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Perturbation analysis for the generalized inverses with prescribed idempotents in Banach algebras PDF

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Preview Perturbation analysis for the generalized inverses with prescribed idempotents in Banach algebras

Perturbation analysis for the generalized inverses with prescribed idempotents in Banach algebras Jianbing Cao∗ 3 1 Department of Mathematics, East China Normal University, 0 2 Shanghai 200241, P.R. China n Department of mathematics, Henan Institute of Science and Technology a J Xinxiang, Henan, 453003, P.R. China 8 1 Yifeng Xue† ] Department of Mathematics, East China Normal University, A F Shanghai 200241, P.R. China . h t a m Abstract [ 1 In this paper, we first study the perturbations and expressions for the gen- v eralized inverses a(2), a(1,2), a(2,l) and a(l) with prescribed idempotents p and p,q p,q p,q p,q 4 q. Then, we investigate the general perturbation analysis and error estimate 1 3 for some of these generalized inverses when p, q and a also have some small 4 perturbations. . 1 0 2010 Mathematics Subject Classification: 15A09; 46L05 3 Key words: gap function, idempotent element, generalized inverse, perturba- 1 : tion v i X r 1 Introduction a Let R be a unital ring and let R• denote the set of all idempotent elements in R. Given p, q ∈ R•. Recall that an element a ∈ R has the (p,q)–outer generalized inverse b = a(2) ∈ R if bab = b, ba = p and 1 − ab = q. If b = a(2) also satisfies p,q p,q the equation aba = a, then we say a has the (p,q)–generalized inverse b, in this case, written b = a(1,2). If an outer generalized inverse with prescribed idempotents p,q exists, it is necessarily unique (cf. [6]). According to this definition, obviously, we see that the Moore–Penrose inverses in a C∗-algebra and (generalized) Drazin inverses in a Banach algebra can be expressed by some (p,q)–outer generalized inverses (cf. [6, 5, 1]). ∗Email: [email protected] †Email: [email protected]; Corresponding author 1 Based on some results of Djordjevi´c and Wei in [6], Ilic, Liu and Zhong gave some equivalent conditions for the existence of the (p,q)–outer generalized inverse in a Banach algebra in [5]. But in our recent paper [1], we find that Theorem 1.4 of [5] is wrong. In [1], we first present a counter–example to [5, Theorem 1.4], then based on our counter–example, we define a new type of generalized inverse with prescribed idempotents in a Banach algebra as follows: Definition 1.1 (see [1]). Let a ∈ A and p, q ∈ A•. An element b ∈ A satisfying bab = b, R (b) = R (p), K (b) = R (q), r r r r will be called the (p,q,l)–outer generalized inverse of a, written as a(2,l) = b. p,q In addition, if a(2,l) satisfies a = aa(2,l)a, we call a(2,l) is the (p,q,l)–generalized p,q p,q p,q inverse of a, denoted by a(l). p,q Perturbation analysis of the generalized inverses is very important in both theory and applications. In recent years, there are many fruitful results concerning the perturbation analysis for various types generalized inverses of operators on Hilbert