# The Expression of the Generalized Drazin Inverse ofA−CB

Authors: Xiaoji Liu; Dengping Tu; Yaoming Yu
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We investigate the generalized Drazin inverse ofA−CB over Banach spaces stemmed from the Drazin inverse of a modified matrix and present its expressions under some conditions.


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## Body

## 1. Introduction

Let𝒳 and 𝒴 be Banach spaces. We denote the set of all bounded linear operators from 𝒳 to 𝒴 by ℬ(𝒳,𝒴). In particular, we write ℬ(𝒳) instead of ℬ(𝒳,𝒳).For anyA∈ℬ(𝒳,𝒴), ℛ(A) and 𝒩(A) represent its range and null space, respectively. If A∈ℬ(𝒳), the symbols σ(A) and acc(σ(A)) stand for its spectrum and the set of all accumulation points of σ(A), respectively.Recall the concept of the generalized Drazin inverse introduced by Koliha [1] that the element Td∈ℬ(𝒳) is called the generalized Drazin inverse of T∈ℬ(𝒳) provided it satisfies(1.1)TTd=TdT,TdTTd=Td,T-T2Tdis quasinilpotent.
If it exists then it is unique. The Drazin index Ind(T) of T is the least positive integer k if (T-T2Td)k=0, and otherwise Ind(T)=+∞.From the definition of the generalized Drazin inverse, it is easy to see that ifT is a quasinilpotent operator, then Td exists and Td=0. It is well known that the generalized Drazin inverse of T∈ℬ(𝒳) exists if and only if 0∉acc(σ(T)) (see [1,Theorem  4.2]). If T is generalized Drazin invertible, then the spectral idempotent Tπ of T corresponding to 0 is given by Tπ=I-TTd.The generalized Drazin inverse is widely investigated because of its applications in singular differential difference equations, Markor chains, (semi-) iterative method numerical analysis (see, for example, [1–5, 7], and references therein).In this paper, we aim to discuss the generalized Drazin inverse ofA-CB over Banach spaces. This question stems from the Drazin inverse of a modified matrix (see, e.g., [6]). In [3], Deng studied the generalized Drazin inverse of A-CB. Here we research the problem under more general conditions than those in [3]. Our results extend the relative results in [3, 4].In this section, we will list some lemmas. In next section, we will present the expressions of the generalized Drazin inverse ofA-CB. In final section, we illustrate a simple example.Lemma 1.1 (see [4, Theorem  2.3]).
LetA,B∈ℬ(𝒳) be the generalized Drazin invertible. If  AB=0, then  A+B is generalized Drazin invertible and
(1.2)(A+B)d=Bπ∑n=0∞BnAdn+1+(∑n=0∞Bdn+1An)Aπ.Lemma 1.2 (see [7, Theorem  5.1]).
IfA∈ℬ(𝒳) and  B∈ℬ(𝒴) are generalized Drazin invertible and  C∈ℬ(𝒴,𝒳), then
(1.3)M=(AC0B)
is also generalized Drazin invertible and
(1.4)Md=(AdSOBd),
where
(1.5)S=Ad2(∑n=0∞AdnCBn)Bπ+Aπ(∑n=0∞AnCBdn)Bd2-AdCBd.


