The Improved Arithmetic-Geometric Mean Inequalities for Matrix Norms

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1 Applied Mathematical Sciences, Vol 7, 03, no 9, HIKARI Ltd, wwwm-hikaricom The Improved Arithmetic-Geometric Mean Inequalities for Matrix Norms I Halil Gumus Adıyaman University, Faculty of Arts and Sciences Department of Mathematics, 0040, Adıyaman, Turkey gumusibo@hotmailcom Necati Taskara Selcuk University, Science Faculty Department of Mathematics 4075, Kampus, Konya, Turkey ntaskara@selcukedutr Copyright c 03 I Halil Gumus and Necati Taskara This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited Abstract In this paper, we prove a scalar inequality such that this inequality is an improvement of the classical arithmetic-geometric mean inequality We obtain its matrix version and investigate Hilbert-Schmidt and trace norm of this matrix version Mathematics Subject Classification: 5A45; 5A60 Keywords: Positive semidefinite matrix, Unitarily invariant norm, arithmeticgeometric mean inequality Introduction Let M n (C be the space of all n n complex matrices For Hermitian matrices A,B M n (C, we write A B to mean A B is positive semidefinite,

2 440 I Halil Gumus and N Taskara particularly, A 0 indicates that A is positive semidefinite Let denote unitarily invariant norm on M n (C So, for all A M n (C and all unitary matrices U,V M n (C, we can write UAV A ( If A [a ij ] M n (C, then the Hilbert-Schmidt norm of A is given by A ( i, a ij ( and the trace norm of A is given by A s j (A tr A, (3 where s (A s n (A are the singular values of A, that is, the eigenvalues of the positive semidefinite matrix A (A A It is known that these norms are unitarily invariant The Young inequality says that if a, b 0 and 0 v, then a v b v va +( vb Specifically, for a b, the above inequality turns into the equality If v, then we obtain the arithmetic-geometric mean inequality ab a + b (4 In [5], for A, B, X M n (C such that A and B are positive semidefinite, Kosaki proved that A v XB v vax +( vxb, where 0 v In [4], for a, b 0 and 0 v, Manasrah and Kittaneh proved that a v b v + r 0 ( a b va +( vb, (5 where r 0 min{v, v} Also, by using (5, they showed that A v B v + r 0 ( A B va +( vb

3 Arithmetic-geometric mean inequalities 44 In [3], for a, b 0 and p, q > with + p q proved that, Hirzallah and Kittaneh ( a p p + bq q r (ap b q + a b By using this inequality, for A, B are positive semidefinite matrices, they obtained that p Ap X + q XBq r Ap X XB q + AXB, where r max(p, q In [], for A, B M n (C such that A, B are positive definite and A B, we proved that 8 s j ( A (A B A s j ( A + B A# B ( 8 s j B (A B B for j,,, n, where A# B is the geometric mean of A and B and s j (X,are the singular values of an n n matrix X for j,,, n In this study, we give a scalar inequality such that this inequality is an improvement of (4 Then we write its matrix version and prove its trace norm inequality By rearranging this scalar inequality, we obtain the Hilbert- Schmidt norm inequality for its matrix version Main Results The following inequality was given by Mitrinovic in [6]: (a b 8 a a + b ab (a b (6 8 b It is easily seen that the left side of this inequality is an improvement of (4 In (6, if we take m,n R + such that a m, b n and m n>0, then we get (m n m + n mn (m n 8 m 8 n If we refine this inequality, then we have (m n (m n (m n 4 m 4 n

4 44 I Halil Gumus and N Taskara Then, we can generalize this inequality for k natural numbers as ( m k+ n k+ ( ( m k n k m k+ n k+ 4 m 4 n Now, in the following lemmas and theorems, we give the proof of the above scalar inequalities and obtain their matrix norm inequalities Lemma If a b>0, then we have (a b 8 a a + b ab (a b 8 b Proof Firstly, let s prove the left side of inequality If b a, then we have b +a a By multiplying both sides of this inequality with a b, we get a b a a b If we take the square of both sides and rearrange this inequality, it is obtained (a b 8 a a + b ab Now, we prove the right side of the inequality If b a, then we get b a Hence we can write b b + a Similarly we have as required a + b ab (a b 8 b Lemma [] Let A, B M n (CThen s j (AB s j (As j (B (7 Theorem 3 For A B>0, we can write the following inequality s j(a+ s j(b A B + A B A + B 8 A Proof By using the left side of Lemma, we can write (s j (A s j (B 8 s j (A + s j (As j (B s j(a+s j (B, (8

