Singular Value and Norm Inequalities Associated with 2 x 2 Positive Semidefinite Block Matrices

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1 Electronic Journal of Linear Algebra Volume 32 Volume 32 (2017) Article Singular Value Norm Inequalities Associated with 2 x 2 Positive Semidefinite Block Matrices Aliaa Burqan Zarqa University, aliaaburqan@zu.edu.jo Fuad Kittaneh University of Jordan, fkitt@ju.edu.jo Follow this additional works at: Part of the Algebra ommons Recommended itation Burqan, Aliaa Kittaneh, Fuad. (2017), "Singular Value Norm Inequalities Associated with 2 x 2 Positive Semidefinite Block Matrices", Electronic Journal of Linear Algebra, Volume 32, pp DOI: This Article is brought to you for free open access by Wyoming Scholars Repository. It has been accepted for inclusion in Electronic Journal of Linear Algebra by an authorized editor of Wyoming Scholars Repository. For more information, please contact scholcom@uwyo.edu.

2 SINGULAR VALUE AND NORM INEQUALITIES ASSOIATED WITH 2 2 POSITIVE SEMIDEFINITE BLOK MATRIES ALIAA BURQAN AND FUAD KITTANEH Abstract. This paper aims to give singular value norm inequalities associated with 2 2 positive semidefinite block matrices. Key words. Singular value, Positive semidefinite matrix, Block matrix, Unitarily invariant norm, Inequality. AMS subject classifications. 15A45, 15A Introduction. Let M n () denote the space of n n complex matrices. A norm. on M n () is called unitarily invariant if UAV = A for all A M n () for all unitary matrices U, V M n (). For Hermitian matrices A, B M n (), we write A B to mean A B is positive semidefinite, particularly, A 0 indicates that A is positive semidefinite. Likewise, if A is positive definite, we write A > 0. For A M n (), the singular values of A, denoted by s 1 (A), s 2 (A),..., s n (A), are the eigenvalues of the positive semidefinite matrix A = (A A) 1 2, arranged in decreasing order repeated according to multiplicity as s 1 (A) s 2 (A) s n (A). If A is Hermitian, we label its eigenvalues as λ 1 (A) λ 2 (A) λ n (A). Several relations between eigenvalues of Hermitian matrices can be obtained by Weyl s monotonicity principle [2, which asserts that if A, B are Hermitian A B, then λ j (A) λ j (B) for j = 1,..., n. [ A 0 The direct sum of A B, denoted by A B, is defined to be the block diagonal matrix. 0 B Positive semidefinite block matrices play an important role in deriving matrix inequalities. A survey of results about 2 2 positive semidefinite block matrices related inequalities can be found in Section 7.7 of [8. [ A B It is evident that if A, B, M n () are such that B 0, then A are positive semidefinite. [ A B A norm inequality due to Horn Mathias [7 says that if A, B, M n () are such that B 0, then for all p > 0 for every unitarily invariant norm, we have (1.1) B p 2 A p p. Received by the editors on November 19, Accepted for publication on March 17, Hling Editor: Roger A. Horn. Department of Mathematics, Zarqa University, Zarqa, Jordan (aliaaburqan@zu.edu.jo). Department of Mathematics, The University of Jordan, Amman, Jordan (fkitt@ju.edu.jo).

3 117 Singular Value Norm Inequalities Recently, Audeh Kittaneh [1 proved that if A, B, M n () are such that B 0, then (1.2) s j (B) s j (A ) for j = 1,..., n. Bhatia Kittaneh [3 proved that if A, B are positive semidefinite, then (1.3) s j A B 2 A + B s j 2 for j = 1,..., n. On the other h, Hirzallah Kittaneh have proved in [6 that if A, B M n (), then A + B (1.4) s j s j (A B) for j = 1,..., n. 2 In this paper, we are interested in finding singular value versions of the inequality (1.1). More singular value inequalities norm inequalities involving products, sums, direct sums of matrices based on the positivity of certain block matrices will be considered. 2. Lemmas. The following lemmas are essential in our analysis. The first lemma is a consequence of the min-max principle (see [2, p. 75). The second third lemmas have been proved in [7. The fourth lemma [ [ [ A + B 0 I I follows from the unitary equivalence of. In fact, if U = 1 0 A B 2, I I [ [ A + B 0 then U is unitary = U U (see [5). The fifth lemma has been proved in 0 A B [4. The sixth lemma is a weak log majorization result that is known as the Weyl majorant theorem (see [2, p. 42). Lemma 2.1. Let A, B, M n (). Then s j (AB) s 1 (A) s j () s 1 (B) for j = 1,..., n. [ A B Lemma 2.2. Let A, B, M n () be such that A > 0 B A 1 B. Lemma 2.3. Let A, B, M n () be such that B 0. Then B B 1 2 U AU 1 2 for some unitary matrix U. Lemma 2.4. Let A, B M n () be Hermitian. Then ±B A if only if 0.

