Open Access

Some trace inequalities for matrix means

Journal of Inequalities and Applications20162016:283

https://doi.org/10.1186/s13660-016-1236-4

Received: 8 September 2016

Accepted: 9 November 2016

Published: 17 November 2016

Abstract

In this short note, we present some trace inequalities for matrix means. Our results are generalizations of the ones shown by Bhatia, Lim, and Yamazaki.

Keywords

positive definite matrices matrix means trace inequalities

MSC

47A63

1 Introduction

Let \(M_{n} \) be the space of \(n\times n\) complex matrices. Let \(A,B\in M_{n} \) be positive definite, the weighted geometric mean of A and B, denoted by \(A\# B\), is defined as
$$A\#_{t}B = A^{1/2} \bigl( {A^{ - 1/2} BA^{ - 1/2} } \bigr) ^{t} A^{1/2}. $$
When \(t=\frac{1}{2}\), this is the geometric mean, denoted by \(A\# B\). For \(A\in M_{n}\), we denote the vector of eigenvalues by \(\lambda ( A ) = ( {\lambda_{1} ( A ),\lambda_{2} ( A ), \ldots,\lambda_{n} ( A ) } ) \), and we assume that the components of \(\lambda ( A ) \) are in descending order. Let \(\Vert \cdot \Vert \) denote any unitarily invariant norm on \(M_{n} \).
Recently, Bhatia, Lim, and Yamazaki proved in [1] that if \(A,B\in M _{n} \) are positive definite, then
$$ \operatorname{tr} \bigl( {A + B + 2 ( {A\# B} ) } \bigr) \le \operatorname{tr} \bigl( { \bigl( {A^{1/2} + B^{1/2} } \bigr) ^{2} } \bigr) $$
(1.1)
and
$$ \operatorname{tr} \bigl( { \bigl( {A + B + 2 ( {A\# B} ) } \bigr) ^{2} } \bigr) \le \operatorname{tr} \bigl( { \bigl( {A^{1/2} + B^{1/2} } \bigr) ^{4} } \bigr). $$
(1.2)
These authors also have shown in [1] that if \(A,B\in M_{n} \) are positive definite and \(0 < t < 1\), then
$$ \operatorname{tr} ( {A\# _{t} B + B\# _{t} A} ) \le \operatorname{tr} \bigl( {A^{1 - t} B ^{t} + A^{t} B^{1 - t} } \bigr) $$
(1.3)
and
$$ \operatorname{tr} \bigl( { ( {A\# _{t} B + B\# _{t} A} ) ^{2} } \bigr) \le \operatorname{tr} \bigl( { \bigl\vert { \bigl( {A^{1 - t} B^{t} + A^{t} B^{1 - t} } \bigr) } \bigr\vert ^{2} } \bigr). $$
(1.4)

In this short note, we first obtain a trace inequality, which is similar to inequality (1.1). Meanwhile, we also obtain generalizations of inequalities (1.1), (1.2), (1.3), and (1.4).

2 Main results

In this section, we first give a trace inequality, which is similar to inequality (1.1). To do this, we need the following lemmas.

Lemma 2.1

[2]

Let \(A,B \in M_{n}\) be positive definite. Then
$$\prod_{j = 1}^{k} {\lambda_{j} ( {AB} ) } \le \prod_{j = 1}^{k} { \lambda_{j}^{1/2} \bigl( {A^{2} B^{2} } \bigr) },\quad 1 \le k \le n. $$

Lemma 2.2

[3]

Let \(A,B \in M_{n}\). If \(\lambda ( A ), \lambda ( B ) > 0\) such that
$$\prod_{j = 1}^{k} {\lambda_{j} ( A ) } \le \prod_{j = 1}^{k} { \lambda_{j} ( B ) },\quad 1 \le k \le n, $$
then
$$\det ( {I + A} ) \le \det ( {I + B} ). $$

