# Is $E^T X E \preceq X$ if $X \succ 0$ and $\|E\|_F = 1$?

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 $$c = \lambda_{\max}(X)/\lambda_{\min}(X)$$.

• 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 $E$. – Rahul Sep 21 '18 at 11:16