bot }f_p \end{aligned}$$

implying

$$\begin{aligned}{}[pc(1-q)a]^\#pc(1-q)= & {} [pc(1-q)a]^\#f_p^{-1} \left[ \begin{array}{cc}c_{1}&{}0\\ 0&{}0\end{array}\right] _{p^\bot \times (1-q^\bot )}f_q\\= & {} f_p^{-1}\left[ \begin{array}{cc} (a_{1})^{(2)}_{p^\bot ,q^\bot }(c_{1})^{(2)}_{1-q^\bot ,1-p^\bot }c_{1}&{} 0\\ 0&{}0\end{array}\right] _{p^\bot \times (1-q^\bot )}f_q\\= & {} f_p^{-1}\left[ \begin{array}{cc}(a_{1})^{(2)}_{p^\bot ,q^\bot }&{} 0\\ 0&{}0\end{array}\right] _{p^\bot \times (1-q^\bot )}f_q\\= & {} a^{\times }_{p,q}. \end{aligned}$$

From

$$\begin{aligned} pc(1-q)ap= & {} f_p^{-1}\left[ \begin{array}{cc}c_{1}a_{1}&{}c_{1}a_{2}\\ 0&{}0\end{array}\right] _{p^\bot \times p^\bot }f_p f_p^{-1}\left[ \begin{array}{cc} p^\bot &{}0\\ 0&{}0\end{array}\right] _{p^\bot \times p^\bot }f_p\\= & {} f_p^{-1}\left[ \begin{array}{cc}c_{1}a_{1}&{}0\\ 0&{}0 \end{array}\right] _{p^\bot \times p^\bot }f_p, \end{aligned}$$

we deduce that $pc(1-q)ap\in {\mathcal {R}}^\#$ and

$$\begin{aligned}{}[pc(1-q)ap]^\#=f_p^{-1} \left[ \begin{array}{cc}(a_{1})^{(2)}_{p^\bot ,q^\bot }(c_{1})^{(2)}_{1-q^\bot ,1-p^\bot }&{}0\\ 0&{}0\end{array}\right] _{p^\bot \times p^\bot }f_p. \end{aligned}$$

Thus,

$$\begin{aligned} (1-q)a[pc(1-q)ap]^\#= & {} f_q^{-1}\left[ \begin{array}{cc}a_{1}&{}a_{2}\\ 0&{}0\end{array}\right] _{(1-q^\bot )\times p^\bot }f_p[pc(1-q)ap]^\#\\= & {} f_q^{-1}\left[ \begin{array}{cc}a_{1}(a_{1})^{(2)}_{p^\bot ,q^\bot } (c_{1})^{(2)}_{1-q^\bot ,1-p^\bot }&{} 0\\ 0&{}0\end{array}\right] _{(1-q^\bot )\times p^\bot }f_p\\= & {} f_q^{-1}\left[ \begin{array}{cc}(c_{1})^{(2)}_{1-q^\bot ,1-p^\bot }&{} 0\\ 0&{}0\end{array}\right] _{(1-q^\bot )\times p^\bot }f_p\\ {}= & {} c^{\times }_{1-q,1-p} \end{aligned}$$

and

$$\begin{aligned}{}[pc(1-q)ap]^\#pc(1-q)= & {} f_p^{-1}\left[ \begin{array}{cc}(a_{1})^{(2)}_{p^\bot ,q^\bot } (c_{1})^{(2)}_{1-q^\bot ,1-p^\bot }c_{1}&{} 0\\ 0&{}0\end{array}\right] _{p^\bot \times (1-q^\bot )}f_q\\= & {} f_p^{-1}\left[ \begin{array}{cc}(a_{1})^{(2)}_{p^\bot ,q^\bot }&{} 0\\ 0&{}0\end{array}\right] _{p^\bot \times (1-q^\bot )}f_q\\ {}= & {} a^{\times }_{p,q}. \end{aligned}$$

The proof of part (ii) is similar to the proof of (i), so it is omitted. $\square$

By the properties of the group inverse and the idempotency of p and $1-q$, we get the next consequence of Theorem 2.6.

Corollary 2.1

Let $p,q\in {\mathcal {R}}^\bullet$ such that $p^\bot$ and $q^\bot$ exist and $a,c\in {\mathcal {R}}$ such that $a^{\times }_{p,q}$ and $c^{\times }_{1-q,1-p}$ exist. Then, $pc(1-q)ap, (1-q)apc(1-q)\in {\mathcal {R}}^\#$,

$$\begin{aligned}&a^{\times }_{p,q}= pc(1-q)[(1-q)apc(1-q)]^\#\\&c^{\times }_{1-q,1-p}= (1-q)ap[pc(1-q)ap]^\# \end{aligned}$$

In a ring ${\mathcal {R}}$ with involution and the GN-property, we can omit the assumptions $p^\bot$ and $q^\bot$ exist in the previous results.

Let ${\mathcal {A}}$ be a $C^*$-algebra with unit 1. An element $x\in {\mathcal {A}}$ is positive (denoted by $x\ge 0$) if $x = x^*$ and $\sigma (x)\subseteq [0,+\infty )$, where $\sigma (x)$ denotes the spectrum of x. We check the next theorem.

Theorem 2.7

Let $a,b\in {\mathcal {A}}$ and $p,q\in {\mathcal {A}}^\bullet$. If $a^{\times }_{p,q}$ exists and $a^{\times }_{p,q}\ge 0$, then $(a+bb^*)^{\times }_{p,q}$ exists and

$$\begin{aligned} (a+bb^*)^{\times }_{p,q}=a^{\times }_{p,q} -a^{\times }_{p,q}b(1+b^*a^{\times }_{p,q}b)^{-1}b^*a^{\times }_{p,q} \end{aligned}$$

(8)

Proof

The hypothesis $a^{\times }_{p,q}\ge 0$ implies that $a^{\times }_{p,q}=s^*s$, for some $s\in {\mathcal {A}}$. So, $b^*a^{\times }_{p,q}b=b^*s^*sb=(sb)^*sb\ge 0$ which yields that $1+b^*a^{\times }_{p,q}b$ is invertible. Denote by $x=a+bb^*$ and by y the right side of (8). Then, from $py=y$, $yq=0$,

$$\begin{aligned} yxp= & {} a^{\times }_{p,q}ap+a^{\times }_{p,q}bb^*p-a^{\times }_{p,q}b(1+b^*a^{\times }_{p,q}b)^{-1}(b^*a^{\times }_{p,q}ap+b^*a^{\times }_{p,q}bb^*p)\\= & {} p+a^{\times }_{p,q}bb^*p-a^{\times }_{p,q}b(1+b^*a^{\times }_{p,q}b)^{-1}(1+b^*a^{\times }_{p,q}b)b^*p\\= & {} p \end{aligned}$$

and

$$\begin{aligned} (1-q)xy= & {} (1-q)aa^{\times }_{p,q}+(1-q)bb^*a^{\times }_{p,q}\\&-\,(1-q)(aa^{\times }_{p,q}b+bb^*a^{\times }_{p,q}b)(1+b^*a^{\times }_{p,q}b)^{-1}b^*a^{\times }_{p,q}\\= & {} (1-q)+(1-q)bb^*a^{\times }_{p,q}\\&-\,(1-q)b(1+b^*a^{\times }_{p,q}b)(1+b^*a^{\times }_{p,q}b)^{-1}b^*a^{\times }_{p,q}\\= & {} 1-q. \end{aligned}$$

We conclude that $x^{\times }_{p,q}$ exists and $x^{\times }_{p,q}=y$. $\square$