International Journal of Mathematical Analysis Vol. 11, 2017, no. 8, 389-396 HIKARI Ltd, www.m-hikari.com https//doi.org/10.12988/ijma.2017.7342 Moore-Penrose Inverses of Operators in Hilbert C -Modules Fugen Gao College of Mathematics and Information Science Henan Normal University, Henan, China Guoqing Hong College of Mathematics and Information Science Henan Normal University, Henan, China Copyright c 2017 Fugen Gao and Guoqing Hong. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract For two given bounded adjointable operators T and S between Hilbert C*-modules, it is well known that an operator Moore-Penrose inverse exists iff the operator has closed range. In this paper, we give some formulas for the Moore-Penrose inverses of products T S. Mathematics Subject Classification 47A05, 46L08, 15A09 Keywords Hilbert C*-module, bounded adjointable operator, Moore- Penrose inverse, product operators 1 Introduction and preliminaries Throughout the paper, we assume that E, F, G are Hilbert A-modules, where A is a C*-algebra. For an operator T, let R(T be the range of T and K(T be the kernel of T. The notation [T, S] denotes the commutator T S ST of T and S. We abbreviate Moore-Penrose inverse to MP-inverse.
390 Fugen Gao and Guoqing Hong For two given invertible operator T, S in Hilbert C*-modules, the equality (T S 1 = S 1 T 1 is called the reverse order law. If T and S are invertible operator then the reverse order law effective but this case does not validate for the MP-inverse in general. The problem first studied by Greville [5] and then reconsidered by Bouldin [1] and Izumino [6]. A number of researchers discussed the problem such that reverse order law holds from different angles [2, 3, 4, 8, 15]. We also refer to another interesting Sharifi and Bonakdar [15] of this type. In this paper we continue and supplement this research by using the space decompositions and operator matrix representations in C*-modules. We specialize the investigations to the MP-inverses of T S and give some formulas for the MP-inverses of T S. The notion of C*-module has been presented by Kaplansky [7] and Paschke [13]. A Hilbert C*-module E is right A-module with an inner product, E E A satisfying (1 x, αy + βz = α x, y + β x, z, (2 x, ya = x, y a, (3 y, x = x, y, (4 x, x 0 and x, x = 0 iff x = 0, for all x, y, z E, a A, α, β C and such that E is complete corresponding to x = x, x. Hilbert C*-module can be regard as a generalization of the Hilbert space. In spite of this promotion is natural, some basic properties of Hilbert spaces are not applicable in Hilbert C*-modules. Therefore, when studying a certain problem in C*-modules, it is useful to find conditions to gain the results homologous to those for Hilbert spaces. The elements of L(E, F is said to be bounded adjointable operators, if there is an operator T F E satisfying T (x, y = x, T (y for each x E and y F. The operator T is selfadjoint if T = T and T L(E, F is MP invertible if there is X L(F, E such that T XT = T, XT X = X, (T X = T X, (XT = XT. There is at most one element X satisfying the above formula, if T L(E, F is MP-invertible, then the unique X is called MP inverse of T. In symbols this is denoted by T. Xu and Sheng [16] show that an operator permits a MP-inverses iff the operator has closed range. Moreover, Sharifi [14] show that T S has closed range iff the kernel of T is orthogonally complemented with the range of S, iff the kernel of S is orthogonally complemented with the range of T. If a Hilbert A-submodule W of a Hilbert A-module E and its orthogonal complement W yield E = W W, then W is orthogonal complemented. We associate the matrix representation of an operator T L(E, F with respect
Moore-Penrose inverses of operators 391 to some natural decompositions of C*-modules. If E = K K, F = H H then operator T has the following matrix form T1 T T = 2 T 3 T 4 where T 1 L(K, H, T 2 L(K, H, T 3 L(K, H, T 4 L(K, H. 2 Main results First, we need following three auxiliary results. Lemma 2.1. [9] Let T L(E, F and R(T be closed. Then T has the following matrix form T1 0 R(T T = R(T K(T K(T, where T 1 is invertible. Moreover T T 1 1 0 = R(T K(T R(T. K(T Lemma 2.2. [15] 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. If T L(E, F has closed range, then T has the following matrix form (i T = T1 T 2 [ E1 E 2 ] R(T K(T. Moreover T = T 1 D 1 0 T2 D 1, 0 where D = T 1 T1 + T 2 T2 L(R(T is positive and invertible. (ii T1 0 R(T T = F1. T 2 0 K(T F 2 Moreover T = [ D 1 T 1 D 1 T 2 ],
