# Further Properties of PPT and Hyponormal Matrices

Hamid Reza Moradi, Ibrahim Halil Gümüş, and Mohammad Sababheh

###### Abstract.

This paper discusses further properties of positive partial transpose
matrices, with applications towards hyponormal, semi-hyponormal, and
$(\alpha,\beta)$-normal matrices. The obtained results present extensions and
improvements of many results in the literature.

###### Key words and phrases:

positive partial transpose, arithmetic-geometric mean inequality, hyponormal
matrices

###### 2010 Mathematics Subject Classification:

Primary 15A45, 47A30; Secondary 15A42, 47A63, 47B65

##  1\. Introduction and Preliminaries

In the sequel, $\mathcal{M}_{n}$ denotes the algebra of all $n\times n$
complex matrices. When $A\in\mathcal{M}_{n}$ is such that
$\left<Ax,x\right>\geq 0$ for all $x\in\mathbb{C}^{n}$, we say that $A$ is
positive semidefinite, and we write $A\geq O$. On the other hand, if
$\left<Ax,x\right>>0$ for all nonzero $x\in\mathbb{C}^{n}$, $A$ is said to be
positive definite, and we write $A>O$. Here, $O$ denotes the zero element of
$\mathcal{M}_{n}$.

Given $A,B,X\in\mathcal{M}_{n}$, the matrix $\left[\begin{matrix}A&X^{*}\\\
X&B\end{matrix}\right]$ is in $\mathcal{M}_{2n}.$ It is well known that if
$\left[\begin{matrix}A&{{X}^{*}}\\\ X&B\\\ \end{matrix}\right]\geq O$, then

(1.1) $\left\|\left[\begin{matrix}A&{{X}^{*}}\\\ X&B\\\
\end{matrix}\right]\right\|\leq\left\|A\right\|+\left\|B\right\|,$

where $\|\cdot\|$ is the usual operator norm. Indeed, (1.1) follows from the
following useful decomposition [5, Lemma 3.4]: For every matrix
$\left[\begin{matrix}A&{{X}^{*}}\\\ X&B\\\ \end{matrix}\right]\geq O$, we have

$\left[\begin{matrix}A&{{X}^{*}}\\\ X&B\\\
\end{matrix}\right]=U\left[\begin{matrix}A&O\\\ O&O\\\
\end{matrix}\right]{{U}^{*}}+V\left[\begin{matrix}O&O\\\ O&B\\\
\end{matrix}\right]{{V}^{*}}$

for some unitaries $U,V$.

However, if $\left[\begin{matrix}A&{{X}^{*}}\\\ X&B\\\ \end{matrix}\right]\geq
O$ and the off-diagonal block $X$ is Hermitian, Hiroshima [12] showed a
stronger inequality than (1.1), as follows

(1.2) $\left\|\left[\begin{matrix}A&X\\\ X&B\\\
\end{matrix}\right]\right\|\leq\left\|A+B\right\|.$

The block matrix $\left[\begin{matrix}A&{{X}^{*}}\\\ X&B\\\
\end{matrix}\right]$ has received the attention of numerous researchers in the
literature due to its usability and applications. For example,
$\left[\begin{matrix}A&{{X}^{*}}\\\ X&B\\\ \end{matrix}\right]$ is said to be
positive partial transpose (PPT) if both $\left[\begin{matrix}A&X\\\
{{X}^{*}}&B\\\ \end{matrix}\right]$ and $\left[\begin{matrix}A&{{X}^{*}}\\\
X&B\\\ \end{matrix}\right]$ are positive semidefinite.

If $\left[\begin{matrix}A&{{X}^{*}}\\\ X&B\\\ \end{matrix}\right]$ is PPT, Lee
[15, Theorem 2.1] showed that

(1.3) $\left|X\right|\leq\frac{A\sharp B+{{U}^{*}}(A\sharp B)U}{2},$

where $X=U\left|X\right|$ is the polar decomposition of $X$, and $\sharp$ is
the matrix geometric mean. Recall, here, that if $A,B>O$ and $0<t<1$, the
weighted geometric mean of $A$ and $B$ is defined by

$\displaystyle
A\sharp_{t}B=A^{\frac{1}{2}}\left(A^{-\frac{1}{2}}BA^{-\frac{1}{2}}\right)^{t}A^{\frac{1}{2}}.$

When $t=\frac{1}{2},$ we simply write $A\sharp B$ instead of
$A\sharp_{\frac{1}{2}}B$. The geometric mean $\sharp$ was first introduced by
Pusz and Woronowicz [16], which was further developed into a general theory of
operator means by Kubo and Ando [14].

Fu et al. [6, Theorem 2.3] improved (1.3) as follows

(1.4) $\left|X\right|\leq\left(A\sharp
B\right)\sharp\left({{U}^{*}}\left(A\sharp B\right)U\right).$

The fact that (1.4) improves (1.3) follows from the arithmetic-geometric mean
inequality that states

(1.5) $\displaystyle A\sharp B\leq\frac{A+B}{2},$

for any $A,B\in\mathcal{M}_{n}$ with $A,B>O.$

Among those useful characterizations of the geometric mean, we have [4,
(4.15)])

(1.6) $X\sharp Y=\max\left\\{Z:\text{ }Z={{Z}^{*}},\left[\begin{matrix}X&Z\\\
Z&Y\\\ \end{matrix}\right]\geq O\right\\};X,Y>O.$

Recently, the following lemma has been shown in [10, Theorem 2.1].

######  Lemma 1.1.

If $\left[\begin{matrix}A&X\\\ {{X}^{*}}&B\\\ \end{matrix}\right]$ is PPT,
then so is $\left[\begin{matrix}A{{\sharp}_{t}}B&X\\\
{{X}^{*}}&A{{\sharp}_{1-t}}B\\\ \end{matrix}\right]$, $0\leq t\leq 1$.

Further, it has been shown in [15, Corollary 2.2] that

(1.7)
${{\lambda}_{j}}\left(2\left|X\right|-A{{\sharp}_{t}}B\right)\leq{{\lambda}_{j}}\left(A{{\sharp}_{1-t}}B\right),$

when $\left[\begin{matrix}A&{{X}^{*}}\\\ X&B\\\ \end{matrix}\right]$ is PPT.
Here $\lambda_{j}$ denotes the $j^{\text{th}}$ largest eigenvalue.

