If a matrix is positive-definite, is its projection on a subspace positive-definite as well?
The question concerns linear operators in general. Let $O$ be a positive-definite operator and $P$ the operator that projects into a given subspace. Is $P\circ O \circ P$ also positive-definite?
Under the assumption of positive-definiteness of $O$, we cannot guarantee that $POP$ is positive-definite. For example, consider
$$ O=\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} , \qquad P=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} $$
However we can show a slightly weaker statement:
If $O$ is positive-semidefinite and $P$ is a projection, then $POP$ is positive-semidefinite.
To see this, let $V$ be an inner product space with inner product $\langle\cdot,\cdot\rangle$, let $O\in L(V)$ be positive-semidefinite, and let $P$ be a projection (i.e. a self-adjoint idempotent). Then for any $x\in V$ we have $$ \langle x,POPx\rangle=\langle P^*x,OPx\rangle =\langle Px,OPx\rangle=\langle(Px),O(Px)\rangle\geq 0.$$ and thus $POP$ is positive-semidefinite.
EDIT In fact,the same argument shows a slightly stronger statment. Since idempotency of the projection wasn't necessary, we have
If $O$ is positive-semidefinite and $T$ is self-adjoint, then $TOT$ is positive-semidefinite.
If $P$ is symmetric (orthogonal projector), then $POP$ is non-negative definite since $a' POP a = (Pa)'O (Pa) \geq 0$.