Theorem 2.1

Let A and B be positive definite. Then
$$\operatorname{tr} \bigl( {\log \bigl( {A^{1/2} + B^{1/2} } \bigr) ^{2} } \bigr) \le \operatorname{tr} \bigl( {\log \bigl( {A + B + 2 ( {A\# B} ) } \bigr) } \bigr). $$

Proof

By Lemma 2.1, we have
$$\begin{aligned} \prod_{j = 1}^{k} {\lambda_{j} \bigl( {A^{ - 1/2} B^{1/2} } \bigr) } &= \prod _{j = 1}^{k} {\lambda_{j} \bigl( {B^{1/2} A ^{ - 1/2} } \bigr) } \\ &\le \prod_{j = 1}^{k} {\lambda_{j} \bigl( { \bigl( {A^{ - 1/2} BA^{ - 1/2} } \bigr) ^{1/2} } \bigr) },\quad 1 \le k \le n. \end{aligned}$$
Using Lemma 2.2, we get
$$ \det \bigl( {I + A^{ - 1/2} B^{1/2} } \bigr) \le \det \bigl( {I + \bigl( {A^{ - 1/2} BA^{ - 1/2} } \bigr) ^{1/2} } \bigr) $$
(2.1)
and
$$ \det \bigl( {I + B^{1/2} A^{ - 1/2} } \bigr) \le \det \bigl( {I + \bigl( {A^{ - 1/2} BA^{ - 1/2} } \bigr) ^{1/2} } \bigr). $$
(2.2)
It follows from (2.1) and (2.2) that
$$\begin{aligned} \det \bigl( {I + A^{ - 1/2} B^{1/2} } \bigr) \det \bigl( {I + B^{1/2} A^{ - 1/2} } \bigr) \le \det \bigl( {I + 2 \bigl( {A^{ - 1/2} BA^{ - 1/2} } \bigr) ^{1/2} + A^{ - 1/2} BA^{ - 1/2} } \bigr), \end{aligned}$$
which is equivalent to
$$\begin{aligned} &\det \bigl( {I + A^{ - 1/2} B^{1/2} + B^{1/2} A^{ - 1/2} + A^{ - 1/2} BA^{ - 1/2} } \bigr) \\ &\quad \le \det \bigl( {I + 2 \bigl( {A^{ - 1/2} BA^{ - 1/2} } \bigr) ^{1/2} + A^{ - 1/2} BA^{ - 1/2} } \bigr). \end{aligned}$$
(2.3)
Multiplying \(\det A^{1/2} \) both sides in inequality (2.3), we have
$$ \det \bigl( {A + B + A^{1/2} B^{1/2} + B^{1/2} A^{1/2} } \bigr) \le \det \bigl( {A + B + 2A^{1/2} \bigl( {A^{ - 1/2} BA^{ - 1/2} } \bigr) ^{1/2} A^{1/2} } \bigr). $$
(2.4)
Note that \(\log \det X = \operatorname{tr}\log X\), inequality (2.4) implies
$$\operatorname{tr} \bigl( {\log \bigl( {A^{1/2} + B^{1/2} } \bigr) ^{2} } \bigr) \le \operatorname{tr} \bigl( {\log \bigl( {A + B + 2 ( {A\# B} ) } \bigr) } \bigr). $$
This completes the proof. □

Next, we show generalizations of inequalities (1.1), (1.2), (1.3), and (1.4). To do this, we need the following lemma.

Lemma 2.3

[2]

Let \(A,B \in M_{n}\) and \(\frac{1}{p} + \frac{1}{q} = 1\), \(p,q > 0\). Then
$$\Vert {AB} \Vert \le \bigl\Vert {\vert A \vert ^{p} } \bigr\Vert ^{1/p} \bigl\Vert {\vert B \vert ^{q} } \bigr\Vert ^{1/q}. $$

This is the Hölder inequality of unitary invariant norms for matrices. For more information on this inequality and its applications the reader is referred to [4] and the references therein.

