Throughout this paper, R denotes an associative ring with 1. We write E(R) and J(R) to denote the set of all idempotents and the Jacobson radical of R, respectively.

An element a∈R is said to be group invertible if there exists a#∈R such that

aa#a=a, a#aa#=a#, aa#=a#a.

The element a# is called the group inverse of a, which is uniquely determined by the above equations [1, 9].

An involution *:a⟼a* in a ring R is an anti-isomorphism of degree 2, that is,

(a*)*=a, (a+b)*=a*+b*, (ab)*=b*a*.

An element a in R is called normal if aa*=a*a.

An element a† in R is called the Moore-Penrose inverse (MP-inverse) of a [6, 10], if

aa†a=a, a†aa†=a†, (aa†)*=aa†, (a†a)*=a†a.

If such a† exists, then it is unique [6]. Denote by R# and R† the set of group invertible elements of R and the set of all MP-invertible elements of R, respectively.

An element a is said to be EP if a∈R#∩R† and satisfies a#=a† [3, 6]. We denote by REP the set of all EP elements of R. Note that if a∈R† is normal, then a∈REP, see [6]. An element a∈R is called normal EP if a is normal and a∈R†. Denote by RNEP the set of all normal EP elements of R.

An element a is called a partial isometry if a†=a* and a is called a strongly EP element if a∈REP is a partial isometry. We denote the sets of all partial isometry elements and strongly EP elements of R by RPI and RSEP, respectively.

In [7], D. Mosić and D. S. Djordjević presented some characterizations of EP elements in rings with involution. In addition, some equivalent conditions for the element a in a ring with involution to be a partial isometry are given. Recent researches on EP elements in rings with involution have produced some interesting results, see [4, 9, 12]. The necessary and sufficient conditions for the existence of a common solution and the general common solution of the equation axb=c (a,b are regular elements) were given for rings with involution in [2]. In [13, 15], a new kind of characterizations of generalized inverse elements has been studied by means of the solution of constructed equations recently.

Motivated by these articles above, this paper is intended to provide, by using certain equations admitting solutions in a definite set, further sufficient and necessary conditions for an element in a ring with involution to be an EP element, partial isometry, normal EP element, and strongly EP element. This is a new way to study generalized inverses in rings.

Lemma 2.1.[5, 7] Let a∈R#∩R†. If a*aa#(1-aa†)=0, then a∈REP.

Proof. Pre-multiplying the equality a*aa#(1-aa†)=0 by (a†)*, we have aa#(1-aa†)=0. That is aa#=aa†. Hence a∈REP by [7, Theorem 1.6] or [5].

Lemma 2.2. Let a∈R#∩R†. Then a∈RPI if and only if a*a†=a†a†.

Proof. ⇒ The equality obviously holds since a*=a†.

⇐ Post-multiplying a*a†=a†a† by a, one has a*a†a=a†a†a. Applying the involution to the last equality, we have a†a2=a†a(a†)*, it follows that a2=a(a†)*. Post-multiplying the equality by a*, we get a2a*=a2a†. Pre-multiplying a2a*=a2a† by a#, one has aa*=aa†. Thus a∈RPI by [7, Theorem 2.1].

In order to prove the theorems given in this paper more clearly, we briefly review the following existing conclusions:

Lemma 2.3.[11, Lemma 2.2] Let a∈R#. Then (a#)*R=a*R and R(a#)*=Ra*.

Lemma 2.4.[11, Lemma 2.3] Let a∈R†. Then

(1) aR=aa†R=aa*R and Ra=Ra†a=Ra*a.

(2) a*R=a†R=a*aR=a†aR and Ra*=Ra†=Raa*=Raa†.}

Lemma 2.5.[12, Theorem 3.9] Let a∈R. Then the following are equivalent:

(1) a∈REP;

(2) a∈R# and aR⊆a*R;

(3) a∈R# and Ra⊆Ra*;

(4) a∈R# and a*R⊆aR;

(5) a∈R# and Ra*⊆Ra.}

Lemma 2.6.[14, Lemma 2.1] Let a∈R#∩R†. Then the following conditions are satisfied:

(1) a*R=a*a2R=a*aa#R=(a#)*R;

(2) Ra=Ra#=Raa*a#=Ra*a=Ra*a*a=Ra†a*a;

(3) (a#)*aa†R=(a#)*a#a†R=(a#)*a#a*R;

(4) a#R=aR and Ra*=Ra†.}

In [14, Theorem 2.4], the authors proved that an element a∈R#∩R† can be an EP element if and only if the equation axa#+axa*=xaa†+a*ax has at least one solution in the set χa={a,a#,a†,a*,(a#)*,(a†)*}.

