# Formulae for the Generalized Drazin Inverse of a Block Matrix in Banach Algebras

Authors: Xiaoji Liu; Xiaolan Qin
Journal: Journal of Function Spaces
 (2015)

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## Abstract

We present some new representations for the generalized Drazin inverse of a block matrix in a Banach algebra under conditions weaker than those used in resent papers on the subject.


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## Body

## 1. Introduction

The Drazin inverse has applications in a number of areas such as control theory, Markov chains, singular differential and difference equations, and iterative methods in numerical linear algebra. Representations for the Drazin inverse of block matrices under certain conditions were given in the literature [1–5]. Generalized inverses of block matrices have important applications in automatics, probability, statistics, mathematical programming, numerical analysis, game theory, econometrics, control, and so on [6, 7]. In 1979, Campbell and Meyer proposed the problem of finding a formula for the Drazin inverse of a 2×2 matrix in terms of its various blocks, where the blocks on the diagonal are required to be square matrices [7]. At the present time, there is no known complete solution to this problem.In this paper, we present the representation for the Drazin inverse ofa+b under the conditions that akb=ab and baπ=b and that akb=ab and ba2=ab (cf. Theorem 4). And we also give several representations for the Drazin inverse of a+b under some weaker conditions.LetA be a complex Banach algebra with the unit1. The sets of all Drazin invertible and quasinilpotent elements (σ(a)={0}) of A will be denoted by Ad and Aqnil, respectively.The generalized Drazin inverse ofa∈A (introduced by Koliha in [8]) is the element x∈A which satisfies(1)xax=x,ax=xa,a-a2x∈Aqnil.If there exist the generalized Drazin inverse, then the generalized Drazin inverse of a is unique and is denoted by ad.Leta∈A and let p∈A be an idempotent (p=p2). We denote p¯=1-p. Then we can write (2)a=pap+pap¯+p¯ap+p¯ap¯.Every idempotentp∈A induces a representation of an arbitrary element a∈A given by the following matrix: (3)a=pappap¯p¯app¯ap¯p.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 [9, Ch. 2]:(4)a=a100a2p,where p=aad, 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 (5)ad=a1d000p.Obviously, if a∈Aqnil, then a is generalized Drazin invertible and ad=0.


## 2. Preliminary Results

First we start the following result which is proved in [10] for matrices, extended in [2] for a bounded linear operator and in [11] for arbitrary elements in a Banach algebra.Lemma 1 (see [11, Theorem  2.3]).
LetA be a Banach algebra, x,y∈A, and p∈A an idempotent. Assume that x and y are represented as (6)x=a0cbp,y=bc0ap.(i)
Ifa∈(pAp)d and b∈(p¯Ap¯)d, then x and y are generalized Drazin invertible, and(7)xd=ad0ubdp,yd=bdu0adp,where(8)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 (7) and (8).Lemma 2.
Letb∈Ad, a∈Aqnil.(1)
[11, Corollary  3.4] If ab=0, then a+b∈Ad and (9)a+bd=∑n=0∞bdn+1an.(2)
[12, Theorem  2.2] If ba=0, then a+b∈Ad and (10)a+bd=∑n=0∞anbdn+1.Lemma 3.
LetA be a Banach algebra and let a∈Ad have the representation (4), where p=aad. If x∈pA and a1x=b for some b∈A, then x=adb (in particular, if a1x=0, then x=0). If y∈Ap and ya1=c for some c∈A, then y=cad (in particular, if ya1=0, then y=0).Proof.
Froma∈Ad we get a1∈Ad, a1d=ad, and aad=a1a1d. There exists u∈A such that x=pu. Since b=a1x, we obtain adb=ada1x=aadx=px=ppu=pu=x. The proof for y is similar.


