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1 ISSN X (Print ISSN (Online Bulletin of the Iranian Mathematical Society Vol 42 (2016, No 1, pp Title The reverse order law for Moore-Penrose inverses of operators on Hilbert C*-modules Author(s K Sharifi and B A Bonakdar Published by Iranian Mathematical Society http//bimsimsir
2 Bull Iranian Math Soc Vol 42 (2016, No 1, pp Online ISSN THE REVERSE ORDER LAW FOR MOORE-PENROSE INVERSES OF OPERATORS ON HILBERT C*-MODULES K SHARIFI AND B A BONAKDAR (Communicated by Behzad Djafari-Rouhani Abstract Suppose T and S are bounded adjointable operators between Hilbert C*-modules admitting bounded Moore-Penrose inverse operators Some necessary and sufficient conditions are given for the reverse order law (T S S T to hold In particular, we show that the equality holds if and only if Ran(T T S Ran(S and Ran(SS T Ran(T, which was studied first by Greville SIAM Rev 8 ( for matrices Keywords Bounded adjointable operator, Hilbert C*-module, Moore- Penrose inverse, reverse order law MSC(2010 Primary 47A05; Secondary 46L08, 15A09 1 Introduction and preliminaries It is well-known that for invertible operators (or nonsingular matrices T, S and T S, (T S 1 S 1 T 1 However, this so-called reverse order law is not necessarily true for other kind of generalized inverses An interesting problem is, for given operators (or matrices T S with T S meaningful, under what conditions, (T S S T? The problem first studied by Greville 7 and then reconsidered by Bouldin and Izumino 2, 9 Many authors discussed the problem like this, see eg 3 5,11,13 and references therein An special case, when S T, was given by Moslehian et al 14 for a Moore-Penrose invertible operator T on Hilbert C*-modules The later paper and the work of 5,7 motivated us to study the problem in the framework of Hilbert C*-modules The notion of a Hilbert C*-module is a generalization of the notion of a Hilbert space However, some well known properties of Hilbert spaces like Pythagoras equality, self-duality, and even decomposition into orthogonal complements do not hold in the framework of Hilbert modules The first use of such Article electronically published on February 22, 2016 Received 16 August 2012, Accepted 26 October 2014 Corresponding author 53 c 2016 Iranian Mathematical Society
3 The reverse order law for Moore-Penrose inverses 54 objects was made by I Kaplansky 10 and then studied more in the work of W L Paschke 15 Let us quickly recall the definition of a Hilbert C*-module Suppose that A is an arbitrary C*-algebra and E is a linear space which is a right A-module and the scalar multiplication satisfies λ(xa x(λa (λxa for all x E, a A, λ C The A-module E is called a pre-hilbert A-module if there exists an A-valued map, E E A with the following properties (i x, y + λz x, y + λ x, z ; for all x, y, z E, λ C, (ii x, ya x, y a; for all x, y E and a A, (iii x, y y, x ; for all x, y E, (iv x, x 0 and x, x 0 if and only if x 0 The A-module E is called a Hilbert C*-module if E is complete with respect to the norm x x, x 1/2 For any pair of Hilbert C*-modules E 1 and E 2, we define E 1 E 2 {(e 1, e 2 e 1 E 1 and e 2 E 2 } which is also a Hilbert C*-module whose A-valued inner product is given by (x 1, y 1, (x 2, y 2 x 1, x 2 + y 1, y 2, for x 1, x 2 E 1 and y 1, y 2 E 2 If V is a (possibly non-closed A-submodule of E, then V {y E x, y 0, for all x V } is a closed A-submodule of E and V V A Hilbert A-submodule V of a Hilbert