REPRESENTATIONS FOR THE GENERALIZED DRAZIN INVERSE IN A BANACH ALGEBRA (COMMUNICATED BY FUAD KITTANEH)

Bulletin of Mathematical Analysis an Applications ISSN: 1821-1291, UL: http://www.bmathaa.org Volume 5 Issue 1 (2013), ages 53-64 EESENTATIONS FO THE GENEALIZED DAZIN INVESE IN A BANACH ALGEBA (COMMUNICATED BY FUAD KITTANEH) J. BENÍTEZ 1, X LIU 2 AND Y. IN 2 Abstract. In this paper, we investigate aitive properties for the generalize Drazin inverse in a Banach algebra A. We give some representations for the generalize Drazin inverse of a + b, a an b are elements of A uner some new conitions, extening some known results. 1. Introuction The Drazin inverse has important applications in matrix theory an fiels such as statistics, probability, linear systems theory, ifferential an ifference equations, Markov chains, an control theory (1, 2, 11). In 9, Koliha extene the Drazin invertibility in the setting of Banach algebras with applications to boune linear operators on a Banach space. In this paper, Koliha was able to euce a formula for the generalize Drazin inverse of a+b when ab ba 0. The general question of how to express the generalize Drazin inverse of a+b as a function of a, b, an the generalize Drazin inverses of a an b without sie conitions, is very ifficult an remains open..e. Hartwig, G.. Wang, an Y. Wei stuie in 8 the Drazin inverse of a sum of two matrices A an B when AB 0. In the papers 3, 4, 5, 7, some new conitions uner which the generalize Drazin inverse of the sum a+b in a Banach algebra is explicitly expresse in terms of a, b, an the generalize Drazin inverses of a an b. In this paper we introuce some new conitions an we exten some known expressions for the generalize Drazin inverse of a+b, a an b are generalize Drazin invertible in a unital Banach algebra. Throughout this paper we will enote by A a unital Banach algebra with unity 1. Let A 1 an A qnil enote the sets of all invertible an quasinilpotent elements in A, respectively. Explicitly, A 1 {a A : x A : ax xa 1}, A qnil {a A : lim n + an 1/n 0}. 0 2000 Mathematics Subject Classification: 15A09, 46H05. Keywors an phrases. generalize Drazin inverse, Banach algebra. c 2012 Universiteti i rishtinës, rishtinë, Kosovë. Submitte Mar 25 2011. ublishe February 16, 2013. 53

54 J. BENÍTEZ1, X LIU 2 AND Y. IN 2 If B is a subalgebra of the unital algebra A, for an element b B 1, we shall enote by b 1 B the inverse of b in B. Let us observe that in general B 1 A 1 (for example, if p A is a nontrivial iempotent an B is the subalgebra pap, then the unity of B is p, an therefore, p B 1 \ A 1 ). Let a A, if there exists b A such that bab b, ab ba, a(1 ab) is nilpotent, (1.1) thenbisthedrazin inverseofa, enotebya D anitisunique. Ifthelastconition in (1.1) is replace by a(1 ab) is quasinilpotent, then b is the generalize Drazin inverse, enote by a an is also unique. Eviently aa is an iempotent, an it is customary to enote a π 1 aa. We shall enote A {a A : a }. In particular, if a(1 ab) 0 then b is calle the group inverse ofa. It was provein 9, Lemma 2.4 that a exists if an only if an only if exists an iempotent q A such that aq qa, aq is quasinilpotent, an a+q is invertible. The following simple remark will be useful. emark 1.1. If the subalgebra B A has unity, then B 1 A an if b B 1, then b b 1 B. In fact, let e be the unity of B, since bb 1 B b 1 B b e, it is easy to see bb 1 B b b, b 1 B bb 1 B b 1 B, an b 1 B b bb 1 B. Following 4, we say that {p 1,p 2,...,p n } is a total system of iempotents in A if p 2 i p i for all i, p i p j 0 if i j, an p 1 + +p n 1. Given a total system of iempotents in A, we consier the set M n (A, ) consisting of all matrices A a ij n i,j1 with elements in A such that a ij p i Ap j for all i,j {1,...,n}. Let us recall that p i Ap i is a subalgebra of A with unity p i. In 4, Lemma 2.1 it was prove that φ : A M n (A, ) given by p 1 xp 1 p 1 xp 2 p 1 xp n p 2 xp 1 p 2 xp 2 p 2 xp n φ(x)...... p n xp 1 p n xp 2 p n xp n is an isometric algebra isomorphism. In the sequel, we shall ientify x φ(x) for x A. Another useful (although trivial) ientity is x n p i xp j x A. i,j1 If a A is generalize Drazin invertible, then we have the following matrix representations: a1 0 a, a 0 a a1 1 pap 0 0 0, a 2 0 0 π 0 1 p, (1.2) p aa, {p,1 p}, a 1 pap 1, an a 2 (1 p)a(1 p) qnil. Let us remark that if a has the above representation, then a a 1 1 pap a 1. The following lemmas are neee in what follows.

