# Tracial and majorisation Heinz mean-type inequalities for matrices

## Abstract

The Heinz mean for every nonnegative real numbers a, b and every $$0\le\nu\le1$$ is $$H_{\nu}(a , b)=\frac{a^{\nu}b^{1-\nu} +a^{1-\nu}b^{\nu}}{2}$$. In this paper we present tracial Heinz mean-type inequalities for positive definite matrices and apply it to prove a majorisation version of the Heinz mean inequality.

## 1 Introduction

The arithmetic-geometric mean inequality for two positive real numbers a, b is $$\sqrt{ab}\le\frac{a+b}{2}$$, where equality holds if and only if $$a=b$$. Heinz means, introduced in [1], are means that interpolate in a certain way between the arithmetic and geometric mean. For every nonnegative real numbers a, b and $$0\le\nu\le1$$, the Heinz mean is defined as

$$H_{\nu}(a , b)=\frac{a^{\nu}b^{1-\nu}+a^{1-\nu}b^{\nu}}{2} .$$

The function $$H_{\nu}$$ is symmetric about the point $$\nu=\frac{1}{2}$$. Note that $$H_{0}(a , b)=H_{1}(a , b)=\frac{a+b}{2}$$, $$H_{\frac{1}{2}}(a , b)=\sqrt{ab}$$, and

$$H_{\frac{1}{2}}(a , b)\le H_{\nu}(a , b)\le H_{1}(a , b)$$
(1)

for every $$0\le\nu\le1$$, and equality holds if and only if $$a=b$$.

Let $$M_{n}(\mathbb {C})$$ denote the space of all $$n\times n$$ matrices. We shall denote the eigenvalues and singular values of a matrix $$A\in M_{n}(\mathbb {C})$$ by $$\lambda_{j}(A)$$ and $$\sigma_{j}(A)$$, respectively. We assume that singular values are sorted in non-increasing order. For two Hermitian matrices $$A,B\in M_{n}(\mathbb {C})$$, $$A\ge B$$ means that $$A-B$$ is positive semi-definite. In particular, $$A\ge0$$ means A is positive semi-definite. Let us write $$A>0$$ when A is positive definite. $$|A|$$ shall denote the modulus $$|A|=(A^{*}A)^{\frac{1}{2}}$$ and $$\operatorname{tr}(A)=\sum_{j=1}^{n} \lambda_{j}(A)$$.

The basic properties of singular values and trace function that some of them are used to establish the matrix inequalities in this paper are collected in the following theorems.

### Theorem 1.1

Assume that $$X, Y \in M_{n}(\mathbb {C})$$, $$A, B\in M_{n}(\mathbb {C})^{+}$$, $$\alpha\in \mathbb {C}$$, and $$j=1,2,\ldots, n$$.

1. (1)

$$\sigma_{j}(X)= \sigma_{j}(X^{*})= \sigma_{j}(|X|)=$$ and $$\sigma_{j}(\alpha X)=|\alpha|\sigma_{j}(X)$$.

2. (2)

If $$A\le B$$, then $$\sigma_{j}(A) \le\sigma_{j}(B)$$.

3. (3)

$$\sigma_{j}(X^{r}) = ( \sigma_{j}(X) )^{r}$$, for every positive real number r.

4. (4)

$$\sigma_{j}(XY^{*}) = \sigma_{j}(YX^{*})$$.

5. (5)

$$\sigma_{j}(XY) \le\|X\| \sigma_{j}(Y)$$.

6. (6)

$$\sigma_{j}(YXY^{*}) \le \|Y\|^{2} \sigma_{j}(X)$$.

### Theorem 1.2

Assume that $$X, Y \in M_{n}(\mathbb {C})$$, $$\alpha\in \mathbb {C}$$.

1. (1)

$$\operatorname{tr}(X+Y)=\operatorname{tr}(X)+\operatorname{tr}(Y)$$.

2. (2)

$$\operatorname{tr}(XY)=\operatorname{tr}(YX)$$.

3. (3)

$$\operatorname{tr}(X) \ge0$$, and for $$A\in M_{n}(\mathbb {C})^{+}$$, $$\operatorname{tr}(A)=0$$ only if $$A=0$$.

