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.

Leave a Reply

Your email address will not be published. Required fields are marked *