This paper discusses extensions of (1.4) and (1.7), where we extend both
inequalities to the weighted geometric mean. We will also extend [6, Theorem
2.2], where we show that when $\left[\begin{matrix}A&X\\\ {{X}^{*}}&B\\\
\end{matrix}\right]\geq O$, then
$\left|X\right|\leq\left(A{{\sharp}_{t}}{{U}^{*}}BU\right)\sharp\left(A{{\sharp}_{1-t}}{{U}^{*}}BU\right)$,
for example. Further, we give an improvement of (1.1) that is different from
(1.2). Many other consequences for PPT matrices and positive semidefinite
matrices of the form $\left[\begin{matrix}A&X\\\ {{X}^{*}}&B\\\
\end{matrix}\right]$ will be presented.

After that, we present some applications that include semi-hyponormal and
$(\alpha,\beta)$-normal matrices. Here we recall that if $T\in\mathcal{M}_{n}$
is such that $|T^{*}|^{2}\leq|T|^{2}$, then $T$ is said to be hyponormal. If
$T$ satisfies the weaker condition $|T^{*}|\leq|T|$, then $T$ is said to be
semi-hyponormal. More generally, if $0\leq\alpha\leq 1\leq\beta$ are such that
$\alpha^{2}|T|^{2}\leq|T^{*}|^{2}\leq\beta^{2}|T|^{2}$, then $T$ is said to be
$(\alpha,\beta)$-normal. For example, we will show that

$\left[\begin{matrix}\left|T\right|&{{T}^{*}}\\\ T&\left|T\right|\\\
\end{matrix}\right]\geq O$

if and only if $T$ is semi-hyponormal. Many other results and consequences
will be shown for these classes. As a consequence, we will be able to present
a possible reverse for the inequality $\|T^{2}\|\leq\|T\|^{2}$, when $T$ is
$(\alpha,\beta)$-normal.

##  2\. Main Results for positive and PPT block matrices

Our first result is an extension of [6, Theorem 2.2], which states that if
$\left[\begin{matrix}A&X\\\ {{X}^{*}}&B\\\ \end{matrix}\right]\geq O$, then

$\left|X\right|\leq A\sharp{{U}^{*}}BU\text{ and }\left|{{X}^{*}}\right|\leq
UA{{U}^{*}}\sharp B.$

Once this has been shown, we use it to extend some results about PPT matrices.

######  Theorem 2.1.

Let $\left[\begin{matrix}A&X\\\ {{X}^{*}}&B\\\ \end{matrix}\right]\geq O$ with
$A,B,X\in{{\mathcal{M}}_{n}}$ and let $X=U\left|X\right|$ be the polar
decomposition of $X$. Then for any $0\leq t\leq 1$,

(2.1)
$\left|X\right|\leq\left(A{{\sharp}_{t}}{{U}^{*}}BU\right)\sharp\left(A{{\sharp}_{1-t}}{{U}^{*}}BU\right)$

and

(2.2)
$\left|{{X}^{*}}\right|\leq\left(UA{{U}^{*}}{{\sharp}_{t}}B\right)\sharp\left(UA{{U}^{*}}{{\sharp}_{1-t}}B\right).$

In particular,

$\left|X\right|\leq A\sharp{{U}^{*}}BU\text{ and }\left|{{X}^{*}}\right|\leq
UA{{U}^{*}}\sharp B.$

###### Proof.

We prove (2.1). Since $X=U\left|X\right|$ is the polar decomposition of $X$,
we have

$\left[\begin{matrix}I&O\\\ O&{{U}^{*}}\\\
\end{matrix}\right]\left[\begin{matrix}A&{{X}^{*}}\\\ X&B\\\
\end{matrix}\right]\left[\begin{matrix}I&O\\\ O&U\\\
\end{matrix}\right]=\left[\begin{matrix}A&{{X}^{*}}U\\\
{{U}^{*}}X&{{U}^{*}}BU\\\ \end{matrix}\right]\geq O,$

which implies

$\left[\begin{matrix}A&\left|X\right|\\\ \left|X\right|&{{U}^{*}}BU\\\
\end{matrix}\right]\geq O,$

since ${{X}^{*}}U={{U}^{*}}X=\left|X\right|$. Lemma 1.1 gives

$\left[\begin{matrix}A{{\sharp}_{t}}{{U}^{*}}BU&\left|X\right|\\\
\left|X\right|&A{{\sharp}_{1-t}}{{U}^{*}}BU\\\ \end{matrix}\right]\geq O.$

Now, the result follows from (1.6). To prove (2.2), we have

$\displaystyle\left|{{X}^{*}}\right|$
$\displaystyle=U\left|X\right|{{U}^{*}}\quad\text{(by
\cite[cite]{[\@@bibref{}{9}{}{}, p. 58]})}$ $\displaystyle\leq
U\left(\left(A{{\sharp}_{t}}{{U}^{*}}BU\right)\sharp\left(A{{\sharp}_{1-t}}{{U}^{*}}BU\right)\right){{U}^{*}}$
$\displaystyle=U\left(A{{\sharp}_{t}}{{U}^{*}}BU\right){{U}^{*}}\sharp
U\left(A{{\sharp}_{1-t}}{{U}^{*}}BU\right){{U}^{*}}$
$\displaystyle=\left(UA{{U}^{*}}{{\sharp}_{t}}BU\right)\sharp\left(UA{{U}^{*}}{{\sharp}_{1-t}}B\right).$

This proves (2.2). Letting $t=\frac{1}{2}$ in (2.1) and (2.2), and noting that
$T\sharp T=T,$ when $T>O$, imply $\left|X\right|\leq A\sharp{{U}^{*}}BU\text{
and }\left|{{X}^{*}}\right|\leq UA{{U}^{*}}\sharp B.$ This completes the
proof. ∎

Using Theorem 2.1, we present the following extension of (1.4).

######  Corollary 2.1.

Let $A,B,X\in{{\mathcal{M}}_{n}}$ be such that
$\left[\begin{matrix}A&{{X}^{*}}\\\ X&B\\\ \end{matrix}\right]$ is PPT, and
let $X=U\left|X\right|$ be the polar decomposition of $X$. Then for any $0\leq
t\leq 1$,

$\left|X\right|\leq\left(A{{\sharp}_{t}}B\right)\sharp\left({{U}^{*}}\left(A{{\sharp}_{1-t}}B\right)U\right),$

and

$\left|{{X}^{*}}\right|\leq\left(U\left(A{{\sharp}_{t}}B\right){{U}^{*}}\right)\sharp\left(A{{\sharp}_{1-t}}B\right).$

###### Proof.

Lemma 1.1 ensures that $\left[\begin{matrix}A{{\sharp}_{t}}B&{{X}^{*}}\\\
X&A{{\sharp}_{1-t}}B\\\ \end{matrix}\right]$ is also PPT, for any $0\leq t\leq
1$. By Theorem 2.1, we conclude the first desired result.

