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 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 for every \(0\le\nu\le1\), and equality holds if and only if \(a=b\).
$$H_{\nu}(a , b)=\frac{a^{\nu}b^{1-\nu}+a^{1-\nu}b^{\nu}}{2} . $$
$$ H_{\frac{1}{2}}(a , b)\le H_{\nu}(a , b)\le H_{1}(a , b) $$
(1)
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 \(\operatorname{tr}(A)=\sum_{j=1}^{n} \lambda_{j}(A)\).
Anzeige
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)
\(\sigma_{j}(X)= \sigma_{j}(X^{*})= \sigma_{j}(|X|)=\)
and
\(\sigma_{j}(\alpha X)=|\alpha|\sigma_{j}(X)\).
(2)
If
\(A\le B\), then
\(\sigma_{j}(A) \le\sigma_{j}(B)\).
(3)
\(\sigma_{j}(X^{r}) = ( \sigma_{j}(X) )^{r}\), for every positive real number
r.
(4)
\(\sigma_{j}(XY^{*}) = \sigma_{j}(YX^{*})\).
(5)
\(\sigma_{j}(XY) \le\|X\| \sigma_{j}(Y)\).
(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)
\(\operatorname{tr}(X+Y)=\operatorname{tr}(X)+\operatorname{tr}(Y)\).
(2)
\(\operatorname{tr}(XY)=\operatorname{tr}(YX)\).
(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:
Anzeige
If \(Y=U|Y|\) is the polar decomposition of Y, with unitary U, then and
$$ \bigl|XY^{*}\bigr|=U\bigl|\bigl(|X|\cdot|Y|\bigr)\bigr| U^{*} $$
(2)
$$ \lambda_{j} \bigl(\bigl| XY^{*}\bigr| \bigr)= \sigma _{j}\bigl( |X|\cdot|Y|\bigr) . $$
(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
where equality holds if and only if
\(A^{p}=B^{q}\).
$$ \sigma_{j}(AB)\le\sigma_{j} \biggl( \frac{1}{p}A^{p}+\frac{1}{q}B^{q} \biggr)\quad \textit{for } j=1, 2,\ldots, n , $$
(4)
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
where equality holds if and only if
\(A^{p}=B^{q}\).
$$ \operatorname{tr}\bigl(|AB|\bigr)\le\frac{1}{p}\operatorname{tr}\bigl(A^{p} \bigr)+ \frac{1}{q}\operatorname{tr}\bigl(B^{q} \bigr) , $$
(5)
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 Or do we have
$$ \sqrt{|AB|} \le\bigl|H_{\nu}(A , B)\bigr|\le H_{1}(A , B) ? $$
(7)
$$ \sqrt{\sigma_{j}(AB)} \le\sigma_{j} \bigl(H_{\nu}(A , B) \bigr)\le\lambda_{j}\bigl(H_{1}(A , B)\bigr) ? $$
(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: 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.
$$ \bigl|\!\bigl|\!\bigl|A^{\frac{1}{2}}B^{\frac{1}{2}}\bigr|\!\bigr|\!\bigr|\le\bigl|\!\bigl|\!\bigl|H_{\nu}(A , B)\bigr|\!\bigr|\!\bigr|\le\biggl|\!\biggl|\!\biggl|\frac{A+B}{2}\biggr|\!\biggr|\!\biggr|, $$
(9)
Theorem 1.6
Let
A, B
be two positive semi-definite matrices in
\(M_{n}(\mathbb {C})\). Then
Equality holds if and only if
\(A=B\).
$$\operatorname{tr}\bigl(\sqrt{|AB|}\bigr) \le \operatorname{tr}\bigl(H_{1} \bigl(\bigl|A^{\nu }B^{1-\nu}\bigr| ,\bigl|A^{1-\nu}B^{\nu}\bigr| \bigr) \bigr) \le \operatorname{tr}\bigl(H_{1}(A , B) \bigr) . $$
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
X is majorised by Y, in symbols \(X \prec Y\), if X is submajorised by Y and
$$\sum_{j=1}^{k} x_{j}^{\downarrow} \leq\sum_{j=1}^{k} y_{j}^{\downarrow} ,\quad1\leq k\leq n . $$
$$\sum_{j=1}^{n} x_{j}^{\downarrow} = \sum_{j=1}^{n} y_{j}^{\downarrow}. $$
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 If \(A \prec_{w} B\) and then we say that A is majorised by B, in symbols \(A \prec B \).