## 2. Main Results

We start with our main result.Theorem 2.1.
LetA∈ℬ(𝒳) be the generalized Drazin invertible,  C∈ℬ(𝒳,𝒴), and  B∈ℬ(𝒴,𝒳). Suppose that there exists a  P∈ℬ(𝒳) such that  AP=PAP and  BP=0. If  R=(I-P)(A-CB) and  AP are generalized Drazin invertible, then  A-CB is generalized Drazin invertible and
(2.1)(A-CB)d=[∑n=0∞(AP)dn+1(Rn+VRn-1+V2Rn-2)]Rπ-(AP)d[VRd+V2Rd2+(AP)dV2Rd]+(AP)π∑n=0∞(AP)n(Rdn+1+VRdn+2+V2Rdn+3),
where V=PA-PCB-AP and the symbols ViRj=0,i=1,2, if j<0.Proof.
LetS:=AP and T:=(A-CB)(I-P). Then
(2.2)TS=(A-CB)(I-P)AP=0,(2.3)RP=(I-P)(A-CB)P=0,(2.4)A-CB=AP+A(I-P)-CB(I-P)=S+T
since AP=PAP and BP=0. So, by Lemma 1.1,
(2.5)(T+S)d=Sπ∑n=0∞SnTdn+1+∑n=0∞Sdn+1TnTπ.
Next, we will give the representations ofTd, Tn, and Tdn. In order to obtain the expression of Td, rewrite T as
(2.6)T=R+PA-PCB-PAP=R+V.
Since VP=PAP-AP2=PAP(I-P),
(2.7)V2P=(PA-PCB-AP)PAP(I-P)=(PAPAP-APPAP)(I-P)=0,
and then Vn=0 for n>2 since V=PA-CB-AP. So Vd exists and Vd=0. By (2.3), RV=RP(A-CB-AP)=0 and then RdV=RdRdRV=0. So, by Lemma 1.1,
(2.8)Td=(R+V)d=Rd+VRd2+V2Rd3,
and then
(2.9)TTd=RRd+VRd+V2Rd2.
Since R(R+V)k=Rk+1 and V2(R+V)k=V2Rk for k≥1,
(2.10)Tn=(R+V)n=(R2+VR+V2)(R+V)n-2=Rn+VRn-1+V2Rn-2,n≥2.
From RdV=0, it is easy to verify that
(2.11)Tdn=(Rd+VRd2+V2Rd3)n=Rdn+VRdn+1+V2Rdn+2.
Hence,
(2.12)(∑n=0∞Sdn+1Tn)Tπ=(AP)d[I+(AP)d(R+V)+(AP)d2(R2+VR+V2)]×(Rπ-VRd-V2Rd2)+∑n=3∞(AP)dn+1(Rn+VRn-1+V2Rn-2)Rπ=(AP)d[I+(AP)d(R+V)+(AP)d2(R2+VR+V2)]Rπ-(AP)d(VRd+V2Rd2+(AP)dV2Rd)+∑n=3∞(AP)dn+1(Rn+VRn-1+V2Rn-2)Rπ,Sπ∑n=0∞SnTdn+1=(AP)π∑n=0∞(AP)n(Rdn+1+VRdn+2+V2Rdn+3).
Therefore, we reach (2.1).WhenInd(AP),Ind(R)<+∞, we have the following corollary.Corollary 2.2.
LetA∈ℬ(𝒳) be generalized Drazin invertible.  C∈ℬ(𝒳,𝒴), and  B∈ℬ(𝒴,𝒳). Suppose that there exists a  P∈ℬ(𝒳) such that  AP=PAP and  BP=0. If  R=(I-P)(A-CB) and  AP are generalized Drazin invertible and  Ind(R)=k<+∞ and  Ind(AP)=h<+∞, then  A-CB is generalized Drazin invertible and
(2.13)(A-CB)d=[∑n=0k-1(AP)dn+1(Rn+VRn-1+V2Rn-2)]Rπ-(AP)d[VRd+V2Rd2+(AP)dV2Rd]+(AP)π∑n=0h-1(AP)n(Rdn+1+VRdn+2+V2Rdn+3),
where V=PA-PCB-AP and the symbols ViRj=0,i=1,2, if j<0.If an operatorT is quasinilpotent, Td=0 and Tπ=I. So, the following corollary follows from Theorem 2.1.Corollary 2.3.
LetA∈ℬ(𝒳) be generalized Drazin invertible,  C∈ℬ(𝒳,𝒴), and  B∈ℬ(𝒴,𝒳). Suppose that there exists a  P∈ℬ(𝒳) such that  AP=PAP and  BP=0. If  R=(I-P)(A-CB) is generalized Drazin invertible and  AP is a quasinilpotent operator, then  A-CB is generalized Drazin invertible and
(2.14)(A-CB)d=∑n=0∞(AP)n(Rdn+1+VRdn+2+V2Rdn+3),
where V=PA-PCB-AP.Theorem 2.4.
LetA∈ℬ(𝒳) be generalized Drazin invertible,  C∈ℬ(𝒳,𝒴), and  B∈ℬ(𝒴,𝒳). Suppose that there exists an idempotent  P∈ℬ(𝒳) such that  PA=PAP and  BP=B. If  R=P(A-CB) is generalized Drazin invertible, then  A-CB is generalized Drazin invertible and
