EP Elements in Rings with Involution

Abstract

Let R be a unital ring with involution. We first show that the EP elements in R can be characterized by three equations. Namely, let $a\in R$, then a is EP if and only if there exists $x\in R$ such that $(xa)^{*}=xa$, $xa^{2}=a$ and $ax^{2}=x.$ Any EP element in R is core invertible and Moore–Penrose invertible. We give more equivalent conditions for a core (Moore–Penrose) invertible element to be an EP element. Finally, any EP element is characterized in terms of the n-EP property, which is a generalization of the bi-EP property.

1 Introduction

Throughout this paper, R will denote a unital ring with involution, i.e., a ring with a mapping $a\mapsto a^*$ satisfying $(a^*)^*=a$, $(ab)^*=b^*a^*$ and $(a+b)^*=a^*+b^*$, for all $a,b\in R$. The notion of core inverse for a complex matrix was introduced by Baksalary and Trenkler [1]. In [13], Rakić et al. generalized the core inverse of a complex matrix to the case of an element in R. More precisely, let $a,x\in R$, if

$$\begin{aligned} axa=a,~xR=aR\quad \text {and}\quad Rx=Ra^{*}, \end{aligned}$$

then x is called a core inverse of a. If such an element x exists, then it is unique and denoted by $a^{\tiny {\textcircled {\tiny {\#}}}}$. The set of all core invertible elements in R will be denoted by $R^{\tiny {\textcircled {\tiny {\#}}}}$. Also, in [13] the authors defined a related inner inverse in a ring with an involution. If $a \in R$, then $x \in R$ is called a dual core inverse of a if

$$\begin{aligned} a x a = a, \ x R = a^* R \quad \text {and} \quad R x = Ra. \end{aligned}$$

If such an element x exists, then it is unique and denoted by ${a}_{\tiny {\textcircled {\tiny {\#}}}}$. The set of all dual core invertible elements in R will be denoted by ${R}_{\tiny {\textcircled {\tiny {\#}}}}$. It is elemental to prove that $a \in R^{\tiny {\textcircled {\tiny {\#}}}}$ if and only if $a^* \in {R}_{\tiny {\textcircled {\tiny {\#}}}}$, and in this case, one has $(a^{\tiny {\textcircled {\tiny {\#}}}})^* = {(a^*)}_{\tiny {\textcircled {\tiny {\#}}}}$. This last observation permits to get results concerning dual core inverses from the corresponding results on core inverses.

Let $a,x\in R$. If

$$\begin{aligned} axa=a, \quad xax=x, \quad (ax)^{*}=ax \quad \text {and}\quad (xa)^{*}=xa, \end{aligned}$$

then x is called a Moore–Penrose inverse of a. If such an element x exists, then it is unique and denoted by $a^{\dagger }$. The set of all Moore–Penrose invertible elements will be denoted by $R^{\dagger }$.

Let $a\in R$. It can be easily proved that the set of elements $x\in R$ such that

$$\begin{aligned} axa=a, \quad xax=x\quad \text {and}\quad ax=xa \end{aligned}$$

is empty or a singleton. If this set is a singleton, its unique element is called the group inverse of a and denoted by $a^\#$. The set of all group invertible elements will be denoted by $R^\#$. The subset of R composed of all invertible elements of R will be denoted by $R^{-1}$.

A matrix $A\in \mathbb {C}^{n\times n}$ is called an EP (range-Hermitian) matrix if $\mathcal {R}(A)=\mathcal {R}(A^{*})$, where $\mathbb {C}^{n\times n}$ denotes the set of all $n\times n$ matrices over the field of complex numbers and $\mathcal {R}(A)$ stands for the range (column space) of $A\in \mathbb {C}^{n\times n}$. This concept can be found in [8]. An element $a\in R$ is said to be an EP element if $a\in R^{\dagger }\cap R^\#$ and $a^{\dagger }=a^\#$ [7]. The set of all EP elements will be denoted by $R^{\mathrm {EP}}$. Mosić et al. in [10, Theorem 2.1] gave several equivalent conditions such that an element in R to be an EP element. Patrício and Puystjens in [12, Proposition 2] proved that for a Moore–Penrose invertible element $a\in R$, $a\in R^{\mathrm {EP}}$ if and only if $aR=a^*R$. As for a Moore–Penrose invertible element $a\in R$, $a\in R^{\mathrm {EP}}$ if and only if $aa^{\dagger }=a^{\dagger }a$, thus we deduce that $aa^{\dagger }=a^{\dagger }a$ if and only if $aR=a^*R$. In [13, Theorem 3.1], Rakić et al. investigated some equivalent conditions such that a (dual) core invertible element in R to be an EP element. Also, they showed that $R^{\dagger }\cap R^{\tiny {\textcircled {\tiny {\#}}}} = R^{\dagger }\cap R^\#$. Motivated by [4, 6, 10, 12, 13], in this paper, we will give new equivalent characterizations such that an element in R to be an EP element.

We first show that the EP elements in R can be characterized by three equations. That is, let $a\in R$, then $a\in R^{\mathrm {EP}}$ if and only if there exists $x\in R$ such that $(xa)^{*}=xa$, $xa^{2}=a$ and $ax^{2}=x.$ In [13], Rakić et al. proved that $a\in R^{\dagger }$ if and only if there exists $x\in R$ such that $axa=a$, $xR=a^{*}R$ and $Rx=Ra^{*}$. Inspired by this result, we show that $a\in R^{\mathrm {EP}}$ if and only if there exists $x\in R$ such that

$$\begin{aligned} axa=a,~xR=aR\quad \text {and}\quad Rx^{*}=Ra. \end{aligned}$$

In [4, Theorem 16], for an operator $T\in L(X)$, where X is a Banach space, Boasso proved that for a Moore–Penrose invertible operator T, T is an EP operator if and only there exists an invertible operator $P\in L(X)$ such that $T^{\dagger }=PT$. We generalize this result to the ring case. Moreover, for $a\in R^{\dagger }$, we show that $a\in R^{\mathrm {EP}}$ if and only if there exists a (left) invertible element v such that $a^{\dagger }=va$. Similarly, for $a\in R^{\tiny {\textcircled {\tiny {\#}}}}$, then $a\in R^{\mathrm {EP}}$ if and only if there exists a (left) invertible element s such that $a^{\tiny {\textcircled {\tiny {\#}}}}=sa$.

In [13], Rakić et al. proved that $a\in R^{\mathrm {EP}}$ if and only if $a\in R^{\dagger }\cap R^\#$ with $a^{\dagger }=a^{\tiny {\textcircled {\tiny {\#}}}}$. Also, it is proved that $a\in R^{\mathrm {EP}}$ if and only if $a\in R^{\tiny {\textcircled {\tiny {\#}}}}$ with $a^\#=a^{\tiny {\textcircled {\tiny {\#}}}}$. In [11, Theorem 2.1], Mosić and Djordjević proved that $a\in R^{\mathrm {EP}}$ if and only if $a\in R^\#\cap R^{\dagger }$ with $a^{n}a^{\dagger }=a^{\dagger }a^{n}$ for all choices $n\geqslant 1$. This result also can be found in [5, Theorem 2.4] by Chen. Motivated by [11, 13], we will give more new equivalent conditions under which a core invertible element is an EP element. And we define the concept of n-EP as a generalization of bi-EP. As an application, we will use n-EP property to give an equivalent characterization of the EP elements in R.

2 New Characterizations of EP Elements by Equations

In this section, we first show that any EP element in R can be characterized by three equations. Let us begin with an auxiliary lemma.

Lemma 2.1

[9, Theorem 7.3] Let $a\in R$. Then $a\in R^{\mathrm {EP}}$ if and only if $a\in R^\#$ with $(aa^\#)^{*}=aa^\#$.

It is well known that the group inverse of an element in a ring can be characterized by three equations and the Moore–Penrose inverse of an element in a ring can be characterized by four equations. In the following theorem, we show that an EP element in a ring can be described by three equations.

Theorem 2.2

Let $a\in R$. Then $a\in R^{\mathrm {EP}}$ if and only if there exists $x\in R$ such that

$$\begin{aligned} (xa)^{*}=xa,~~xa^{2}=a~~\text {and}~~ax^{2}=x. \end{aligned}$$

(2.1)

Proof

Suppose $a\in R^{\mathrm {EP}}$. Let $x=a^{\dagger }=a^\#$, then $(xa)^{*}=(a^{\dagger }a)^{*}=a^{\dagger }a=xa$, $xa^{2}=a^\#a^{2}=a$ and $ax^{2}=a(a^\#)^{2}=a^\#=x.$ Conversely, if there exists $x\in R$ such that $(xa)^{*}=xa$, $xa^{2}=a$ and $ax^{2}=x,$ then $a(x^2a) = (ax^2)a = xa = x(xa^2) = (x^2a)a$, $a(x^2a)a = (xa)a = xa^2 = a$, and $(x^2a)a(x^2a) = (xa)(x^2a) = x(ax^2)a = x^2a$. These three equalities prove that $a \in R^\#$, $a^\# = x^2a$, and $aa^\# = xa$. By Lemma 2.1, we get $a \in R^\mathrm{EP}$. $\square$

For an idempotent p in a ring R, every $a\in R$ can be written as

$$\begin{aligned} a = pap+pa(1-p)+(1-p)ap+(1-p)a(1-p) \end{aligned}$$

or in the matrix form

$$\begin{aligned} a = \left[ \begin{matrix} a_{11} &{}\quad a_{12} \\ a_{21} &{}\quad a_{22} \end{matrix} \right] , \end{aligned}$$

where $a_{11}=pap$, $a_{12}=pa(1-p)$, $a_{21}=(1-p)ap$ and $a_{22}=(1-p)a(1-p)$.

Let us observe that pRp and $(1-p)R(1-p)$ are rings whose unities are p and $1-p$, respectively. Also, we notice that if $p=p^{*}$, then the above matrix representation preserves the involution. The term projection will be reserved for a Hermitian idempotent.