EESENTATIONS FO THE GENEALIZED DAZIN INVESE 55 Lemma 1.1. Let {p,1 p} be a total system of iempotents in A an let a,b A have the following representation x 0 x t a, b. z y 0 y Then there exist (z n ) n0 (1 p)ap an (t n) n0 pa(1 p) such that a n x n 0 z n y n an b n x n t n 0 y n n N. The proof of this lemma is trivial by inuction an we will not give it. Lemma1.2. 4, Theorem3.3 Let b A, a A qnil, an let ab π a an b π ab 0. Then a+b A an (a+b) b + (b ) n+2 a(a+b) n. n0 The following Lemma is a generalization of Theorem 1 in 6. Although it was state for boune linear operators in a Banach space, its proof remains vali for Banach algebras. Lemma 1.3. Let a,b A such that ab ba. Then a + b A if an only if 1+a b A. In this case we have (a+b) a (1+a b)bb +b π ( b) n (a ) n+1 + (b ) n+1 ( a) n a π. n0 Lemma 1.4. 4, Example 4.5 Let a,b A be generalize Drazin invertible an ab 0, then a+b is generalize Drazin invertible an (a+b) b π b n (a ) n+1 + (b ) n+1 a n a π. n0 Lemma 1.5. 4, Theorem 2.3 Let x,y A, p an iempotent of A an let x an y have the representation a 0 b c x, y. (1.3) c b 0 a {p,1 p} {1 p,p} n0 n0 (i) If a pap an b (1 p)a(1 p), then x,y A an a x 0 b u b, y u 0 a {p,1 p} n0 {1 p,p} (1.4) u (b ) n+2 ca n a π + b π b n c(a ) n+2 b ca. (1.5) n0 (ii) If x A an a pap, then b (1 p)a(1 p), an x an y are given by (1.4) an (1.5). Lemma 1.6. 5, Lemma 2.1 Let a,b A qnil an let ab ba or ab 0, then a+b A qnil.