392 Fugen Gao and Guoqing Hong where D = T 1 T 1 + T 2 T 2 L(R(T is positive and invertible. Lemma 2.3. Let N L(E, F have a closed range, and let M L(F be selfadjoint and invertible. Then R(MN = R(N iff [M, NN ] = 0. Proof. Suppose N have a closed range, M be selfadjoint and invertible. We consider the orthogonal direct sums E = R(N + K(N and F = R(N + K(N. Then the operators N and M have the corresponding matrix forms as follows N = N1 0 where N 1 is invertible, and M1 M M = 2 M 3 M 4 R(N R(N K(N K(N, R(N R(N K(N K(N, where M 3 = M2. It follows that M1 N MN = 1 0 R(N R(N M 3 N 1 0 K(N K(N. M1 0 Hence, R(MN = R(N implies M 3 = 0 and M 2 = 0, so M =. 0 M 4 Since M is selfadjoint and invertible, we [ obtain ] that M 1 and M 4 are also N selfadjoint and invertible. Since N 1 1 0 =, we obtain that MNN = NN M holds. Conversely, if M is invertible and MNN = NN M, then R(MN = R(MNN = R(NN M = R(NN = R(N. In the following, we give some formulas about (T S. Theorem 2.4. Suppose S L(E, F, T L(F, G, TS have closed ranges. Then (T S = (T T S T iff R(T T T S = R(T S. Proof. By Lemma (2.1, the operator S and its MP-inverse S have the following matrix forms S1 0 R(S S = R(S K(S K(S, S S 1 1 0 R(S R(S = K(S. K(S From Lemma 2.2 it follows that the T and its MP-inverse T have the following matrix forms
Moore-Penrose inverses of operators 393 T = T1 T 2 R(S R(T K(S K(T, T = T 1 D 1 0 T2 D 1, 0 where D = T 1 T1 + T 2 T2 L(R(T is positive and invertible. Then we have the following products T1 S T S = 1 0, (T S (T1 S = 1 0 S, S T 1 1 T1 = D 1 0. Notice that R((T T S = R(S T T = S R(T T = S R(T = R((T S is closed, so R(T T S is closed. First, we find the equivalent conditions for our statements. (1. Let us denote C = T T S. Using [[11], Corollary 2.4] we find C as follows [ C = (C C C = ] (S 1 T1 D 1 T 1 S 1 S1T 1 D 1 T 1 (S1T 1 D 1 T 1 S 1 S1T 1 D 1 T 2. Now, we can see that (T S = (T T S T is equivalent with (T 1 S 1 = (S1T 1 D 1 T 1 S 1 S1T 1 D 1 = (D 1 2 T 1 S 1 D 1 2. (2. It is obvious that T T DT1 S T S = 1 0, so R(T T T S = R(T S holds iff R(DT 1 S 1 = R(T 1 S 1. (1 (2 From the third Moore-Penrose equation for (T 1 S 1 = (D 1 2 T 1 S 1 D 1 2, we see that T 1 S 1 (D 1 2 T 1 S 1 D 1 2 is selfadjoint. So we have the following equivalents Now, T 1 S 1 (D 1 2 T 1 S 1 D 1 2 is selfadjoint D 1 2 T 1 S 1 (D 1 2 T 1 S 1 D 1 is selfadjoint [D, D 1 2 T 1 S 1 (D 1 2 T 1 S 1 ] = 0 D 1 2 T 1 S 1 (D 1 2 T 1 S 1 = D 1 2 T 1 S 1 (D 1 2 T 1 S 1 D DT 1 S 1 (D 1 2 T 1 S 1 = T 1 S 1 (D 1 2 T 1 S 1 D. R(DT 1 S 1 = R(DT 1 S 1 (T 1 S 1 = R(T 1 S 1 (T 1 S 1 D = R(T 1 S 1. (2 (1 If R(DT 1 S 1 = R(T 1 S 1, then we apply Lemma 2.3 to obtain [D, T 1 S 1 (T 1 S 1 ]=0.
394 Fugen Gao and Guoqing Hong Now, from the previous argument it follows that T 1 S 1 (D 1 2 T 1 S 1 D 1 2 is selfadjoint. Noticed that (D 1 2 T 1 S 1 D 1 2 T 1 S 1 is the orthogonal projection onto R((T 1 S 1 D 1 2 = R((T 1 S 1, so T 1 S 1 (D 1 2 T 1 S 1 D 1 2 T 1 S 1 = T 1 S 1. Finally, it is not difficult to examine that (T 1 S 1 = (D 1 2 T 1 S 1 D 1 2 holds. Similarly to Theorem 2.4 we have Theorem 2.5. Suppose S L(E, F, T L(F, G, TS have closed ranges. Then (T S = S (T SS iff R(S S(T S = R((T S. Proof. According to Theorem 2.4, we have the following equivalences (S T = (T (S T T R(T T T S = R(T S Now,take T = S and S = T, (T S = S (T S S R(S S S T = R(S T. Theorem 2.6. Suppose S L(E, F, T L(F, G, TS have closed ranges. Then S = (T S T iff R(T T S = R(S. Proof. We keep the matrix forms of T and S as in previous theorems. (1 We can obatin that S = (T S T iff I = (T 1 S 1 T 1 S 1 and (T 1 S 1 T 2 = 0. Hence, S = (T S T is equivalent to the following two conditions T 1 is injective with closed range and (T 1 S 1 T 2 = 0. (2 R(T T S = R(S iff R(T 1 T 1 S 1 = R(S 1 and T 2 T 1 S 1 = 0. Hence, R(T T S = R(S is equivalent to the following two conditions T 1 is injective with closed range and T 1 T 2 = 0. To prove the equivalence (1 (2, we have the following (T 1 S 1 T 2 = 0 R(T 2 K((T 1 S 1 = K((T 1 S 1 (T 1 S 1 T 2 = 0 T 1 T 2 = 0. We also prove the following result. Theorem 2.7. Suppose S L(E, F, T L(F, G, TS have closed ranges. Then T = S(T S iff R(SS T = R(T. Proof. From the Theorem 2.6 it follows that S = T (S T iff R(T T S = R(S. Now replace, T and S by S and T, to obtain theorem holds.
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