The second inequality can be shown using the method we used to show (2.2).
This completes the proof. ∎

Interestingly, Corollary 2.1 can be used to present a weighted version of
(1.7), as follows.

######  Corollary 2.2.

Let $A,B,X\in{{\mathcal{M}}_{n}}$ be such that
$\left[\begin{matrix}A&{{X}^{*}}\\\ X&B\\\ \end{matrix}\right]$ is PPT. Then
for any $0\leq t\leq 1$,

${{\lambda}_{j}}\left(2\left|X\right|-A{{\sharp}_{t}}B\right)\leq{{\lambda}_{j}}\left(A{{\sharp}_{1-t}}B\right)$

and

${{\lambda}_{j}}\left(2\left|{{X}^{*}}\right|-A{{\sharp}_{1-t}}B\right)\leq{{\lambda}_{j}}\left(A{{\sharp}_{t}}B\right)$

for all $j=1,2,\ldots,n$.

###### Proof.

Corollary 2.1 means that

$\displaystyle\left|X\right|$
$\displaystyle\leq\left(A{{\sharp}_{t}}B\right)\sharp\left({{U}^{*}}(A{{\sharp}_{1-t}}B)U\right)$
$\displaystyle\leq\frac{A{{\sharp}_{t}}B+{{U}^{*}}(A{{\sharp}_{1-t}}B)U}{2}.$

Thus,

$2\left|X\right|-A{{\sharp}_{t}}B\leq{{U}^{*}}(A{{\sharp}_{1-t}}B)U.$

Therefore,

${{\lambda}_{j}}\left(2\left|X\right|-A{{\sharp}_{t}}B\right)\leq{{\lambda}_{j}}\left(A{{\sharp}_{1-t}}B\right)$

as desired. ∎

Related to the discussion, we employ (1.2) to obtain another refinement of
(1.1), as follows.

######  Theorem 2.2.

Let $A,B,X\in{{\mathcal{M}}_{n}}$ be such that
$\left[\begin{matrix}A&{{X}^{*}}\\\ X&B\\\ \end{matrix}\right]\geq O$ and let
$X=U\left|X\right|$ be the polar decomposition of $X$. Then

$\left\|\left[\begin{matrix}A&{{X}^{*}}\\\ X&B\\\
\end{matrix}\right]\right\|\leq\left\|A+{{U}^{*}}BU\right\|.$

###### Proof.

If $X=U\left|X\right|$ is the polar decomposition of $X$, then

$\left[\begin{matrix}I&O\\\ O&{{U}^{*}}\\\
\end{matrix}\right]\left[\begin{matrix}I&O\\\ O&U\\\
\end{matrix}\right]=\left[\begin{matrix}I&O\\\ O&U\\\
\end{matrix}\right]\left[\begin{matrix}I&O\\\ O&{{U}^{*}}\\\
\end{matrix}\right]=\left[\begin{matrix}I&O\\\ O&I\\\ \end{matrix}\right],$

since $U$ is unitary. The fact that $\left[\begin{matrix}A&{{X}^{*}}\\\ X&B\\\
\end{matrix}\right]\geq O$ implies

$\displaystyle\left\|\left[\begin{matrix}A&{{X}^{*}}\\\ X&B\\\
\end{matrix}\right]\right\|$ $\displaystyle=\left\|\left[\begin{matrix}I&O\\\
O&{{U}^{*}}\\\ \end{matrix}\right]\left[\begin{matrix}A&{{X}^{*}}\\\ X&B\\\
\end{matrix}\right]\left[\begin{matrix}I&O\\\ O&U\\\
\end{matrix}\right]\right\|$
$\displaystyle=\left\|\left[\begin{matrix}A&\left|X\right|\\\
\left|X\right|&{{U}^{*}}BU\\\ \end{matrix}\right]\right\|$
$\displaystyle\leq\left\|A+{{U}^{*}}BU\right\|,$

where we have used (1.2) to obtain the last inequality. This completes the
proof. ∎

The following is an interesting characterization related to PPT matrices.

######  Theorem 2.3.

Let $A,B,X\in\mathcal{M}_{n}$ be such that $\left[\begin{matrix}A&X\\\
X^{*}&B\end{matrix}\right]$ is PPT and let $0\leq t\leq 1$. Then there are
some isometries $\widetilde{U},\widetilde{V}\in{{\mathcal{M}}_{2n,n}}$
(depending on $t$), such that

$\left[\begin{matrix}A{{\sharp}_{t}}B&X\\\ {{X}^{*}}&A{{\sharp}_{1-t}}B\\\
\end{matrix}\right]=\widetilde{U}\left(A{{\sharp}_{t}}B\right){{\widetilde{U}}^{*}}+\widetilde{V}\left(A{{\sharp}_{1-t}}B\right){{\widetilde{V}}^{*}}.$

###### Proof.

By Lemma 1.1, $\left[\begin{matrix}A{{\sharp}_{t}}B&X\\\
X&A{{\sharp}_{1-t}}B\\\ \end{matrix}\right]$ is PPT. From [5, Lemma 3.4],
there are two unitaries $U,V\in{{\mathcal{M}}_{2n}}$ partitioned into equally
sized matrices,

$U=\left[\begin{matrix}{{U}_{11}}&{{U}_{12}}\\\ {{U}_{21}}&{{U}_{22}}\\\
\end{matrix}\right]\text{ and }V=\left[\begin{matrix}{{V}_{11}}&{{V}_{12}}\\\
{{V}_{21}}&{{V}_{22}}\\\ \end{matrix}\right]$

such that

$\left[\begin{matrix}A{{\sharp}_{t}}B&X\\\ {{X}^{*}}&A{{\sharp}_{1-t}}B\\\
\end{matrix}\right]=U\left[\begin{matrix}A{{\sharp}_{t}}B&O\\\ O&O\\\
\end{matrix}\right]{{U}^{*}}+V\left[\begin{matrix}O&O\\\
O&A{{\sharp}_{1-t}}B\\\ \end{matrix}\right]{{V}^{*}}.$

Hence,

$\left[\begin{matrix}A{{\sharp}_{t}}B&X\\\ {{X}^{*}}&A{{\sharp}_{1-t}}B\\\
\end{matrix}\right]=\widetilde{U}\left(A{{\sharp}_{t}}B\right){{\widetilde{U}}^{*}}+\widetilde{V}\left(A{{\sharp}_{1-t}}B\right){{\widetilde{V}}^{*}}$

where

$\widetilde{U}=\left[\begin{matrix}{{U}_{11}}\\\ {{U}_{21}}\\\
\end{matrix}\right]\text{ and }\widetilde{V}=\left[\begin{matrix}{{V}_{12}}\\\
{{V}_{22}}\\\ \end{matrix}\right]$

are isometries. This completes the proof. ∎

Theorem 2.3 implies the following remarkable result.

