Moore-Penrose Inverse and Operator Inequalities

Transcription

1 E extracta mathematicae Vol. 30, Núm. 1, 9 39 (015) Moore-Penrose Inverse and Operator Inequalities Ameur eddik Department of Mathematics, Faculty of cience, Hadj Lakhdar University, Batna, Algeria seddikameur@hotmail.com Presented by Manuel González Received October 4, 014 Abstract: In this note, we shall give complete characterizations of the class of all normal operators with closed range, and the class of all selfadjoint operators with closed range multiplied by scalars in terms of some operator inequalities. Key words: Closed range operator, Moore-Penrose inverse, selfadjoint operator, unitary operator, normal operator, operator inequality. AM ubject Class. (010): 47A30, 47A05, 47B Introduction and preliminaries results Let B(H) be the C*-algebra of all bounded linear operators acting on a complex Hilbert space H, and let N (H), and (H) denote the class of all normal operators, and the class of all selfadjoint operators in B(H), respectively. We denote by I(H), the group of all invertible elements in B(H), 0 (H) = (H) I(H), the set of all invertible selfadjoint operators in B(H), N 0 (H) = N (H) I(H), the set of all invertible normal operators in B(H), R(H), the set of all operators with closed ranges in B(H), x y (where x, y H), the one rank operator on H defined by (x y) z = z, y x, for every z H, the positive square root of the positive operator (where B(H)), {} = {X B(H) : X = X} the commutant of (where B(H)). 9

2 30 a. seddik For B(H), let R() and ker denote the range and the kernel of, respectively. It is known that R(H) if and only if there exits a unique operator R(H) satisfying the following four equations =, =, ( ) =, ( ) =. Then, the operator is called the Moore-Penrose inverse of, and it satisfies that and are orthogonal projections onto R() and R( ), respectively. It is clear that if I(H), then = 1, and if B(H) is a surjective operator (resp. injective with closed range), then = I (rep. = I). [ For every ] in R(H), we associate the matrix representation = 1 on R() ker. The operator is called an EP operator if R( ) = [ ] R(), or equivalently = 0 and 1 is invertible; in this case 1 = 1 0 (see []). Any normal operator with a closed range in B(H) is an EP operator (see[4]). One of the most essential inequalities in operator theory is the arithmeticgeometric mean inequality given by (see [1, 1]) A, B, X B(H), A AX XBB AXB. (1) From this inequality, we deduce immediately that for every 0 (H) the following inequality holds X 1 1 X X. () The inequality () was proved by Corach-Porta-Recht [6] with another motivation and independently of inequality (1). In [14], we give some characterization of some distinguished classes of operators in terms of operator inequalities. We proved that the class of all operators I(H) satisfying () is exactly the class C 0 (H) (that is the class of all rotations of invertible selfadjoint operators). o, the class C 0 (H) is characterized by the following property X 1 1 X X, ( I(H)). (3) In [15], we have found two other characterizations of this last class given by

3 moore-menrose inverse and operator inequalities 31 X 1 1 X = X 1 1 X, ( I(H)). (4) X 1 1 X X 1 1 X, ( I(H)). (5) For the class of all invertible normal operators N 0 (H), we have showed in [15] that this class is characterized by each of the following three properties X 1 1 X X, ( I(H)). (6) X 1 1 X = X 1 1 X, ( I(H)). (7) X 1 1 X X 1 1 X, ( I(H)). (8) In this note, we consider the following extensions of the above six properties from the domain I(H) to the domain R(H) X X X, ( R(H)). (9) X X = X X, ( R(H)). (10) X X X X, ( R(H)). (11) X X X, ( R(H)). (1) X X = X X, ( R(H)). (13) X X X X, ( R(H)). (14) We consider also in this note the following two properties X X X, ( R(H)). (15) X X X, ( R(H)). (16)

4 3 a. seddik In the finite dimensional case, we showed in [13] that each of properties (9) and (15) characterizes the class C(H) and each of properties (1) and (16) characterizes the class N (H) (here R(H) = B(H)). In this note and in general situation, we shall show that any one of properties (9), (10), (11) and (15) characterizes the class C ((H) R(H)) and any one of properties (1), (13), (14) and (16) characterizes the class N (H) R(H).. Characterizations and Moore-Penrose inverse To achieve our new characterizations, we need the following lemma. Lemma 1. Let B(H). If is injective with a closed range (or is surjective) and satisfies property (16), then it is normal. Proof. Assume that is injective. It is clear that is also injective with a closed range. Hence, = I = ( ). Put P =, Q = and R =. ince is injective with a closed range, then ker P = ker = {0}, and R(P ) = R( ) is closed (since R( ) is also closed). Thus ker P = {0} and R(P ) = (ker P ) = H. o, P is invertible. On the other hand, since is also injective with a closed range, using the same argument as used before R is invertible. The proof is given in six steps. tep 1. {R} = {P }. From (16), it follows that X X ( ) X ( ). ince = 1, the last inequality shows that X X ( ) X ( ). Thus the following inequality holds X X ( ) X. By taking the polar decomposition of each of the two operators and in this last inequality, we obtain X RXR 1 P XP 1.