56 J. BENÍTEZ1, X LIU 2 AND Y. IN 2 2. Main results In this section, for a,b A, we will investigate some formulas of (a + b) in terms of a, b, a, an b. Theorem2.1. Let a,b A be generalize Drazin invertible an satisfying b π a π ba 0, b π aa baa 0, ab π a. Then n0 (a+b) b +u+b π v, v a + (a ) n+2 b(a+b) n, u (b ) n+2 a(a+b) n (a+b) π b av. roof. Letp bb an {p,1 p}. Letaanbhavethefollowingrepresentation b1 0 a3 a b, a 1, (2.1) 0 b 2 a 4 a 2 b 1 is invertible in pap an b 2 is quasinilpotent in (1 p)a(1 p). Since ab π a an ab π a3 a 1 0 0, a 4 a 2 0 1 p we have a 3 a 4 0. Hence b1 0 b, a 0 b 2 0 a1 0 a 2 n0, a+b b1 a 1 0 a 2 +b 2. (2.2) By observing the representation of a given in (2.2) an a simple appealing of Lemma 1.5 yiel a 0 a1 (a 2) 2 0 a (2.3) 2 an p 0 0 a π 1 aa a1 0 a1 (a 2 )2 p a1 a 0 1 p 0 a 2 0 a 2 2 0 a π 2 p. (2.4) To explain better the southeast block of a π in the above relation, let us permit say that a π 2 is efine as 1 a 2 a 2, the element 1 p a 2 a 2 a π 2 p belongs to (1 p)a(1 p), but a π 2 oes not nee belong to (1 p)a(1 p). In fact, since a π 2 p (1 p)a(1 p), one has (aπ 2 p)p p(aπ 2 p) 0, or equivalently, a π 2p pa π 2 p. In view of the last representation in (2.2), we shall apply Lemma 1.5 to fin an expression of (a + b). To this en, we nee prove b 1 pap an a 2 + b 2 (1 p)a(1 p). The fact b 1 pap follows from b A an the representation of b in (2.2), in fact, we have b b 1 1 pap b 1. We shall stuy (a 2 +b 2 ) in the following lines. Let us represent a 2 an b 2 as follows: a 2 a11 0 0 a 22, b 2 b11 b 12 b 21 b 22, (2.5) q a 2 a 2 an {q,1 p q} (which is a total system of iempotents in the subalgebra (1 p)a(1 p)). Observe that since q (1 p)a(1 p) an 1 p is the unity of (1 p)a(1 p), then q(1 p) (1 p)q q, or equivalently,

EESENTATIONS FO THE GENEALIZED DAZIN INVESE 57 qp pq 0. ecall that in the above representation of a 2 in (2.5), the element a 11 is invertible in qaq an a 22 is quasinilpotent. Since b π a π ba 0 an by (2.2), (2.4), we have 0 b π a π ba 0 0 0 1 p 0 0 0 (a π 2 p)b 2a 2 (a π 2 p)b 2a 2. p a1 a 2 0 a π 2 p b1 0 0 b 2 0 a1 0 a 2 But observe that b 2 (1 p)a(1 p), an thus, pb 2 0. Therefore, 0 a π 2 b 2a 2 hols. It seems that we can use (2.5) an 0 a π 2 b 2a 2 to get some information on b ij, but observe that we cannot represent a π 2 in the total system of iempotents since in general a π 2 / (1 p)a(1 p). To avoi this situation, let us efine {p,q,1 p q}, which in view of pq qp 0, it is trivial to see that is a total system of iempotents in A. Since a π 2 1 a 2a 2 1 q, 0 a π 2 b 2a 2 p 0 0 0 0 0 0 0 1 p q 0 0 0 0 0 0 0 b 21 a 11 b 22 a 22. 