######  Corollary 2.3.

Let $A,B,X\in{{\mathcal{M}}_{n}}$ be such that
$\left[\begin{matrix}A&{{X}^{*}}\\\ X&B\\\ \end{matrix}\right]$ is PPT. Then
for any $0\leq t\leq 1$,

$\left\|\left[\begin{matrix}A{{\sharp}_{t}}B&X\\\
{{X}^{*}}&A{{\sharp}_{1-t}}B\\\
\end{matrix}\right]\right\|\leq\left\|A{{\sharp}_{t}}B\right\|+\left\|A{{\sharp}_{1-t}}B\right\|.$

In particular

$\left\|\left[\begin{matrix}A\sharp B&X\\\ {{X}^{*}}&A\sharp B\\\
\end{matrix}\right]\right\|\leq 2\left\|A\sharp B\right\|.$

######  Remark 2.1.

Ando [1, Theorem 3.3] proved that if $\left[\begin{matrix}A&{{X}^{*}}\\\
X&B\\\ \end{matrix}\right]$ is PPT, then

$\left\|X\right\|\leq\left\|A\sharp B\right\|.$

We know that [18] if $\left[\begin{matrix}A&{{X}^{*}}\\\ X&B\\\
\end{matrix}\right]\geq O$, then

$2\left\|X\right\|\leq\left\|\left[\begin{matrix}A&{{X}^{*}}\\\ X&B\\\
\end{matrix}\right]\right\|.$

Consequently, if $\left[\begin{matrix}A&{{X}^{*}}\\\ X&B\\\
\end{matrix}\right]$ is PPT, then

$\left\|X\right\|\leq\frac{1}{2}\left\|\left[\begin{matrix}A\sharp B&X\\\
{{X}^{*}}&A\sharp B\\\ \end{matrix}\right]\right\|\leq\left\|A\sharp
B\right\|.$

######  Remark 2.2.

It is well-known that

${{\left|\left\langle
T{{\left|T\right|}^{\alpha+\beta-1}}x,y\right\rangle\right|}^{2}}\leq\left\langle{{\left|T\right|}^{2\alpha}}x,x\right\rangle\left\langle{{\left|{{T}^{*}}\right|}^{2\beta}}y,y\right\rangle;\left(\alpha,\beta\in\left[0,1\right],\alpha+\beta\geq
1\right)$

for any $T$ [8]. By [13, Lemma 1], we infer that

$\left[\begin{matrix}{{\left|T\right|}^{2\alpha}}&{{\left|T\right|}^{\alpha+\beta-1}}{{T}^{*}}\\\
T{{\left|T\right|}^{\alpha+\beta-1}}&{{\left|{{T}^{*}}\right|}^{2\beta}}\\\
\end{matrix}\right]\geq O.$

So,

$\displaystyle 2\left\|T{{\left|T\right|}^{\alpha+\beta-1}}\right\|$
$\displaystyle\leq\left\|\left[\begin{matrix}{{\left|T\right|}^{2\alpha}}&{{\left|T\right|}^{\alpha+\beta-1}}{{T}^{*}}\\\
T{{\left|T\right|}^{\alpha+\beta-1}}&{{\left|{{T}^{*}}\right|}^{2\beta}}\\\
\end{matrix}\right]\right\|$
$\displaystyle\leq\left\|{{\left|T\right|}^{2\alpha}}+{{U}^{*}}{{\left|{{T}^{*}}\right|}^{2\beta}}U\right\|$
$\displaystyle=2\left\|{{\left|T\right|}^{2\alpha}}+{{\left|T\right|}^{2\beta}}\right\|$

i.e.,

$\left\|T{{\left|T\right|}^{\alpha+\beta-1}}\right\|\leq\frac{1}{2}\left\|\left[\begin{matrix}{{\left|T\right|}^{2\alpha}}&{{\left|T\right|}^{\alpha+\beta-1}}{{T}^{*}}\\\
T{{\left|T\right|}^{\alpha+\beta-1}}&{{\left|{{T}^{*}}\right|}^{2\beta}}\\\
\end{matrix}\right]\right\|\leq\left\|{{\left|T\right|}^{2\alpha}}+{{\left|T\right|}^{2\beta}}\right\|.$

In particular,

$\left\|T\right\|=\frac{1}{2}\left\|\left[\begin{matrix}\left|{{T}^{*}}\right|&T\\\
{{T}^{*}}&\left|T\right|\\\ \end{matrix}\right]\right\|.$

######  Theorem 2.4.

Let $A,B,X\in{{\mathcal{M}}_{n}}$ be such that
$\left[\begin{matrix}A&{{X}^{*}}\\\ X&B\\\ \end{matrix}\right]\geq O$. Then

$\left\|\left[\begin{matrix}A&{{X}^{*}}\\\ X&B\\\
\end{matrix}\right]\right\|\leq\left\|A+B+\left|X\right|+\left|{{X}^{*}}\right|\right\|.$

###### Proof.

Indeed,

$\displaystyle\left\|\left[\begin{matrix}A&{{X}^{*}}\\\ X&B\\\
\end{matrix}\right]\right\|$ $\displaystyle=\left\|\left[\begin{matrix}A&O\\\
O&B\\\ \end{matrix}\right]+\left[\begin{matrix}O&{{X}^{*}}\\\ X&O\\\
\end{matrix}\right]\right\|$
$\displaystyle\leq\left\|\left[\begin{matrix}A&O\\\ O&B\\\
\end{matrix}\right]+\left[\begin{matrix}\left|X\right|&O\\\
O&\left|{{X}^{*}}\right|\\\ \end{matrix}\right]\right\|$
$\displaystyle=\left\|\left[\begin{matrix}A+\left|X\right|&O\\\
O&B+\left|{{X}^{*}}\right|\\\ \end{matrix}\right]\right\|$
$\displaystyle\leq\left\|A+B+\left|X\right|+\left|{{X}^{*}}\right|\right\|$

where the first inequality follows from the fact that
$\left[\begin{matrix}O&{{X}^{*}}\\\ X&O\\\ \end{matrix}\right]$ is Hermitian,
and for any Hermitiam matrix $T$ , we have $T\leq\left|T\right|$. The second
inequality is also obtained from (1.2). This completes the proof. ∎

##  3\. Applications towards hyponormal and $(\alpha,\beta)$-normal matrices

This section presents several results on hyponormal, semi-hyponormal, and
$(\alpha,\beta)$-normal matrices. While some of these results can be
considered as applications of the results of the previous section, other
results are related but independent.