5 moore-menrose inverse and operator inequalities 33 Hence from [14, Lemma 3.], {R} {P }. o, from this last inequality, we obtain also the following inequality P 1 XP ( RP 1 ) X ( RP 1) 1 X. where RP 1 is a positive invertible operator. o, from [14, Theorem 3.3], {P } = { RP 1}. Hence {R} = {P }. tep. ( ) =. From the property (16) the following inequality holds X X X. ( ) It is known that is the unique solution of the following four equations: X =, XX = X, (X) = X, (X) = X. It is easy to see that ( ) satisfies the first three equations. Now we prove that ( ) also satisfies the last equation. ince the operator ( ) is a projection, it suffices to prove that its norm is less than or equal to one. By taking X = ( ) in ( ), we obtain Hence ( ) ( ) ( ). ( ) 1. Therefore ( ) =. tep 3. ( ) = ( ). ince ( ) = ( ), then ( ) = ( ). o from tep, we obtain ( ) = ( ). ince is injective, we have ( ) = ( ). tep 4. P and R are diagonal matrices with respect to the orthogonal direct sum H = R() ker. All matrices given here are given [ with] respect to the[ orthogonal ] direct sum H = R() ker 1. Put =. Then = 1 1. ince ( ) = ( ), then the operators and commute (see [3, 10]). [ ] [ ] Hence P = P o that P =, where P 0 P 1 = 1 and P =. From tep 1 and { }, we deduce that ( ) and

6 34 a. seddik [( ) commute. Therefore R ] [ ] = 1 1 ( ) 0 R Hence R =, 0 R where R 1 = 1 and R =. tep 5. ker = {0}. ince is injective, ker = {0} if and only if = 0. [ Assume that ker {0}. It is easy to see that 1 = 1 ] 0. [ ] Q1 0 Then Q =, where Q 1 = ( 1 1 ). Using the polar decomposition of the operators, and in ( ) we obtain the following inequality XQ RXP 1 P X. ( ) By putting X = 0 Y P (where Y B(ker )) in inequality ( ), we obtain Y B(ker ), R Y P Y P. Hence = R P =. o that. Thus = 0, which is a contradiction with ker {0}. Therefore ker = {0}. tep 6. is normal. From tep 5, is surjective. o that is invertible and satisfies property (16). Thus satisfies property (6). Hence is normal. With the second assumption surjective, is injective with a closed range satisfying also property (16), so that is normal. Hence is normal. Theorem 1. Let R(H). Then the following properties are equivalent: (i) N (H), (ii) X X = X X, (iii) X X X X, (iv) X X X, (v) X X X. Proof. The proof is trivial if = 0. Assume now that 0.

7 moore-menrose inverse and operator inequalities 35 (i) (ii). Assume N (H). Then the equality X = X holds for every X B(H). Hence the equalities X = X and X = X hold for every X B(H). o, we obtain (ii). The implication (ii) (iii) is trivial. (iii) (vi). This implication follows immediately using [11, Theorem.4]. (iv) (v). Assume (iv) holds. Then the following inequality holds X X X. From this inequality and since = = 1, property (v) follows immediately. [ ] [ ] 1 (v) (i). Assume (v) holds. Let = R() ker and let = [ ] [ T1 T R( ] [ ] [ ] ) X1 0 R(). Put X = ker ker. By a simple computation, we obtain [ ] [ X = 1 X 1 0, X X1 = ] [ ] 1 X X, X = X 1. [ ] [ ] Y1 Y Put Y = R() ker, where Y denotes one of the above three opera- [ tors. Then Y = Y Y = Y1 Y1 Y ] Y 0 = Y 1 Y1 Y Y. Hence we have X = 1 X 1, X = (X1 1)(X 1 1) (X 1 1 )(X 1 1 ) and = X 1 1 K (X 1 1 ) = X 1 1 K, X = ( 1 X 1 1 )( 1 X 1 1 ) ( 1 X 1 )( 1 X 1 ) = (1 X 1 )K ( 1 X 1 ) = 1 X 1 K, (where K is the positive square root of the positive operator 1 1 ). Then using (v), we obtain the following inequality X 1 B (R()), 1 X 1 X1 1 K 1 X 1 K (*) On the other hand, if we put X = x y (for x, y H) in (v), we obtain x, y H, y x ( x ) y x y (**)