Suppose in this paragraph that $a \in R$ is an EP element. If we denote $p=aa^\dagger =a^{\dagger } a$, since $ap=pa=a$ and $a^{\dagger } p = p a^{\dagger } = a^\dagger$, then the matrix representations of a and $a^\dagger$ with respect to the Hermitian idempotent p are

$$\begin{aligned} a = \left[ \begin{array}{cc} a &{}\quad 0 \\ 0 &{}\quad 0 \end{array} \right] \quad \text {and} \quad a^{\dagger } = \left[ \begin{array}{cc} a^{\dagger } &{}\quad 0 \\ 0 &{}\quad 0 \end{array} \right] , \end{aligned}$$

(2.2)

respectively.

Recall that a ring R is prime if for any two elements a and b of R, $aRb = 0$ implies that either $a = 0$ or $b = 0$ and a ring R is semiprime if for any element a in R, $aRa = 0$ implies that $a = 0$.

In the next theorem, the set of elements $x \in R$ satisfying (2.1) is described.

Theorem 2.3

Let $a \in R$. If a is EP, then $\{ x \in R: (xa)^* = xa, xa^2 =a, ax^2 = x \} = \{a^{\dagger } +aa^{\dagger } y (1-aa^{\dagger }): y \in R\}$. Moreover, if R is a prime ring, then $\{ x \in R: (xa)^* = xa, xa^2 =a, ax^2 = x \} = \{ a^{\dagger }\}$ if and only if $a=0$ or a is invertible.

Proof

Suppose a that is an EP element. We use the matrix representations of a and $a^{\dagger }$ with respect to the projection $p=aa^\dagger$ given in (2.2). Let $x=\left[ {\begin{matrix} u &{} v \\ w &{} z \end{matrix}} \right]$ be the representation of any $x\in \{ x \in R: (xa)^* = xa, xa^2 =a, ax^2 = x \}$ with respect to p. From $xa^{2}=a$, we get

$$\begin{aligned} \left[ \begin{matrix} a &{}\quad 0 \\ 0 &{}\quad 0 \end{matrix} \right] =\left[ \begin{matrix} u &{}\quad v \\ w &{}\quad z \end{matrix} \right] \left[ \begin{matrix} a^{2} &{}\quad 0 \\ 0 &{}\quad 0 \end{matrix} \right] =\left[ \begin{matrix} ua^{2} &{}\quad 0 \\ wa^{2} &{}\quad 0 \end{matrix} \right] . \end{aligned}$$

Since a is EP and $aa^{\dagger }=p=a^{\dagger }a$, then a is invertible in pRp and its inverse is $a^{\dagger }$. Hence from $a=ua^{2}$ and $0=wa^{2}$, we obtain $u=a^{\dagger }$ and $0=w$, respectively. Now, from $x=ax^{2}$ we have

$$\begin{aligned} \left[ \begin{matrix} a^{\dagger } &{}\quad v \\ 0 &{}\quad z \end{matrix} \right] = \left[ \begin{matrix} a &{}\quad 0 \\ 0 &{}\quad 0 \end{matrix} \right] \left[ \begin{matrix} a^{\dagger } &{}\quad v \\ 0 &{}\quad z \end{matrix} \right] x =\left[ \begin{matrix} p &{}\quad av \\ 0 &{}\quad 0 \end{matrix} \right] \left[ \begin{matrix} a^{\dagger } &{}\quad v \\ 0 &{}\quad z \end{matrix} \right] =\left[ \begin{matrix} a^{\dagger } &{}\quad v+avz \\ 0 &{}\quad 0 \end{matrix} \right] , \end{aligned}$$

which implies $z=0$. Therefore,

$$\begin{aligned} x=\left[ \begin{matrix} a^{\dagger } &{}\quad v \\ 0 &{}\quad 0 \end{matrix} \right] =a^{\dagger }+v, \end{aligned}$$

that is $\{ x \in R: (xa)^* = xa, xa^2 =a, ax^2 = x \}\subseteq \{a^{\dagger }+aa^{\dagger }y(1-aa^{\dagger }):y\in R\}$.

Let us prove the opposite inclusion. We have that $aa^{\dagger }=a^{\dagger }a$ since a is EP. Let $x=a^{\dagger }+aa^{\dagger }y(1-aa^{\dagger })$, then

$$\begin{aligned} xa&= [a^{\dagger }+aa^{\dagger }y(1-aa^{\dagger })]a=a^{\dagger }a \text { is Hermitian},\\ xa^{2}&= [a^{\dagger }+aa^{\dagger }y(1-aa^{\dagger })]a^{2}=a^{\dagger }a^{2}=a,\\ ax^{2}&= a[a^{\dagger }+aa^{\dagger }y(1-aa^{\dagger })]^{2} =[aa^{\dagger }+ay(1-aa^{\dagger })][a^{\dagger }+aa^{\dagger }y(1-aa^{\dagger })]\\&=a^{\dagger }+aa^{\dagger }y(1-aa^{\dagger })=x. \end{aligned}$$

Suppose that R is a prime ring. If $a=0$, then $\{ x \in R: (xa)^* = xa, xa^2 =a, ax^2 = x \}=\{0\}$. If a is invertible, then $\{ x \in R: (xa)^* = xa, xa^2 =a, ax^2 = x \} = \{a^{-1}\}$. If $\{ x \in R: (xa)^* = xa, xa^2 =a, ax^2 = x \}$ is a singleton, then $aa^{\dagger } y(1-aa^{\dagger })=0$ for all $y \in R$, by using that R is prime, then $aa^{\dagger }=0$ or $1-aa^\dagger =0$. The first of the previous alternatives is equivalent to $a=0$ and the second one (since a is EP) is equivalent to the invertibility of a. $\square$

We will also use the following notations: $aR=\{ax : x\in R\}$, $Ra=\{xa : x\in R\}$, ${}^{\circ }a = \{x\in R : xa=0\}$ and $a^{\circ }=\{x\in R : ax=0\}$ . The following lemma will be useful in the sequel.

Lemma 2.4

[14, Lemma 8] Let $a, b\in R$. Then:

• (1) $aR\subseteq bR$ implies ${}^{\circ }b\subseteq {}^{\circ }a$ and the converse is valid whenever b is regular;

• (2) $Ra\subseteq Rb$ implies $b^{\circ }\subseteq a^{\circ }$ and the converse is valid whenever b is regular.

Theorem 2.5

Let $a\in R$. Then the following are equivalent:

• (1) $a\in R^{\mathrm {EP}}$;

• (2) there exists $x\in R$ such that $axa=a$, $xR=aR$ and $Rx^{*}=Ra$;

• (3) there exists $x\in R$ such that $axa=a$, $xR=aR$ and $Rx^{*}\subseteq Ra$;

• (4) there exists $x\in R$ such that $xax=x$, $xR=aR$ and $Rx^{*}=Ra$;

• (5) there exists $x\in R$ such that $xax=x$, $xR=aR$ and $Ra\subseteq Rx^{*}$;

• (6) there exists $x\in R$ such that $axa=a$, ${}^{\circ }x={}^{\circ } a$ and $(x^{*})^{\circ }=a^{\circ }$;

• (7) there exists $x\in R$ such that $axa=a$, ${}^{\circ }x={}^{\circ }a$ and $a^{\circ }\subseteq (x^{*})^{\circ }$;

• (8) there exists $x\in R$ such that $xax=x$, ${}^{\circ }x={}^{\circ } a$ and $(x^{*})^{\circ }=a^{\circ }$;

• (9) there exists $x\in R$ such that $xax=x$, ${}^{\circ }x={}^{\circ }a$ and $(x^{*})^{\circ }\subseteq a^{\circ }$.

Proof

$(1)\Rightarrow (2)$: Let $x=a^{\dagger }=a^\#$, then $axa=a$, $x=a(a^\#)^{2}$, $x^{*}=(a^{\dagger })^{*}=(a^{\dagger } aa^{\dagger })^{*}=(a^{\dagger })^{*}a^{\dagger }a,$ and $xa^{2}=a^\#a^{2}=a=aa^{\dagger }a=aa^{*}(a^{\dagger })^{*}=aa^{*}x^{*}.$ Thus $xR=aR$ and $Rx^{*}=Ra$.

$(2)\Rightarrow (3)$ and $(6)\Rightarrow (7)$ are clear.

$(2)\Rightarrow (6)$ and $(3)\Rightarrow (7)$ are obvious by Lemma 2.4.

$(7)\Rightarrow (1)$: Suppose that there exists $x\in R$ such that $axa=a$, $^{\circ }x=^{\circ }a$ and $a^{\circ }\subseteq (x^{*})^{\circ }$. Since $(1-ax)a=0,$ then $1-ax\in \ {}^{\circ }a={}^{\circ }x$, hence $(1-ax)x=0$. Since $a(1-xa)=0$, then $1-xa\in a^{\circ }\subseteq (x^{*})^{\circ }$, hence $x^{*}(1-xa)=0$, i.e., $x=(xa)^{*}x$. We get $xa=(xa)^{*}xa$, hence xa is Hermitian. Finally, $x=xax$ implies $1-xa\in \ ^{\circ }x={}^{\circ }a$, whence $xa^{2}=a$. Therefore $a\in R^{\mathrm {EP}}$ by Theorem 2.2.

The implications $(1)\Leftrightarrow (4)\Leftrightarrow (5) \Leftrightarrow (8)\Rightarrow (9)$ are similar to $(1)\Leftrightarrow (2)\Leftrightarrow (3)\Leftrightarrow (6)\Rightarrow (7)$.

