We correct a mistake which affect our main results, namely the proof of Lema 1. The main results of the article remain unchanged.peerReviewe

Seddik, Ameur

Extracta Mathematicae

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E extracta mathematicae Vol. 32, Núm. 2, 209 – 211 (2017)Corrigendum to“Moore-Penrose Inverse and Operator Inequalities”Extracta Mathematicae 30 (2015), 29 – 39Ameur SeddikDepartment of Mathematics, Faculty of Mathematics and Computer Science,University of Batna 2, Batna, Algeriaseddikameur@hotmail.comPresented by Manuel González Received January 18, 2017Abstract : We correct a mistake which affect our main results, namely the proof of Lema 1.The main results of the article remain unchanged.Key words: Closed range operator, Moore-Penrose inverse, normal operator, operator in-equality.AMS Subject Class. (2010): 47A30, 47A03, 47B15.The paper mentioned in the title includes the following result as Lemma 1:Lemma 1. Let S ∈ B(H). If S is surjective or injective with closed rangeand satisfies the following inequality∀X ∈ B(H),∥∥S2X∥∥+ ∥∥XS2∥∥ ≥ 2 ∥SXS∥ , (∗)then S is normal.In the proof of this lemma, the matrix representation of the operator R2was computed incorrectly obtaining[S∗1S1 00 S∗2S2], while the correct form ofthis matrix is R2 =[S∗1S1 00 (S1S2)∗ (S1S2)]. Since all the results of the paperare based on this lemma, we shall give here a correct proof of it.The original proof is given in two cases. The second case follows imme-diately from the first one. The proof of the first case is divided in six steps.The mistake is in the fourth step.For S ∈ B(H) with closed range, S+ denotes the Moore-Penrose inverseof S.209210 a. seddikProof of Lemma 1. Assume that S ̸= 0 and that all 2×2 matrices used in thisproof are given with respect to the orthogonal direct sum H = R(S)⊕ kerS∗.Then S =[S1 S20 0].We put P = |S| , Q = |S∗| , P1 = |S1| , P2 = |S2| , Q1 = (S1S∗1 + S2S∗2)12 .So we have S∗S = P 2 =[P 21 S∗1S2S∗2S1 P22], SS∗ = Q2 =[Q2100 0]. It is clearthat Q1 is invertible and Q+ =[Q−11 00 0].Case 1. Assume that S is injective with closed range and satisfies (∗). ThenS+S = I, kerP = kerS = {0}, and R(P ) = R(S∗S) is closed (since R(S∗) isalso closed). Thus kerP = {0} and R(P ) = (kerP )⊥ = H. So, P is invertible.Note that inequality (∗) implies the following inequality:∀X ∈ B(H),∥∥S2S+XS+∥∥+ ∥∥S+XS∥∥ ≥ 2∥∥SS+X∥∥ . (1)The proof is given in four steps.Step 1.(S2)+S = S+. See Step 2 of the original proof.Step 2.(S2)+= (S+)2. See Step 3 of the original proof.Step 3. kerS∗ = {0} . Since S is injective, then kerS∗ = {0} if and only ifS2 = 0. Assume that S2 ̸= 0.Since(S2)+= (S+)2, then the two operators S∗S and SS+ commute (see[1, 2]). Thus P 2 =[P 2100 P 22], hence P =[P1 00 P2].Since kerS∗ ̸= {0}, then σ(Q2) = σ(Q21)∪{0}. From the fact that σ(P 2) =σ(Q2) − {0}, we have σ(P 2) = σ(Q21). Then σ(P 21 ) ∪ σ(P22 ) = σ(Q21). Henceσ(P 21) ⊂ σ(Q21). Thus σ(P1) ⊂ σ(Q1).Using the polar decomposition of S and S∗ in inequality (1), we obtainthe following inequality:∀X ∈ B(H),∥∥S2S+XP−1∥∥+ ∥∥Q+XQ∥∥ ≥ 2∥∥SS+X∥∥ .By taking X =[X1 00 0](resp. X =[0 X20 0]), where X1 ∈ B(R(S))(resp. X2 ∈ B(kerS∗, R(S))) in the last inequality, and since S2S+ =[S1 00 0], we deduce the two following inequalities∀X1 ∈ B(R(S)),∥∥P1X1P−11 ∥∥+ ∥∥Q−11 X1Q1∥∥ ≥ 2 ∥X1∥ , (2)corrigendum to “moore-penrose inverse and . . . ” 211∀X2 ∈ B(kerS∗, R(S)),∥∥P1X2P−12 ∥∥ ≥ 2 ∥X2∥ . (3)By taking X2 = x⊗ y (where x ∈ (R(S))1, y ∈ kerS∗) in (3), we obtain∀x ∈ (R(S))1,∀y ∈ kerS∗, ∥P1x∥∥∥P−12 y∥∥ ≥ 2 ∥y∥ .So we have∀x ∈ (R(S))1,∀y ∈ (kerS∗)1 , ∥P1x∥ ≥ 2 ∥P2y∥ .Thus ∥P2y∥ ≤ k2 , for every y ∈ (kerS∗)1 (where k = inf∥x∥=1 ∥P1x∥ > 0),and then⟨P 22 y, y⟩≤ k2/4, for every y ∈ (kerS∗)1. So we obtain σ(P 22 ) ⊂(0, k24]and σ(P 21 ) ⊂ [k2,∞).Since σ(P1) ⊂ σ(Q1), and P1 , Q1 satisfy the inequality (2), then usinga variation of [3, Theorem 3.6] (in that paper Theorem 3.6 is stated withequality between the spectra but the proof is the same for inclusion betweenthe spectra), we obtain P1 = Q1 . Hence σ(Q21) = σ(P 21 ) = σ(P21) ∪ σ(P 22 ).Then σ(P 22 ) ⊂ σ(P 21 ), that is impossible since(0, k24]∩ [k2,∞) = ∅. ThereforekerS∗ = {0}.Step 4. S is normal. Since kerS∗ = {0}, we obtain R(S) = H. So thatS is invertible and satisfies the inequality (∗). Hence S satisfies the followinginequality∀X ∈ B(H),∥∥SXS−1∥∥+ ∥∥S−1XS∥∥ ≥ 2 ∥X∥ .Therefore S is normal (using [4]).Case 2. Assume that S is surjective and satisfies (∗). Then S∗ is injectivewith closed range and satisfies inequality (∗). From Case 1, S∗ is normal.Hence S is normal, and the proof is finished.References[1] R. Bouldin, The pseudo-inverse of a product, SIAM J. Appl. Math. 24 (1973),489 – 495.[2] S. Izumino, The product of operators with closed range and an extension ofthe reverse order law, Tôhoku Math. J. (2) 34 (1) (1982), 43 – 52.[3] A. Seddik, Some results related to the Corach-Porta-Recht inequality, Proc.Amer. Math. Soc. 129 (2001), 3009 – 3015.[4] A. Seddik, On the injective norm and characterization of some subclasses ofnormal operators by inequalities or equalities, J. Math. Anal. Appl. 351 (1)(2009), 277 – 284.

Corrigendum to “Moore-Penrose inverse and operator inequalities” Extracta Mathematicae 30 (2015), 29 - 39

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Corrigendum to “Moore-Penrose inverse and operator inequalities” Extracta Mathematicae 30 (2015), 29 - 39

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