Recall that an element a is said to be EP if a∈R#∩R† and satisfies a#=a†. Thus, we can modify the above existing theorem in [14] and construct the following equation, with the help of which we can explore a new kind of characterization of EP elements:

(2.1)axa†+axa*=xaa#+a*ax.

Theorem 2.7. Let a∈R#∩R†. Then a∈REP if and only if equation (2.1) has at least one solution in χa={a,a#,a†,a*,(a#)*,(a†)*}.

Proof. ⇒ Obviously, x=a† is a solution because a†=a#.

⇐ (1) If x=a is a solution, then a2a†+a2a*=a2a#+a*a2=a+a*a2. By Lemma 2.6, we have

a*R=a*a2R=(a2a†+a2a*-a)R⊆aR.

Therefore a∈REP by Lemma 2.5.

(2) If x=a# is a solution, then aa#a†+aa#a*=a#aa#+a*aa#=a#+a*aa#. By Lemma 2.6, we obtain that

a*R=a*aa#R=(aa#a†+aa#a*-a#)R⊆aR.

The fact that a∈REP follows from Lemma 2.5.

(3) If x=a† is a solution, then aa†a†+aa†a*=a†aa#+a*aa†=a†aa#+a*. Pre-multiplying it by aa#, we obtain aa†a†+aa†a*=a#+a#aa*. By Lemma 2.6, we have

Ra#=R(aa†a†+aa†a*-a#aa*)⊆Ra†+Ra*=Ra†.

Since Ra#=Ra,Ra†=Ra* by Lemma 2.6, we get Ra⊆Ra*. From Lemma 2.5, a∈REP.

(4) If x=a* is a solution, then aa*a†+aa*a*=a*aa#+a*aa*. Post-multiplying it by (1-aa†), we have a*aa#(1-aa†)=0. By Lemma 2.1, we get a∈REP.

(5) If x=(a#)* is a solution, then a(a#)*a†+a(a#)*a*=(a#)*aa#+a*a(a#)*. Post-multiplying it by (1-aa†), we get (a#)*aa#(1-aa†)=0. Pre-multiplying the last equation by (a2)*, we get a*aa#(1-aa†)=0. Therefore a∈REP by Lemma 2.1.

(6) If x=(a†)* is a solution, then a(a†)*a†+a(a†)*a*=(a†)*aa#+a*a(a†)*. Taking involution of the above equality, we obtain that

(a†)*a†a*+aa†a*=(a#)*a*a†+a†a*a.

Lemma 2.6 now leads to

Ra=Ra†a*a=R((a†)*a†a*+aa†a*-(a#)*a*a†)⊆Ra*+Ra†=Ra†=Ra*.

From Lemma 2.5, a∈REP.

Multipying the equation (2.1) on the right by a, we obtain:

(2.2)axa†a+axa*a=ax+a*axa.

Theorem 2.8. Let a∈R#∩R†. Then a∈REP if and only if the equation (2.2) has at least one solution in χa.

Proof. ⇒Obviously, x=a† is a solution.

⇐ (1) If x=a is a solution, then a2a†a+a2a*a=a2+a*a3. It is immediate that a2a*a=a*a3. By Lemma 2.4, we obtain that

a*R=a*a3R=a2a*aR⊆aR.

Hence a∈REP by Lemma 2.5.

(2) If x=a# is a solution, then aa#a†a+aa#a*a=aa#+a*aa#a. That is aa#a*a=a*a. Post-multiplying it by a, we obtain that aa#a*a2=a*a2.

By Lemma 2.6, we get

a*R=a*a2R=aa#a*a2R⊆aR.

Therefore, a∈REP by Lemma 2.5.