spacesorBanachspaces. TheconceptofstableperturbationofanoperatoronHilbert spaces and Banach spaces is introduced by Chen and Xue in [1]. Later the notation is generalized to the set of Banach algebras by the second author in [13] and to the set of Hilbert C∗–modules by Xu, Wei and Gu in [16]. Using the notation “stable perturbation”, many important results in perturbation analyses for Moore–Penrose inverses onHilbertspacesandDrazininverses onBanachspacesorinBanachalgebras have been obtained. Please see [2, 3, 4, 14, 13, 15] for detail. Let X,Y be Banach spaces over complex field C. Let T (resp. S) be a given closed space in X (resp, Y). Let A be a bounded linear operator from X to Y such that A(2) exists. The perturbation analysis of A(2) for small perturbation of T, S T,S T,S and A has been done in [7, 8]. Motivated by some recent results concerning the per- turbation analysis for the generalized inverses of operators, in this paper, we mainly study the perturbations and expressions for various types of generalized inverses with prescribed idempotents in Banach algebras. We first consider the stable perturbation characterizations for a(2), a(1,2), a(2,l) and a(l) with prescribed idempotents p and q. p,q p,q p,q p,q Then, by using stable perturbation characterizations, we can investigate the general perturbation analysis and error estimate for some of these generalized inverses when p, q and a also have some small perturbations. The results obtained in this paper extend and improve many recent results in this area. 2 Preliminaries In this section, we give some notations in this paper, we also list some preliminary results which will be frequently used in our main sections. Throughout the paper, A is always a complex Banach algebra with the unit 1. Let a ∈ A. If there is b ∈ A such that aba = a and bab = b, then a is called to be generalized invertible and b is called the generalized inverse of a, denoted by 2 b = a+. Let Gi(A) denote the set of all generalized invertible elements in A\{0}. Let A• denote the set of all idempotent elements in A. If a ∈ Gi(A), then a+a and 1−aa+ are all idempotent elements. For a ∈ A, set K (a) = {x ∈ A | ax = 0}, R (a) = {ax | x ∈ A}; r r K (a) = {x ∈ A | xa = 0}, R (a) = {xa | x ∈ A}. l l Clearly, if p ∈ A•, then A has the direct sum decompositions: A = K (p)∔R (p) or A = K (p)∔R (p). r r l l The following useful and well–known lemma can be easily proved. Lemma 2.1. Let x ∈ A and p ∈ A•. Then (1) K (p) and R (p) are all closed and K (p) = R (1−p), R (p)A ⊂ R (p); r r r r r r (2) px = x if and only if R (x) ⊂ R (p) or K (p) ⊂ K (x); r r l l (3) xp = x if and only if K (p) ⊂ K (x) or R (x) ⊂ R (p). r r l l We list some of the necessary and sufficient conditions for the existence of a(2,l) p,q in the following lemma, which will be frequently used in the paper. Here we should indicate that a(2,l) is unique if it exists. Please see [1] for the proofs and more p,q information. Lemma 2.2. Leta ∈ A and p,q ∈ A•. Then the followingstatements are equivalent: (1) a(2,l) exists; p,q (2) There exists b ∈ A such that bab = b, R (b) = R (p) and K (b) = R (q); r r r r (3) K (a)∩R (p) = {0} and A = aR (p)∔R (q); r r r r (4) There exists b ∈ A satisfying b = pb, p = bap, b(1−q) = b, 1−q = (1−q)ab; (5) p ∈ R ((1−q)ap) = {x(1−q)ap | x ∈ A} and 1−q ∈ R ((1−q)ap); l r (6) There exist some s,t ∈ A such that p = t(1−q)ap, 1−q = (1−q)aps. The following lemma gives some equivalent conditions about the existence of a(l) . p,q See [1] for more information. Lemma 2.3. Let a ∈ A and p,q ∈ A•. Then the following conditions are equivalent: (1) a(l) exists, i.e.,there exists some b ∈ A such that p,q aba = a, bab = b, R (b) = R (p), K (b) = R (q), r r r r (2) A = R (a)∔R (q) = K (a)∔R (p), r r r r (3) A = aR (p)∔R (q), R (a)∩R (q) = {0}, K (a)∩R (p) = {0}. r r r r r r Let X be a complex Banach space. Let M, N be two closed subspaces in X. Set sup{dist(x,N)|x ∈ M, kxk = 1}, M 6= {0} δ(M,N) = , (0 M = {0} where dist(x,N) = inf{kx − yk|y ∈ N}. The gap δˆ(M,N) of M, N is given by δˆ(M,N) = max{δ(M,N),δ(N,M)}. For convenience, we list some properties about δ(M,N) and δˆ(M,N) which come from [10] as follows. 3 Proposition 2.4 ([10]). Let M, N be closed subspaces in a Banach space X. Then (1) δ(M,N) = 0 if and only if M ⊂ N; (2) δˆ(M,N) = 0 if and only if M = N; (3) δˆ(M,N) = δˆ(N,M); (4) 0 ≤ δ(M,N) ≤ 1, 0 ≤ δˆ(M,N) ≤ 1. 3 Stable perturbations for the (p, q)–generalized inverses Let a ∈ Gi(A) and let a¯ = a + δa ∈ A. Recall from [14] that a¯ is a stable perturbation of a if R (a¯) ∩ K (a+) = {0}. Obviously, we can define the stable r r perturbation for various kind of generalized inverses. In this section, we concern the stable perturbation problem for various types of (p,q)–generalized inverses in a Banach algebra. Lemma 3.1 ([9, Lemma 2.2]). Let a, b ∈ A. If 1+ ab is left invertible, then so is 1+ba. Lemma 3.2. Let a,δa ∈ A and p, q ∈ A• such that a(2,l) exists. Put a¯ = a+δa. If p,q 1+δaa(2,l) is invertible, w = a(2,l)(1+δaa(2,l))−1. Then a¯(2,l) exists and w = a¯(2,l) . p,q p,q p,q p,q p,q Proof. We prove our result by showing that waw = w,R (w) = R (p),K (w) = r r r R (q). It is easy to check that r w = a(2,l)(1+δaa(2,l))−1 = (1+a(2,l)δa)−1a(2,l). p,q p,q p,q p,q Then, by using these two equalities, we can show R (w) = R (a(2,l)) = R (p) and r r p,q r K (w) = K (a(2,l)) = R (q). We can also compute r r p,q r wa¯w = a(2,l)(1+δaa(2,l))−1a¯a(2,l)(1+δaa(2,l))−1 p,q p,q p,q p,q = a(2,l)(1+δaa(2,l))−1[(aa(2,l) −1)+(1+δaa(2,l))](1+δaa(2,l))−1 p,q p,q p,q p,q p,q = a(2,l)(1+δaa(2,l))−1(aa(2,l) −1)(1+δaa(2,l))−1 +w p,q p,q p,q p,q = w. By Definition 1.1 and the uniqueness of a(2,l), we see a¯(2,l) exists and w = a¯(2,l) . p,q p,q p,q Obviously, from the proof of Lemma 3.2, we see that if a(2,l) exists and 1+a(2,l)δa p,q p,q isinvertible, set v = (1+a(2,l)δa)−1a(2,l), thenwealso havev = a¯(2,l). Inorder toprove p,q p,q p,q the main results about the stable perturbation, we need one more characterizations of the existence of a(2,l). p,q For an element a ∈ A and p, q ∈ A•. Let R : A → A be the right multiplier on a A(i.e., R (x) = xa for any x ∈ A). Then it easy to see that a(2,l) exists in A if and a p,q only if (R )(2) exists inthe Banach algebra B(A). So fromtheequivalences a A(1−q),A(1−p) of(1),(2)and(3)inLemma2.2, dually, wecangetthefollowingequivalent