New Characterizations of SEP Elements in a Ring with Involution

Abstract

The characterization of SEP elements is a classical problem in generalized inverse theory. Most existing characterizations are formulated in terms of specific algebraic identities. This paper proposes a new approach based on polynomial equations with parameters to characterize SEP elements. This framework provides an alternative characterization and, more importantly, naturally clarifies the structural relationships between SEP elements and three other types of elements: square rootable elements, invertible elements, and involutional projection elements.

1. Introduction

Let R be an associative ring with 1, and let $a\in R$. Then a is called group invertible if there is $a^{\#}\in R$ such that

$$a{a}^{\#}a=a,a^{\#}a{a}^{\#}=a^{\#},a{a}^{\#}=a^{\#}a;$$

$a^{\#}$ is a group inverse of a and it is uniquely determined by these equations [1]. We use $R^{\#}$ to denote the set of all group invertible elements of R. Recent work can be referenced in [2].

An involution ∗:$a\mapsto a^{*}$ in R is an anti-isomorphism of drgee 2, that is,

$${\left(a^{*}\right)}^{*}=a,{\left(ab\right)}^{*}=b^{*}{a}^{*},{(a+b)}^{*}=a^{*}+b^{*}.$$

A ring R with involution is called a ∗-ring. An element $a\in R$ satisfying $a{a}^{*}=a^{*}a$ is called normal. In this article, R is always a ∗-ring.

An element $a\in R$ is said to be Moore-Penrose invertible if the following equations:

$$axa=a,xax=x,{\left(ax\right)}^{*}=ax,{\left(xa\right)}^{*}=xa$$

have a common solution. Such solution is unique if it exists [3]. We call it the Moore-Penrose inverse of a and denote it by $a^{†}$ (see, e.g., [1,4,5,6]). The set of all Moore-Penrose invertible elements of R is denoted by $R^{†}$.

We call an element $a\in R^{\#}\cap R^{†}$ EP if $a^{\#}=a^{†}$ (see [3]). The equivalent conditions under which the elements of R is EP can be found in [7,8,9,10,11,12,13,14,15]. The set of all EP elements of R is denoted by $R^{EP}$. An element $a\in R^{†}$ is called partial isometry if $a^{†}=a^{*}$ (see [12,16,17]). Mosić and Djordjević provide numerous equivalence characterizations of EP elements that are partial isometry in [17,18]. The second author and Zhao [19] termed these elements strongly EP elements (abbreviated as SEP), characterizing them using equations. Guan and Wei [20] characterized the SEP elements via the existence of solutions to equations within a given set. Cao et al. [21] constructed additional equations characterizing SEP elements and further characterized them using w-core inverses. Li et al. [22] presented numerous new properties and equivalent characterizations of SEP elements. For further related results, see [23,24]. The set of SEP elements of R will be denoted by $R^{SEP}$. For a ring R, $E\left(R\right)$ denotes the set of all idempotent elements of R. We call $e\in E\left(R\right)$ projection if $e^{*}=e$. The set of all projections of R will be denoted by $PE\left(R\right)$.

The study of SEP elements has traditionally relied on definitions that explicitly involve various types of inverses. This paper establishes a new perspective by showing that a simple linear equation provides a complete characterization of these elements. Moving beyond a mere reformulation, this equation-based approach proves to be a powerful tool. As a direct application of this perspective, we demonstrate that the analysis of the equation’s structure naturally leads to the introduction of three key types of elements: involutional projection elements, square-rootable elements, and invertible elements. We then show that these concepts yield a novel characterization of SEP elements, revealing their fundamental properties through these equivalences. Furthermore, we extend this approach to multivariate and nonlinear settings, obtaining a unified framework for understanding SEP elements through systems of equations.

2. The Construction of Projection Elements

The following lemma can be found in [25].

Lemma 1.

Let $a\in R^{\#}\cap R^{†}$. Then $a\in R^{SEP}$ if and only if ${\left(a^{\#}\right)}^2{a}^{†}=a^{*}{a}^{†}{a}^{\#}$.

Observing the equality ${\left(a^{\#}\right)}^2{a}^{†}=a^{*}{a}^{†}{a}^{\#}$ in Lemma 1, we can construct the following Equation (1).

$${\left(a^{\#}\right)}^{2}x=a^{*}x{a}^{\#}.$$

(1)

Let $a\in R^{\#}\cap R^{†}$ and set ${\varsigma }_a=\{a,a^{\#},a^{†},{\left(a^{†}\right)}^{*},{\left(a^{\#}\right)}^{*},a^{†}{a}^3{a}^{†},{\left(a{a}^{\#}\right)}^{*}a{\left(a{a}^{\#}\right)}^{*}\}$.

Theorem 1.

Let $a\in R^{\#}\cap R^{†}$. Then $a\in R^{SEP}$ if and only if the Equation (1) has at least one solution in ${\varsigma }_a$.

Proof. 

⇒ It is obvious by Lemma 1.

⇐ Applying the assumption, we have:

(1)

If $x=a$ is a solution, then $a^{\#}=a^{*}a{a}^{\#}$. We get $a^{\#}a=a^{*}a$. By [23] (Theorem 1.5.3), we have $a\in R^{SEP}$.

(2)

If $x=a^{\#}$ is a solution, then ${\left(a^{\#}\right)}^3=a^{*}{\left(a^{\#}\right)}^2$. Multiplying the equation on the right by $a^2$, we obtain $a^{\#}=a^{*}a{a}^{\#}$. Hence $a\in R^{SEP}$ by (1).

(3)

If $x=a^{†}$ is a solution, then ${\left(a^{\#}\right)}^2{a}^{†}=a^{*}{a}^{†}{a}^{\#}.$ Thus $a\in R^{SEP}$ by [23] (Theorem 1.5.3).

(4)

If $x={\left(a^{†}\right)}^{*}$ is a solution, then ${\left(a^{\#}\right)}^2{\left(a^{†}\right)}^{*}=a^{†}a{a}^{\#}.$ Multiplying the equation on the left by $a^2$, we obtain $a^{*}=a^{†}$. Noting that $a^{*}a=a^{†}a=a^{†}a{a}^{\#}a={\left(a^{\#}\right)}^2{\left(a^{†}\right)}^{*}a={\left(a^{\#}\right)}^2{a}^2=a^{\#}a$. By [23] (Theorem 1.5.3), we have $a\in R^{SEP}$.

(5)

If $x={\left(a^{\#}\right)}^{*}$ is a solution, then ${\left(a^{\#}\right)}^2{\left(a^{\#}\right)}^{*}=a^{*}{\left(a^{\#}\right)}^{*}{a}^{\#}.$ Multiplying the equation on the left by $a^2{a}^{†}$, we obtain $a^{\#}{\left(a^{\#}\right)}^{*}=a{a}^{\#}$. By [23] (Theorem 1.5.3), $a\in R^{EP}$. It follows that $a^{\#}=a^{*}{\left(a^{\#}\right)}^{*}{a}^{\#}={\left(a^{\#}\right)}^2{\left(a^{\#}\right)}^{*}$ and $a^{\#}{a}^{*}={\left(a^{\#}\right)}^2{\left(a^{\#}\right)}^{*}{a}^{*}={\left(a^{\#}\right)}^2.$ Hence $a\in R^{SEP}$.

(6)

If $x=a^{†}{a}^3{a}^{†}$ is a solution, then ${\left(a^{\#}\right)}^2{a}^{†}{a}^3{a}^{†}=a^{*}{a}^{†}{a}^3{a}^{†}{a}^{\#}.$ That is $a^{\#}a{a}^{†}=a^{*}{a}^{†}a.$ Hence $a\in R^{SEP}$ by [23] (Theorem 1.5.3).

(7)

If $x={\left(a{a}^{\#}\right)}^{*}a{\left(a{a}^{\#}\right)}^{*}$ is a solution, then

$${\left(a^{\#}\right)}^2{\left(a{a}^{\#}\right)}^{*}a{\left(a{a}^{\#}\right)}^{*}=a^{*}{\left(a{a}^{\#}\right)}^{*}a{\left(a{a}^{\#}\right)}^{*}{a}^{\#}=a^{*}a{\left(a{a}^{\#}\right)}^{*}{a}^{\#}.$$

It follows that

$$\begin{array}{cc}\hfill {\left(a^{\#}\right)}^2{\left(a{a}^{\#}\right)}^{*}a& ={\left(a^{\#}\right)}^2{\left(a{a}^{\#}\right)}^{*}a{a}^{†}a={\left(a^{\#}\right)}^2{\left(a{a}^{\#}\right)}^{*}a{\left(a{a}^{\#}\right)}^{*}{a}^{†}a\hfill \\ \hfill & =a^{*}a{\left(a{a}^{\#}\right)}^{*}{a}^{\#}{a}^{†}a=a^{*}a{\left(a{a}^{\#}\right)}^{*}{a}^{\#}={\left(a^{\#}\right)}^2{\left(a{a}^{\#}\right)}^{*}a{\left(a{a}^{\#}\right)}^{*}\hfill \end{array}$$

and

$$a^{†}a=a^{†}{a}^{†}{a}^3{\left(a^{\#}\right)}^2{\left(a{a}^{\#}\right)}^{*}a=a^{†}{a}^{†}{a}^3{\left(a^{\#}\right)}^2{\left(a{a}^{\#}\right)}^{*}a{\left(a{a}^{\#}\right)}^{*}={\left(a{a}^{\#}\right)}^{*}.$$

Hence $a\in R^{EP}$, which implies

$$a^{\#}=a^{\#}{a}^{\#}a={\left(a^{\#}\right)}^2{\left(a{a}^{\#}\right)}^{*}a{\left(a{a}^{\#}\right)}^{*}=a^{*}a{\left(a{a}^{\#}\right)}^{*}{a}^{\#}=a^{*}a{a}^{\#}=a^{*}.$$

Thus $a\in R^{SEP}.$ □

Noting that for any $x\in {\chi }_a=\{a,a^{\#},a^{†},{\left(a^{\#}\right)}^{*},{\left(a^{†}\right)}^{*},{\left(a^{†}\right)}^{\#}\}$, ${\left(a^{*}x{a}^{\#}\right)}^{†}=a^{†}{a}^3{a}^{†}$$x^{\#}{\left(a^{†}\right)}^{*}$. Hence, if $a\in R^{EP}$, then ${\left(a^{*}x{a}^{\#}\right)}^{†}=a{x}^{+}{\left(a^{+}\right)}^{*}$ for any $x\in {\tau }_a=\{a,a^{\#},{\left(a^{+}\right)}^{\#}\}$. Therefore, by Theorem 1, we naturally consider characterizing elements of SEP via projection elements.

Theorem 2.

Let $a\in R^{\#}\cap R^{†}$. Then $a\in R^{SEP}$ if and only if ${\left(a^{\#}\right)}^{2}xa{x}^{†}{\left(a^{†}\right)}^{*}$ is projection for some $x\in {\tau }_a$.

Proof. 

⇒ Since $a^{\#}=a^{†}=a^{*}$, ${\left(a^{\#}\right)}^2{a}^2{a}^{†}{\left(a^{†}\right)}^{*}=a^{\#}a{a}^{†}a=a^{†}a$. Obviously, it is projection.

