HYPO-EP OPERATORS 1. (Received 21 May 2013; after final revision 29 November 2014; accepted 7 October 2015)

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1 Indian J. Pure Appl. Math., 47(1): 73-84, March 2016 c Indian National Science Academy DOI: /s x HYPO-EP OPERATORS 1 Arvind B. Patel and Mahaveer P. Shekhawat Department of Mathematics, Sardar Patel University, Vallabh Vidyanagar , Gujarat, India Department of Mathematics, M. P. Shah Arts & Science College, Surendranagar , Gujarat, India, s: abp1908@gmail.com; mahaveer.maths@gmail.com (Received 21 May 2013; after final revision 29 November 2014; accepted 7 October 2015) In this paper we study closed range operators on a Hilbert space such that the range is contained in the range of its adjoint. Some results pertaining to these operators and operator matrices are discussed. Key words : Hypo-EP operator; EP operator; operator matrices. 1. INTRODUCTION Let H be a separable Hilbert space and BL(H) be the space of all bounded linear operators from H to H. We say that A BL(H) is hypo-ep if the range R(A) of A is closed, and R(A) R(A ) equivalently N (A) N (A ), where A denotes the adjoint of A. For A BL(H) with R(A) closed, there is a unique A BL(H) such that AA A = A, A AA = A, (AA ) = AA, (A A) = A A. The operator A is known as the Moore-Penrose inverse of A [11]. Infact AA is the orthogonal projection onto R(A) and A A is the orthogonal projection onto R(A ). 1 The research work is supported by UGC-SAP-DRS-II Grant No. F.510/3/DRS/2009/SAP-1 provided to the Department of Mathematics, Sardar patel University. The second author is also supported by UGC-JRF Grant

2 74 ARVIND B. PATEL AND MAHAVEER P. SHEKHAWAT An operator A BL(H) is called an EP operator if A A = AA [11]. In fact A BL(H) is EP if and only if R(A) is closed and R(A) = R(A ). It is well known that EP stands for equiprojection. An operator A BL(H) is normal if A A = AA and hyponormal if A x Ax for each x H. It is observed in [9], that A is hypo-ep if and only if A Ax AA x for each x H. It is also proved that A is hypo-ep if and only if A A 2 A = AA. It is observed in ([10], Corollary 2.5), that for closed range operators A and B, AB is a closed ranged operator if and only if N (A) + R(B) is closed. We note that the unilateral right shift operator on the Hilbert space l 2 is hypo-ep but it is not EP. For A BL(H), it is easy to observe the following: (1) Both A and A are hypo-ep if and only if A is EP. (2) A hypo-ep operator on a finite dimensional Hilbert space is EP. (3) A hyponormal operator with closed range is hypo-ep. Infinite dimensional EP operators have been studied by several authors. Hypo-EP operators have been introduced by Itoh [9] presumably motivated by hyponormal operators. In the present paper, we discuss powers and product of hypo-ep operators. We also discuss hypo-ep operator matrices. 2. POWERS AND PRODUCT OF HYPO-EP OPERATORS For a hyponormal A BL(H), A 2 need not be hyponormal [6]; however for a hypo-ep operator, we have the following. Theorem 2.1 If A BL(H) is hypo-ep, then A n is hypo-ep for each n. We shall need a couple of lemmas. It is well known that for a hyponormal operator A BL(H), N (A) = N (A n ) for each n, where N (A) = {x H : Ax = 0} is the null space of A. The following is a hypo-ep analogue of this. Lemma 2.2 If A BL(H) is hypo-ep, then N (A) = N (A n ) for each n. PROOF : We prove this result by induction on n. First we prove it for n = 2. Clearly, N (A) N (A 2 ). Let x N (A 2 ). Then Ax N (A). Since A is hypo-ep, A Ax = 0. So, Ax 2 = A Ax, x = 0. Thus x N (A). Hence N (A) = N (A 2 ). Now suppose that the result holds for n = k. Let x N (A k+1 ). Then Ax N (A k ) = N (A) N (A ). Thus A Ax = 0, which implies that Ax = 0. So, x N (A). Thus N (A k+1 ) N (A). Hence N (A) = N (A k+1 ).