conditions for the existence of a(2,l). p,q 4 Proposition 3.3. Let a ∈ A and p, q ∈ A•. Then the following statements are equivalent: (1) a(2,l) exists; p,q (2) There exists c ∈ A such that cac = c, R (c) = R (1−q) and K (c) = R (1−p); l l l l (3) K (a)∩R (1−q) = {0} and A = R (1−q)a∔R (1−p). l l l l Proof. (1) ⇔ (2) Suppose that a(2,l) exists. Let c = a(2,l). Then from Definition 1.1, p,q p,q we know that cac = c, and then ca, ac ∈ A•, R (ca) = R (c) = R (p), K (ac) = r r r r K (c) = R (q). Thus, it follows from Lemma 2.1 that r r cap = p, pca = ca, ac(1−q) = ac, (1−q)ac = 1−q. Then, by using Lemma 2.1 again, we have K (ca) ⊂ K (p) ⊂ K (ca), R (ac) ⊂ R (1−q) ⊂ R (ac). (3.1) l l l l l l By using cac = c, we have K (ca) = K (c) and R (ac) = R (c). Thus from Eq. (3.1) l l r r we see that (2) holds. If (2) holds, similarly, by using Definition 1.1 and Lemma 2.1, we can obtain a(2,l) exists. p,q (2) ⇔ (3) By our remark above this lemma, we see these hold simply from the equivalences of (2) and (3) in Lemma 2.2. Note that we can also prove these equiva- lences directly by using the right multiplier R on A. Here we omit the detail. a Now we can present one of our main results about the stable perturbation of the generalized inverse a(2,l). p,q Theorem 3.4. Let a, δa ∈ A and p, q ∈ A• such that a(2,l) exists. Put a¯ = a+δa. p,q Then the following statements are equivalent: (1) 1+δaa(2,l) is invertible; p,q (2) 1+a(2,l)δa is invertible; p,q (3) a¯(2,l) exists. p,q In this case, we have a¯(2,l) = a(2,l)(1+δaa(2,l))−1 = (1+a(2,l)δa)−1a(2,l). p,q p,q p,q p,q p,q Proof. (1) ⇔ (2) follows from the well-known spectral theory in Banach algebras. (2) ⇒ (3) We prove our result by using Lemma 2.2. Let x ∈ K (a¯)∩R (p) = {0}. r r Since R (p) = R (a(2,l)), thenthere exists some t ∈ A satisfying x = a(2)tanda¯t = 0. r r p,q p,q Thus we have (1+a(2,l)δa)a(2,l)t = a(2,l)t+a(2,l)δaa(2,l)t p,q p,q p,q p,q p,q = a(2,l)(a+δa)a(2,l)t p,q p,q = aa¯t = 0. Since 1+a(2,l)δa is invertible, it follows that x = a(2,l)t = 0. Therefore, p,q p,q K (a¯)∩R (p) = {0} (3.2) r r 5 Let s ∈ a¯R (p) ∩R (q). Since R (p) = R (a(2,l)) and R (q) = K (a(2,l)), then there r r r r p,q r r p,q exists some z ∈ A such that s = a¯a(2,l)z and a(2,l)s = 0. Similar to the proof of p,q p,q Eq.(3.2), we can get s = a(2,l)t = 0, i.e, a¯R (p) ∩R (q) = {0}. Since 1+ a(2,l)δa is p,q r r p,q invertible, then for any w ∈ A there is some v ∈ A such that a(2,l)w = (1+a(2,l)δa)v. p,q p,q From a¯ = a+δa, we have (1−a(2,l)a)v = a(2,l)(w−a¯v) ∈ R (a(2,l)a)∩K (a(2,l)a) = {0}. p,q p,q r p,q r p,q Thus, w−a¯v ∈ K (a(2,l)) and v = a(2,l)av ∈ R (a(2,l)). Since for w ∈ A, we also have r p,q p,q r p,q w = a¯v +(w−a¯v) ∈ a¯R (p)∔R (q). Thus, we have r r A = a¯R (p)∔R (q). (3.3) r r Now, from Eqs. (3.2) and (3.3), by using Lemma 2.2, we see that a¯(2,l) exists. p,q (3) ⇒ (1) Suppose that a¯(2,l) exists, we want to prove 1+δaa(2,l) is both left and p,q p,q right invertible. Since a¯(2,l) exists, then from Lemma 2.2, A = a¯R (p) ∔ R (q) = p,q r r a¯R (a(2,l))∔K (a(2,l)). Thus, for any x ∈ A, we can write x = a¯a(2,l)t +t , where r p,q r