⇐ From assumption, there exists $x\in {\tau }_a$ such that ${\left(a^{\#}\right)}^{2}xa{x}^{†}{\left(a^{†}\right)}^{*}$ is projection, this gives

$${\left(a^{\#}\right)}^{2}xa{x}^{†}{\left(a^{†}\right)}^{*}={\left(a^{\#}\right)}^{2}xa{x}^{†}{\left(a^{†}\right)}^{*}{\left(a^{\#}\right)}^{2}xa{x}^{†}{\left(a^{†}\right)}^{*},$$

(2)

and

$${\left(a^{\#}\right)}^{2}xa{x}^{†}{\left(a^{†}\right)}^{*}={\left({\left(a^{\#}\right)}^{2}xa{x}^{†}{\left(a^{†}\right)}^{*}\right)}^{*}.$$

(3)

Multiplying the equality (2) on the left by $x^{†}{a}^2$ and noting that $x^{†}a{a}^{\#}x=a^{†}a$, one gets

$$a^{†}{a}^2{x}^{†}{\left(a^{†}\right)}^{*}=a^{†}{a}^2{x}^{†}{\left(a^{†}\right)}^{*}{\left(a^{\#}\right)}^{2}xa{x}^{†}{\left(a^{†}\right)}^{*}.$$

(4)

Multiplying the equality (4) on the left by $x{a}^{\#}$ and noting that $xa{a}^{\#}{x}^{†}=a{a}^{†}$, one gets

$${\left(a^{†}\right)}^{*}={\left(a^{†}\right)}^{*}{\left(a^{\#}\right)}^{2}xa{x}^{†}{\left(a^{†}\right)}^{*}.$$

(5)

Multiplying the equality (5) on the left and on the right all by $a^{*}$ and noting that $x^{†}a{a}^{†}=x^{†}$, one gets

$$a^{*}=a^{†}{a}^{\#}xa{x}^{†}.$$

(6)

If $x=a$ or $x=a^{\#}$, then (6) changes $a^{*}=a^{†}$; If $x={\left(a^{†}\right)}^{*}$, then (6) changes $a^{*}=a^{†}{a}^{\#}{\left(a^{†}\right)}^{*}a{a}^{*}$. This gives

$$a^2=a^2{a}^{†}a=a^2{a}^{*}{\left(a^{†}\right)}^{*}=a^2{a}^{+}{a}^{\#}{\left(a^{†}\right)}^{*}a{a}^{*}{\left(a^{†}\right)}^{*}={\left(a^{†}\right)}^{*}a.$$

Hence $a=a^2{a}^{\#}={\left(a^{†}\right)}^{*}a{a}^{\#}={\left(a^{†}\right)}^{*}$. Thus, in any case, one yields $a^{*}=a^{†}$. Multiplying the equality (3) on the right by $a{a}^{†}$, one has

$${\left(a^{\#}\right)}^{2}xa{x}^{†}{\left(a^{†}\right)}^{*}=\left({\left(a^{\#}\right)}^{2}xa{x}^{†}{\left(a^{†}\right)}^{*}\right)a{a}^{†}.$$

(7)

Multiplying the equality (7) on the left by $x{a}^{\#}{x}^{†}{a}^2$, one has

$${\left(a^{†}\right)}^{*}={\left(a^{†}\right)}^{*}a{a}^{†}.$$

Hence $a\in R^{EP}$. Therefore $a\in R^{SEP}$. □

Theorem 3.

Let $a\in R^{\#}\cap R^{†}$. Then $a\in R^{SEP}$ if and only if ${\left(a^{\#}\right)}^{2}xa{x}^{†}{\left(a^{\#}\right)}^{*}{a}^{†}a$ is projection for some $x\in {\omega }_a=\{a^{+},{\left(a^{\#}\right)}^{*},a^{+}{a}^3{a}^{+}\}$.

Proof. 

⇒ It is easy to check.

⇐ From the assumption, one gets

$$\begin{array}{cc}\hfill {\left(a^{\#}\right)}^{2}xa{x}^{†}{\left(a^{\#}\right)}^{*}{a}^{†}a& ={\left({\left(a^{\#}\right)}^{2}xa{x}^{†}{\left(a^{\#}\right)}^{*}{a}^{†}a\right)}^{*}\hfill \\ \hfill & =a^{†}a{a}^{\#}{\left(x^{†}\right)}^{*}{a}^{*}{x}^{*}{\left(a^{\#}\right)}^{*}{\left(a^{\#}\right)}^{*}\mathrm{for}\mathrm{some}x\in {\omega }_a.\hfill \end{array}$$

Multiplying the equality on the right by $a{a}^{†}$ and notice that ${\left(a^{\#}\right)}^{*}a{a}^{†}={\left(a^{\#}\right)}^{*}$, one yields

$${\left(a^{\#}\right)}^{2}xa{x}^{†}{\left(a^{\#}\right)}^{*}{a}^{†}a={\left(a^{\#}\right)}^{2}xa{x}^{†}{\left(a^{\#}\right)}^{*}{a}^{†}{a}^2{a}^{†}.$$

Multiplying the last equality on the left by $x^{†}{a}^2$ and notice that $x^{†}a{a}^{\#}x=a{a}^{†}$ for any $x\in {\omega }_a$, then one obtains

$$a{x}^{†}{a}^{\#}x=a{x}^{†}{\left(a^{\#}\right)}^{*}{a}^{†}{a}^2{a}^{†}.$$

Since $x{a}^{\#}a{x}^{†}=a^{†}a$ for any $x\in {\omega }_a$, one has

$$\begin{array}{cc}\hfill {\left(a^{\#}\right)}^{*}{a}^{†}a& =a^{†}a{\left(a^{\#}\right)}^{*}{a}^{†}a\hfill \\ \hfill & =x{a}^{\#}a{x}^{†}{\left(a^{\#}\right)}^{*}{a}^{†}a\hfill \\ \hfill & =x{a}^{\#}a{x}^{†}{\left(a^{\#}\right)}^{*}{a}^{†}{a}^2{a}^{†}\hfill \\ \hfill & ={\left(a^{\#}\right)}^{*}{a}^{†}{a}^2{a}^{†}.\hfill \end{array}$$

It follows that

$$a^{†}a=a^{*}a{\left(a^{\#}\right)}^{*}{a}^{†}a=a^{*}{\left(a^{\#}\right)}^{*}{a}^{†}{a}^2{a}^{†}=a^{†}{a}^2{a}^{†}.$$

Hence $a\in R^{EP}$, which implies ${\left(a^{\#}\right)}^{*}{a}^{†}a={\left(a^{†}\right)}^{*}{a}^{†}a={\left(a^{†}\right)}^{*}$, ${\tau }_a={\omega }_a$ and ${\left(a^{\#}\right)}^{2}xa{x}^{†}{\left(a^{\#}\right)}^{*}{a}^{†}a={\left(a^{\#}\right)}^{2}xa{x}^{†}{\left(a^{†}\right)}^{*}.$ By Theorem 2, we have $a\in R^{SEP}$. □

Theorem 4.

Let $a\in R^{\#}\cap R^{†}$. Then $a\in R^{SEP}$ if and only if $a^{*}x{a}^{\#}{x}^{†}{a}^3{a}^{†}$ is projection for some $x\in {\tau }_a\cup {\omega }_a$.

Proof. 

⇒ It is easy to check.

⇐ If there exists $x\in {\tau }_a\cup {\omega }_a,$ such that $a^{*}x{a}^{\#}{x}^{†}{a}^3{a}^{†}$ is projection, then

$$a^{*}x{a}^{\#}{x}^{†}{a}^3{a}^{†}=\left(a^{*}x{a}^{\#}{x}^{†}{a}^3{a}^{†}\right)=a{a}^{†}{\left(a^{*}\right)}^2{\left(x^{†}\right)}^{*}{\left(a^{\#}\right)}^{*}{x}^{*}a.$$

Multiplying the equality on the right by $a^{†}a$, one gets

$$a^{*}x{a}^{\#}{x}^{†}{a}^3{a}^{†}=a^{*}x{a}^{\#}{x}^{†}{a}^3{a}^{†}{a}^{†}a$$

and

$$\begin{array}{cc}\hfill x^{\#}a{a}^{†}x{a}^{\#}{x}^{†}{a}^3{a}^{†}& =x^{\#}{\left(a^{†}\right)}^{*}{a}^{*}x{a}^{\#}{x}^{†}{a}^3{a}^{†}\hfill \\ \hfill & =x^{\#}{\left(a^{†}\right)}^{*}{a}^{*}x{a}^{\#}{x}^{†}{a}^3{a}^{†}{a}^{†}a\hfill \\ \hfill & =x^{\#}a{a}^{†}{\left(a^{\#}\right)}^{*}x{a}^{\#}{x}^{†}{a}^3{a}^{†}{a}^{†}a.\hfill \end{array}$$

Notice that $x^{\#}a{a}^{†}x=\left\{\begin{array}{c}a{a}^{\#},x\in {\tau }_a\hfill \\ {\left(a{a}^{\#}\right)}^{*},x\in {\omega }_a\hfill \end{array}\right.$, this leads to

(1)

If $x\in {\tau }_a$, then $a^{\#}{x}^{†}{a}^3{a}^{†}=a^{\#}{x}^{†}{a}^3{a}^{†}{a}^{†}a.$

(2)

If $x\in {\omega }_a$, then ${\left(a{a}^{\#}\right)}^{*}{a}^{\#}{x}^{†}{a}^3{a}^{†}={\left(a{a}^{\#}\right)}^{*}{a}^{\#}{x}^{†}{a}^3{a}^{†}{a}^{†}a.$

It follows that

$$\begin{array}{cc}\hfill a^{\#}{x}^{†}{a}^3{a}^{†}& =a{a}^{†}{a}^{\#}{x}^{†}{a}^3{a}^{†}\hfill \\ \hfill & =a{a}^{†}{\left(a{a}^{\#}\right)}^{*}{a}^{\#}{x}^{†}{a}^3{a}^{†}\hfill \\ \hfill & =a{a}^{†}{\left(a{a}^{\#}\right)}^{*}{a}^{\#}{x}^{†}{a}^3{a}^{†}{a}^{†}a\hfill \\ \hfill & =a^{\#}{x}^{†}{a}^3{a}^{†}{a}^{†}a.\hfill \end{array}$$

Hence, one always has

$$a^{\#}{x}^{†}{a}^3{a}^{†}=a^{\#}{x}^{†}{a}^3{a}^{†}{a}^{†}a\mathrm{for}\mathrm{some}x\in {\tau }_a\cup {\omega }_a.$$

Since $xa{a}^{\#}{x}^{†}=\left\{\begin{array}{c}a{a}^{†},x\in {\tau }_a\hfill \\ a^{†}a,x\in {\omega }_a\hfill \end{array}\right.$, one has

(1)

If $x\in {\tau }_a$, then $a^3{a}^{†}=a{a}^{†}{a}^3{a}^{†}=xa{a}^{\#}{x}^{†}{a}^3{a}^{†}{a}^{†}a=xa{a}^{\#}{x}^{†}{a}^3{a}^{†}{a}^{†}a=a^3{a}^{†}{a}^{†}a$ and $a^{†}=a^{†}{a}^{\#}{a}^{\#}{a}^3{a}^{†}=a^{†}{a}^{\#}{a}^{\#}{a}^3{a}^{†}{a}^{†}a=a^{†}{a}^{†}a.$ Hence $a\in R^{EP}$.

(2)

If $x\in {\omega }_a$, then $a^{†}{a}^4{a}^{†}=xa{a}^{\#}{x}^{†}{a}^3{a}^{†}=xa{a}^{\#}{x}^{†}{a}^3{a}^{†}{a}^{†}a=a^{†}{a}^4{a}^{†}{a}^{†}a.$ It follows that $a^3{a}^{†}=a^{\#}a{a}^{†}{a}^4{a}^{†}=a^{\#}a{a}^{†}{a}^4{a}^{†}{a}^{†}a.$ By (1), $a\in R^{EP}$.

