Let $P$ be a positive semidefinite matrix and $A$ any matrix. Is $X=APA^\intercal$ positive semidefinite, if so give a proof, if not a counterexample. (not sure if the context of the problem means that A should be square)
I doubt X is positive semidefinite, but need help with a counterexample in a 2x2 case.
Example that does not work: $\left(\begin{matrix}-1&1\\0&0\end{matrix}\right)\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}-1&0\\1&0\end{matrix}\right)=\left(\begin{matrix}-1&1\\0&0\end{matrix}\right)\left(\begin{matrix}-1&0\\1&0\end{matrix}\right)=\left(\begin{matrix}1&0\\0&0\end{matrix}\right)$
Maybe if you got a positive semidefinite matrix without zeros in the off diagonals and tried something it could work. This problem is only worth a few points on my homework but I feel bothered by it.
For real matrices, this is true, but for complex matrices, this is false!
Consider first the (counter)example for $2\times 2$ complex matrices: Let $P=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ and $A=\begin{pmatrix} i & 0 \\ 0 & i \end{pmatrix}$. Then, $APA^T=\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$.
Now for real matrices, for any vector $v$, $\langle APA^T v, v \rangle=\langle P(A^T v), (A^T v) \rangle \geq 0$, as $P$ is positive semidefinite.
For the result to remain true for complex matrices as well, one must replace the transpose by the adjoint.