3 HYPO-EP OPERATORS 75 The converse of above lemma need not be true as is seen from the following example. Consider A : C C C C define by A(x, y) = (x, ix). Then A (x, y) = (x iy, 0). Since A is idempotent, N (A) = N (A n ) for each n. Also N (A) = {(0, y)/y C} and N (A ) = {(x, y)/x = iy, x, y C}. Thus A is not hypo-ep. We note that the right unilateral shift operator A on l 2 is hypo-ep but R(A) R(A n ) for any n > 1. Corollary 2.3 If A BL(H) is hypo-ep and nilpotent, then A = 0. Lemma 2.4 If A BL(H) is hypo-ep, then R(A n ) is closed for each n. PROOF : First we prove that N (A) is invariant under A j for each j. Let x N (A). Since A is hypo-ep, x N (A ) and so A x N (A). Thus A (N (A)) N (A). Hence A(N (A) ) N (A). Thus A j (N (A) ) N (A). To see that R(A n ) is closed, let x N (A n ). Since A is hypo-ep, by Lemma 2.2, N (A) = N (A j ) for each j. Thus x N (A). So A j x N (A) for each j. In particular A n 1 x N (A). Since R(A) is closed, there exists α 0 such that Ay α y for each y N (A). So A(A n 1 x) α A n 1 x. Now again using the fact that A n 2 x N (A), we get A n x α 2 A n 2 x. Continuing this process we get A n x α n x for x N (A n ). Hence R(A n ) is closed. PROOF OF THEOREM 2.1 : By Lemma 2.4, R(A n ) is closed for each n. So it is enough to prove that N (A n ) N (A n ). Now by Lemma 2.2, N (A) = N (A n ) for each n. Thus N (A n ) = N (A) N (A ) N (A n ). Hence N (A n ) N (A n ). Thus A n is hypo-ep. Hartwig [8] and Basket and Katz [1], discussed the product of two EP operators on finite dimensional spaces. Necessary and sufficient conditions for the product of two EP operators to be an EP operator have been discussed by Djordjevic [4]. It is shown that the product of two commuting EP operators is EP. We discuss the case of hypo-ep operators. Theorem 2.5 Let A, B BL(H) be hypo-ep operators. (a) If R(AB) is closed, R(AB) R(A) R(B ) and N (AB) N (A) + N (B), then AB is hypo-ep. (b) If A is injective and R(AB) R(A) R(B ), then AB is hypo-ep. (c) If AB is hypo-ep, then R(AB) R(A) R(B ).

4 76 ARVIND B. PATEL AND MAHAVEER P. SHEKHAWAT PROOF : (a) Since R(AB) is closed, it is enough to prove that R(AB) R((AB) ). Now, R(AB) R(A) R(B ) = (R(A) R(B )) = (R(A) + R(B ) ) = (N (A ) + N (B)) (N (A) + N (B)) R((AB) ) = R((AB) ). Thus AB is hypo-ep. (b) Suppose A is injective and R(AB) R(A) R(B ). As A is injective and B is hypo-ep, N (A) + R(B) = R(B) is closed. Hence R(AB) is closed. Since N (AB) = N (B), by part (a), AB is hypo-ep. (c) Suppose AB is a hypo-ep operator. Since A is hypo-ep, R(A ) is closed. Thus H = R(A ) N (A). Consider the following decomposition of A: A = A : R(A ) N (A) R(A ) N (A) Since B is hypo-ep, R(B ) is closed. Thus H = R(B ) N (B). Let x H. Then x = x 1 + x 2, for some x 1 R(B ), x 2 N (B). Since Bx H = R(A ) N (A), Bx = y 1 + y 2, for some y 1 R(A ), y 2 N (A). Define B 1 : R(B ) R(A ) and B 2 : R(B ) N (A) by B 1 (x 1 ) = y 1 and B 2 (x 1 ) = y 2. Clearly B 1, B 2 are well defined. Thus we have following decomposition for B B = B 1 0 B 2 0 : R(B ) N (B) R(A ) N (A). Since B 1 : R(B ) R(A ), R(B ) = N (B 1 ) (N (B 1 ) R(B )). Now AB = A 1B 1 0 : R(B ) R(A ). 0 0 N (B) N (A) Clearlly N (B 1 ) N (B) N (AB). Since A 1 is injective, N (AB) N (B 1 ) N (B). Thus N (AB) = N (B) N (B 1 ). Therefore N (AB) = (N (B 1 ) N (B)) = N (B 1 ) N (B) = N (B 1 ) R(B ). Since AB is hypo-ep, R(AB) R(AB) = N (AB) R(B ). Therefore R(AB) R(A) R(B ). Corollary 2.6 Let A, B BL(H) be hypo-ep with A injective and AB = BA, then AB is hypo-ep. The following example shows that we cannot drop the condition AB = BA in Corollary 2.6. We

