# Representations for the Generalized Drazin Inverse of the Sum in a Banach Algebra and Its Application for Some Operator Matrices

Authors: Xiaoji Liu; Xiaolan Qin
Journal: The Scientific World Journal
 (2015)

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## Abstract

We investigate additive properties of the generalized Drazin inverse in a Banach algebraA. We find explicit expressions for the generalized Drazin inverse of the sum a+b, under new conditions on a,b∈A. As an application we give some new representations for the generalized Drazin inverse of an operator matrix.


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## Body

## 1. Introduction

LetA be a complex Banach algebra with unite 1. We use σ(a) to denote the spectrum of a∈A. The sets of all nilpotent and quasinilpotent elements (σ(a)={0}) of A will be denoted by Anil and Aqnil, respectively.The generalized Drazin inverse ofa∈A (introduced by Koliha in [1]) is the element b∈A which satisfies
(1)xax=x,ax=xa,a-a2x∈Aqnil.
If there exists the generalized Drazin inverse, then the generalized Drazin inverse of a is unique and is denoted by ad. The set of all generalized Drazin invertible elements of A is denoted by Ad. For interesting properties of the generalized Drazin inverse see [2–6]. For a complete treatment of the generalized Drazin inverse, see [7, Chapter 2].Ifp=p2∈A is an idempotent, we denote p¯=1-p. We can represent element a∈A as
(2)a=a11a12a21a22p,
where a11=pap, a12=pap¯, a21=p¯ap, and a22=p¯ap¯.Leta∈Ad and aπ=1-aad be the spectral idempotent of a corresponding to {0}. It is well known that a∈A can be represented in the following matrix form ([7, Chapter 2]):
(3)a=a100a2p,
relative to p=aad, where a1 is invertible in the algebra pAp, ad is its inverse in pAp, and a2 is quasinilpotent in the algebra p¯Ap¯. Thus, the generalized Drazin inverse of a can be expressed as
(4)ad=a1d000p.
Obviously, if a∈Aqnil, then a is generalized Drazin invertible and ad=0.In this paper, we first give the formulas of(a+b)d under the conditions ab=babπ and ab=aπbabπ, respectively. Then we will apply these formulas to provide some representations for the generalized Drazin inverse of the operator matrix M=ABCD under some conditions.


