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Let $E, X \in M_n(\mathbb R)$. $X \succ 0$ is positive definite and $\|E\|_F = 1$ where $\|\cdot\|_F$ denotes the Frobenius norm and no particular structure is assumed for $E$. I am trying to determine whether \begin{align*} E^T X E \preceq X, \end{align*} where the order is understood in the positive semdefinite cone.

I can only determine $\lambda_{\max}(E^T X E) \le \lambda_{\max}(X)$. Let $B_1$ denote the closed unit ball in $\mathbb R^n$. Then $\mathcal E = \{E x: \|x\| \le 1 \}$ is a subset of $B_1$ for fixed $E$ since $\|Ex\|_2 \le \|E\|_2 \|x\|_2 \le \|E\|_F \|x\|_2 \le 1$. So we have $$ \lambda_{\max}(E^TXE) = \sup_{\|x\| \in B_1}(x^T E^T X E X ) = \sup_{y \in \mathcal E}(y^T X y) \le \lambda_{\max}(X).$$


I am not sure if this should go to a separate question. But in the event the inequality is not true in general, what is smallest constant $c$ making the inequality hold, i.e., $E^TXE \preceq cX$? Obviously we could take $c = \lambda_{\max}(X)/\lambda_{\min}(X)$.

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    $\begingroup$ Counterexample: Let $E=\begin{bmatrix}0&1\\0&0\end{bmatrix}$ and $X=\begin{bmatrix}100&0\\0&1\end{bmatrix}$. This also shows that your bound is tight if you want a constant $c$ independent of $E$. $\endgroup$ – Rahul Sep 21 '18 at 11:16

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