$(9)\Rightarrow (1)$: There exists $x\in R$ such that $xax=x$, ${}^{\circ }x={}^{\circ }a$ and $(x^{*})^{\circ }\subseteq a^{\circ }$. It is obvious that $(x^{*})^{\circ }\subseteq a^{\circ }$ is equivalent to $^{\circ }x\subseteq {}^{\circ }(a^{*})$. We have $a=xa^{2}$ since $(1-xa)x=0$ implies $(1-xa)a=0$. Similarly, we have $a^{*}=xaa^{*}$ since $(1-xa)x=0$ implies $(1-xa)a^{*}=0$. Thus $(xa)^{*}=a^{*}x^{*}=xaa^{*}x^{*}=xa(xa)^{*}$, that is $(xa)^{*}=xa.$ By $a^{*}=xaa^{*}$ and $(xa)^{*}=xa$, we have $a^{*}=xaa^{*}=(xa)^{*}a^{*}=(axa)^{*}$, that is $a=axa.$ Hence by ${}^{\circ }x={}^{\circ }a$, we have $(1-ax)a=0$ implies $(1-ax)x=0$, which gives $x=ax^{2}$. Therefore $a\in R^{\mathrm {EP}}$ by Theorem 2.2. $\square$

Theorem 2.6

Let $a\in R^\mathrm{EP}$ and denote $p=aa^{\dagger }$. Then the following sets are the same

• (1) $\{x\in R: axa=a, xR\subseteq aR\}=\{a^{\dagger }+py(1-p):y\in R\}$;

• (2) $\{x\in R: xax=x, xR=aR\}=\{a^{\dagger }+py(1-p):y\in R\}$;

• (3) $\{x\in R: axa=a, {}^{\circ }a\subseteq {}^{\circ }x\} = \{a^{\dagger }+py(1-p):y\in R\}$;

• (4) $\{x\in R: xax=x, {}^{\circ }a = {}^{\circ }x\} = \{a^{\dagger }+py(1-p):y \in R\}$.

Furthermore, if R is prime, then any of the above subsets is a singleton if and only if $a=0$ or a is invertible.

Proof

We use the matrix representations of a and $a^{\dagger }$ with respect to the projection $p=aa^\dagger$ given in (2.2). Let $x=\left[ {\begin{matrix} u &{} v \\ w &{} z \end{matrix}} \right]$ be the representation of any x with respect to p.

• (1) Let x satisfy $axa=a$ and $xR \subseteq aR$. From $axa=a$, we have

  $$\begin{aligned} \left[ \begin{matrix} a &{}\quad 0 \\ 0 &{}\quad 0 \end{matrix} \right] =\left[ \begin{matrix} a &{}\quad 0 \\ 0 &{}\quad 0 \end{matrix} \right] \left[ \begin{matrix} u &{}\quad v \\ w &{}\quad z \end{matrix} \right] \left[ \begin{matrix} a &{}\quad 0 \\ 0 &{}\quad 0 \end{matrix} \right] =\left[ \begin{matrix} au &{}\quad av \\ 0 &{}\quad 0 \end{matrix} \right] \left[ \begin{matrix} a &{}\quad 0 \\ 0 &{}\quad 0 \end{matrix} \right] =\left[ \begin{matrix} aua &{}\quad 0 \\ 0 &{}\quad 0 \end{matrix} \right] . \end{aligned}$$

  Since $a\in R^{\mathrm {EP}}$, we have that a is invertible in pRp and its inverse is $a^{\dagger }.$ Hence $a=aua$ gives $u=a^{\dagger }.$ Since $xR \subseteq aR$, we can write

  $$\begin{aligned} \left[ \begin{matrix} u &{}\quad v \\ w &{}\quad z \end{matrix} \right] =\left[ \begin{matrix} a &{}\quad 0 \\ 0 &{}\quad 0 \end{matrix} \right] \left[ \begin{matrix} \xi _{1} &{}\quad \xi _{2} \\ \xi _{3} &{}\quad \xi _{4} \end{matrix} \right] =\left[ \begin{matrix} a\xi _{1} &{}\quad a\xi _{2} \\ 0 &{}\quad 0 \end{matrix} \right] . \end{aligned}$$

  Therefore, $w=z=0$. Hence

  $$\begin{aligned} x=\left[ \begin{matrix} a^{\dagger } &{}\quad v \\ 0 &{}\quad 0 \end{matrix} \right] =a^{\dagger } +px(1-p)\in \left\{ a^{\dagger }+py(1-p):y\in R\right\} . \end{aligned}$$

  The opposite inclusion is trivial.

• (2) Let $x \in R$ satisfy $xax=x$ and $xR = aR$. Since $x \in a R$, by the proof of (1), we have $w=z=0$. Now, since $a \in xR$, we can write

  $$\begin{aligned} \left[ \begin{matrix} a &{}\quad 0 \\ 0 &{}\quad 0 \end{matrix} \right] =\left[ \begin{matrix} u &{}\quad v \\ 0 &{}\quad 0 \end{matrix} \right] \left[ \begin{matrix} \delta _{1} &{}\quad \delta _{2} \\ \delta _{3} &{}\quad \delta _{4} \end{matrix} \right] , \end{aligned}$$

  which implies

  $$\begin{aligned} a=u \delta _{1}+v \delta _{3},\qquad 0=u \delta _{2}+v \delta _{4}. \end{aligned}$$

  (2.3)

  Now, we use $xax=x$:

  $$\begin{aligned} \left[ \begin{matrix} u &{}\quad v \\ 0 &{}\quad 0 \end{matrix} \right] =\left[ \begin{matrix} u &{}\quad v \\ 0 &{}\quad 0 \end{matrix} \right] \left[ \begin{matrix} a &{}\quad 0 \\ 0 &{}\quad 0 \end{matrix} \right] \left[ \begin{matrix} u &{}\quad v \\ 0 &{}\quad 0 \end{matrix} \right] =\left[ \begin{matrix} ua &{}\quad 0 \\ 0 &{}\quad 0 \end{matrix} \right] \left[ \begin{matrix} u &{}\quad v \\ 0 &{}\quad 0 \end{matrix} \right] =\left[ \begin{matrix} uau &{}\quad uav \\ 0 &{}\quad 0 \end{matrix} \right] . \end{aligned}$$

  Therefore

  $$\begin{aligned} u=uau,\qquad v=uav. \end{aligned}$$

  (2.4)

  Post-multiply the first equality of (2.4) by $\delta _{1}$ and the second equality of (2.4) by $\delta _{3}$ to obtain

  $$\begin{aligned} u \delta _{1}=uau \delta _{1}, \qquad v \delta _{3}=uav \delta _{3}. \end{aligned}$$

  From (2.3),

  $$\begin{aligned} a=u \delta _{1}+v \delta _{3}=uau \delta _{1}+uav \delta _{3}=ua(u \delta _{1}+v \delta _{3})=ua^{2}. \end{aligned}$$

  We get $u=a^{\dagger }$ because a is invertible in pRp and its inverse is $a^{\dagger }$. Therefore

  $$\begin{aligned} x=\left[ \begin{matrix} a^{\dagger } &{}\quad v \\ 0 &{}\quad 0 \end{matrix}\right] = a^{\dagger }+v=a^{\dagger } +px(1-p)\in \left\{ a^{\dagger }+py(1-p):y\in R\right\} . \end{aligned}$$

  For the opposite inclusion, it is easy to check that $[a^{\dagger } +py(1-p)]a[a^{\dagger } + py(1-p)] = a^{\dagger } + py(1-p)$ and $a^{\dagger } +py(1-p) \in a R$ in view of $a \in R^\mathrm{EP}$. From $a =[a^{\dagger } +py(1-p)]a^2$, we deduce that $a \in [a^{\dagger } + py(1-p)]R$.

Since $a\in R^{\mathrm {EP}}$, a is regular. Hence, from Lemma 2.4, $xR\subseteq aR$ is equivalent to ${}^{\circ }a\subseteq {}^{\circ }x$ and $xR=aR$ is equivalent to $^{\circ }a = {}^{\circ }x$. Therefore, (1) is equivalent to (3) and (2) is equivalent to (4).

The proof of the last affirmation of this theorem has the same proof as the corresponding part of Theorem 2.3. $\square$

By considering that a is EP if and only if $a^{*}$ is EP and having in mind Theorems 2.2, 2.3, 2.5 and 2.6, we get the following four theorems.

Theorem 2.7

Let $a\in R$. Then $a\in R^{\mathrm {EP}}$ if and only if there exists $y\in R$ such that

$$\begin{aligned} (ay)^{*}=ay,~~a^{2}y=a~~\text {and}~~y^{2}a=y. \end{aligned}$$

Theorem 2.8

Let $a \in R$. If a is EP, then $\{ y \in R: (ay)^* = ay, a^2y =a, y^2a = y \} = \{ a^{\dagger } +(1-aa^{\dagger })xaa^{\dagger } : x \in R\}$. Moreover, if R is a prime ring, then $\{ y \in R: (ay)^* = ay, a^2y =a, y^2a = y \} = \{ a^{\dagger }\}$ if and only if $a=0$ or a is invertible.

Theorem 2.9

Let $a\in R$. Then the following are equivalent:

• (1) $a\in R^{\mathrm {EP}}$;

• (2) there exists $y\in R$ such that $aya=a$, $Ry=Ra$ and $y^{*}R=aR$;

• (3) there exists $y\in R$ such that $aya=a$, $Ry=Ra$ and $y^{*}R\subseteq aR$;

• (4) there exists $y\in R$ such that $yay=y$, $Ry=Ra$ and $y^{*}R=aR$;

• (5) there exists $y\in R$ such that $yay=y$, $Ry=Ra$ and $aR\subseteq y^{*}R$;

• (6) there exists $y\in R$ such that $aya=a$, $y^{\circ }=a^{\circ }$ and ${}^{\circ }(y^{*})={}^{\circ }a$;

• (7) there exists $y\in R$ such that $aya=a$, $y^{\circ }=a^{\circ }$ and $^{\circ }a\subseteq {}^{\circ }(y^{*})$;

• (8) there exists $y\in R$ such that $yay=y$, $y^{\circ }=a^{\circ }$ and $^{\circ }(y^{*}) = {}^{\circ }a$;

• (9) there exists $y\in R$ such that $yay=y$, $y^{\circ }=a^{\circ }$ and $^{\circ }(y^{*})\subseteq {}^{\circ }a$.