## 2. Main Results

First we start the following result which is proved in [8] for matrices, extended in [9] for a bounded linear operator and in [10] for arbitrary elements in a Banach algebra.Lemma 1 (see [10, Theorem 2.3]).
Letx,y∈A and p∈A be an idempotent. Assume that x and y are represented as
(5)x=a0cbp,y=bc0ap.(i)
Ifa∈(pAp)d and b∈(p¯Ap¯)d, then x and y are generalized Drazin invertible, and
(6)xd=ad0ubdp,yd=bdu0adp,
where
(7)u=∑n=0∞‍bdn+2canaπ+∑n=0∞‍bπbncadn+2-bdcad.(ii)
Ifx∈Ad and a∈(pAp)d, then b∈(p¯Ap¯)d and xd and yd are given by (6) and (7).Lemma 2 (see [11, Lemma 2.1]).
Leta,b∈Aqnil. If ab=ba or ab=0, then a+b∈Aqnil.The following result is a generalization of [10, Corollary 3.4].Theorem 3.
Ifa∈Aqnil, b∈Ad, and ab=babπ, then a+b∈Ad and
(8)a+bd=bd+∑n=0∞‍bdn+2aa+bn.Proof.
First, suppose thatb∈Aqnil. Therefore, bπ=1 and from ab=babπ we obtain ab=ba. Using Lemma 2, a+b∈Aqnil and (8) holds.
Now we assumeb is not quasinilpotent, using matrix representations of a and b relative to p=bbd. We have
(9)b=b100b2p,bd=b1d000p,
where b1∈(pAp)-1, b2∈(p¯Ap¯)qnil.
Let us represent(10)a=a1a2a3a4p.
From ab=babπ and
(11)ab=a1b1a2b2a3b1a4b2p,babπ=0b1a20b2a4p,
we obtain a1b1=0 and a3b1=0. Since b1 is invertible, we have a1=0 and a3=0.
Hence we have(12)a+b=b1a20a4+b2p.
The conditionab=babπ implies that a4b2=b2a4. Hence, using Lemma 2, we get a4+b2∈Aqnil. By Lemma 1, we obtain that a+b∈Ad and
(13)a+bd=b1du00p,
where
(14)u=∑n=0∞‍b1dn+2a2a4+b2n.
Now from (14), using the matrix representation of bd, a, and a+b, we easily obtain formula (8) of the theorem.The next result is a generalization of [12, Theorem 2.2] and [10, Example 4.5].Theorem 4.
Leta,b∈Ad. If ab=aπbabπ, then a+b∈Ad and
(15)a+bd=bπad+bdaπ+∑n=0∞‍bdn+2aa+bnaπ+bπ∑n=0∞‍a+bnbadn+2-∑n=0∞‍∑k=0∞‍bdk+1aa+bn+kbadn+2-∑n=0∞‍bdn+2aa+bnbad.Proof.
Ifa is quasinilpotent, we can apply Theorem 3 and we obtain (15) for this particular case. Now we assume that a is neither invertible nor quasinilpotent and consider the following matrix representations of a, ad, and b relative to the p=aad:
(16)a=a100a2p,ad=a1d000p,b=b1b2b3b4p.
The conditionab=aπbabπ implies that a1b1=0 and a1b2=0. Since a1 is invertible, we have b1=0 and b2=0.
Thus,b can be represented as
(17)b=00b3b4p.
Therefore, b4∈(p¯Ap¯)d and, from Lemma 1, we have
(18)bd=00b4d2b3b4dp,bπ=p0-b4db3b4πp.
Fromab=aπbabπ and
(19)ab=00a2b3a2b4p,aπbabπ=00b3a1-b4a2b4db3b4a2b4πp,
we obtained a2b4=b4a2b4π. From Theorem 3, we get a2+b4∈Ad and
(20)a2+b4d=b4d+∑n=0∞‍b4dn+2a2a2+b4n.
Further, applying Lemma 1 to a+b, we get
(21)a+bd=a1d0ua2+b4d,
where
(22)u=∑n=0∞‍a2+b4dn+2b3a1na1π+∑n=0∞‍a2+b4πa2+b4nb3a1dn+2-a2+b4db3a1d.
Observe that since a1∈(pAp)-1, then a1π=0.
Hence, the expression ofu reduces to
(23)u=∑n=0∞‍a2+b4πa2+b4nb3a1dn+2-a2+b4db3a1d.
From a2b4=b4a2b4π we get =a2b4(b4d)2=b4a2b4π(b4d)2=0.
Hence, from formula (20) and a2b4d=0, we have
(24)a2+b4π=p¯-a2+b4·b4d+∑n=0∞‍b4dn+2a2a2+b4n=p¯-b4b4d+∑n=0∞‍b4dn+2a2a2+b4n=b4π-∑n=0∞‍b4dn+1a2a2+b4n.
Then substituting (20) and (24) in (22), we get
(25)u=∑n=0∞‍b4πa2+b4nb3a1dn+2-∑n=0∞‍∑k=0∞‍b4dk+1a2a2+b4n+kb3a1dn+2-b4db3a1d-∑n=0∞‍b4dn+2a2a2+b4nb3a1d.
Now, replacingu by the above expression and considering matrix representations of a and b, after direct computations, we obtain the formula (15) for (a+b)d.


