# Can the second term of the Schur complement of a symmetric matrix be undefined?

Given the next symmetric matrix conformably partitioned

$$\begin{bmatrix} A &B \\ B^T &C \end{bmatrix}$$

I know that $A$ and $C$ are positive definite matrices.

The Schur complement is $S=C-B^TA^{-1}B$

What can I say about $B^TA^{-1}B$? Is this undefined in general? or it is positive/negative (semi)definite.

Probably it is an easy question, but I do not see why. Thanks in advance.

If $A \succ 0$, $S$ is positive semidefinite iff the block matrix is positive semidefinite.
• I asked the wrong question sorry. I meant $B^TA^{-1}B$ – user51196 May 23 '13 at 22:02
• That is always positive semidefinite. Furthermore, if $B$ has full column rank, it is positive definite. – Michael Grant May 23 '13 at 22:09
• $A$ being positive definite means $A^{-1}$ is positive definite. Then define $Bx \triangleq y$, so we have $x^TB^TA^{-1}Bx = y^TA^{-1}y$, so positive semidefiniteness follows from positive definiteness of $A^{-1}$. The degenerate case possibly occurs if $B$ does not have full column rank ($y$ could be zero even if $x$ is not). – Ross B. May 24 '13 at 13:05