(3) If x=a† is a solution, then aa†a†a+aa†a*a=aa†+a*aa†a=aa†+a*a. Post-multiplying it by a, we have aa†a†a2+aa†a*a2=a+a*a2. We thus get

a*R=a*a2R=(aa†a†a2+aa†a*a2-a)R⊆aR

by Lemma 2.6.

And then it follows from Lemma 2.5 that a∈REP.

(4) If x=a* is a solution, then aa*a†a+aa*a*a=aa*+a*aa*a. We conclude from Lemma 2.4 that

Ra*=Raa*=R(aa*a†a+aa*a*a-a*aa*a)⊆Ra.

Hence a∈REP by Lemma 2.5.

(5) If x=(a#)* is a solution, then a(a#)*a†a+a(a#)*a*a=a(a#)*+a*a(a#)*a. Pre-multiplying it by a†, we get (a#)*a†a+(a#)*a*a=(a#)*+a†a*a(a#)*a. Then from Lemma 2.3, we obtain that

Ra*=R(a#)*=R((a#)*a†a+(a#)*a*a-a†a*a(a#)*a)⊆Ra,

which yields a∈REP by Lemma 2.5.

(6) If x=(a†)* is a solution, then a(a†)*a†a+a(a†)*a*a=a(a†)*+a*a(a†)*a. That is a2=a*a(a†)*a. Post-multiplying it by a#, we obtain that a=a*a(a†)*aa#. Then

aR=a*a(a†)*aa#R⊆a*R.

Therefore, a∈REP by Lemma 2.5.

Further, we revised the equation (2.2) as follows:

(2.3)axa†a+xaa*a=ax+a*axa.

Theorem 2.9. Let a∈R#∩R†. Then a∈REP if and only if the equation (2.3) has at least one solution in χa.

Proof. ⇒ x=a† is a solution since aa†=a†a.

⇐ (1) If x=a is a solution, then a2a†a+a2a*a=a2+a*a3. It is immediate from the proof of Theorem 2.8(1) that a∈REP.

(2) If x=a# is a solution, then aa#a†a+a#aa*a=aa#+a*aa#a. Then a∈REP by the proof of Theorem 2.8(2) since aa#=a#a.

(3) If x=a† is a solution, then aa†a†a+a†aa*a=aa†+a*aa†a=aa†+a*a. That is aa†a†a=aa†. Applying the involution, one has aa†=a†a2a†. By Lemma 2.4 and Lemma 2.5, we have a∈REP.

(4) If x=a* is a solution, then aa*a†a+a*aa*a=aa*+a*aa*a. That is aa*a†a=aa*. From Lemma 2.4, we obtain that

Ra*=Raa*=Raa*a†a⊆Ra.

By Lemma 2.5, a∈REP.

(5) If x=(a#)* is a solution, then a(a#)*a†a+(a#)*aa*a=a(a#)*+a*a(a#)*a. Pre-multiplying it by a†, we have (a#)*a†a+a†(a#)*aa*a=(a#)*+a†a*a(a#)*a. By Lemma 2.3, we have

Ra*=R(a#)*=R((a#)*a†a+a†(a#)*aa*a-a†a*a(a#)*a)⊆Ra,

which gives a∈REP by Lemma 2.5.

(6) If x=(a†)* is a solution, then a(a†)*a†a+(a†)*aa*a=a(a†)*+a*a(a†)*a. That is (a†)*aa*a=a*a(a†)*a.

Pre-multiplying it by (1-aa†), we have

(1-aa†)a*a(a†)*a=0.

Hence 0=(1-aa†)a*a(a†aa†)*a=(1-aa†)a*a(a†)*a†a2.

Multiplying the last equality by a# on the right, we obtain that 0=(1-aa†)a*a(a†)*a†a=(1-aa†)a*a(a†)*. Post-multiply it by a* and then we have

(1-aa†)a*a2a†=0.

Post-multiplying it by aa#a†, we get (1-aa†)a*=0, which implies a=a2a†. Consequently, a∈REP.

Theorem 2.10. Let a∈R#∩R†. Then a∈REP if and only if the equality a†xa=a* has at least one solution.

Proof. ⇒ Since a∈REP, we have aa†=a†a. Hence x=aa*a† is a solution of the equation a†xa=a*.

