The Schur Complement of a block matrix $A$ in $X$ where $$ X = \begin{bmatrix} A & B \\ B^T& C \end{bmatrix} $$ is defined as $S = C-B^TA^{-1}B$ whenever $A$ is non-singular. For one, I know that if $A = 0$, then for $X$ to be positive semidefinite, $B = 0$ and $C \succeq 0$ must hold true. (Here $\succeq$ denotes positive semi-definiteness). But I don't know how to prove it here (even for a $2 \times 2$ block matrix).
Here is my attempt at a proof (when $A = 0$)
Consider $P = \begin{bmatrix} I & 0 \\ -B^T & I\end{bmatrix}$. Then, since $X$ is congruent to $Y = PXP^T$, so $X \succeq 0 \Leftrightarrow Y \succeq 0$ because $P$ is non-singular. Expanding $Y$ we get $Y = \begin{bmatrix} 0 & 0 \\ 0 & C-B^TB \end{bmatrix}$. Now, for $Y$ to be PSD, all we need is $C-B^TB \succeq 0$.
I don't see how that puts a restriction on $C$ and $B$ individually.
Any help is greatly appreciated. Thanks in advance!!!