The Drazin inverses of products and differences of orthogonal projections

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1 J Math Anal Appl wwwelseviercom/locate/jmaa The Drazin inverses of products and differences of orthogonal projections Chun Yuan Deng School of Mathematics Science, South China Normal University, Guangzhou 51631, PR China Received 13 September 6 Available online 3 January 7 Submitted by JH Shapiro Abstract In this note, the Drazin inverses of products and differences of orthogonal projections on a Hilbert space are established 7 Elsevier Inc All rights reserved Keywords: Subspace; Projection; Drazin inverse 1 Introduction Let BH be the set of all linear bounded operators on a Hilbert space H An operator A BH is said to be positive if Ax,x for all x H An operator P BH is said to be an orthogonal projection if P = P = P Clearly, an orthogonal projection is a positive operator [] Throughout this paper, the range, the null space and the adjoint of A BH are denoted by RA, N A and A, respectively If A BH, then the Drazin inverse [1,4,5,7, 13 16] of A, denoted by A D, is the unique operator X BH satisfying the relations A k+1 X = A k, XAX= X, AX = XA, where k = IndA, theindexofa, is the smallest nonnegative integer for which N A k = N A k+1 and RA k = RA k+1 If no such k exists, we say that A is not Drazin invertible If IndA = 1, the operator X is called the group inverse of A IfIndA =, then it can be easily seen that A is invertible and A D = A 1 Partial support was provided by the National Natural Science Foundation Grants of China No address: cy-deng@63net -47X/$ see front matter 7 Elsevier Inc All rights reserved doi:1116/jjmaa7156

2 CY Deng / J Math Anal Appl The concept of a Drazin inverse was shown to be very useful in various applied mathematical settings For example, applications to Markov Chains, singular differential and difference equation, iterative method or multibody system dynamics can be found in Refs [8,1 1], respectively This paper is concerned with the Drazin inverses of products and differences of orthogonal projections on a Hilbert space This problem was first considered by Drazin in 1958 in his celebrated paper [4] Herein, it was proved that P + Q D = P D + Q D provided PQ= QP =, and AB D = B D A D provided AB = BA The general question of how to express P + Q D as a function of P,Q,P D,Q D, without side conditions, is very difficult and remains open [7] The aim of this paper is to present some formulae for the Drazin inverses of products and differences of orthogonal projections on a Hilbert space We shall assume familiarity with the theory of Drazin inverse as given in [7,13 16] We start by discussing some lemmas which are well known For the sake of convenience, we state them without proof Lemma 11 See [5] Let T BH be Drazin invertible with IndT = k Then RT k is closed and RT k = RT D Lemma 1 Let A and B be two elements of BH If there exists an invertible operator G BH such that A = BG or A = GB, then RA is closed if and only if RB is closed Lemma 13 See [3] Let A BH be a positive operator and k> be an arbitrary positive integer Then the following statements hold: 1 RA RA 1 k and RA = RA 1 k, where K denotes the closure of K; RA is closed if and only if RA = RA 1 k ; 3 RA = H if and only if A is invertible Lemma 14 See [3,6,9] Let P and Q are orthogonal projections in BH Denote H 1 = RP RQ, H = RP N Q, H 3 = N P RQ, H 4 = N P N Q, H 5 = RP H 1 H and H 6 = H 5 i=1 H i Then P and Q have the following operator matrices: I P = I I, Q Q = I I Q 1 I Q 1 D 1 D Q 1 I Q 1 D I Q D with respect to the space decomposition H = 6 i=1 H i Where Q is a positive contraction on H 5 with that neither nor 1 belong to the point spectrum of Q,Dis a unitary operator from H 6 onto H 5, and I and are the identity operator and the zero operator in the subspace under consideration Main results In this section, we first study the Drazin invertibility of the products of two orthogonal projections, and get the following results

