# Some trace inequalities for matrix means

## 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.

## 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.â€ƒâ–¡

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 HÃ¶lder 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 Youngâ€™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.â€ƒâ–¡

### 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 Youngâ€™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.â€ƒâ–¡

### 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).

## References

1. Bhatia, R, Lim, Y, Yamazaki, T: Some norm inequalities for matrix means. Linear Algebra Appl. 501, 112-122 (2016)

2. Bhatia, R: Matrix Analysis. Springer, New York (1997)

3. Lin, M: On a determinantal inequality arising from diffusion tensor imaging. Commun. Contemp. Math. (2016). doi:10.1142/S0219199716500449

4. Hu, X: Some inequalities for unitarily invariant norms. J. Math. Inequal. 6, 615-623 (2012)

## 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).

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Correspondence to Limin Zou.

### Competing interests

The authors declare that they have no competing interests.

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All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

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Zou, L., Peng, Y. Some trace inequalities for matrix means. J Inequal Appl 2016, 283 (2016). https://doi.org/10.1186/s13660-016-1236-4