5 HYPO-EP OPERATORS 77 do not know whether theinjectivity assumptioncan be omitted. Example : Let A = 1 0 and B = 1 1. Then A is invertible and B is EP. Then AB = 1 1 with N (AB) = {(x, x) : x R} and N ((AB) ) = {( 2x, x) : x R}. 2 2 Thus N (AB) is not contained in N ((AB) ). Hence AB is not hypo-ep. Theorem 2.7 If every operator on a Hilbert space H of rank n is hypo-ep, then dim H = n. PROOF : Suppose that dim H > n. Then there exist orthonormal vectors u 1, u 2,.... u n and orthonormal vectors w 1, w 2,.... w n such that Span{u 1, u 2,..... u n } Span{w 1, w 2,.... w n }. Define T : H H by T x = n x, u i w i, x H. Thus for x, y H, T x, y = n x, u i w i, y = i=1 n i=1 i=1 x, y, w i u i = x, n y, w i u i = x, T y. Therefore T y = n y, w i u i. Thus we have N (T ) = {u 1, u 2,... u n } and N (T ) = {w 1, w 2,... w n }. i=1 Since T is hypo-ep, {u 1, u 2,... u n } {w 1, w 2,... w n }. Therefore Span{w 1, w 2,... w n } Span{u 1, u 2,... u n }. Thus Span{w 1, w 2,.... w n } = Span{u 1, u 2,.... u n }. This contradiction gives dim H = n. Theorem 2.8 Let A BL(H) be hypo-ep and B BL(H) be unitarily equivalent to A. Then B is hypo-ep. PROOF : Since R(A) is closed and B is unitarily equivalent to A, R(B) is closed. Since B is unitarily equivalent to A, B = U AU for some unitary operator U. Let x N (B). Then U AUx = 0, so that AUx = 0. Since A is hypo-ep, A Ux = 0. Thus B x = 0. Hence N (B) N (B ). Therefore B is hypo-ep. The following example shows that in above theorem unitary equivalence cannot be replaced by similarity Example : Let A = and B = R}. Thus B is not hypo-ep. and C = i= Take B = C 1 AC. Then A is EP with N (B) = {(0, 0, α) : α R} and N (B ) = {(0, α, α) : α

6 78 ARVIND B. PATEL AND MAHAVEER P. SHEKHAWAT 3. HYPO-EP OPERATOR MATRICES Hartwig [7] has discussed when a block matrix A C is EP. In this section, we discuss hypo- B D EP operator matrices. We generalize some results of Hartwig to infinite dimensional spaces. Theorem 3.1 Let A, X BL(H) and M = A AX. Then M is hypo-ep if X A X AX and only if A is hypo-ep. PROOF : Suppose M is a hypo-ep operator. Let x N (A). Then A AX x = 0. Since M is a hypo-ep operator, A A X X A X AX 0 0 X A X A X x = 0. Therefore A x = 0. Thus N (A) N (A ). Next we prove that R(A) is closed. 0 0 Suppose A(x n ) u, for some u H. Then A AX x n u. Since X A X AX 0 X u R(M) is closed, there exist x, y H such that u = A AX x = X u X A X AX y A(x + Xy). Thus u = A(x + Xy). Therefore R(A) is closed. Hence A is hypo-ep. X A(x + Xy) Conversely suppose A is a hypo-ep operator. Let x A(x + Xy) N (M). Then = y X A(x + Xy) 0. Thus x + Xy N (A). Since A is a hypo-ep operator, x + Xy N (A ). Therefore M x = A (x + Xy) 0 = 0. Thus N (M) N (M ). To see that y X A (x + Xy) 0 R(M) is closed, let M x n u. Then A(x n + Xy n ) u. Thus y n v X A(x n + Xy n ) v A(x n + Xy n ) u and X A(x n + Xy n ) v = X u. Since R(A) is closed, u = Ax and v = X u = X Ax. Thus u = v Hence M is hypo-ep. A AX X A X AX x. Therefore R(M) is closed. 0