The absolute value for matrices does not satisfy $$|XY|=|X|\cdot|Y|$$; however, a weaker version of this is the following:

If $$Y=U|Y|$$ is the polar decomposition of Y, with unitary U, then

$$\bigl|XY^{*}\bigr|=U\bigl|\bigl(|X|\cdot|Y|\bigr)\bigr| U^{*}$$
(2)

and

$$\lambda_{j} \bigl(\bigl| XY^{*}\bigr| \bigr)= \sigma _{j}\bigl( |X|\cdot|Y|\bigr) .$$
(3)

The Young inequality is among the most important inequalities in matrix theory. We present here the following theorem from [2, 3].

### Theorem 1.3

Let $$A, B\in M_{n}(\mathbb {C})$$ be positive semi-definite. If $$p, q>1$$ with $$\frac{1}{p}+\frac{1}{p}=1$$, then

$$\sigma_{j}(AB)\le\sigma_{j} \biggl( \frac{1}{p}A^{p}+\frac{1}{q}B^{q} \biggr)\quad \textit{for } j=1, 2,\ldots, n ,$$
(4)

where equality holds if and only if $$A^{p}=B^{q}$$.

### Corollary 1.4

Let $$A, B\in M_{n}(\mathbb {C})$$ be positive semi-definite. If $$p, q>1$$ with $$\frac{1}{p}+\frac{1}{p}=1$$, then

$$\operatorname{tr}\bigl(|AB|\bigr)\le\frac{1}{p}\operatorname{tr}\bigl(A^{p} \bigr)+ \frac{1}{q}\operatorname{tr}\bigl(B^{q} \bigr) ,$$
(5)

where equality holds if and only if $$A^{p}=B^{q}$$.

Another interesting inequality is the following version of the triangle inequality for the matrix absolute value [1, 4].

### Theorem 1.5

Let X and Y be $$n\times n$$ matrices, then there exist unitaries U, V such that

$$|X+Y|\le U|X|U^{*}+V|Y|V^{*} .$$
(6)

We are interested to find what types of inequalities (1) hold for positive semi-definite matrices A, B? For example, do we have

$$\sqrt{|AB|} \le\bigl|H_{\nu}(A , B)\bigr|\le H_{1}(A , B) ?$$
(7)

Or do we have

$$\sqrt{\sigma_{j}(AB)} \le\sigma_{j} \bigl(H_{\nu}(A , B) \bigr)\le\lambda_{j}\bigl(H_{1}(A , B)\bigr) ?$$
(8)

Here

$$H_{\nu}(A , B)=\frac{A^{\nu}B^{1-\nu}+A^{1-\nu}B^{\nu}}{2} .$$

Bhatia and Davis [5] extended inequality (1) to the matrix case, they showed that it holds for positive semi-definite matrices, in the following form:

$$\bigl|\!\bigl|\!\bigl|A^{\frac{1}{2}}B^{\frac{1}{2}}\bigr|\!\bigr|\!\bigr|\le\bigl|\!\bigl|\!\bigl|H_{\nu}(A , B)\bigr|\!\bigr|\!\bigr|\le\biggl|\!\biggl|\!\biggl|\frac{A+B}{2}\biggr|\!\biggr|\!\biggr|,$$
(9)

where $$|\!|\!|\cdot|\!|\!|$$ is any invariant unitary norm. An example shows that the first inequality in (9), to singular values, does not hold [6]. One of the results in the present article is a version of Heinz mean-type inequalities for matrices in the following theorem.

### Theorem 1.6

Let A, B be two positive semi-definite matrices in $$M_{n}(\mathbb {C})$$. Then

$$\operatorname{tr}\bigl(\sqrt{|AB|}\bigr) \le \operatorname{tr}\bigl(H_{1} \bigl(\bigl|A^{\nu }B^{1-\nu}\bigr| ,\bigl|A^{1-\nu}B^{\nu}\bigr| \bigr) \bigr) \le \operatorname{tr}\bigl(H_{1}(A , B) \bigr) .$$

Equality holds if and only if $$A=B$$.