Hence, in any case, one has $a\in R^{EP}$, which implies ${\tau }_a={\omega }_a$ and $a^{*}x{a}^{\#}{x}^{†}{a}^2=a^{*}x{a}^{\#}{x}^{†}{a}^3{a}^{†}$ is projection for some $x\in {\tau }_a.$

This gives

$$a^{*}x{a}^{\#}{x}^{†}{a}^2=a^{*}x{a}^{\#}{x}^{†}{a}^2{a}^{*}x{a}^{\#}{x}^{†}{a}^2$$

and

$$\begin{array}{cc}\hfill a^{\#}{x}^{†}{a}^2& =a^{\#}a{a}^{\#}{x}^{†}{a}^2\hfill \\ \hfill & =\left(x^{\#}a{a}^{†}x\right)a^{\#}{x}^{†}{a}^2\hfill \\ \hfill & =x^{\#}{\left(a^{†}\right)}^{*}\left(a^{*}x{a}^{\#}{x}^{†}{a}^2\right)\hfill \\ \hfill & =x^{\#}{\left(a^{†}\right)}^{*}\left(a^{*}x{a}^{\#}{x}^{†}{a}^2{a}^{*}x{a}^{\#}{x}^{†}{a}^2\right)\hfill \\ \hfill & =a^{\#}{x}^{†}{a}^2{a}^{*}x{a}^{\#}{x}^{†}{a}^2.\hfill \end{array}$$

It follows that

$$a^2=a{a}^{†}{a}^2=xa{a}^{\#}{x}^{†}{a}^2=xa{a}^{\#}{x}^{†}{a}^2=a^2{a}^{*}x{a}^{\#}{x}^{†}{a}^2$$

and

$$a^{\#}={\left(a^{\#}\right)}^2{a}^2{a}^{\#}={\left(a^{\#}\right)}^2{a}^2{a}^{*}x{a}^{\#}{x}^{†}{a}^2{a}^{\#}=a^{*}x{a}^{\#}{x}^{†}{a}^2{a}^{\#}.$$

(1)

If $x=a$, then $a^{\#}=a^{*}a{a}^{\#}{a}^{†}a=a^{*}$;

(2)

If $x=a^{\#}$, then $a^{\#}=a^{*}{a}^{\#}{a}^{\#}{a}^2=a^{*}$;

(3)

If $x={\left(a^{†}\right)}^{*}$, then $a^{\#}=a^{*}{\left(a^{†}\right)}^{*}{a}^{\#}{a}^{*}a=a^{\#}{a}^{*}a$, and $a^{\#}=a{a}^{\#}{a}^{\#}=a{a}^{\#}{a}^{*}a{a}^{\#}=a^{*}.$

Thus, in any case, $a\in R^{SEP}$. □

3. The Group-Inverse and MP-Inverse of Variable Equations

According to Equation (1), a natural question arises as to the form of the MP inverse and group inverse in this general expression. Readers may find such questions addressed in the literature [19,22,24]. In this section, we shall focus primarily on presenting the generalized inverse form of Equation (1).

Lemma 2.

Let $a\in R^{\#}\cap R^{†}$ and take ${\psi }_a=\{a^{+},a^{*},{\left(a^{\#}\right)}^{*},a^{+}{a}^3{a}^{+}\}$. Then

(1) 

${\left({\left(a^{\#}\right)}^{2}x\right)}^{\#}=\left\{\begin{array}{c}{x}^{\#}{a}^{†}{a}^3,x\in {\tau }_a\hfill \\ x^{†}{a}^3{a}^{†},x\in {\psi }_a\hfill \end{array}\right.$;

(2) 

${\left({\left(a^{\#}\right)}^{2}x\right)}^{†}=x^{†}{a}^3{a}^{†}$ for $x\in {\tau }_a\cup {\psi }_a$;

(3) 

${\left(a^{*}x{a}^{\#}\right)}^{\#}={\left(a^{*}x{a}^{\#}\right)}^{†}=a^{†}{a}^3{a}^{†}{x}^{\#}{\left(a^{†}\right)}^{*}$ for $x\in {\tau }_a\cup {\psi }_a$.

Proof. 

(1) Noting that $x{x}^{\#}=a{a}^{\#}$ and $x^{\#}{a}^{†}a=x^{\#}$ for any $x\in {\tau }_a$, then

$$\left({\left(a^{\#}\right)}^{2}x\right)\left(x^{\#}{a}^{†}{a}^3\right)\left({\left(a^{\#}\right)}^{2}x\right)={\left(a^{\#}\right)}^{2}a{a}^{\#}{a}^{†}ax={\left(a^{\#}\right)}^{2}x,$$

$$\left(x^{\#}{a}^{†}{a}^3\right)\left({\left(a^{\#}\right)}^{2}x\right)\left(x^{\#}{a}^{†}{a}^3\right)=x^{\#}{a}^{†}ax{x}^{\#}{a}^{†}{a}^3=x^{\#}{a}^{†}aa{a}^{\#}{a}^{†}{a}^3=x^{\#}{a}^{†}{a}^3,$$

$$\left({\left(a^{\#}\right)}^{2}x\right)\left(x^{\#}{a}^{†}{a}^3\right)={\left(a^{\#}\right)}^{2}a{a}^{\#}{a}^{†}{a}^3=a^{\#}{a}^{\#}{a}^{†}{a}^3=a^{\#}a,$$

$$\left(x^{\#}{a}^{†}{a}^3\right)\left({\left(a^{\#}\right)}^{2}x\right)=x^{\#}{a}^{+}ax=x^{\#}x=a{a}^{\#}.$$

Hence if $x\in {\tau }_a$, then ${\left({\left(a^{\#}\right)}^{2}x\right)}^{\#}=x^{\#}{a}^{†}{a}^3$.

Now assume $x\in {\psi }_a$, we have $x{x}^{†}=a^{†}a$ and $x^{†}a{a}^{\#}x=a{a}^{†}$. Thus

$$\begin{array}{cc}\hfill \left({\left(a^{\#}\right)}^{2}x\right)\left(x^{†}{a}^3{a}^{†}\right)& ={\left(a^{\#}\right)}^2{a}^{†}a{a}^3{a}^{†}\hfill \\ \hfill & ={\left(a^{\#}\right)}^2{a}^{†}{a}^4{a}^{†}\hfill \\ \hfill & =a{a}^{†};\hfill \\ \hfill \left(x^{†}{a}^3{a}^{†}\right)\left({\left(a^{\#}\right)}^{2}x\right)& =x^{†}a{a}^{\#}x\hfill \\ \hfill & =a{a}^{†};\hfill \\ \hfill \left({\left(a^{\#}\right)}^{2}x\right)\left(x^{†}{a}^3{a}^{†}\right)\left({\left(a^{\#}\right)}^{2}x\right)& =a{a}^{†}{\left(a^{\#}\right)}^{2}x\hfill \\ \hfill & ={\left(a^{\#}\right)}^{2}x;\hfill \\ \hfill \left(x^{†}{a}^3{a}^{†}\right)\left({\left(a^{\#}\right)}^{2}x\right)\left(x^{†}{a}^3{a}^{†}\right)& =a{a}^{†}{x}^{†}{a}^3{a}^{†}\hfill \\ \hfill & =x^{†}{a}^3{a}^{†}.\hfill \end{array}$$

Hence if $x\in {\psi }_a$, then ${\left({\left(a^{\#}\right)}^{2}x\right)}^{\#}=x^{†}{a}^3{a}^{†}.$

(2) It is obvious that ${\left({\left(a^{\#}\right)}^{2}x\right)}^{†}=x^{†}{a}^3{a}^{†}$ for $x\in {\psi }_a$ by (1).

For $x\in {\tau }_a$, we have $x{x}^{†}=a{a}^{†}$ and $a^{†}a{x}^{†}=x^{†}$. Then

$$\begin{array}{cc}\hfill \left({\left(a^{\#}\right)}^{2}x\right)\left(x^{†}{a}^3{a}^{†}\right)& =a^{\#}{a}^{†}{a}^3{a}^{†}\hfill \\ \hfill =a{a}^{†}={\left(\left({\left(a^{\#}\right)}^{2}x\right)\left(x^{†}{a}^3{a}^{†}\right)\right)}^{*};\end{array}$$

For $x\in {\tau }_a$, we have $x^{†}x=a^{†}a$ and $x^{†}a{a}^{\#}x=a^{†}a$. Then

$$\left(x^{†}{a}^3{a}^{†}\right)\left({\left(a^{\#}\right)}^{2}x\right)=x^{†}a{a}^{\#}x=a^{†}a={\left(\left(x^{†}{a}^3{a}^{†}\right)\left({\left(a^{\#}\right)}^{2}x\right)\right)}^{*};$$

$$\left({\left(a^{\#}\right)}^{2}x\right)\left(x^{†}{a}^3{a}^{†}\right)\left({\left(a^{\#}\right)}^{2}x\right)=a{a}^{†}{\left(a^{\#}\right)}^{2}x={\left(a^{\#}\right)}^{2}x;$$

$$\left(x^{†}{a}^3{a}^{†}\right)\left({\left(a^{\#}\right)}^{2}x\right)\left(x^{†}{a}^3{a}^{†}\right)=a^{†}a{x}^{†}{a}^3{a}^{†}=x^{†}{a}^3{a}^{†}.$$

Thus ${\left({\left(a^{\#}\right)}^{2}x\right)}^{†}=x^{†}{a}^3{a}^{†}$ for $x\in {\tau }_a$.

(3) For $x\in {\tau }_a\cup {\psi }_a$, we have

$$\left(a^{*}x{a}^{\#}\right)\left(a^{†}{a}^3{a}^{†}{x}^{\#}{\left(a^{†}\right)}^{*}\right)=a^{*}xa{a}^{†}{x}^{\#}{\left(a^{†}\right)}^{*},$$

$$\left(a^{†}{a}^3{a}^{†}{x}^{\#}{\left(a^{†}\right)}^{*}\right)\left(a^{*}x{a}^{\#}\right)=a^{†}{a}^3{a}^{†}{x}^{\#}a{a}^{†}x{a}^{\#}.$$

If $x\in {\tau }_a$, then $x{x}^{†}=a{a}^{†}$ and $x{x}^{\#}=a{a}^{\#}$. Hence

$$a^{*}xa{a}^{†}{x}^{\#}{\left(a^{†}\right)}^{*}=a^{*}xx{x}^{†}{x}^{\#}{\left(a^{†}\right)}^{*}=a^{*}x{x}^{\#}{\left(a^{†}\right)}^{*}=a^{*}a{a}^{\#}{\left(a^{†}\right)}^{*}=a^{†}a,$$

$$a^{†}{a}^3{a}^{†}{x}^{\#}a{a}^{†}x{a}^{\#}=a^{†}{a}^3{a}^{†}{x}^{\#}x{x}^{†}x{a}^{\#}=a^{†}{a}^3{a}^{†}a{\left(a^{\#}\right)}^2=a^{†}a.$$

According to the definition, it is easy to get

$${\left(a^{*}x{a}^{\#}\right)}^{\#}={\left(a^{*}x{a}^{\#}\right)}^{†}=a^{†}{a}^3{a}^{†}{x}^{\#}{\left(a^{†}\right)}^{*},forx\in {\tau }_a.$$

If $x\in {\psi }_a$, then $a{a}^{†}=x^{†}x$ and $x{x}^{\#}=a^{*}{\left(a^{\#}\right)}^{*}$. Therefore

$$a^{*}xa{a}^{†}{x}^{\#}{\left(a^{†}\right)}^{*}=a^{*}x{x}^{\#}{\left(a^{†}\right)}^{*}={\left(a^{*}\right)}^2{\left(a^{\#}\right)}^{*}{\left(a^{†}\right)}^{*}=a^{†}a,$$

$$a^{†}{a}^3{a}^{†}{x}^{\#}a{a}^{†}x{a}^{\#}=a^{†}{a}^3{a}^{†}{x}^{\#}{x}^{†}{x}^2{a}^{\#}=a^{†}{a}^3{a}^{†}{a}^{*}{\left(a^{\#}\right)}^{*}{a}^{\#}=a^{†}a.$$

Hence we have

$${\left(a^{*}x{a}^{\#}\right)}^{\#}={\left(a^{*}x{a}^{\#}\right)}^{†}=a^{†}{a}^3{a}^{†}{x}^{\#}{\left(a^{†}\right)}^{*}forx\in {\psi }_a.$$

□

Based on the expanded form of the generalized inverse given in Lemma 2, we provide the following equivalent characterization of SEP elements:

Theorem 5.

