Consider a complex symmetric matrix $$ C= C\_R + i C\_I $$ with $C\_R,C\_I \in \text{Mat}\_{n\times n}(\mathbb R)$ symmetric, and assume that the eigenvalues of $C\_R$ are all strictly positive. Then, $C$ is invertible, and the real part of its inverse is $$ (C^{-1})\_R = \Big(C\_R + C\_I (C\_R)^{-1} C\_I\Big)^{-1}. $$ Let now $S\in \text{Mat}\_{n\times n}(\mathbb R)$ be symmetric matrix with nonnegative eigenvalues and diagonal entries all equal to $1$. Let $$\tilde C = C \circ S $$ be the Hadamard (i.e. entrywise) product of $C$ and $S$. It is easy to show that the eigenvalues of its real part $\tilde C\_R = C\_R\circ S $ are convex combinations of the eigenvalues of $C\_R $, and so $\tilde C $ is invertible with real part $$ (\tilde C^{-1})\_R = \Big(\tilde C\_R + \tilde C\_I (\tilde C\_R)^{-1} \tilde C\_I\Big)^{-1}. $$ Here, $\tilde C\_I=C\_I\circ S $. Based on numerical simulations, and exact computations for special cases (such as $C\_R= 1$ or $S $ a block matrix), I expect that the diagonal values $$ \Big(C\_R + C\_I (C\_R)^{-1} C\_I\Big)\_{ii} \geq\Big(\tilde C\_R + \tilde C\_I (\tilde C\_R)^{-1} \tilde C\_I\Big)\_{ii}. $$ (Note that the $()^{-1}$ is gone). Since $(C\_R)\_{ii} = (\tilde C\_R)\_{ii} $ (as $S$ has $1$'s on the diagonal), this is really a statement about the diagonal values of $C\_IC\_R^{-1}C\_I $. Numerically, it is even suggested that $$\Big( C\_I^aC\_R^{-b}C\_I^a\Big)\_{ii}\geq \Big( \tilde C\_I^a\tilde C\_R^{-b}\tilde C\_I^a\Big)\_{ii} $$ for any $a,b\in \mathbb N $. 


I know this is a rather specific question, but any hints or suggestions from people with background in such problems would be greatly appreciated! This Lemma would very much simplify part of my research, which is in mathematical physics.



Inequalities for Hadamard products of complex symmetric matrices
The first inequality is **true** while the second one is **false**. The proof below suggests a version of the second inequality that does hold.


To ease notation, let $A = C\_R$, and $B=C\_I$. We wish to show that


\begin{equation\*}
 (BA^{-1}B)\_{ii} \ge [(S\circ B)(S\circ A)^{-1}(S\circ B)]\_{ii}.
\end{equation\*}


Since $A \succ 0$ (symmetric, positive definite), and $B$ is symmetric, using Schur complements it follows that
\begin{equation\*}
 \begin{bmatrix}
 BA^{-1}B & B\\
 B & A 
 \end{bmatrix} \succeq 0.
\end{equation\*}


Since $S$ is a correlation matrix, it is semidefinite; also $
\begin{bmatrix}
 S & S\\
 S & S
\end{bmatrix} = 11^T \otimes S \succeq 0$, whereby using the Schur product theorem we see
\begin{equation\*}
 \begin{bmatrix}
 S & S\\
 S & S
\end{bmatrix} \circ \begin{bmatrix}
 BA^{-1}B & B\\
 B & A 
 \end{bmatrix} = \begin{bmatrix}
 S \circ BA^{-1}B & S \circ B\\
 S \circ B & S\circ A 
 \end{bmatrix} \succeq 0.
\end{equation\*}
Now using Schur complements again, this implies that (in Loewner order)
\begin{equation\*}
 S \circ BA^{-1}B \succeq (S\circ B)(S\circ A)^{-1}(S\circ B).
\end{equation\*}
Since $S$ is a correlation matrix, it has 1s on its diagonals, so that $[S \circ BA^{-1}B]\_{ii} = (BA^{-1}B)\_{ii}$, which yields
\begin{equation\*}
 (BA^{-1}B)\_{ii} \ge [(S\circ B)(S\circ A)^{-1}(S\circ B)]\_{ii},
\end{equation\*}
as desired. 


To prove the other inequality, we proceed similarly, by starting from (using $a, b \in \mathbb{N}$),
\begin{equation\*}
 \begin{bmatrix}
 B^aA^{-b}B^a & B^a\\
 B^a & A^b
 \end{bmatrix} \succeq 0.
\end{equation\*}


This ultimately leads to the inequality
\begin{equation\*}
 (B^aA^{-b}B^a)\_{ii} \ge [(S\circ B^a)(S\circ A^b)^{-1}(S\circ B^a)]\_{ii}.
\end{equation\*}




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**NOTE:** The original version written by the OP, i.e.,
\begin{equation\*}
 (B^aA^{-b}B^a)\_{ii} \ge [(S\circ B)^a(S\circ A)^{-b}(S\circ B)^a]\_{ii}
\end{equation\*}
is false (Matlab quickly yields a counterexample; I can include it here in case anyone needs the gory details).