For a real vector $$X=(x_{1} , x_{2} , \ldots,x_{n})$$, let $$X^{\downarrow}=(x_{1}^{\downarrow} , x_{2}^{\downarrow} ,\ldots ,x_{n}^{\downarrow})$$ be the decreasing rearrangement of X. Let X and Y are two vectors in $$\mathbb {R}^{n}$$, we say X is (weakly) submajorised by Y, in symbols $$X \prec_{w} Y$$, if

$$\sum_{j=1}^{k} x_{j}^{\downarrow} \leq\sum_{j=1}^{k} y_{j}^{\downarrow} ,\quad1\leq k\leq n .$$

X is majorised by Y, in symbols $$X \prec Y$$, if X is submajorised by Y and

$$\sum_{j=1}^{n} x_{j}^{\downarrow} = \sum_{j=1}^{n} y_{j}^{\downarrow}.$$

### Definition 1.7

If $$A, B\in M_{n}(\mathbb {C})$$, then we write $$A \prec_{w} B$$ to denote that A is weakly majorised by B, meaning that

$$\sum_{j=1}^{k}\sigma_{j}(A) \le \sum_{j=1}^{k}\sigma_{j}(B), \quad\mbox{for all } 1\le k\le n .$$

If $$A \prec_{w} B$$ and

$$\operatorname{tr}\bigl(|A|\bigr) = \operatorname{tr}\bigl(|B|\bigr) ,$$

then we say that A is majorised by B, in symbols $$A \prec B$$.

Let $$S(A)$$ denote the n-vector whose coordinates are the singular values of A. Then we write $$A \prec_{w} B$$ ($$A \prec B$$) when $$S(A) \prec_{w} S(B)$$ ($$S(A) \prec S(B)$$) .

The following theorem has been proved in [1].

### Theorem 1.8

If X and Y are two matrices in $$M_{n}(\mathbb {C})$$, then

$$S^{r}(XY)\prec_{w} S^{r}(X) S^{r}(Y) \quad\textit{for all } r>0 .$$
(10)

## 2 Main results

We present here the matrix inequalities that we will use in the proof of our main results. The next theorem has been proved in [6].

### Theorem 2.1

For positive semi-definite matrices A and B and for all $$j=1, 2,\ldots, n$$

$$\sigma_{j} \bigl(H_{\nu}(A , B) \bigr)\le \sigma_{j} \bigl(H_{1}(A , B) \bigr) ,$$

for every $$\nu\in[0 , 1]$$.

Thus, this proves that the second inequality in (8) holds. The arithmetic-geometric mean inequality

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

is used in the matrix setting, much of this is associated with Bhatia and Kittaneh. They established the next inequality in [7]:

$$\sigma_{j} \bigl(A^{*}B \bigr)\le\lambda_{j} \biggl(\frac{AA^{*}+BB^{*}}{2} \biggr) ,$$
(11)

where A and B are two matrices in $$M_{n}(\mathbb {C})$$. They also studied many possible versions of this inequality in [8], and put a lot of emphasis on what they described as level three inequalities [9]. Drury [10] answered to the key question in this area in the following theorem.

### Theorem 2.2

For positive semi-definite matrices A and B in $$M_{n}(\mathbb {C})$$ and for all $$j=1, 2,\ldots, n$$

$$\sqrt{\sigma_{j}(AB)}\le\lambda_{j} \bigl(H_{1}(A , B) \bigr) .$$

We will show that in both Theorems 2.1 and 2.2 equality holds if and only if $$A=B$$. It is still unknown whether

$$\sqrt{\sigma_{j}(AB)}\le\sigma_{j} \bigl(H_{\nu}(A , B) \bigr)$$

for every $$\nu\in(0 , 1)$$. However, by using Theorems 2.1 and 2.2, we present a different version of this inequality.

### Lemma 2.3

For positive semi-definite matrices A and B in $$M_{n}(\mathbb {C})$$ and for all $$j=1, 2,\ldots, n$$

$$\sqrt{\sigma_{j}(AB)}\le\lambda_{j} \bigl(H_{1} \bigl(\bigl|A^{\nu}B^{1-\nu}\bigr| ,\bigl|A^{1-\nu }B^{\nu}\bigr| \bigr) \bigr)$$
(12)

for every $$\nu\in(0 , 1)$$.

