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					Journal of Inequalities and
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                                     Inequalities for eigenvalues of matrices
        Journal of Inequalities and Applications 2013, 2013:6                               doi:10.1186/1029-242X-2013-6

                                               Xiaozeng Xu (cquxuxz@163.com)
                                              Chuanjiang He (cquxuxz@163.com)




                                            ISSN        1029-242X

                                  Article type          Research

                         Submission date                5 July 2012

                         Acceptance date                14 December 2012

                          Publication date              4 January 2013

                                  Article URL           http://www.journalofinequalitiesandapplications.com/content/2013/1/


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         Inequalities for eigenvalues of matrices
                                        a, b 1                     a
                        Xiaozeng Xu              , Chuanjiang He
              a. College of Mathematics and Statistics, Chongqing University,
                             Chongqing, 401331, P. R. China
        b. School of Mathematics and Statistics,Chongqing University of Technology,
                             Chongqing, 400054, P. R. China




Abstract

The purpose of the paper is to present some inequalities for eigenvalues of
positive semidefinite matrices.
Keywords: singular values; eigenvalues; unitarily invariant norm
Subject Classification: MSC (2010) 15A18, 15A60


1. Introduction

       Throughout this paper, Mn denotes the space of n × n complex matrices
and Hn denotes the set of all Hermitian matrices in Mn . Let A, B ∈ Hn ; the
order relation A ≥ B means, as usual, that A − B is positive semidefinite.
We always denote the singular values of A by s1 (A) ≥ · · · ≥ sn (A). If A
has real eigenvalues, we label them as λ1 (A) ≥ · · · ≥ λn (A). Let · denote
any unitarily invariant norm on Mn . We denote by |A| the absolute value
                                          1
operator of A, that is, |A| = (A∗ A) 2 , where A∗ is the adjoint operator of A.
       For positive real number a, b, the arithmetic-geometric mean inequality

   1
    Corresponding author.
    E-mail address: cquxuxz@163.com (X. Xu).



Preprint submitted to Journal of Inequalities and Applications         December 27, 2012
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says that
                                   √            a+b
                                       ab ≤         .
                                                 2
It is equivalent to
                                            2m
                                  a+b
                      (ab)m ≤                     , m = 1, 2, · · · .   (1.1)
                                   2

    Let A, B ∈ Mn be positive semidefinite. Bhatia and Kittaneh [1] proved
that for all m = 1, 2, · · ·,
                                                              2m
                                                   A+B
                         λj ((AB) ) ≤ λj
                                   m
                                                                   .    (1.2)
                                                    2

This is a matrix version of (1.1). For more information on matrix versions
of the arithmetic-geometric mean inequality, the reader is referred to [1-11]
and the references therein.
    It is easy to see that the arithmetic-geometric mean inequality is also
equivalent to
                                            2/3       a+b
                                a3/4 b3/4         ≤       .             (1.3)
                                                       2
As pointed out in [10, p.198], although the arithmetic-geometric mean in-
equalities can be written in different ways and each of them may be obtained
the other, the matrix versions suggested by them are different.
    In this note, we obtain a refinement of (1.2) and a log-majorization in-
equality for eigenvalues. As an application of our result, we give a matrix
version of (1.3).


2. Main results

    We begin this section with the following lemma, which is a question posed
by Bhatia and Kittaneh [1](see also [8, 10]) and settled in the affirmative by

                                            2
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Drury in [2].
Lemma 2.1. Let A, B ∈ Mn be positive semidefinite. Then
                                                                 2
                                                       A+B
                              sj (AB) ≤ sj                           .
                                                        2

As a consequence of Lemma 2.1, we have

                                                   1
                                |AB|1/2 ≤            A+B .                                 (2.1)
                                                   2

It is a matrix version of the arithmetic-geometric mean inequality. By prop-
erties of the matrix square function, we know that this last inequality is
stronger than the assertion
                                                             2
                                                 A+B
                                AB ≤                             ,
                                                  2

which is due to Bhatia and Kittaneh [1] and is also a matrix version of (1.1).
Theorem 2.1. Let A, B ∈ Mn be positive semidefinite. Then for all
m = 1, 2, · · ·,
                                                                                  2m
                                   A + B + A1/2 B 1/2 + B 1/2 A1/2
                 λj ((AB) ) ≤ λj
                         m
                                                                                       .   (2.2)
                                                4

