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Let $X$ be an $n\times n$ real positive definite matrix and consider the following matrix $$ Y=SX, $$ where $S$ is an $n\times n$ real skew-symmetric matrix. Let $\{\lambda_i\}_{i=1}^n$, $\lambda_1\ge \lambda_2\ge\cdots\ge \lambda_n$, be the ordered eigenvalues of $X$ and let $\{\mu_i\}_{i=1}^n$, $\Re[\mu_1]\ge \Re[\mu_2]\ge\cdots\ge \Re[\mu_n]$ be the ordered (w.r.t. real parts) eigenvalues of $Y$.

My question. Does there exist a skew-symmetric $S$ such that the real part of the eigenvalues of $Y$ are greater than those of $X$, that is, $$ \Re[\mu_i] \ge \lambda_i, \quad i=1,\dots,n \ \ ? $$ If not, does there exist a skew-symmetric $S$ such that the largest real part of the eigenvalues of $Y$ is greater than the largest eigenvalue of $X$, that is, $$ \Re[\mu_1] \ge \lambda_1 \ \ ? $$

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$Y=SX$ is similar to the skew-symmetric matrix $X^{1/2}SX^{1/2}$. Therefore the real parts of the eigenvalues of $Y$ are always zero and they cannot possibly be greater than any eigenvalue of $X$ (which is positive).

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