(2.15)(A-CB)d=Rd+Ad(I-P)+∑n=0∞Adn+2(I-P)(A-CB)P(A-CB)nRπ+Aπ∑n=0∞An(I-P)(A-CB)PRdn+2-Ad(I-P)(A-CB)Rd.Proof.
SinceP2=P, we have 𝒳=ℛ(P)⨁𝒩(P) and can write P in the following matrix form:
(2.16)P=(I000).
The condition PA=PAP, therefore, yields the matrix form of A as follows:
(2.17)A=(A10A3A2).
From σ(A)=σ(A1)∪σ(A2) and the hypothesis that Ad exists, A1∈ℬ(ℛ(P)) and A2∈ℬ(𝒩(P)) are generalized Drazin invertible since 0∉acc(σ(A)) if and only if 0∉acc(σ(A1)) and 0∉acc(σ(A2)). And, by Lemma 1.2,
(2.18)Ad=(A1d0WA2d),
where W is some operator. Since
(2.19)A(I-P)=(000A2),(A(I-P))d exists and
(2.20)(A(I-P))d=(000A2d)=Ad(I-P).
To use Theorem2.1 to complete the proof, let Q=(I-P). So R=(I-Q)(A-CB) and AQ are generalized Drazin invertible. And from the conditions PA=PAP and BP=B, we can obtain AQ=QAQ and BQ=0. Thus, by Theorem 2.1, we have
(2.21)(A-CB)d=(AQ)dRπ+(AQ)d2(R+V)Rπ+[∑n=2∞(AQ)dn+1(Rn+VRn-1+V2Rn-2)]Rπ-(AQ)d[VRd+V2Rd2+(AQ)dV2Rd]+(AQ)π(Rd+VRd2+V2Rd3)+(AQ)π∑n=1∞(AP)n(Rdn+1+VRdn+2+V2Rdn+3),
where V=QA-QCB-AQ.
SinceP2=P and Q2=Q and then VQ=0 and V=QV. So V2=0. Note that QR=0 and then QRd=0 and (AQ)dR=0. Thus it follows from (2.21) that
(2.22)(A-CB)d=(AQ)d+(AQ)d2VRπ+[∑n=2∞(AQ)dn+1VRn-1]Rπ-(AQ)dVRd+Rd+(AQ)πVRd2+(AQ)π∑n=1∞(AQ)nVRdn+2=(AQ)d+[∑n=0∞(AQ)dn+2VRn]Rπ-(AQ)dVRd+Rd+(AQ)π∑n=0∞(AQ)nV(Rd)n+2.
Since V=Q(A-CB)-(A-CB)Q=(A-CB)(I-Q)-(I-Q)(A-CB), VR=Q(A-CB)R and QV=Q(A-CB)(I-Q). Note that Rn=P(A-CB)n and (AQ)n=AnQ. Substituting V and Q=I-P in (2.22) yields (2.15).Adding the conditionPC=C in Theorem 2.4 yields a result below.Corollary 2.5.
LetA∈ℬ(𝒳) be generalized Drazin invertible,  C∈ℬ(𝒳,𝒴), and  B∈ℬ(𝒵,𝒳). Suppose that there exists an idempotent  P∈ℬ(𝒳) such that  PA=PAP,  BP=B, and  PC=C. If  R=P(A-CB) is generalized Drazin invertible, then  A-CB is generalized Drazin invertible and
(2.23)(A-CB)d=Rd+Ad(I-P)+∑n=0∞Adn+2(I-P)AP(A-CB)nRπ+Aπ∑n=0∞An(I-P)APRdn+2-Ad(I-P)ARd.Adding the conditionPC=0 in Theorem 2.4 yields R=PA. So similar to the proof of (A(I-P))d=Ad(I-P) in Theorem 2.4, we can gain (PA)d=PAd.Corollary 2.6.
LetA∈ℬ(𝒳) be generalized Drazin invertible,  C∈ℬ(𝒳,𝒴), and  B∈ℬ(𝒵,𝒳). Suppose that there exists an idempotent  P∈ℬ(𝒳) such that  PA=PAP,  BP=B, and  PC=0; then  A-CB is generalized Drazin invertible and
(2.24)(A-CB)d=Ad+∑n=0∞Adn+2(I-P)(A-CB)PAnAπ+Aπ∑n=0∞An(I-P)(A-CB)PAdn+2-Ad(I-P)(A-CB)PAd.Analogously, we can deduce Theorem2.7 and Corollary 2.9 below.Theorem 2.7.
LetA∈ℬ(𝒳) be generalized Drazin invertible,  C∈ℬ(𝒳,𝒴), and  B∈ℬ(𝒴,𝒳). Suppose that there exists an idempotent  P∈ℬ(𝒳) such that  AP=PAP and  PC=C. If  R=(A-CB)P is generalized Drazin invertible, then  A-CB is generalized Drazin invertible and