######  Theorem 3.1.

Let $T\in\mathcal{M}_{n}$. Then
$\left[\begin{matrix}\left|T\right|&{{T}^{*}}\\\ T&\left|T\right|\\\
\end{matrix}\right]\geq O$ if and only if $T$ is semi-hyponormal.

###### Proof.

It is easy to show that [17] if $T$ is semi-hyponormal, then

$\left[\begin{matrix}\left|T\right|&{{T}^{*}}\\\ T&\left|T\right|\\\
\end{matrix}\right]\geq O.$

We show that if $\left[\begin{matrix}\left|T\right|&{{T}^{*}}\\\
T&\left|T\right|\\\ \end{matrix}\right]\geq O$, then $T$ is semi-hyponormal.
Indeed, if $T=U\left|T\right|$ is the polar decomposition of $T$, then

$\left[\begin{matrix}\left|T\right|&{{T}^{*}}U\\\
{{U}^{*}}T&{{U}^{*}}\left|T\right|U\\\
\end{matrix}\right]=\left[\begin{matrix}I&O\\\ O&{{U}^{*}}\\\
\end{matrix}\right]\left[\begin{matrix}\left|T\right|&{{T}^{*}}\\\
T&\left|T\right|\\\ \end{matrix}\right]\left[\begin{matrix}I&O\\\ O&U\\\
\end{matrix}\right]\geq O.$

Since ${{T}^{*}}U={{U}^{*}}T=\left|T\right|$, we get

$\left[\begin{matrix}\left|T\right|&\left|T\right|\\\
\left|T\right|&{{U}^{*}}\left|T\right|U\\\ \end{matrix}\right]\geq O.$

From (1.6), we get

$\left|T\right|\leq\left|T\right|\sharp{{U}^{*}}\left|T\right|U.$

By the definition of geometric mean, we have

$\left|T\right|\leq{{\left|T\right|}^{\frac{1}{2}}}{{\left({{\left|T\right|}^{-\frac{1}{2}}}{{U}^{*}}\left|T\right|U{{\left|T\right|}^{-\frac{1}{2}}}\right)}^{\frac{1}{2}}}{{\left|T\right|}^{\frac{1}{2}}}.$

Multiplying both sides by ${{\left|T\right|}^{-\frac{1}{2}}}$, we infer that

$I\leq{{\left({{\left|T\right|}^{-\frac{1}{2}}}{{U}^{*}}\left|T\right|U{{\left|T\right|}^{-\frac{1}{2}}}\right)}^{\frac{1}{2}}}.$

This implies

$\left|T\right|\leq{{U}^{*}}\left|T\right|U.$

Thus,

$\left|{{T}^{*}}\right|=U\left|T\right|{{U}^{*}}\leq\left|T\right|,$

which shows that $T$ is semi-hyponormal. This completes the proof. ∎

In [5, (2.11)], it has been shown that if $A,B$ are normal, then

(3.1) $|A+B|\leq\frac{|A|+|B|+U^{*}(|A|+|B|)U}{2},$

where $U$ is the unitary matrix in the polar decomposition of $A+B.$ Since
every normal matrix is necessarily semi-hyponormal, and because of (1.5), the
following result significantly improves [5, (2.11)].

######  Corollary 3.1.

Let $A,B\in\mathcal{M}_{n}$ be semi-hyponormal and let $U$ be the unitary part
in the polar decomposition $A+B=U\left|A+B\right|$. Then

$\left|A+B\right|\leq\left(\left|A\right|+\left|B\right|\right)\sharp\left({{U}^{*}}\left(\left|A\right|+\left|B\right|\right)U\right).$

###### Proof.

Using Theorem 3.1, we see that

(3.2)
$\left[\begin{matrix}\left|A\right|+\left|B\right|&{{A}^{*}}+{{B}^{*}}\\\
A+B&\left|A\right|+\left|B\right|\\\ \end{matrix}\right]\geq O.$

By (3.2) and Theorem 2.1, we get the desired result. ∎

######  Corollary 3.2.

Let $A,B\in\mathcal{M}_{n}$ be semi-hyponormal. Then

$\left\|A+B\right\|\leq\left\|\left|A\right|+\left|B\right|\right\|.$

######  Remark 3.1.

We highlight that Corollary 3.2 is well-known for normal operators $A,B$ [3,
(1.42)].

######  Theorem 3.2.

Let $T\in\mathcal{M}_{n}$. Then the following assertions are equivalent.

  * (i)

$T$ is $(\alpha,\beta)$-normal.

  * (ii)

$\left[\begin{matrix}\frac{1}{\alpha}{{\left|{{T}^{*}}\right|}^{2}}&{{\left|T\right|}^{2}}\\\
{{\left|T\right|}^{2}}&\frac{1}{\alpha}{{\left|T\right|}^{2}}\\\
\end{matrix}\right]\geq O$ and
$\left[\begin{matrix}\beta{{\left|T\right|}^{2}}&{{\left|{{T}^{*}}\right|}^{2}}\\\
{{\left|{{T}^{*}}\right|}^{2}}&\beta{{\left|{{T}^{*}}\right|}^{2}}\\\
\end{matrix}\right]\geq O$.

  * (iii)

$\left[\begin{matrix}\frac{1}{{{\alpha}^{2}}}{{\left|{{T}^{*}}\right|}^{2}}&{{\left|T\right|}^{2}}\\\
{{\left|T\right|}^{2}}&\frac{1}{{{\alpha}^{2}}}{{\left|{{T}^{*}}\right|}^{2}}\\\
\end{matrix}\right]\geq O$ and
$\left[\begin{matrix}{{\beta}^{2}}{{\left|T\right|}^{2}}&{{\left|{{T}^{*}}\right|}^{2}}\\\
{{\left|{{T}^{*}}\right|}^{2}}&{{\beta}^{2}}{{\left|T\right|}^{2}}\\\
\end{matrix}\right]\geq O$.