## 3. Applications

In this section, we give some formulas for the generalized Drazin inverse of a2×2 operator matrix under some conditions.Finding an explicit representation for the generalized Drazin inverse of an operator matrixM=ABCD in terms of A, B, C, D and related generalized Drazin inverse has been studied by several authors [9, 13–15]. Djordjević and Stanimirović [9] generalize the well-known result in [8, 16] concerning the Drazin inverse of block 2×2 upper triangular matrices to the generalized Drazin inverse for a block triangular operator matrix. Further, they consider the case where BC=0, BD=0, and DC=0.This section is devoted to the generalized Drazin inverse of2×2 operator matrix:
(26)M=ABCD,
where A∈B(X) and D∈B(Y) are generalized Drazin invertible.Next we will state some auxiliary lemmas.Lemma 5 (see [2, 3]).
LetA and D be generalized Drazin invertible and let M be matrix of form (26). If BC=0 and BD=0, then
(27)Md=AdAd2BX0Dd+X1B,
where
(28)Xn=∑i=0∞‍Ddi+n+2CAiAπ+Dπ∑i=0∞‍DiCAdi+n+2-∑i=0n‍Ddi+1CAdn-i+1,n≥0.Lemma 6 (see [17, Lemma 3.1]).
IfM is matrix of form (26), such that A is generalized Drazin invertible, D is quasinilpotent, and BDnC=0 for any nonnegative integer n, then M is generalized Drazin invertible and
(29)Md=AdΓΔΔAΓ,
where
(30)Γ=∑n=0∞‍Adn+2BDn,Δ=∑n=0∞‍DnCAdn+2.Lemma 7.
LetA∈Cn×n. Then (AAπ)d=0, (A2Ad)d=Ad, (A2Ad)π=Aπ, and Ind(AAπ)=Ind(A) and Ind(A2Ad)=1.Proof.
The Jordan canonical form ofX permits us to write A=S(C⊕N)S-1, where S and C are nonsingular, and N is nilpotent with index Ind(A). Thus Ad=S(C-1⊕0)S-1. Now, it is evident that A2Ad=S(C⊕0)S-1 and AAπ=S(0⊕N)S-1, which lead to the affirmations of this lemma.In [9, Theorem 5.3] authors gave an explicit representation for Md under conditions BC=0, DC=0, and BD=0. Here we replace the last two conditions by the two weaker conditions DC=DπCAAπ and BD=AAπB.Theorem 8.
LetA and D be generalized Drazin invertible and let M be matrix of form (26). If AAπB=BD, DC=DπCAAπ and BC=0. Then
(31)Md=AdAd2B+∑n=0∞‍AnBDdn+2CAd2Dd+CAd3B+∑n=1∞∑i=1n‍Di-1CAn-iBDdn+2‍.Proof.
We can split matrixM as M=P+Q, where
(32)P=AAπ00D,Q=A2AdBC0,Pd=000Dd,Pπ=I00Dπ.
SinceDC=DπCAAπ and AAπB=BD, we have
(33)DdC=Dd2DC=Dd2DπCAAπ=0,DCAd=DπCAAπAd=0,AdBD=AdAAπB=0.
FromBC=0 and applying Lemma 5 to Q, we obtain
(34)Qdi=AdiAdi+1BXi-1XiB,Qπ=Aπ-AdB-CAdI-CAd2B,
where Xn is defined in (28). From DdC=0 and DCAd=0, we get Xn=C(Ad)n+2.
SinceAAπB=BD and DC=DπCAAπ, we obtain PQ=PπQPQπ. Applying Theorem 4, we get
(35)P+Qd=QπPd+QdPπ+∑n=0∞‍Qdn+2PP+QnPπ+Qπ∑n=0∞‍P+QnQPdn+2-∑n=0∞‍∑k=0∞‍Qdk+1PP+Qn+kQPdn+2-∑n=0∞‍Qdn+2PP+QnQPd.
FromAdBD=0, we have
(36)QπPd=Pd,QdPπ=Qd,QdP=0.
Hence from (35), we obtain