### Proof

We first aim to show that

$$\sigma_{j}(AB) \le\sigma_{j} \bigl(A^{1-\nu}A^{\nu}B^{1-\nu}B^{\nu} \bigr).$$

We have

\begin{aligned} \sigma_{j}(AB)&=\sigma_{j} \bigl(A^{1-\nu}A^{\nu}B^{1-\nu}B^{\nu }A^{1-\nu}A^{\nu-1} \bigr) \\ &\le\|A\|^{1-\nu}\sigma_{j} \bigl(A^{\nu}B^{1-\nu}B^{\nu}A^{1-\nu} \bigr)\|A \|^{\nu -1} \quad\bigl(\mbox{by part (5) Theorem }1.1 \bigr). \end{aligned}
(13)

As $$\nu-1<0$$, the matrix $$A^{\nu-1}$$ exists only if A is invertible. Therefore, to prove (13) we shall assume that A is invertible. This assumption entails no loss in generality, for if A were not invertible, then we could replace A by $$A+\varepsilon I$$, which is invertible and which satisfies $$\sigma_{j}((A+\varepsilon I)B)\rightarrow \sigma_{j}(AB)$$ for every $$B\in M_{n}(\mathbb {C})$$ and $$j=1, 2,\ldots, n$$. Thus, (13) is achieved for noninvertible A as a limiting case of (13) using the invertibility of A.

By using equation (3), we get

$$\sigma_{j} \bigl(A^{\nu}B^{1-\nu}B^{\nu}A^{1-\nu} \bigr) =\sigma_{j} \bigl(\bigl|A^{\nu}B^{1-\nu}\bigr| \cdot\bigl|A^{1-\nu}B^{\nu}\bigr| \bigr).$$

Hence, by using TheoremÂ 2.2,

$$\sqrt{\sigma_{j}(AB)} \le\sqrt{\sigma_{j} \bigl(\bigl|A^{\nu}B^{1-\nu}\bigr|\cdot\bigl|A^{1-\nu}B^{\nu }\bigr| \bigr)}\le\lambda_{j} \bigl(H_{1} \bigl(\bigl|A^{\nu}B^{1-\nu }\bigr| ,\bigl|A^{1-\nu }B^{\nu}\bigr| \bigr) \bigr) .$$

â€ƒâ–¡

### Remark 2.4

Note that LemmaÂ 2.3 generalizes TheoremÂ 2.2, in fact, it is the special case with $$\nu=1$$ of LemmaÂ 2.3.

### Theorem 2.5

Let A, B be two positive semi-definite matrices in $$M_{n}(\mathbb {C})$$. Then

$$\operatorname{tr}\bigl(\sqrt{|AB|}\bigr) \le \operatorname{tr}\bigl(H_{1} \bigl(\bigl|A^{\nu }B^{1-\nu}\bigr| ,\bigl|A^{1-\nu}B^{\nu}\bigr| \bigr) \bigr) \le \operatorname{tr}\bigl(H_{1}(A , B) \bigr) .$$

### Proof

By the definition of the trace, we have

\begin{aligned} \operatorname{tr}\bigl(\sqrt{|AB|}\bigr) =& \sum_{j=1}^{n} \lambda_{j}\sqrt{|AB|}\\ =& \sum_{j=1}^{n} \sqrt{\sigma_{j}(AB)} \quad \bigl(\mbox{by part (3) Theorem }1.1\bigr) \\ \le& \sum_{j=1}^{n} \lambda_{j}\bigl(H_{1}\bigl(\bigl|A^{\nu }B^{1-\nu}\bigr| ,\bigl|A^{1-\nu }B^{\nu}\bigr|\bigr)\bigr)\\ =& \operatorname{tr}\bigl(H_{1}\bigl(\bigl|A^{\nu}B^{1-\nu}\bigr| , \bigl|A^{1-\nu}B^{\nu}\bigr|\bigr)\bigr) \quad\bigl(\mbox{using inequality } (12)\bigr)\\ =& \frac{1}{2}\operatorname{tr}\bigl(A^{\nu}B^{1-\nu}\bigr)+\frac{1}{2} \operatorname{tr}\bigl(A^{1-\nu}B^{\nu}\bigr)\\ \le& \frac{1}{2} \bigl( \operatorname{tr}\bigl(\nu A+(1-\nu)B\bigr) +\operatorname{tr}\bigl(\nu B+(1-\nu)A\bigr) \bigr). \end{aligned}