Let $a\in R^{\#}\cap R^{†}$. Then $a\in R^{SEP}$ if and only if ${\left({\left(a^{\#}\right)}^{2}x\right)}^{\#}=\left\{\begin{array}{c}{x}^{\#}{a}^{†}{a}^2{\left(a^{\#}\right)}^{*},x\in {\tau }_a\hfill \\ x^{†}{a}^2{\left(a^{†}\right)}^{*}{a}^{\#},x\in {\psi }_a\hfill \end{array}\right.$.

Proof. 

⇒ Assume that $a\in R^{SEP}$. Then ${\left(a^{+}\right)}^{*}=a={\left(a^{\#}\right)}^{*}$. By Lemma 2, we are done.

⇐ If $x\in {\tau }_a$, then by Lemma 2, we have $x^{\#}{a}^{†}{a}^3=x^{\#}{a}^{†}{a}^2{\left(a^{\#}\right)}^{*}$. Multiplying the equality on the left by x and noting that $x{x}^{\#}=a{a}^{\#}$, we get $a^2=a{\left(a^{\#}\right)}^{*}$. Thus by [23] (Theorem 1.5.3), $a\in R^{SEP}$.

If $x\in {\psi }_a$, then by Lemma 2, we have $x^{†}{a}^3{a}^{†}=x^{†}{a}^2{\left(a^{†}\right)}^{*}{a}^{\#}$. Multiplying the equality on the left by x and noting that $x{x}^{†}=a^{†}a$, we get $a^{†}{a}^4{a}^{†}=a^{†}{a}^3{\left(a^{†}\right)}^{*}{a}^{\#}$. Multiplying the equality on the left by ${\left(a^{\#}\right)}^2$ and the equality on the right by a, we have $a^{*}=a^{†}$. Thus $a{a}^{\#}={\left(a^{\#}\right)}^2{a}^{†}{a}^4{a}^{\#}={\left(a^{\#}\right)}^2{a}^{†}{a}^3{\left(a^{†}\right)}^{*}{a}^{\#}={\left(a^{\#}\right)}^2{a}^{†}{a}^4{a}^{†}=a{a}^{†}=a{a}^{*}.$ So $a\in R^{SEP}$ by [23] (Theorem 1.5.3). □

Theorem 6.

Let $a\in R^{\#}\cap R^{†}$. Then $a\in R^{SEP}$ if and only if ${\left({\left(a^{\#}\right)}^{2}x\right)}^{†}=x^{†}{\left(a^{†}\right)}^{*}a$ for $x\in {\tau }_a\cup {\psi }_a$.

Proof. 

⇒ It is an immediate result of Lemma 2 because $a^3{a}^{+}={\left(a^{+}\right)}^{*}{a}^2{a}^{\#}={\left(a^{+}\right)}^{*}a$.

⇐ By the assumption and Lemma 2, we have $x^{†}{a}^3{a}^{†}=x^{†}{\left(a^{†}\right)}^{*}a$. Multiplying the equality on the left by x and using that

$$x{x}^{†}=a{a}^{†},x\in {\tau }_a$$

and

$$x{x}^{†}=a^{†}a,x\in {\psi }_a.$$

we have

$$a^3{a}^{†}={\left(a^{†}\right)}^{*}a\mathrm{for}x\in {\tau }_a,$$

(8)

and

$$a^{†}{a}^4{a}^{†}=a^{†}a{\left(a^{†}\right)}^{*}a\mathrm{for}x\in {\psi }_a,$$

(9)

Multiplying the equality (8) on the right by $a^{\#}$, we get $a^{*}=a^{†}$. Thus $a^{\#}a={\left(a^{\#}\right)}^2{a}^2={\left(a^{\#}\right)}^2{\left(a^{†}\right)}^{*}a={\left(a^{\#}\right)}^2{a}^3{a}^{†}=a{a}^{†}=a{a}^{*}$.

Similarly, multiplying equality (9) by $a^{\#}$ on the right and by $a{a}^{\#}$ on the left, we get $a^{*}=a^{†}$. Then $a^{\#}a={\left(a^{\#}\right)}^2{a}^{†}{a}^3={\left(a^{\#}\right)}^2{a}^{†}a{\left(a^{†}\right)}^{*}a={\left(a^{\#}\right)}^2{a}^{†}{a}^4{a}^{†}=a{a}^{†}=a{a}^{*}.$ So, in any case, $a\in R^{SEP}$ by [23] (Theorem 1.5.3). □

Theorem 7.

Let $a\in R^{\#}\cap R^{†}$. Then $a\in R^{SEP}$ if and only if ${\left(a^{*}x{a}^{\#}\right)}^{\#}=a^{*}{a}^3{a}^{†}{x}^{\#}{\left(a^{\#}\right)}^{*}$ for $x\in {\tau }_a\cup {\psi }_a$.

Proof. 

⇒ Suppose that $a\in R^{SEP}$. Then $a^{†}=a^{*}$ and ${\left(a^{†}\right)}^{*}={\left(a^{\#}\right)}^{*}$. By Lemma 2, we are done.

⇐ Under the hypothesis and using Lemma 2, we have

$$a^{†}{a}^3{a}^{†}{x}^{\#}{\left(a^{†}\right)}^{*}=a^{*}{a}^3{a}^{†}{x}^{\#}{\left(a^{\#}\right)}^{*}\mathrm{for}x\in {\tau }_a\cup {\psi }_a$$

(10)

Multiplying the Equation (10) on the right by $a{a}^{†}$, one gets

$$a^{†}{a}^3{a}^{†}{x}^{\#}{\left(a^{†}\right)}^{*}=a^{†}{a}^3{a}^{†}{x}^{\#}{\left(a^{†}\right)}^{*}a{a}^{†}$$

and

$$a{a}^{†}{x}^{\#}{\left(a^{†}\right)}^{*}=a^{\#}{a}^{†}{a}^3{a}^{†}{x}^{\#}{\left(a^{†}\right)}^{*}=a^{\#}{a}^{†}{a}^3{a}^{†}{x}^{\#}{\left(a^{†}\right)}^{*}a{a}^{†}=a^{†}{a}^3{a}^{†}{x}^{\#}{\left(a^{†}\right)}^{*}a{a}^{†}.$$

Noting that $xa{a}^{†}{x}^{\#}=\left\{\begin{array}{c}a{a}^{\#},x\in {\tau }_a\hfill \\ {\left(a{a}^{\#}\right)}^{*},x\in {\psi }_a,\hfill \end{array}\right..$ Then we have

$${\left(a^{†}\right)}^{*}={\left(a^{†}\right)}^{*}a{a}^{†}\mathrm{for}x\in {\tau }_a,$$

(11)

and

$${\left(a^{\#}\right)}^{*}{a}^{†}a={\left(a^{\#}\right)}^{*}{a}^{†}{a}^2{a}^{†}\mathrm{for}x\in {\psi }_a.$$

(12)

Multiplying the Equations (11) and (12) on the left by $a^{*}$, one gets $a^{†}a=a^{†}{a}^2{a}^{†}$. Hence $a\in R^{EP}.$

Now Equation (10) changes

$$a{x}^{\#}{\left(a^{†}\right)}^{*}=a^{*}{a}^2{x}^{\#}{\left(a^{†}\right)}^{*}\mathrm{for}x\in {\tau }_a\cup {\psi }_a.$$

This gives $a{x}^{\#}a{a}^{†}x=a{x}^{\#}{\left(a^{†}\right)}^{*}{a}^{*}x=a^{*}{a}^2{x}^{\#}{\left(a^{†}\right)}^{*}{a}^{*}x=a^{*}{a}^2{x}^{\#}a{a}^{†}x.$

If $x\in {\tau }_a$, then $x^{\#}a{a}^{†}x=a{a}^{\#}$, this infers $a=a^{*}{a}^2$;

If $x\in {\psi }_a$, then $x^{\#}a{a}^{†}x={\left(a{a}^{\#}\right)}^{*},$ this infers $a{\left(a{a}^{\#}\right)}^{*}=a^{*}{a}^2{\left(a{a}^{\#}\right)}^{*}.$

It follows that $a=a{a}^{†}a=a{\left(a{a}^{\#}\right)}^{*}{a}^{†}a=a^{*}{a}^2{\left(a{a}^{\#}\right)}^{*}{a}^{†}a=a^{*}{a}^2$. Hence, in any case, we have $a=a^{*}{a}^2$. Thus $a\in R^{SEP}$ by [23] (Theorem 1.5.3). □

By Lemma 2 and Theorem 1, we obtain the following conclusion:

Theorem 8.

Let $a\in R^{\#}\cap R^{†}$. Then the following conditions are equivalent:

(1) 

$a\in R^{SEP}$;

(2) 

$x^{\#}{a}^{†}{a}^3=a^{†}{a}^3{a}^{†}{x}^{\#}{\left(a^{†}\right)}^{*}$ for some $x\in {\tau }_a$;

(3) 

$x^{†}{a}^3{a}^{†}=a^{†}{a}^3{a}^{†}{x}^{\#}{\left(a^{†}\right)}^{*}$ for some $x\in {\tau }_a\cup {\psi }_a$.

4. Compatibility of Equations Involving Parametric Variables

In the preceding sections, we established that the SEP property can be characterized by a specific linear equation, ${\left(a^{\#}\right)}^{2}x=a^{*}x{a}^{\#}$, which provides a remarkably simple description. This equation, however, is univariate and fixed. A natural question arises: what happens when we introduce parameters into such equations? This section investigates the compatibility of equations involving parametric, variables that is, we ask under what conditions do there exist solutions to families of related equations. This shift in perspective, from solving a single equation to understanding a system of equations parameterized by elements in the ring, is not merely a technical generalization. Instead, it serves as the crucial conceptual bridge that guides us from the linear characterization of SEP elements to the discovery of the novel constructs of square-rootable and involutional projection elements, which will be introduced in Section 5 and Section 7.

Theorem 9.

Let $a\in R^{\#}\cap R^{†}$. Then $a\in R^{SEP}$ if and only if $a^{*}x{a}^{\#}{x}^{†}=$$\left\{\begin{array}{c}{a}^{\#}{a}^{†},x\in {\tau }_a\hfill \\ {\left(a^{\#}\right)}^2,x\in {\eta }_a=\{a^{†},{\left(a^{\#}\right)}^{*},a^{†}{a}^3{a}^{†},{\left(a{a}^{\#}\right)}^{*}a{\left(a{a}^{\#}\right)}^{*}\}\hfill \end{array}\right..$

Proof. 

