Hi I have the following problem. 
Let the symmetric matrix M of the form:
\begin{bmatrix} 
 A & B \newline
 B^T & C \newline
\end{bmatrix}


We have that $C$ is positive semidefinite. Is there a way to transform the constraint $A-BC^{-1}B^T \leq 0$ to a constraint using matrix $M$? 


I know that in case the constraint was $A-BC^{-1}B^T \geq 0$ the answer is $M \geq 0$.
But can we say something similar for the negative definite constraint? (I have already checked that $A-BC^{-1}B^T \leq 0 \Leftrightarrow M \leq 0$ Does not hold.) Is there another way to transform this constraint to a linear over the matrix B?


Thank you very much!



Schur complement and "negative definite"!
$M$ is congruent to ${\rm diag}(A-BC^{-1}B^T,C)$. Therefore the condition $A-BC^{-1}B^T\le0$ amounts to saying that the maximal dimension of positive subspace is the size $p$ of $C$.If $A-BC^{-1}B^T<0$, this is saying that the signature of $M$ is $(p,0,q)$, where $q$ is the size of $A$.


