If $0\leq X \leq \text{Id}$ and $0\leq A$, then $XAX \leq A$?

Question

Let $X,A\in\mathbb{C}^{n\times n}$ and suppose $$ 0 \leq X \leq \text{Id}, \quad 0\leq A, $$ where $\text{Id}\in\mathbb{C}^{n\times n}$ denotes the identity matrix. Is it true that $$ XAX \leq A, $$ or can you give a counterexample? For a proof I would imagine $0\leq \text{Id} - X$ is important, however, I have not been able to show the inequality.

Any help is appreciated!

Answer 1

Here seems to be a counterexample. Let $$ A= \begin{pmatrix} 1 & 1\\ 1 & 1 \end{pmatrix},\ X= \begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix}. $$ We have $0\leq A$, $0\leq X\leq Id$, $XAX=X$, and $$ B=A-XAX=A-X= \begin{pmatrix} 1 & 1\\ 1 & 0 \end{pmatrix}. $$ Matrix $B$ is not positive definite matrix.