⇒ It is a routine verification because $a^{\#}=a^{†}=a^{*}$ and ${\tau }_a=\{a^{†},{\left(a^{\#}\right)}^{*},a^{†}{a}^3{a}^{†},{\left(a{a}^{\#}\right)}^{*}a{\left(a{a}^{\#}\right)}^{*}\}$.

⇐ If $x\in {\tau }_a$, then $a^{*}x{a}^{\#}{x}^{†}=a^{\#}{a}^{†}$. Multiplying the equality on the left by $a^{†}a$, one gets $a^{\#}{a}^{†}=a^{†}a{a}^{\#}{a}^{†}$ and $a^{\#}=a^{\#}{a}^{†}a=a^{†}a{a}^{\#}{a}^{†}a=a^{†}a{a}^{\#}$. Hence $a\in R^{EP}$ by [23] (Theorem 1.2.1). If $x\in {\eta }_a$, then

$${\left(a^{\#}\right)}^2=a^{*}x{a}^{\#}{x}^{†}=a^{†}a{a}^{*}x{a}^{\#}{x}^{†}=a^{†}a{\left(a^{\#}\right)}^2=a^{†}{a}^{\#}.$$

Hence $a\in R^{EP}$ by [23] (Theorem 1.2.1).

Thus, in any case, $a\in R^{EP}$, which infers ${\tau }_a={\eta }_a$ and $a^{*}x{a}^{\#}{x}^{†}=a^{\#}{a}^{†}={\left(a^{\#}\right)}^2$ for any $x\in {\tau }_a$.

If $x=a$, then $a^{\#}{a}^{\#}=a^{*}a{a}^{\#}{a}^{†}=a^{*}{a}^{\#}$.

If $x=a^{\#}$, then $a^{\#}{a}^{\#}=a^{*}{a}^{\#}{a}^{\#}{\left(a^{\#}\right)}^{†}=a^{*}{a}^{\#}{a}^{\#}{\left(a^{†}\right)}^{†}=a^{*}{a}^{\#}$.

If $x={\left(a^{†}\right)}^{*}$, then $a^{\#}{a}^{\#}=a^{*}{\left(a^{†}\right)}^{*}{a}^{\#}{\left({\left(a^{†}\right)}^{*}\right)}^{†}=a^{†}a{a}^{\#}{a}^{*}=a^{\#}{a}^{*}$.

In any case, we have $a\in R^{SEP}$ by [23] (Theorem 1.5.3). □

Theorem 10.

Let $a\in R^{\#}\cap R^{†}$. Then $a\in R^{SEP}$ if and only if ${\left(a^{\#}\right)}^{2}xa{x}^{†}=\left\{\begin{array}{c}{a}^{*},x\in {\tau }_a\hfill \\ a^{*}{a}^{†}a,x\in {\eta }_a\hfill \end{array}\right..$

Proof. 

⇒ It is obvious.

⇐ Under the assumption, one has

(1)

If $x\in {\tau }_a$, then $a^{*}={\left(a^{\#}\right)}^{2}xa{x}^{†}=a{a}^{†}{\left(a^{\#}\right)}^{*}xa{x}^{†}=a{a}^{†}{a}^{*}$;

(2)

If $x\in {\eta }_a$, then $a^{*}{a}^{†}a={\left(a^{\#}\right)}^{2}xa{x}^{†}=a{a}^{†}{\left(a^{\#}\right)}^{*}xa{x}^{†}=a{a}^{†}{a}^{*}{a}^{†}a$.

It follows that

$$a^{*}=a^{*}{a}^{†}a{\left(a{a}^{\#}\right)}^{*}=a{a}^{†}{a}^{*}{a}^{†}a{\left(a{a}^{\#}\right)}^{*}=a{a}^{†}{a}^{*}.$$

Thus, in any case, we have $a\in R^{EP}$. This gives ${\tau }_a={\eta }_a$ and $a^{*}={\left(a^{\#}\right)}^{2}xa{x}^{†}$.

If $x=a$, then $a^{*}={\left(a^{\#}\right)}^2{a}^2{a}^{†}={\left(a^{\#}\right)}^2{a}^2{a}^{\#}=a^{\#}.$

If $x=a^{\#}$, then $a^{*}={\left(a^{\#}\right)}^2{a}^{\#}a{\left(a^{\#}\right)}^{†}={\left(a^{\#}\right)}^3{a}^2=a^{\#}.$

If $x={\left(a^{†}\right)}^{*}$, then $a^{*}={\left(a^{\#}\right)}^2{\left(a^{†}\right)}^{*}a{a}^{*}$ and $a^{\#}a=a^{†}a=a^{*}{\left(a^{†}\right)}^{*}={\left(a^{\#}\right)}^2{\left(a^{†}\right)}^{*}a{a}^{*}{\left(a^{†}\right)}^{*}={\left(a^{\#}\right)}^2{\left(a^{†}\right)}^{*}a.$ It follows that $a^{\#}=a^{\#}a{a}^{\#}={\left(a^{\#}\right)}^2{\left(a^{†}\right)}^{*}a{a}^{\#}={\left(a^{\#}\right)}^2{\left(a^{†}\right)}^{*}$ and $a=a^2{a}^{\#}=a^2{\left(a^{\#}\right)}^2{\left(a^{†}\right)}^{*}={\left(a^{†}\right)}^{*}$. So $a^{*}=a^{†}=a^{\#}.$ Thus, in any case, $a^{\#}=a^{*}$ and $a\in R^{SEP}$. □

Theorem 11.

Let $a\in R^{\#}\cap R^{†}$. Then $a\in R^{SEP}$ if and only if $x^{†}{a}^2{a}^{*}x{a}^{\#}=\left\{\begin{array}{c}a{a}^{†},x\in {\tau }_a\hfill \\ a^{†}a,x\in {\eta }_a\hfill \end{array}\right..$

Proof. 

⇒ Assume that $a\in R^{SEP}$. Then ${\eta }_a={\tau }_a$ and $a^{†}=a^{\#}=a^{*}$, it follows that

$x^{†}{a}^2{a}^{*}x{a}^{\#}=x^{†}ax{a}^{\#}=\left\{\begin{array}{c}{a}^{†}aa{a}^{\#}=a^{†}a=a{a}^{†},x=a\hfill \\ {\left(a^{\#}\right)}^{†}a{a}^{\#}{a}^{\#}={\left(a^{†}\right)}^{†}{a}^{\#}=a{a}^{\#}=a{a}^{†}=a^{†}a,x=a^{\#}\hfill \\ {\left({\left(a^{†}\right)}^{*}\right)}^{†}a{\left(a^{†}\right)}^{*}{a}^{\#}=a^{†}{a}^2{a}^{\#}=a^{†}a,x={\left(a^{†}\right)}^{*}\hfill \end{array}\right..$ As desired.

⇐ According to the assumption, one has

(1)

If $x\in {\tau }_a$, then $a{a}^{†}=x^{†}{a}^2{a}^{*}x{a}^{\#}=\left(x^{†}{a}^2{a}^{*}x{a}^{\#}\right)a^{†}a=a{a}^{†}{a}^{†}a,$ which infers $a\in R^{EP}.$

(2)

If $x\in {\eta }_a$, then $a^{†}a=x^{†}{a}^2{a}^{*}x{a}^{\#}=\left(a{a}^{†}{x}^{†}\right)a^2{a}^{*}x{a}^{\#}=a{a}^{†}{a}^{†}a,$ this induces $a\in R^{EP}.$

Hence, in any case, we have $a\in R^{EP}$ and ${\tau }_a={\eta }_a,$ it follows that $x^{†}{a}^2{a}^{*}x{a}^{\#}=a{a}^{†}$ for $x\in {\tau }_a={\eta }_a$.

(1)

If $x=a$, then $a{a}^{†}=a^{†}{a}^2{a}^{*}a{a}^{\#}=a{a}^{*}$. Hence $a\in R^{SEP}$.

(2)

If $x=a^{\#}$, then $a{a}^{†}={\left(a^{\#}\right)}^{†}{a}^2{a}^{*}{a}^{\#}{a}^{\#}=a^3{a}^{*}{a}^{\#}{a}^{\#},$ this gives $a=a^{\#}{a}^2=a^{\#}a{a}^{†}{a}^2=a^{\#}\left(a^3{a}^{*}{a}^{\#}{a}^{\#}\right)a^2=a^2{a}^{*}.$ Hence $a\in R^{SEP}$ by [23] (Theorem 1.5.3).

(3)

If $x={\left(a^{†}\right)}^{*}$, then $a{a}^{†}={\left({\left(a^{†}\right)}^{*}\right)}^{†}{a}^2{a}^{*}{\left(a^{†}\right)}^{*}{a}^{\#}=a^{*}{a}^2{a}^{†}a{a}^{\#}=a^{*}a.$ Hence $a\in R^{SEP}$.

□

Theorem 12.

Let $a\in R^{\#}\cap R^{†}$. Then $a\in R^{SEP}$ if and only if $x^{\#}{\left(a^{†}\right)}^{*}{\left(a^{\#}\right)}^{2}x=$$\left\{\begin{array}{c}{a}^{†},x\in {\tau }_a\hfill \\ {\left(a{a}^{\#}\right)}^{*}{a}^{\#},x\in {\eta }_a\hfill \end{array}\right..$

Proof. 

⇒ Since $a\in R^{SEP}$, ${\eta }_a={\tau }_a$ and $x^{\#}{\left(a^{†}\right)}^{*}{\left(a^{\#}\right)}^{2}x=x^{\#}a{\left(a^{\#}\right)}^{2}x=x^{\#}{a}^{\#}x=\left\{\begin{array}{c}{a}^{\#}{a}^{\#}a,x=a\hfill \\ {\left(a^{\#}\right)}^{\#}{a}^{\#}{a}^{\#},x=a^{\#}\hfill \\ {\left({\left(a^{†}\right)}^{*}\right)}^{\#}{a}^{\#}{\left(a^{†}\right)}^{*}{a}^{\#}=a^{*}{a}^{\#}{\left(a^{†}\right)}^{*},x={\left(a^{†}\right)}^{*}\hfill \end{array}\right.=a^{\#}=a^{†}={\left(a^{\#}a\right)}^{*}{a}^{\#}.$ As desired.

⇐ Applying the hypothesis, one obtains:

If $x\in {\tau }_a$, then $a^{†}=x^{\#}{\left(a^{†}\right)}^{*}{\left(a^{\#}\right)}^{2}x=\left(a{a}^{†}{x}^{\#}\right){\left(a^{†}\right)}^{*}{\left(a^{\#}\right)}^{2}x=a{a}^{†}{a}^{†}$. Hence $a\in R^{EP}$;

If $x\in {\eta }_a$, then

$${\left(a{a}^{\#}\right)}^{*}{a}^{\#}=x^{\#}{\left(a^{†}\right)}^{*}{\left(a^{\#}\right)}^{2}x=x^{\#}{\left(a^{†}\right)}^{*}{\left(a^{\#}\right)}^2\left(xa{a}^{†}\right)={\left(a{a}^{\#}\right)}^{*}{a}^{\#}a{a}^{†}$$

and

$$a^{\#}=a{a}^{†}{a}^{\#}=a{a}^{†}{\left(a{a}^{\#}\right)}^{*}{a}^{\#}=a{a}^{†}{\left(a{a}^{\#}\right)}^{*}{a}^{\#}a{a}^{†}=a^{\#}a{a}^{†}.$$

Hence $a\in R^{EP}$.