$$\sum_{j=1}^{k}\sigma_{j}(A) \le \sum_{j=1}^{k}\sigma_{j}(B), \quad\mbox{for all } 1\le k\le n . $$
$$\operatorname{tr}\bigl(|A|\bigr) = \operatorname{tr}\bigl(|B|\bigr) , $$
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
\(j=1, 2,\ldots, n\)
for every
\(\nu\in[0 , 1]\).
$$\sigma_{j} \bigl(H_{\nu}(A , B) \bigr)\le \sigma_{j} \bigl(H_{1}(A , B) \bigr) , $$
Thus, this proves that the second inequality in (8) holds. The arithmetic-geometric mean inequality is used in the matrix setting, much of this is associated with Bhatia and Kittaneh. They established the next inequality in [7]: 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.
$$\sqrt{ab}\le\frac{a+b}{2} $$
$$ \sigma_{j} \bigl(A^{*}B \bigr)\le\lambda_{j} \biggl(\frac{AA^{*}+BB^{*}}{2} \biggr) , $$
(11)
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 for every \(\nu\in(0 , 1)\). However, by using Theorems 2.1 and 2.2, we present a different version of this inequality.
$$\sqrt{\sigma_{j}(AB)}\le\sigma_{j} \bigl(H_{\nu}(A , B) \bigr) $$
Lemma 2.3
For positive semi-definite matrices
A
and
B
in
\(M_{n}(\mathbb {C})\)
and for all
\(j=1, 2,\ldots, n\)
for every
\(\nu\in(0 , 1)\).
$$ \sqrt{\sigma_{j}(AB)}\le\lambda_{j} \bigl(H_{1} \bigl(\bigl|A^{\nu}B^{1-\nu}\bigr| ,\bigl|A^{1-\nu }B^{\nu}\bigr| \bigr) \bigr) $$
(12)
Proof
We first aim to show that We have 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.
$$\sigma_{j}(AB) \le\sigma_{j} \bigl(A^{1-\nu}A^{\nu}B^{1-\nu}B^{\nu} \bigr). $$
$$\begin{aligned} \sigma_{j}(AB)&=\sigma_{j} \bigl(A^{1-\nu}A^{\nu}B^{1-\nu}B^{\nu }A^{1-\nu}A^{\nu-1} \bigr) \\ &\le\|A\|^{1-\nu}\sigma_{j} \bigl(A^{\nu}B^{1-\nu}B^{\nu}A^{1-\nu} \bigr)\|A \|^{\nu -1} \quad\bigl(\mbox{by part (5) Theorem }1.1 \bigr). \end{aligned}$$
(13)
By using equation (3), we get Hence, by using Theorem 2.2, □
$$\sigma_{j} \bigl(A^{\nu}B^{1-\nu}B^{\nu}A^{1-\nu} \bigr) =\sigma_{j} \bigl(\bigl|A^{\nu}B^{1-\nu}\bigr| \cdot\bigl|A^{1-\nu}B^{\nu}\bigr| \bigr). $$
$$\sqrt{\sigma_{j}(AB)} \le\sqrt{\sigma_{j} \bigl(\bigl|A^{\nu}B^{1-\nu}\bigr|\cdot\bigl|A^{1-\nu}B^{\nu }\bigr| \bigr)}\le\lambda_{j} \bigl(H_{1} \bigl(\bigl|A^{\nu}B^{1-\nu }\bigr| ,\bigl|A^{1-\nu }B^{\nu}\bigr| \bigr) \bigr) . $$
Remark 2.4
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, □
$$\begin{aligned} \operatorname{tr}\bigl(\sqrt{|AB|}\bigr) \le& \operatorname{tr}\bigl(H_{1}\bigl(\bigl|A^{\nu }B^{1-\nu}\bigr| , \bigl|A^{1-\nu}B^{\nu}\bigr|\bigr)\bigr)\\ \le& \frac{1}{2} \bigl( \nu \operatorname{tr}(A)+(1-\nu)\operatorname{tr}(B)+(1-\nu)\operatorname{tr}(A)+\nu \operatorname{tr}(B) \bigr)\\ = & \frac{1}{2}\operatorname{tr}(A+B)=\operatorname{tr}\bigl(H_{1}(A , B)\bigr). \end{aligned}$$
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)
\(\operatorname{tr}(\sqrt{|AB|}) = \operatorname{tr}(H_{1}(A , B))\).
