On the invertibility of the operator A XB

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1 NUMERICAL LINEAR ALGEBRA WITH APPLICATIONS Numer. Linear Algebra Appl. 29; 16: Published online 8 April 29 in Wiley InterScience On the invertibility of the operator A XB Chun Yuan Deng, School of Mathematics Science, South China Normal University, Guangzhou 51631, People s Republic of China SUMMARY For a given pair of A, B) an arbitrary operator X, expressions for the inverse, the Moore Penrose inverse the generalized Drazin inverse of the operator A XB are derived under some conditions. Copyright q 29 John Wiley & Sons, Ltd. Received 13 February 28; Revised 23 February 29; Accepted 24 February 29 KEY WORDS: inverse; Moore Penrose inverse; generalized Drazin inverse; projection; spectrum 1. INTRODUCTION Let H be a separable, complex Hilbert space, denote by BH) the Banach space of all bounded linear operators on H. For an operator A BH),σA),RA),NA) A denote the spectrum, the range, the null space the adjoint of A, respectively. If M N are two closed subspaces of H, thenm denotes the closure of subspace M, M {x H: x y for all y M} M NM N. In addition, if M N, we write M+NM N. It is easy to see that HM M N N. An operator P BH) is said to be a projection if P 2 P P.The projection onto a subspace U is denoted by P U. The identity on a Hilbert space H is denoted by I H or I if there does not exist confusion. The Moore Penrose inverse of T is denoted by T +, it is the unique solution to the following four operator equations: TXT T, XTX X, TXTX), XT XT) 1) Correspondence to: Chun Yuan Deng, School of Mathematics Science, South China Normal University, Guangzhou 51631, People s Republic of China. cydeng@scnu.edu.cn Contract/grant sponsor: National Natural Science Foundation Grants of China; contract/grant number: Copyright q 29 John Wiley & Sons, Ltd.

2 818 C. Y. DENG It is well known that T has a Moore Penrose inverse if only if RT ) is closed. The Moore Penrose inverse of T is unique T + T TT ) + T T ) + T see [1]). We use T {1} to denote one arbitrary solution to the operator equation TXT T. In general, T {1} is not unique. The concept of the generalized Drazin inverse was introduced by Koliha [2], which is the element T d BX) such that TT d T d T, T d TT d T d, T T 2 T d is quasi-nilpotent 2) If there exists an integer k such that T T 2 T d ) k, then least such k is the index of T, denoted by indt )k. Otherwise, we say indt )+. When indt )1, then the Drazin inverse T d is called the group inverse. When indt ), then the Drazin inverse T d T 1.Itiswellknown that for T BX), T d exists if only if / acc[σt )] in that case T d is unique. If T is the generalized Drazin invertible, then the spectral idempotent T π of T corresponding to {} is given by T π I TT d. The operator matrix form of T with respect to the space decomposition X NT π ) RT π ) is given by T T 1 T 2, where T 1 is invertible T 2 is quasi-nilpotent see [3 ]). The inverse, the Moore Penrose inverse the related generalized Drazin inverse formulae of operator A XB have been used in a wide variety of applications. An excellent review by Hager [12] described some of the applications to statistics, networks, structural analysis, asymptotic analysis, optimization partial differential equations see [2, 8 25]). The objectives of this paper are to derive the properties of an operator A XB establish some interesting results. The paper is organized as follows. We study the invertibility the representation of the inverse of the operator A XB in Section 2. In Section 3, we study the generalized invertibility the representation of the generalized inverse of the operator A XB. Finally in Section 4, we study the representation of the generalized Drazin inverse of the operator A XB. 2. THE INVERSE OF A XB First we consider the invertibility of operator A XB. Let M be a closed subspace of H with orthocomplement M. According to the orthogonal decomposition HM M, every operator T BH) is written in a block-form A ) B T C D 3) If A BM) is invertible, the Schur complement of A in T is the matrix D CA 1 B see [1, 19, 2, 26 28]). The idea of the Schur complement goes back to Sylvester in Emilie Haynsworth was the first author to introduce the notion of Schur complement but it had already been used by Schur [2]. The following lemma is well known can be found in stard textbooks on linear algebra. Lemma 2.1 see Schur [2]; Ando [27] Corach et al. [28]) Let T BH) be given by 3) such that A is invertible. Then T is invertible if only if the Schur complement D CA 1 B is invertible. Let T BK,H). Say that T is right invertible if there is an S BH,K) with TS I H.We have the following result.