Theorem 2.10

Let $a\in R^{EP}$ and denote $p=aa^{\dagger }$. Then the following sets are the same

• (1) $\{y\in R: aya=a, Ry \subseteq Ra\}=\{a^{\dagger }+(1-p)xp:x\in R\}$;

• (2) $\{y\in R: yay=y, Ry=Ra\}=\{a^{\dagger }+(1-p)xp:x\in R\}$;

• (3) $\{y\in R: aya=a, a^{\circ }\subseteq y^{\circ }\} = \{a^{\dagger }+(1-p)xp : x\in R\}$;

• (4) $\{y\in R: yay=y, a^{\circ } = y^{\circ }\} = \{a^{\dagger }+(1-p)xp : x\in R\}$.

Furthermore, if R is prime, then any of the above subsets is a singleton if and only if $a=0$ or a is invertible.

We will characterize when $a \in R$ is EP by another three equations.

Theorem 2.11

Let $a\in R$. Then $a\in R^{\mathrm {EP}}$ if and only if there exists $x\in R$ such that

$$\begin{aligned} a^{2}x=a, ~~ ax=xa ~~ \text {and} ~~ (ax)^{*}=ax. \end{aligned}$$

(2.5)

Proof

If a is EP, by taking $x=a^{\dagger }=a^\#$, we get (2.5). Conversely, assume that there exists $x \in R$ such that (2.5) is satisfied. We shall show that $a\in R^{\mathrm {EP}}$ and $a^\#=ax^{2}.$ Since $ax=xa$, we get $a(ax^{2})=(ax^{2})a$, but in addition, $a(ax^{2})=(a^{2}x)x=ax$, which leads to $a(ax^{2})a=a^{2}x=a$ and $(ax^{2})a(ax^{2})=(ax^{2})ax =(a^{2}x)x^{2}=ax^{2}.$ Since $aa^\#=a^{2}x^{2}=ax$ is Hermitian, the conclusion follows from Lemma 2.1. $\square$

We have seen that if $a\in R$ is EP, then $\{ x \in R: a^2x=a, ax=xa, (ax)^*=ax \}$ is not empty. In next theorem, we describe this set.

Theorem 2.12

Let $a \in R$. If a is EP, then $\{ x \in R: a^{2}x=a, ax=xa, (ax)^{*}=ax \} = \{ a^{\dagger } +(1-aa^{\dagger }) y (1-aa^{\dagger }): y \in R\}$. Moreover, if R is a semiprime ring, then $\{ x \in R: a^{2}x=a, ax=xa, (ax)^{*}=ax \} = \{ a^{\dagger }\}$ if and only if a is invertible.

Proof

Suppose that a is an EP element. We use the matrix representations of a and $a^{\dagger }$ with respect to the projection $p=aa^\dagger$ given in (2.2). Let $x=\left[ {\begin{matrix} u &{} v \\ w &{} z \end{matrix}} \right]$ be the representation of any $x\in R$ with respect to p.

Let $x \in R$ satisfy $a^2x=a$, $ax=xa$ and $(ax)^*=ax$. From $a^{2}x=a$, we get

$$\begin{aligned} \left[ \begin{matrix} a &{}\quad 0 \\ 0 &{}\quad 0 \end{matrix} \right] ^{2} \left[ \begin{matrix} u &{}\quad v \\ w &{}\quad z \end{matrix} \right] =\left[ \begin{matrix} a &{}\quad 0 \\ 0 &{}\quad 0 \end{matrix} \right] , \end{aligned}$$

which leads to $a^2u=a$ and $a^2 v=0$. Since a is EP and $aa^{\dagger }=p=a^{\dagger }a$, then a is invertible in pRp and its inverse is $a^{\dagger }$. Hence from $a=a^{2}u$ and $0=a^{2}v$, we obtain $u=a^{\dagger }$ and $0=v$, respectively. Now, from $ax=xa$ we have

$$\begin{aligned} \left[ \begin{matrix} a &{}\quad 0 \\ 0 &{}\quad 0 \end{matrix} \right] \left[ \begin{matrix} a^{\dagger } &{}\quad 0 \\ w &{}\quad z \end{matrix} \right] = \left[ \begin{matrix} a^{\dagger } &{}\quad 0 \\ w &{}\quad z \end{matrix} \right] \left[ \begin{matrix} a &{}\quad 0 \\ 0 &{}\quad 0 \end{matrix} \right] , \end{aligned}$$

which implies $0=wa$, and taking into account that a is invertible in pRp, we have $0=w$. Therefore,

$$\begin{aligned} x=\left[ \begin{matrix} a^{\dagger } &{}\quad 0 \\ 0 &{}\quad z \end{matrix} \right] =a^{\dagger }+z, \end{aligned}$$

that is $\{ x \in R: a^{2}x=a, ax=xa, (ax)^{*}=ax \}\subseteq \{a^{\dagger }+(1-aa^{\dagger })y(1-aa^{\dagger }):y\in R\}.$

Let us prove the opposite inclusion. We have $aa^{\dagger }=a^{\dagger }a$ since a is EP. Now,

$$\begin{aligned} a[a^{\dagger }+(1-aa^{\dagger })y(1-aa^{\dagger })]= & {} aa^{\dagger } ~\text {is Hermitian},\\ a^{2}[a^{\dagger }+(1-aa^{\dagger })y(1-aa^{\dagger })]= & {} a^{2}a^{\dagger }=a,\\ a[a^{\dagger }+(1-aa^{\dagger })y(1-aa^{\dagger })]= & {} aa^{\dagger }=a^{\dagger }a =[a^{\dagger }+(1-aa^{\dagger })y(1-aa^{\dagger })]a. \end{aligned}$$

Suppose that R is a semiprime ring. If a is invertible, then $\{ x \in R: a^{2}x=a, ax=xa, (ax)^{*}=ax \} = \{a^{-1}\}$. If $\{ x \in R: a^{2}x=a, ax=xa, (ax)^{*}=ax \}$ is a singleton, then $(1-aa^{\dagger })y(1-aa^{\dagger })=0$ for all $y \in R$. By using that R is semiprime, we get $1-aa^\dagger =0$, which (since a is EP) is equivalent to the invertibility of a. $\square$

Theorem 2.13

Let $a\in R$. Then the following are equivalent:

• (1) $a\in R^{\mathrm {EP}}$;

• (2) $^{\circ }(a^{2})\subseteq {}^{\circ }a$ and there exists $x\in R$ such that $xa^{2}=a$ and $(xa)^{*}=xa$;

• (3) $(a^{2})^{\circ }\subseteq a^{\circ }$ and there exists $x\in R$ such that $a^{2}x=a$ and $(ax)^{*}=ax$.

Furthermore, under these equivalences one has that the set of elements x satisfying (2) is $\{a^{\dagger }+y(1-aa^{\dagger }): y\in R\}$ and the set of elements x satisfying (3) is $\{a^{\dagger }+(1-aa^{\dagger })z: z\in R\}$.

Proof

(1) $\Rightarrow$ (2): The inclusion $^{\circ }(a^{2})\subseteq {}^{\circ }a$ is evident from $a\in R^\#$. For the remaining, it is sufficient to take $x=a^\dagger =a^\#$.

(2) $\Rightarrow$ (1): Since $(ax-1)a^2 = a(xa^2)-a^2 = a^2-a^2=0$, we get $ax-1 \in \ ^{\circ }(a^{2})\subseteq {}^{\circ }a$, hence $axa=a$. From

$$\begin{aligned} ax^2a^2 = ax(xa^2)=axa=a=xa^2 \end{aligned}$$

we get $ax^2-x \in {}^{\circ }(a^{2})\subseteq {}^{\circ }a$, hence $ax^2a=xa$. Now, we prove $a^\# = x^2a$ by the definition of the group inverse,

$$\begin{aligned} a(x^2a)= & {} ax^2a = xa, \qquad (x^2a)a = x(xa^2)=xa; \\ a(x^2a)a= & {} xa^2 = a; \qquad (x^2a)a(x^2a) = x(xa^2)x^2a =xax^2a=x(ax^2a)=x^2a. \end{aligned}$$

Now, $a^\# = x^2a$ and $aa^\# =ax^2a=xa$ is Hermitian. Hence (1) follows from Lemma 2.1.

The proof of $(1) \Leftrightarrow (3)$ is similar to the proof of $(1)\Leftrightarrow (2)$.

Now, let us prove the last part of the theorem. Recall that a is EP. If $x \in R$ satisfies $xa^2=a$, then $(x-a^{\dagger })aa^{\dagger } = xaa^{\dagger } -a^{\dagger } = xa^2(a^{\dagger })^2-a^{\dagger } = a(a^{\dagger })^2-a^{\dagger } = 0$. Hence $x-a^{\dagger } = (x-a^{\dagger })(1-aa^{\dagger })$, which yields $x \in \{ a^{\dagger } + y(1-aa^{\dagger }): y \in R\}$. Reciprocally, it is evident that for any $y \in R$ one has that $[a^{\dagger } + y(1-aa^{\dagger })]a^2 = a$ and $[a^{\dagger } + y(1-aa^{\dagger })]a$ is Hermitian. In a similar manner we can prove that x satisfies (3) if and only if $x \in \{ a^{\dagger } + (1-aa^{\dagger })z: z \in R\}$. $\square$

3 When a Core Invertible Element is an EP Element

Any EP element is core invertible, but when a core invertible element is EP? In this section, we answer this question. Let us start this section with a lemma.

Lemma 3.1

[13, Theorem 2.14] An element $a \in R$ is core invertible if and only if there exists $x \in R$ such that

$$\begin{aligned} axa=a,~~xax=x,~~(ax)^{*}=ax,~~xa^{2}=a~~\text {and}~~ax^{2}=x. \end{aligned}$$

Under this equivalence, one has that $x=a^{\tiny {\textcircled {\tiny {\#}}}}$.

Let us recall the following result.

Theorem 3.2

Let $a \in R$.