(2)
\(\operatorname{tr}(H_{1}(|A^{\nu}B^{1-\nu}| ,|A^{1-\nu}B^{\nu}|)) = \operatorname{tr}(H_{1}(A , B))\).
(3)
\(\operatorname{tr}(|H_{\nu}(A , B)|) = \operatorname{tr}(H_{1}(A , B))\).
(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 If the equation in part (2) holds, then from what was proved in the last theorem we conclude that Thus, By Corollary 1.4, this equality holds if and only if and therefore \(A^{1-\nu}=B^{\nu}\), \(B^{1-\nu}=A^{\nu}\), which implies \(A=B\). It is clear that \((4)\Longrightarrow(1)\).
$$\operatorname{tr}\bigl(H_{1} \bigl(\bigl|A^{\nu}B^{1-\nu}\bigr| ,\bigl|A^{1-\nu}B^{\nu}\bigr| \bigr) \bigr) = \operatorname{tr}\bigl(H_{1}(A , B) \bigr) . $$
$$\begin{aligned} \operatorname{tr}\bigl(H_{1}(A , B)\bigr) = & \operatorname{tr}\bigl(H_{1}\bigl(\bigl|A^{\nu}B^{1-\nu}\bigr| , \bigl|A^{1-\nu}B^{\nu}\bigr|\bigr)\bigr)\\ = & \frac{1}{2}\operatorname{tr}\bigl(\bigl|A^{\nu}B^{1-\nu}\bigr| + \bigl|A^{1-\nu}B^{\nu}\bigr|\bigr)\\ \le& \frac{1}{2}\bigl( \operatorname{tr}\bigl(\nu A+(1-\nu)B\bigr)+\operatorname{tr}\bigl(\nu B+(1-\nu)A\bigr) \bigr)=\operatorname{tr}\bigl(H_{1}(A , B)\bigr) . \end{aligned}$$
$$ \operatorname{tr}\bigl(\bigl|A^{\nu}B^{1-\nu}\bigr| \bigr) + \operatorname{tr}\bigl(\bigl|A^{1-\nu}B^{\nu}\bigr| \bigr) = \operatorname{tr}\bigl(\nu A+(1-\nu)B \bigr)+ \operatorname{tr}\bigl(\nu B+(1-\nu)A \bigr) . $$
(14)
$$\operatorname{tr}\bigl(\bigl|A^{\nu}B^{1-\nu}\bigr| \bigr) = \operatorname{tr}\bigl(\nu A+(1-\nu )B\bigr)\quad\mbox{and}\quad \operatorname{tr}\bigl(\bigl|A^{1-\nu}B^{\nu}\bigr| \bigr) = \operatorname{tr}\bigl(\nu B+(1-\nu)A \bigr), $$
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 for some unitaries U and \(V\in M_{n}(\mathbb {C})\).
