Additive results for the generalized Drazin inverse in a Banach algebra

Additive results for the generalized Drazin inverse in a Banach algebra Dragana S. Cvetković-Ilić Dragan S. Djordjević and Yimin Wei* Abstract In this aer we investigate additive roerties of the generalized Drazin inverse in a Banach algebra. We find some new conditions under which the generalized Drazin inverse of the sum a + b could be exlicitly exressed in terms of a a d b b d. Also some recent results of Castro and Koliha (Proc. Roy. Soc. Edinburgh Sect. 34 (2004) 1085-1097) are extended. 2000 Mathematics Subject Classification: 15A09 46H3046H05 15A33. Key words: Banach algebra; generalized Drazin inverse; Drazin inverse; additive result; 1 Introduction Let A be a comlex Banach algebra with the unit 1. By A 1 A nil A qnil we denote the sets of all invertible nilotent and quasinilotent elements in A resectively. Let us recall that the Drazin inverse of a A [1] is the element x A (denoted by a D ) which satisfies xax x ax xa a k+1 x a k (1) * Corresonding author. Suorted by Grant No. 1232 of the Ministry of Science Technology and Develoment Reublic of Serbia and Natural National Science Foundation of China under grant 10471027 1

for some nonnegative integer k. The least such k is the index of a denoted by ind(a). When ind(a) 1 then the Drazin inverse a D is called the grou inverse and it is denoted by a #. The conditions (1) are equivalent to xax x ax xa a a 2 x A nil. (2) The concet of the generalized Drazin inverse in a Banach algebra was introduced by Koliha [2]. The condition a a 2 x A nil was relaced by a a 2 x A qnil. Hence the generalized Drazin inverse of a is the element x A (written a d ) which satisfies xax x ax xa a a 2 x A qnil. (3) We mention that an alternative definition of the generalized Drazin inverse in a ring is also given in [3 4 5]. These two concets of generalized Drazin inverse are equivalent in the case when the ring is actually a comlex Banach algebra with a unit. It is well-known that a d is unique whenever it exists [2]. The set A d consists of all a A such that a d exists. For interesting roerties of Drazin inverse see [6 7 8]. Let a A and let A be a idemotent ( 2 ). Then we write and use the notations a a + a(1 ) + (1 )a + (1 )a(1 ) a 11 a a 12 a(1 ) a 21 (1 )a a 22 (1 )a(1 ). Every idemotent A induces a reresentation of an arbitrary element a A given by the following matrix [ ] a a(1 ) a11 a a 12. (4) (1 )a (1 )a(1 ) a 21 a 22 Let a π be the sectral idemotent of a corresonding to {0}. It is wellknown that a A d can be reresented in the following matrix form: a11 0 a 0 a 22 relative to aa d 1 a π where a 11 is invertible in the algebra A and a 22 is quasinilotent in the algebra (1 )A(1 ). Then the generalized Drazin inverse is given by a d a 1 11 0. 0 0 2

The motivation for this aer was the aer of Djordjević and Wei [9] and the aer of Castro and Koliha [10]. In both of these aers the conditions under which the generalized Drazin inverse (a + b) d could be exressed in terms of a a d b b d were considered. In [9] this roblem is investigated for a bounded linear oerator on an arbitrary comlex Banach sace under assumtion that AB 0 and these results are the generalizations of the results from [11] where the same roblem was considered for matrices. Castro and Koliha [10] considered the same roblem for the elements of the Banach algebra with unit under some weaker conditions. They generalized the results from [9]. In the resent aer we investigate additive roerties of the generalized Drazin inverse in a Banach algebra and find an exlicit exression for the generalized Drazin inverse of the sum a + b under various conditions. In the first art of the aer we find some new conditions which are nonequivalent to the conditions from [10] allowing for the generalized Drazin inverse of a + b to be exressed in terms of a a d b b d. It is interesting to note that in some cases we obtain the same exression for (a + b) d as in [10]. In the rest of the aer we generalize recent results from [10]. 2 Results First we state the following result which is roved in [12] for matrices extended in [13] for a bounded linear oerator and in [10] for arbitrary elements in a Banach algebra. Theorem 2.1 Let x A and let x relative to the idemotent A. [ a c 0 b (1) If a (A) d and b ((1 )A(1 )) d then x is generalized Drazin invertible and x d a d u 0 b d (5) where u (a d ) n+2 cb n b π + a π a n c(b d ) n+2 a d cb d. (2) If x A d and a (A) d then b ((1 )A(1 )) d and x d is given by (5). 3 ]