7 HYPO-EP OPERATORS 79 Corollary 3.2 Let A, X BL(H) and M = only if A is EP. A AX X A X AX. Then M is EP if and PROOF : Suppose M is EP. Then M is also EP. Therefore M and M are hypo-ep. Thus by Theorem 3.1, A and A are hypo-ep. Therefore A is EP. Conversely suppose A is EP. Therefore A and A are hypo-ep. Thus by Theorem 3.1, M = A AX and M = A A X are hypo-ep. Hence M is EP. X A X AX X A X A X The following results are observed by Hartwig [7], for EP matrices. Next we discuss similar results for hypo-ep and EP operator matrices. Theorem 3.3 An operator A BL(H) is hypo-ep if and only if M = A 0 is hypo-ep. A A PROOF : Suppose A is a hypo-ep operator. Let A 0 x n u. Then Ax n A A y n v u and Ay n v u. Since R(A) is closed, there exist x, y H such that u = Ax and v u = Ay. Thus v = A(x + y). Therefore u = v A 0 A A x. Thus R(M) is closed. Now y let x y N (M). Then Ax = 0 = Ay. Since A is hypo-ep, A x = 0 = A y. Therefore x y N (M ). Hence M is hypo-ep. Conversely suppose M is a hypo-ep operator. Let Ax n u. Then A 0 x n A A 0 u. Since R(M) is closed, there exist x, y H such that A 0 x = u. u A A y u Thus Ax = u and Ay = 0. Thus R(A) is closed. Now let x N (A). Then x N (M) 0 A A N (M ). Thus x = 0. Therefore A x = 0. Thus N (A) N (A ). 0 A 0 0 Hence A is hypo-ep.

8 80 ARVIND B. PATEL AND MAHAVEER P. SHEKHAWAT For A, C, D BL(H) with R(A) and R(D) closed, A C 0 D and only if N (D) N (C) and R(C) R(A) [3]. = A A CD 0 D if Theorem 3.4 Let A, C, D BL(H) with N (D) N (C) and R(C) R(A). Then M = A C is hypo-ep if and only if A and D are hypo-ep. 0 D PROOF : Suppose M is hypo-ep. Let x N (A), then x 0 N (M). Thus x 0 N (M ). Therefore x N (A ). Hence N (A) N (A ). Suppose Ax n u, for some u H. Then A C x n u. Since R(M) is closed, there exists x, y H such that 0 D 0 0 u = A C x. Thus Dy = 0 and Ax + Cy = u. Since N (D) N (C), 0 0 D y Ax = u. Thus R(A) isclosed. Hence A is hypo-ep. Let y N (D). Then y N (C). Thus A C 0 = 0. Since M is hypo-ep, A 0 0 = 0. Thus 0 D y 0 C D y 0 y N (D ). Hence N (D) N (D ). To Prove R(D) is closed, it is enough to prove that R(D ) is closed. Let D y n v. Then A Since R(M ) is closed, there C D y n v exist x, y H such that 0 = A 0 x. Therefore A x = 0 and C x + D y = v C D y v. Since R(C) R(A), D y = v. Therefore R(D ) is closed. Conversely suppose that A and D are hypo-ep. Thus R(A) and R(D) are closed. Since N (D) N (C) and R(C) R(A), M = A A CD. Thus M M 2 M 0 D = AA AA CD + A ACD = AA 0 = MM. Therefore M 0 DD 0 DD is hypo-ep. Corollary 3.5 Let A, D, X BL(H). If A and D are hypo-ep, then A AXD is hypo-ep. 0 D

9 HYPO-EP OPERATORS 81 Corollary 3.6 Let A BL(H). Then A is hypo-ep if and only if M = hypo-ep. A A 0 A is Theorem 3.7 Let A, C, D BL(H) with N (D) N (C) and R(C) R(A). Then M = A C is EP if and only if A and D are EP. 0 D PROOF : Suppose M is EP. In view of Theorem 3.4, to prove A and D are EP, it is enough to prove N (A ) N (A) and N (D ) N (D). Let x N (A ). Since R(C) R(A), x N (C ). Therefore A 0 x = 0. Since M is EP, A C x = 0. C D D 0 0 Thus Ax = 0. Therefore N (A ) N (A). Let y N (D ). Then A 0 0 = C D y 0. Since M is EP, A C 0 = 0. Thus Dy = 0. Therefore N (D ) 0 0 D y 0 N (D). Conversely suppose that A and D are EP operators. Thus R(A) and R(D) are closed. Therefore under the assumption of hypothesis we have M = A A CD. Thus 0 D M M MM = A A AA 0 = 0 0. Therefore M is EP. 0 D D DD 0 0 We study similar results to above for another type of operator matrices in the following. Theorem 3.8 Let A, B, C BL(H) with N (B) = N (C) N (A) and R(A) R(C). Then M = A C is hypo-ep if and only if C and B are hypo-ep. B 0 PROOF : Suppose M is hypo-ep. Let x N (C). Since N (B) = N (C) N (A), x N (A) and x N (B). Therefore A C x = 0. Since M is hypo-ep, A B x = B C Thus x N (C ). Hence N (C) N (C ). Suppose Cx n u, for some u H. 0

