SINGULAR VALUE INEQUALITIES FOR COMPACT OPERATORS

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1 SINGULAR VALUE INEQUALITIES FOR OMPAT OPERATORS WASIM AUDEH AND FUAD KITTANEH Abstract. A singular value inequality due to hatia and Kittaneh says that if A and are compact operators on a complex separable Hilbert space such that A is self-adjoint, 0, and A ; then s j (A) s j ( ) We give an equivalent inequality, which states that if A; ; and are compact operators such that 0; then s j () s j (A ) Moreover, we give a sharper inequality and we prove that this inequality is equivalent to three equivalent inequalities considered by Tao. In particular, we show that if A and are compact operators such that A is self-adjoint, 0, and A ; then 2s j (A) s j (( + A) ( A)) Some applications of these results will be given. 1. Introduction Let (H) denote the space of all bounded linear operators on a complex separable Hilbert space H; and let K(H) denote the twosided ideal of compact operators in (H): For T 2 K(H); the singular values of T, denoted by s 1 (T ); s 2 (T ); ::: are the eigenvalues of the positive operator jt j = (T T ) 1=2 ; enumerated as s 1 (T ) s 2 (T ) ::: and repeated according to multiplicity: Note that s j (T ) = s j (T ) = s j (jt j) It follows by Weyl s monotonicity principle (see, e.g., [1; p: 63] or [4; p: 26] ) that if S; T 2 K(H) are positive and S T, then s j (S) s j (T ) Moreover, for S; T 2 K(H); s j (S) s j (T ) if and only if s j (S S) s j (T T ) The 2000 Mathematics Subject lassi cation. 15A18, 15A42, 15A60, 47A63, 4705, Key words and phrases. Singular value, compact operator, inequality, positive operator, self-adjoint operator. 1

2 2 WASIM AUDEH AND FUAD KITTANEH 0 T singular values of S T and are the same, and they consist S 0 of those of S together with those of T. Here, we use the direct sum S 0 notation S T for the block-diagonal operator de ned on 0 T H H. hatia and Kittaneh have proved in [3] that if A; 2 K(H) such that A is self-adjoint, 0, and A ; then s j (A) s j ( ) (1.1) We will give a new equivalent form of (1.1): If A; ; 2 K(H) such that 0; then s j () s j (A ) (1.2) The well-known arithmetic-geometric mean inequality for singular values, due to hatia and Kittaneh [2], says that if A; 2 K(H), then 2s j (A ) s j (A A + ) (1.3) On the other hand, Zhan has proved in [8] that if A; 2 K(H) are positive, then s j () s j (A ) (1.4) Moreover, Tao has proved in [7] that if A; ; 2 K(H) such that 0; then 2s j () s j (1.5) It has been pointed out in [7] that the three inequalities (1.3)-(1.5) are equivalent. It should be mentioned here that while the inequalities in [3], [7], and [8] are formulated for matrices, they can be extended in a natural way to compact operators on a complex separable Hilbert space. In this paper, we will give a new inequality, which is equivalent to these inequalities: If A; 2 K(H) such that A is self-adjoint, 0, and A ; then 2s j (A) s j (( + A) ( A)) (1.6)

3 SINGULAR VALUE INEQUALITIES FOR OMPAT OPERATORS 3 This inequality is sharper than (1.1). 2. Main results Our rst singular value inequality is equivalent to the inequality (1.1). Theorem 2.1. Let A; ; 2 K(H) such that 0: Then s j () s j (A ) Proof. Since 0; it follows that 0. In I 0 fact, if U = ; then U is unitary and 0 I = A 0 0 U U 0. Thus, 0 ; and so by 0 applying the inequality (1.1), we get s j ( ) s j ((A)(A)) : Thus, s j ( ) s j ((A)(A)) This is equivalent to saying that s j () s j (A ) Remark 1. While the proof of the inequality (1.2), given in Theorem 2.1 is based on the inequality (1.1), it can be obtained by employing the inequality (1.5) as follows: If 0; then 0; and so A 0 + = 2 : y Weyl s 0 monotonicity principle, we have A 0 s j 2s j = 2s 0 j (A ) haining this with the inequality (1.5), yields the inequality (1.2). Now, we prove that the inequalities (1.1) and (1.2) are equivalent. Theorem 2.2. The following statements are equivalent: (i) Let A; 2 K(H); where A is self-adjoint, 0; and A : Then s j (A) s j ( )

4 4 WASIM AUDEH AND FUAD KITTANEH A (ii) Let A; ; 2 K(H) such that s j () s j (A ) 0: Then Proof. (i))(ii) This implication follows from the proof of Theorem 2.1. (ii))(i) Let A; 2 K(H); where A is self-adjoint, 0; and A A : Then the matrix 0: In fact, if U = p I I 1 2 ; then I I U is unitary and A + A 0 = U U 0. Thus, by (ii) we have 0 A s j (A) s j ( ) Our second singular value inequality is equivalent to the inequalities (1.3)-(1.5). Theorem 2.3. Let A; 2 K(H); where A is self-adjoint, 0; and A. Then 2s j (A) s j (( + A) ( A A Proof. Let X = : Since is unitarily equivalent to + A 0, and since A ; it follows that X is positive. 0 A Now, applying the inequality (1.5) to the operator X; we get + A 0 2s j (A) s j (X) = s j = s 0 A j (( + A) ( A)) Remark 2. While the proof of the inequality (1.6), given in Theorem 2.3 is based on the inequality (1.5), it can be obtained by applying the inequality (1.4) to the positive operators + A and A: The inequality (1.6) is sharper than (1.1). In fact, if A; 2 K(H); where A is self-adjoint, 0; and A, then by Weyl s monotonicity principle, we have + A 0 s j (( + A) ( A)) = s j 0 A 2s j ( ) A)) Theorem 2.4. The following statements are equivalent: s j =