0 0 0 0 b 11 b 12 0 b 21 b 22 0 0 0 0 a 11 0 0 0 a 22 Thus, b 21 a 11 0. Since a 11 is invertible in qaq an b 21 (1 p q)aq (this last assertion follows from the representation of b 2 given in (2.5)), we get Let us calculate b π aa baa. b π aa baa 0 0 0 a1 a 2 0 1 p 0 a 2 a 2 0 0 0 a 2 a 2 b 2a 2 a. 2 b 21 0. (2.6) b1 0 0 b 2 0 a1 a 2 0 a 2 a 2 Thus, b π aa baa a 2 a 2 b 2a 2 a 2 qb 2q; hence representation (2.5) entails b 11 0. Since (2.6) hols, then b 2 a π 2 0 0 0 0 0 b 12 p 0 0 0 0 0 0 0 0 0 0 b 12 b 2. 0 0 b 22 0 0 1 p q 0 0 b 22 Thus, the following conitions (i) a 2 A (ii) b 2 is quasinilpotent (iii) b 2 a π 2 b 2 (iv) a π 2b 2 a 2 0. are satisfie. Hence, we can apply Lemma 1.2 to get an expression of (b 2 + a 2 ) obtaining (a 2 +b 2 ) a 2 + (a 2) n+2 b 2 (a 2 +b 2 ) n. n0

58 J. BENÍTEZ1, X LIU 2 AND Y. IN 2 By Lemma 1.5 applie to the representation of a+b given in (2.2) we obtain (a+b) b 1 u 0 (a 2 +b 2 ), (2.7) u (b 1) n+2 a 1 (a 2 +b 2 ) n (a 2 +b 2 ) π + b π 1b n 1a 1 (a 2 +b 2 ) n+2 b 1a 1 (a 2 +b 2 ). n0 n0 (2.8) ecall that b 1 b. Easily we have bb a a 1, bb b b 1, an b π b b 2. From (2.3) we get b π a a 2. By Lemma 1.1 an the representations of a+b an a given in (2.2) an (2.3), respectively, we have b π (a+b) k (a 2 +b 2 ) k, an b π (a ) k (a 2 )k for any positive integer k, Hence (a 2 +b 2 ) b π a + b π (a ) n+2 b π bb π (a+b) n. n0 This last expression can be simplifie by observing that b π bb π b π b an that by Lemma 1.1, there exists a sequence z n A such that (a ) n+2 b π 0 zn 0 0 0 zn 0 (a 2 )n+2 0 1 p 0 (a (a ) n+2. 2 )n+2 Therefore, (a 2 +b 2 ) b π (a + ) (a ) n+2 b(a+b) n. (2.9) n0 Now we will simplify the expression of u given in (2.8). Observe that for any n 0, one has n 0 (b 1) n+2 a 1 (b ) n+2 bb a (b ) n+2 a. (2.10) Moreover, by (2.2) we have (a+b) π 1 (a+b)(a+b) p 0 b1 a 1 b 1 u 0 1 p 0 a 2 +b 2 0 (a 2 +b 2 ) p b1 b 1 b 1 u a 1 (a 2 +b 2 ) 0 1 p (a 2 +b 2 )(a 2 +b 2 ) Thus, an by using Lemma 1.1 an (2.7), there exists a sequence (x n ) n1 in A such that b π (a+b) n (a+b) π 0 0 0 1 p b n 1 x n 0 (a 2 +b 2 ) n (a 2 +b 2 ) n (a 2 +b 2 ) π p,. p b1 b 1 b 1 u a 1 (a 2 +b 2 ) 0 (a 2 +b 2 ) π p but recall that a 2 + b 2 (1 p)a(1 p), an thus, if n > 0, then (a 2 + b 2 ) n (a 2 +b 2 ) π p (a 2 +b 2 ) n (a 2 +b 2 ) π. Thus n > 0 b π (a+b) n (a+b) π (a 2 +b 2 ) n (a 2 +b 2 ) π. (2.11) Now we can prove that (b 1) n+2 a 1 (a 2 +b 2 ) n (a 2 +b 2 ) π (b ) n+2 a(a+b) n (a+b) π (2.12)

EESENTATIONS FO THE GENEALIZED DAZIN INVESE 59 hols for any n N. Since, as is easy to see, ab π a, then we have for any n > 0 that (2.10) an (2.11) lea to (b 1) n+2 a 1 (a 2 +b 2 ) n (a 2 +b 2 ) π (b ) n+2 ab π (a+b) n (a+b) π (b ) n+2 a(a+b) n (a+b) π. Now, we will prove that (2.12) hols for n 0: (b ) 2 a(a+b) π (b 1 ) 2 0 0 a1 p b1 b 1 b 1 u a 1 (a 2 +b 2 ) 0 0 0 a 2 0 (a 2 +b 2 ) π p 0 (b 1 ) 2 a 1 (a 2 +b 2 ) π p (b 0 0 1 )2 a 1 (a 2 +b 2 ) π p. But observe that a 1 pa(1 p), an hence a 1 p 0. Thus, we have prove (b ) 2 a(a+b) π (b 1) 2 a 1 (a 2 +b 2 ) π. An thus, (2.12) hols for any n N. Now, we aregoing to simplify the