Thus, in any case, we have $a\in R^{EP}$ and ${\tau }_a={\eta }_a$.

Now we have $x^{\#}{\left(a^{†}\right)}^{*}{\left(a^{\#}\right)}^{2}x=a^{\#}$ for $x\in {\tau }_a$.

If $x=a$, then $a^{\#}=a^{\#}{\left(a^{†}\right)}^{*}{\left(a^{\#}\right)}^{2}a=a^{\#}{\left(a^{†}\right)}^{*}{a}^{\#}$ and $a=a{a}^{\#}a=a{a}^{\#}{\left(a^{†}\right)}^{*}{a}^{\#}a={\left(a^{†}\right)}^{*}.$ Hence $a\in R^{SEP}.$

If $x=a^{\#}$, then $a^{\#}={\left(a^{\#}\right)}^{\#}{\left(a^{†}\right)}^{*}{\left(a^{\#}\right)}^3=a{\left(a^{†}\right)}^{*}{\left(a^{\#}\right)}^3$ and $a=a^{\#}{a}^{\#}{a}^3=a^{\#}\left(a{\left(a^{†}\right)}^{*}{\left(a^{\#}\right)}^3\right)a^3={\left(a^{†}\right)}^{*}.$ Hence $a\in R^{SEP}.$

If $x={\left(a^{†}\right)}^{*}={\left(a^{\#}\right)}^{*}$, then

$$a^{\#}={\left({\left(a^{\#}\right)}^{*}\right)}^{\#}{\left(a^{†}\right)}^{*}{\left(a^{\#}\right)}^2{\left(a^{†}\right)}^{*}=a^{*}{\left(a^{†}\right)}^{*}{\left(a^{\#}\right)}^2{\left(a^{†}\right)}^{*}={\left(a^{\#}\right)}^2{\left(a^{†}\right)}^{*}$$

and

$$a=a^2{a}^{\#}=a^2{\left(a^{\#}\right)}^2{\left(a^{†}\right)}^{*}={\left(a^{†}\right)}^{*}.$$

Hence $a\in R^{SEP}$. □

5. Square Rootable Elements

Let R be a ring and $a\in R$. Then a is called a square rootable element of R if there exists $b\in R$ such that $a=b^2$. In this case, b is called the square root of a.

The set of all square rootable elements of R will be denoted by $\sqrt{R}$. Obviously, $0,1\in \sqrt{R}$ and $E\left(R\right)\subseteq \sqrt{R}$. Let $a\in \sqrt{R}$. Write $\sqrt{a}=\{b\in R|b^2=a\}$. By Theorem 9, we find that square rootable elements can be utilised to characterize SEP elements.

Theorem 13.

Let $a\in R^{\#}\cap R^{†}$. Then $a\in R^{SEP}$ if and only if $a^{*}{a}^{†}{a}^{\#}a\in \sqrt{R}$ and $a^{\#}\in \sqrt{a^{*}{a}^{†}{a}^{\#}a}$.

Proof. 

⇒ Since $a\in R^{SEP}$, by Lemma 1, $a^{*}{a}^{†}{a}^{\#}=a^{\#}{a}^{\#}{a}^{†}$. Thus

$${\left(a^{\#}\right)}^2={\left(a^{\#}\right)}^2{a}^{†}a=a^{*}{a}^{†}{a}^{\#}a.$$

⇐ Using the assumption, one gets

$$a^{*}{a}^{†}{a}^{\#}a={\left(a^{\#}\right)}^2.$$

Then

$$a^{†}{a}^{\#}=a^{†}a{\left(a^{\#}\right)}^2=a^{†}a{a}^{*}{a}^{†}{a}^{\#}a=a^{*}{a}^{†}{a}^{\#}a={\left(a^{\#}\right)}^2.$$

Hence $a\in R^{EP}$ and

$${\left(a^{\#}\right)}^2=a^{*}{a}^{†}{a}^{\#}a=a^{*}{a}^{\#}.$$

Thus $a\in R^{SEP}$. □

By Theorem 13, and [23] (Theorem 1.5.3), we get the following corollary.

Corollary 1.

Let $a\in R^{\#}\cap R^{†}$. Then the following conditions are equivalent:

(1) 

$a\in R^{SEP}$;

(2) 

$a^{†}\in \sqrt{a^{*}{a}^{†}{a}^{\#}a}$;

(3) 

$a^{†}\in \sqrt{a^{*}{a}^{\#}}$;

(4) 

$a^{\#}\in \sqrt{a^{*}{a}^{†}}$.

Theorem 14.

Let $a\in R^{\#}\cap R^{†}$. Then $a\in R^{SEP}$ if and only if $a^{\#}+1-a^{†}a\in \sqrt{a^{*}{a}^{†}{a}^{\#}a+1-a{a}^{†}}$.

Proof. 

⇒ It follows from Theorem 13.

⇐ Under the hypothesis, we have

$${\left(a^{\#}\right)}^2+a^{\#}-a^{†}a+1-a^{†}a{a}^{\#}={(a^{\#}+1-a^{†}a)}^2=a^{*}{a}^{†}{a}^{\#}a+1-a{a}^{†}.$$

$${\left(a^{\#}\right)}^2+a^{\#}-a^{†}a+a{a}^{†}-a^{†}a{a}^{\#}=a^{*}{a}^{†}{a}^{\#}a.$$

(13)

Multiplying the equality (13) by $a^{†}a$ on the right, we get

$$a{a}^{†}=a{a}^{†}{a}^{†}a.$$

Then $a\in R^{EP}$, which leads to (13) change ${\left(a^{\#}\right)}^2=a^{*}{a}^{\#}.$ Hence $a\in R^{SEP}$. □

Theorem 15.

Let $a\in R^{\#}\cap R^{†}$. Then $a\in R^{SEP}$ if and only if $a^{*}{a}^{†}{\left(a^{\#}\right)}^{*}\in \sqrt{a^{\#}{a}^{†}{a}^{†}{\left(a^{\#}\right)}^{*}}$.

Proof. 

⇒ Since $a\in R^{SEP}$, $a^{*}{a}^{+}{\left(a^{\#}\right)}^{*}=a^{\#}{a}^{\#}a=a^{\#}$ and $a^{\#}{a}^{+}{a}^{+}{\left(a^{\#}\right)}^{*}=a^{*}{a}^{+}{a}^{\#}a$. By Theorem 13, we are done.

⇐ From the assumption, one gets

$${\left(a^{*}{a}^{†}{\left(a^{\#}\right)}^{*}\right)}^2=a^{\#}{a}^{†}{a}^{†}{\left(a^{\#}\right)}^{*}.$$

Then

$$a^{*}{\left(a^{†}\right)}^2{\left(a^{\#}\right)}^{*}=a^{\#}{\left(a^{†}\right)}^2{\left(a^{\#}\right)}^{*}.$$

(14)

Multiplying the equality (14) by $a^{*}a{a}^{*}{\left(a^{\#}\right)}^{*}$ on the right, we get $a^{*}{a}^{†}=a^{\#}{a}^{†}$ and $a\in R^{SEP}$ by [23] (Theorem 1.5.3). □

6. Characterization Using Reversible Elements

Let $U\left(R\right)$ denote the set of invertible elements in R. Based on the study of SEP elements in Section 3, we find that certain special invertible elements in ring R can be used to characterize SEP elements. Furthermore, it is evident from our proof that the inverses of these invertible elements can be explicitly written out.

Theorem 16.

Let $a\in R^{\#}\cap R^{†}$. Then $a\in R^{SEP}$ if and only if $a^{*}{a}^{†}{a}^{\#}=u{a}^{\#}{a}^{†}$, where $u\in U\left(R\right)$ and $a^{†}au=a^{†}a{a}^{\#}$.

Proof. 

⇒ Since $a\in R^{SEP}$, by Lemma 1, $a^{*}{a}^{†}{a}^{\#}={\left(a^{\#}\right)}^2{a}^{†}$. Take $u=a^{\#}+1-a{a}^{\#}$. Then $u\in U\left(R\right)$ with

$$u^{-1}=a+1-a{a}^{\#}.$$

Clearly, $a^{†}au=a^{†}a{a}^{\#}$ and $u{a}^{\#}{a}^{†}=a^{*}{a}^{†}{a}^{\#}$.

⇐ If there exists $u\in U\left(R\right)$ such that $a^{*}{a}^{†}{a}^{\#}=u{a}^{\#}{a}^{†}$ and $a^{†}au=a^{†}a{a}^{\#}$, then

$$a^{†}{a}^{\#}{a}^{†}=a^{†}au{a}^{\#}{a}^{†}=a^{†}a{a}^{*}{a}^{†}{a}^{\#}=a^{*}{a}^{†}{a}^{\#}=a^{*}{a}^{†}{a}^{\#}{a}^{†}a=a^{†}{a}^{\#}{a}^{†}{a}^{†}a.$$

It follows that $a{a}^{†}=a^3{a}^{†}{a}^{\#}{a}^{†}=a^3{a}^{†}{a}^{\#}{a}^{†}{a}^{†}a=a{a}^{†}{a}^{†}a.$ Hence $a\in R^{EP}$. This gives

$${\left(a^{\#}\right)}^2{a}^{†}=a^{†}{a}^{\#}{a}^{†}=a^{*}{a}^{†}{a}^{\#}.$$

By Lemma 1, $a\in R^{SEP}$. □

Theorem 17.

Let $a\in R^{\#}\cap R^{†}$. Then $a\in R^{SEP}$ if and only if $a^{*}{a}^{†}{a}^{\#}=a^{\#}u{a}^{†}$, where $u\in U\left(R\right)$ and $u{a}^{*}=a^{†}{a}^{*}$.

Proof. 

⇒ Assume that $a\in R^{SEP}$. Then $a^{*}{a}^{†}{a}^{\#}={\left(a^{\#}\right)}^2{a}^{†}$ by Lemma 1. Clearly, $a^{\#}u{a}^{†}=a^{\#}{a}^{†}{a}^{†}={\left(a^{\#}\right)}^2{a}^{†}=a^{*}{a}^{†}{a}^{\#}$ and $u{a}^{*}=a^{†}{a}^{*}$.

⇐ From the assumption, one gets

$$a^{*}{a}^{†}{a}^{\#}=a^{\#}u{a}^{†}\mathrm{and}u{a}^{*}=a^{†}{a}^{*}\mathrm{for}\mathrm{some}u\in U\left(R\right).$$

This gives

$$a^{*}{a}^{†}{a}^{\#}=a^{\#}u{a}^{*}{\left(a^{\#}\right)}^{*}{a}^{†}=a^{\#}{a}^{†}{a}^{*}{\left(a^{\#}\right)}^{*}{a}^{†}=a^{\#}{a}^{†}{a}^{†}=\left(a{a}^{†}{a}^{\#}\right)a^{†}{a}^{†}=a{a}^{†}{a}^{*}{a}^{†}{a}^{\#}.$$

Multiplying the equality by $a^2{\left(a{a}^{\#}\right)}^{*}$ on the right, we get $a^{*}=a{a}^{†}{a}^{*}$.

Hence $a\in R^{EP}$, which implies $a^{*}{a}^{†}{a}^{\#}=a^{\#}{a}^{†}{a}^{†}={\left(a^{\#}\right)}^2{a}^{†}.$

By Lemma 1, $a\in R^{SEP}$. □

Theorem 18.