A-module E is orthogonally complemented if V and its orthogonal complement V yield E V V, in this case, V and its biorthogonal complement V coincide For the basic theory of Hilbert C*-modules we refer to the book by E C Lance 12 Note that every Hilbert space is a Hilbert C-module and every C*-algebra A, can be regarded as a Hilbert A-module via a, b a b when a, b A Throughout this paper we assume that A is an arbitrary C*-algebra We use, for commutator of two elements The notations Ker( and Ran( stand for kernel and range of operators, respectively Suppose E and F are Hilbert A-modules, L(E, F denotes the set of all bounded adjointable operators from E to F, that is, all operator T E F for which there exists T F E such that T x, y x, T y, for all x E and y F Closed submodules of Hilbert modules need not to be orthogonally complemented at all, however we have the following well known results Suppose T in L(E, F, the operator T has closed range if and only if T has In this case, E Ker(T Ran(T and F Ker(T Ran(T, cf 12, Theorem 32 In view of 16, Lemma 21, Ran(T is closed if and only if Ran(T T is, and in this case, Ran(T Ran(T T Let T L(E, F The Moore Penrose inverse T of T (if it exists is an element X L(F, E which satisfies (1 T XT T, (2 XT X X, (3 (T X T X, (4 (XT XT
4 55 Sharifi and Bonakdar If θ {1, 2, 3, 4}, and X satisfies the equation (i for all i θ, then X is an θ-inverse of T The set of all θ-inverses of T is denoted by T {θ} In particular, T {1, 2, 3, 4} {T } The properties (1 to (4 imply that T is unique and T T and T T are orthogonal projections Moreover, Ran(T Ran(T T, Ran(T Ran(T T, Ker(T Ker(T T and Ker(T Ker(T T which lead us to E Ker(T T Ran(T T Ker(T Ran(T and F Ker(T Ran(T We also have Ran(T Ran(T and Ker(T Ker(T Xu and Sheng in 19 have shown that a bounded adjointable operator between two Hilbert C*-modules admits a bounded Moore-Penrose inverse if and only if the operator has closed range The reader should be aware of the fact that a bounded adjointable operator may admit an unbounded operator as its Moore-Penrose, see 6, 8, 16, 18 for more detailed information It is a classical result of Greville 7, that (T S S T if and only if T T SS T SS T and SS T T S T T S (or equivalently, Ran(SS T Ran(T and Ran(T T S Ran(S for Moore-Penrose invertible matrices T and S The present paper is an extension of some results of 5,7,14 to Hilbert C*-modules settings Indeed, we give some necessary and sufficient conditions for reverse order law for the Moore Penrose inverse by using the matrix form of bounded adjointable module maps These enable us to derive Greville s result for bounded adjointable module maps The matrix form of a bounded adjointable operator T L(E, F is induced by some natural decompositions of Hilbert C*-modules If F M M, E K K then T can be written as the following 2 2 matrix T1 T (11 T 2 T 3 T 4 with operator entries, T 1 L(K, M, T 2 L(K, M, T 3 L(K, M and T 4 L(K, M Lemma 11 Let T L(E, F have a closed range Then T has the following matrix decomposition with respect to the orthogonal decompositions of submodules E Ran(T Ker(T and F Ran(T Ker(T T1 0 Ran(T T Ran(T Ker(T Ker(T, where T 1 is invertible Moreover, T T Ran(T Ker(T Ran(T Ker(T Proof The operator T and its adjoint T have the following representations T1 T T 2 Ran(T Ran(T T 3 T 4 Ker(T Ker(T,
