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.

  • $\begingroup$ I asked the wrong question sorry. I meant $B^TA^{-1}B$ $\endgroup$ – user51196 May 23 '13 at 22:02
  • 1
    $\begingroup$ That is always positive semidefinite. Furthermore, if $B$ has full column rank, it is positive definite. $\endgroup$ – Michael Grant May 23 '13 at 22:09
  • $\begingroup$ Why is that? I can not see it directly. $\endgroup$ – user51196 May 24 '13 at 6:00
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    $\begingroup$ $A$ 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). $\endgroup$ – Ross B. May 24 '13 at 13:05

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