$$\begin{aligned} \operatorname{tr}\bigl(H_{1}(A , B)\bigr) = & \operatorname{tr}\bigl(\bigl|H_{\nu}(A , B)\bigr|\bigr)\\ = & \frac{1}{2}\operatorname{tr}\bigl(\bigl|A^{\nu}B^{1-\nu} + A^{1-\nu}B^{\nu}\bigr|\bigr)\\ \le& \frac{1}{2}\bigl[ \operatorname{tr}\bigl(U\bigl|A^{\nu}B^{1-\nu}\bigr|U^{*}\bigr) + \operatorname{tr}\bigl(V^{*}\bigl|A^{1-\nu}B^{\nu}\bigr|V\bigr)\bigr]\ \ \bigl(\mbox{by the triangle inequality} (6)\bigr) \end{aligned}$$
Thus, thereby proving (2). \((2)\Longrightarrow(4)\) was shown in the first part. It is clear that \((4)\Longrightarrow(3)\). □
$$\begin{aligned} \operatorname{tr}\bigl(H_{1}(A , B)\bigr) \le& \frac{1}{2}\operatorname{tr}\bigl(\bigl|A^{\nu }B^{1-\nu}\bigr| + \bigl|A^{1-\nu}B^{\nu}\bigr|\bigr) \\ = & \operatorname{tr}\bigl(H_{1}\bigl(\bigl|A^{\nu}B^{1-\nu}\bigr| , \bigl|A^{1-\nu}B^{\nu}\bigr|\bigr)\bigr) \\ \le& \operatorname{tr}\bigl(H_{1}(A , B)\bigr) \quad(\mbox{by Theorem }2.5), \end{aligned}$$
The following two corollaries are almost immediate from Theorem 2.6.
Corollary 2.7
For positive semi-definite matrices
A
and
B
in
\(M_{n}(\mathbb {C})\)
and for all
\(j=1, 2,\ldots, n\)
if and only if
\(A = B\).
$$\sqrt{\sigma_{j}(AB)} = \lambda_{j} \bigl(H_{1}(A , B) \bigr) , $$
Corollary 2.8
For positive semi-definite matrices
A
and
B
in
\(M_{n}(\mathbb {C})\)
and for all
\(j=1, 2,\ldots, n\)
for
\(\nu\in[0 , 1]\)
if and only if
\(A = B\).
$$\sigma_{j} \bigl(H_{\nu}(A , B) \bigr) = \lambda_{j} \bigl(H_{1}(A , B) \bigr) , $$
We do not know whether for every \(\nu\in[0 , 1]\).
$$\sqrt{\sigma_{j}(AB)} \le\sigma_{j} \bigl(H_{\nu}(A , B) \bigr) \le\lambda_{j} \bigl(H_{1}(A , B) \bigr) $$
To answer this question, just we need to know whether for every \(\nu\in[0 , 1]\).
$$\sqrt{\sigma_{j}(AB)} \le\sigma_{j} \bigl(H_{\nu}(A , B) \bigr) $$
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 Theorem 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, Thus, which implies \(S^{\frac{1}{2}}(AB) \prec_{w} \frac{1}{2}(S(A) + S(B)) \). □
$$\sum_{j=1}^{k} \sigma_{j}(AB)^{\frac{1}{2}} \le\sum_{j=1}^{k}\frac{1}{2} \lambda_{j}(A) + \sum_{j=1}^{k} \frac{1}{2} \lambda_{j}(B) \quad\mbox{for every } 1\le k\le n . $$
$$\sum_{j=1}^{k} \sigma_{j}(AB)^{\frac{1}{2}} \le\sum_{j=1}^{k}\frac{1}{2}( \lambda_{j}(A) + \lambda_{j}(B) \quad\mbox{for every } 1\le k\le n , $$
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 Theorem 2.2. □
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 Theorem 2.1. □
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 Lemma 2.3. □
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)
\(S^{\frac{1}{2}}(AB) \prec\frac{1}{2}(S(A) + S(B)) \).
(2)
\(\sqrt{|AB|} \prec(H_{1}(A , B))\).
(3)
\(H_{\nu}(A , B) \prec H_{1}(A , B) \).
(4)
\(\sqrt{|AB|} \prec_{w} H_{1}(|A^{\nu}B^{1-\nu}| ,|A^{1-\nu }B^{\nu }|) \).
(5)
\(A = B\).
Acknowledgements
This work was supported by the Department of Mathematical Sciences at Isfahan University of Technology, Iran.
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.
Competing interests
The author declares that he has no competing interests.