4 Aliaa Burqan Fuad Kittaneh 118 Lemma 2.5. Let A, B M n () be Hermitian such that ±B A. Then (2.5) s j (B) s j (A A) for j = 1,..., n (2.6) B A for every unitarily invariant norm. Note that the inequality (2.5) is equivalent to saying that s j (B) s [ j+1 2 (A) for j = 1,..., n, where [k denotes the integer part of k. In view of Lemma 2.4, this inequality also follows from the inequality (1.2). Lemma 2.6. Let A M n () have eigenvalues λ 1 (A), λ 2 (A),..., λ n (A) arranged in such a way that λ 1 (A) λ 2 (A) λ n (A). Then with equality for k = n. λ j (A) s j (A) for k = 1,..., n 1, 3. Main results. In view of the inequality (1.1), one might wonder if the singular value inequality s 2 j (B) s j (A)s j () for j = 1,..., n holds true. However, this is refuted by considering A = case, [ [ 2 1, = 1 1, B = A In this [ A A A 1 2 = [ A [ A s 2 2 A > 0.14 s 2 (A)s 2 (). [ A B A weaker inequality follows from the fact that if B 0, then B = A 1 2 K 1 2 matrix K (see [2, p. 269). Using this Lemma 2.1, we have for some contraction s 2 j (B) min {s j (A)s 1 (), s j ()s 1 (A)} for j = 1,..., n. Our next singular value inequality, which involves a commutativity condition, can be stated as follows.

5 119 Singular Value Norm Inequalities [ A B Theorem 3.1. Let A, B, M n () be such that B 0 AB = BA. Then s j (B) s j A for j = 1,..., n. Proof. First assume that A > 0. The general case follows by a continuity argument. Using Lemma 2.2 the commutativity of A B, we have B B = B A 1 AB = A 1 2 B A 1 BA 1 2 A 1 2 A 1 2. Now, by the Weyl monotonicity principle, we have s j (B) = λ 1 2 j (B B) λ 1 2 j A A 2 ( ) = λ 1 2 j (A )(A 2 2 ) = s j A for j = 1,..., n. If AB = BA, the singular value inequality in Theorem 3.1 is a refinement of the inequality (1.2). By the inequalities (1.3) (1.4), we have s j (B) s j A s j (A + ) s j (A ) for j = 1,..., n. Theorem [ 3.1 is not true[ if the hypothesis of commutativity of A B is omitted. To see this, let A = [ A B = B =. Then B 0, but s 2 (B) 0.70 > 0.30 s 2 (A ). [ A B Theorem 3.2. Let A, B,, X, Y M n () be such that (3.7) X AX + Y Y ±(X BY + Y B X). Proof. Since [ X AX + Y B X + X BY + Y Y [ X AX Y B X X BY + Y Y = = [ X 0 Y 0 [ X 0 Y 0 B B [ X 0 Y 0 [ X 0 Y 0 0 0,

6 Aliaa Burqan Fuad Kittaneh 120 it follows that X AX + Y B X + X BY + Y Y 0 X AX Y B X X BY + Y Y 0. Thus, X AX + Y Y ±(X BY + Y B X). Using Lemma 2.5 Theorem 3.2, we have the following singular value norm inequalities involving sums direct sums of matrices. [ A B orollary 3.3. Let A, B,, X, Y M n () be such that (3.8) s j (X BY + Y B X) s j ((X AX + Y Y ) (X AX + Y Y )) for j = 1,..., n (3.9) X BY + Y B X X AX + Y Y for every unitarily invariant norm. Letting X = Y = I in orollary 3.3, we have the following result. [ A B orollary 3.4. Let A, B, M n () be such that (3.10) s j (B + B ) s j ((A + ) (A + )) for j = 1,..., n (3.11) B + B A + for every unitarily invariant norm. The inequality (3.11) has been recently obtained in [9 using a different argument. [ [ [ A B B A + B + B If B 0, then 0, so B 0. Thus, applying Theorem , we have the following result. [ A B orollary 3.5. Let A, B, M n () be such that B 0, AB = BA, B = B. Then s j (B + B ) s j (A + ) for j = 1,..., n. As an application of orollary 3.5, we have the following singular value inequality for normal matrices. orollary 3.6. Let A, B M n () be normal such that AB = BA. Then s j (A B + B A) s j (A A + B B) for j = 1,..., n. [ [ A A A [ 0 A 0 Proof. First observe that B A B = 0. Since AB = BA, it follows by B B 0 B 0 the Fuglede-Putnam theorem (see [2, p. 235) that A B commutes with A A B B. Thus, by orollary 3.5, we have s j (A B + B A) s j (A A + B B) for j = 1,..., n.