Proof. By Lemma 2.1, we have
                                       m                                 m
                         λj   A2 B 2         =         λj A2 B 2
                                                                         m
                                             =         λj AB 2 A
                                             = (sj (AB))2m
                                                                         4m                (2.3)
                                                    A+B
                                             ≤ sj
                                                      2
                                                                         4m
                                                    A+B
                                             = λj                             .
                                                      2


                                               3
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Replacing A, B by A1/2 , B 1/2 in (2.3), we have
                                                                                     2m
                                         A + B + A1/2 B 1/2 + B 1/2 A1/2
                 λj ((AB) ) ≤ λj
                         m
                                                                                          .
                                                      4

This completes the proof.
Remark 2.1. Let A, B ∈ Mn be positive semidefinite. Note that
                                2
           A1/2 − B 1/2               A + B A + B + A1/2 B 1/2 + B 1/2 A1/2
        0≤                          =      −                                .
                2                       2                4

Therefore, the inequality (2.2) is a refinement of the inequality (1.2).
Remark 2.2. For m = 1, by (1.2), we have
                                                                  2
                                                          A+B
                                λj (AB) ≤ λj                          .                               (2.4)
                                                           2

For m = 1, by (2.2), we have
                                                                      4
                                                          A+B
                               λj A B 2    2
                                               ≤ λj                       .                           (2.5)
                                                           2

In view of the inequalities (2.4) and (2.5), one may ask whether it is true
that
                                                                      2m
                                                          A+B
                              λj (A B ) ≤ λj
                                     m     m
                                                                                                      (2.6)
                                                           2
for all m = 1, 2, · · ·. The answer is no. For m = 3, the inequality (2.6) is
refuted by the following example:
                        ⎡         ⎤      ⎡      ⎤
                           5 −1            6 −4
                    A=⎣           ⎦, B = ⎣      ⎦.
                          −1 9             −4 5

Theorem 2.2. Let A, B ∈ Mn be positive semidefinite. Then
             k                                                        k                       3
                             Av B 1−v + A1−v B v                                   A+B
                  λj A                                    B   ≤               λj                  .
           j=1
                                      2                           j=1
                                                                                    2

                                                  4
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Proof. By Weyl’s inequality, Horn’s inequality and Lemma 2.1, we have
                   k                             k
                        |λj (AXB)| =                 |λj (XAB)|
                  j=1                        j=1
                                              k
                                         ≤           sj (XAB)
                                             j=1
                                              k                                                              (2.7)
                                         ≤           sj (X) sj (AB)
                                             j=1
                                              k                 k                       2
                                                                              A+B
                                         ≤           sj (X)          sj                     .
                                             j=1               j=1
                                                                               2

Putting
                                  Av B 1−v + A1−v B v
                             X=                       , 0 ≤ v ≤ 1,
                                           2
in (2.7) gives
 k                                                        k                                              k                 2
                 Av B 1−v + A1−v B v                                 Av B 1−v + A1−v B v                             A+B
      λj A                                   B       ≤         sj                                             sj               .
j=1
                          2                              j=1
                                                                              2                         j=1
                                                                                                                      2
                                                                                                             (2.8)
In response to a conjecture by Zhan [11], Audenaert [3] proved that if 0 ≤
v ≤ 1, then
                             Av B 1−v + A1−v B v                         A+B
                        sj                                    ≤ sj                  .                        (2.9)
                                      2                                   2
                                     1
The special case where v =           2
                                         was obtained earlier in [6, 12] and the special
                    1
case where v =      4
                        was obtained earlier in [15]. It follows from (2.8) and (2.9)
that
             k                                                            k                     3
                             Av B 1−v + A1−v B v                                    A+B
                 λj A                                     B          ≤         λj                   .
           j=1
                                      2                                  j=1
                                                                                     2

This completes the proof.
Remark 2.3. As an application of Theorem 2.2, we now present a matrix