Theorem 2.2

Let A and B be positive definite and \(1 \le r \le 2\). Then
$$ \operatorname{tr} \bigl( { \bigl( {A + B + 2 ( {A\# B} ) } \bigr) ^{r} } \bigr) \le ( {2 - r} ) \operatorname{tr} \bigl( {A^{1/2} + B^{1/2} } \bigr) ^{2} + ( {r - 1} ) \operatorname{tr} \bigl( {A^{1/2} + B^{1/2} } \bigr) ^{4}. $$
(2.5)

Proof

Let
$$p = \frac{1}{{2 - r}},\qquad q = \frac{1}{{r - 1}}, $$
then
$$\frac{1}{p} + \frac{1}{q} = 1,\quad p,q > 0. $$
By Lemma 2.3, we obtain
$$ \begin{aligned}[b] &\operatorname{tr} \bigl( \bigl( A + B + 2 ( A \# B ) \bigr)^{r} \bigr) \\ &\quad = \operatorname{tr} \bigl( { \bigl( {A + B + 2 ( {A\# B} ) } \bigr) ^{2 - r} \bigl( {A + B + 2 ( {A\# B} ) } \bigr) ^{2r - 2} } \bigr) \\ &\quad \le \bigl( {\operatorname{tr} \bigl( {A + B + 2 ( {A\# B} ) } \bigr) ^{p ( {2 - r} ) } } \bigr) ^{1/p} \bigl( {\operatorname{tr} \bigl( {A + B + 2 ( {A\# B} ) } \bigr) ^{q ( {2r - 2} ) } } \bigr) ^{1/q} \\ &\quad = \bigl( {\operatorname{tr} \bigl( {A + B + 2 ( {A\# B} ) } \bigr) } \bigr) ^{2 - r} \bigl( {\operatorname{tr} \bigl( {A + B + 2 ( {A\# B} ) } \bigr) ^{2} } \bigr) ^{r - 1}. \end{aligned} $$
(2.6)
It follows from (1.1), (1.2), and (2.6) that
$$\operatorname{tr} \bigl( { \bigl( {A + B + 2 ( {A\# B} ) } \bigr) ^{r} } \bigr) \le \bigl( {\operatorname{tr} \bigl( {A^{1/2} + B^{1/2} } \bigr) ^{2} } \bigr) ^{2 - r} \bigl( { \operatorname{tr} \bigl( {A^{1/2} + B^{1/2} } \bigr) ^{4} } \bigr) ^{r - 1}. $$
By Young’s inequality, we have
$$\operatorname{tr} \bigl( { \bigl( {A + B + 2 ( {A\# B} ) } \bigr) ^{r} } \bigr) \le ( {2 - r} ) \operatorname{tr} \bigl( {A^{1/2} + B^{1/2} } \bigr) ^{2} + ( {r - 1} ) \operatorname{tr} \bigl( {A^{1/2} + B^{1/2} } \bigr) ^{4}. $$
This completes the proof. □

Remark 2.1

Putting \(r=1\) in (2.5), we get (1.1). Putting \(r=2\) in (2.5), we get (1.2). Therefore, inequality (2.5) is a generalization of inequalities (1.1) and (1.2).

Remark 2.2

Let A and B be positive definite. By the concavity of \(f ( x ) = x^{r}\), \(x \ge 0\), \(0 < r<1\), then we have
$$n^{r - 1} \operatorname{tr}f ( X ) \le f ( {\operatorname{tr}X} ), $$
where X is positive definite. It follows from this last inequality and inequality (1.1) that
$$\begin{aligned} n^{r - 1} \operatorname{tr} \bigl( {A + B + 2 ( {A\# B} ) } \bigr) ^{r} & \le \bigl( {\operatorname{tr} \bigl( {A + B + 2 ( {A\# B} ) } \bigr) } \bigr) ^{r} \\ &\le \bigl( {\operatorname{tr} \bigl( { \bigl( {A^{1/2} + B^{1/2} } \bigr) ^{2} } \bigr) } \bigr) ^{r}. \end{aligned}$$
Meanwhile, we also have
$$f ( {\operatorname{tr}X} ) \le \operatorname{tr}f ( X ), $$
which implies
$$n^{r - 1} \operatorname{tr} \bigl( {A + B + 2 ( {A\# B} ) } \bigr) ^{r} \le \operatorname{tr} \bigl( { \bigl( {A^{1/2} + B^{1/2} } \bigr) ^{2r} } \bigr). $$
This is a complement of (1.1) for \(0 < r<1\).