8 36 a. seddik We shall prove (i) in three steps. tep 1. 1 or T 1 is bounded below. Assume that it is not the case. With the condition 1 is not bounded below, we may choose a sequence (u n ) in H such that u n 0 and u n = 1, for n 1. For every n 1, there exist x n R( ) and z n ker such that u n = x n z n. Thus, we obtain x n = u n 0, x n =u n =1, x n = u n, for n 1 With the second condition T 1 is not bounded below, by the same argument, we may choose a bounded sequence (y n ) in H satisfying ( ) y n 0, y n = 1, for n 1. Applying (**) for x = x n and y = y n, we obtain n 1, y n x n xn ( ) y n. Letting n, we have 0, which is impossible. Therefore 1 or T 1 is bounded below. tep. 1 or T 1 is surjective. Assume that T 1 is bounded below. Then there exists a constant k > 0 such that x H, ( ) x k x. o we have ( ) k. From [7], we obtain R( ) R(). Thus R( ) = R(). o 1 is surjective. Also, if 1 is bounded below, then by the same argument, we deduce that T 1 is surjective. tep 3. is normal. Assume that 1 is surjective (on R()). Then 1 is also surjective on R(). ince R() {0}, thus 1 1 = I 1 = 1 ( ) 1 (where I1 is the identity operator on R()), 1 1 and ( 1) 1 are nonzero orthogonal projections. By putting X 1 = ( 1) in (*), we obtain 1 ( ) ( ) 1 ( ) 1 1 K 1 1 K

9 moore-menrose inverse and operator inequalities 37 Hence, 1 ( 1 ) = 1, (1 ) 1 K = (1 ) 1 1 K 1 K, and 1 (1 ) K 1 1 ( 1 ) K = 1 K. Thus 1 1 K. Hence 1 1 K = 1 K ( 1 ) = 1 1 ( 1 )( 1 ) 1 1 = 1. Hence 1 1 ( 1 )( 1 ) = 1. ince 1 1 is an orthogonal projection, by a simple computation, we obtain that = 0. Hence = = 0. o, we obtain that 1 is a surjective operator (as element in B (R()) and satisfies the following inequality X B(R()), 1 X X 1 1 X 1 Utilizing Lemma 1, we obtain that 1 is normal. Hence is normal. With the second assumption T 1 surjective, and since satisfies (v), by using the same argument as used with the first assumption, we obtain also that is normal. Thus is normal. Corollary 1. Assume dim H <. The class N (H) is characterized by each of the following properties X X X, ( B(H)), X X = X X, ( B(H)), X X X X, ( B(H)), X X X, ( B(H)). Theorem. Let R(H). Then the following properties are equivalent: (i) C(H), (ii) X X = X X, (iii) X X X X, (iv) X X X, (v) X X X. Proof. The implications (i) (ii) and (ii) (iii) are trivial. The implication (iii) (iv) follows immediately from [11, Theorem.4]. The implication (i) (v) follows immediately from (1). Assume now that (iv) or (v) holds. Applying the triangular inequality in (iv) or (v), we obtain from Theorem 1, that is normal (with a closed range).

10 38 a. seddik [ ] [ ] 1 0 R() o that is an EP operator satisfying (iv) or (v). o = ker where 1 is invertible on R(). Hence we obtain the following inequality X B(R()), 1 X X 1 X. Hence 1 is a selfadjoint operator in B(R()) multiplied by a nonzero scalar. Thus C(H). Corollary. Assume that dim H <. The class C(H) is characterized by each of the four following properties X X X, ( B(H)), X X = X X, ( B(H)), X X X X, ( B(H)), X X X, ( B(H)). Remarks. 1. Let the following extension of the property (15) to the domain B(H): X X X, ( B(H)). (17) The restriction of this property to the domain R(H) characterizes the class C((H) R(H)) (by using Theorem ). On the other hand, by using the inequality (1), the property (17) is satisfied for every C(H). o, does property (17) characterize the class C(H)?. Let the following extension of the property (16) to the domain B(H): X X X, ( B(H)). (18) The restriction of this property to the domain R(H) characterizes the class N (H) R(H) (by using Theorem 1). The property (18) is satisfied for every N (H). Indeed, if N (H), then X X = X X X X X, for every X B(H). o, does property (18) characterize the class N (H)? Acknowledgements The author would like to thank the referee for valuable suggestions and comments.

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