3 ON THE INVERTIBILITY OF THE OPERATOR A XB 819 Theorem 2.2 Let A BH) B BH,K) be given. If A, B ) is right invertible dimna) dimna ), then there always exists an operator X BK,H) such that A XB is invertible. Without loss of generality, we can assume A 1. Let A QU be the polar decomposition of A. As dimna)dimna ), we can suppose that U is a unitary operator. As A, B ) is right invertible, HRA, B ))RA, B )A, B ) )RA A+ B B). Hence, positive operator A A+ B B is invertible. From U A+ B B U QU + B B U Q 2 U + B B A A+ B B we have U A+ B B is invertible. Put X UB,then A XB A+UB B UU A+ B B) is invertible. Note that ) ) ) ) A XB I X A X I I I B I B I ) Then A XB is invertible if only if A X B I is invertible. Moreover, we have the following result. Theorem 2.3 see Lu [18]) 1) Let A BH) B BH,K) be given such that A is invertible. Then for every X BK,H), A XBis invertible if only if the Schur complement I BA 1 X of A is invertible. In this case ) 1 A X A 1 + A 1 X I BA 1 X) 1 BA 1 A 1 X I BA 1 X) 1 ) B I I BA 1 X) 1 BA 1 I BA 1 X) 1 A XB) 1 A 1 + A 1 X I BA 1 X) 1 BA 1 2) Let A BH) B BH,K) be given such that B is invertible. Then for every X BK,H), A XB is invertible if only if the Schur complement X AB 1 of B is invertible. In this case ) 1 A X B 1 X AB 1 ) 1 B 1 + B 1 X AB 1 ) 1 AB 1 ) B I X AB 1 ) 1 X AB 1 ) 1 AB 1 A XB) 1 B 1 X AB 1 ) 1

4 82 C. Y. DENG Theorem 2.4 Let A, B BH) be given normal operators RA) be closed with NA)RB). Then for every X BH), A XB is invertible if only if SXS is invertible. In this case where S P NA) T P NB). A XB) 1 T B + )S A + )I +TXS)[SXS) 1 T ] If A, B BH) are normal operators RA) is closed with NA)RB), ThenNA) NB) X X 12 ) A X A 22 X 21 X 22 B I B I I with respect to the space decomposition H HNA) NA) NB) NB), where A 22 B are invertible. Note that I X X 12 I I A 22 X 21 X 22 I I B I I I I I B I A 22 X 22 X 21 I X 12 X Denote by S P NA) T P NB). We know A XBis invertible if only if X is invertible, that is, X P NA) XP NB) P NA) XP NA) SXS is invertible. From 4) 5) we have B 1 ) 1 X 1 B 1 B 1 X 1 X 12 A X A 1 22 X 21 X 1 A 1 22 A 1 22 X 21X 1 X 12 A 1 B I X 1 X 1 X 12 I 22 X 22 4) 5)

5 ON THE INVERTIBILITY OF THE OPERATOR A XB 821 Hence, ) ) ) A XB) 1 1 ) I A X I X I B I B I I B 1 X 1 A 1 22 X 21X 1 A 1 22 I I B 1 X 1 A XB) 1 A 1 22 X 21 X 1 A 1 22 ) B 1 I I ) A 1 22 ) ) I X 1 I X 21 I P NB) B + )P NA) A + )I + P NB) XP NA) ) P NA) XP NA) ) 1 P NB) ) T B + )S A + )I +TXS)[SXS) 1 T ] ) Remark Let A A XB be invertible. From A XB AI A 1 XB) we have 1 / σa 1 XB). As σa 1 XB)\{}σBA 1 X)\{}σXBA 1 )\{}, I A 1 XB, I BA 1 X I XBA 1 are all invertible. Then A XB) 1 A 1 I XBA 1 ) 1 I A 1 XB) 1 A 1 A 1 + A 1 X I BA 1 X) 1 BA 1 6) Thisresultis calledsherman Morrison Woodburyformula see [21, 22]). Infact, ifra) is closed I A + XB is invertible, we still have A + I XBA + ) 1 I A + XB) 1 A + A + + A + X I BA + X) 1 BA + 3. THE GENERALIZED INVERSE OF A XB Second, we consider the generalized invertibility of operator A XB. We first give some lemmas. Lemma 3.1 see Chen et al. [16]) Let Q be an idempotent. Then P RQ) QQ + Q I ) 1.