expression n0 bπ 1b n 1a 1 (a 2 +b 2 ) n+2 appearing in (2.8). ecall that b π 1 bπ an a 1 bb a were obatine. Observe that if n > 0, then b π 1 bn 1 bπ b n 1 (1 p)bn 1 0 since b 1 pap. Furthermore, b π 1 a 1 b π bb a 0. Thus b π 1 bn 1 a 1(a 2 +b 2 ) n+2 0. (2.13) n0 From (2.9), b 1 b, a 1 bb a, an ab π a we get b 1a 1 (a 2 +b 2 ) ) ( ) b bb ab (a π + (a ) n+2 b(a+b) n b a a + (a ) n+2 b(a+b) (2.14) n. n0 n0 Now, (2.7), (2.8), (2.9), (2.12), (2.13), an (2.14) prove the theorem. emark 2.1. If b is group invertible, then the conition b π a π ba 0 implies b π aa baa 0. In fact, since bb π 0, then b π aa baa b π (1 a π )baa b π a π baa 0. emark 2.2. Theorem 2.1 extens Lemma 1.2. If b is quasinilpotent, then b 0 an b π 1. Notice that ba π b clearly implies aa baa 0. Theorem 2.2. Let a,b A be generalize Drazin invertible. Assume that b π ab b π ba an ab π a. Then (a+b) b + +b π ( 1) n (a ) n+1 b n + (b ) n+2 a(a+b) n (a+b) π b a ( 1) n (a ) n+1 b n. n0 n0 roof. As in the proof of Theorem 2.1, if we set p bb an by using ab π a, then the representations given in (2.2) are vali, {p,1 p}, b 1 is invertible in pap an b 2 is quasinilpotent. Since 0 0 0 b π a1 b1 0 0 0 ab, 0 1 p 0 a 2 0 b 2 0 a 2 b 2 b π ba 0 0 0 1 p b1 0 0 b 2 0 a1 0 a 2 n0 0 0 0 b 2 a 2,

60 J. BENÍTEZ1, X LIU 2 AND Y. IN 2 anb π ab b π ba,thena 2 b 2 b 2 a 2. Lemma1.3guaranteesthatb 2 +a 2 isgeneralize Drazin invertible if an only if 1+b 2a 2 is generalize Drazin invertible; but observe that b 2 is quasinilpotent, an therefore b 2 0. So, b 2 + a 2 is generalize Drazin invertible. Also, b π 2 1 b 2b 2 1 an Lemma 1.3 lea to (b 2 +a 2 ) (a 2) n+1 ( b 2 ) n. n0 By (2.3) an Lemma 1.1, there exist a sequence (x n ) n1 in pa(1 p) such that (a ) n x n +(a 2) n for any n N, thus b π (a ) n+1 b n 0 0 0 xn b n 1 0 0 1 p 0 (a 2) n+1 0 b n 2 0 0 0 (a 2 )n+1 b n (a 2) n+1 b n 2. 2 Thus, (a 2 +b 2 ) b π ( 1) n (a ) n+1 b n. (2.15) By employing Lemma 1.5 for the representation of a+b given in (2.2) we get n0 n0 (a+b) b 1 +(b 2 +a 2 ) +u, (2.16) u (b 1) n+2 a 1 (a 2 +b 2 ) n (a 2 +b 2 ) π + b π 1b n 1a 1 (a 2 +b 2 ) n+2 b 1a 1 (a 2 +b 2 ). n0 As in the proof of Theorem 2.1, we have that (2.12) an b π 1 bn 1 a 1 0 for any n 0 hol. Furthermore, since a 1 bb a an ab π a, then b 1 a 1(a 2 +b 2 ) b bb ab π ( 1) n (a ) n+1 b n b a ( 1) n (a ) n+1 b n. Therefore, n0 n0 u (b ) n+2 a(a+b) n (a+b) π b a ( 1) n (a ) n+1 b n. (2.17) n0 n0 Expressions (2.15), (2.16), an (2.17) permit finish the proof. Theorem 2.3. Let a, b A be generalize Drazin invertible. Assume that they satisfy aba 0 an ab 2 0. Then a+b A a π (a+b) A b π a π (a+b) A a π b π (a+b) A. Furthermore, if b π ab 0 or b π ba 0, or b π ab b π ba, then a+b A an (a+b) b +v +b π a +u+b π (a ) 2 b+ua b, v u b π a π b π (a+ba π ) n bb π (a ) n+2, n0 (b ) n+2 a(a+b) n 1 s(a+b) b aa, n0 s b π a +u+b π (a ) 2 b+ua b.