p,q p,q 1 2 t ∈ A and t ∈ K (a(2,l)). Set s = aa(2,l)t +t , then 1 2 r p,q p,q 1 2 (1+δaa(2,l))s = (1+δaa(2,l))(aa(2,l)t +t ) p,q p,q p,q 1 2 = a¯a(2,l)t +t = x. p,q 1 2 Since x ∈ A is arbitrary, let x = 1, then we see that 1+δaa(2,l) is right invertible. p,q Now we prove that 1+δaa(2,l) is also left invertible. In fact, from Proposition 3.3, we p,q also have A = R (1−q)a¯∔R (1−p) = R (a(2,l))a¯∔K (a(2,l)) for a¯(2,l) exists. Then for l l l p,q l p,q p,q any z ∈ A, we can write z = s a(2,l)a¯+s , where s ∈ R (a(2,l)a) and s ∈ K (a(2,l)). 1 p,q 2 1 l p,q 2 l p,q Let t = s +s , then we have 1 2 t(1+a(2,l)δa) = (s +s )(1+a(2,l)δa) p,q 1 2 p,q = s +s +s a(2,l)(a¯−a) 1 2 1 p,q = s a(2,l)a¯+s +s (1−a(2,l)a) 1 p,q 2 1 p,q = z. Since z ∈ A is arbitrary, let z = 1, then we get that 1 + a(2,l)δa is left invertible. p,q But from Lemma 3.1 we see 1 + δaa(2,l) is also left invertible. Thus, 1 + δaa(2,l) is p,q p,q invertible. Now, from Lemma 3.2, a¯(2,l) = a(2,l)(1 + δaa(2,l))−1 = (1 + a(2,l)δa)−1a(2,l). This p,q p,q p,q p,q p,q completes the proof. Lemma 3.5. Let a, δa ∈ A and p, q ∈ A• such that a(2,l) exists. If 1 + a(2,l)δa is p,q p,q invertible. Put a¯ = a+δa and f = (1+a(2,l)δa)−1(1−a(2,l)a). Then p,q p,q (1) f ∈ A• with K (a¯) ⊂ R (f); r r (2) K (a¯) = R (f) if and only if R (a¯)∩R (q) = {0}. r r r r 6 Proof. (1) Since (1−a(2,l)a)(1+a(2,l)δa) = 1−a(2,l)a and 1+a(2,l)δa is invertible, we p,q p,q p,q p,q have 1−a(2,l)a = (1−a(2,l)a)(1+a(2,l)δa)−1. Thus, p,q p,q p,q f2 = (1+a(2,l)δa)−1(1−a(2,l)a)(1+a(2,l)δa)−1(1−a(2,l)a) = f. p,q p,q p,q p,q Now for any x ∈ A, from (1−aa(2,l))x = (1+a(2,l)δa)x−a(2,l)a¯x, we have p,q p,q p,q fx = (1+a(2,l)δa)−1(1−a(2,l)a)x = x−(1+a(2,l)δa)−1a(2,l)a¯x. (3.4) p,q p,q p,q p,q Eq. (3.4) implies that K (a¯) ⊂ R (f). r r (2) (⇒) Let t ∈ R (a¯)∩R (q) = {0}. Since R (q) = K (a(2,l)), then there is some r r r r p,q x ∈ A such that t = a¯x and a(2,l)a¯x = a(2,l)t = 0. Thus, x = fx by Eq. (3.4). So, p,q p,q x ∈ R (f) = K (a¯) and t = a¯x = 0, i.e., R (a¯)∩R (q) = {0}. r r r r (⇐) Thanks to (1), we need only to prove R (f) ⊂ K (a¯). Let t ∈ R (f). Since r r r f ∈ A•, we have t = ft. So by Eq. (3.4), we get (1+a(2,l)δa)−1a(2,l)a¯t = 0 and then p,q p,q a(2,l)a¯t = 0. Hence a¯t ∈ R (a¯)∩R (q) = {0} and K (a¯) = R (f). p,q r r r r Similar to [13, Proposition 2.2] or [14, Theorem 2.4.7], but by using some of our characterizations for a(2,l) and a(l) , we can obtain the following results about the p,q p,q stable perturbations for these two kinds of generalized inverses. Theorem 3.6. Let a, δa ∈ A and p, q ∈ A• such that a(2,l) exists. Suppose that p,q 1 + a(2,l)δa is invertible. Put a¯ = a + δa and w = (1 + a(2,l)δa)−1a(2,l). Then the p,q p,q p,q following statements are equivalent: (1) w = a¯(l) , i.e., a¯(2,l) = a¯(l) ; p,q p,q p,q (2) R (a¯)∩R (q) = {0}, i.e., a¯ is a stable perturbation of a; r r (3) a¯(1+a(2,l)δa)−1(1−a(2,l)a) = 0; p,q p,q (4) (1−aa(2,l))(1+δaa(2,l))−1a¯ = 0. p,q p,q Proof. The implication (1) ⇔ (2) comes from Lemma 2.2, Lemma 2.3 and Theorem 3.4. The implication (2) ⇔ (3) comes from Lemma 3.5. (3) ⇔ (4) we can compute in the following