Singular Value Inequalities for Real and Imaginary Parts of Matrices

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Filomat 3:1 16, 63 69 DOI 1.98/FIL16163C Published by Faculty of Sciences Mathematics, University of Niš, Serbia Available at: http://www.pmf.ni.ac.rs/filomat Singular Value Inequalities for Real Imaginary Parts of Matrices Dongjun Chen a, Yun Zhang a a School of Mathematical Sciences, Huaibei Normal University, Huaibei 35, China Abstract. Let A = Re A + i Im A be the Cartesian decomposition of square matrix A of order n with Re A = A+A Im A = A A. Fan-Hoffman s result asserts that i λ j ReA s j A, j = 1,..., n, where λ j M s j M st for the jth largest eigenvalue of M the jth largest singular value of M, respectively. We investigate singular value inequalities for real imaginary parts of matrices prove the following inequalities: s j Re A 1 4 s j [ A + A A + A ] [ A + A + A + A ], s j Im A 1 4 s j [ A + A ia A] [ A + A + ia A], j = 1,..., n. In particular, we have s j Re A 1 s j A + A A + A, s j Im A 1 s j A + A A + A, j = 1,..., n. Moreover, we also show that these inequalities are sharp. 1. Introduction Let M n denote the vector space of all complex n n matrices let H n be the set of all Hermitian matrices of order n. We always denote the eigenvalues of A H n in decreasing order by λ 1 A λ A λ n A. For A, B H n, we use the notation A B or B A to mean that B A is positive semidefinite. Clearly, 1 Mathematics Subject Classification. 47A63; 47B5; 47B15; 15A6; 47A3; 15A4; 15A18 Keywords. Hermitian matrices, Positive semidefinite matrices, Singular values, Cartesian decomposition Received: 14 April 14; Accepted: 11 June 14 Communicated by Mohammad Sal Moslehian Corresponding author: Yun Zhang Research supported by the Anhui Provincial Natural Science Foundation14885MA8,15885MA14 the Natural Science Foundation of Anhui Higher Education Institutions of China KJ16A633,KJ16A633,KJ16B1,KJ15A35. Email address: zhangyunmaths@163.com Yun Zhang

D. Chen, Y. Zhang / Filomat 3:1 16, 63 69 64 define two partial orders on H n, each of which is called Löwner partial order. In particular, B res. B > means that B is positive semidefinite res. B is positive definite. For T M n, the singular values of T, denoted by s 1 T, s T,..., s n T are the eigenvalues of the positive semidefinite matrix T = T T 1, enumerated as s 1 T s T s n T repeated according to multiplicity. It follows that the singular values of a normal matrix are just the moduli of its eigenvalues. In particular, if T M n is positive semidefinite, then singular values eigenvalues of T are the same. For more information on this related topic, we refer to [1, 6, 7]. Let A M n. Then A = Re A + i Im A, where Re A = A+A Im A = A A i. This is called the Cartesian decomposition of A. It is clear that both Re A Im A are Hermitian. Here we A denote the block matrix by A B. B Fan Hoffman [] asserts that for A M n, λ j Re A s j A 1 for j = 1,..., n. In the book [4] of page 37, it is said that 1 implies that λ j Re A s j A for j = 1,..., n. Since the singular values of a Hermitian matrix are just the moduli of its eigenvalues., it seems that is presented singular value inequalities between the real part Re A A. But, there exists a gap in. We shall point out that through an example. Consider the square matrix A = i. Then Re A =. It is obvious that λ 1 Re A =, λ Re A = s 1 A =, s A = 1. However, λ Re A = > 1 = s A. This contradicts. More details on the monograph [4] review, we refer to the helpful paper by Zhang [8]. In this paper, our main consideration is singular value inequalities involving real imaginary parts of matrices themselves. We prove the following inequalities, i.e., s j Re A 1 4 s j [ A + A A + A ] [ A + A + A + A ] s j Im A 1 4 s j [ A + A ia A] [ A + A + ia A], for all j = 1,..., n. In particular, the following inequalities hold: s j Re A 1 s j A + A A + A s j Im A 1 s j A + A A + A, for all j = 1,..., n. Furthermore, we show that these inequalities are sharp. Some applications of these results other related inequalities will be also obtained. Finally, we give new revision form of between the absolute values of the eigenvalues of Re A the singular values of A.