6 822 C. Y. DENG Since Q + Q I ) 2 I +UU, where U Q Q, Q + Q I is always invertible. Since P RQ) Q Q P RQ) Q P RQ),wehaveP RQ) Q + Q I ) Q, thatis, P RQ) QQ + Q I ) 1 Lemma 3.2 see Deng Du [29]) Let A BH) have a closed range. If A {1} is one solution to the operator equation AXA A, then P RA) AA A) {1} A, A + P RA ) A {1} P RA) where P RA) A + are irrespective of the choice of A A) {1} A {1}, respectively. Since RA) is closed, A has the operator matrix form ) A1 RA ) A : NA) ) ) RA) NA ) where A 1 as an operator from RA ) onto RA) is invertible. It is clear that A + A 1 1, P RA ) I RA ), P RA) I RA). The general solution to the operator equation AXA A is A 1 ) ) ) X X 12 X 21 X 22 : RA) NA ) RA ) NA) where X 12, X 21, X 22 are arbitrary elements. Hence, if A {1} is one solution to the operator equation AXA A, wehave P RA ) A {1} P RA) A 1 1 A+ Similarly, the general solution to the operator equation A A)X A A)A A) has the form A A) 1 ) S X 12 RA ) ) RA ) ) : : S 21 S 22 NA) NA) where S 12, S 21, S 22 are arbitrary elements. Hence, if A A) {1} is one solution to the operator equation A A)X A A)A A), wehave AA A) {1} A I RA) P RA) The first part of the following results was first given by Ton Steerneman Frederieke van Perlo-ten Kleij see [23]) in the case of matrices A, B X. We will show that it also holds in an infinite-dimensional Hilbert space. Theorem 3.3 Let B BH). ForeveryX BH), 1) if BX I,thenB +, X + exist I XB) + I XX + )I B + B) 2) if RI XB)NB), thenb + exists B + P NB) XP RB).

7 ON THE INVERTIBILITY OF THE OPERATOR A XB 823 1) Put Q I XB.SinceBX I, X B are bounded below. So that RX) RB ) are closed. Then B + X + exist. From BX I, we have QX BQ. Put M I XX + )I B + B). Then QM I XB)I XX + )I B + B)I XB)I B + B) I B + B MQ I XX + )I B + B)I XB)I XX + )I XB) I XX + Hence, QMQI B + B)I XB)I XB) Q MQMI XX + )I XX + ) I XB) M. Hence I XB) + I XX + )I B + B) 2) Since RQ)NB), BXB BI Q) B. SoBX is an idempotent RB)RBXB) RBX) RB), i.e. RB)RBX) RB) is closed. Put N P NB) XP RB).Then BN BP NB) XP RB) BXP RB) P RB) NB P NB) XP RB) B P NB) XB P NB) I Q) P NB) Obviously, we have BNB P RB) B B NBN P NB) P NB) XP RB) N. Hence, B + P NB) XP RB). Theorem 3.4 Let A, B BH) RA) be closed. For every X BH), 1) if 1 / σxba + ) I AA + )I XBA + ) 1 A XB), then A XB) {1} I A + XB) 1 A + 2) if I AA + )A XB), then A XB) + A XB) [I XBA + )AA I XBA + ) +XBI AA + )B X ] + 3) if I AA + )A XB) A XB)I AA + ), then A XB) + I A + XB) + A + A + I XBA + ) + A + + A + X I BA + X) + BA + Since RA) is closed, A + exists. Let A, XB A XB as operators from RA ) NA) into RA) NA ) have the form A A 1, ) ) N1 N 2 A1 N 1 N 2 XB A XB N 3 N 4 N 3 N 4