• (1) [3, Proposition 8.24] a is group invertible if and only if there exists an idempotent $p \in R$ such that $ap=pa=0$ and $a+p$ is invertible. Under this equivalence, we have $p=1-aa^\#$ and $a^\# = (a+p)^{-1}-p$.

• (2) [2, Theorem 2.1] a is EP if and only if there exists a projection $p \in R$ such that $ap=pa=0$ and $a+p$ is invertible. Under this equivalence, we have $p=1-aa^{\dagger }$ and $a^{\dagger } = (a+p)^{-1}-p$.

In fact, the second item of the previous result was stated for unital $C^*$-algebras, but as one can easily check, its proof remains valid for unital rings with an involution.

In the next theorem, we give analogous equivalences for the core invertibility.

Theorem 3.3

Let $a \in R$. Then following are equivalent:

• (1) a is core invertible;

• (2) there exists a projection p such that $pa=0$ and $a(1-p)$ is invertible in the ring $(1-p)R(1-p)$;

• (3) there exists a projection p such that $pa=0$ and $a(1-p)+p$ is invertible.

Under this equivalence, one has that this projection p is unique and $p=1-a a^{\tiny {\textcircled {\tiny {\#}}}}$. In addition,

$$\begin{aligned} (a(1-p))^{-1}_{(1-p)R(1-p)} = a^{\tiny {\textcircled {\tiny {\#}}}}, \quad (a(1-p)+p)^{-1} = p+a^{\tiny {\textcircled {\tiny {\#}}}}. \end{aligned}$$

Proof

(1) $\Rightarrow$ (2): Let $p=1-aa^{\tiny {\textcircled {\tiny {\#}}}}$. Then p is a projection with $pa=0$. Put $\overline{p}=1-p=aa^{\tiny {\textcircled {\tiny {\#}}}}$. Now, using Lemma 3.1, we get $a^{\tiny {\textcircled {\tiny {\#}}}} =aa^{\tiny {\textcircled {\tiny {\#}}}}a^{\tiny {\textcircled {\tiny {\#}}}} =aa^{\tiny {\textcircled {\tiny {\#}}}}a^{\tiny {\textcircled {\tiny {\#}}}} aa^{\tiny {\textcircled {\tiny {\#}}}}\in \overline{p}R \overline{p}$, $a^{\tiny {\textcircled {\tiny {\#}}}} a \overline{p}= a^{\tiny {\textcircled {\tiny {\#}}}}a^2 a^{\tiny {\textcircled {\tiny {\#}}}} =aa^{\tiny {\textcircled {\tiny {\#}}}}=\overline{p}$ (Since $a^{\tiny {\textcircled {\tiny {\#}}}}a^2=a$) and $a\overline{p}a^{\tiny {\textcircled {\tiny {\#}}}} = a a^{\tiny {\textcircled {\tiny {\#}}}} = \overline{p}$. Therefore, $a \overline{p}$ is invertible in the ring $\overline{p}R \overline{p}$ and its inverse is $a^{\tiny {\textcircled {\tiny {\#}}}}$.

(2) $\Leftrightarrow$ (3): Let $p \in R$ be a projection such that $pa=0$. It is easy to verify that

$$\begin{aligned} a(1-p) = \left[ \begin{matrix} 0 &{}\quad 0 \\ 0 &{}\quad a (1-p) \end{matrix} \right] , \quad a(1-p) + p = \left[ \begin{matrix} p &{}\quad 0 \\ 0 &{}\quad a (1-p) \end{matrix} \right] . \end{aligned}$$

(3.1)

Taking into account that p is invertible in the ring pRp (in fact, p is the unity), evidently we have that $a (1-p) \in ((1-p)R (1-p))^{-1} \Leftrightarrow a(1-p) + p \in R^{-1}$.

(2) $\Rightarrow$ (1): Let $x \in (1-p) R (1-p)$ be the inverse of $a (1-p)$ in $(1-p) R (1-p)$. This means that $a(1-p)x = xa(1-p) = 1-p$. Observe that $x \in (1-p) R (1-p)$ implies $(1-p)x=x(1-p)=x$, and therefore, $ax = a(1-p)x = xa(1-p) = 1-p$. We will prove that $x=a^{\tiny {\textcircled {\tiny {\#}}}}$ by using Lemma 3.1. Let us recall that we can use $pa=0$ and $(1-p)a=a$ by hypothesis.

$$\begin{aligned} axa= & {} (ax)a = (1-p)a = a. \\ xax= & {} x(ax) = x(1-p) = x. \\ ax= & {} 1-p \text { is Hermitian}. \\ xa^2= & {} xa(1-p)a = (1-p)a = a. \\ ax^2= & {} (ax)x = (1-p)x = x. \end{aligned}$$

Now, we shall prove the uniqueness of the projection p. Assume that q is another projection such that $qa = 0$ and $a(1-q)$ is invertible in $(1-q)R(1-q)$. By the proof of (2) $\Rightarrow$ (1) we get that the inverse of $a(1-q)$ in $(1-q)R(1-q)$ is $a^{\tiny {\textcircled {\tiny {\#}}}}$, in particular $a^{\tiny {\textcircled {\tiny {\#}}}} \in (1-q) R (1-q)$ and $a(1-q)a^{\tiny {\textcircled {\tiny {\#}}}} = 1-q$. By using also Lemma 3.1 we get

$$\begin{aligned} (1-q) a a^{\tiny {\textcircled {\tiny {\#}}}} =a(1-q)a^{\tiny {\textcircled {\tiny {\#}}}} a a^{\tiny {\textcircled {\tiny {\#}}}} = a(1-q) a^{\tiny {\textcircled {\tiny {\#}}}} = 1-q. \end{aligned}$$

(3.2)

Since $a^{\tiny {\textcircled {\tiny {\#}}}} \in (1-q)R(1-q)$, exists $u \in R$ such that $a^{\tiny {\textcircled {\tiny {\#}}}}=u(1-q)$. Now,

$$\begin{aligned} a a^{\tiny {\textcircled {\tiny {\#}}}} (1-q) = au(1-q)^2 =au(1-q)=aa^{\tiny {\textcircled {\tiny {\#}}}}. \end{aligned}$$

Apply involution and use Lemma 3.1 in this last equality to get $(1-q) aa^{\tiny {\textcircled {\tiny {\#}}}} =aa^{\tiny {\textcircled {\tiny {\#}}}}$. From this last equality and (3.2) we obtain $a a^{\tiny {\textcircled {\tiny {\#}}}} =1-q$, which implies $p=q$.

Finally, $(a(1-p))^{-1}_{(1-p)R(1-p)} =a^{\tiny {\textcircled {\tiny {\#}}}}$ holds in view of the proof of (2) $\Leftrightarrow$ (1). Let us represent a and $a^{\tiny {\textcircled {\tiny {\#}}}}$ with respect to the projection $p=1-aa^{\tiny {\textcircled {\tiny {\#}}}}$. From the matrix representation of $a(1-p)+p$ given in (3.1), we easily get $(a(1-p)+p)^{-1} = p+ (a(1-p))^{-1}_{(1-p)R(1-p)} =p+a^{\tiny {\textcircled {\tiny {\#}}}}$. The proof is finished. $\square$

The following lemmas are necessary to prove next Theorem 3.6. We will use the notation $[a,b]=ab-ba$.

Lemma 3.4

[13, Theorem 3.1] Let $a\in R$. Then the following are equivalent:

• (1) $a\in R^{\mathrm {EP}}$;

• (2) $a\in R^{\dagger }$ and $[a,a^{\dagger }]=0$;

• (3) $a\in R^{\tiny {\textcircled {\tiny {\#}}}}$ and $[a,a^{\tiny {\textcircled {\tiny {\#}}}}]=0$;

• (4) $a\in R^{\tiny {\textcircled {\tiny {\#}}}}$ and $a^\#=a^{\tiny {\textcircled {\tiny {\#}}}}$;

• (5) $a\in R^{\dagger }\cap R^\#$ and $a^{\dagger }=a^{\tiny {\textcircled {\tiny {\#}}}}$.

Lemma 3.5

Let $a\in R^{\tiny {\textcircled {\tiny {\#}}}}$. Then

• (1) [13, Theorem 2.19] $a\in R^\#$ and $a^\#=(a^{\tiny {\textcircled {\tiny {\#}}}})^{2}a$;

• (2) [13, Theorem 2.18] $a^{\tiny {\textcircled {\tiny {\#}}}} \in R^{\mathrm {EP}}$ and $(a^{\tiny {\textcircled {\tiny {\#}}}})^{\tiny {\textcircled {\tiny {\#}}}} = (a^{\tiny {\textcircled {\tiny {\#}}}})^{\dagger } = (a^{\tiny {\textcircled {\tiny {\#}}}})^\# = a^2 a^{\tiny {\textcircled {\tiny {\#}}}}$. Moreover, if $a\in R^{\dagger }$, then $(a^{\dagger })^{\tiny {\textcircled {\tiny {\#}}}} = (a^{\tiny {\textcircled {\tiny {\#}}}}a)^{*}a$.

Theorem 3.6

Let $a\in R$. Then the following are equivalent:

• (1) $a \in R^{\mathrm {EP}}$;

• (2) $a \in R^{\tiny {\textcircled {\tiny {\#}}}}$ and $(a^{\tiny {\textcircled {\tiny {\#}}}}a)^{*} = a^{\tiny {\textcircled {\tiny {\#}}}}a$;

• (3) $a \in R^{\tiny {\textcircled {\tiny {\#}}}}$ and $(a^{\tiny {\textcircled {\tiny {\#}}}})^{\tiny {\textcircled {\tiny {\#}}}} = a$;

• (4) $a \in R^{\tiny {\textcircled {\tiny {\#}}}}$ and $(a^{\tiny {\textcircled {\tiny {\#}}}})^{\dagger }=a$;

• (5) $a \in R^{\tiny {\textcircled {\tiny {\#}}}}$ and $(a^{\tiny {\textcircled {\tiny {\#}}}})^\#=a$;

• (6) $a\in R^{\dagger }\cap R^\#$ and $(a^{\dagger })^{\tiny {\textcircled {\tiny {\#}}}} = a$;

• (7) $a\in R^{\dagger }\cap R^\#$ and $(a^{\dagger })^{\tiny {\textcircled {\tiny {\#}}}} = (a^{\tiny {\textcircled {\tiny {\#}}}})^{\dagger }$;

• (8) $a\in R^{\tiny {\textcircled {\tiny {\#}}}}$ and $ap=0$, where $p = 1-aa^{\tiny {\textcircled {\tiny {\#}}}}$.