Theorem 2.3

Let A and B be positive definite and \(1 \le r \le 2\). Then
$$ \operatorname{tr} \bigl( { ( {A\# _{t} B + B\# _{t} A} ) ^{r} } \bigr) \le ( {2 - r} ) \operatorname{tr} \bigl( {A^{1 - t} B^{t} + A^{t} B^{1 - t} } \bigr) + ( {r - 1} ) \operatorname{tr} \bigl( { \bigl\vert { \bigl( {A^{1 - t} B^{t} + A^{t} B^{1 - t} } \bigr) } \bigr\vert ^{2} } \bigr). $$
(2.7)

Proof

Let
$$p = \frac{1}{{2 - r}},\qquad q = \frac{1}{{r - 1}}, $$
then
$$\frac{1}{p} + \frac{1}{q} = 1,\quad p,q > 0. $$
By Lemma 2.3, we obtain
$$ \begin{aligned}[b] \operatorname{tr} \bigl( { ( {A\# _{t} B + B\# _{t} A} ) ^{r} } \bigr) &= \operatorname{tr} \bigl( { ( {A\# _{t} B + B\# _{t} A} ) ^{2 - r} ( {A\# _{t} B + B\# _{t} A} ) ^{2r - 2} } \bigr) \\ &\le \bigl( {\operatorname{tr} ( {A\# _{t} B + B\# _{t} A} ) ^{p ( {2 - r} ) } } \bigr) ^{1/p} \bigl( {\operatorname{tr} ( {A\# _{t} B + B\# _{t} A} ) ^{q ( {2r - 2} ) } } \bigr) ^{1/q} \\ &= \bigl( {\operatorname{tr} ( {A\# _{t} B + B\# _{t} A} ) } \bigr) ^{2 - r} \bigl( {\operatorname{tr} ( {A\# _{t} B + B\# _{t} A} ) ^{2} } \bigr) ^{r - 1}. \end{aligned} $$
(2.8)
It follows from (1.3), (1.4), and (2.8) that
$$\operatorname{tr} \bigl( { ( {A\# _{t} B + B\# _{t} A} ) ^{r} } \bigr) \le \bigl( {\operatorname{tr} \bigl( {A^{1 - t} B^{t} + A^{t} B^{1 - t} } \bigr) } \bigr) ^{2 - r} \bigl( {\operatorname{tr} \bigl( { \bigl\vert { \bigl( {A^{1 - t} B^{t} + A^{t} B ^{1 - t} } \bigr) } \bigr\vert ^{2} } \bigr) } \bigr) ^{r - 1}. $$
By Young’s inequality, we have
$$\operatorname{tr} \bigl( { ( {A\# _{t} B + B\# _{t} A} ) ^{r} } \bigr) \le ( {2 - r} ) \operatorname{tr} \bigl( {A^{1 - t} B^{t} + A^{t} B^{1 - t} } \bigr) + ( {r - 1} ) \operatorname{tr} \bigl( { \bigl\vert { \bigl( {A^{1 - t} B^{t} + A^{t} B^{1 - t} } \bigr) } \bigr\vert ^{2} } \bigr). $$
This completes the proof. □

Remark 2.3

Putting \(r=1\) in (2.7), we get (1.3). Putting \(r=2\) in (2.7), we get (1.4). Therefore, inequality (2.7) is a generalization of inequalities (1.3) and (1.4).

Declarations

Acknowledgements

The authors wish to express their heartfelt thanks to the referees and Professor Sin E Takahasi for their detailed and helpful suggestions for revising the manuscript. This research was supported by Scientific and Technological Research Program of Chongqing Municipal Education Commission (Grant No. KJ1501004).

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
School of Mathematics and Statistics, Chongqing Three Gorges University

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Copyright

© Zou and Peng 2016