(2.25)(A-CB)d=Rd+(I-P)Ad+∑n=0∞Rdn+2P(A-CB)(I-P)AnAπ+Rπ∑n=0∞(A-CB)nP(A-CB)(I-P)Adn+2-Rd(A-CB)(I-P)Ad.Remark 2.8 (see [4, Theorem  2.4]).
It is a special case of Theorem2.7.Corollary 2.9.
LetA∈ℬ(𝒳) be generalized Drazin invertible,  C∈ℬ(𝒳,𝒴), and  B∈ℬ(𝒵,𝒳). Suppose that there exists an idempotent  P∈ℬ(𝒳) such that  AP=PAP,  PC=C, and  BP=0; then  A-CB is generalized Drazin invertible and
(2.26)(A-CB)d=Ad+∑n=0∞Adn+2P(A-CB)(I-P)AnAπ+Aπ∑n=0∞AnP(A-CB)(I-P)Adn+2-AdP(A-CB)(I-P)Ad.Similar to Theorem2.1 and Corollary 2.2, we can show the following two results.Theorem 2.10.
LetA∈ℬ(𝒳) be generalized Drazin invertible,  C∈ℬ(𝒳,𝒴), and  B∈ℬ(𝒴,𝒳). Suppose that there exists a  P∈ℬ(𝒳) such that  PA=PAP and  PC=0. If  R=(A-CB)(I-P) and  PA are generalized Drazin invertible, then  A-CB is generalized Drazin invertible and
(2.27)(A-CB)d=Rπ∑n=0∞(Rn+Rn-1V+Rn-2V2)(PA)dn+1-[RdV+Rd2V2+RdV2(PA)d](PA)d+[∑n=0∞(Rdn+1+Rdn+2V+Rdn+3V2)(PA)n](PA)π,
where V=AP-CBP-PA and the symbols RiVj=0,j=1,2, if i<0.Corollary 2.11.
LetA∈ℬ(𝒳) be generalized Drazin invertible.  C∈ℬ(𝒳,𝒴), and  B∈ℬ(𝒴,𝒳). Suppose that there exists a  P∈ℬ(𝒳) such that  PA=PAP and  PC=0. If  R=(A-CB)(I-P) and  PA are generalized Drazin invertible and  Ind(R)=k<+∞ and  Ind(PA)=h<+∞, then  A-CB is generalized Drazin invertible and
(2.28)(A-CB)d=Rπ∑n=0k-1(Rn+Rn-1V+Rn-2V2)(PA)dn+1-[RdV+Rd2V2+RdV2(PA)d](PA)d+[∑n=0h-1(Rdn+1+Rdn+2V+Rdn+3V2)(PA)n](PA)π,
where V=AP-CBP-PA and the symbols RiVj=0,j=1,2, if i<0.WhenPA=AP and P2=P in Theorem 2.10, we can obtain the following result since Rn=(A-CB)n(I-P).Corollary 2.12 (see [3, Theorem  4.3]).
LetA∈ℬ(𝒳) be the generalized Drazin invertible,  C∈ℬ(𝒳,𝒴), and  B∈ℬ(𝒴,𝒳). Suppose that there exists an idempotent  P∈ℬ(𝒳) commuting with  A such that  PC=0. If  R=(A-CB)(I-P) is generalized Drazin invertible, then  A-CB is the generalized Drazin invertible and
(2.29)(A-CB)d=Rd+PAd-RdVAd+Rπ∑n=0∞(A-CB)nVAdn+2+∑n=0∞Rdn+2VAnAπ,
where V=-CBP.


## 3. Example

Before ending this paper, we give an example as follows.Example 3.1.
Let(3.1)A=(12410-1100-1100000),B=(0001),C=(1-100).
Then
(3.2)CB=(0001000-100000000),A-CB=(12400-1110-1100000).
We will compute the Drazin inverse of A-CB. To do this, we choose the matrix
(3.3)P=(100001000-1200000).
Apparently, P is not idempotent and PA≠AP. But BP=0 and
(3.4)AP=PAP=(1-2800-2200-2200000).
Obviously, Ind(AP)=2. Computing
(3.5)R=(I-P)(A-CB)=(0000000000010000),Rd=(0000000000000000),(3.6)V=PA-PCB-AP=(04-4001-1101-1-10000),
we have Ind(R)=2. So, by Corollary 2.2,
(3.7)(A-CB)d=(1-410-4000000000000).



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*Source: 412872-2012-02-16.xml*