5 The reverse order law for Moore-Penrose inverses 56 T T 1 T3 Ran(T Ran(T T2 T4 Ker(T Ker(T From T (Ker(T {0} we obtain T3 0 and T4 0, thus T 3 0 and T 4 0 Since T (Ker(T {0}, T 2 0 therefore T T 1 0 Moreover since Ran(T is closed, T 1 possesses a bounded adjointable inverse from Ran(T onto Ran(T Now, it is easy to check that the matrix T is the Moore Penrose inverse of T T 1 0 Lemma 12 Let T L(E, F have a closed range Let E 1, E 2 be closed submodules of E and F 1, F 2 be closed submodules of F such that E E 1 E 2 and F F 1 F 2 Then the operator T has the following matrix representations with respect to the orthogonal sums of submodules E Ran(T Ker(T and F Ran(T Ker(T (12 T T1 T 2 E1 E 2 Ran(T Ker(T where D T 1 T1 + T 2 T2 L(Ran(T is positive and invertible Moreover, (13 T T 1 D 1 0 T2 D 1 0 (14 T T1 0 T 2 0 Ran(T Ker(T F1 where D T1 T 1 + T2 T 2 L(Ran(T is positive and invertible Moreover, (15 T D 1 T1 D 1 T2 Proof We prove only the matrix representations (12 and (13, the proof of (14 and (15 are analogous The operator T has the following representation T1 T T 2 E1 Ran(T T 3 T 4 E 2 Ker(T, which yields T T 1 T3 T2 T4 Ran(T Ker(T F 2 E1,, E 2 From T (Ker(T {0} we obtain T3 0 and T4 0 Then T 3 0 and T 4 0 which yield the matrix form (12 of T Consequently, the adjoint operator T has the matrix representation T T 1 0 Ran(T E1 T2 0 Ker(T E 2
6 57 Sharifi and Bonakdar We therefore have (16 T T D 0 Ran(T Ker(T Ran(T Ker(T where D T 1 T1 + T 2 T2 Ran(T Ran(T From Ker(T T Ker(T, it follows that D is injective From Ran(T T Ran(T it follows that D is surjective Hence, D is invertible Now using 14, Corollary 24 and (16 we obtain T T (T T T 1 0 T2 0 D The reverse order law, T 1 D 1 0 T2 D 1 0 We begin this section with the following useful facts about the product of module maps with closed range Suppose E, F and G are Hilbert C*-modules and S L(E, F and T L(F, G are bounded adjointable operators with closed ranges Then T S has closed range if and only if T T SS has, if and only if Ker(T + Ran(S is an orthogonal summand in F, if an only if Ker(S + Ran(T is an orthogonal summand in F For the proofs of the results and historical notes about the problem we refer to 17 and references therein Theorem 21 Suppose E, F and G are Hilbert C*-modules and S L(E, F, T L(F, G and T S L(E, G have closed ranges Then following statements are equivalent (i T S(T S T SS T, (ii T T S SS T T S, (iii S T (T S{1, 2, 3} Proof Using Lemma 11, the operator S and its Moore-Penrose inverse S have the following matrix forms S1 0 Ran(S Ran(S S Ker(S Ker(S, S S Ran(S Ran(S Ker(S Ker(S From Lemma 12 it follows that the operator T and T have the following matrix forms T1 T 2 Ran(S Ran(T T Ker(S Ker(T, T T 1 D 1 0 T2 D 1, 0 where D T 1 T1 + T 2 T2 is invertible and positive in L(Ran(T Then we have the following products T1 S 1 0 T S, (T S (T1 S 1 0, S T S 1 1 T 1 D 1 0
7 The reverse order law for Moore-Penrose inverses 58 It is easy to check that the following three expressions in terms of T 1, T 2 and S 1 are equivalent to our statements (1 T 1 S 1 (T 1 S 1 T 1 T 1 D 1, which is equivalent to (i (2 T 2 T 1 0, which is equivalent to (ii (3 T 1 T 1 D 1 T 1 T 1 and T 1 T 1, D 1 0, which are equivalent to (iii Note that T 1 T1, D 1 0, since T 1 S 1 (T 1 S 1 is self-adjoint We show that (3 (2 (1 (3 To prove (1 (2, we observe that T 1 S 1 (T 1 S 1 T 1 T1 D 1 if and only if (T 1 S 1 (T 1 S 1 T 1 T1 D 1 The last statement is obtained by multiplying the first expression by (T 1 S 1 from the left side, or multiplying the second expression by T 1 S 1 from the left side, and using T 1 T1 T 1 S 1 