⇐ Assume that a†xa=a* have a solution x0. Then Ra*=Ra†x0a⊆Ra, it follows that a∈REP by Lemma 2.5.

Theorem 2.11. Let a∈R#∩R†. Then a∈REP if and only if the equation a†xa=aa#-aa† has at least one solution.

Proof. ⇒ Since a∈REP, we have aa#-aa†=0. Then x=aa#-aa† is a solution of the equation a†xa=aa#-aa†.

⇐ Assume that a†xa=aa#-aa† has a solution. Then, by [2, Theorem 2.1], we get

(2.4)aa#−aa†=a†a(aa#−aa†)a†a.

Pre-multiplying (2.4) by a, we obtain that

a-a2a†=a(aa#-aa†)a†a=a-a2a†a†a.

That is a2a†=a2a†a†a.

On the other hand, post-multiply (2.4) by a, we have

(2.5)a†a(aa#−aa†)a†a2=0.

Pre-multiplying (2.5) by a and then post-multiplying the last equation by a#, we obtain that

a(aa#-aa†)a†a=0.

That is a=a2a†a†a. Obviously, we can deduce that a=a2a†. Consequently, a∈REP.

Let a∈R. Write a0={x∈R|ax=0}. Clearly, a0 is a right ideal of R, which is called the right annihilator of a. Similarly, we can define  0a. Then, we have the following theorem.

Theorem 2.12. Let a∈R#∩R†. Then a∈REP if and only if a0=(a†)0.

Proof. ⇒ Since a∈REP, we have a#=a†. Therefore (a#)o=(a†)0. Note that a0=(a#)0. Then a0=(a†)0.

⇐ Assume that a0=(a†)0. Note that 1-a†a∈(a#)0. Then 1-a†a∈(a†)0, which implies a†=a†a†a. Hence Ra†⊆Ra, one obtains a∈REP by Lemma 2.4 and Lemma 2.5.

Recall that an element c∈R is semi-idempotent if c-c2∈J(R). Using the semi-idempotent elements of R, we have the following theorem.

Theorem 3.1. Let a∈R†. Then a∈RPI if and only if the following two conditions hold:

(1) aa* is a semi-idempotent;

(2) a†-a*∈R#.}

Proof. The equality a*=a† implies that

aa*-aa*aa*=0∈J(R) and a†-a*=0∈E(R)⊆R#.

On the contrary, assume that aa* is semi-idempotent and a†-a*∈R#. Take aa*-aa*aa*=x∈J(R). Then, by the proof of [14, Theorem 2.6], one has a†-a*=a†(a†)*a†x∈J(R). Set z=(a†-a*)# because a†-a*∈R#. Then a†-a*=(a†-a*)z(a†-a*). Thus, we get

(a†-a*)(1-z(a†-a*))=(a†-a*)-(a†-a*)z(a†-a*)=0.

Since z(a†-a*)∈J(R), we obtain that 1-z(a†-a*) is invertible. Hence a†-a*=0, so a∈RPI.

Theorem 3.2. Let a∈R†. Then the following conditions are equivalent:

(1) a†(a†)*∈E(R);

(2) (a†)*a†∈E(R);

(3) a∈RPI;

(4) a†(a†)* is a semi-idempotent and a*-a†∈E(R);

(5) a†a-a†(a†)*∈E(R);

(6) aa†-(a†)*a†∈E(R);

(7) (a†)*a† is a semi-idempotent and a*-a†∈E(R).

Proof. (1)⇒(2) From the assumption, we know that a†(a†)*=a†(a†)*a†(a†)*. Pre-multiplying it by a and then post-multiplying the last equality by a†, we get (a†)*a†=(a†)*a†(a†)*a†.

(2)⇒(3) From (2), we obtain that (a†)*a†=(a†)*a†(a†)*a†. Post-multiplying it by aa*, we get aa†=(a†)*a†. Multiply the equality by a on the right and then we get a=(a†)*. Applying involution to a=(a†)*, we get a*=a†. Consequently, a∈RPI.

(3)⇒(4) Since a∈RPI, a*=a†. Then we know that a*-a†=0∈E(R) and a†(a†)*-a†(a†)*a†(a†)*=0∈J(R). It is immediate that a†(a†)* is a semi-idempotent and a*-a†∈E(R).