Proof

$(1)\Leftrightarrow (2)$: Suppose $a\in R^{\mathrm {EP}}$. Then by Lemma 3.4, we have $a^{\dagger } =a^{\tiny {\textcircled {\tiny {\#}}}}$. Thus $(a^{\dagger }a)^{*}=a^{\dagger }a$ implies $(a^{\tiny {\textcircled {\tiny {\#}}}}a)^{*} =a^{\tiny {\textcircled {\tiny {\#}}}}a$. Conversely, suppose $a\in R^{\tiny {\textcircled {\tiny {\#}}}}$ and $(a^{\tiny {\textcircled {\tiny {\#}}}}a)^{*} =a^{\tiny {\textcircled {\tiny {\#}}}}a$. By Lemma 3.1, we have $a a^{\tiny {\textcircled {\tiny {\#}}}} a = a$, $a^{\tiny {\textcircled {\tiny {\#}}}} a a^{\tiny {\textcircled {\tiny {\#}}}} =a^{\tiny {\textcircled {\tiny {\#}}}}$ and $(aa^{\tiny {\textcircled {\tiny {\#}}}})^{*} =aa^{\tiny {\textcircled {\tiny {\#}}}}$. Thus by the definition of Moore–Penrose inverse, we have $a^{\dagger } =a^{\tiny {\textcircled {\tiny {\#}}}}$. Hence by Lemma 3.4, we have $a\in R^{\mathrm {EP}}$.

$(1)\Leftrightarrow (3)$: Suppose $a\in R^{\mathrm {EP}}$. Then by Lemma 3.4, we have $[a,a^{\tiny {\textcircled {\tiny {\#}}}}]=0$. By Lemma 3.5, we have $(a^{\tiny {\textcircled {\tiny {\#}}}})^{\tiny {\textcircled {\tiny {\#}}}} = a^2 a^{\tiny {\textcircled {\tiny {\#}}}}$. Thus $(a^{\tiny {\textcircled {\tiny {\#}}}})^{\tiny {\textcircled {\tiny {\#}}}} = a^2 a^{\tiny {\textcircled {\tiny {\#}}}} =a(aa^{\tiny {\textcircled {\tiny {\#}}}}) =a(a^{\tiny {\textcircled {\tiny {\#}}}}a)=a$. Conversely, suppose $(a^{\tiny {\textcircled {\tiny {\#}}}})^{\tiny {\textcircled {\tiny {\#}}}} = a$. By Lemma 3.5, we have $a =(a^{\tiny {\textcircled {\tiny {\#}}}})^{\tiny {\textcircled {\tiny {\#}}}} = a^2 a^{\tiny {\textcircled {\tiny {\#}}}}$. Thus

$$\begin{aligned} a^{\tiny {\textcircled {\tiny {\#}}}}a =a^{\tiny {\textcircled {\tiny {\#}}}} a^2 a^{\tiny {\textcircled {\tiny {\#}}}} =aa^{\tiny {\textcircled {\tiny {\#}}}}. \end{aligned}$$

Therefore $a\in R^{\mathrm {EP}}$ by Lemma 3.4.

$(3)\Leftrightarrow (4)\Leftrightarrow (5)$ is clear by Lemma 3.5.

$(1)\Rightarrow (6)$: By Lemmas 3.4 and 3.5, we have $(a^\dagger )^{\tiny {\textcircled {\tiny {\#}}}} = (a^{\tiny {\textcircled {\tiny {\#}}}}a)^{*}a =(a^{\dagger }a)^{*}a=a^{\dagger }a^{2}=a^\#a^2=a$.

$(6)\Rightarrow (7)$: Suppose that $a\in R^{\tiny \textcircled {\tiny \#}}$ and $(a^{\dagger })^{\tiny \textcircled {\tiny \#}}=a$. Then by Lemma  3.1, we have $a^{\dagger }a^{2}=a$ and $a(a^{\dagger })^{2}=a^{\dagger }$. Since $a^{\dagger } a$ is Hermitian, an appeal to Theorem 2.2 leads to $a \in R^\mathrm{EP}$. Hence $a^{\tiny {\textcircled {\tiny {\#}}}}=a^\#$. By Lemma 3.5, we have $(a^{\tiny {\textcircled {\tiny {\#}}}})^{\dagger }=a^2 a^{\tiny {\textcircled {\tiny {\#}}}}=a^2 a^\#=a$. Thus by $(a^{\dagger })^{\tiny {\textcircled {\tiny {\#}}}}=a$, we have $(a^{\dagger })^{\tiny {\textcircled {\tiny {\#}}}} =(a^{\tiny {\textcircled {\tiny {\#}}}})^{\dagger }$.

$(7)\Rightarrow (1)$: Suppose that $a\in R^{\tiny {\textcircled {\tiny {\#}}}}$ and $(a^{\dagger })^{\tiny {\textcircled {\tiny {\#}}}} =(a^{\tiny {\textcircled {\tiny {\#}}}})^{\dagger }$. By Lemma 3.5, we have $(a^{\tiny {\textcircled {\tiny {\#}}}})^{\dagger } = a^2 a^{\tiny {\textcircled {\tiny {\#}}}}$ and $(a^{\dagger })^{\tiny {\textcircled {\tiny {\#}}}} =(a^{\tiny {\textcircled {\tiny {\#}}}}a)^{*}a$. Thus by $(a^{\dagger })^{\tiny {\textcircled {\tiny {\#}}}} =(a^{\tiny {\textcircled {\tiny {\#}}}})^{\dagger }$, we have

$$\begin{aligned} a^{2} a^{\tiny {\textcircled {\tiny {\#}}}} =(a^{\tiny {\textcircled {\tiny {\#}}}} a)^{*}a. \end{aligned}$$

(3.3)

Taking involution on (3.3), we have $a^{*} a^{\tiny {\textcircled {\tiny {\#}}}}a =(aa^{\tiny {\textcircled {\tiny {\#}}}})^{*}a^{*} =aa^{\tiny {\textcircled {\tiny {\#}}}}a^{*}$. Thus by Lemma 3.1, we have $a^{\tiny {\textcircled {\tiny {\#}}}}a = a (a^{\tiny {\textcircled {\tiny {\#}}}})^2 a = (a a^{\tiny {\textcircled {\tiny {\#}}}})a^{\tiny {\textcircled {\tiny {\#}}}}a = (a a^{\tiny {\textcircled {\tiny {\#}}}})^* a^{\tiny {\textcircled {\tiny {\#}}}}a =(a^{\tiny {\textcircled {\tiny {\#}}}})^{*}a^{*} a^{\tiny {\textcircled {\tiny {\#}}}}a =(a^{\tiny {\textcircled {\tiny {\#}}}})^* a a^{\tiny {\textcircled {\tiny {\#}}}} a^* =(a^{\tiny {\textcircled {\tiny {\#}}}})^* (a a^{\tiny {\textcircled {\tiny {\#}}}})^* a^* = (a^2 (a^{\tiny {\textcircled {\tiny {\#}}}})^2)^* = (a a^{\tiny {\textcircled {\tiny {\#}}}})^* =aa^{\tiny {\textcircled {\tiny {\#}}}}$. That is, $[a,a^{\tiny {\textcircled {\tiny {\#}}}}]=0$, therefore $a\in R^{\mathrm {EP}}$ by Lemma 3.4.

$(8)\Leftrightarrow (1)$ follows by Theorems 3.2 and 3.3. $\square$

By the following example, we show that the relation $ap=0$ in (8) of Theorem 3.6 cannot be replaced by $pa=0$. We also show that we cannot consider other projections that the projection $p=1-aa^{\tiny {\textcircled {\tiny {\#}}}}$.

Example 3.7

Let R be the ring of all complex $2\times 2$ matrices. Let

$$\begin{aligned} A=\left[ \begin{matrix} 0 &{}\quad 1 \\ 0 &{}\quad 1 \end{matrix} \right] \quad \text {and} \quad P=\left[ \begin{matrix} 1 &{}\quad 0 \\ 0 &{}\quad 0 \end{matrix} \right] . \end{aligned}$$

It is simple to prove that

$$\begin{aligned} A^\#=\left[ \begin{matrix} 0 &{}\quad 1 \\ 0 &{}\quad 1 \end{matrix} \right] , \quad A^{\tiny {\textcircled {\tiny {\#}}}}=\left[ \begin{matrix} 1/2 &{}\quad 1/2 \\ 1/2 &{}\quad 1/2 \end{matrix}\right] , \quad AP=\left[ \begin{matrix} 0 &{}\quad 0 \\ 0 &{}\quad 0 \end{matrix} \right] , \end{aligned}$$

and $P^{2}=P=P^{*}$. But $A^\# \ne A^{\tiny {\textcircled {\tiny {\#}}}}$, therefore A is not EP. The projection given by Theorem 3.3 is $I-AA^{\tiny {\textcircled {\tiny {\#}}}} = \left[ {\begin{matrix} 1/2 &{} -1/2 \\ -1/2 &{} 1/2 \end{matrix}} \right]$. This projection satisfies $(I-AA^{\tiny {\textcircled {\tiny {\#}}}})A=0$ and $A(I-AA^{\tiny {\textcircled {\tiny {\#}}}}) \ne 0$.

In the following Theorem 3.9, we show that the equality $aR=a^{*}R$ in Lemma 3.8 can be replaced by the weaker inclusions $aR\subseteq a^{*}R$ or $a^{*}R\subseteq aR$.