## 3. Main Results

In [13, Theorem  3.2] authors gave an explicit representation for (a+b)d under conditions aπb=b and a2baπ=ab2aπ=0. Here we replace the last condition by the condition akb=ab; we will get a much simpler expression for (a+b)d.Theorem 4.
LetA be a Banach algebra and let a,b∈Ad. If there exist k∈N, k>1, such that akb=ab and baπ=b, then a+b∈Ad and(11)a+bd=ad+aπ∑n=0∞bdn+1an-adb∑n=0∞bdn+1an+∑n=0∞adn+2ba+bnbπ-∑n=0∞∑k=0∞adn+2ba+bnbdk+1ak+1,(12)a+bd-ad≤aπ+adb∑n=0∞bdn+1an+∑n=0∞adn+2ba+bnbπ+∑n=0∞∑k=0∞adn+2ba+bnbdk+1ak+1.Proof.
Letp=aad. We can represent a as in (4), where a1 is invertible in the subalgebra pAp and a2 is quasinilpotent. Hence,(13)ad=a1d000p.Let us write(14)b=b1b2b3b4.
Frombaπ=b and (15)b1-aπ=b10b30p=0000p,we get b1=b3=0. Thus, b and a+b can be represented as(16)b=0b20b4p,a+b=a1b20a2+b4p.By Lemma 1 and b∈Ad, we get b4∈(p¯Ap¯)d and(17)bd=0b2b4d20b4dp.An elementary computation from (17) and bπ=1-bbd leads to (18)bπ=p-b2b4d0b4πp.Since akb=ab, (19)akb=0a1kb20a2kb4p,ab=0a1b20a2b4p,we get a2kb4=a2b4. An induction argument proves a2r(k-1)+1b4=a2b4 for any r∈N. Let us denote mr=r(k-1)+1 and observe that since k>1, then {a2mr}r=1∞ is a subsequence of {a2r}r=1∞. Since a2 is quasinilpotent, we get (20)a2b41/mr=a2mrb41/mr≤a2mr1/mrb41/mr→r→∞0.Hence a2b4=0. By Lemma 2, we have that a2+b4∈Ad, because b4∈Ad, a2∈Aqnil, and a2b4=0. Also, Lemma 2 allows us to obtain(21)a2+b4d=∑n=0∞b4dn+1a2n.Again, from Lemma 1 and (16), we get that a+b∈Ad and(22)a+bd=a1du0a2+b4d,where (23)u=∑n=0∞a1dn+2b2a2+b4na2+b4π+∑n=0∞a1πa1nb2a2+b4dn+2-a1db2a2+b4d.Observe that since a1∈(pAp)-1, then a1π=0. Hence, the expression of u reduces to (24)u=∑n=0∞a1dn+2b2a2+b4na2+b4π-a1db2a2+b4d.From a2b4=0 we get a2b4d=a2b4(b4d)2=0. Hence (25)a2+b4π=p¯-a2+b4a2+b4d=p¯-a2+b4∑n=0∞b4dn+1a2n=p¯-b4b4d+b4d2a2+b4d3a22+⋯=p¯-b4b4d-b4b4d2a2+b4b4d3a22+⋯=b4π-b4da2+b4d2a22+⋯.
So, we get(26)u=∑n=0∞a1dn+2b2a2+b4nb4π-∑n=0∞∑k=0∞a1dn+2b2a2+b4nb4dk+1a2k+1-a1db2∑n=0∞b4dn+1a2n.Observe that (22) and a1d=ad yield(27)a+bd=ad+a2+b4d+u.
We will express(a2+b4)d using (21) in terms only of a and b. We have from (17) (28)aπbdn+1an=000p¯p0b2b4dn+20b4dn+1pa1n00a2np=000b4dn+1a2np=b4dn+1a2n.Hence from (21) we obtain(29)a2+b4d=aπ∑n=0∞bdn+1an.Also we have (we have written with an asterisk any entry whose exactly expression is not necessary)(30)adn+2ba+bnbπ=a1dn+2000p0b20b4pa1n∗0a2+b4npp-b2b4d0b4πp=0a1dn+2b2a2+b4nb4π00p=a1dn+2b2a2+b4nb4π,adbbdn+1an=a1d000p0b20b4p0∗0b4dn+1pa1n00a2np=0a1db2b4dn+1a2n00p=a1db2b4dn+1a2n,a1dn+2ba+bnbdk+1ak+1=a1dn+2000p0b20b4pa1n∗0a2+b4np0∗0b4dk+1pa1k+100a2k+1p=0a1dn+2b2a2+b4nb4dk+1a2k+100p=a1dn+2b2a2+b4nb4dk+1a2k+1.
From (27), (29), and (30), it follows (11). The inequality (12) trivially follows from (11).IfA is a Banach algebra, then we can define another multiplication in A by a⊙b=ba. It is trivial that (A,⊙) is a Banach algebra. If we apply Theorem 4 to this new algebra, we can immediately establish the following result.Theorem 5.
LetA be a Banach algebra and let a,b∈Ad. If there exist k∈N, k>1, such that bak=ba and aπb=b, then a+b∈Ad and(31)a+bd=ad+∑n=0∞anbdn+1aπ-∑n=0∞anbdn+1bad+bπ∑n=0∞a+bnbadn+2-∑n=0∞∑k=0∞ak+1bdk+1a+bnbadn+2,a+bd-ad≤∑n=0∞anbdn+1aπ+bad+bπ∑n=0∞a+bnbadn+2+∑n=0∞∑k=0∞ak+1bdk+1a+bnbadn+2.Observe that we can obtain paired results as Theorems4 and 5. We will not write explicitly these (trivially obtained) results.Theorem 6.
LetA be a Banach algebra and let a,b∈Ad be such that w=(a+b)aad∈Ad. If there exist k∈N, k>1, such that akb=ab and ba=ab2, then a+b∈Ad and(32)a+bd=wd+aπbd+∑n=0∞wdn+2baπa+bnbπ+∑n=0∞wπwnbaπbdn+2-wdbaπbd.Proof.
Let us consider the matrix representations ofa, ad, and b given in (4), (13), and (14) relative to the idempotent p=aad. As in the proof of Theorem 4, from akb=ab, we get a2b3=a2b4=0.
Fromba=ab2 we have (33)ab2=a1b12+a1b2b3a1b1b2+a1b2b4a2b3b1+a2b4b3a2b3b2+a2b42p=a1b12+a1b2b3a1b1b2+a1b2b400p,ba=b1a1b2a2b3a1b4a2p.Therefore, b3a1=0 and a2b4=b4a2=0. From Lemma 3 we get b3=0. Hence (34)b=b1b20b4p,a+b=a1+b1b20a2+b4p,w=a+baad=a1+b1b20a2+b4pp000p=a1+b1000p=a1+b1.
Lemma2 implies that a2+b4∈Ad because a∈Aqnil, b∈Ad, and a2b4=0. Also, we obtain(35)a2+b4d=∑n=0∞b4dn+1a2n.
In this situation, we obtain(36)a+bd=wdu0a2+b4dp=wd+u+a2+b4d,(37)u=∑n=0∞a1+b1dn+2b2a2+b4na2+b4π+∑n=0∞a1+b1πa1+b1nb2a2+b4dn+2-a1+b1db2a2+b4d.
Sincea2b4=b4a2=0, we have b4da2=(b4d)2b4a2=0. Thus (35) reduces to(38)a2+b4d=b4d=aπbd.From a2b4=0 we have a2b4d=a2b4(b4d)2=0. Therefore, (39)a2+b4π=p¯-a2+b4a2+b4d=p¯-a2b4d-b4b4d=b4π,so we get (40)u=∑n=0∞a1+b1dn+2b2a2+b4nb4π+∑n=0∞a1+b1πa1+b1nb2b4dn+2-a1+b1db2b4d.
We have(41)wdn+2baπa+bnbπ=a1+b1dn+2000pb1b20b4p000p¯pb1n∗0a2+b4npb1π∗0b4πp=0a1+b1dn+2b2a2+b4nb4π00=a1+b1dn+2b2a2+b4nb4π.
Also,(42)wπwnbaπbdn+2=a1+b1π00p¯pa1+b1n000pb1b20b4p000p¯pb1dn+2∗0b4dn+2p=0a1+b1πa1+b1nb2b4dn+200p=a1+b1πa1+b1nb2b4dn+2,wdbaπbd=a1+b1d000pb1b20b4p000p¯pb1d∗0b4dp=0a1+b1db2b4d00=a1+b1db2b4d.
From (36), (38), (41), and (42), it follows (32).
The proof is finished.As we have commented before, we can obtain a paired result by considering the Banach algebraA with the product a⊙b=ba. The key hypothesis of this new result will be bak=ba and b2a=ab.Theorem 7.