Now we state an auxiliary result. Lemma 2.1 Let a b A qnil and let ab ba or ab 0 then a + b A qnil. Proof. If ab ba we have that r(a + b) r(a) + r(b) which gives that a + b A qnil. The case when ab 0 follows from the equation (λ a)(λ b) λ(λ (a + b)) Considering the revious lemma the first idea was to relace the basic condition ab 0 which was used in the aers [11] [9] by the condition ab ba. As we exected this condition wasn t enough to derive a formula for (a + b) d. Hence to this aim we assume the following three conditions for a b A d : a ab π b π ba π b π b and b π a π ba b π a π ab. (6) Instead of the condition ab ba we assume the weaker condition b π a π ba b π a π ab. Notice that a ab π ab d 0 Aa Ab π (7) b π ba π b π b b π ba d 0 Ab π b Aa π (8) b π a π ba b π a π ab (ba ab)a (b π a π ) (9) where for u A u {x A : ux 0}. For matrices and bounded linear oerators on a Banach sace the conditions (7) (8) (9) are equivalent to N (b π ) N (a) N(a π ) N (b π b) R(ba ab) N (b π a π ). Remark that conditions (6) are not symmetric in a b like the conditions (3.1) from [10] so our exression for (a + b) d is not symmetric in a b at all. In the next theorem under the assumtion that for a b A d the conditions (6) hold we offer the following exression for (a + b) d. 4

Theorem 2.2 Let a b A d be such that (6) is satisfied. Then a + b A d and (a + b) d (b d + (b d ) n+2 a(a + b) n )a π (b d ) n+2 a(a + b) n (a d ) k+2 b(a + b) k+1 k0 + (b d ) n+2 a(a + b) n a d b b d a(a d ) n+2 b(a + b) n (10) Before roving Theorem 2.2 we have to rove the following result which is a secial case of this theorem: Theorem 2.3 Let a A qnil b A d are such that b π ab b π ba and a ab π. Then (6) is satisfied a + b A d and (a + b) d b d + (b d ) n+2 a(a + b) n. (11) Proof. First suose that b A qnil. Then b π 1 and from b π ab b π ba we obtain that ab ba. Using Lemma 2.1 a + b A qnil and (11) holds. Now we assume that b is not quasinilotent and we consider the matrix reresentation of a and b relative to the 1 b π. We have b1 0 a11 a b a 12 0 b 2 a 21 a 22 where b 1 (A) 1 and b 2 ((1 )A(1 )) qnil A qnil. From a ab π it follows that a 11 0 and a 21 0. We denote a 1 a 12 and a 2 a 22. Hence b1 a a + b 1. 0 a 2 + b 2 The condition b π ab b π ba imlies that a 2 b 2 b 2 a 2. Hence using Lemma 2.1 we get a 2 + b 2 ((1 )A(1 )) qnil. Now by Theorem 2.1 we obtain that a + b A d and (a + b) d [ b 1 1 b (n+2) 1 a 1 (a 2 + b 2 ) n 0 0 b d + (b d ) n+2 a(a + b) n 5 ]

Let us observe that the exressions for (a + b) d in (11) and in (3.6) Theorem 3.3 [10] are exactly the same. If we assume that ab ba instead of b π ab b π ba we will get a much simler exression for (a + b) d. Corollary 2.1 Let a A qnil b A d are such that ab ba and a ab π then a + b A d and (a + b) d b d. Proof. From the condition a ab π as we mentioned before it follows that ab d 0. Now because the Drazin inverse b d is double commutant of a we have that (b d ) n+2 a(a + b) n a(b d ) n+2 (a + b) n 0 Proof of the Theorem 2.2: If b is quasinilotent we can aly Theorem 2.3. Hence we assume that b is neither invertible nor quasinilotent and consider the following matrix reresentation of a and b relative to the 1 b π : b1 0 a11 a b a 12 0 b 2 a 21 a 22 where b 1 (A) 1 and b 2 ((1 )A(1 [ )) qnil. ] As in the roof of 0 Theorem 2.3 from a ab π a1 it follows that a and 0 a 2 a + b [ b1 a 1 0 a 2 + b 2 From the conditions b π a π ba b π a π ab and b π ba π b π b we obtain that a π 2 b 2a 2 a π 2 a 2b 2 and b 2 b 2 a π 2. Now by Theorem 2.3 it follows that (a 2 + b 2 ) ((1 )A(1 )) d and (a 2 + b 2 ) d a d 2 + (a d 2) n+2 b 2 (a 2 + b 2 ) n. (12) By Theorem 2.1 we get (a + b) d ] [ b 1 1 u 0 (a 2 + b 2 ) d 6. ]

