Huihui ZHU Qing-Wen WANG

Abstract In this paper,the authors derive the existence criteria and the formulae of the weighted Moore-Penrose inverse,the e-core inverse and the f-dual core inverse in rings.Also,new characterizations between weighted Moore-Penrose inverses and one-sided inverses along an element are given.

Keywords Weighted Moore-Penrose inverses,One-sided inverses along an element,Inverses along an element,e-Core inverses,f-Dual core inverses

1 Introduction

Suppose that R is a unital*-ring,that is a ring with unity 1 and an involution aa*satisfying(a*)*=a,(ab)*=b*a*and(a+b)*=a*+b*for all a,b∈R.

Throughout this paper,we assume that R is a unital*-ring.Recall that an element a∈R is called(von Neumann)regular if there exists some x∈R such that a=axa.Such an x is called an inner inverse or{1}-inverse of a,and is denoted by a−.An element x∈R is called Hermitian if x=x*.In what follows,let e,f∈R be invertible Hermitian elements.

We say that a∈R has a weighted Moore-Penrose inverse with weights e,f if there exists x∈R such that

In this paper,we mainly investigate weighted Moore-Penrose inverses,e-core inverses,f-dual core inverses and one-sided inverses along an element in rings.The paper is organized as follows.In Section 2,several characterizations and expressions for{e,1,3}-inverses and{f,1,4}-inverses of elements are derived.Also,the existence criterion of the weighted Moore-Penrose inverse is given.Moreover,it is proved that a∈R is weighted Moore-Penrose invertible if and only if it is both{e,1,3}-invertible and{f,1,4}-invertible.In Section 3,we present the existence criterion of both e-core invertible and f-dual core invertible elements.In Section 4,it is shown that a∈R is weighted Moore-Penrose invertible if and only if a∈R is left invertible along f−1a*e if and only if a∈R is right invertible along f−1a*e,extending[1,Theorem 3.2].Also,it is proved that a∈R is weighted Moore-Penrose invertible if and only if f−1a*e is left invertible along a if and only if f−1a*e is right invertible along a.Under the assumptionwe further prove that a∈R is e-core invertible if and only if it is invertible along af−1a*e,and a is f-dual core invertible if and only if it is invertible along f−1a*ea.

2 Characterizations for Weighted Moore-Penrose Inverses

We begin this section with several characterizations for{e,1,3}-inverses and{f,1,4}-inverses of an element in a ring.

Proposition 2.1Let a∈R and let e∈R be an invertible Hermitian element.Then a is{e,1,3}-invertible if and only if a∈Ra*ea.Moreover,if a=xa*ea for some x∈R,then x*e is an{e,1,3}-inverse of a.

ProofSuppose that a is{e,1,3}-invertible.Then we have a=*a*ea∈Ra*ea.

Conversely,if a∈Ra*ea,then a=xa*ea for some x∈R,and hence ax*=xa*eax*=xa*e(xa*)*.So,ax*is Hermitian.

It follows ax*ea=(ax*)*ea=xa*ea=a and(eax*e)*=exa*e=e(ax*)*e=eax*e that x*e is an{e,1,3}-inverse of a.

Proposition 2.2Let a∈R and let f∈R be an invertible Hermitian element.Then a is{f,1,4}-invertible if and only if a∈af−1a*R.Moreover,if a=af−1a*y for some y∈R,then f−1y*is an{f,1,4}-inverse of a.

It is well known that a∈R†if and only if a∈aa*R∩Ra*a.Motivated by this,we derive the characterization of the weighted Moore-Penrose inverse.

Theorem 2.1Let a∈R and let e,f∈R be invertible Hermitian elements.Thenif and only if a∈af−1a*R∩Ra*ea.Moreover,if a=xa*ea=af−1a*y for some x,y∈R,then

ProofApplying Propositions 2.1–2.2,it is obvious thatimplies a∈af−1a*R∩Ra*ea.

Suppose that a=xa*ea=af−1a*y for some x,y∈R.We next show that z=f−1y*ax*e is the weighted Moore-Penrose inverse of a.

Note that f−1y*and x*e are inner inverses of a.Then af−1y*a=a=ax*ea,and consequently aza=af−1y*ax*ea=a and zaz=z.

Also,eaz=eaf−1y*ax*e=eax*e=eaawhich implies eaz=(eaz)*.

Analogously,fza=ff−1y*ax*ea=y*ax*ea=y*a.As y*a=y*af−1a*y,we get fza=(fza)*.

We next characterize the weighted Moore-Penrose inverse by ideals.Herein,a lemma is given.

Lemma 2.1Let a∈R and let e,f∈R be invertible Hermitian elements.We have

(i)If a=af−1a*eax for some x∈R,then f−1(eax)*is both an{e,1,3}-inverse and an{f,1,4}-inverse of a.