We applied (1.4) with $$p=\frac{1}{\nu}$$ and $$q=\frac{1}{1-\nu}$$ for the first summand, and $$q=\frac{1}{\nu}$$ and $$p=\frac{1}{1-\nu}$$ for the second one.

Therefore,

\begin{aligned} \operatorname{tr}\bigl(\sqrt{|AB|}\bigr) \le& \operatorname{tr}\bigl(H_{1}\bigl(\bigl|A^{\nu }B^{1-\nu}\bigr| , \bigl|A^{1-\nu}B^{\nu}\bigr|\bigr)\bigr)\\ \le& \frac{1}{2} \bigl( \nu \operatorname{tr}(A)+(1-\nu)\operatorname{tr}(B)+(1-\nu)\operatorname{tr}(A)+\nu \operatorname{tr}(B) \bigr)\\ = & \frac{1}{2}\operatorname{tr}(A+B)=\operatorname{tr}\bigl(H_{1}(A , B)\bigr). \end{aligned}

â€ƒâ–¡

### Theorem 2.6

If $$A, B\in M_{n}(\mathbb {C})$$ are two positive semi-definite matrices and $$0\le\nu\le1$$. Then the following conditions are equivalent:

1. (1)

$$\operatorname{tr}(\sqrt{|AB|}) = \operatorname{tr}(H_{1}(A , B))$$.

2. (2)

$$\operatorname{tr}(H_{1}(|A^{\nu}B^{1-\nu}| ,|A^{1-\nu}B^{\nu}|)) = \operatorname{tr}(H_{1}(A , B))$$.

3. (3)

$$\operatorname{tr}(|H_{\nu}(A , B)|) = \operatorname{tr}(H_{1}(A , B))$$.

4. (4)

$$A = B$$.

### Proof

We shall show that $$(1)\Longrightarrow (2)\Longrightarrow(4)\Longrightarrow(1)$$ and $$(3)\Longrightarrow (2)\Longrightarrow(4)\Longrightarrow(3)$$.

Let $$\operatorname{tr}(\sqrt{|AB|}) = \operatorname{tr}(H_{1}(A , B))$$. Then the arguments of the proof of the above theorem implies

$$\operatorname{tr}\bigl(H_{1} \bigl(\bigl|A^{\nu}B^{1-\nu}\bigr| ,\bigl|A^{1-\nu}B^{\nu}\bigr| \bigr) \bigr) = \operatorname{tr}\bigl(H_{1}(A , B) \bigr) .$$

If the equation in part (2) holds, then from what was proved in the last theorem we conclude that

\begin{aligned} \operatorname{tr}\bigl(H_{1}(A , B)\bigr) = & \operatorname{tr}\bigl(H_{1}\bigl(\bigl|A^{\nu}B^{1-\nu}\bigr| , \bigl|A^{1-\nu}B^{\nu}\bigr|\bigr)\bigr)\\ = & \frac{1}{2}\operatorname{tr}\bigl(\bigl|A^{\nu}B^{1-\nu}\bigr| + \bigl|A^{1-\nu}B^{\nu}\bigr|\bigr)\\ \le& \frac{1}{2}\bigl( \operatorname{tr}\bigl(\nu A+(1-\nu)B\bigr)+\operatorname{tr}\bigl(\nu B+(1-\nu)A\bigr) \bigr)=\operatorname{tr}\bigl(H_{1}(A , B)\bigr) . \end{aligned}