8 824 C. Y. DENG It is clear that A + A ) Since 1 / σxba + ), I XBA + I A + XB are invertible A1 I XBA + ) 1 A 1 N 1 ) 1 ) A1, I A + XB) 1 A + N 1 ) 1 ) N 3 A 1 N 1 ) 1 I From I AA + )I XBA + ) 1 A XB) wegetn 3 A 1 N 1 ) 1 N 2 + N 4. So A XB)I A + XB) 1 A + A XB) ) A1 N 1 N 2 A1 N 1 ) 1 ) ) A1 N 1 N 2 N 3 N 4 N 3 N 4 ) ) A1 N 1 N 2 A1 N 1 N 2 N 3 N 3 A 1 N 1 ) 1 N 2 N 3 N 4 A XB 2) If I AA + )A XB), then N 3 N 4. Note that ) ) N1 XBA + N2 A XBI A + A) We get that A XB) + A XB) [A XB)A XB) ] + [ A1 A XB) N 1 ) N 2 A1 N 1 ) )] + N2 [A1 A XB) N 1 )A 1 N 1 ) + N 2 N2 ]+ ) A XB) [I XBA + )AA I XBA + ) + XBI AA + )B X ] + 3) If I AA + )A XB) A XB)I AA + ), then N 2, N 3 N 4. The result follows directly by item 2). Remark 1) In Theorem 3.4, if 1 / σxba + ) A is invertible, the formulae change into Sherman Morrison Woodbury formula 6). Hence, Theorem 3.4 generalizes formula 6). 2) In the proof of Theorem 3.4, we can see the condition I AA + )A XB) implies that I AA + )I XBA + ) 1 A XB).

9 ON THE INVERTIBILITY OF THE OPERATOR A XB 825 Theorem 3.5 Let A, B BH) A be invertible. For every X BH), ifba 1 X 1, then A XB) + exists A XB) + [I A 1 XBA 1 XB+A 1 XB) I ) 1 ] A 1 I XBA 1 ) + I XBA 1 )[I XBA 1 XBA 1 ) ] 1 Since BA 1 X 1,I XBA 1 )I XBA 1 ) I XBA 1. This indicates that RI XBA 1 ) is closed. So RA XB)RI XBA 1 )A)RI XBA 1 ) is closed. Hence, A XB) + exists. An easy calculation yields A XB) {1} A 1 I XBA 1 ) +. Hence, by Lemma 3.2, A XB) + P NA XB) A 1 I XBA 1 ) + P RA XB) Note that XBA 1 A 1 XB are idempotent. By Lemma 3.1 P NA XB) I P NA XB) I P RA 1 XB) I A 1 XB[A 1 XB+A 1 XB) I ] 1 hence, P RA XB) P RI XBA 1 ) I XBA 1 )[I XBA 1 XBA 1 ) ] 1 A XB) + [I A 1 XBA 1 XB+A 1 XB) I ) 1 ] A 1 I XBA 1 ) + I XBA 1 )[I XBA 1 XBA 1 ) ] 1 If RA),RB) RAB) are closed, then the rule AB) + B + A + is called the reverse order rule for the Moore Penrose inverse it does not hold in general). In [3] it is shown that if RA),RB) RAB) are closed, then AB) + B + A + if only if RA AB) RB) RBB A ) RA ). Hence, we have the following result. Theorem 3.6 Let A, B BH) RA )+RB ) be closed. For every X BH), if RI, X ) ) RA, B ) ),then A XB) + A A+ B B)A I + XX ) 1 A A+ B B)B X I + XX ) 1 Denote by Then M ) I X N ) A B ) ) ) M A XB I A XB MN X A XB) X