###### Proof.

$\left(i\right)\Leftrightarrow\left(ii\right)$ Since $T$ is
$(\alpha,\beta)$-normal, we have

$\displaystyle{{\left|T\right|}^{2}}\leq\frac{1}{{{\alpha}^{2}}}{{\left|{{T}^{*}}\right|}^{2}}$
$\displaystyle\text{
}\Leftrightarrow{{\left|T\right|}^{2}}{{\left(\frac{1}{\alpha}{{\left|T\right|}^{2}}\right)}^{-1}}{{\left|T\right|}^{2}}\leq\frac{1}{\alpha}{{\left|{{T}^{*}}\right|}^{2}}$
$\displaystyle\Leftrightarrow\left[\begin{matrix}\frac{1}{\alpha}{{\left|{{T}^{*}}\right|}^{2}}&{{\left|T\right|}^{2}}\\\
{{\left|T\right|}^{2}}&\frac{1}{\alpha}{{\left|T\right|}^{2}}\\\
\end{matrix}\right]\geq O\quad\text{(by \cite[cite]{[\@@bibref{}{11}{}{},
Theorem 1.3.3]})}.$

Again, since $T$ is $(\alpha,\beta)$-normal, we have

$\displaystyle{{\left|{{T}^{*}}\right|}^{2}}\leq{{\beta}^{2}}{{\left|T\right|}^{2}}$
$\displaystyle\text{
}\Leftrightarrow{{\left|{{T}^{*}}\right|}^{2}}{{\left(\beta{{\left|{{T}^{*}}\right|}^{2}}\right)}^{-1}}{{\left|{{T}^{*}}\right|}^{2}}\leq\beta{{\left|T\right|}^{2}}$
$\displaystyle\Leftrightarrow\left[\begin{matrix}\beta{{\left|T\right|}^{2}}&{{\left|{{T}^{*}}\right|}^{2}}\\\
{{\left|{{T}^{*}}\right|}^{2}}&\beta{{\left|{{T}^{*}}\right|}^{2}}\\\
\end{matrix}\right]\geq O\quad\text{(by \cite[cite]{[\@@bibref{}{11}{}{},
Theorem 1.3.3]})}.$

$\left(i\right)\Leftrightarrow\left(iii\right)$ See [17, Theorem 2.2]. ∎

We can establish the following theorem by employing the same arguments as in
the proof of $\left(i\right)\Leftrightarrow\left(ii\right)$ in Theorem 3.2,
but we present another proof.

######  Theorem 3.3.

Let $T\in\mathcal{M}_{n}$ be $\left(\alpha,\beta\right)$-normal. Then

$\left[\begin{matrix}\frac{1}{\sqrt{\alpha}}\left|{{T}^{*}}\right|&\left|T\right|\\\
\left|T\right|&\frac{1}{\sqrt{\alpha}}\left|T\right|\\\
\end{matrix}\right]\geq O\text{ and
}\left[\begin{matrix}\sqrt{\beta}\left|T\right|&\left|{{T}^{*}}\right|\\\
\left|{{T}^{*}}\right|&\sqrt{\beta}\left|{{T}^{*}}\right|\\\
\end{matrix}\right]\geq O.$

###### Proof.

According to the assumption,

$\left\langle\left|T\right|x,x\right\rangle\left\langle\left|{{T}^{*}}\right|y,y\right\rangle\leq\frac{1}{\alpha}\left\langle\left|{{T}^{*}}\right|x,x\right\rangle\left\langle\left|{{T}^{*}}\right|y,y\right\rangle$

and

$\left\langle\left|T\right|x,x\right\rangle\left\langle\left|{{T}^{*}}\right|y,y\right\rangle\leq\beta\left\langle\left|T\right|x,x\right\rangle\left\langle\left|T\right|y,y\right\rangle$

for any $x,y\in\mathcal{H}$. On the other hand, we know that (see, e.g., [9,
p. 216])

${{\left|\left\langle
Tx,y\right\rangle\right|}^{2}}\leq\left\langle\left|T\right|x,x\right\rangle\left\langle\left|{{T}^{*}}\right|y,y\right\rangle.$

Consequently,

${{\left|\left\langle
Tx,y\right\rangle\right|}^{2}}\leq\frac{1}{\alpha}\left\langle\left|{{T}^{*}}\right|x,x\right\rangle\left\langle\left|{{T}^{*}}\right|y,y\right\rangle$

and

${{\left|\left\langle
Tx,y\right\rangle\right|}^{2}}\leq\beta\left\langle\left|T\right|x,x\right\rangle\left\langle\left|T\right|y,y\right\rangle.$

The last two inequalities are equivalent to

(3.3)
$\left[\begin{matrix}\frac{1}{\sqrt{\alpha}}\left|{{T}^{*}}\right|&{{T}^{*}}\\\
T&\frac{1}{\sqrt{\alpha}}\left|{{T}^{*}}\right|\\\ \end{matrix}\right]\geq
O\text{ and }\left[\begin{matrix}\sqrt{\beta}\left|T\right|&{{T}^{*}}\\\
T&\sqrt{\beta}\left|T\right|\\\ \end{matrix}\right]\geq O,$

thanks to [13, Lemma 1].

Now assume that $T=U\left|T\right|$ is the polar decomposition of $T$. Then

(3.4) $\left[\begin{matrix}I&O\\\ O&{{U}^{*}}\\\
\end{matrix}\right]\left[\begin{matrix}\frac{1}{\sqrt{\alpha}}\left|{{T}^{*}}\right|&{{T}^{*}}\\\
T&\frac{1}{\sqrt{\alpha}}\left|{{T}^{*}}\right|\\\
\end{matrix}\right]\left[\begin{matrix}I&O\\\ O&U\\\
\end{matrix}\right]=\left[\begin{matrix}\frac{1}{\sqrt{\alpha}}\left|{{T}^{*}}\right|&{{T}^{*}}U\\\
{{U}^{*}}T&\frac{1}{\sqrt{\alpha}}{{U}^{*}}\left|{{T}^{*}}\right|U\\\
\end{matrix}\right]\geq O.$

On the other hand,

$\left[\begin{matrix}\sqrt{\beta}\left|T\right|&{{T}^{*}}\\\
T&\sqrt{\beta}\left|T\right|\\\ \end{matrix}\right]\geq O\text{
}\Leftrightarrow\text{ }\left[\begin{matrix}\sqrt{\beta}\left|T\right|&T\\\
{{T}^{*}}&\sqrt{\beta}\left|T\right|\\\ \end{matrix}\right]\geq O.$

So,

(3.5) $\left[\begin{matrix}I&O\\\ O&U\\\
\end{matrix}\right]\left[\begin{matrix}\sqrt{\beta}\left|T\right|&T\\\
{{T}^{*}}&\sqrt{\beta}\left|T\right|\\\
\end{matrix}\right]\left[\begin{matrix}I&O\\\ O&{{U}^{*}}\\\
\end{matrix}\right]=\left[\begin{matrix}\sqrt{\beta}\left|T\right|&T{{U}^{*}}\\\
U{{T}^{*}}&\sqrt{\beta}U\left|T\right|{{U}^{*}}\\\ \end{matrix}\right]\geq O.$

One can easily check that ${{U}^{*}}T={{T}^{*}}U=\left|T\right|$. Meanwhile,
$\left|{{T}^{*}}\right|=U\left|T\right|{{U}^{*}}$ (see [9, p. 58]), so
${{U}^{*}}\left|{{T}^{*}}\right|U=\left|T\right|$. Hence, by (3.4) and (3.5),
we obtain

$\left[\begin{matrix}\frac{1}{\sqrt{\alpha}}\left|{{T}^{*}}\right|&\left|T\right|\\\
\left|T\right|&\frac{1}{\sqrt{\alpha}}\left|T\right|\\\
\end{matrix}\right]\geq O\text{ and
}\left[\begin{matrix}\sqrt{\beta}\left|T\right|&\left|{{T}^{*}}\right|\\\
\left|{{T}^{*}}\right|&\sqrt{\beta}\left|{{T}^{*}}\right|\\\
\end{matrix}\right]\geq O,$

as desired. ∎

The matrices in (3.3) are P.P.T. Therefore, by Corollary 2.2, we have the
following eigenvalue inequalities.