where u b (n+2) 1 a 1 (a 2 + b 2 ) n (a 2 + b 2 ) π b 1 1 a 1(a 2 + b 2 ) d and by b 1 1 we denote the inverse of b 1 in the algebra A. Using (12) we have that u b (n+2) 1 a 1 (a 2 + b 2 ) n a π 2 b (n+2) 1 a 1 (a 2 + b 2 ) n a d 2b 2 (b 1 ) (n+2) a 1 (a 2 + b 2 ) n (a d 2) k+2 b 2 (a 2 + b 2 ) k+1 b 1 1 a 1a d 2 k0 b 1 1 a 1(a d 2) n+2 b 2 (a 2 + b 2 ) n By a straightforward comutation we obtain that (10) holds. Corollary 2.2 Let a b A d are such that ab ba a ab π and b π ba π b π b then a + b A d and (a + b) d b d. Let us also observe that if a b are such that a is invertible and b is grou invertible than the conditions (8) and (9) are satisfied so we have to assume just that a ab π. In the oosite case when b is invertible we get a 0. As we mentioned before Hartwig et al. in [11] for matrices and Djordjević and Wei [9] for oerators used the condition AB 0 to derive the formula (a + b) d. Castro and Koliha [10] relaxed this hyothesis by assuming the following three conditions symmetric in a b A d a π b b ab π a b π aba π 0. (13) It is easy to see that ab 0 imlies (13) but the converse is not true (see Ex. 3.1 [10]). It is interesting to remark that the conditions (13) and (6) are indeendent neither of them imlies the other one but in some cases we obtain the same exressions for (a + b) d. If we consider the algebra A of all comlex 3 3 matrices and a b A which are given in the Examle 3.1 [10] we can see that the conditions (13) are satisfied but the conditions (6) are not satisfied. In the following examle we have the oosite case. We construct matrices a b in the algebra A of all comlex 3 3 matrices such that (6) is satisfied but (13) is not satisfied. If we assume that ab ba in Theorem 2.2 the exression for 7

(a + b) d will be exactly the same as in the Theorem 3.5 [10] (in this aer Corollary 2.4). Examle. Then Let a 1 0 0 0 0 0 0 0 0 a π b 0 0 0 0 1 0 0 0 1 0 1 0 0 0 0 0 0 0 and b π 1. Now we can see that a ab π a π ab a π ba and ba π b i.e. (6) is satisfied. Also a π b 0 b so (13) is not satisfied. In the rest of the aer we will resent a generalization of the results from [10]. We will use some weaker conditions than in [10]. For examle in the next theorem which is the generalization of Theorem 3.3 [10] we will assume that e (1 [ b π )(a + b)(1 ] b π ) A d instead of ab π a. If ab π a then e (1 b π b1 0 )b for 1 b 0 0 π and e d b d. Theorem 2.4 Let b A d a A qnil be such that. e (1 b π )(a + b)(1 b π ) A d and b π ab 0 then a + b A d and (a + b) d e d + (e d ) n+2 ab π (a + b) n. Proof. The case when b A qnil follows from Lemma 2.1. Hence we assume that b is not quasinilotent b1 0 a11 a b and a 12 0 b 2 a 21 a 22 where 1 b π. From b π ab 0 we have that b π a(1 b π ) 0 i.e. a 21 0. Denote a 1 a 11 a 22 a 2 and a 12 a 3. Then a1 + b a + b 1 a 3. 0 a 2 + b 2 8

Also b π ab 0 imlies that a 2 b 2 0 so a 2 + b 2 ((1 )A(1 )) qnil according to Lemma 2.1. Now alying Theorem 2.1 we obtain that (a + b) d (a1 + b 1 ) d u 0 0 where u ((a 1 + b 1 ) d ) (n+2) a 3 (a 2 + b 2 ) n. By a direct comutation we verify that (a + b) d e d + (e d ) n+2 ab π (a + b) n. Now as a corollary we obtain Theorem 3.3 from [10]. Corollary 2.3 Let b A d a A qnil and let ab π a b π ab 0. Then a + b A d and (a + b) d b d + (b d ) n+2 a(a + b) n. The next result is a generalization of Theorem 3.5 in [10]. For simlicity we use the following notation: e (1 b π )(a + b)(1 b π ) A d f (1 a π )(a + b)(1 a π ) (1 a π )A(1 a π ) A 2 (1 b π )A(1 b π ) where a b A d are given. We also rove the next result which is the generalization of Theorem 3.5 [10]. Theorem 2.5 Let a b A d be such that (1 a π )b(1 a π ) A d f A 1 1 and e A d 2. If (1 a π )ba π 0 b π aba π 0 a π a(1 b π )a π 0 then a + b A d and (a + b) d (b d + (b d ) n+2 a(a + b) n )a π + (b d ) k+1 a(a + b) n+k a π b(f) (n+2) k0 (b d ) n+2 a(a + b) n a π b(f) 1 + (f) 1 9 b π (a + b) n a π b(f) (n+2) b d a π b(f) 1

