Matrix semi-positive definite If $n \times n$ matrix $A \succeq 0$, and one $ n \times q$ column orthogonal matrix $U$, does this inequality hold? $$A- UU^{T} A UU^{T} \succeq 0$$
 A: The conclusion seem not to be true. Here is my argument. First, since $U$ is column orthogonal we have that $P=U U^T $ is a projection matrix (we do assume $q<n$, the case $q=n$ is uninteresting since then $P=I$ and the result is trivial).  Let $\mathcal{L}=\text{kernel of $P $}^\perp$, that is, the subspace where $P$ is projecting, $\mathcal{M}=\text{the kernel of $A$}^\perp$. The critical case seems to be when the nonzero vector $x$ is in the nullspace of 
$A$ but is projected by  $P$ out of the nullspace of $A$, then we can calculate 
$$
   x^T(A - PAP)x = x^TAx - x^T PAPx = 0 - \text{possibly positive}
$$
which would make a counterexample to the claim. The possibly positive term would be zero 
if $\mathcal{L} \subseteq  \mathcal{M}^\perp$, which gives a condition we can use to make a refined conjecture:
CONJECTURE
We have that $A - PAP$ is positive semidefinite if $P$ projects into the kernel of $A$.
COUNTEREXAMPLE TO ORIGINAL CLAIM
$n=2, q=1$ $A=\left(\begin{smallmatrix} 0 & 0 \\ 0 & 0.7\end{smallmatrix}\right)$, $P=\frac{1}{2}\left(\begin{smallmatrix} 1&1 \\1 & 1 \end{smallmatrix}\right)$, and $x= \left(\begin{smallmatrix} 1 \ 0 \end{smallmatrix}\right)$.  With this definitions, calculate
$$
   x^T \left( A-PAP\right) x = -7/40
$$
completing the counterexample.
What makes the counterexample work? First, if the column making up tha matrix $U$ are eigenvectors of $A$, the OP conjecture is true, ew van find no counterexample. This is easy to see. So the point is finding a simple matrix $U$ where the columns are not eigenvectors of $A$, and then using the idea of projecting vectors in the kernel of $A$ out of the kernel. Then, drawing a simple figure gives the example.
