An element a in a Banach algebra A has p-Drazin inverse provided that there exists b2 comm(a) such that b = b2a, ak ↋ ak+1b 2 J(A) for some k 2 N. In this paper, we present new conditions for a block operator matrix to have p-Drazin inverse. As applications, we prove the p-Drazin invertibility of the block operator matrix under certain spectral conditions. © 2020, University of Nis. All rights reserved

Calci, Tuğçe Pekacar

Chen, Huanyin

Köse, Handan

Zou, Honglin

English

Kırşehir Ahi Evran University Institutional Repository

Filomat 34:14 (2020), 4597–4605https://doi.org/10.2298/FIL2014597CPublished by Faculty of Sciences and Mathematics,University of Niš, SerbiaAvailable at: http://www.pmf.ni.ac.rs/filomatThe p-Drazin Inverse for Operator Matrix over Banach AlgebrasHuanyin Chena, Honglin Zoub, Tugce Pekacar Calcic, Handan KosedaDepartment of Mathematics, Hangzhou Normal University, Hangzhou, ChinabDepartment of Mathematics, Hubei Normal University, Huangshi, ChinacDepartment of Mathematics, Ankara University, Ankara, TurkeydDepartment of Mathematics, Ahi Evran University, Kirsehir, TurkeyAbstract. An element a in a Banach algebraA has p-Drazin inverse provided that there exists b ∈ comm(a)such that b = b2a, ak − ak+1b ∈ J(A) for some k ∈ N. In this paper, we present new conditions for a blockoperator matrix to have p-Drazin inverse. As applications, we prove the p-Drazin invertibility of the blockoperator matrix under certain spectral conditions.1. IntroductionLet A be a Banach algebra with an identity. The commutant of a ∈ A is defined by comm(a) = {x ∈A | xa = ax}. An element a in a Banach algebraA has p-Drazin inverse provided that there exists b ∈ comm(a)such that b = b2a, ak − ak+1b ∈ J(A) for some k ∈ N. The preceding b is unique if exists, and we denote itby a‡. We refer the reader to [11, 13, 17, 18] and [20] for more properties of p-Drazin inverse in a Banachalgebra.Recall that a ∈ A has g-Drazin inverse provided that there exists b ∈ comm(a) such that b = b2a, a − a2b ∈Aqnil (see [6]). More results on g-Drazin inverse can be found in [1–5, 10, 19, 21]. As is well known, everyp-Drazin inverse is just the g-Drazin inverse. The p-Drazin inverse should be expressed as that of theg-Drazin inverse if exists. We will suffice to investigate the existence for p-Drazin inverse. This motivatesus to present new conditions for a block operator matrix to have p-Drazin inverse.Let M =(a bc d)∈M2(A). Let a, d ∈ A‡. Ifbd‡ = 0 or d‡c = 0, and bdic = 0 for all i ≥ 0,in Section 2, we prove that M has p-Drazin inverse.In Section 3, we determine the p-Drazin invertibility of the block operator matrix M under certainspectral conditions. If (bc)πabc = 0 or bca(bc)π = 0, and bd = 0, then M has p-Drazin inverse.2010 Mathematics Subject Classification. 15A09, 47A11, 16U99Keywords. p-Drazin inverse; operator matrix; Banach algebra.Received: 24 January 2020; Accepted: 20 April 2020Communicated by Dragana S. Cvetković-IlićCorresponding author: Tugce Pekacar CalciResearch supported by the Natural Science Foundation of Zhejiang Province, China (No. LY21A010018). Thethird author thanks the Scientific and Technological Research Council of Turkey (TUBITAK) for the financial support.Email addresses: huanyinchen@aliyun.com (Huanyin Chen), honglinzou@163.com (Honglin Zou), tcalci@ankara.edu.tr (TugcePekacar Calci), handankose@gmail.com (Handan Kose)H. Chen et al. / Filomat 34:14 (2020), 4597–4605 4598Throughout the paper, all Banach algebras are complex with an identity. We use J(A) denotes theJacobson radical of A.