Let $a\in R^{\#}\cap R^{†}$. Then $a\in R^{SEP}$ if and only if $a^{*}{a}^{†}{a}^{\#}=a^{\#}{a}^{†}u$, where $u\in U\left(R\right)$ and $u{\left(a^{†}\right)}^{*}=a^{†}{\left(a^{†}\right)}^{*}$.

Proof. 

⇒ Suppose that $a\in R^{SEP}$. Then, $a^{\#}=a^{†}$ and by Lemma 1, $a^{*}{a}^{†}{a}^{\#}={\left(a^{\#}\right)}^2{a}^{†}$. Set $u=a^{†}+1-a{a}^{†}$. Then $u\in U\left(R\right)$ with $u^{-1}=a+1-a{a}^{†}$. Clearly, $a^{\#}{a}^{†}u=a^{\#}{a}^{†}{a}^{†}={\left(a^{\#}\right)}^2{a}^{†}=a^{*}{a}^{†}{a}^{\#}$ and $u{\left(a^{†}\right)}^{*}=a^{†}{\left(a^{†}\right)}^{*}$.

⇐ If there exists $u\in U\left(R\right)$ such that $a^{*}{a}^{†}{a}^{\#}=a^{\#}{a}^{†}u$ and $u{\left(a^{†}\right)}^{*}=a^{†}{\left(a^{†}\right)}^{*}.$ Then $a^{*}{a}^{†}{a}^{\#}{\left(a^{†}\right)}^{*}=a^{\#}{a}^{†}u{\left(a^{†}\right)}^{*}=a^{\#}{a}^{†}{a}^{†}{\left(a^{†}\right)}^{*}=a{a}^{†}{a}^{\#}{a}^{†}{a}^{†}{\left(a^{†}\right)}^{*}=a{a}^{†}{a}^{*}{a}^{†}{a}^{\#}{\left(a^{†}\right)}^{*}.$

Multiplying the equality by $a^{*}{a}^2{\left(a{a}^{\#}\right)}^{*}$ on the right, one gets $a^{*}=a{a}^{†}{a}^{*}$.

Hence $a\in R^{EP}$. It follows that

$$a^{*}{a}^{†}{a}^{\#}=a^{*}{a}^{†}{a}^{†}=a^{*}{a}^{†}{a}^{†}{\left(a^{\#}\right)}^{*}{a}^{*}=a^{*}{a}^{†}{a}^{\#}{\left(a^{\#}\right)}^{*}{a}^{*}=a^{\#}{a}^{†}{a}^{†}{\left(a^{†}\right)}^{*}{a}^{*}={\left(a^{\#}\right)}^2{a}^{†}.$$

By Lemma 1, $a\in R^{SEP}$. □

Theorem 19.

Let $a\in R^{\#}\cap R^{†}$. Then $a\in R^{SEP}$ if and only if $a^{*}{a}^{†}{a}^{\#}=a^{\#}u$, where $u\in U\left(R\right)$ and $a^{*}u=a^{*}{a}^{\#}{a}^{†}$.

Proof. 

⇒ Since $a\in R^{SEP}$, $a^{\#}=a^{†}$ and $a^{*}{a}^{†}{a}^{\#}={\left(a^{\#}\right)}^2{a}^{†}$. Let $u=a^{\#}{a}^{†}+1-a{a}^{\#}.$ Then $u\in U\left(R\right)$ and $u^{-1}=a^2+1-a{a}^{\#}.$ Clearly, $a^{\#}u=a^{\#}{a}^{\#}{a}^{†}=a^{*}{a}^{†}{a}^{\#}$ and $a^{*}u=a^{*}{a}^{\#}{a}^{†}+a^{*}-a^{*}a{a}^{\#}=a^{*}{a}^{\#}{a}^{†}.$

⇐ Under the hypothesis, one gets $a^{*}{a}^{†}{a}^{\#}=a^{\#}u$ and $a^{*}u=a^{*}{a}^{\#}{a}^{†}$ for some $u\in U\left(R\right)$. This gives

$$a^{*}{a}^{†}{a}^{\#}=a^{\#}u=a{a}^{†}{a}^{\#}u=a{a}^{†}\left(a^{*}{a}^{†}{a}^{\#}\right).$$

Multiplying the equality by $a^2{\left(a{a}^{\#}\right)}^{*}$ on the right, one yields $a^{*}=a{a}^{†}{a}^{*}$.

Hence $a\in R^{EP}$ and

$$a^{*}{a}^{†}{a}^{\#}=a^{\#}u=a^{†}u=a^{†}{\left(a^{\#}\right)}^{*}{a}^{*}u=a^{†}{\left(a^{\#}\right)}^{*}{a}^{*}{a}^{\#}{a}^{†}=a^{†}{a}^{\#}{a}^{†}={\left(a^{\#}\right)}^2{a}^{†}.$$

Thus, by Lemma 1, $a\in R^{SEP}$. □

7. The Involutional Projection Elements

Let R be a $\ast$-ring and $u\in R$. Then u is called an involutional projection element if $u^2=1$ and $u^{*}=u$. Denote the set of all involutional projection elements of R by $IP\left(R\right)$. Let $p\in PE\left(R\right)$. Then, clearly, $2p-1\in IP\left(R\right)$. Indeed, from the conclusions drawn in Section 4 and the structural characterization of SEP elements presented in Section 6, it follows naturally that involutional projection elements should be employed to characterize SEP elements.

Theorem 20.

Let $a\in R^{\#}\cap R^{†}$. Then $a\in R^{SEP}$ if and only if $u{a}^{*}{a}^{†}{a}^{\#}={\left(a^{\#}\right)}^2{a}^{†}$, where $u\in IP\left(R\right)$ and $a^{†}u=a^{†}$.

Proof. 

⇒ Assume that $a\in R^{SEP}$. Then $a^{+}=a^{\#}$ and $a^{*}{a}^{†}{a}^{\#}={\left(a^{\#}\right)}^2{a}^{†}$ by Lemma 1. Take $u=2a{a}^{†}-1$. Then $u\in IP\left(R\right)$. It can be readily verified that $u{a}^{*}{a}^{†}{a}^{\#}={\left(a^{\#}\right)}^2{a}^{†}$ and $a^{†}u=a^{†}$.

⇐ Under the assumption, one gets

$$a^{†}{\left(a^{\#}\right)}^2{a}^{†}=a^{†}u{a}^{*}{a}^{†}{a}^{\#}=a^{†}{a}^{*}{a}^{†}{a}^{\#},$$

and

$$a^{†}{\left(a^{\#}\right)}^2{a}^{†}=a^{†}{a}^{*}{a}^{†}{a}^{\#}=\left(a^{†}{a}^{*}{a}^{†}{a}^{\#}\right)a{a}^{\#}=a^{†}{\left(a^{\#}\right)}^2{a}^{†}a{a}^{\#}=a^{†}{\left(a^{\#}\right)}^3.$$

Multiplying equality by $a^2$ on the left, one gets

$$a^{\#}{a}^{†}={\left(a^{\#}\right)}^2.$$

Hence $a\in R^{EP}.$ This induces

$$a^{\#}{a}^{†}={\left(a^{\#}\right)}^2={\left(a^{\#}\right)}^5{a}^3=a^{†}{\left(a^{\#}\right)}^2{a}^{†}{a}^3{a}^{†}=a^{†}{a}^{*}{a}^{†}{a}^{\#}{a}^3{a}^{†}=a^{†}{a}^{*}.$$

One has $a\in R^{SEP}$. □

In fact, both the above theorem and the following theorem are inspired by Equation (1).

Theorem 21.

Let $a\in R^{\#}\cap R^{†}$. Then $a\in R^{SEP}$ if and only if $a^{*}{a}^{†}{a}^{\#}={\left(a^{\#}\right)}^2{a}^{†}u$, where $u\in IP\left(R\right)$ and $a^{\#}au=2a{a}^{†}-a{a}^{\#}$.

Proof. 

⇒ Assume that $a\in R^{SEP}$. Then $a\in R^{EP}$ and $a^{*}{a}^{†}{a}^{\#}={\left(a^{\#}\right)}^2{a}^{†}$ by Lemma 1. Take $u=2a{a}^{†}-1$, then $u\in IP\left(R\right)$. By a simple verification, one has $a^{*}{a}^{†}{a}^{\#}={\left(a^{\#}\right)}^2{a}^{†}u$ and $a^{\#}au=2a{a}^{†}-a{a}^{\#}$.

⇐ Applying the hypothesis, one gets

$$u{\left(a{a}^{\#}\right)}^{*}=u^{*}{\left(a{a}^{\#}\right)}^{*}={\left(a{a}^{\#}u\right)}^{*}={(2a{a}^{†}-a{a}^{\#})}^{*}=2a{a}^{†}-{\left(a{a}^{\#}\right)}^{*}.$$

It follows that

$$a^{*}{a}^{†}{a}^{\#}{\left(a{a}^{\#}\right)}^{*}={\left(a^{\#}\right)}^2{a}^{†}u{\left(a{a}^{\#}\right)}^{*}={\left(a^{\#}\right)}^2{a}^{†}{(2a{a}^{†}-a{a}^{\#})}^{*})={\left(a^{\#}\right)}^2{a}^{†}$$

and

$${\left(a^{\#}\right)}^2{a}^{†}=a^{*}{a}^{†}{a}^{\#}{\left(a{a}^{\#}\right)}^{*}=a^{†}a{a}^{*}{a}^{†}{a}^{\#}{\left(a{a}^{\#}\right)}^{*}=a^{†}a{\left(a^{\#}\right)}^2{a}^{†}=a^{†}{a}^{\#}{a}^{†}.$$

So $a^{\#}a={\left(a^{\#}\right)}^2{a}^{†}{a}^3=a^{†}{a}^{\#}{a}^{†}{a}^3=a^{†}a$. Hence $a\in R^{EP}$, this implies

$${\left(a^{\#}\right)}^2{a}^{†}=a^{*}{a}^{†}{a}^{\#}{\left(a{a}^{\#}\right)}^{*}=a^{*}{a}^{†}{a}^{\#}a{a}^{\#}=a^{*}{a}^{†}{a}^{\#}.$$

Thus $a\in R^{SEP}$ by Lemma 1. □

Theorem 22.

Let $a\in R^{\#}\cap R^{†}$. Then $a\in R^{SEP}$ if and only if $a^{*}{a}^{†}u{a}^{\#}={\left(a^{\#}\right)}^2{a}^{†}$, where $u\in PE\left(R\right)$ and $u{\left(a^{\#}\right)}^{*}={\left(a^{†}\right)}^{*}$.

Proof. 

