Inequalities for Hadamard products of complex symmetric matrices

Question

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.

Answer 1

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).