EESENTATIONS FO THE GENEALIZED DAZIN INVESE 61 roof. Since ab 2 0, then a an b have the matrix representation given in (2.2). Now, we use 0 aba. 0 a1 b1 0 0 a1 0 a1 b aba 2 a 2. 0 a 2 0 b 2 0 a 2 0 a 2 b 2 a 2 Thus a 1 b 2 a 2 a 2 b 2 a 2 0. Let q a 2 a 2 an {q,1 p q} (a total system of iempotents in the algebra (1 p)a(1 p)). We represent a11 0 b11 b a 2, b 0 a 2 12, (2.18) 22 b 21 b 22 a 11 is invertible in the subalgebra qaq an a 22 is quasinilpotent. We use a 2 b 2 a 2 0 a11 0 b11 b a 2 b 2 a 2 12 a11 0 a11 b 11 a 11 a 11 b 12 a 22 0 a 22 b 21 b 22 0 a 22 a 22 b 21 a 11 a 22 b 22 a 22 Thus, 0 a 11 b 11 a 11 an 0 a 11 b 12 a 22. The invertibility of a 11 in the subalgebra qaq an b 11 qaq ensure b 11 0. In a similar way we have b 12 a 22 0. Using ab 2 0 leas to a 2 b 2 2 0. Hence a 11 b 12 b 21 0 an a 11 b 12 b 22 0. The invertibility of a 11 (in qaq) leas to b 12 b 21 0 an b 12 b 22 0. Now let us efine 0 b12 a11 0 x an y (2.19) 0 0 b 21 a 22 +b 22 From (2.18) an b 11 0 one trivially gets a 2 + b 2 x + y. From b 12 b 21 0, b 12 a 22 0, an b 12 b 22 0 we have xy 0. Let us prove a+b A a 22 +b 22 (1 p q)a(1 p q). (2.20) : Assume that a+b A, then by the representations given in (2.2) we have that a 2 +b 2 (1 p)a(1 p), i.e., a 2 +b 2 x+y (1 p)a(1 p). We can apply Lemma 1.4 to y x+(x+y) because x A (since ( x) 2 0) an x(x+y) 0 obtaining y A. Lemma 1.5 an the representation of y in (2.19) ensure that a 22 +b 22 is generalize Drazin invertible. : Assume in this paragraph that a 22 + b 22 is generalize Drazin invertible. By recalling that a 11 is invertible in the subalgebra qaq, the representation of y in (2.19) leas to y (1 p)a(1 p). Since x 2 0, xy 0, an x + y a 2 +b 2, Lemma 1.4 yiels a 2 +b 2 (1 p)a(1 p). Now, Lemma 1.5 an the representation of a+b in (2.2) ensure that a+b A. Our next goal is to express the right sie of the equivalence (2.20) in terms of a an b. To this en, let us efine {p,q,1 q p}, which is easy to see that it is a total system of iempotents in A. First we notice that a π 2 (a 2 +b 2 ) p 0 0 0 0 0 0 0 1 p q 0 0 0 0 0 0 0 (1 p q)a 21 a 22 +b 22 0 0 0 0 a 11 +b 11 b 12 0 b 21 a 22 +b 22, (2.21) which, in view of Lemma 1.5, ensures that a 22 +b 22 is generalize Drazin invertible if an only if a π 2(a 2 +b 2 ) is generalize Drazin invertible. Now, we shall use (2.4),.