7 121 Singular Value Norm Inequalities We remark here that [ orollary 3.6 is not [ true if the hypothesis of commutativity of A B is omitted To see this, let A = B =. Then A B are normal, but s 2 (A B + B A) = > 1 = s 2 (A A + B B). Moreover, orollary 3.6 is not true if the hypothesis of normality of A B is [ [ omitted. To see this, let A = B =. Then AB = BA, but s 2 (A B + B A) 0.83 > s 2 (A A + B B). In spite of the failure of Theorem 3.1 without the hypothesis of commutativity of A B, using Lemma 2.3 the Weyl monotonicity principle, one can prove the following related result. [ A B Theorem 3.7. Let A, B, M n () be such that 0. Then s j (B) s j (A 1 2 U 1 2 ) for some unitary matrix U for j = 1,..., n. B Employing Theorem 3.7 the inequalities (1.3) (1.4), we can give a different proof of the inequality (1.2) as follows: s j (B) s j (A U 2 ) = s j ( A U 2 ) 1 2 s j(u AU + ) s j (U AU ) = s j (A ) for j = 1,..., n. 4. On the inequality ±B A. If A, B M n () are Hermitian if ±B A, then the inequality (4.12) s j (B) s j (A) for j = 1,..., n, which is stronger than the inequalities (2.5) (2.6), need not be true. To see this, let A = [ 0 1 B =. Then ±B A, but s 2 (B) = 1 > 0.70 s 2 (A). 1 0 [ In spite of the failure of the inequality (4.12), the inequality (2.6) can be strengthened to the following weak log majorization relation, which can also be concluded from the inequality (2.4) in [7. Theorem 4.1. Let A, B M n () be Hermitian such that ±B A. Then s j (B) s j (A) for k = 1,..., n. Proof. By Lemma 2.4, 0. Thus, B = A 1 2 KA 1 2 for some contraction matrix K, so for

8 Aliaa Burqan Fuad Kittaneh 122 k = 1,..., n, we have s j (B) = λ j (B) = = λ j A KA 2 λ j (AK) s j (AK) (by Lemma 2.6) s j (A)s 1 (K) (by Lemma 2.1) s j (A). As an application of Theorem 4.1, in view of the inequality (3.7), we have the following result, which is stronger than the inequality (3.9). [ A B orollary 4.2. Let A, B,, X, Y M n () be such that s j (X BY + Y B X) s j (X AX + Y Y ) for k = 1,..., n. Letting X = Y = I in orollary 4.2, we have the following corollary (see [9), which is stronger than the inequality (3.11). [ A B orollary 4.3. Let A, B, M n () be such that s j (B + B ) s j (A + ) for k = 1,..., n. We conclude the paper by observing that the inequalities (2.5) (3.10) are equivalent. Theorem 4.4. The following statements are equivalent. (i) Let A, B M n () be Hermitian such that ±B A. Then s j (B) s j (A A) for j = 1,..., n. (ii) Let A, B, M n () be such that B 0.Then s j (B + B ) s j ((A + ) (A + )) for j = 1,..., n.

9 123 Singular Value Norm Inequalities Proof. (i) = (ii). Let A, B, M n () be such that B (3.7) that ±(B + B ) A Then it follows by the inequality Thus, by (i), we have s j (B + B ) s j ((A + ) (A + )) for j = 1,..., n. (ii)= (i). Let A, B M n () be Hermitian such that ±B A. Then, by Lemma 2.4, [ A B 0. Thus, by (ii), we have s j (2B) s j (2A 2A), so s j (B) s j (A A) for j = 1,..., n. In a similar fashion, one can prove the following theorem, which asserts that the inequalities (2.6) (3.11) are equivalent. Theorem 4.5. The following statements are equivalent. (i) Let A, B M n () be Hermitian such that ±B A. Then B A for every unitarily invariant norm. (ii) Let A, B, M n () be such that B 0.Then B + B A + for every unitarily invariant norm. Acknowledgment. The authors are grateful to the referee Roger Horn for their valuable comments suggestions. The work of A. Burqan was funded by the Deanship of Research Graduate Studies at Zarqa University, Jordan. REFERENES [1 W. Audeh F. Kittaneh. Singular value inequalities for compact operators. Linear Algebra Appl., 437: , [2 R. Bhatia. Matrix Analysis. Springer-Verlag, New York, [3 R. Bhatia F. Kittaneh. On the singular values of a product of operators. SIAM J. Matrix Anal. Appl., 11: , [4 R. Bhatia F. Kittaneh. The matrix arithmetic-geometric mean inequality revisited. Linear Algebra Appl., 428: , [5.H. Fitzgerald R.A. Horn. On the structure of Hermitian-symmetric inequalities. J. London Math. Soc., 15: , 1977.

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