D. Chen, Y. Zhang / Filomat 3:1 16, 63 69 65. Main Results We start with some lemmas. Lemma 1 If A 1, A, B 1, B M n such that s j A 1 s j B 1, s j A s j B, for all j = 1,..., n, then Moreover, if only if s j A 1 A s j B 1 B, for all j = 1,..., n. s j S s j T, for all j = 1,..., n, s j S S s j T T, for all j = 1,..., n. Proof. Note that singular values are unitarily invariant: For any A M n unitary U, V M n, suav = sa. In particular, for positive semidefinite matrices, singular values eigenvalues are the same. Let Hence A = diags 1 A 1, s A 1,, s n A 1, s 1 A, s A,, s n A B = diags 1 B 1, s B 1,, s n B 1, s 1 B, s B,, s n B. sa 1 A = sa = λa, sb 1 B = sb = λb. Since s j A 1 s j B 1 s j A s j B, 1 j n,, it follows that A B. By Weyl s Monotonicity Theorem, λ j A λ j B, 1 j n, i.e., s j A 1 A = λ j A λ j B = s j B 1 B, 1 j n. The left part of the lemma is trivial. This completes the proof. The following useful result can be founded in [1, 6, 7]. A B Lemma The partitioned block matrix B is positive semidefinite if only if both A C are positive C semidefinite there exists a contraction W such that B = A 1 WC 1. The following useful singular inequality was given by Zhan [5]. Lemma 3 Let A, B H n. If A B, then s j A B s j A B, 3 for j = 1,,..., n. The following inequalities are due to Hirzallah Kittaneh[3]. Lemma 4 Let X, Y M n. Then s j X + Y As a consequence, s j X Y, j = 1,,..., n. 4 s j Re X s j X X, j = 1,,..., n. 5

D. Chen, Y. Zhang / Filomat 3:1 16, 63 69 66 Next, we shall prove our main results about singular value inequalities involving real imaginary parts of matrices themselves. Theorem 5 Let A M n. Then s j Re A 1 4 s j [ A + A A + A ] [ A + A + A + A ] 6 s j Im A 1 4 s j [ A + A ia A] [ A + A + ia A], 7 for all j = 1,..., n. Proof. Note that A = A 1 U A 1 with unitary U. By Lemma, we have A A A A A, A A. A Then A + A A + A A + A A + A Using Lemma again, there exists a contraction W M n such that. A + A = A + A 1 W A + A 1. Next, we shall show that we can choose W such that W is Hermitian. We divide into two cases. First, consider the case that A is invertible. Then both A A are invertible, i.e., A >, A >. Thus W = A + A 1 A + A A + A 1 is Hermitian. In general, we have A + A + m 1 I A + A A + A A + A + m 1 >, I for any positive integer m. By proved case above, for each m there is a Hermitian contraction W m M n such that A + A = A + A + m 1 I 1 W m A + A + m 1 I 1. 8 Since M n is a finite-dimensional space, the unit ball {X M n : X 1} of the spectral norm is compact. By Bolzano-Weierstrass theorem, {W m } m=1 has a convergent subsequence {W m k }, i.e., lim = W. Since k=1 W mk are Hermitian, it follows that In 8, letting k yields W = limw mk = lim W mk = lim k k k W m k = W. A + A = A + A 1 W A + A 1, where W is a Hermitian contraction. Since W is Hermitian contraction, it follows that ±W I. Then we have ±A + A = A + A 1 ±W A + A 1 A + A. k W m k Note that A + A + A + A, A + A A + A

D. Chen, Y. Zhang / Filomat 3:1 16, 63 69 67 Re A = 1 4 {[ A + A + A + A ] [ A + A A + A ]}. By Lemma 3, for each j = 1,..., n, we have s j Re A 1 4 s j [ A + A A + A ] [ A + A + A + A ]. Note that Im A = Re ia. Replacing A by ia, we obtain 7. This completes the proof. Remark 1 In the proof of Theorem 5, we know that ±A + A A + A, ±ia A A + A. Using Theorem 5 the above remark, we can obtain the following inequality. Theorem 6 Let A M n. Then s j Re A 1 s j A + A A + A 9 s j Im A 1 s j A + A A + A, 1 for all j = 1,..., n.. Proof. Note that We have A + A + A + A, A + A A + A A + A ia A, A + A + ia A. A + A ± A + A A + A A + A ± ia A A + A. Using the fact that for positive semidefinite matrices, singular values eigenvalues are the same Weyl s Monotonicity Principle, we have s j A + A ± A + A s j A + A s j A + A ± ia A s j A + A for all j = 1,..., n. By Lemma 1 Theorem 5, 9 1 hold. This completes the proof. Remark It should be mentioned here that the inequality s j Re A 1 s j A + A, is false for j = 1,..., n. To see this, consider the matrix A = s ReA = 1 > 1 i i. Then = 1 s A + A.

