Representations of the Drazin inverse involving idem potents in a ring

（Department of Mathematics，Southeast University，Nanjing 211189，China）

Representations of the Drazin inverse involving idem potents in a ring

Zhu Huihui Chen Jianlong

（Department of Mathematics，Southeast University，Nanjing 211189，China）

An element a of a ring R is called Drazin invertible if there exists b∈R such that ab＝ba，bab＝b，and a－a2b is nilpotent．The element b above is unique if it exists and is denoted as aD．The equivalent conditionsof the Drazin inverse involving idempotents in R are established．As applications，some formulae for the Drazin inverse of the difference and the product of idempotents in a ring are given．Hence，a number of results of bounded linear operators in Banach spaces are extended to the ring case．

idempotent；Drazin inverse；spectral idempotent

L et R be an associative ring w ith unity 1．The symbols R－1，RDand Rnildenote the sets of invertible，Drazin invertible and nilpotent elements of R，respectively．The commutant of an element a∈R is defined as comm（a）＝｛x∈R：xa＝ax｝．Recall that an element a∈R is said to have a Drazin inverse［1］if there is b∈R such that b∈comm（a），bab＝b，a－a2b∈Rnil．The element b∈R above is unique if it exists and is denoted by aD．In this case，we call aπ＝1－aaDthe spectral idempotent of a．The nilpotency index of a－a2b is called the Drazin index of a，denoted by ind（a）．By RDwemean the setof all Drazin invertible elements in R．It is well known that a∈RDimplies that a2∈RDand（a2）D＝（aD）2．

Groβand Trenkler［2］considered the invertibility of pq for general matrix projectors p，q．Koliha and Rakocevic［3］studied the invertibility of the sum p＋q and described the relationship between the invertibility of p－q and p＋q for idempotents p and q in a ring．Later，Koliha and Rakocevic［4］obtained the equivalent conditions for the invertibility of p－q in a ring．

Many authors considered Drazin invertibility in different sets．For example，Deng［5］considered the Drazin inverse of the difference and the product of projections in Hilbert spaces．Deng and Wei［6］presented the formulae for the Drazin inverse involving idempotent bounded linear operators in Banach spaces．More results on the Drazin inverse of the difference and the product of idempotents can be found in Refs．［7- 9］．

In this paper，we present the formulae for the Drazin inverse of the difference and the product of idempotents in a ring．Moreover，the equivalent relationships of Drazin inverse involving idempotents are established．Hence，the results in Refs．［5- 6］are extended to a general ring case．Note that dimensional analysis and spectral decompositions cannot be used in a ring case．The results in this paper are proved by a purely algebraic method．

1 Some Lemmas

In what follows，p，q always mean any two idempotents in a ring R．We first state several known results in the form of lemmas．

Lemm a 1［10］Let S＝｛p－q，1－pq，p－pq，p－qp，p－pqp，1－qp，q－pq，q－qp，p＋q－pq｝．If one of the elements in the set S is Drazin invertible，then all elements in S are Drazin invertible．

Lemma 2［10］The following statements are equivalent：

1）pq∈RD；

2）1－p－q∈RD；

3）（1－p）（1－q）∈RD．

Lemm a 3［11］（Cline’s formula） Let a，b∈RD．Then（ba）D＝b（（ab）D）2a．

Lemm a 4［1］Let a，b∈RDw ith ab＝ba．Then（ab）D＝bDaD＝aDbD．

Lemma 5［12］Let a，b∈R．If 1－ab∈RDwith ind（1－ab）＝k，then 1－ba∈RDw ith ind（1－ba）＝k and

2 M ain Results

In this section，we present some formulae on the Drazin inverse of the difference and the product of idempotents of ring R．

Definition 1 Let p－q∈RD．Define F，G and H as F＝p（p－q）D，G＝（p－q）Dp，and H＝（p－q）D（p－q）．

Theorem 1 Let p－q∈RD．Then F，G and H above are idempotents and

1）F＝（p－q）D（1－q）；

2）G＝（1－q）（p－q）D．

Proof Since p，q are idempotents，we obtain p（pq）2＝（p－q）2p＝p－pqp．Note that a∈RDand ab＝ba imply aDb＝baDby Theorem 1 of Ref．［1］．It follows that p∈comm（（p－q）D）2．Hence，we have

