Schur complement and "negative definite"! 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!
 A: $M$ 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 $A$.
A: Its not completely clear what you are asking: what does it mean to "transform this constraint to a linear over the matrix B?" But perhaps it will be helpful to note that the set of matrices of your form satisfying $$C \geq 0, ~~A - B C^{-1} B^T \leq 0$$ is not convex: just consider what happens when you take the average of two $2 \times 2$ matrices, one with $A=2,C=1/2,B=1$ and the other with $A=1/2, C=2, B=1$. So it is impossible to express these two constraints in terms of linear inequalities on matrices $A,B,C,D$. 
A: No, that's not quite the generalization that you'll get when you extend the Schur complement theorem for positive definite matrices to negative definite matrices.  Try using the fact that a matrix $X$ is negative definite if and only if $-X$ is positive definite and then follow the sign changes through- you'll end up with a slightly different $M$ matrix.  