Thus,

$$\operatorname{tr}\bigl(\bigl|A^{\nu}B^{1-\nu}\bigr| \bigr) + \operatorname{tr}\bigl(\bigl|A^{1-\nu}B^{\nu}\bigr| \bigr) = \operatorname{tr}\bigl(\nu A+(1-\nu)B \bigr)+ \operatorname{tr}\bigl(\nu B+(1-\nu)A \bigr) .$$
(14)

By CorollaryÂ 1.4, this equality holds if and only if

$$\operatorname{tr}\bigl(\bigl|A^{\nu}B^{1-\nu}\bigr| \bigr) = \operatorname{tr}\bigl(\nu A+(1-\nu )B\bigr)\quad\mbox{and}\quad \operatorname{tr}\bigl(\bigl|A^{1-\nu}B^{\nu}\bigr| \bigr) = \operatorname{tr}\bigl(\nu B+(1-\nu)A \bigr),$$

and therefore $$A^{1-\nu}=B^{\nu}$$, $$B^{1-\nu}=A^{\nu}$$, which implies $$A=B$$. It is clear that $$(4)\Longrightarrow(1)$$.

Now, we try to show that $$(3)\Longrightarrow(2)\Longrightarrow (4)\Longrightarrow(3)$$. Therefore assume (3): $$\operatorname{tr}(|H_{\nu}(A , B)|) = \operatorname{tr}(H_{1}(A , B))$$. Then

\begin{aligned} \operatorname{tr}\bigl(H_{1}(A , B)\bigr) = & \operatorname{tr}\bigl(\bigl|H_{\nu}(A , B)\bigr|\bigr)\\ = & \frac{1}{2}\operatorname{tr}\bigl(\bigl|A^{\nu}B^{1-\nu} + A^{1-\nu}B^{\nu}\bigr|\bigr)\\ \le& \frac{1}{2}\bigl[ \operatorname{tr}\bigl(U\bigl|A^{\nu}B^{1-\nu}\bigr|U^{*}\bigr) + \operatorname{tr}\bigl(V^{*}\bigl|A^{1-\nu}B^{\nu}\bigr|V\bigr)\bigr]\ \ \bigl(\mbox{by the triangle inequality} (6)\bigr) \end{aligned}

for some unitaries U and $$V\in M_{n}(\mathbb {C})$$.

Thus,

\begin{aligned} \operatorname{tr}\bigl(H_{1}(A , B)\bigr) \le& \frac{1}{2}\operatorname{tr}\bigl(\bigl|A^{\nu }B^{1-\nu}\bigr| + \bigl|A^{1-\nu}B^{\nu}\bigr|\bigr) \\ = & \operatorname{tr}\bigl(H_{1}\bigl(\bigl|A^{\nu}B^{1-\nu}\bigr| , \bigl|A^{1-\nu}B^{\nu}\bigr|\bigr)\bigr) \\ \le& \operatorname{tr}\bigl(H_{1}(A , B)\bigr) \quad(\mbox{by Theorem }2.5), \end{aligned}

thereby proving (2). $$(2)\Longrightarrow(4)$$ was shown in the first part. It is clear that $$(4)\Longrightarrow(3)$$.â€ƒâ–¡

The following two corollaries are almost immediate from TheoremÂ 2.6.

### Corollary 2.7

For positive semi-definite matrices A and B in $$M_{n}(\mathbb {C})$$ and for all $$j=1, 2,\ldots, n$$

$$\sqrt{\sigma_{j}(AB)} = \lambda_{j} \bigl(H_{1}(A , B) \bigr) ,$$

if and only if $$A = B$$.

### Corollary 2.8

For positive semi-definite matrices A and B in $$M_{n}(\mathbb {C})$$ and for all $$j=1, 2,\ldots, n$$

$$\sigma_{j} \bigl(H_{\nu}(A , B) \bigr) = \lambda_{j} \bigl(H_{1}(A , B) \bigr) ,$$

for $$\nu\in[0 , 1]$$ if and only if $$A = B$$.

We do not know whether

$$\sqrt{\sigma_{j}(AB)} \le\sigma_{j} \bigl(H_{\nu}(A , B) \bigr) \le\lambda_{j} \bigl(H_{1}(A , B) \bigr)$$

for every $$\nu\in[0 , 1]$$.

