Inequalities for eigenvalues of matrices

Abstract

The purpose of the paper is to present some inequalities for eigenvalues of positive semidefinite matrices.

MSC:15A18, 15A60.

1 Introduction

Throughout this paper, $M_n$ denotes the space of $n\times n$ complex matrices and $H_n$ denotes the set of all Hermitian matrices in $M_n$. Let $A,B\in H_n$; the order relation $A\ge B$ means, as usual, that $A-B$ is positive semidefinite. We always denote the singular values of A by $s_1(A)\ge \cdots \ge s_n(A)$. If A has real eigenvalues, we label them as ${\lambda }_1(A)\ge \cdots \ge {\lambda }_n(A)$. Let $\parallel \cdot \parallel$ denote any unitarily invariant norm on $M_n$. We denote by $|A|$ the absolute value operator of A, that is, $|A|={(A^{\ast }A)}^{\frac{1}{2}}$, where $A^{\ast }$ is the adjoint operator of A.

For positive real number a, b, the arithmetic-geometric mean inequality says that

$$\sqrt{ab}\le \frac{a+b}{2}.$$

It is equivalent to

$${(ab)}^m\le {\left(\frac{a+b}{2}\right)}^{2m},m=1,2,\dots .$$

(1.1)

Let $A,B\in M_n$ be positive semidefinite. Bhatia and Kittaneh [1] proved that for all $m=1,2,\dots$ ,

$${\lambda }_j\left({(AB)}^m\right)\le {\lambda }_j{\left(\frac{A+B}{2}\right)}^{2m}.$$

(1.2)

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

$${\left(a^{3/4}{b}^{3/4}\right)}^{2/3}\le \frac{a+b}{2}.$$

(1.3)

As pointed out in [10], p.198], although the arithmetic-geometric mean inequalities can be written in different ways and each of them may be obtained from the other, the matrix versions suggested by them are different.

In this note, we obtain a refinement of (1.2) and a log-majorization inequality 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 Drury in [2].

Lemma 2.1 Let $A,B\in M_n$ be positive semidefinite. Then

$$s_j(AB)\le s_j{\left(\frac{A+B}{2}\right)}^2.$$

As a consequence of Lemma  2.1, we have

$$\parallel |AB{|}^{1/2}\parallel \le \frac{1}{2}\parallel A+B\parallel .$$

(2.1)

It is a matrix version of the arithmetic-geometric mean inequality. By properties of the matrix square function, we know that this last inequality is stronger than the assertion

$$\parallel AB\parallel \le \parallel {\left(\frac{A+B}{2}\right)}^2\parallel ,$$

which is due to Bhatia and Kittaneh [1]and is also a matrix version of (1.1).

Theorem 2.1 Let $A,B\in M_n$ be positive semidefinite. Then for all $m=1,2,\dots$ ,

$${\lambda }_j\left({(AB)}^m\right)\le {\lambda }_j{\left(\frac{A+B+A^{1/2}{B}^{1/2}+B^{1/2}{A}^{1/2}}{4}\right)}^{2m}.$$

(2.2)

Proof By Lemma 2.1, we have

$$\begin{array}{rl}{\lambda }_j\left({\left(A^2{B}^2\right)}^m\right)& ={\left({\lambda }_j\left(A^2{B}^2\right)\right)}^m\\ ={\left({\lambda }_j\left(A{B}^{2}A\right)\right)}^m\\ ={(s_j(AB))}^{2m}\\ \le s_j{\left(\frac{A+B}{2}\right)}^{4m}\\ ={\lambda }_j{\left(\frac{A+B}{2}\right)}^{4m}.\end{array}$$

(2.3)

Replacing A, B by $A^{1/2}$, $B^{1/2}$ in (2.3), we have

$${\lambda }_j\left({(AB)}^m\right)\le {\lambda }_j{\left(\frac{A+B+A^{1/2}{B}^{1/2}+B^{1/2}{A}^{1/2}}{4}\right)}^{2m}.$$

This completes the proof. □

Remark 2.1 Let $A,B\in M_n$ be positive semidefinite. Note that

$$0\le \frac{{(A^{1/2}-B^{1/2})}^2}{2}=\frac{A+B}{2}-\frac{A+B+A^{1/2}{B}^{1/2}+B^{1/2}{A}^{1/2}}{4}.$$

Therefore, the inequality (2.2) is a refinement of the inequality (1.2).

Remark 2.2 For $m=1$, by (1.2), we have

$${\lambda }_j(AB)\le {\lambda }_j{\left(\frac{A+B}{2}\right)}^2.$$

(2.4)

For $m=1$, by (2.2), we have

$${\lambda }_j\left(A^2{B}^2\right)\le {\lambda }_j{\left(\frac{A+B}{2}\right)}^4.$$

(2.5)

In view of the inequalities (2.4) and (2.5), one may ask whether it is true that

$${\lambda }_j\left(A^m{B}^m\right)\le {\lambda }_j{\left(\frac{A+B}{2}\right)}^{2m}$$

(2.6)

for all $m=1,2,\dots$ . The answer is no. For $m=3$, the inequality (2.6) is refuted by the following example:

$$A=\left[\begin{array}{ll}5& -1\\ -1& 9\end{array}\right],B=\left[\begin{array}{ll}6& -4\\ -4& 5\end{array}\right].$$

