PSD Matrix Inequality proof

Question

Is it true that for any two matrices $A \preceq B \iff C^T A C \preceq C^T B C$ for any conformable matrix $C$?

This related question proves ($\Rightarrow$) We have $x^TC^T(B-A)C x = (Cx)^T(B-A)C x \ge 0$ for any conformable vector $x$ so that $C^T(B-A)C \ge 0.$

What about the other direction?

Answer 1

The converse is clearly false. Consider $C=0$ for instance.

For another counterexample we have $\pmatrix{1&0}\pmatrix{1&0\\ 0&-1}\pmatrix{1\\ 0}\succeq0$ but $\pmatrix{1&0\\ 0&-1}$ is indefinite.