Moore-Penrose inverses and positive semidefinite-ness. For $X\in\mathbb{R}^{n\times p}$ with $\operatorname{rank}(X)=p$, and positive definite $V\in\mathbb{R}^{n\times n}$, consider the following matrix
$$A=X^+V(X^+)^T-(X^TV^{-1}X)^{-1}$$
Where $X^+$ denotes the Moore-Penrose pseudoinverse of $X$, i.e. $(X^TX)^{-1}X^T$.
I know that both terms are positive definite and that $A=0$ when $X$ is square, but I'm not sure how to show that $A$ is positive semi-definite (if it even is). 

It is positive semidefinite, because it is congruent to
\begin{align}
(X^TX)A(X^TX)
&=(X^TX)\left[X^+V(X^+)^T - (X^TV^{-1}X)^{-1}\right](X^TX)\\
&=X^TVX - (X^TX)(X^TV^{-1}X)^{-1}(X^TX)
\end{align}
and the expression on last line is the Schur complement of $X^TV^{-1}X$ in the positive semidefinite matrix
$$
\pmatrix{X^TVX&X^TX\\ X^TX&X^TV^{-1}X}
=\pmatrix{X^TV^{1/2}\\ X^TV^{-1/2}}\pmatrix{V^{1/2}X& V^{-1/2}X}.
$$