Next，we prove that F is idempotent．From p（p－q）D＝（p－q）D（1－q），we have

Sim ilarly，G2＝G＝（1－q）（p－q）D．It is clear that H is idempotent and

Sim ilarly，we obtain more relationships among F，G and H．

Corollary 1 Let p－q∈RD．Then

1）q（p－q）D＝（p－q）D（1－p）；

2）（p－q）Dq＝（1－p）（p－q）D；

3）qH＝Hq；

4）G（1－q）＝（1－q）F．

Proof 1）and 2）can be obtained by a sim ilar way of Theorem 1．

3）Since H＝（p－q）D（p－q），we have

4）By Theorem 1，we have

Proposition 1 Let p－q∈RD．Then

1）Fp＝pG＝pH＝Hp；

2）qHq＝qH＝Hq＝HqH．

Proof 1）It is clear that Fp＝pG，we only need to show pG＝pH and pH＝Hp．

According to Theorem 1，we obtain

Hence，1）holds．

2）Note that qH＝Hq in 3）of Corollary 1．We obtain that qHq＝（Hq）q＝Hq．Since H is idempotent，HqH＝H2q＝Hq．

Thus，qHq＝qH＝Hq＝HqH．

The follow ing theorems，themain results of this paper，give the formulae of the Drazin inverses of the difference and the product of idempotents in a ring R．

Theorem 2 Let p－q∈RD．Then

1）（1－pqp）D＝［（p－q）D］2p＋1－p；

2）（p－pqp）D＝［（p－q）D］2p＝p［（p－q）D］2；

3）（p－pq）D＝p［（p－q）D］3；

4）（p－qp）D＝［（p－q）D］3p；

5）If ind（p－q）＝k，then

Proof 1）As 1－pqp＝（p－q）2p＋1－p，［（pq）2］D＝［（p－q）D］2and（p－q）2p（1－p）＝（1－p）（pq）2p＝0；then（1－pqp）D＝［（p－q）D］2p＋1－p by Corollary 1 of Ref．［1］．

2）Observing that p－pqp＝p（p－q）2＝（p－q）2p，we obtain（p－pqp）D＝［（p－q）D］2p＝p［（p－q）D］2from Lemma 4．

3）Let x＝p［（p－q）D］3．We prove that x is the Drazin inverse of p－pq by show ing the follow ing conditions hold．

①From p（p－q）2＝（p－q）2p＝（p－pq）p，it follows that

②Note that（p－pq）x＝p（p－q）D．We have

③Since（p－pq）x＝p（p－q）D，we obtain that

According to pH＝Hp and qH＝Hq，it follows that p（p－q）（p－q）π＝（p－q）πp（p－q）．By induction，one can obtain［p（p－q）］m＝p（p－q）2m－1．Take m≥ind（p－q），then［p（p－q）（p－q）π］m＝p（p－q）2m－1（p－q）π＝0．This implies that（p－pq）－（p－pq）2x is nilpotent．

Therefore，（p－pq）D＝p［（p－q）D］3．

4）Use a similar proof of 3）．

5）It follows from Lemma 1 that 1－pq∈RD．Lemma 5 guarantees that

Substituting Eq．（2）into Eq．（1），we have

Theorem 3 Let 1－p－q∈RD．Then

1）（pqp）D＝［（1－p－q）D］2p＝p［（1－p－q）D］2；

2）（pq）D＝［（1－p－q）D］4pq．

Proof 1）By pqp＝p（1－p－q）2＝（1－p－q）2p and Lemma 4，it follows that（pqp）D＝［（1－p－q）D］2p＝p［（1－p－q）D］2．

2）From pq＝ppq and Lemma 3，we have（pq）D＝p［（pqp）D］2pq＝［（pqp）D］2pq．According to Eq．（1），we obtain（pq）D＝［（pqp）D］2pq＝［（1－p－q）D］4pq．

Deng［5］and Li［13］considered the following result for projections in Hilbert spaces，C*-algebras，respectively．Indeed，they still hold for idempotents in a ring．