5 Arithmetic-geometric mean inequalities 443 where j,,, n From (3 and (8, we have A + B tr(a + B s j (As j (B+ 8 s j (A+s j (B s j(a s j (As j (B+s j(b s j (A s j (A sj (B + s j(a+ s j(b 8 s j(a s j(as j (B By considering Cauchy-Schwarz inequality and (7, we obtain A + B s j (A B + 8 A B + 8 Therefore we have the required inequality s j (A+ s j (B ( n s j(a( n s j(b tra s j (A+ s j (B A B A Theorem 4 For A B>0, we have s j (A+ s j (B A B + A B A + B 8 B Proof By the same operations, the proof is easily checked Lemma 5 For m n>0 and k natural numbers, we can write (m k+ n k+ ( m k n k (m k+ n k+ 4 m 4 n Proof We prove the left side of this inequality with mathematical induction method For k, we show that the inequality is true It is clear that m+n If m this inequality is multiplied with conjugate of m + n, then we have ( m n (m n m Assume that it is true for all positive integers a, that is, m a+ n a+ m a n a m

6 444 I Halil Gumus and N Taskara If we refine this inequality, then we get n a+ mn a m a+ Hence we have to show that it is true for a + Then, for m n>0, we have n a+ nm a+ mn a+ mm a+ mn a+ m a+ mn a+ If it is arranged by adding m + to both side of inequality, we obtain m a+ n a+ m a+ mn a+ m a+ n a+ m a+ n a+ m ( m a+ n a+ ( m a+ n a+ m which ends up the induction Also, the proof of rigth side can be seen by using the same method Theorem 6 Let A, B M n (C be positive semidefinite matricesif all eigenvalues of A are bigger than all eigenvalues of B, then we get A (A k+ X XB k+ 4 ( A k X XB k A k+ X XB k+ B 4 Proof Let s prove the left side of the inequality Since every positive semidefinite matrix is unitarily diagonalizable, it is seen that there are unitary matrices U, V M n (C such that A UλU and B VMV, where λ diag(λ,λ,, λ n,m diag(μ,μ,, μ n and all λ i,μ i are nonnegative If Y U XV [y ij ], then we have 4 A (A k+ X XB k+ 4 Uλ U (Uλ k+ U UY V UY V VM k+ V 4 Uλ U (Uλ k+ YV UY M k+ V 4 Uλ (λ k+ Y YM k+ V U( λk+ i μ k+ j 4λ i y ij V

7 Arithmetic-geometric mean inequalities 445 Similarly, we have A k X XB k Uλ k U UY V UY V VM k V U(λ k Y YM k V U( ( λ k i μk j yij V By using (, ( and applying Lemma 5 to the nonnegative numbers λ i,μ j for i,, n, we obtain A (A k+ X XB k+ 4 i, i, ( λk+ i μ k+ j y ij λ i ( λ k i μk j yij A k X XB k Thus, this completes the proof of the left side Similarly, we show the right side of theorem We have 4 (Ak+ X XB k+ B 4 (Uλk+ U UY V UY V VM k+ V VM V 4 (Uλk+ YV UY M k+ V VM V 4 U(λk+ Y YM k+ M V U( λk+ i μ k+ j y ij V 4μ i Also, we write AX XB i, i, ( λ k i μ k j yij ( λk+ i μ k+ j μ i y ij ( A k+ X XB k+ B 4 and so the proof of theorem is completed References [] Bhatia, R: Matrix Analysis Springer-Verlag 997 [] Gumus, IH, Hirzallah, O, Taskara, N: Singular value inequalities for the arithmetic, geometric and Heinz means of matrices Linear and Multilinear Algebra Vol 59, No, (0

8 446 I Halil Gumus and N Taskara [3] Hirzallah, O, Kittaneh, F: Matrix Young inequalities for the Hilbert- Schmidt norm Linear Algebra Appl 308, (000 [4] Kittaneh, F, Manasrah Y: Improved Young and Heinz inequalities for matrices Journal of Mathematical Analysis and Applications 36, 6-69 (00 [5] Kosaki, H: Arithmetic geometric mean and related inequalities for operators J Funct Anal 56, (998 [6] Mitrinovic, DS: Analytic Inequalities New York Springer Verlag 970 Received: January, 03