To answer this question, just we need to know whether

$$\sqrt{\sigma_{j}(AB)} \le\sigma_{j} \bigl(H_{\nu}(A , B) \bigr)$$

for every $$\nu\in[0 , 1]$$.

In the rest of this paper, we apply the results of singular value inequalities for the means to present a new majorisation version of the means.

### Lemma 2.9

Let A and B be two positive semi-definite matrices. Then

$$S^{\frac{1}{2}}(AB) \prec_{w} \frac{1}{2} \bigl(S(A) + S(B) \bigr) .$$

### Proof

By TheoremÂ 1.8,

$$\sum_{j=1}^{k} \sigma_{j}(AB)^{\frac{1}{2}} \le\sum_{j=1}^{k} \lambda_{j}(A)^{\frac{1}{2}} \lambda_{j}(B)^{\frac{1}{2}} \quad\mbox{for every } 1\le k\le n .$$

By using an arithmetic-geometric mean inequality for singular values of A and B,

$$\sum_{j=1}^{k} \sigma_{j}(AB)^{\frac{1}{2}} \le\sum_{j=1}^{k}\frac{1}{2} \lambda_{j}(A) + \sum_{j=1}^{k} \frac{1}{2} \lambda_{j}(B) \quad\mbox{for every } 1\le k\le n .$$

Thus,

$$\sum_{j=1}^{k} \sigma_{j}(AB)^{\frac{1}{2}} \le\sum_{j=1}^{k}\frac{1}{2}( \lambda_{j}(A) + \lambda_{j}(B) \quad\mbox{for every } 1\le k\le n ,$$

which implies $$S^{\frac{1}{2}}(AB) \prec_{w} \frac{1}{2}(S(A) + S(B))$$.â€ƒâ–¡

### Lemma 2.10

If A and $$B \in M_{n}(\mathbb {C})$$, then

$$\sqrt{|AB|} \prec_{w} H_{1}(A , B) .$$

### Proof

It is direct result of the definition of the majorisation and TheoremÂ 2.2.â€ƒâ–¡

### Lemma 2.11

If A and B are positive semi-definite $$\in M_{n}(\mathbb {C})$$, then

$$H_{\nu}(A , B) \prec_{w} H_{1}(A , B) .$$

### Proof

It is direct result of definition of the majorisation and TheoremÂ 2.1.â€ƒâ–¡

It is interesting to know whether

$$\sqrt{|AB|} \prec_{w} H_{j}(A , B) .$$

### Lemma 2.12

If A and B are positive semi-definite $$\in M_{n}(\mathbb {C})$$, then

$$\sqrt{|AB|} \prec_{w} H_{1} \bigl(\bigl|A^{\nu}B^{1-\nu}\bigr| ,\bigl|A^{1-\nu}B^{\nu}\bigr| \bigr) .$$

### Proof

It is direct result of definition of the majorisation and LemmaÂ 2.3.â€ƒâ–¡

The results to this point lead to the following theorem about majorisation for positive definite matrices.

### Theorem 2.13

For every two positive matrices A and B in $$M_{n}(\mathbb {C})$$, the following conditions are equivalent:

1. (1)

$$S^{\frac{1}{2}}(AB) \prec\frac{1}{2}(S(A) + S(B))$$.

2. (2)

$$\sqrt{|AB|} \prec(H_{1}(A , B))$$.

3. (3)

$$H_{\nu}(A , B) \prec H_{1}(A , B)$$.

4. (4)

$$\sqrt{|AB|} \prec_{w} H_{1}(|A^{\nu}B^{1-\nu}| ,|A^{1-\nu }B^{\nu }|)$$.

5. (5)

$$A = B$$.

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## Acknowledgements

This work was supported by the Department of Mathematical Sciences at Isfahan University of Technology, Iran.

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Correspondence to Seyed Mahmoud Manjegani.

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The author declares that he has no competing interests.

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Mahmoud Manjegani, S. Tracial and majorisation Heinz mean-type inequalities for matrices. J Inequal Appl 2016, 23 (2016). https://doi.org/10.1186/s13660-016-0965-8