S1 1 T 1 We therefore have (T 1S 1 (T 1S 1 T 1T 1 D 1 (T 1S 1 (T 1T 1 + T 2T 2 (T 1S 1 T 1T 1 (T 1 S 1 T 2 T 2 0 Ran(T 2 T 2 Ker((T 1 S 1 Ker((T 1 S 1 S 1 T 1 T 2T 2 0 T 2T 2 T 1 0 Ran(T 1 Ker(T 2T 2 Ker(T 2 T 2 T 1 0 To demonstrate (1 (3, we multiply T 1 S 1 (T 1 S 1 T 1 T1 D 1 by T 1 S 1 from the right side, we find T 1 T1 D 1 T 1 T 1, ie (3 holds Finally, we prove (3 (2 If T 1 T1 D 1 T 1 T 1 and T 1 T1, D 1 0, then T 1 T1 T 1 DT 1 T 1 T1 T 1 + T 2 T2 T 1 Consequently, T 2 T2 T 1 0 which implies T 2 T1 0, since Ran(T 1 Ker(T 2 T2 Ker(T2 Theorem 22 Suppose E, F and G are Hilbert C*-modules and S L(E, F, T L(F, G and T S L(E, G have closed ranges Then following statements are equivalent (i (T S T S S T T S, (ii T SS T SS T T, (iii S T (T S{1, 2, 4} Proof The operators T, S and T S and their Moore-Penrose inverses have the same matrix representations as in the previous theorem To prove the assertions, we first find the equivalent expressions for our statements in terms of T 1, T 2 and S 1 (1 (T 1 S 1 T 1 S 1 S 1 1 T 1 D 1 T 1 S 1, which is equivalent to (i (2 T 1 S 1 S 1T 1 D 1 T 1 T 1 S 1 S 1 and T 1 S 1 S 1T 1 D 1 T 2 0, which are equivalent to (ii (3 T 1 T 1 D 1 T 1 T 1 and S 1 S 1, T 1 D 1 T 1 0, which are equivalent to (iii
8 59 Sharifi and Bonakdar Note that S 1 S1, T1 D 1 T 1 0, since (T 1 S 1 T 1 S 1 is self-adjoint We show that (1 (3 (2 (1 Suppose (1 holds If we multiply (T 1 S 1 T 1 S 1 S1 1 T 1 D 1 T 1 S 1 by T 1 S 1 from the left side, we obtain T 1 T 1 T1 D 1 T 1 Furthermore, S 1 S1, T1 D 1 T 1 0, ie (3 holds Suppose (3 holds Obviously, T 1S 1S1 T1 D 1 T 1 T 1T1 D 1 T 1S 1S1 T 1S 1S1, that is, the first equality of (2 holds According to the fact that (T 1 T1 + T 2 T2 D 1 T 1 T 1 and the assumption T 1 T1 D 1 T 1 T 1, we have T2 D 1 T 1 0 Consequently, Ran(D 1 T 1 Ker(T 2 T 2 Ker(T 2, which yields T2 D 1 T 1 0 Therefore, T1 D 1 T 2 0 which establishes the second equality of (2 In order to prove (2 (1, we multiply T 1 S 1 S1T 1 D 1 T 1 T 1 S 1 S1 by (T 1 S 1 from the left side In view of S 1 S1, T1 D 1 T 1 0, we find S 1T 1 D 1 T 1 (T 1 S 1 T 1 S 1 S 1 (T 1 S 1 T 1 S 1 S 1T 1 D 1 T 1 (S 1 1 (T 1 S 1 T 1 S 1 S 1 1 T 1 D 1 T 1 S 1 Now we are ready to derive Greville s result, which also gives an answer to a problem of 17 about the reverse order law for Moore-Penrose inverses of modular operators The operators SS and T T are orthogonal projections onto Ran(S and Ran(T Ran(T, respectively These facts together with Theorems 21 and 22 lead us to the following result Corollary 23 Suppose E, F and G are Hilbert C*-modules and S L(E, F, T L(F, G and T S L(E, G have closed ranges Then following statements are equivalent (i (T S S T, (ii T S(T S T SS T and (T S T S S T T S, (iii SS T T S T T S and T SS T T T SS, (iv Ran(T T S Ran(S and Ran(SS T Ran(T Acknowledgments This research was done while the first author stayed at Mathematisches Institut of the Westfälische Wilhelms-Universität in Münster, Germany He would like to express his thanks to Professor Wend Werner and his colleagues in functional analysis and operator algebras group for warm hospitality and great scientific atmosphere The first author gratefully acknowledges financial support from the Alexander von Humboldt Foundation This research was partially supported by a grant from Shahrood University of Technology The authors are grateful to the referee for his/her careful reading and useful comments
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