Theorem 2.2 Let $A,B\in M_n$ be positive semidefinite. Then

$$\prod _{j=1}^k\left|{\lambda }_j\left(A\left(\frac{A^v{B}^{1-v}+A^{1-v}{B}^v}{2}\right)B\right)\right|\le \prod _{j=1}^k{\lambda }_j{\left(\frac{A+B}{2}\right)}^3.$$

Proof By Weyl’s inequality, Horn’s inequality and Lemma 2.1, we have

$$\begin{array}{rl}\prod _{j=1}^k|{\lambda }_j(AXB)|& =\prod _{j=1}^k|{\lambda }_j(XAB)|\\ \le \prod _{j=1}^k{s}_j(XAB)\\ \le \prod _{j=1}^k{s}_j(X)s_j(AB)\\ \le \prod _{j=1}^k{s}_j(X)\prod _{j=1}^k{s}_j{\left(\frac{A+B}{2}\right)}^2.\end{array}$$

(2.7)

Putting

$$X=\frac{A^v{B}^{1-v}+A^{1-v}{B}^v}{2},0\le v\le 1,$$

in (2.7) gives

$$\prod _{j=1}^k\left|{\lambda }_j\left(A\left(\frac{A^v{B}^{1-v}+A^{1-v}{B}^v}{2}\right)B\right)\right|\le \prod _{j=1}^k{s}_j\left(\frac{A^v{B}^{1-v}+A^{1-v}{B}^v}{2}\right)\prod _{j=1}^k{s}_j{\left(\frac{A+B}{2}\right)}^2.$$

(2.8)

In response to a conjecture by Zhan [11], Audenaert [3] proved that if $0\le v\le 1$, then

$$s_j\left(\frac{A^v{B}^{1-v}+A^{1-v}{B}^v}{2}\right)\le s_j\left(\frac{A+B}{2}\right).$$

(2.9)

The special case where $v=\frac{1}{2}$ was obtained earlier in [6, 12] and the special case where $v=\frac{1}{4}$ was obtained earlier in [13]. It follows from (2.8) and (2.9) that

$$\prod _{j=1}^k\left|{\lambda }_j\left(A\left(\frac{A^v{B}^{1-v}+A^{1-v}{B}^v}{2}\right)B\right)\right|\le \prod _{j=1}^k{\lambda }_j{\left(\frac{A+B}{2}\right)}^3.$$

This completes the proof. □

Remark 2.3 As an application of Theorem 2.2, we now present a matrix version of (1.3). Taking $v=\frac{1}{2}$ in this last inequality, we have

$$\prod _{j=1}^k\left|{\lambda }_j\left(A^{3/2}{B}^{3/2}\right)\right|\le \prod _{j=1}^k{s}_j{\left(\frac{A+B}{2}\right)}^3$$

and so

$$\prod _{j=1}^k{s}_j\left(A^{3/4}{B}^{3/4}\right)\le \prod _{j=1}^k{s}_j{\left(\frac{A+B}{2}\right)}^{3/2},$$

which is equivalent to

$$\prod _{j=1}^k{s}_j\left(\right|A^{3/4}{B}^{3/4}{|}^{2/3})\le \prod _{j=1}^k{s}_j\left(\frac{A+B}{2}\right).$$

Since weak log-majorization is stronger than weak majorization, we have

$$\sum _{j=1}^k{s}_j\left(\right|A^{3/4}{B}^{3/4}{|}^{2/3})\le \sum _{j=1}^k{s}_j\left(\frac{A+B}{2}\right).$$

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

$$\parallel |A^{3/4}{B}^{3/4}{|}^{2/3}\parallel \le \frac{1}{2}\parallel A+B\parallel .$$

(2.10)

This is a matrix version of (1.3).

Next, we give another proof of the inequality (2.10). Araki [14] (also see [15]) obtained the following log-majorization inequality:

$$\prod _{j=1}^k{s}_j\left({\left(A^{p/2}{B}^p{A}^{p/2}\right)}^{q/p}\right)\le \prod _{j=1}^k{s}_j\left(A^{q/2}{B}^q{A}^{q/2}\right),0<p\le q.$$

(2.11)

Putting

$$p=\frac{3}{2},q=2$$

in (2.11) gives

$$\prod _{j=1}^k{s}_j\left({\left(A^{3/4}{B}^{3/2}{A}^{3/4}\right)}^{1/3}\right)\le \prod _{j=1}^k{s}_j{\left(A{B}^{2}A\right)}^{1/4},$$

and so

$$\sum _{j=1}^k{s}_j\left(\right|A^{3/4}{B}^{3/4}{|}^{2/3})\le \sum _{j=1}^k{s}_j(|AB{|}^{1/2}).$$

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

$$\parallel |A^{3/4}{B}^{3/4}{|}^{2/3}\parallel \le \parallel |AB{|}^{1/2}\parallel .$$

(2.12)

It follows from (2.1) and (2.12) that

$$\parallel |A^{3/4}{B}^{3/4}{|}^{2/3}\parallel \le \frac{1}{2}\parallel A+B\parallel .$$