Then D. Chen, Y. Zhang / Filomat 3:1 16, 63 69 68 Next, we shall show that 6, 7, 9 1 are sharp. 1 Example Consider the square matrix A =. Let Note that H 1 = A + A A + A, H = A + A + A + A H 3 = A + A. H 1 = Re A = 1 1 1 1 1 1, H =, Im A = 1 1 1 1 1 i 1 i, H 3 = 1 1 s 1 ReA = 1 4 s 1H 1 H = 1 s 1H 3 H 3 = 1 s ReA = 1 4 s H 1 H = 1 s H 3 H 3 = 1. Similarly, let H 4 = A + A ia A, H 5 = A + A + ia A, we also have. s 1 ImA = 1 4 s 1H 4 H 5 = 1 s 1H 3 H 3 = 1 s ImA = 1 4 s H 4 H 5 = 1 s H 3 H 3 = 1. This example shows that the inequalities 6, 7, 9 1 are sharp. On the other h, using this example we could see that 6 9 seem sharper than 5 in Lemma 4, since s 1 A A = s A A = 1. In [3, Corollary.4], Hirzallah Kittaneh show that let X, Y M n. Then s j XY + YX s j X + Y X + Y, j = 1,..., n. Replacing A in 9 of Corollary 6 by XY, we have following related inequality. Corollary 7 Let X, Y M n. Then s j XY + YX s j XY + YX XY + YX, j = 1,..., n. 11 An immediate consequence of Theorem 6 gives the following inequality related to normal matrices. Corollary 8 Let T M n be normal matrix. Then s j Re T s j T T, s j Im T s j T T, j = 1,..., n. In the end, we shall give a new revision of. Theorem 9 Let A M n let j be positive integer with 1 j n. If j n+1, then λ j Re A s j A, λ j Im A s j A. Otherwise, we have λ j Re A s n j+1 A, λ j Im A s n j+1 A. 1 13 14 15

D. Chen, Y. Zhang / Filomat 3:1 16, 63 69 69 Proof. Replacing A by A in 1 using the fact s j A = s j A, j = 1,..., n, we have λ n j+1 Re A s n j+1 A, j = 1,..., n. Note that λ j Re A = λ n j+1 Re A, j = 1,..., n. Then λ j Re A = λ n j+1 Re A s n j+1 A, j = 1,..., n. By 1 due to Fan Hoffman, we have Therefore s j A λ j Re A s n j+1 A, j = 1,..., n. λ j Re A max {s j A, s n j+1 A}. Note that Im A = Re ia. Replacing A by ia using the fact s j ia = s j A, the inequality 1 holds. Comparing the value between j n j + 1, this completes the proof. Acknowledgement The authors are grateful to Professor Xingzhi Zhan for his helpful suggestions the authors also thank Professor Fuzhen Zhang to provide them with the interesting paper [8]. References [1] R. Bhatia, Matrix Analysis, Springer-Verlag, 1997. [] K. Fan A.J. Hoffman, Some metric inequalities in the space of matrices, Proc. Amer. Math. Soc., 6 1955 111-116. [3] O. Hirzallah, F. Kittaneh, Inequalities for sums direct sums of Hilbert space operators, Linear Algebra Appl., 44 7, 71C8. [4] A.W. Marshall, I. Olkin B.C. Arnold, Inequalities: theory of majorization its applications. Second edition. Springer Series in Statistics. Springer, New York, 11. [5] X. Zhan, Singular values of differences of positive semidefinite matrices, SIAM J. Matrix Anal. Appl.,, no.3, 819-83. [6] X. Zhan, Matrix Inequalities, Lecture Notes in Mathematics 179, Springer-Verlag, Berlin,. [7] X. Zhan, Matrix Theory, Graduate Studies in Mathematics 147, American Mathematical Society, Providence, R.I., 13. [8] F. Zhang, Inequalities: Theory of Majorization Its Applications Springer Series in Statistics by Albert W. Marshall, Ingram Olkin Barry C. Arnold, nd edition, Springer 11 xxvii+99 pp, Hardback, ISBN 978--387-487-7; e-isbn 978--387-6876-1, Linear Algebra Appl., 436 1 1535-154.