                                                     5
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                                                  1
version of (1.3). Taking v =                      2
                                                      in this last inequality, we have
                             k                                               k                      3
                                                                                            A+B
                                      λj A      3/2
                                                      B   3/2
                                                                     ≤            sj
                         j=1                                                j=1
                                                                                             2

and so
                             k                                          k                         3/2
                                                                                       A+B
                                  sj A3/4 B 3/4 ≤                            sj                         ,
                         j=1                                         j=1
                                                                                        2
which is equivalent to
                         k                                                       k
                                                        3/4 2/3                              A+B
                                 sj       A 3/4
                                                    B                   ≤             sj         .
                        j=1                                                  j=1
                                                                                              2

Since weak log-majorization is stronger than weak majorization, we have
                         k                                                       k
                                                        3/4 2/3                              A+B
                                 sj       A 3/4
                                                    B                   ≤              sj        .
                        j=1                                                  j=1
                                                                                              2

By Fan’s dominance theorem [4, p.93], we get
                                                              2/3            1
                                          A3/4 B 3/4                    ≤      A+B .                                  (2.10)
                                                                             2
This is a matrix version of (1.3).
    Next, we give another proof of the inequality (2.10). Araki [13] (also see
[14]) obtained the following log-majorization inequality:
         k                                                     k
                                            q/p
             sj   Ap/2 B p Ap/2                       ≤             sj Aq/2 B q Aq/2 , 0 < p ≤ q.                     (2.11)
       j=1                                                    j=1

Putting
                                                            3
                                                          p= , q=2
                                                            2
in (2.11) gives
                    k                                                                 k
                                                           3/4 1/3                                          1/4
                        sj            A   3/4
                                                B   3/2
                                                          A                  ≤             sj AB 2 A              ,
                  j=1                                                                j=1


                                                                    6
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and so
                   k                                        k
                                         3/4 2/3
                        sj   A 3/4
                                     B                 ≤         sj |AB|1/2 .
                  j=1                                      j=1

By Fan’s dominance theorem [4, p.93], we get

                                            2/3
                             A3/4 B 3/4                ≤ |AB|1/2 .              (2.12)

It follows from (2.1) and (2.12) that

                                            2/3            1
                             A3/4 B 3/4                ≤     A+B .
                                                           2



References


 [1] R. Bhatia, F. Kittaneh. Notes on matrix arithmetic-geometric mean
      inequalities. Linear Algebra Appl. 308 (2000) 203-211.

 [2] S. W. Drury. On a question of Bhatia and Kittaneh. Linear Algebra
      Appl. 437 (2012) 1955-1960.

 [3] K. M. R. Audenaert. A singular value inequality for Heinz means. Linear
      Algebra Appl. 422 (2007) 279-283.

 [4] R. Bhatia. Matrix Analysis. Springer-Verlag, New York, 1997.

 [5] R. Bhatia, C. Davis. More matrix forms of the arithmetic-geometric
      mean inequality. SIAM J. Matrix Anal. Appl. 14 (1993) 132-136.

 [6] R. Bhatia, F. Kittaneh. On the singular values of a product of operators.
      SIAM J. Matrix Anal. Appl. 11 (1990) 272-277.

                                                   7
JIA_182_edited                  [12/27 09:37]                           8/8


 [7] R. Bhatia. Interpolating the arithmetic-geometric mean inequality and
      its operator version. Linear Algebra Appl. 413 (2006) 355-363.

 [8] R. Bhatia, F. Kittaneh. The matrix arithmetic–geometric mean inequal-
      ity revisited. Linear Algebra Appl. 428 (2008) 2177-2191.

 [9] H. Kosaki. Arithmetic-geometric mean and related inequalities for op-
      erators. J. Funct. Anal. 156 (1998) 429-451.

[10] R. Bhatia. Positive Definite Matrices. Princeton University Press,
      Princeton, 2007.

[11] X. Zhan. Matrix Inequalities. Lecture Notes in Mathematics, vol.1790,
      Springer-Verlag, Berlin, 2002.

[12] T. Ando. Matrix Young inequalities. Oper. Theory Adv. Appl. 75 (1995)
      33-38.

[13] H. Araki. On an inequality of Lieb and Thirring. Lett. Math. Phys. 19
      (1990) 167-170.

[14] F. Hiai. Matrix Analysis: Matrix Monotone Functions, Matrix Means,
      and Majorization. Interdisciplinary Information Sciences. 16 (2010) 139-
      248.

[15] Y. Tao. More results on singular value inequalities of matrices. Linear
      Algebra Appl. 416 (2006) 724-729.




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