######  Corollary 3.3.

Let $T\in\mathcal{M}_{n}$ be $\left(\alpha,\beta\right)$-normal. Then

${{\lambda}_{j}}\left(2\sqrt{\alpha}\left|T\right|-\left|{{T}^{*}}\right|\right)\leq{{\lambda}_{j}}\left(\left|{{T}^{*}}\right|\right)$

and

${{\lambda}_{j}}\left(\frac{2}{\sqrt{\beta}}\left|{{T}^{*}}\right|-\left|T\right|\right)\leq{{\lambda}_{j}}\left(\left|T\right|\right)$

for $j=1,2,\ldots,n$.

######  Remark 3.2.

It is well-known that for any $T\in\mathcal{M}_{n}$,

$\left\|\;\left|T\right|-\left|{{T}^{*}}\right|\;\right\|\leq\left\|T\right\|.$

From Corollary 3.3, we infer that if $T$ is
$\left(\alpha,\beta\right)$-normal, then

$\left\|\;2\sqrt{\alpha}\left|T\right|-\left|{{T}^{*}}\right|\;\right\|\leq\left\|T\right\|,$

and

$\left\|\;\frac{2}{\sqrt{\beta}}\left|{{T}^{*}}\right|-\left|T\right|\;\right\|\leq\left\|T\right\|.$

The inequality (1.5) is usually referred to as the operator arithmetic-
geometric mean inequality. It is of great interest in the literature to find
possible reverses for this inequality. Usually, such reverses are found under
additional conditions, as seen in [7, 11]. In the following, we present a
reverse of (1.5) for $(\alpha,\beta)$-normal matrices.

######  Proposition 3.1.

Let $T\in\mathcal{M}_{n}$ be $\left(\alpha,\beta\right)$-normal. Then

$\frac{\left|T\right|+\left|{{T}^{*}}\right|}{2}\leq\min\left\\{\frac{1}{\sqrt{\alpha}},\sqrt{\beta}\right\\}\left(\left|T\right|\sharp\left|{{T}^{*}}\right|\right).$

###### Proof.

Theorem 3.3 implies

(3.6)
$\left|T\right|\leq\frac{1}{\sqrt{\alpha}}\left(\left|T\right|\sharp\left|{{T}^{*}}\right|\right)\text{
and
}\left|{{T}^{*}}\right|\leq\sqrt{\beta}\left(\left|T\right|\sharp\left|{{T}^{*}}\right|\right)$

due to (1.6). Further,

(3.7)
$\left|{{T}^{*}}\right|\leq\frac{1}{\sqrt{\alpha}}\left(\left|T\right|\sharp\left|{{T}^{*}}\right|\right)\text{
and
}\left|T\right|\leq\sqrt{\beta}\left(\left|T\right|\sharp\left|{{T}^{*}}\right|\right)$

by utilizing the same approach as in the proof of inequality (2.2).
Inequalities (3.6) and (3.7) say that

$\left|T\right|\leq\min\left\\{\frac{1}{\sqrt{\alpha}},\sqrt{\beta}\right\\}\left(\left|T\right|\sharp\left|{{T}^{*}}\right|\right)\text{
and
}\left|{{T}^{*}}\right|\leq\min\left\\{\frac{1}{\sqrt{\alpha}},\sqrt{\beta}\right\\}\left(\left|T\right|\sharp\left|{{T}^{*}}\right|\right).$

Adding the above two inequalities together implies the desired result. ∎

######  Remark 3.3.

Inequalities (3.6) and (3.7) can be shown in another way. Since
$f\left(t\right)=\sqrt{t}$ is operator monotone on $\left(0,\infty\right)$,
and since $\alpha^{2}|T|^{2}\leq|T^{*}|^{2},$ we infer that

$\left|T\right|\leq\frac{1}{\alpha}\left|{{T}^{*}}\right|.$

This implies

$\left|T\right|\leq\frac{1}{\sqrt{\alpha}}\left(\left|T\right|\sharp\left|{{T}^{*}}\right|\right),$

where we have used the fact that if $A,B,C,D>O$ are such that $A\leq B$ and
$C\leq D$, then $A\sharp C\leq B\sharp D.$

For the following result, we remind the reader of positive linear maps. A
linear map $\Phi:\mathcal{M})_{n}\to\mathcal{M}_{n}$ is said to be positive if
$\Phi(A)\geq O$ whenever $A\geq O.$

######  Theorem 3.4.

Let $T\in\mathcal{M}_{n}$ be $\left(\alpha,\beta\right)$-normal and let $\Phi$
be a positive linear map. If
$\Phi\left(T\right)=U\left|\Phi\left(T\right)\right|$ is the polar
decomposition of $\Phi\left(T\right)$, then

$\left|\Phi\left(T\right)\right|\leq\frac{1}{\sqrt{\alpha}}\left(\Phi\left(\left|{{T}^{*}}\right|\right)\sharp{{U}^{*}}\Phi\left(\left|{{T}^{*}}\right|\right)U\right)$

and

$\left|\Phi\left(T\right)\right|\leq\sqrt{\beta}\left(\Phi\left(\left|T\right|\right)\sharp{{U}^{*}}\Phi\left(\left|T\right|\right)U\right).$

###### Proof.

First notice that every positive linear map is adjoint-preserving; i.e.,
${{\Phi}^{*}}\left(T\right)=\Phi\left({{T}^{*}}\right)$ for all $T$ [3, Lemma
2.3.1]. It follows from (3.3) that

$\left[\begin{matrix}\frac{1}{\sqrt{\alpha}}\Phi\left(\left|{{T}^{*}}\right|\right)&{{\Phi}^{*}}\left(T\right)\\\
\Phi\left(T\right)&\frac{1}{\sqrt{\alpha}}\Phi\left(\left|{{T}^{*}}\right|\right)\\\
\end{matrix}\right]\geq O\text{ and
}\left[\begin{matrix}\sqrt{\beta}\Phi\left(\left|T\right|\right)&{{\Phi}^{*}}\left(T\right)\\\
\Phi\left(T\right)&\sqrt{\beta}\Phi\left(\left|T\right|\right)\\\
\end{matrix}\right]\geq O$

thanks to [3, Exercise 3.2.2]. We get the desired result by mimicking the
technique of the proof of Theorem 2.1. ∎