(ii)If a=yaf−1a*ea for some y∈R,then(yaf−1)*e is both an{e,1,3}-inverse and an{f,1,4}-inverse of a.

Proof(i)By Proposition 2.2,we know that f−1(eax)*is an{f,1,4}-inverse of a.To show that f−1(eax)*is also an{e,1,3}-inverse of a,it is sufficient to prove that eaf−1(eax)*is Hermitian.

By calculations,we have

Hence,f−1(eax)*is an{e,1,3}-inverse of a.

(ii)It can be proved similarly.

Theorem 2.2Let a∈R and let e,f∈R be invertible Hermitian elements.Then the following conditions are equivalent:

Hence,a∈af−1a*eaR.

(ii)⇔(iii)Assume that a∈af−1a*eaR.Then there exists x∈R such that a=af−1a*eax,and hence a*=x*a*eaf−1a*.Also,we have(eax)*a=(eax)*af−1a*eax,which implies that(eax)*a is Hermitian.

We obtain

Thus,a∈Raf−1a*ea.

Conversely,if a∈Raf−1a*ea,then we can similarly obtain a∈af−1a*eaR.

(iii)⇒(i)As a∈Raf−1a*ea,and consequently a∈af−1a*eaR,then a∈af−1a*R∩Ra*ea.It follows from Theorem 2.1 that a∈

By Lemma 2.1,we get that f−1(eax)*is both an{e,1,3}-inverse and an{f,1,4}-inverse of a.

Applying Theorem 2.1,we have

Set e=f=1 in Theorem 2.2,then we get the characterization for the Moore-Penrose inverse.

Corollary 2.1(see[9,Theorem 2.16]Let a∈R.Then the following conditions are equivalent:

(i)a∈R†;

(ii)a∈aa*aR;

(iii)a∈Raa*a.

In this case,a†=(ax)*=(ya)*,where x,y∈R satisfy a=aa*ax=yaa*a.

Lemma 2.2Let a,b∈R.

(i)If there exists c∈R such that(1+ab)c=1,then(1+ba)(1−bca)=1.

(ii)If there exists d∈R such that d(1+ab)=1,then(1−bda)(1+ba)=1.

It follows from Lemma 2.2 that 1+ab is(left,right)invertible if and only if 1+ba is(left,right)invertible.Moreover,(1+ba)−1=1−b(1+ab)−1a.The formula above is known as Jacobson’s lemma.

Theorem 2.3Let a∈R be regular and let e,f∈R be invertible Hermitian elements.Then the following conditions are equivalent:

Corollary 2.2Let a∈R be regular and let e,f∈R be invertible Hermitian elements.Then the following conditions are equivalent:

Corollary 2.3(see[10,Theorem 3.3])Let a∈R be regular and let e,f∈R be invertible Hermitian elements.Then the following conditions are equivalent:

(i)a∈R†;

(ii)u=aa*+1−aa−is invertible;

(iii)v=a*a+1−a−a is invertible.

In this case,a†=(u−1a)*=(av−1)*.

3 Characterizations of e-Core Inverses and f-Dual Core Inverses

Recall that an element a∈R is group invertible if there exists b∈R such that aba=a,bab=b and ab=ba.Such a b is called a group inverse of a.It is unique if it exists,and is denoted by a#.By R#we denote the set of all group invertible elements in R.It is well known that a∈R#if and only if a∈a2R∩Ra2if and only if a∈anR∩Ranfor any integer n≥2.In particular,if a=a2x=ya2for some x,y∈R,then a#=yax=y2a=ax2.

Next,we mainly investigate e-core inverses and f-dual core inverses by the intersection of ideals and units.

Lemma 3.1Let a∈R be regular.Then the following conditions are equivalent:

(i)a∈R#;

(ii)a+1−aa−is invertible;

(iii)a+1−a−a is invertible.

In this case,a#=(a+1−aa−)−2a=a(a+1−a−a)−2.

Lemma 3.2(see[6,Theorem 2.1])Let a∈R and let e∈R be an invertible Hermitian

element.Then the following conditions are equivalent:

(i)a is e-core invertible;

(ii)a∈R#∩

(iii)there exists x∈R such that(eax)*=eax,xa2=a and ax2=x;

(iv)there exists x∈R such that(eax)*=eax,xa2=a,ax2=x,xax=x and axa=a.In this case,

Lemma 3.3(see[6,Theorem 2.2])Let a∈R and let f∈R be an invertible Hermitian element.Then the following conditions are equivalent:

Lemma 3.4Let a∈R and let e,f∈R be invertible Hermitian elements.Suppose that n≥2 is an integer.Then

(i)a∈af−1a*R∩Ranif and only if a∈af−1(a*)nR.