(37)P+Qd=Pd+Qd+Qπ∑n=0∞‍P+QnQPdn+2.
Since AdBD=0, we have
(38)QπQPd2=BDd2.
The conditionsBC=0 and BD=AAπB  imply that BDnC=0. From BDnC=0 and AdBD=0, we get
(39)Qπ∑n=1∞‍P+QnQPdn+2=0∑n=1∞‍AnBDdn+20∑n=1∞∑i=1n‍Di-1CAn-iBDdn+2‍.
From (36), (38), and (39) it follows (31).
The proof is finished.Since(40)ABCD=0InIm0DCBA0ImIn0,
we can obtain the following result, applying Theorem 8 to DCBA.Theorem 9.
LetA and D be generalized Drazin invertible and let M be matrix of form (26). If DDπC=CA, AB=AπBDDπ and CB=0. Then
(41)Md=Ad+BDd3C+∑n=1∞‍∑i=1n‍An-iBDi-1CAdn+2BDd2Dd2C+∑n=0∞‍DnCAdn+2Dd.Theorem 10.
LetA, D, and BC be generalized Drazin invertible and let M be matrix of form (26). If AB=AπBD, DC=DπCA and BC=0. Then
(42)Md=Ad∑n=0∞‍AnBDdn+2∑n=0∞‍DnCAdn+2Dd+∑n=1∞∑i=0n-1‍DiCAn-i-1BDdn+2‍.Proof.
We can split matrixM as M=P+Q, where
(43)P=A00D,Q=0BC0,(44)Pd=Ad00Dd,Pπ=Aπ00Dπ.
Since
(45)Q2=BC00CB,Q3=0BCBCBC0,
from BC=0, it is easy to get Q3=0. Since Q is nilpotent, we have Qd=0. Applying Theorem 4 to the particular case, we get
(46)P+Qd=Pd+∑n=0∞‍P+QnQPdn+2.
The conditionsAB=AπBD and BC=0  imply that BDnC=0, for n≥0, so we get
(47)∑n=0∞‍P+QnQPdn+2=0∑n=0∞‍AnBDdn+2∑n=0∞‍DnCAdn+2∑n=1∞∑i=0n-1‍DiCAn-i-1BDdn+2‍.
From (44) and (47) it follows (42).
The proof is finished.Theorem 11.
LetA and D be generalized Drazin invertible and let M be matrix of form (26). If AAπB=BD2Dd, D2DdC=DπCAAπ and BDnC=0 for any nonnegative integer n. Then
(48)Md=AdΓ+∑n=0∞‍AnBDdn+2ΔDd+ΔAΓ+∑n=1∞∑i=0n-1‍DiCAn-i-1BDdn+2‍,
where Γ and Δ are defined in (30).Proof.
We can split matrixM as M=P+Q, where
(49)P=AAπ00D2Dd,Q=A2AdBCDDπ,Pd=000Dd,Pπ=I00Dπ.
FromAAπB=BD2Dd and D2DdC=DπCAAπ, we have
(50)DdC=Dd3D2C=Dd2D2DdC=Dd2DπCAAπ=0,(51)AdBDd=AdBD2Dd3=AdBD2DdDd2=AdAAπBDd2=0,
so we get DπC=C.
Note thatDDπ is quasinilpotent, DπC=C, and B(DDπ)nC=BDnDπC=BDnC=0 for any nonnegative integer n; we can apply Lemma 6 to Q with D replaced by DDπ; we have
(52)Qd=AdΓ′Δ′Δ′AΓ′,
where
(53)Γ′=∑n=0∞‍Adn+2BDnDπ,Δ′=∑n=0∞‍DnDπCAdn+2.
Observe that (50) and (51) yield
(54)Γ=∑n=0∞‍Adn+2BDn,Δ=∑n=0∞‍DnCAdn+2,
so we get
(55)Qd=AdΓΔΔAΓ.
The condition BDnC=0 implies that
(56)BΔ=B∑n=0∞‍DnCAdn+2=0.
Hence we have
(57)QQd=AAd+BΔA2AdΓ+BΔAΓCAd+DDπΔCΓ+DDπΔAΓ=AAdAΓCAd+DΔCΓ+DΔAΓ,Qπ=Aπ-AΓ-CAd-DΔI-CΓ-DΔAΓ.
FromAAπB=BD2Dd and D2DdC=DπCAAπ, we obtain PQ=PπQPQπ. Applying Theorem 4, we get
(58)P+Qd=QπPd+QdPπ+∑n=0∞‍Qdn+2PP+QnPπ+Qπ∑n=0∞‍P+QnQPdn+2-∑n=0∞‍∑k=0∞‍Qdk+1PP+Qn+kQPdn+2-∑n=0∞‍Qdn+2PP+QnQPd,
where
(59)ΓD2Dd=∑n=0∞‍Adn+2BDnDπD2Dd=0,ΔAAπ=∑n=0∞‍DnDπCAdn+2AAπ=0,
so we get
(60)QdP=AdΓΔΔAΓAAπ00D2Dd=AdAAπΓD2DdΔAAπΔAΓD2Dd=0000.
Hence from (58) and (60) we obtain
(61)P+Qd=QπPd+QdPπ+Qπ∑n=0∞‍P+QnQPdn+2.
By direct computation we verify that(62)QπPd=Pd,QdPπ=Qd.