10 826 C. Y. DENG ) ) ) AA XB) NN M A A XB) BA XB) B ))R )) From R I A X B we have RM MN) RN) RNN M ) RM ). Hence, the reverse order rule MN) + N + M + holds, i.e. A XB) + ) ) + ) + A I X B Since RA )+RB )RN )RN N) 1/2 )RA A+ B B) 1/2 ) RA )+RB ) is closed, it follows that RN ) RN N) are closed. Hence, N + N N) + exist. From N + N N) + N, we obtain A N + A+ B B) + ) A B ) A A+ B B) + A A A+ B B) + B ) Similarly, we have Hence, A XB) + ) I + XX M + ) 1 ) X I + XX ) 1 A A+ B B) + A A A+ B B) + B ) I + XX ) 1 ) X I + XX ) 1 A A+ B B) + A I + XX ) 1 A A+ B B) + B X I + XX ) 1 ) Hence, A XB) + A A+ B B) + A I + XX ) 1 A A+ B B) + B X I + XX ) 1 4. THE GENERALIZED DRAZIN INVERSE OF A XB In this section, we will obtain the generalized Drazin inverse for A XB, where A BH) is the generalized Drazin invertible, B BH,K) X BK,H). We will prove that Sherman Morrison Woodbury formula see Equation 6)) has the analogous result concerning the generalized Drazin inverse the generalized Drazin Schur complement.

11 ON THE INVERTIBILITY OF THE OPERATOR A XB 827 Lemma 4.1 see Djordjević Wei [6]) Let P Q BH) be the generalized Drazin invertible. If PQ, then P + Q is the generalized Drazin invertible [ [ P + Q) d I QQ d ) Q n P d ) ]+ n+1 Let Z I BA d X. Then we have the following result. Q d ) n+1 P ]I n PP d ) Theorem 4.2 Let A BH) be the generalized Drazin invertible, B BH,K) X BK,H). IfI AA d )X, BI AA d ) X I ZZ d )B, then A XB) d A d + A d X I BA d X) d BA d Let X A d + A d X I BA d X) d BA d. We prove the theorem by checking four conditions in 2). Note that A XB)X A XB)[A d + A d X I BA d X) d BA d ] AA d + AA d XZ d BA d XBA d XBA d XZ d BA d AA d + XZ d BA d XBA d X I Z)Z d BA d AA d XBA d + XZZ d BA d AA d Similarly, X A XB) AA d, X A XB)X X A XB) A XB) 2 X A XB)[I A XB)X] A XB)[I AA d ] AI AA d ) is a quasinilpotent operator Remark In particular 1) If indi BA d X) inda) in Theorem 4.2, then I BA d X) d I BA d X) 1 A d A 1, the formulae change into Sherman Morrison Woodbury formula 6). Hence, Theorem 4.2 generalizes the formula 6). 2) If I BA d X A are normal operators such that indi BA d X)1 inda)1 in Theorem 4.2, then I BA d X) d I BA d X) + A d A +, Theorem 4.2 becomes Theorem 3.4, item 3).

12 828 C. Y. DENG Theorem 4.3 Let A BH) be the generalized Drazin invertible, B BH,K) X BK,H). If there exists an idempotent P such that AP PA PX, then A XB) d R d + PA d + R d XBPA d R d ) n+2 XBPA n I AA d ) where R A XB)I P). I RR d ) A XB) n XBPA d ) n+2 Let S AP T I P)A XB). ThenST. From AP PA PX wehave A XB AP+I P)A I P)XB S +T By Lemma 4.1, we have [ [ S +T ) d I TT d ) T n S d ) ]+ n+1 T d ) n+1 S ]I n SS d ) Next, we will give the representations SS d, TT d, T n S d ) n+1 T d ) n+1 S n, respectively. Since P is an idempotent PA AP,wehaveP d P, S d AP) d PA d SS d PAA d. Note that T I P)A XB) AI P) XBI P)+ XBI P) XB R XBP with XBP) 2 [I P)XBP] 2 XBPR XBPA XB)I P) XBPI P)A XBI P)) Hence, by Lemma 4.1 again, we get that T d XBP+ R) d R d R d ) 2 XBP Since XBPR, XBPR d XBPRR d ) 2 TT d R XBP)[R d R d ) 2 XBP]RR d R d XBP Since RSA XB)I P)APA XB)I P)PA, T d S R d R d ) 2 XBP)S R d ) 2 XBS T d ) 2 S R d R d ) 2 XBP)R d ) 2 XBS R d ) 3 XBS. T d ) n+1 S n R d ) n+2 XBS n n 1)