(ii)a∈Ra*ea∩anR if and only if a∈R(a*)nea.

Proof(i)“⇒”If a∈af−1a*R∩Ran,then there exist some s,t∈R such that a=af−1a*s=tan,and hence a=af−1(tan)*s=af−1(a*)nt*s∈af−1(a*)nR.

(ii)It can be proved by a similar way as(i).

Theorem 3.1Let a∈R and let e,f∈R be invertible Hermitian elements.Suppose that n≥2 is an integer.Then the following conditions are equivalent:

The following result gives the characterization of both e-core invertible and f-dual core invertible elements by units in a ring R.

Theorem 3.2Let a∈R be regular and let e,f∈R be invertible Hermitian elements.Then the following conditions are equivalent:

Proof(i)⇔(ii)It follows from Theorem 2.1 and Lemmas 3.2–3.3.

Analogously,we can prove(i)⇔(v)⇔(vi).

As a#=(u−1af−1a*e)2a and a=u−1af−1a*ea2,by applying Lemma 3.2,we get

Also,as a=a2(f−1a*eas−1),we obtain a#=a(f−1a*eas−1)2and

The proof is completed.

If e=1,then the e-core inverse is just the core inverse.If f=1,then the f-dual core inverse is the dual core inverse.By Rand Rwe denote the sets of all core invertible and dual core invertible elements in R.

Corollary 3.1(see[2,Theorem 5.6])Let a∈R be regular.Then the following conditions are equivalent:

By Corollary 3.1,we know that the core and dual core inverses of a are characterized by the invertibility of a2a*+1−aa−.In[3,Theorem 4.1],Li and Chen proved that the result is true when the quadratic component a2a*in a2a*+1−aa−is changed to a(a*)2.More precisely,a∈R∩Rif and only if a(a*)2+1−aa−is invertible if and only if(a*)2a+1−a−a is invertible.

For the case of the e-core inverse and the f-dual core inverse,one can also get their similar characterizations.

Theorem 3.3Let a∈R be regular and let e,f∈R be invertible Hermitian elements.Then the following conditions are equivalent:

ProofIt follows from Lemma 2.2 that(ii)⇔(iii).

So,u is right invertible and s is a right inverse of u.

4 Relations with(one-sided)Inverses along an Element

Given a,d∈R,a is left invertible along d(see[7])if there exists b∈R such that bad=d and b∈Rd.Such b is called a left inverse of a along d,and is denoted byDually,we call a is right invertible along d(see[7])if there exists b∈R satisfying dab=b and b∈dR.A right inverse of a along d is denoted by

Lemma 4.1(see[9,Theorems 2.3–2.4])Let a,d∈R.Then

(i)a is left invertible along d if and only if d∈Rdad.

(ii)a is right invertible along d if and only if d∈dadR.

Theorem 4.1Let a∈R and let e,f∈R be invertible Hermitian elements.Then the following conditions are equivalent:

Theorem 4.2Let a∈R and let e,f∈R be invertible Hermitian elements.Then the following conditions are equivalent:

It follows from Theorem 3.2 that a∈if and only if a2f−1a*e+1−aa−is invertible.Hence,a is both e-core and f-dual core invertible if and only if af−1a*e is invertible along a.

It is natural to consider whether we can characterize the e-core inverse(resp.the f-dual core inverse)by the inverse of an element.We next show the fact that a is e-core invertible if and only if it is invertible along af−1a*e,and a is f-dual core invertible if and only if it is invertible along f−1a*ea,under the assumption a∈

Theorem 4.3LetThen a is e-core invertible if and only if it is invertible along af−1a*e.In this case,

ProofSuppose that a is invertible along af−1a*e with x=a‖af−1a*e.Then,we have

By a direct calculation,it follows

which implies eax=(eax)*.

As e is an invertible Hermitian element,and hence axa=a.Since x∈af−1a*eR,there exists some y∈R such that x=af−1a*ey=axaf−1a*ey=ax2.

Similarly,we get

Therefore,x=a‖af−1a*eis the e-core inverse of a.

which gives z∈dR.

Also,note the equality z=zaz,we can obtain z∈Rd.

Therefore,a is invertible along af−1a*e.

Corollary 4.1(see[8,Theorem 4.3])Let a∈R†.Then

(i)a is core invertible if and only if it is invertible along aa*.In this case,=a‖aa*.

(ii)a is dual core invertible if and only if it is invertible along a*a.In this case,=a‖a*a.

AcknowledgementThe authors are highly grateful to the referees for their valuable comments and suggestions which led to improvements of this paper.In particular,Theorem 3.3 is suggested to the authors by one referee.