LetA be a Banach algebra and let a,b∈Ad be such that w=aad(a+b)∈Ad. If there exist k∈N, k>1, such that bak=ba and b2a=ab, then a+b∈Ad and (43)a+bd=wd+bdaπ-bdaπbwd+∑n=0∞bdn+2aπbwnwπ+bπ∑n=0∞a+bnaπbwdn+2.Theorem 8.
LetA be a Banach algebra and let a,b∈Ad be such that w=aad(a+b)∈Ad. If there exist k,n,m∈N, k>1, such that anb=bam and akb=ab, then a+b∈Ad and (44)a+bd=wd+∑n=0∞bdn+1anaπ.Proof.
Let us consider the matrix representations ofa, ad, and b given in (4), (13), and (14) relative to the idempotent p=aad. From anb=bam, we have(45)a1nb1=b1a1m,a1nb2=b2a2m,a2nb3=b3a1m,a2nb4=b4a2m.
The second equality of (45) implies a1nkb2=b2a2mk for any k∈N. By Lemma 3 we get b2=(ad)nkb2a2mk. Hence (46)b21/mk≤adnk1/mkb21/mka2mk1/mk.Therefore, if β denotes limk→∞b21/(mk) and α denotes limk→∞(ad)nk1/(mk), then 0≤β≤βαr(a2). Recall that a2 is quasinilpotent, and thus r(a2)=0. We have just proved β=0 or, equivalently, b2=0. Similarly we obtain b3=0.
Thus,b and a+b can be represented as(47)b=b100b4p,a+b=a1+b100a2+b4p.
Sinceakb=ab, by the proof of Theorem 4, we get a2b4=0. From Lemma 2, we have a2+b4∈Ad, because b4∈Ad, a2∈Aqnil, and a2b4=0. Also, Lemma 2 allows us to obtain (48)a2+b4d=∑n=0∞b4dn+1a2n.
We havea+b∈Ad if and only if a1+b1∈Ad. But w=aad(a+b)=a1+b1. If w∈Ad, then (49)a+bd=a1+b1d00a2+b4dp=wd+∑n=0∞bdn+1anaπ.The theorem is just proved.Theorem 9.
LetA be a Banach algebra and let a,b∈Ad. If there exist k∈N, k>1, such that akb=abaπ and ba=ab2, then a+b∈Ad and(50)a+bd=ad+aπbd+ad3ba+b,(51)a+bd-ad≤aπbd+ad3ba+b.Proof.
Let us consider the matrix representations ofa, ad, and b given in (4), (13), and (14) relative to the idempotent p=aad. Since akb=abaπ, (52)akb=a1kb1a1kb2a2kb3a2kb4p,abaπ=0a1b20a2b4p,we get a1kb1=0 and a2kb4=a2b4; as in the proof of Theorem 4, we have a2b4=0. From Lemma 3 we get b1=0.
Fromba=ab2 we have (53)ab2=a1b12+a1b2b3a1b1b2+a1b2b4a2b3b1+a2b4b3a2b3b2+a2b42p=a1b2b3a1b2b40a2b3b2p,ba=b1a1b2a2b3a1b4a2p=0b2a2b3a1b4a2p.Therefore, b3a1=0, b2a2=a1b2b4, and b4a2=a2b3b2. From Lemma 3 we get b3=0 and b4a2=a2b3b2=0. Hence (54)b=0b20b4p,a+b=a1b20a2+b4p.
Lemma2 implies that a2+b4∈Ad because a∈Aqnil, b∈Ad, and a2b4=b4a2=0; as in the proof of Theorem 6, we obtain(55)a2+b4d=b4d,a2+b4π=b4π.
In this situation, we obtain(56)a+bd=a1du0a2+b4dp=a1d+u+a2+b4d,(57)u=∑n=0∞a1dn+2b2a2+b4na2+b4π+∑n=0∞a1πa1nb2a2+b4dn+2-a1db2a2+b4d.Observe that since a1∈(pAp)-1, then a1π=0. Hence, the expression of u reduces to (58)u=∑n=0∞a1dn+2b2a2+b4nb4π-a1db2b4d.