62 J. BENÍTEZ1, X LIU 2 AND Y. IN 2 b 1 pap, a 1 pa(1 p), an a 2,b 2 (1 p)a(1 p), a π p a1 a (a+b) 2 b1 a 1 b1 a 0 a π 2 p 1 a 1 a 2 (a 2 +b 2 ) 0 a 2 +b 2 0 a π 2 (a 2 +b 2 ) b π a π 0 0 b1 a (a+b) 1 a 1 a 2 (a 2 +b 2 ) 0 0 0 1 p 0 a π 2 (a 2 +b 2 ) 0 a π 2(a 2 +b 2 ) an a π b π (a+b) an appealing to Lemma 1.5, leas to p a1 a 2 0 0 0 a π 2 p 0 1 p 0 a1 a 2 (a 2 +b 2 ) 0 a π, 2(a 2 +b 2 ) b1 a 1 0 a 2 +b 2 a π 2(a 2 +b 2 ) A a π (a+b) A b π a π (a+b) A a π b π (a+b) A. We shall prove the secon part of the Theorem. Since b π 0 0 0 a1 b1 0 0 0 ab 0 1 p 0 a 2 0 b 2 0 a 2 b 2 an then b π ba 0 0 0 1 p b1 0 0 b 2 0 a1 0 a 2 0 0 0 b 2 a 2 a 2 b 2 b 2 a 2, b π ab 0 or b π ba 0 or b π ab b π ba a 2 b 2 0 or b 2 a 2 0 or a 2 b 2 b 2 a 2. The representations given in (2.18) lea to a 2 b 2 0 or b 2 a 2 0 or a 2 b 2 b 2 a 2 a 22 b 22 0 or b 22 a 22 0 or a 22 b 22 b 22 a 22. Since a 22 an b 22 are quasinilpotent, then the above implications an Lemma 1.6 lea to b π ab 0 or b π ba 0 or b π ab b π ba a 22 +b 22 is quasinilpotent. Inparticular, byemployingequivalence(2.20) wegetthat a+b A. Furthermore, by using Lemma 1.4, x 2 0, an xy 0, one gets (a 2 +b 2 ) (x+y) y π y n (x ) n+1 + (y ) n+1 x n x π y +(y ) 2 x. n0 From (2.19) an Lemma 1.5 we get y a 11 0 u (a 22 +b 22 ) So (y ) 2 x a 11 0 u 0 u n0 a 11 0 u 0, (a 22 +b 22 ) n b 21 (a 11 )n+2. (2.22) n0 a 11 0 u 0 0 b12 0 0, 0 (a 11 ) 2 b 12 0 ua. (2.23) 11b 12,

EESENTATIONS FO THE GENEALIZED DAZIN INVESE 63 In view of (2.2) an (2.18) it is simple to obtain b π a a 2 a 11 an bπ (a ) 2 (a 2) 2 (a 11) 2. We have a+ba π 0 a1 b1 0 p a1 a + 2 b1 a 0 a 2 0 b 2 0 a π 2 p 1 b 1 a 1 a 2 0 a 2 +b 2 a π. 2 By Lemma 1.1, there exists a sequence (w n ) n0 in A such that (a+ba π ) n b n 1 w n 0 (a 2 +b 2 a π 2) n an thus, b π (a+ba π ) n (a 2 +b 2 a π 2 )n. But another appealing to Lemma 1.1 an some computations as in (2.21) lea to (1 p q)(a 2 + b 2 a π 2 )n (a 22 + b 22 ) n. Observe that (2.4) yiels b π a π a π 2 p 1 q p. Thus b π a π b π (a + ba π ) n (a 22 +b 22 ) n. In view of (2.18) an b 11 0 we get b 21 b 2 q. But, it is simple to prove b π aa a 2 a 2 q an b 2 bb π. Hence b 21 bb π b π aa bb π aa. Moreover, (a 11) k (a 2) k b π (a ) k hols for any k N in view of Lemma 1.1. If we take into account that a b π a hols, then (2.22) becomes u b π a π b π (a+ba π ) n bb π a(a ) n+3 b π a π b π (a+ba π ) n bb π (a ) n+2. (2.24) n0 From b 11 0, (2.18), an a b π a we have b 12 qb 2 b π aa b π b b π aa b. This observation allows us to simplify the entries of (y ) 2 x given in (2.23): an Therefore, ua 11 b 12 n0 (a 11 )2 b 12 b π (a ) 2 b π aa b b π (a ) 2 b b b π a π b π (a+ba π ) n bb π (a ) n+2 π a b π aa b ua b. n0 (a 2 +b 2 ) y +(y ) 2 x a 11 +u+(a 11 )2 b 12 +ua 11 b 12 b