10 82 ARVIND B. PATEL AND MAHAVEER P. SHEKHAWAT Thus A C 0 u. Since R(M) is closed, there exist x, y H such that B 0 x n 0 u = A C x. Thus Bx = 0 and Ax + Cy = u. Since N (B) N (A), 0 B 0 y Cy = u. Thus R(C) is closed. Hence C is hypo-ep. Next we prove that B is a hypo-ep operator. Let y N (B). Since N (B) = N (C), y N (C). Thus A C 0 = 0. Since M is hypo-ep, A B 0 = 0. Thus B 0 y 0 C 0 y 0 y N (B ). Hence N (B) N (B ). To prove R(B) is closed, it is enough to prove that R(B ) is closed. Let B y n u. Therefore A B 0 u. Since R(M ) C 0 y n 0 is closed, there exist x, y H such that u = x. Thus C x = 0 and 0 C 0 y A x + B y = u. Since R(A) R(C), B y = u. Therefore R(B ) is closed. Hence B is hypo-ep. Conversely suppose that B and C are hypo-ep. Let x N (M). Then A C x y B 0 y A B = 0 0. Thus Bx = 0 and Ax+Cy = 0. Since N (B) N (A), Cy = 0. Since N (B) = N (C), Cx = 0 = By. As C and B arehypo-ep, C x = 0 = B y. Since R(A) R(C), A x = 0. Thus A B C 0 x = 0. Thus N (M) N (M ). y 0 Let A C x n u. Thus Ax n + Cy n u and Bx n v. Since R(B) is B 0 y n v closed, there exists x H such that Bx = v. As R(A) R(C), for each n there exists z n H such that Ax n = Cz n. Thus C(z n +y n ) u. Since R(C) is closed, there exists z H such that u = Cz. As R(A) R(C), Cz Ax R(C). Thus there exists y H such that Cz Ax = Cy. Thus A C x Ax + Cy = = u. Thus R(M) is closed. Hence M is hypo-ep. B 0 y Bx v Theorem 3.9 Let A, B, C BL(H) with N (B) = N (C) N (A) and R(A) R(C). Then M = A C is EP if and only if C and B are EP. B 0

11 HYPO-EP OPERATORS 83 PROOF : Suppose M is EP. In view of Theorem 3.8, to prove C and B are EP, it is enough to prove that N (C ) N (C) and N (B ) N (B). Let x N (C ). Since R(A) R(C), x N (A ). Thus A B x = 0. Since M is EP, 0 = A C x = C B 0 0 Ax. Thus Bx = 0. Since N (B) = N (C), x N (C). Therefore N (C ) N (C). Let y Bx N (B ). Then A B 0 = 0. Since M is EP, 0 = A C 0 = C 0 y 0 0 B 0 y Cy. Thus Cy = 0. Since N (B) = N (C), y N (B). Therefore N (B ) N (B). 0 Conversely suppose C and B are EP. Inview of the Theorem 3.8, to prove M is EP, it is enough to prove that N (M ) N (M). Let x N (M ). Thus x = 0. y C 0 y 0 Therefore C x = 0 and A x + B y = 0. Since R(A) R(C), A x = 0. Therefore C x = 0 = A B y. Since C and B are EP, Cx = 0 = By. Since N (B) = N (C) N (A), Ax = 0 = Bx = Cy. Therefore A C x = 0. Thus N (M ) N (M). B 0 y 0 Theorem 3.10 Let A BL(H) be hypo-ep and B BL(H) be invertible. If M = A 0 is hypo-ep, then M is injective. B D PROOF : Suppose M is hypo-ep. Let x Ax N (M). Then = 0. y Bx + Dy 0 Therefore x N (A) and Bx + Dy = 0. Since A is hypo-ep, A x = 0. Also, since M is hypo-ep, A B x = 0. Thus A x+b y = 0. So B y = 0. Since B is injective, y = 0. 0 D y 0 As Bx + Dy = 0, Bx = 0. So x = 0. Hence N (M) = {0}. B ACKNOWLEDGEMENT The authors wish to thank the referee for useful suggestions and detailed comments, which greatly improve the manuscript. Authors also thank Prof. S. J. Bhatt for encouragement and suggestions.

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