13 ON THE INVERTIBILITY OF THE OPERATOR A XB 829 Since AP PA PX, T n S d ) n+1 [I P)A XB)] n PA d ) n+1 Hence, [ ] T d ) n+1 S n [I P)A XB] n PA d ) n+1 [I P)A XB] n 1 [I P)A XB]PA d ) n+1 [I P)A XB] n 1 XBPA d ) n+1 A XB) n 1 XBPA d ) n+1 n 1) XBPA XB) n XB XBPA XB)A XB) n 1 XB XBAPA XB) n 1 XB XBA 2 PA XB) n 2 XB XBA n PXB I SS d ) [ ] T d ) n+1 S n T d ) n+1 S n SS d R d [ R d ) n+2 XBPA n R d ] R d ) n+2 XBPA n R d R d ) n+2 XBPA n I AA d ) [ ] I TT d ) T n S d ) n+1 [I RR d + R d I P)XBP] [PA d A XB) n XBPA d ) n+2 ] The result is obtained. PAA d PA d + R d XBPA d I RR d ) A XB) n XBPA d ) n+2 In addition, if B also satisfies that BP B in Theorem 4.3, then R d [A XB)I P)] d A d I P) I RR d I AA d + AA d P. Theorem 4.3 can be simplified as: Corollary 4.4 Let A BH) be the generalized Drazin invertible, B BH,K) X BK,H). If there exists an idempotent P such that AP PA, PX BP B, then A XB) d A d + A d XBA d A d ) n+2 XBA n I AA d ) I AA d ) A n XBA d ) n+2

14 83 C. Y. DENG In particular, if P I AA d,then A XB) d A d A d ) n+2 XBA n If P AA d,then A XB) d A d A n XBA d ) n+2 5. CONCLUDING REMARKS This paper is devoted to presenting various representation formulae for the inverse, the Moore Penrose inverse the generalized Drazin inverse of the operator A XB under certain circumstances. These results can be applied to update infinite Markov chains. It is of interest to derive the most general formulae, this would be rather complicated. ACKNOWLEDGEMENTS We thank two anonymous referees for their valuable comments detailed suggestions which are helpful in obtaining the shortened paper improving the readability of the paper significantly. Many thanks also go to Yimin Wei for his comments some helpful discussions on parts of this work. REFERENCES 1. Ben-Israel A, Greville TNE. Generalized Inverses: Theory Applications 2nd edn). Springer: New York, Koliha JJ. A generalized Drazin inverse. Glasgow Mathematical Journal 1996; 38: Deng CY, Du HK. The reduced minimum modulus of Drazin inverses of linear operators on Hilbert spaces. Proceedings of the American Mathematical Society 26; 134: Deng CY. The Drazin inverse of bounded operators with commutativity up to a factor. Applied Mathematics Computation 28; 26: Deng CY. The Drazin inverses of sum difference of idempotents. Linear Algebra its Applications 29; 43: Djordjević DS, Wei Y. Additive results for the generalized Drazin inverse. Journal of the Australian Mathematical Society 22; 73: Djordjević DS. Characterizations of normal, hyponormal EP operators. Journal of Mathematical Analysis Applications 27; 329: Wei Y. The Drazin inverse of a modified matrix. Applied Mathematics Computation 22; 125: Wei Y, Li X. An improvement on perturbation bounds for the Drazin inverse. Numerical Linear Algebra with Applications 23; 1: Wei Y, Li X, Bu F. A perturbation bound of the Drazin inverse of a matrix by separation of simple invariant subspaces. SIAM Journal on Matrix Analysis Applications 25; 27: Wei Y, Qiao S. The representation approximation of the Drazin inverse of a linear operator in Hilbert space. Applied Mathematics Computation 23; 138: Hager WW. Updating the inverse of a matrix. SIAM Review 1989; 31: Choi M-D, Wu PY. Convex combinations of projections. Linear Algebra its Applications 199; 13615): Du H. Operator matrix forms of positive operator matrices. Chinese Quarterly Journal of Mathematics 1992; 7: Du H. Another generalization of Anderson s theorem. Proceedings of the American Mathematical Society 1995; 1239):

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