π a +u+b π (a ) 2 b+ua b. (2.25) By Lemma 1.5, (a+b) b 1 v 0 (a 2 +b 2 ), (2.26) v (b 1) n+2 a 1 (a 2 +b 2 ) n a π 2 + b π 1b n 1a 1 (a 2 +b 2 ) n+2 b 1a 1 (a 2 +b 2 ). n0 n0 Since b 1 b, a 1 bb a, (a 2 + b 2 ) n b π (a + b) n, a 2,b 2 (1 p)a(1 p), a π 2 p+b π a π, ab π a, a b π a an ub π u (this last equality is obtaine from (2.24)) we have an (a 2 +b 2 ) π 1 (a 2 +b 2 ) (a 2 +b 2 ) 1 (b π a +u+b π (a ) 2 b+ua b)b π (a+b) 1 (b π a +u+b π (a ) 2 b+ua b)(a+b) (b 1) n+2 a 1 (a 2 +b 2 ) n (a 2 +b 2 ) π (b ) n+2 bb ab π (a+b) n (a 2 +b 2 ) π (b ) n+2 a(a+b) n (a 2 +b 2 ) π.

64 J. BENÍTEZ1, X LIU 2 AND Y. IN 2 Asiseasytosee,b π 1 bn 1 0foranyn 1. Moreover,bπ 1 a 1 b π (bb a) b π (1 b π )a 0, an b 1a 1 a 2 (b )(bb a)(b π a ) b aa. Thus, v reuces to v (b ) n+2 a(a+b) n 1 s(a+b) b aa, (2.27) n0 s b π a + u + b π (a ) 2 b + ua b. Expressions (2.25) (2.27) allow finish the proof. Acknowlegements The first author was supporte by Spanish roject MTM2010-18539. eferences 1 Ben-Israel, A. an Greville T.N.E.: Generalize Inverses: Theory an Applications, Wiley- Interscience New York, 1974, secon e. Springer, New York, 2002. 2 Campbell S.L. an Meyer C.D.: Generalize Inverses of Linear Transformations, itman, Lonon, 1979. 3 Castro González N.: Aitive perturbation results for the Drazin inverse, Linear Algebra Appl. Vol.397, (2005) 279-297. 4 Castro González N an Koliha J.J.: New aitive results for the g-drazin inverse, roceeings of the oyal Society of Einburgh: Section A Mathematics, Vol.134, (2004) 1085-1097. 5 Cvetković-Ilić D.S., Djorjević D.S. an Wei Y.: Aitive results for the generalize Drazin inverse in a Banach algebra, Linear Algebra Appl. Vol.418, (2006) 53-61. 6 Deng C. an Wei Y.: New aitive results for the generalize Drazin inverse, Math. Anal. Appl. Vol.370, (2010) 313-321. 7 Djorjević D.S. an Wei Y.: Aitive results for the generalize Drazin inverse, J. Aust. Math. Soc. Vol.73, (2002) 115-126. 8 Hartwig.E., Wang G.. an Wei Y.: Some aitive results on Drazin inverse, Linear Algebra Appl. Vol.322, (2001) 207-217. 9 Koliha J.J.: A generalize Drazin inverse, Glasg. Math. J. Vol.38, (1996) 367-381. 10 Martínez-Serrano M.F. an Castro-González N.: On the Drazin inverse of blocks matrices an generalize schur complement, App. Math. Comp. Vol.215, (2009) 2733-2740. 11 Yang H. an Liu X.: The Drazin inverse of the sum of two matrices an its applications, J. Comput. Appl. Math. Vol.235, (2011), 1412-1417. 1 Departamentoe MatemáticaAplicaa, Universia olitécnica e Valencia, Camino e Vera s/n. 46022, Valencia, España. E-mail aress: jbenitez@mat.upv.es 2 College of Mathematics an Computer Science, Guangxi University for Nationalities, Nanning 530006,.. China. E-mail aress: xiaojiliu72@yahoo.com.cn, yonghui1676@163.com