Lemma 3.8

[12, Proposition 2] Let $a\in R$. Then the following are equivalent:

• (1) $a\in R^{\mathrm {EP}}$;

• (2) $a\in R^\#$ and $aR=a^{*}R$;

• (3) $a\in R^\#$ and $Ra=Ra^{*}$;

• (4) $a\in R^{\dagger }$ and $aR=a^{*}R$;

• (5) $a\in R^{\dagger }$ and $Ra=Ra^{*}$.

Theorem 3.9

Let $a\in R$. Then the following are equivalent:

• (1) $a\in R^{\mathrm {EP}}$;

• (2) $a\in R^\#$ and $aR\subseteq a^{*}R$;

• (3) $a\in R^\#$ and $Ra\subseteq Ra^{*}$;

• (4) $a\in R^\#$ and $a^{*}R\subseteq aR$;

• (5) $a\in R^\#$ and $Ra^{*}\subseteq Ra$.

Proof

$(1)\Rightarrow (2)$–(5) is obvious by Lemma 3.8.

$(2)\Rightarrow (1)$: By $aR\subseteq a^{*}R$, we have $a=a^{*}r$ for some $r\in R$, then $a=(aa^\#a)^{*}r=(a^\#a)^{*}a^{*}r=(a^\#a)^{*}a$. Thus $a^\#a=aa^\#=(a^\#a)^{*}aa^\#=(a^\#a)^{*}a^\#a$, which gives $(a^\#a)^{*}=a^\#a.$ Therefore $a\in R^\mathrm{EP}$ by the definition of EP element.

(3)–$(5)\Rightarrow (1)$ is similar to $(2)\Rightarrow (1)$. $\square$

Corollary 3.10

Let $a\in R$. Then the following are equivalent:

• (1) $a\in R^{\mathrm {EP}}$;

• (2) $a\in R^{\tiny {\textcircled {\tiny {\#}}}}$ and $aR\subseteq a^{*}R$;

• (3) $a\in R^{\tiny {\textcircled {\tiny {\#}}}}$ and $Ra\subseteq Ra^{*}$;

• (4) $a\in R^{\tiny {\textcircled {\tiny {\#}}}}$ and $a^{*}R\subseteq aR$;

• (5) $a\in R^{\tiny {\textcircled {\tiny {\#}}}}$ and $Ra^{*}\subseteq Ra$.

Proof

It is obvious by Lemma 3.5 and Theorem 3.9. $\square$

Theorem 3.11

Let $a\in R^{\tiny {\textcircled {\tiny {\#}}}}$. Then $a\in R^{\mathrm {EP}}$ if and only if $[a^{\tiny {\textcircled {\tiny {\#}}}}, (a^{\tiny {\textcircled {\tiny {\#}}}}a)^{*}a]=0$.

Proof

If a is EP, then $a\in R^{\tiny {\textcircled {\tiny {\#}}}}$ and $a^\# = a^{\dagger } = a^{\tiny {\textcircled {\tiny {\#}}}}$. Thus $[a^{\tiny {\textcircled {\tiny {\#}}}},(a^{\tiny {\textcircled {\tiny {\#}}}}a)^{*}a] =[a^\#,(a^{\dagger }a)^{*}a]=[a^\#,a^{\dagger }a^{2}]=[a^\#,a^\#a^{2}]=[a^\#,a]=0.$

Conversely, if $[a^{\tiny {\textcircled {\tiny {\#}}}}, (a^{\tiny {\textcircled {\tiny {\#}}}}a)^{*}a]=0$, then

$$\begin{aligned} a^{\tiny {\textcircled {\tiny {\#}}}}(a^{\tiny {\textcircled {\tiny {\#}}}}a)^{*}a = (a^{\tiny {\textcircled {\tiny {\#}}}}a)^{*}aa^{\tiny {\textcircled {\tiny {\#}}}}. \end{aligned}$$

(3.4)

Taking involution $*$ on (3.4), in view of Lemma 3.1, we get $a^* a^{\tiny {\textcircled {\tiny {\#}}}}a(a^{\tiny {\textcircled {\tiny {\#}}}})^* = a (a^{\tiny {\textcircled {\tiny {\#}}}})^2 a = a^{\tiny {\textcircled {\tiny {\#}}}} a$, Thus, $a^*a^{\tiny {\textcircled {\tiny {\#}}}} a(a^{\tiny {\textcircled {\tiny {\#}}}})^*a = a^{\tiny {\textcircled {\tiny {\#}}}}a^2=a$. Hence, $aR\subseteq a^{*}R$. Therefore, $a\in R^{\mathrm {EP}}$ by Corollary 3.10. $\square$

In [4, Theorem 16] and [6, Proposition 4.3], for an operator $T\in L(X)$, where X is a Banach space, the authors proved that for a Moore–Penrose invertible operator T, T is an EP operator if and only if there exists an invertible operator $P\in L(X)$ such that $T^{\dagger }=PT.$ Inspired by this result, we get the following theorem.

Theorem 3.12

Let $a\in R^{\tiny \textcircled {\tiny \#}}$. Then the following are equivalent:

• (1) $a\in R^{\mathrm {EP}};$

• (2) there exists a unit $u\in R$ such that $a^{\tiny {\textcircled {\tiny {\#}}}} = ua;$

• (3) there exists an element $b\in R$ such that $a^{\tiny {\textcircled {\tiny {\#}}}}=ba.$

Proof

$(1)\Rightarrow (2)$: If $a\in R^{\mathrm {EP}}$, then $a\in R^{\tiny {\textcircled {\tiny {\#}}}}$ and $a^{\tiny {\textcircled {\tiny {\#}}}}=a^\#$. Let $u=(a^\#)^{2}+1-aa^\#$. Since $u(a^{2}+1-aa^\#)=(a^{2}+1-aa^\#)u=1$, we get that u is a unit. Furthermore, we have $ua=((a^\#)^{2}+1-aa^\#)a=a^\#=a^{\tiny \textcircled {\tiny \#}}$.

$(2)\Rightarrow (3)$ is clear.

$(3)\Rightarrow (1)$: We know that $R a^{\tiny {\textcircled {\tiny {\#}}}} = Ra^*$ by the definition of the core inverse. From $a^{\tiny {\textcircled {\tiny {\#}}}} = ba$ we get $R a^{\tiny {\textcircled {\tiny {\#}}}} \subseteq Ra$. Thus $Ra^* = Ra^{\tiny \textcircled {\tiny \#}} \subseteq Ra$. Therefore, we deduce that $a\in R^{\mathrm {EP}}$ by Corollary 3.10. $\square$

By the previous result, we know that if $a\in R$ is an EP element, then the equation $a^{\tiny {\textcircled {\tiny {\#}}}} = xa$ has at least one solution. In fact, one solution is $x=(a^{\tiny {\textcircled {\tiny {\#}}}})^2+1-aa^{\tiny {\textcircled {\tiny {\#}}}}$, as one can see in the proof of (1) $\Rightarrow$ (2) of the previous result. We will describe the set of solutions in the next theorem.

Theorem 3.13

If $a \in R^\mathrm{EP}$, then $\{ x \in R: a^{\tiny {\textcircled {\tiny {\#}}}} = xa \} = \{ (a^{\tiny {\textcircled {\tiny {\#}}}})^2+w(1-aa^{\tiny {\textcircled {\tiny {\#}}}}): w \in R \}$.

Proof

Let $x\in R$ such that $a^{\tiny {\textcircled {\tiny {\#}}}}=xa$. Since $x = x-xaa^{\tiny {\textcircled {\tiny {\#}}}} +(a^{\tiny {\textcircled {\tiny {\#}}}})^2 =x(1-aa^{\tiny {\textcircled {\tiny {\#}}}})+(a^{\tiny {\textcircled {\tiny {\#}}}})^2$ we have proved the “$\subseteq$” inclusion of the statement of the theorem.

Let us prove the opposite inclusion. Since $a \in R^\mathrm{EP}$, by Theorem 3.12 there exists $b \in R$ such that $a^{\tiny {\textcircled {\tiny {\#}}}}=ba$. Now, $(a^{\tiny {\textcircled {\tiny {\#}}}})^2 a = ba a^{\tiny {\textcircled {\tiny {\#}}}} a=ba=a^{\tiny {\textcircled {\tiny {\#}}}}$. Finally, if w is any element of R, then $\left[ (a^{\tiny {\textcircled {\tiny {\#}}}})^2 +w (1-aa^{\tiny {\textcircled {\tiny {\#}}}})\right] a = a^{\tiny {\textcircled {\tiny {\#}}}}$. $\square$

4 When a Moore–Penrose Invertible Element is an EP Element

Since any EP element is Moore–Penrose invertible, it is natural to ask when a Moore–Penrose invertible element is an EP element. The concept of bi-EP was introduced by Hartwig and Spindelböck in [8] for complex matrices. They proved that for $A\in \mathbb {C}^{n\times n}$, if A is group invertible, then A is an EP matrix if and only if A is bi-EP. We give a generalization of this result in Theorem 4.3. In this section, we give the definition of n-EP, which is a generalization of bi-EP. We show that any n-EP element is an EP element whenever this element is group invertible.

Definition 4.1

[8] An element $a\in R$ is called bi-EP if $a\in R^{\dagger }$ and $[aa^{\dagger },a^{\dagger }a]=0.$

Definition 4.2

Let n be a positive integer. An element $a\in R$ is called n-EP if $a\in R^{\dagger }$ and $[a^na^{\dagger },a^{\dagger }a^n]=0.$

Note that 1-EP is coincide with bi-EP.

In [11, Theorem 2.1], Mosić and Djordjević proved that $a\in R^{\mathrm {EP}}$ if and only if $a\in R^\#\cap R^{\dagger }$ and $a^{n}a^{\dagger }=a^{\dagger }a^{n}$ for some $n\geqslant 1$. This result also can be found in [5, Theorem 2.4] by Chen. In the following theorem, we give a generalization of this result.

Theorem 4.3

Let $a\in R$ and n be a positive integer. Then $a\in R^{\mathrm {EP}}$ if and only if $a\in R^{\dagger }\cap R^\#$ and a is n-EP.

