Let $E$ be a $d \times d$ matrix of full rank. Assume that $E \le A$ in the Loewner ordering over positive semidefinite matrices and further assume that $A$ is also full rank. If $P$ is a change of basis matrix, is the following true?
$$P^{-1} E P \le P^{-1} A P$$
If $P$ were orthogonal, I think this is immediate from the conjugation rule (if $X \le Y$ then $C^T X C \le C^T Y C$ and the converse if $C$ is full rank)on the Loewner ordering as then we have $P^{-1} = P^T$.
This question arose as I have a product $EH$ and know how to control $E$ under the Loewner ordering. However, I would like to understand its product with $H$ so I applied a change of variables by pre/post multiplying by $H^{1/2},H^{-1/2}$, respectively (assume $H$ is full rank). The only rule I am aware of is the conjugate rule given above.