# Schur Complement and Positive Definite Matrix

For a matrix $M = \begin{pmatrix} A &B\\C&D\end{pmatrix}$, we call $S\triangleq D -CA^{-1}B$ the Schur complement of $A$ in $M$. Note that $S$ naturally appear in block matrix inversion.

Note that when $M$ is symmetric and $A$ is positive definite, $M$ is positive definite if and only if $S$ is also positive definite. The proof is rather straightforward. Consider the function

$f(u,v)=[u^t,v^t]M\begin{pmatrix} u \\ v\end{pmatrix}$

$= u^tAu+u^tBv+v^tCu+v^tDv = u^tAu+2u^tBv+v^tDv$.

Let’s try to minimize $f(u,v)$ w.r.t. $u$,

$\frac{\partial f(u,v)}{\partial u} =u^tA+u^tA^t+2v^tB^t=u^tA+v^tB^t$.

Set the derivative to zero and we get

$u*=-A^{-1}Bv$. And the minimum is

$v^t(D-B^t A^{-1}B)v = v^tSv\triangleq g(v)$. Now, we see that for $M$ to be positive definite, $f(u,v)>0$ for all $u$ and $v$. This holds if and only if $g(v) > 0$ for all $v$, which in turn means that $S$ has to be positive definite. The argument works for positive semidefinite as well.