⇒ Since $a\in R^{SEP}$ and Lemma 1, we have $a^{\#}=a^{†}$ and $a^{*}{a}^{†}{a}^{\#}={\left(a^{\#}\right)}^2{a}^{†}$. Set $u=2{\left(a{a}^{\#}\right)}^{*}-1.$ Clearly, $u\in IP\left(R\right)$ with $a^{*}{a}^{†}u{a}^{\#}=a^{*}{a}^{†}{a}^{\#}={\left(a^{\#}\right)}^2{a}^{†}$ and $u{\left(a^{\#}\right)}^{*}={\left(a^{\#}\right)}^{*}={\left(a^{†}\right)}^{*}.$

⇐ Using the assumption, one gets

$${\left(a^{\#}\right)}^2{a}^{†}=a^{*}{a}^{†}u{a}^{\#}=a^{†}a{a}^{*}{a}^{†}u{a}^{\#}=a^{†}a{\left(a^{\#}\right)}^2{a}^{†}=a^{†}{a}^{\#}{a}^{†}.$$

By the proof of Theorem 21, $a\in R^{EP}.$ Noting that $a^{†}u=a^{\#}{u}^{*}={\left(u{\left(a^{\#}\right)}^{*}\right)}^{*}={\left({\left(a^{†}\right)}^{*}\right)}^{*}=a^{†}.$ Then one obtains ${\left(a^{\#}\right)}^2{a}^{†}=a^{*}{a}^{†}u{a}^{\#}=a^{*}{a}^{†}{a}^{\#}.$ Hence $a\in R^{SEP}$ by Lemma 1. □

8. Representation of the General Solution to a Bivariate Equation

From the perspective of self-contained research, characterizing the generalized inverse using multivariable equations constitutes a natural extension of characterizing the generalized inverse using univariable equations. We now analyze the general solution to the bivariate Equation (15), which naturally arises from (1). The parametric representation of its solution set will be instrumental in the derivations that follow.

$${\left(a^{\#}\right)}^{2}x=a^{*}y{a}^{\#}.$$

(15)

Theorem 23.

Let $a\in R^{\#}\cap R^{†}$. Then the general solution to Equation (15) is given by the following expression:

$$\left\{\begin{array}{c}x=a^2{a}^{†}p{a}^{\#}+u-a^{†}au,\hfill \\ y={\left(a^{†}\right)}^{*}{a}^{†}p+v-a{a}^{†}va{a}^{†}.\hfill \end{array}\right.p,u,v\in R\mathrm{and}{a}^{†}p=a{a}^{†}{a}^{†}p.$$

(16)

Proof. 

First, it is easy to check (16) is the solution of Equation (15). Next, let $\left\{\begin{array}{c}x=x_0\hfill \\ y=y_0\hfill \end{array}\right.$ be any solution of the Equation (15). Then

$${\left(a^{\#}\right)}^2{x}_0=a^{*}{y}_0{a}^{\#}.$$

Take

$$\begin{array}{c}p=a{a}^{*}{y}_{0}a{a}^{†},\hfill \\ u=x_0-a^2{a}^{†}p{a}^{\#},\hfill \\ v=y_0.\hfill \end{array}$$

We have

$$\begin{array}{cc}\hfill a{a}^{†}{a}^{†}p& =a{a}^{†}{a}^{*}{y}_{0}a{a}^{†}\hfill \\ & =a{a}^{†}{\left(a^{\#}\right)}^2{x}_0{a}^2{a}^{†}\hfill \\ & ={\left(a^{\#}\right)}^2{x}_0{a}^2{a}^{†}\hfill \\ & =a^{*}{y}_{0}a{a}^{†}\hfill \\ & =a^{†}p,\hfill \\ \hfill a^{†}au& =a^{†}a{x}_0-a^{†}{a}^3{a}^{†}p{a}^{\#}\hfill \\ & =a^{†}{a}^3{\left(a^{\#}\right)}^2{x}_0-a^{†}{a}^3{a}^{†}a{a}^{*}{y}_{0}a{a}^{†}{a}^{\#}\hfill \\ & =a^{†}{a}^3{a}^{*}{y}_0{a}^{\#}-a^{†}{a}^3{a}^{*}{y}_{0}a{a}^{†}{a}^{\#}\hfill \\ & =0.\hfill \end{array}$$

Thus

$$x=a^2{a}^{†}p{a}^{\#}+u-a^{†}au.$$

Similarly $a{a}^{†}va{a}^{†}=0$ and

$$y={\left(a^{†}\right)}^{*}{a}^{†}p+v-a{a}^{†}va{a}^{†}.$$

□

Theorem 24.

Let $a\in R^{\#}\cap R^{†}$. Then $a\in R^{SEP}$ if and only if the general solution to Equation (15) is given by the following expression:

$$\left\{\begin{array}{c}x=a^2{a}^{†}p{a}^{\#}+u-a^{†}au,\hfill \\ y=a{a}^{†}p+v-a{a}^{†}va{a}^{†}.\hfill \end{array}\right.p,u,v\in R.$$

(17)

Proof. 

⇒ Suppose $a\in R^{SEP}$, then $a^{†}p=a{a}^{†}{a}^{†}p$ and $a={\left(a^{+}\right)}^{*}.$ From Theorem 23, (17) is the general solution of Equation (15).

⇐ Assume that (17) is the general solution of Equation (15), one gets

$$\begin{array}{c}{\left(a^{\#}\right)}^2(a^2{a}^{†}p{a}^{\#}+u-a^{†}au)=a^{*}(a{a}^{†}p+v-a{a}^{†}va{a}^{†})a^{\#}.\end{array}$$

That is $a^{\#}a{a}^{†}p{a}^{\#}=a^{*}p{a}^{\#}.$ Take $p=a$, we have $a^{\#}=a^{*}a{a}^{\#}.$ By Theorem 1, $a\in R^{SEP}$. □

Observing Theorem 23, we can construct the following equation:

$$a^{*}{a}^{\#}x=a^{*}y{a}^{\#}.$$

(18)

Theorem 25.

Let $a\in R^{\#}\cap R^{†}$. Then the general solution to Equation (18) is given by (17).

Proof. 

The proof is similar to Theorem 23. □

From Theorem 23 and Theorem 25, we have the following result.

Theorem 26.

Let $a\in R^{\#}\cap R^{†}$. Then $a\in R^{SEP}$ if and only if Equations (15) and (18) have the same solution.

9. Representation of the General Solution to a Bivariate Nonlinear Equation

The characterization of SEP elements using linear equations in Section 8 is complemented by the natural characterization of nonlinear elements in this section. Moreover, this indeed represents a novel formulation distinct from Mosić’s characterization of SEP elements in Reference [23].

Equation (15) prompts us to consider the following nonlinear equation:

$$a^{*}{a}^{\#}x-a^{*}y{a}^{\#}=a^{†}.$$

(19)

Theorem 27.

Let $a\in R^{\#}\cap R^{†}$. The general solution to Equation (19) is given by the following expression:

$$\left\{\begin{array}{c}x=a^2{a}^{†}p{a}^{\#}+u-a^{†}au+a{\left(a^{†}\right)}^{*}{a}^{†}+(1-a^{†}a)w,\hfill \\ y=a{a}^{†}p+v-a{a}^{†}va{a}^{†}.\hfill \end{array}\right.p,u,v,w\in R.$$

(20)

Proof. 

First, it is easy to check (20) is the solution of Equation (19).

Next let $\left\{\begin{array}{c}x=x_0\hfill \\ y=y_0\hfill \end{array}\right.$ be any solution of Equation (19). Then

$$a^{*}{a}^{\#}{x}_0-a^{*}{y}_0{a}^{\#}=a^{†}.$$

Take

$$\begin{array}{c}p=y_0,\hfill \\ u=x_0-a^2{a}^{†}p{a}^{\#}-a{\left(a^{†}\right)}^{*}{a}^{†},\hfill \\ w=a^{†},\hfill \\ v=y_0-a{a}^{†}p.\hfill \end{array}$$

We have

$$\begin{array}{cc}\hfill (1-a^{†}a)w& =(1-a^{†}a)a^{†}\hfill \\ \hfill & =0,\hfill \\ \hfill a^{†}au& =a^{†}a(x_0-a^2{a}^{†}p{a}^{\#}-a{\left(a^{†}\right)}^{*}{a}^{†})\hfill \\ & =a^{†}a{x}_0-a^{†}{a}^3{a}^{†}{y}_0{a}^{\#}-a^{†}{a}^2{\left(a^{†}\right)}^{*}{a}^{†}\hfill \\ & =a^{†}a{x}_0-a^{†}{a}^2{\left(a^{†}\right)}^{*}{a}^{*}{y}_0{a}^{\#}-a^{†}{a}^2{\left(a^{†}\right)}^{*}{a}^{†}\hfill \\ & =a^{†}a{x}_0-a^{†}{a}^2{\left(a^{†}\right)}^{*}(a^{*}{a}^{\#}{x}_0-a^{†})-a^{†}{a}^2{\left(a^{†}\right)}^{*}{a}^{†}\hfill \\ & =0,\hfill \\ \hfill a^2{a}^{†}p{a}^{\#}+u+a{\left(a^{†}\right)}^{*}{a}^{†}& =a^2{a}^{†}p{a}^{\#}+x_0-a^2{a}^{†}p{a}^{\#}-a{\left(a^{†}\right)}^{*}{a}^{†}+a{\left(a^{†}\right)}^{*}{a}^{†}\hfill \\ & =x_0.\hfill \end{array}$$

Thus

$$x_0=a^2{a}^{†}p{a}^{\#}+u-a^{†}au+a{\left(a^{†}\right)}^{*}{a}^{†}+(1-a^{†}a)w.$$

Similarly $y_0=a{a}^{†}p+v-a{a}^{†}va{a}^{†}$. □

Theorem 28.

Let $a\in R^{\#}\cap R^{†}$. Then $a\in R^{SEP}$ if and only if the general solution to Equation (19) is given by the following expression:

$$\left\{\begin{array}{c}x=a^2{a}^{†}p{a}^{\#}+u-a^{†}au+a+(1-a^{†}a)w,\hfill \\ y=a{a}^{†}p+v-a{a}^{†}va{a}^{†}.\hfill \end{array}\right.p,u,v,w\in R.$$

(21)

Proof. 

⇒ Suppose $a\in R^{SEP}$. Then

$$x=a^2{a}^{†}p{a}^{\#}+u-a^{†}au+a+(1-a^{†}a)w=a^2{a}^{†}p{a}^{\#}+u-a^{†}au+a{\left(a^{†}\right)}^{*}{a}^{†}+(1-a^{†}a)w.$$

From Theorem 27, we know that (21) is the general solution of Equation (19).

⇐ Assume that (21) is the general solution of Equation (19), one has

$$a^{*}{a}^{\#}(a^2{a}^{†}p{a}^{\#}+u-a^{†}au+a+(1-a^{†}a)w)-a^{*}(a{a}^{†}p+v-a{a}^{†}va{a}^{†})a^{\#}=a^{†}.$$

That is $a^{*}{a}^{\#}a=a^{†}$. Hence $a\in R^{SEP}$ by [23] (Theorem 1.5.3). □

Theorem 29.

Let $a\in R^{\#}\cap R^{†}$. Then $a\in R^{SEP}$ if and only if the general solution to Equation (19) is given by the following expression:

$$\left\{\begin{array}{c}x=a^2{a}^{*}p{a}^{\#}+u-a^{†}au+a{\left(a^{†}\right)}^{*}{a}^{\#}+(1-a^{†}a)w,\hfill \\ y=a{a}^{†}p+v-a{a}^{†}va{a}^{†}.\hfill \end{array}\right.p,u,v,w\in R.$$

(22)

Proof. 

The proof is similar to Theorem 28. □

From Theorems 27 and 29, we have the following result.

Theorem 30.

Let $a\in R^{\#}\cap R^{†}$. Then $a\in R^{SEP}$ if and only if the general solution to Equation (19) is given by the following expression:

$$\left\{\begin{array}{c}x=a^2{a}^{†}p{a}^{†}+u-a^{†}au+a^2{a}^{†}+(1-a^{†}a)w,\hfill \\ y=a{a}^{†}p+v-a{a}^{†}va{a}^{†}.\hfill \end{array}\right.p,u,v,w\in R.$$

(23)