The following result presents an interesting reverse of the well-known
inequality $\|T^{2}\|\leq\|T\|^{2},$ for any $T$. We recall that a contraction
$K$ satisfies $KK^{*}\leq I$, the identity. We also recall that the spectral
radius $r(X)$ coincides with the operator norm $\|X\|$ when $X\geq O.$

######  Theorem 3.5.

Let $T\in\mathcal{M}_{n}$ be $\left(\alpha,\beta\right)$-normal. Then

${{\left\|T\right\|}^{2}}\leq\frac{1}{\alpha}\left\|{{T}^{2}}\right\|\text{
and }{{\left\|T\right\|}^{2}}\leq\beta\left\|{{T}^{2}}\right\|.$

Indeed,

${{\left\|T\right\|}^{2}}\leq\min\left\\{\frac{1}{\alpha},\beta\right\\}\left\|{{T}^{2}}\right\|.$

###### Proof.

Ando [2] proved that $\left[\begin{matrix}A&X\\\ {{X}^{*}}&B\\\
\end{matrix}\right]\geq O$ if and only if there exists a contraction $K$ such
that $X={{A}^{\frac{1}{2}}}K{{B}^{\frac{1}{2}}}$. It has been revealed in the
proof of Theorem 3.3 that
$\left[\begin{matrix}\frac{1}{\sqrt{\alpha}}\left|{{T}^{*}}\right|&\left|T\right|\\\
\left|T\right|&\frac{1}{\sqrt{\alpha}}\left|T\right|\\\
\end{matrix}\right]\geq O$. Therefore, there exists a contraction $K$ such
that
$\left|T\right|=\frac{1}{\sqrt{\alpha}}{{\left|{{T}^{*}}\right|}^{\frac{1}{2}}}K{{\left|T\right|}^{\frac{1}{2}}}\geq
O$. So, we have

$\displaystyle\left\|\;\left|T\right|\;\right\|$
$\displaystyle=\left\|T\right\|$
$\displaystyle=\frac{1}{\sqrt{\alpha}}\left\|{{\left|{{T}^{*}}\right|}^{\frac{1}{2}}}K{{\left|T\right|}^{\frac{1}{2}}}\right\|$
$\displaystyle=\frac{1}{\sqrt{\alpha}}r\left({{\left|{{T}^{*}}\right|}^{\frac{1}{2}}}K{{\left|T\right|}^{\frac{1}{2}}}\right)\quad\text{(since
$r\left(X\right)=\left\|X\right\|$ for positive $X$)}$
$\displaystyle=\frac{1}{\sqrt{\alpha}}r\left(K{{\left|T\right|}^{\frac{1}{2}}}{{\left|{{T}^{*}}\right|}^{\frac{1}{2}}}\right)\quad\text{(since
$r\left(XY\right)=r\left(YX\right)$)}$
$\displaystyle\leq\frac{1}{\sqrt{\alpha}}\left\|K{{\left|T\right|}^{\frac{1}{2}}}{{\left|{{T}^{*}}\right|}^{\frac{1}{2}}}\right\|\quad\text{(since
$r\left(X\right)\leq\left\|X\right\|$)}$
$\displaystyle\leq\frac{1}{\sqrt{\alpha}}\left\|K\right\|\left\|{{\left|T\right|}^{\frac{1}{2}}}{{\left|{{T}^{*}}\right|}^{\frac{1}{2}}}\right\|\quad\text{(by
the submultiplicative property of usual operator norm)}$
$\displaystyle\leq\frac{1}{\sqrt{\alpha}}\left\|{{\left|T\right|}^{\frac{1}{2}}}{{\left|{{T}^{*}}\right|}^{\frac{1}{2}}}\right\|\quad\text{(since
$K$ is contraction)}$
$\displaystyle\leq\frac{1}{\sqrt{\alpha}}{{\left\|\left|T\right|\left|{{T}^{*}}\right|\right\|}^{\frac{1}{2}}}\quad\text{(by
\cite[cite]{[\@@bibref{}{4}{}{}, Theorem IX.2.1]})}$
$\displaystyle=\frac{1}{\sqrt{\alpha}}\left\|{{T}^{2}}\right\|^{\frac{1}{2}}.$

The second inequality comes from
$\left[\begin{matrix}\sqrt{\beta}\left|T\right|&\left|{{T}^{*}}\right|\\\
\left|{{T}^{*}}\right|&\sqrt{\beta}\left|{{T}^{*}}\right|\\\
\end{matrix}\right]\geq O$ and the same method as above. This completes the
proof. ∎

######  Remark 3.4.

We will give another method to prove Theorem 3.5. The operator inequality

$\left|T\right|\leq\frac{1}{\sqrt{\alpha}}\left(\left|T\right|\sharp\left|{{T}^{*}}\right|\right)$

implies the following norm inequality

$\left\|T\right\|\leq\frac{1}{\sqrt{\alpha}}\left\|\left|T\right|\sharp\left|{{T}^{*}}\right|\right\|.$

But, for any positive operators $A,B$, we know that

$\left\|A\sharp
B\right\|\leq\left\|{{A}^{\frac{1}{2}}}{{B}^{\frac{1}{2}}}\right\|,$

i.e.,

$\left\|T\right\|\leq\frac{1}{\sqrt{\alpha}}\left\|{{\left|T\right|}^{\frac{1}{2}}}{{\left|{{T}^{*}}\right|}^{\frac{1}{2}}}\right\|=\frac{1}{\sqrt{\alpha}}\left\|{{T}^{2}}\right\|^{\frac{1}{2}}.$

### Declarations

  * •

Availability of data and materials: Not applicable.

  * •

Competing interests: The authors declare that they have no competing
interests.

  * •

Funding: Not applicable.

  * •

Authors’ contributions: Authors declare that they have contributed equally to
this paper. All authors have read and approved this version.

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(H. R. Moradi) Department of Mathematics, Mashhad Branch, Islamic Azad
University, Mashhad, Iran

E-mail address<EMAIL_ADDRESS>
(I. H. Gümüş) Department of Mathematics, Faculty of Arts and Sciences,
Adıyaman University, Adıyaman, Turkey

E-mail address<EMAIL_ADDRESS>
(M. Sababheh) Department of Basic Sciences, Princess Sumaya University for
Technology, Amman, Jordan

E-mail address<EMAIL_ADDRESS>