Theorem 4 Let pq∈RD．Then

1）（pq）D＝（pqp）D－p［（1－q）（1－p）］D；

2）（pq）Dpq＝（pqp）Dpq．

Proof 1）By 4）of Theorem 2，we have（p－qp）D＝［（p－q）D］3p and（q－pq）D＝［（q－p）D］3q＝－［（pq）D］3q．

Hence，

We replace p by 1－p in Eq．（3）to obtain

Multiplying Eq．（4）by p on the left yields

Note that p（pq）D＝p（pq）（pq）D（pq）D＝（pq）Dand Theorem 3．We have

2）By Lemma 3，we have

The proof is completed．

Theorem 5 Let1－pq∈RD．Then p－q∈RDand

Proof By 5）of Theorem 2，we have

Substituting p and q by 1－p and 1－q，respectively，in Eq．（6），we obtain

Multiplying Eq．（6）by p－pq on the right yields

Multiplying Eq．（7）by pq－p on the right yields

From（8）and（9），one can obtain

The proof is complete．

Let p，q be two idempotents in a Banach algebra．Then，p＋q∈RDif and only if p－q∈RD．However，in general，this need not be true in a ring．For example，let R＝Z and p＝q＝1．Then p－q＝0∈RD，but p＋q＝2?RD．Next，we consider what conditions p and q satisfy，and p－q∈RDimplies that p＋q∈RD．

The following result，proved by Deng and Wei［6］for bounded linear operators in Banach spaces，indeed holds in a ring．

Theorem 6 Let p－q∈RD．If F，G and H are given by Definition 1 and（p＋q）（p－q）π∈Rnil，then

1）（p＋q）D＝（p－q）D（p＋q）（p－q）D；

2）（p－q）D＝（p＋q）D（p－q）（p＋q）D；

3）（p－q）π＝（p＋q）π；

4）（p－q）D＝F＋G－H；

5）（p＋q）D＝（2G－H）（F＋G－H）．

Koliha et al．［4］proved that p－q∈R－1implies that p＋q∈R－1for idempotents p and q in a ring R．Hence，we have the follow ing results．

Corollary 2［14］Let p－q∈R－1．If F＝p（p－q）－1and G＝（p－q）－1p，then

1）（p＋q）－1＝（p－q）－1（p＋q）（p－q）－1；

2）（p－q）－1＝（p＋q）－1（p－q）（p＋q）－1；

3）（p－q）－1＝F＋G－1；

4）（p＋q）－1＝（2G－1）（F＋G－1）．

Corollary 3 Let p－qp∈RD，and then（p－q）D＝（p－q）2［（p－qp）D－（q－qp）D］．

Proof Since（p－qp）D＝［（p－q）D］3p and（q－qp）D＝q［（q－p）D］3，we obtain

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環(huán)中涉及冪等元的Drazin逆的表示

朱輝輝 陳建龍

（東南大學(xué)數(shù)學(xué)系，南京211189）

稱環(huán)R中的元素a為Drazin可逆的，如果存在R中的元素b使得ab＝ba，bab＝b，a－a2b是冪零的．上述元素b如果存在則是唯一的，并表示為aD．給出了一些環(huán)中涉及冪等元的Drazin逆的等價條件．作為應(yīng)用，給出了環(huán)中冪等元的積與差的Drazin逆的公式．因此，一些關(guān)于Banach空間中有界線性算子的結(jié)果被推廣到環(huán)上．

冪等元；Drazin逆；譜冪等元

O151．2

10．3969／j．issn．1003－7985．2015．03．023

2013-10-14．

Biographies：Zhu Huihui（1985—），male，graduate；Chen Jianlong（corresponding author），male，doctor，professor，jlchen＠seu．edu．cn．

s：The National Natural Science Foundation of China（No．11371089），the Specialized Research Fund for the Doctoral Program of Higher Education（No．20120092110020），the Scientific Innovation Research of College Graduates in Jiangsu Province（No．CXLX13-072），the Scientific Research Foundation of Graduate School of Southeast University，the Fundamental Research Funds for the Central Universities（No．22420135011）．

：Zhu Huihui，Chen Jianlong．Representations of the Drazin inverse involving idempotents in a ring［J］．Journal of Southeast University（English Edition），2015，31（3）：427- 430．

10．3969／j．issn．1003－7985．2015．03．023