Proof

Suppose $a\in R^{\mathrm {EP}}$. Then $[a,a^{\dagger }]=0$, which gives $[a^{n}a^{\dagger },a^{\dagger }a^{n}]=0.$ That is, a is n-EP.

Conversely, suppose that $a\in R^{\dagger }\cap R^\#$ and a is n-EP. Then we have

$$\begin{aligned} a^{\dagger }a^{2n}a^{\dagger }=a^{n}(a^{\dagger })^{2}a^{n}. \end{aligned}$$

(4.1)

Pre-multiplication and post-multiplication of (4.1) by a respectively now yields $a^{2n}a^{\dagger }=a^{n+1}(a^{\dagger })^{2}a^{n},$ and $a^{\dagger }a^{2n}=a^{n}(a^{\dagger })^{2}a^{n+1}.$ Thus

$$\begin{aligned} a^{2n-1}a^{\dagger }= & {} a^\#a^{2n}a^{\dagger }=a^\#a^{n+1}(a^{\dagger })^{2}a^{n} =a^{n}(a^{\dagger })^{2}a^{n}. \end{aligned}$$

(4.2)

$$\begin{aligned} a^{\dagger }a^{2n-1}= & {} a^{\dagger }a^{2n}a^\#=a^{n}(a^{\dagger })^{2}a^{n+1}a^\# =a^{n}(a^{\dagger })^{2}a^{n}. \end{aligned}$$

(4.3)

By (4.2) and (4.3), we have $a^{2n-1}a^{\dagger }=a^{\dagger }a^{2n-1}.$ Hence by [11, Theorem 2.1], we have $a\in R^{\mathrm {EP}}$. $\square$

Theorem 4.4

Let $a\in R^{\dagger }$. Then the following are equivalent:

• (1) $a\in R^{\mathrm {EP}}$;

• (2) there exists a unit $u\in R$ such that $a^{\dagger }=ua$;

• (3) there exists a left invertible element $v\in R$ such that $a^{\dagger }=va$.

Proof

$(1)\Rightarrow (2)$: If $a\in R^{\mathrm {EP}}$, then $a\in R^{\dagger }$ and $a^{\dagger }=a^\#.$ Let $u=(a^\#)^{2}+1-aa^\#$. Since $u(a^{2}+1-aa^\#)=(a^{2}+1-aa^\#)u=1$ we get that u is a unit. Furthermore, $ua=((a^\#)^{2}+1-aa^\#)a=a^\#=a^{\dagger }$.

$(2)\Rightarrow (3)$ is clear.

$(3)\Rightarrow (1)$ Suppose that there exists a left invertible element $v\in R$ such that $a^{\dagger }=va$. Then $1=tv$ for some $t\in R$ and $ta^{\dagger }=tva=a.$ Thus $Ra^{\dagger }\subseteq Ra$ and $Ra\subseteq Ra^{\dagger }$. Since $Ra^{\dagger }=Ra^{*}$, we deduce that $Ra^{*} = Ra$, that is a is an EP element. $\square$

Remark 4.5

In Theorem 3.12, we proved that for a core invertible element $a\in R$, $a\in R^{\mathrm {EP}}$ if and only if there exists an element $b\in R$ such that $a^{\tiny \textcircled {\tiny \#}}=ba$. The following example shows that this affirmation can not be obtained for a Moore–Penrose invertible element. In Theorem 3.9, we proved that for a group invertible element $a\in R$, $a\in R^{\mathrm {EP}}$ if and only if $aR\subseteq a^{*}R$. The following example also shows that this affirmation can not be obtained for a Moore–Penrose invertible element. Observe that if $a \in R^{\dagger } \cap R^\#$, then $a \in R^\mathrm{EP}$ if and only if there exists $b \in R$ such that $a^{\dagger } = ba$, which follows from Theorem 3.9.

Recall that an infinite matrix M is said to be bi-finite if it is both row-finite and column-finite.

Example 4.6

Let R be the ring of all bi-finite real matrices with transpose as involution and let $e_{i,j}$ be the matrix in R with 1 in the (i, j) position and 0 elsewhere. Let $A=\sum \nolimits _{i=1}^{\infty }e_{i+1,i}$ and $B=A^{*}$, now $AB=\sum \nolimits _{i=2}^{\infty }e_{i,i}$, $BA=I$. So $A^{\dagger }=B$ and $A^{\dagger }=A^{\dagger }BA=B^{2}A$. It is easy to check that $B^{2}$ is not left invertible and A is not EP (since $AB\ne BA$). In addition, A is not group invertible (if $A \in R^\#$, then $AA^\#=A^\#A=BAA^\#A=BA=I$; thus, A is invertible, which is not possible). This example also shows that the equality $aR=a^{*}R$ in the equivalence $(a \in R^\mathrm{EP} \Leftrightarrow a \in R^{\dagger }, aR=a^{*}R)$ cannot be replaced by the inclusions $aR\subseteq a^{*}R$ or $a^{*}R\subseteq aR$.

Proposition 4.7

Let $a\in R^{\dagger }$. Then the following are equivalent:

• (1) $a\in R^{\mathrm {EP}}$;

• (2) $[a^{\dagger }a,a]=[a^{\dagger },aa^{\dagger }]=0$;

• (3) $[a^{\dagger }a,a]=[a,aa^{\dagger }]=0$;

• (4) $[a^{\dagger }a,a^{\dagger }]=[a^{\dagger },aa^{\dagger }]=0$;

• (5) $[a^{\dagger }a,a^{\dagger }]=[a,aa^{\dagger }]=0$.

Proof

$(1)\Rightarrow (2)$–(3): If $a\in R^{\mathrm {EP}}$, then $aa^{\dagger }=a^{\dagger }a$. Thus, (2) and (3) are obvious.

$(2)\Rightarrow (1)$: Observe that $[a^{\dagger }a,a]=0$ implies that $a=a^{\dagger }a^{2}\in a^{*}R$ and $[a^{\dagger },aa^{\dagger }]=0$ implies that $a^{\dagger }=a(a^{\dagger })^{2}\in aR$, that is $a^{*}R\subseteq aR$ since $a^{\dagger }R=a^{*}R.$ Thus, $aR=a^{*}R$, i.e., a is EP.

$(3)\Rightarrow (1)$: Observe that $[a^{\dagger }a,a]=0$ implies that $a=a^{\dagger }a^{2}\in a^{*}R$ and $[a,aa^{\dagger }]=0$ implies that $a=a^{2}a^{\dagger }\in Ra^{\dagger }$, that is $Ra\subseteq Ra^*$ since $Ra^{\dagger }=Ra^{*}$, and therefore, $aR\subseteq a^{*}R$. Thus, $aR=a^{*}R$, i.e., a is EP.

The equivalence between $(1) \Leftrightarrow (4) \Leftrightarrow (5)$ is similar to the proof of the equivalence between $(1) \Leftrightarrow (2) \Leftrightarrow (3)$. $\square$

Example 4.8

The condition $[a^{\dagger }a,a^{\dagger }]=0$ in Proposition 4.7 does not imply that a is an EP element in general. Let R, A and B be the same as Example 4.6, then $AB=\sum \nolimits _{i=2}^{\infty }e_{i,i}$, $BA=I$. So $A^{\dagger }=B$ and $[A^{\dagger }A,A^{\dagger }]=0$. But A is not EP since $AB\ne BA.$

Theorem 4.9

Let $a\in R^{\dagger }$. Then the following are equivalent:

• (1) $a\in R^{\mathrm {EP}}$;

• (2) $aR=a^{2}R$ and $[a^{\dagger }a,a^{\dagger }]=0$;

• (3) $aR=a^{2}R$ and $[a^{\dagger }a,a]=0$;

• (4) $aR=a^{2}R$ and $aR\subseteq a^{\dagger }R$;

• (5) $aR=a^{2}R$ and $aR\subseteq a^{*}R$.

Proof

$(1)\Rightarrow (2)$–(5): For a Moore–Penrose invertible element $a\in R$, we have $a\in R^\mathrm{EP}$ if and only if $aa^{\dagger }=a^{\dagger }a$. Hence (2)–(5) hold.

$(2)\Rightarrow (1)$: Observe that $[a^{\dagger }a,a^{\dagger }]=0$ implies $a^{\dagger }=(a^{\dagger })^{2}a$. Thus $Ra^{*}\subseteq Ra$ (since $Ra^{\dagger }=Ra^{*}$). That is $a^{*}=ra$ for some $r\in R.$ We deduce that $a^{*}=ra=raa^{\dagger }a=a^{*} a^{\dagger }a$, applying involution on that last equality we obtain $a=a^{\dagger }a^{2}$. Therefore $a\in R^{\mathrm {EP}}$ by Lemma 2.4 and Theorem 2.13.

$(3)\Rightarrow (1)$: It is clear that $[a^{\dagger }a,a]=0$ implies $a=a^{\dagger }a^{2}$. Thus $a\in R^{\mathrm {EP}}$ by Lemma 2.4 and Theorem 2.13.

$(5)\Rightarrow (1)$: As $aR\subseteq a^{*}R$ is equivalent to $Ra^{*}\subseteq Ra$, we get $a\in R^{\mathrm {EP}}$ by the proof of $(2)\Rightarrow (1)$.

$(4)\Leftrightarrow (5)$: It is clear by $a^{*}R=a^{\dagger }R$. $\square$

Similarly, we have the following theorem.

Theorem 4.10

Let $a\in R^{\dagger }$. Then the following are equivalent:

• (1) $a\in R^{\mathrm {EP}}$;

• (2) $Ra=Ra^{2}$ and $[aa^{\dagger },a^{\dagger }]=0$;

• (3) $Ra=Ra^{2}$ and $[aa^{\dagger },a]=0$;

• (4) $Ra=Ra^{2}$ and $Ra\subseteq Ra^{\dagger }$;

• (5) $Ra=Ra^{2}$ and $Ra\subseteq Ra^{*}$.