way, a¯(1+a(2,l)δa)−1(1−a(2,l)a) = a¯(1+a(2,l)δa)−1[(1+a(2,l)δa)−a(2,l)a¯] p,q p,q p,q p,q p,q = a¯−a¯a(2,l)(1+δaa(2,l))−1a¯ p,q p,q = a¯−[(1+δaa(2,l))−(1−aa(2,l))](1+δaa(2,l))−1a¯ p,q p,q p,q = (1−aa(2,l))(1+δaa(2,l))−1a¯. p,q p,q This completes the proof. Furthermore, by using the above theorem, we have the following results. Corollary 3.7. Let a, δa ∈ A and p, q ∈ A• such that a(2,l) exists. Put a¯ = a+δa. p,q If 1+a(2,l)δa is invertible. Then the following statements are equivalent: p,q 7 (1) R (a¯)∩R (q) = {0}, i.e., a¯ is stable perturbation of a; r r (2) (1+a(2,l)δa)−1K (a(2,l)a) = K (a¯); p,q r p,q r (3) (1+δaa(2,l))−1R (a¯) = R (aa(2,l)). p,q r r p,q Proof. Notethat we have K (a(2,l)a) = R (1−a(2,l)a) andR (aa(2,l)) = K (1−aa(2,l)). r p,q r p,q r p,q r p,q So we can get the assertions by using Theorem 3.6. Theorem 3.8. Let a, δa ∈ A and p, q ∈ A• such that a(l) exists. Put a¯ = a+δa. p,q Then the following statements are equivalent: (1) 1+a(l) δa is invertible, R (a¯) = K (q) and a¯(l) = a(l) (1+δaa(l) )−1. p,q r r p,q p,q p,q (2) R (a¯)∩R (q) = {0}, K (a¯)∩R (p) = {0} and a¯R (p) = K (q). r r r r r r Proof. (1) ⇒ (2) Suppose that (1) holds. Since a(l) exists, we obtain that a(2,l) exists p,q p,q and a(2,l) = a(l) . Thus, from our assumption, by using Theorem 3.6 and Lemma 2.3, p,q p,q we have R (a¯)∩R (q) = {0}, K (a¯)∩R (p) = {0}. r r r r Now we need to show a¯R (p) = K (q). But since R (a¯) = K (q), so we can prove r r r r our result by showing that a¯R (p) = R (a¯). Obviously, a¯R (p) ⊂ R (a¯). Ontheother r r r r hand, since a¯(l) exists, then by Lemma 2.3 again, we have A = a¯R (p)∔R (q). Now p,q r r for any x ∈ R (a¯), we can write x = x +x with x ∈ a¯R (p) and x ∈ R (q). From r 1 2 1 r 2 r a¯R (p) ⊂ R (a¯), we get x ∈ R (a¯). Thus, r r 1 r x = x−x ∈ R (a¯)∩R (q) = {0}. 2 1 r r Therefore, x = 0 and then x = x ∈ a¯R (p). Hence, a¯R (p) = R (a¯) = K (q). 2 1 r r r r (2) ⇒ (1) Since q ∈ A• and a¯R (p) = K (q), we can write A = K (q)∔R (q) = r r r r a¯R (p)∔R (q). Note that R (a¯)∩R (q) = {0}, K (a¯)∩R (p) = {0}, then by using r r r r r r Lemma 2.3, we get a(l) exists, then a(2,l) also exists and a(2,l) = a(l) . Thus, from p,q p,q p,q p,q Theorem 3.4, we see 1+a(l) δa is invertible and a¯(l) = a(l) (1+δaa(l) )−1. Now, by p,q p,q p,q p,q using Lemma 2.3, we cangetA = a¯R (p)∔R (q). Similarly, asin(1) ⇒ (2), byusing r r R (a¯) ∩ R (q) = {0}, we can show that a¯R (p) = R (a¯) and then R (a¯) = K (q). r r r r r r This completes the proof. The first result in the following lemma has been proved for generalized inverse a+ by the second author (see [13, Proposition 2.5]). By using the same method, we can prove the following results for the general inverse a(l) . p,q Lemma 3.9 ([13, Proposition 2.5]). Let a,δa ∈ A and p, q ∈ A• such that a(l) p,q exists. Put a¯ = a+δa. (1) If δ(R (a¯),R (a)) < k1−aa(l) k−1, then R (a¯)∩R (q) = {0}; r r p,q r r (2) If δ(K (a¯),K (a)) < ka(l) ak−1, then K (a¯)∩R (p) = {0}. r r p,q r r By using Theorem 3.6, Theorem 3.8 and Lemma 3.9, we have the following: 8 Corollary 3.10. Let a,δa ∈ A and p, q ∈ A• such that a(l) exists. Put a¯ = a+δa. p,q If one of the following condition holds, (i) δ(K (a¯),K (a)) < ka(l) ak−1, δ(R (a¯),R (a)) < k1 − aa(l) k−1 and a¯R (p) = r r p,q r r p,q r K (q). r (ii) 1+a(l) δa is invertible and δ(R (a¯),R (a)) < k1−aa(l) k−1. p,q r r p,q Then a¯(l) exists and a¯(l) = a(l) (1+δaa(l) )−1. p,q p,q p,q p,q Finally, we present some perturbation results for a(2). p,q Theorem 3.11. Let a, δa ∈ A and p, q ∈ A• such that a(2) exists. Put a¯ = a+δa. p,q If 1+a(2)δa is invertible. Then the following statements are equivalent: p,q (1) a¯(2) exists and a¯(2) = a(2)(1+δaa(2))−1; p,q p,q p,q p,q (2) a¯p = (1−q)a¯; (3) a¯a(2) = (1−q)a¯a(2) and a(2)a¯ = a(2)a¯p. p,q p,q p,q p,q Proof. (1) ⇔ (2) comes from [6, Theorem 4.1]. We show that (2) and (3) are equiv- alent. If a¯p = (1−q)a¯, then a¯a(2) = a¯pa(2) = (1−q)a¯a(2) and a(2)a¯ = a(2)(1−q)a¯ = a(2)a¯p. p,q p,q p,q p,q p,q p,q Conversely, if (3) holds, then a¯p = a¯a(2)a = (1−q)a¯a(2)a = aa(2)a¯p = (1−q)a¯. p,q p,q p,q Corollary 3.12. Let a, δa ∈ A and p, q ∈ A• such that a(2) exists. Put a¯ = a+δa. p,q If 1+a(2)δa is invertible and δa = (1−q)δa = δap. Then a¯(2) exists and p,q p,q a¯(2) = a(2)(1+δaa(2))−1 = (1+a(2)δa)−1a(2). p,q p,q p,q p,q p,q Proof. If δa = (1 −q)δa = δap, then it is easy to check that a¯p = (1 − q)a¯. Thus, Theorem 3.11 shows that our results hold. 4 Perturbation analysis for the (p,q)–generalized inverses Inthissection, wemainlyinvestigate thegeneralperturbationsproblemforthe(p,q)– generalized inverses a(2,l) and a(1,2). Let κ = kakka(2,l)k, which is the generalized p,q p,q p,q condition number of the generalized inverse a(2,l). p,q Lemma 4.1. Let a ∈ A and p ∈ A• with R (p) = R (a). Let c ∈ A with R (c) r r r 1 closed and δˆ(R (c), R (a)) < . Then A = R (c)∔K (p). r r r r 1+kpk Proof. Let L x = px, ∀x ∈ A. Then L is an idempotent operator on A with p p kL k = kpk and {L x|x ∈ A} = R (p). By [11, Theorem 11] or [14, Lemma 4.4.4], p p r 1 δˆ(R (c), R (a)) < implies that A = R (c)∔K (p). r r r r 1+kL k p 9 Lemma 4.2 ([13,Lemma2.4]). Forany p, q ∈ A• we have δˆ(R (p),R (q)) ≤ kp−qk. r r Lemma 4.3. Let a ∈ A and p, q ∈ A• such that a(2,l) exists. Suppose that p′ ∈ A• p,q 1 satisfying kp−p′k < . Then 1+κ κkp−p′k (1) δˆ(aR (p),aR (p′)) ≤ ; r r 1−(1+κ)kp−p′k (2) aR (p′) ⊂ A is closed and K (a)∩R (p′) = {0}. r r r Proof. (1) Set b = a(2,l). For any t′ ∈ R (p′), we have p,q r dist(at′,aR (p)) = inf kat′ −atk ≤ kak inf kt−t′k r t∈Rr(p) t∈Rr(p) ≤ kakdist(t′,R (p)) r ≤ kakkt′kδˆ(R (p′),R (p)). r r Thus, we get δ(at′,aR (p)) ≤ kakkt′kδˆ(R (p′),R (p)). (4.1) r r r But for any t′ ∈ R (p′) and t ∈ R (p), we have r r kbkkat′k = kbkka(t′ −t+t)k ≥ kbkkatk−kbkkakkt′ −tk ≥ ktk−kbkkakkt′ −tk ≥ kt′k−(1+kbkkak)kt′ −tk. Thus, kt′k−kbkkat′k ≤ (1+kbkkak)kt′ −tk, and then kt′k−kbkkat′k ≤ (1+kbkkak)dist(t′,R (p)) r ≤ (1+kbkkak)kt′kδˆ(R (p′),R (p)). r r Therefore, we have kbkkat′k kt′k ≤ . (4.2) 1−(1+kbkkak)δˆ(R (p′),R (p)) r r Then by Eq. (4.1) and Eq. (4.2), we get kbkkakkat′kδˆ(R (p′),R (p)) δ(at′,aR (p)) ≤ r r . r 1−(1+kbkkak)δˆ(R (p′),R (p)) r r Now, by Lemma 4.2 and the Definition of gap–function, we have κkp−p′k δ(aR (p′),aR (p)) ≤ . (4.3) r r 1−(1+κ)kp−p′k 10

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