Froma2b4=b4a2=0 and b2a2=a1b2b4, we have (59)a2+b4n=a2n+b4n,a1db2b4d=a1d2a1b2b4b4b4d3=a1d2b2a2b4b4d2=0.So we get (60)a1dn+2b2a2+b4nb4b4d=a1dn+2b2a2nb4b4d+a1dn+2b2b4nb4b4d=0,a1dn+2b2a2+b4n=a1dn+2b2a2n+a1dn+2b2b4n=a1dn+2b2a2a2n-1+a1dn+3a1b2b4b4n-1=a1dn+2a1b2b4a2n-1+a1dn+3b2a2b4n-1=0,n≥2.
Hence(61)u=∑n=0∞a1dn+2b2a2+b4nb4π-a1db2b4d=∑n=0∞a1dn+2b2a2+b4n-∑n=0∞a1dn+2b2a2+b4nb4b4d=a1d3b2a2+b4.
Thus (55) reduces to(62)a2+b4d=b4d=aπbd.
We have(63)ad3ba+b=a1d3000p0b20b4pa1b20a2+b4p=0a1d3b2a2+b400=a1d3b2a2+b4.
From (56), (62), and (63), it follows (50). The inequality (51) trivially follows from (50).
The proof is finished.As we have commented before, we can obtain a paired result by considering the Banach algebraA with the product a⊙b=ba. The key hypothesis of this new result will be bak=aπba and b2a=ab.Theorem 10.
LetA be a Banach algebra and let a,b∈Ad. If there exist k∈N, k>1, such that bak=aπba and b2a=ab, then a+b∈Ad and(64)a+bd=ad+bdaπ+a+bbad3,a+bd-ad≤bdaπ+a+bbad3.In the rest of the paper, we will use some weaker conditions than in [11, Theorem  4.1]. For example, if we assume that akbaπ=ab and aπb=ab2 instead of abaπ=0 and aπb=0, we will get a much simpler expression for (a+b)d.Theorem 11.
LetA be a Banach algebra and let a,b∈Ad. If there exist k∈N, k>1, such that akbaπ=ab and ab2=aπb, then a+b∈Ad and(65)a+bd=ad+∑n=0∞adn+2ban,a+bd-ad≤∑n=0∞adn+2ban.Proof.
Let us consider the matrix representations ofa, ad, and b given in (4), (13), and (14) relative to the idempotent p=aad. We will use the condition akbaπ=ab: since (66)akbaπ=0a1kb20a2kb4p,ab=a1b1a1b2a2b3a2b4p,we get a1b1=a2b3=0 and a2kb4=a2b4; as in the proof of Theorem 4, we have a2b4=0. From Lemma 3 we get b1=0.
Fromab2=aπb we have (67)ab2=a1b12+a1b2b3a1b1b2+a1b2b4a2b3b1+a2b4b3a2b3b2+a2b42p=a1b2b3a1b2b400p,aπb=00b3b4p.Therefore, b3=b4=0. Hence (68)b=0b200p,a+b=a1b20a2p,a+bd=a1du00p=a1d+u,where (69)u=∑n=0∞a1dn+2b2a2n=∑n=0∞adn+2ban.
The proof is finished.Theorem 12.
LetA be a Banach algebra and let a,b∈Ad. If there exist k∈N, k>1, such that akbaπ=ab and aπb=aba, then a+b∈Ad and(70)a+bd=ad+ad2b,a+bd-ad≤ad2b.Proof.
Let us consider the matrix representations ofa, ad, and b given in (4), (13), and (14) relative to the idempotent p=aad. As in the proof of Theorem 11, from akbaπ=ab, we get b1=0 and a2b3=a2b4=0.
Sinceaπb=aba, (71)ab=a1b1a1b2a2b3a2b4p=0a1b200p,aba=0a1b200pa100a2p=0a1b2a200p,aπb=00b3b4p,we get a1b2a2=b3=b4=0. As in the proof of Theorem 11, we have (72)a+bd=a1du00p=a1d+u,where (73)u=∑n=0∞a1dn+2b2a2n.From a1b2a2=0, we get (74)u=∑n=0∞a1dn+2b2a2n=a1d2b2=ad2b.
The proof is finished.



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