where by (f) 1 we denote the inverse of f in. Proof. Obviously if a is invertible then the statement of the theorem holds. If a is quasinilotent then the result follows from Theorem 2.4. Hence we assume that a is neither invertible nor quasinilotent. As in the roof of Theorem 2.2 we have that a1 0 b11 b a b 12 0 a 2 b 21 b 22 where 1 a π a 1 (A) 1 and a 2 ((1 )A(1 )) qnil. From (1 a π )ba π 0 we have that b 12 0. Denote b 1 b 11 b 22 b 2 and b 21 b 3. Then a1 + b a + b 1 0. b 3 a 2 + b 2 The condition a π b π aba π 0 exressed in the matrix form yields a π b π aba π 0 0 a1 0 0 0 0 b π 2 0 a 2 0 b 2 0 0 0 0 0 b π 2 a. 2b 2 0 0 Similarly a π a(1 b π ) 0 imlies that a 2 b π 2 a 2. From Corollary 2.3 we get that a 2 + b 2 A d and (a 2 + b 2 ) d b d 2 + (b d 2) n+2 a 2 (a 2 + b 2 ) n. where Now using Theorem 2.1 we obtain that a + b A d and (a + b) d (a1 + b 1 ) d 0 u (a 2 + b 2 ) d u k0 b π 2 (a 2 + b 2 ) n b 3 (f) (n+2) (b d 2) k+1 a 2 (a 2 + b 2 ) n+k b 3 (f) (n+2) (b d 2) n+2 a 2 (a 2 + b 2 ) n b 3 (f) 1 10. b d 2b 3 (f) 1

By a straightforward comutation we obtain that the result holds. Corollary 2.4 Let a b A d satisfy the conditions (13). Then a + b A d and (a + b) d (b d + (b d ) n+2 a(a + b) n )a π + b π (a + b) n b(a d ) (n+2) (b d ) k+1 a(a + b) n+k b(a d ) (n+2) + b π a d k0 (b d ) n+2 a(a + b) n ba d Proof. We have that f (1 a π )a so (f) 1 a d. The authors would like to thank the referees for their helful comments and suggestions that hel to imrove this aer. References [1] M.P. Drazin Pseudoinverse in associative rings and semigrous Amer. Math. Monthly 65 (1958) 506 514. [2] J.J. Koliha A generalized Drazin inverse Glasgow Math. J. 38 (1996) 367-381. [3] R.E. Harte Sectral rojections Irish Math. Soc. Newsletter 11 (1984) 10 15. [4] R.E. Harte Invertibility and Singularity for Bounded Linear Oerators New York Marcel Dekker 1988. [5] R.E. Harte On quasinilotents in rings PanAm. Math. J. 1 (1991) 10 16. [6] N. Castro González J.J. Koliha V. Rakocevic Continuity and general erturbation of the Drazin inverse for closed linear oerators Abstract and A. Anal. 7 (2002) 335 347. [7] J.J. Koliha V. Rakocevic Holomorhic and meromorhic roerties of the g-drazin inverse Demonstratio Mathematica 38 (2005) 657-666. 11

[8] J.J. Koliha V. Rakocevic Differentiability of the g-drazin inverse Studia Mathematica 168 (2005) 193-201. [9] D.S. Djordjević Y. Wei Additive results for the generalized Drazin inverse J. Austral. Math. Soc. 73 (2002) 115 125. [10] N. Castro González J.J. Koliha New additive results for the g-drazin inverse Proc. Roy. Soc. Edinburgh Sect. 34 (2004) 1085-1097. [11] R.E. Hartwig G. Wang Y. Wei Some additive results on Drazin inverse Linear Algebra Al. 322 (2001) 207 217. [12] C.D. Meyer Jr. N.J. Rose The index and the Drazin inverse of block triangular matrices SIAM J. Al. Math. 33(1) (1977) 1 7. [13] D.S. Djordjević P.S. Stanimirović On the generalized Drazin inverse and generalized resolvent Czechoslovak Math. J. 51(126) (2001) 617 634. Address: Dragana S. Cvetković-Ilić: Deartment of Mathematics Faculty of Sciences University of Niš P.O. Box 224 Višegradska 33 18000 Niš Serbia E-mail: dragana@mf.ni.ac.yu gagamaka@tt.yu Dragan S. Djordjević: Deartment of Mathematics Faculty of Sciences University of Niš P.O. Box 224 Višegradska 33 18000 Niš Serbia E-mail: dragan@mf.ni.ac.yu ganedj@eunet.yu Yimin Wei: School of Mathematical Sciences Fudan University Shanghai 200433 P.R. China and Key Laboratory of Mathematics for Nonlinear Sciences (Fudan University) Ministry of Education. E-mail: ymwei@fudan.edu.cn 12