3 66 CY Deng / J Math Anal Appl Theorem 1 Let P and Q be two orthogonal projections in BH Then the following statements are equivalent: 1 PQis Drazin invertible; QP is Drazin invertible; 3 I P I Q is Drazin invertible; 4 I QI Pis Drazin invertible Proof By Lemma 14, we have Q PQ= I Q 1 I Q 1 D Let Q = UP 1 be the polar decomposition Q and P 1 = Q λde λ be the spectral representation of P 1 Defining P + 1 by Q { P + 1 = λ + de λ, where λ + λ 1 if λ, = if λ = Then P + 1 is not necessarily bounded but it is a densely defined and closed operator on H 5 Put Q + = P + 1 U, I S = I I I I I Q 1 Q + I Then Q + Q = I H5 and SPQ S 1 Q = I If PQ is Drazin invertible, then SPQ S 1 is Drazin invertible By Lemma 11, there exists a positive integer k such that RQ k is closed Since Q is a positive operator and is not the point spectrum of Q, it follows that RQ is closed and RQ = H 5 by Lemma 13 Hence Q is invertible on H 5 Conversely, if Q is invertible on H 5, then SPQ S 1 D Q 1 = I Note that SP Q D S 1 = SP Q S 1 D So P Q D = S 1 SPQ S 1 D Q 1 S = I Q 3 I Q 1 D This shows that PQis Drazin invertible if and only if Q is invertible Similarly, we can show that Q QP = I D Q 1, I Q 1

4 CY Deng / J Math Anal Appl I P I Q = I I Q 1 Q 1 I QI P= I I Q 1 D, D Q D D Q D and each of the three operators mentioned above is Drazin invertible if and only if Q BH 5 is invertible, therefore the results are clear Theorem Let P and Q be two orthogonal projections and PQ be Drazin invertible Then the following statements hold: 1 P Q D = QP if and only if PQ= QP ; P Q D = P QP D P I QI P D ; 3 P Q D PQ= P QP D PQ Proof 1 By the proof of Theorem 1, we have P Q D Q = I 1 Q 3 I Q 1 D 3 Note that Q QP = I D Q 1 I Q 1 By Lemma 14, it is clear that PQ= QP if P Q D = QP On the contrary, if PQ= QP, P and Q have the following matrix representation forms: P = I and Q = Q 1 Q 4 with respect to the space decomposition H = RP N P, where Q 1 and Q are orthogonal projections on RP and N P, respectively Hence P Q D = QP = Q 1 Put I Q 1 S = I I I I I Q 1 D I Then Q PQP = I, SI QI PS 1 = I D Q D So D I QI P = S 1 [ SI QI PS 1] D S Q 3 = I I Q 1 D, D Q 1 D Q 1 P QP D = I 5,

5 68 CY Deng / J Math Anal Appl Hence P Q D = P QP D P I QI P D and P Q D PQ= P QP D PQ Let P and Q be two orthogonal projections, S m,p denotes an m-factor product of P,Q with P being the first factor and P,Q occurring alternately and S n,q denotes an n-factor product of P,Q, with Q be the first factor and P,Q occurring alternately Then an immediate consequence of Lemma 14, Theorems 1 and is the following Corollary 3 Let P and Q be two orthogonal projections and PQ be Drazin invertible Then the following statements are equivalent: 1 Sm,P D = SD n,q for some m, n except for the trivial case where simultaneously m = n and P = Q; PQ= QP ; 3 Sm,P D = SD n,q for every m, n Theorem 4 Let P and Q be two orthogonal projections Then the following statements are equivalent: 1 P Q is Drazin invertible; P + Q is Drazin invertible; 3 P QP is Drazin invertible; 4 Q QP is Drazin invertible Proof 1 By Lemma 14, we have I Q P Q = I I Q 1 I Q 1 D, 6 I Q 1 D I Q D I P Q Q = I I D 7 I Q D If P Q is Drazin invertible, then P Q is Drazin invertible and P Q D = P Q D According to Lemmas 11 and 13 we have RI Q is closed Since Q is a positive contraction on H 5 and 1 is not the point spectrum of Q, we have I Q is invertible Conversely, if I Q is invertible, then I Q Q 1 I Q 1 D I Q 1 D I Q D is invertible on H 5 H 6 and the inverse is I Q 1 I Q 1 D I Q 1 I

6 CY Deng / J Math Anal Appl Hence, P Q D I Q 1 = I I I Q 1 D 8 I Q 1 I Note that P + Q and I + Q P + Q = I I I Q 1 I Q 1 D 9 D Q 1 I Q 1 D I Q D If P + Q is Drazin invertible, then RP + Q is closed by Lemmas 11 and 13 Define operators S and T by I S = I I I I I Q 1 I + Q 1 I and I I + Q T = I I I I 1 Q 1 I Q 1 D I It is clear that S,T are invertible and I + Q SP + QT = I I I D I Q I + Q 1 D By Lemma 1, if RP + Q is closed, then RI Q is closed Since Q is a positive contraction on H 5 and 1 is not the point spectrum of Q, we have I Q is an invertible operator on H 5 Conversely, if I Q is invertible, then I + Q Q 1 I Q 1 D D Q 1 I Q 1 D I Q D is invertible on H 5 H 6 and the inverse is I Q 1 I Q 1 D I Q 1 D I + Q I Q 1 D So P + Q D = I I I I Q 1 I Q 1 D I Q 1 D I + Q I Q 1 D 1 Using the same method, we can show that the Drazin invertibility of P QP and Q QP are all equivalent to invertibility of I Q on H 5 So the details are omitted Theorem 5 Let P and Q be two orthogonal projections and P Q be Drazin invertible Then the following statements hold:

7 7 CY Deng / J Math Anal Appl P Q D = P Q if and only if PQ= QP ; P + Q D = P + Q if and only if PQ= QP = ; 3 P Q D = P Q P QP D Q QP D ; 4 P P + Q D P Q D P Q = Proof 1 If P Q D = P Q, then it is clear PQ= QP by 6 and 8 On the other hand, by 4, P Q D = I Q 1 Q = P Q if PQ= QP If P + Q D = P + Q, then PQ= QP = by 9 and 1 On the other hand, if PQ= QP =, then Q 1 = in 4 It is clear that P + Q D = P + Q = I Q by 4 3 Similar to Theorem, we have I Q 1 P QP D = I D Q 1, I Q 3 Q QP D Q 1 = I I Q 3 D D I Q 1 D Hence, by 7, 8 and 1, P Q D = P Q P QP D Q QP D, P P + Q D P Q D P Q = Remark 1 Let P and Q as two orthogonal projections have the form 1 In particular, if PQ= T, where T {,P,Q,QP}, then H 5 H 6 ={} Hence, by 1, 8, 1 and Theorem 5 we obtain immediately the following results: P Q D = P Q for PQ= T, where T {,P,Q,QP}; P + Q D = P + Q 3 PQ for PQ= QP ; P + Q D = 1 P + Q for PQ= P ; P + Q D = P 1 Q for PQ= Q If P and Q satisfy Theorems 1 and 4, then PQ QP is also Drazin invertible and, by 1, 5 and 8, we have P Q QP D Q 1 = I Q 1 D D Q 1 I Q 1 Acknowledgments = P Q D P QP D P QP D P Q D The author would like to thank Hong Ke Du for his comments on some early draft versions Many thanks also go to an anonymous referee for his/her valuable comments and suggestions which help to improve the presentation of this paper

8 CY Deng / J Math Anal Appl References [1] SL Campbell, CD Meyer, Generalized Inverse of Linear Transformations, Pitman, London, 1979 [] S Cheng, Y Tian, Moore Penrose inverses of products and differences of orthogonal projectors, Acta Sci Math Szeged [3] CY Deng, HK Du, Common complements of two subspaces and an answer of Groß s question, Acta Math Sinica [4] MP Drazin, Pseudoinverse in associative rings and semigroups, Amer Math Monthly [5] HK Du, CY Deng, The representation and characterization of Drazin inverses of operators on a Hilbert space, Linear Algebra Appl [6] P Halmos, Two subspaces, Trans Amer Math Soc [7] RE Hartwig, GR Wang, Y Wei, Some additive results on Drazin inverse, Linear Algebra Appl [8] RE Hartwig, J Levine, Applications of the Drazin inverse to the Hill cryptographic system, Part III, Cryptologia [9] S Kruglyak, V Rabanovich, Y Samolenko, Decomposition of a scalar matrix into a sum of orthogonal projections, Linear Algebra Appl [1] CD Meyer Jr, The condition number of a finite Markov chains and perturbation bound for the limiting probabilities, SIAM J Algebraic Discrete Methods [11] CD Meyer Jr, JM Shoaf, Updating finite Markov chains by using techniques of group inversion, J Stat Comput Simul [1] B Simeon, C Rentrop, The Drazin inverse in multibody system dynamics, Numer Math [13] H Tian, On the reverse order laws AB D = B D A D, J Math Res Exposition [14] Y Wei, The Drazin inverse of updating of a square matrix with application to perturbation formula, Appl Math Comput [15] Y Wei, S Qiao, The representation and approximation of the Drazin inverse of a linear operator in Hilbert space, Appl Math Comput [16] Y Wei, G Wang, The perturbation theory for the Drazin inverse and its applications, Linear Algebra Appl

Some additive results on Drazin inverse

Linear Algebra and its Applications 322 (2001) 207 217 www.elsevier.com/locate/laa Some additive results on Drazin inverse Robert E. Hartwig a, Guorong Wang a,b,1, Yimin Wei c,,2 a Mathematics Department,