(4)⇒(5) Write x=a†a-a†(a†)*. Then x-x2=a†(a†)*-a†(a†)*a†(a†)*∈J(R) by hypothesis. Clearly, a(x-x2)a*a=a-(a†)*. Note that a*-a†∈E(R). Then a-(a†)*∈E(R), this gives a(x-x2)a*a∈J(R)∩E(R), so a(x-x2)a*a=0. It follows that a=(a†)*. Hence a†a-a†(a†)*∈E(R).

(5)⇒(6) From (5), we know that a†(a†)*=a†(a†)*a†(a†)*. Pre-multiplying it by a and then multiplying the last equality by a† on the right, we obtain that

(a†)*a†=(a†)*a†(a†)*a†.

Hence aa†-(a†)*a†-(aa†-(a†)*a†)(aa†-(a†)*a†)=(a†)*a†-(a†)*a†(a†)*a†=0. Consequently, aa†-(a†)*a†∈E(R).

(6)⇒(7) From (6), we obtain that (a†)*a†=(a†)*a†(a†)*a†. So (a†)*a† is an idempotent. By (2)⇒(3), we get a*=a†, which gives a*-a†=0∈E(R).

(7)⇒(1) Similar to (4)⇒(5), we obtain that a*=a†. Then

a†(a†)*a†(a†)*=a†aa†(a†)*=a†(a†)*.

Therefore, a†(a†)*∈E(R).

Lemma 3.3. Let a∈R†. Then a∈RPI if and only if a†=a†(a†)*a†.

Proof. ⇒ Since a∈RPI, a*=a†. And then we can easily get

a†(a†)*a†=a†aa†=a†.

⇐ From the assumption, we know that a†=a†(a†)*a†. Pre-multiplying it by a and then post-multiplying the last equality by a, we get a=(a†)*. Taking involution of the above equality, we obtain a*=a†. So a∈RPI.

Lemma 3.4. Let a∈R#∩R†. Then a∈RPI if and only if a2=a(a†)*.

Proof. ⇒ We know that a*=a† according to the assumption. Then we can get the following equality.

a(a†)*=a(a*)*=a2.

⇐ From the assumption, we know that a2=a(a†)*. Post-multiplying it by a*, we have a2a*=a2a†. Hence a∈RPI by [8, Theorem 2.1].

Lemma 3.5. Let a∈R#∩R†. Then a∈RPI if and only if the following equation has at least one solution in χa:

(2.1)x=x(a†)*a†.

Proof. ⇒ Obviously, a* is a solution.

⇐ (1) If x=a is a solution, then a=a(a†)*a†. Post-multiplying it by a, we have a2=a(a†)*. By Lemma 3.4, a∈RPI.

(2) If x=a# is a solution, then a#=a#(a†)*a† is a solution. Pre-multiplying it by a2, we get a=a(a†)*a†. By the proof of (1), we know a∈RPI.

(3) If x=a† is a solution, then a†=a†(a†)*a†. Hence a∈RPI by Lemma 3.3.

(4) If x=a* is a solution, then a*=a*(a†)*a†=a†. It is immediate that a∈RPI.

(5) If x=(a#)* is a solution, then (a#)*=(a#)*(a†)*a†. Post-multiplying it by a, we get (a#)*a=(a#)*(a†)*. Applying involution to the above equality, we have a*a#=a†a#. Then we deduce that a∈RPI by [8, Theorem 2.1].

(6) If x=(a†)* is a solution, then (a†)*=(a†)*(a†)*a†. Firstly, multiply the equality on the right by a, apply involution to the latest equation, and then we get a*a†=a†a†. By Lemma 2.2, a∈RPI.

Similarly, we have the following theorem.

Theorem 3.6. Let a∈R#∩R†. Then a∈RPI if and only if the following equation has at least one solution in χa:

(2.2)x=(a†)*a†x.

Using the symmetricity, we have the following corollary.

Corollary 3.7. Let a∈R#∩R†. Then a∈RPI if and only if the following equation has at least one solution in χa:

(2.3)x=xa†(a†)*.