√J(A) stands for the radical of J(A), i.e.,√J(A) = {x | xm ∈ J(A) for some m ∈ N}.A‡ denotes the set of all elements having p-Drazin inverses in A. For any a ∈ A‡, we use aπ to stand forthe spectral idempotent 1 − aa‡.2. 2 × 2 Operator matricesLet M =(a bc d)∈ M2(A). The aim of this section is to determine when M has p-Drazin inverseunder certain conditions and generalize [9, Theorem 3.2] from g-Drazin inverse to p-Drazin inverse. Thefollowing lemmas are crucial.Lemma 2.1. Let a, b ∈ A and a, b ∈√J(A). If abka = 0 for any k ∈N, then a + b ∈√J(A).Proof. Let am, bn ∈ J(A). Assume that t = m + n. Then at, bt ∈ J(A). Let s = 3t + 1. Then every term of (a + b)sshould beai1 b j1 ai2 b j2 · · · ais b js ,where i1, j1, · · · , is, js ≥ 0, i1 + j1 + · · · + is + js = s. If the term contains abka(k ∈ N), then it is zero. So thenonzero terms should be written in the form bkalbr(k + l + r = s). Then k or l or r is greater than t. Hence,bkalbr ∈ J(A). Therefore (a + b)s ∈ J(A), as asserted.Lemma 2.2. Suppose that a, d ∈√J(A). If bdic = 0 for all i ≥ 0, then M ∈√J(M2(A)).Proof. Let am, dn ∈ J(A) and M = P + Q, where P =(a bc 0), and Q =(0 00 d).Since bc = 0, we see that (a 1bc 0)m+1=(a 10 0)m+1∈ J(M2(A)),and soPm+2 =[ ( a 1c 0) (1 00 b) ]m+2=(a 1c 0) (a 1bc 0)m+1 (1 00 b)∈ J(M2(A)).Clearly, Qn ∈ J(M2(A)). We easily check thatPQkP = 0for any k ∈N. Hence, M = P + Q ∈√J(A) by Lemma 2.1.Lemma 2.3. Let A be a Banach algebra. If a ∈ A‡, d ∈√J(A) and bdic = 0 for all i ≥ 0, then M has p-Drazininverse.Proof. Let N =(a‡ γδ δaγ), whereγ =∞∑i=0(a‡)i+2bdi, δ =∞∑i=0dic(a‡)i+2.H. Chen et al. / Filomat 34:14 (2020), 4597–4605 4599By hypothesis, we haveγdic = 0, bdiδ = 0, γdiδ = 0for any i ≥ 0. We shall prove that N = M‡.Step 1. MN = NM. We compute thatMN =(aa‡ aγca‡ + dδ cγ + dδaγ),NM =(a‡a a‡b + γdδa δb + δaγd).As in the proof of [9, Lemma 3.1], we easily check that MN = NM.Step 2. N = MN2. We haveI −MN =(aπ −aγ−ca‡ − dδ 1 − cγ − dδaγ).As γc = γδ = 0, we haveN(I −MN) =(0 γ − a‡aγδaπ 0)= 0.Hence, N = MN2.Step 3. M −M2N ∈√J(A). Since bdic = 0 for all i ≥ 0, we easily verify thatM(I −MN) =(aaπ b − a2γcaπ − dca‡ − d2δ d − σ),where σ = caγ + dcγ + d2δaγ.Clearly, aaπ ∈√J(A). By hypothesis, we see that σdiσ = 0 for all i ≥ 0. Hence, σ2 = 0, and sod,−σ ∈√J(A). In view of Lemma 2.1, we see that d − σ ∈√J(A). Moreover, we have(b − a2γ)(d − σ)m(caπ − dca‡ − d2δ) = 0for all m ≥ 0. Therefore M −M2N ∈√J(A) by Lemma 2.2.Therefore N = M‡, as asserted.We have accumulated all information necessary to prove the following.Theorem 2.4. Let a, d ∈ A‡. Ifbd‡ = 0, bdic = 0 for all i ≥ 0,then M has p-Drazin inverse.Proof. Clearly, M = P + Q, whereP =(a bc ddπ), Q =(0 00 d2d‡).Obviously, Q has p-Drazin inverse. Clearly, bc = 0. For any k ≥ 0, we have b(ddπ)kc = bdkdπc = bdkc = 0,and so b(ddπ)kc = 0 for all k ≥ 0. Clearly, ddπ = d − d2d‡ ∈√J(A). In light of Lemma 2.3, P has p-Drazininverse. On the other hand,PQ =(a bc ddπ) (0 00 d2d‡)= 0.Therefore M has p-Drazin inverse, by [13, Theorem 5.4].H. Chen et al. / Filomat 34:14 (2020), 4597–4605 4600Corollary 2.5. Let a, d ∈ A‡. Ifca‡ = 0, caib = 0 for all i ≥ 0,M has p-Drazin inverse.Proof. In view of Theorem 2.4, the matrix(d cb a)has p-Drazin inverse. It is easy to check that(a bc d)=(0 11 0) (d cb a) (0 11 0).Since(0 11 0)2= I2, we easily obtain the result.Theorem 2.6. Let a, d ∈ A‡. Ifd‡c = 0, bdic = 0 for all i ≥ 0,M has p-Drazin inverse.Proof. Clearly, M = P + Q, whereP =(0 00 d2d‡), Q =(a bc ddπ).Obviously, P has p-Drazin inverse. As in the proof of Theorem 2.4 we easily check that b(ddπ)kc = bdkdπc =bdkc = 0 for any k ≥ 0. In view of Lemma 2.3, Q has p-Drazin inverse. By hypothesis, we see thatPQ =(0 00 d2d‡) (a bc ddπ)= 0.According to [13, Theorem 5.4], M has p-Drazin inverse, as asserted.Corollary 2.7. Let a, d ∈ A‡. Ifa‡b = 0, caib = 0 for all i ≥ 0,M has p-Drazin inverse.Proof. In view of Theorem 2.6, the matrix(d cb a)has p-Drazin inverse. As in the proof Corollary 2.5, weeasily obtain the result.3. Spectral conditionsIn this section we apply the preceding results and demonstrate the p-Drazin invertibility of the blockmatrix M under certain spectral conditions. We now deriveLemma 3.1. Let a, d ∈ A‡. If abc = 0, bd = 0 and bc ∈√J(A), then M =(a bc d)∈M2(A)‡.Proof. Clearly, we haveM2 =(a2 + bc abca + dc cb + d2).H. Chen et al. / Filomat 34:14 (2020), 4597–4605 4601Clearly, a2 has p-Drazin inverse. By hypothesis, bc has p-Drazin inverse. Since a2(bc) = 0, it follows from [13,Theorem 5.4] that a2 + bc has p-Drazin inverse. In light of [13, Theorem 3.6], cb has p-Drazin inverse. Thenwe easily see that cb + d2 has p-Drazin inverse as (cb)d2 = 0. It is easy to verify thatab(cb + d2)‡ = ab(cb + d2)((cb + d2)‡)2= 0;ab(cb + d2)i(ca + dc) = 0.In view of Theorem 2.4, M2 has p-Drazin inverse. Therefore M has p-Drazin inverse by [20, Lemma 2.8].We come now to generalize [16, Theorem 3.1 and Corollary 3.3] from the generalized Drazin inverse tothe p-Drazin inverse.Theorem 3.2. Let a, d, bc, (bc)πa ∈ A‡. If (bc)πabc = 0 and bd = 0, then M =(a bc d)∈M2(A)‡.Proof. Step 1. Let h = bc,N =(a 1h 0)and e =(hπ 00 0). Then N =(a′ b′c′ d′)e, wherea′ = eNe, b′ = eN(I − e), c′ = (I − e)Ne, d′ = (I − e)N(I − e).Since (bc)πabc = 0, we have hπah = 0, we easily check thata′ =(hπa 00 0), b′ =(0 hπ0 0),c′ =(hh‡ahπ 0hhπ 0), d′ =(ahh‡ hh‡h2h‡ 0).Since hπahh‡ = (bc)πabc(bc)‡ = 0, it follows by [16, Lemma 2.2] that (hπa)d = hπad. This shows that(hπa)(hπad) = (hπad)(hπa),hπad = (hπad)(hπa)(hπad).Since a ∈ A‡, we have ak − ak+1ad ∈ J(A) for some k ∈N. Then we verify that[hπa − (hπa)2(hπad)]k= (hπa)k[1 − (hπa)(hπad)]= (hπa)k − (hπa)k+1ad= hπak − hπak+1ad= hπ(ak − ak+1ad)∈ J(A),and so (hπa)‡ = hπa‡. Hence, we easily verify that(a′)‡ =(hπa‡ 00 0), (d′)‡ =(0 h‡hh‡ −ahh‡).Hence, a′, d′ ∈ A‡. Moreover, we havea′b′c′ =(hπahhπ 00 0);b′d′ =(hπh2h‡ 00 0);b′c′ =(hhπ 00 0).H. Chen et al. / Filomat 34:14 (2020), 4597–4605 4602Therefore a′b′c′ = 0, b′d′ = 0 and b′c′ ∈√J(A). In light of Lemma 3.1, N has p-Drazin inverse.Step 2. It is easy to check thatN =(1 00 