From BDnC=0, we have
(63)∑n=0∞‍P+QnQPdn+2=0∑n=0∞‍AnBDdn+20∑n=1∞‍∑i=0n-1‍DiCAn-i-1BDdn+2.
Observe that (51) and BDnC=0 yield
(64)Qπ∑n=0∞‍P+QnQPdn+2=Aπ-AΓ-CAd-DΔI-CΓ-DΔAΓ0∑n=0∞‍AnBDdn+20∑n=1∞‍∑i=0n-1‍DiCAn-i-1BDdn+2=0Aπ∑n=0∞‍AnBDdn+2-AΓ∑n=1∞‍∑i=0n-1‍DiCAn-i-1BDdn+20-CAd-DΔ∑n=0∞‍AnBDdn+2+I-CΓ-DΔAΓ∑n=1∞‍∑i=0n-1‍DiCAn-i-1BDdn+2=0∑n=0∞‍AnBDdn+20∑n=1∞‍∑i=0n-1‍DiCAn-i-1BDdn+2.
From (62) and (64) it follows (48).
The proof is finished.Theorem 12.
LetA and D be generalized Drazin invertible and let M be matrix of form (26). If AAπBDπ=BD, BC=0, CAd=0, and CBDπ=0. Then
(65)Md=AdAd2BX0Dd+∑n=0∞‍Xn+1BDn,
where
(66)Xn=∑i=0∞‍Ddi+n+2CAi,n≥0.Proof.
We can split matrixM as M=P+Q, where
(67)P=A2AdB00,Q=AAπ0CD,Pd=AdAd2B00,Pπ=Aπ-AdB0I.
Applying Lemma7, we have (AAπ)d=0, so we get
(68)Qdn=00Xn-1Ddn,Qπ=I0-DX0Dπ,
where Xn is defined in (28).
SinceAAπBDπ=BD, BC=0, CBDπ=0, and CA2Ad=0, we obtain PQ=PπQPQπ. Applying Theorem 4, we get
(69)P+Qd=QπPd+QdPπ+∑n=0∞‍Qdn+2PP+QnPπ+Qπ∑n=0∞‍P+QnQPdn+2-∑n=0∞‍∑k=0∞‍Qdk+1PP+Qn+kQPdn+2-∑n=0∞‍Qdn+2PP+QnQPd.
FromCAd=0, we have QPd=0. Hence from (69) we obtain
(70)P+Qd=QπPd+QdPπ+∑n=0∞‍Qdn+2PP+QnPπ,
where XnAd=0, we get
(71)QπPd=Pd,QdPπ=Qd,Qd2PPπ=00X1Dd2A2AdB00Aπ-AdB0I=000X1B.
The conditionsAAπBDπ=BD and BC=0  imply that BDiC=0. So we get
(72)∑n=0∞‍Qdn+2PP+QnPπ=000∑n=1∞‍Xn+1BDn,n≥1.
From (71) and (72) it follows (65).
The proof is finished.Using (40) and Theorem 12, we have the following result.Theorem 13.
IfCA=DπDCAπ, BDd=0, CB=0, and BCAπ=0. Then
(73)Md=Ad+∑n=0∞‍Xn+2CAnX1Dd2CDd,
where
(74)Xn=∑i=0∞‍Adi+n+1BDi,n≥1.Using the case of Theorem3, we get the following results.Theorem 14.
LetA and D be generalized Drazin invertible and let M be matrix of form (26). If DCAπ=CA, ABD=0, BC=0, and CB=0. Then
(75)Md=Ad+∑n=0∞‍BDdn+3CAnAd2B+BDd2∑n=0∞‍Ddn+2CAnDd.Proof.
We can split matrixM as M=P+Q, where
(76)P=00C0,Q=AB0D.
FromABD=0, we have
(77)Qdi=AdiXi0Ddi,Qπ=Aπ-AX1-BDd0Dπ,
where
(78)Xn=Adn+2B+BDdn+2,n≥0.
Note thatP is quasinilpotent; since DCAπ=CA, ABD=0, BC=0, and CB=0, we obtain PQ=QPQπ. Applying Theorem 3, we get
(79)P+Qd=Qd+∑n=0∞‍Qdn+2PP+Qn.
FromBC=0, we have
(80)Qd2P=Ad2X20Dd200C0=X2C0Dd2C0=BDd30Dd2C0.
The conditionsDCAπ=CA and CB=0  imply that CAiB=0. From ABD=0, CAiB=0, and BC=0, we get
(81)∑n=1∞‍Qdn+2PP+Qn=∑n=1∞‍Xn+2CAn∑n=1∞‍Xn+2CAn-1B∑n=1∞‍Ddn+2CAn∑n=1∞‍Ddn+2CAn-1B=∑n=1∞‍BDdn+3CAn0∑n=1∞‍Ddn+2CAn0.
From (77), (80), and (81) it follows (75).Using (40) and Theorem 14, we have the following result.Theorem 15.
LetA and D be generalized Drazin invertible and let M be matrix of form (26). If ABDπ=BD, DCA=0, BC=ABCAd, and CB=0. Then
(82)Md=Ad∑n=0∞‍Adn+2BDnDd2C+CAd2Dd+∑n=0∞‍CAdn+3BDn.



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