For a matrix $latex M = \begin{pmatrix} A &B\\C&D\end{pmatrix}$, we call $latex S\triangleq D -CA^{-1}B$ the Schur complement of $latex A$ in $latex M$. Note that $latex S$ naturally appear in block matrix inversion.
Note that when $latex M$ is symmetric and $latex A$ is positive definite, $latex M$ is positive definite if and only if $latex S$ is also positive definite. The proof is rather straightforward. Consider the function
$latex f(u,v)=[u^t,v^t]M\begin{pmatrix} u \\ v\end{pmatrix}$
$latex = u^tAu+u^tBv+v^tCu+v^tDv = u^tAu+2u^tBv+v^tDv$.
Let’s try to minimize $latex f(u,v)$ w.r.t. $latex u$,
$latex \frac{\partial f(u,v)}{\partial u} =u^tA+u^tA^t+2v^tB^t=u^tA+v^tB^t$.
Set the derivative to zero and we get
$latex u*=-A^{-1}Bv$. And the minimum is
$latex v^t(D-B^t A^{-1}B)v = v^tSv\triangleq g(v)$. Now, we see that for $latex M$ to be positive definite, $latex f(u,v)>0$ for all $latex u$ and $latex v$. This holds if and only if $latex g(v) > 0$ for all $latex v$, which in turn means that $latex S$ has to be positive definite. The argument works for positive semidefinite as well.