Let $\mathbf{Z,R}$ two Hermitian semidefinite positive matrices with all eigenvalues larger than one. Intuition drives me that



> 
> $\mathbf{R}^{-1/2}\mathbf{Z} \left(\mathbf{R}^{-1/2}\right)^H - \mathbf{Z} \preceq \mathbf{0}$
> 
> 
> 


Any idea of how to proceed with the inequality verification?


Thank you.



Matrix inequality
Let $A$ and $B$ be Hermitian positive definite matrices. Then, the following matrix
\begin{equation\*}
\begin{pmatrix}
A & X\\
X^\* & B
\end{pmatrix}
\end{equation\*}
is positive definite *if and only if* $X=A^{1/2}ZB^{1/2}$ for some $Z$ that satisfies $\|Z\| \le 1$ (thus, the range of $X$ is a subspace of the range of $A$, and the range of $X^\*$ is a subspace of the range of $B$).


To this matrix, apply the Schur-complement's lemma to conclude that
\begin{equation\*}
 A \ge X^\* B^{-1}X.
\end{equation\*}


Using $X=R^{-1/2}$ and $B=A^{-1}$, we must therefore have
$R^{-1/2} = A^{1/2}ZA^{-1/2}$ for some contraction $Z$. If $R$ satisfies such an equality, the original claim will be true.


