Some additive results on Drazin inverses

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Abstract

In this paper, some additive results on Drazin inverse of a sum of Drazin invertible elements are derived. Some converse results are also presented.

Section snippets

Background

Our aim is to investigate the existence of the Drazin inverse (p+q)d of the sum p+q, where p and q are either ring elements or matrices. The Drazin inverse is the unique solution to the equationsak+1x=ak,xax=x,ax=xafor some k0, if any. The minimal such k is called the index in(a) of a. If the Drazin inverse exists we shall call the element D-invertible.

An element a is called regular if axa=a for some x, and we denote the set of all such solutions by a{1}.

A ring with 1 is von Neumann (Dedekind)

D-inverses via powering

As a first example where powering can be used, we present the case where a2=0=b2. We have

Proposition 2.1

Suppose a,b and ab are D-invertible and that a2=0=b2. Then

  • 1.

    a+b is D-invertible;

  • 2.

    (a+b)d=a(ba)d+b(ab)d and [(a+b)2]d=(ab)d+(ba)d.

Proof

Using induction it is easily seen that(a+b)2k=(ab)k+(ba)kand(a+b)2k+1=(ab)ka+(ba)kb.It is now straight forward to check that x=a(ba)d+b(ab)d satisfies the necessary equations (a+b)2k+1x=(a+b)2k,x(a+b)x=x and (a+b)x=x(a+b). 

We note in passing that this result takes care of the example

Splittings

As always our starting point for the splitting approach is the factorization a+b=1ba1. Using Cline’s formula [1], we may write(a+b)d=1b(Md)2a1,whereM=a11b=aab1b,M2=a2+aba2b+ab2a+bab+b2andM3=a3+a2b+aba+ab2a3b+a2b2+abab+ab3a2+ab+ba+b2a2b+ab2+bab+b3.There are two approaches that we can take, namely we can compute Md and then square the result, or we can directly compute (M2)d or (M3)d. We shall start by using the second approach.

Our first result is

Theorem 3.1

Suppose that a2+ab and ab+b2 are D-invertible, and

Converse results

We shall now assume that a+b is D-invertible, and examine the D-invertibility of the related elements, a,b,ab and ba. We shall present one local result in addition to one global result.

Proposition 4.1

Let a3=0=b2=a2bab=baba2=0=(ab)3. If a+b has a Drazin inverse then so do a2b and aba.

Proof

Using the notation of Proposition 3.1, we see that nx=0. Now if a+b is D-invertible, then the matrices M and M3 in (5), (6) are D-invertible, so that P+Q is D-invertible. Now P=(P+Q)-Q is a LO splitting because PQ=0=Q2.

Acknowledgement

The authors wish to thank an anonymous referee for his/her remarks and corrections.

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Research supported by CMAT – Centro de Matemática da Universidade do Minho, Portugal, by the Portuguese Foundation for Science and Technology – FCT through the research program POCTI, and by Fundação Luso Americana para o Desenvolvimento (project # 273/2008).

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