b) (a 1c 0),it follows by [13, Theorem 3.6] that(a 1c 0) (1 00 b)has p-Drazin inverse. Therefore(a bc 0)has p-Drazininverse.Step 3. Write M = P + Q, whereP =(0 00 d),Q =(a bc 0).Then QP = 0. Clearly, P has p-Drazin inverse. By the preceding discussion, we have Q has p-Drazininverse. In light of [13, Theorem 5.4], M has p-Drazin inverse, as asserted.Corollary 3.3. Let a, d, cb, (cb)πd ∈ A‡. If (cb)πdcb = 0 and ca = 0, then M =(a bc d)∈M2(A)‡.Proof. In view of [13, Theorem 3.6], cb ∈ A‡. By virtue of Theorem 3.2, we prove that(d cb a)∈ M2(A)‡.We easily check thatM =(0 11 0) (d cb a) (0 11 0).This completes the proof.Corollary 3.4. Let a, d, bc, (bc)πa ∈ A‡. If (bc)πabc = 0, dπdc = 0 and bd‡ = 0, then M =(a bc d)∈M2(A)‡.Proof. Obviously, we have M = P + Q, whereP =(0 00 dπd),Q =(a bc d2d‡).In light of Theorem 3.2, Q has p-Drazin inverse. Since dπdc = 0, we have PQ = 0, and therefore we completethe proof by [13, Theorem 5.4].The following is the symmetric version of Theorem 3.2.Theorem 3.5. Let a, d, bc, (bc)πa ∈ A‡. If bca(bc)π = 0 and bd = 0, then M =(a bc d)∈M2(A)‡.Proof. Step 1. Let h = bc and let N =(a 1h 0). Let e =(hh‡ 00 1). Then N =(a′ b′c′ d′)e, wherea′ = eNe, b′ = eN(1 − e), c′ = (1 − e)Ne, d′ = (1 − e)N(1 − e).By hypothesis, we havea′ =(hh‡a hh‡h2h‡ 0), b′ =(0 0hhπ 0),c′ =(hπahh‡ hπ0 0), d′ =(ahπ 00 0).H. Chen et al. / Filomat 34:14 (2020), 4597–4605 4603By hypothesis, we havea′b′c′ = 0;b′d′ =(0 0hπhahπ 0)= 0;b′c′ =(0 0hhπahh‡ hhπ)∈√J(M2(A)).As in the proof of Theorem 3.2, a′, d′ ∈ A‡. In light of Lemma 3.1, N has p-Drazin inverse.Step 2. Since (a bc1 0)=(0 11 −a)−1 (a 1bc 0) (0 11 −a),we prove that(a bc1 0)has p-Drazin inverse.Step 3. Obviously, we have (a bc1 0)=(a b1 0) (1 00 c).In light of [13, Theorem 3.6],(1 00 c) (a b1 0)has p-Drazin inverse. Therefore(a bc 0)has p-Drazininverse.Step 4. Write M = P + Q, whereP =(0 00 d),Q =(a bc 0).Then QP = 0, and therefore we complete the proof by the discussion above.As in the proof of Corollary 3.3, we now deriveCorollary 3.6. Let a, d, cb, (cb)πd ∈ A‡. If cbd(cb)π = 0 and ca = 0, then M =(a bc d)∈M2(A)‡.Corollary 3.7. Let a, d, bc, (bc)πa ∈ A‡. If bca(bc)π = 0, dπdc = 0 and bd‡ = 0, then M =(a bc d)∈M2(A)‡.Example 3.8. Let C be the field of complex number, and letA ={(a b0 c) ∣∣∣∣a, b, c ∈ C} .Let M =(E I4F 0), whereE =1 1 0 00 −1 0 00 0 0 00 0 0 0 ,F =−1 2 0 10 0 0 00 0 1 00 0 0 −1 ∈M2(A).Then M has p-Drazin inverse.H. Chen et al. / Filomat 34:14 (2020), 4597–4605 4604Proof. We see thatJ(A) ={(0 b0 0) ∣∣∣∣b ∈ C} .Clearly, E and F have p-Drazin inverses. In fact, we haveE‡ = E =1 1 0 00 −1 0 00 0 0 00 0 0 0 ,F‡ = F =−1 2 0 10 0 0 00 0 1 00 0 0 −1 ,(F‡E)‡ =0 −2 0 00 −1 0 00 0 0 00 0 0 0 .Moreover, we haveFπEF =0 2 0 20 1 0 00 0 0 00 0 0 0−1 2 0 00 0 0 00 0 0 00 0 0 0 = 0,and we are done by Theorem 3.2.AcknowledgementThe authors would like to thank the referee for his/her careful reading.References[1] D.S. 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The P-Drazin İnverse for Operator Matrix Over Banach Algebras

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The P-Drazin İnverse for Operator Matrix Over Banach Algebras

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