5 SINGULAR VALUE INEQUALITIES FOR OMPAT OPERATORS 5 (i) Let A; 2 K(H) be positive operators. Then (ii) Let A; 2 K(H): Then s j () s j (A ) 2s j (A ) s j (A A + ) A (iii) Let A; ; 2 K(H) such that 2s j () s j A 0: Then (iv) Let A; 2 K(H) such that A is self-adjoint, 0; and A. Then 2s j (A) s j (( + A) ( Proof. Note that (i), (ii), and (iii) are equivalent by [7]. We will prove that (iii) is equivalent to (iv), and this will complete the proof of the theorem. (iii) ) (iv) This implication follows from the proof of Theorem 2.3. (iv) ) (iii) Assume that 0: Since and are unitarily equivalent; it follows that A 0 0 0, and so 0 : Now, applying (iv), we get 0 2s j ( ) s j Thus, 2s j ( ) s j This is equivalent to saying that 2s j () s j A)) As an application of our results, we present the following theorem, which is sharper and more general than the hatia-kittaneh inequality given in [3], which states that if A; 2 K(H), then s j (A + A ) s j ((AA + ) (AA + )) (2.1)

6 6 WASIM AUDEH AND FUAD KITTANEH For related auchy-schwarz type inequalities, we refer to [5] and references therein. Theorem 2.5. Let A; 2 K(H); and let X 2 (H): Then 2s j (AX + XA ) s j ((A jxj A + jx j + AX + XA ) (A jxj A + jx j AX XA )) (2.2) jxj X Proof. Let Y = and Z = 0 0 X jx. Then Z 0 j A jxj A (see, e.g., [6]), and Y ZY = XA AX + jx j ; which implies that A jxj A + jx j (AX + XA ): (2.3) Now, applying Theorem 2.3 to the operators AX + XA and A jxj A + jx j we get the result. Remark 3. Letting X = I in Theorem 2.5, we have 2s j (A +A ) s j ((AA + +A +A )(AA + A A )) (2.4) In view of Weyl s monotonicity principle and the fact that (A + A ) AA + ; one can see that the inequality (2.4) is sharper than the inequality (2.1). jaj + jj A + If A; 2 (H); then the operator matrix A + ja j + j j is positive, and so if A; 2 K(H); then it follows from the inequalities (1.2) and (1.5), respectively, that s j (A + ) s j ((jaj + jj) (ja j + j j)) (2.5) and jaj + jj A + 2s j (A + ) s j A + ja j + j (2.6) j

7 SINGULAR VALUE INEQUALITIES FOR OMPAT OPERATORS 7 The inequality (2.5), which is a triangle-type inequality has been obtained earlier in [3]. Our nal application is a singular value inequality involving the Jordan parts of compact self-adjoint operators. The Jordan decomposition asserts that every self-adjoint operator can be expressed as the di erence of two positive operators. In fact, if A 2 (H) is self-adjoint, then A = A + A ; where A + and A are the positive operators given by A + = jaj+a and A = jaj A : 2 2 Theorem 2.6. Let A; 2 K(H) be self-adjoint operators. Then s j (A + ) s j ((A ) (A + )) (2.7) Proof. Since A and are self-adjoint, it follows that A jaj and jj ; and so (A + ) jaj + jj : Now, applying Theorem 2.3, we have 2s j (A + ) s j ((jaj + jj + A + ) (jaj + jj (A + )) = s j ((jaj + A + jj + ) (jaj A + jj )) = s j ((2A ) (2A + 2 )) = 2s j ((A ) (A + )) ; and so s j (A + ) s j ((A ) (A + )) It can be easily veri ed that the inequality (2.7) is a generalization of the inequality (1.4). In fact, if A; 2 K(H) are positive, then by the inequality (2.7), we have s j () = s j (A + ( )) s j ((A + + ( ) + ) (A + ( ) )) = s j ((A + 0) (0 + )) = s j (A ) for j = 1; 2; :: Acknowledgments The authors are grateful to the referee for his comments and suggestions. References [1] R. hatia, Matrix Analysis, GTM169, Springer-Verlag, New York, [2] R. hatia, F. Kittaneh, On the singular values of a product of operators, SIAM J. Matrix Anal. Appl. 11 (1990) [3] R. hatia, F. Kittaneh, The matrix arithmetic-geometric mean inequality revisited, Linear Algebra Appl. 428 (2008)

8 8 WASIM AUDEH AND FUAD KITTANEH [4] I.. Gohberg and M. G. Krein, Introduction to the Theory of Linear Nonselfadjoint Operators, Amer. Math. Soc., Providence, RI, [5] O. Hirzallah, F. Kittaneh, Inequalities for sums and direct sums of Hilbert space operators, Linear Algebra Appl. 424 (2007) [6] F. Kittaneh, Some norm inequalities for operators, anad. Math. ull. 42 (1999) [7] Y. Tao, More results on singular value inequalities of matrices, Linear Algebra Appl. 416 (2006) [8] X. Zhan, Singular values of di erences of positive semide nite matrices, SIAM J. Matrix Anal. Appl. 22 (3) (2000) Department of asic Sciences, Petra University, Amman, Jordan address: Department of Mathematics, University of Jordan, Amman, Jordan address: