Question

Hello,

My question regards to the Schur complement lemma. Consider the matrix $M=\left( \begin{array}{cc} A & B\\ B^T & C \end{array}\right) $.

According to the lemma $M\geq0$ iff $C>0$ and $A-BC^{-1}B^T\geq 0$.

In my current research I'm working on an optimization problem over a domain of matrices; I'm trying to convert this optimization problem into it's convex form. In order to do so I need a similar relation for negative definite matrices. Can the Schur complement lemma be extended to the case of negative definite matrices? And if so, how? Namely, is it true that for a matrix $M$ of the same structure we have $M\leq 0$ iff $C<0$ and $A-BC^{-1}B^T\leq 0$?

Another similar but different problem I have regards to the following non-convex (nor linear) constraint: $A-BC^{-1}B^T\leq 0$ and $A\geq 0$. Is there some way such constraints can be converted to an equivalent constraint which is linear in these variables? For example the two constraints $A-BC^{-1}B^T\geq 0$ and $C> 0$ can be simply converted to $M\geq 0$ using the Schur complement lemma. The new constraint $M\geq 0$ is equivalent to the two old ones and is indeed linear in the matrices $A,B,C$. I'm looking for a way to do something similar to this for my case.

Thank you all in advance,

Best regards!

Answer 1

This should be a comment, but I can not yet post comments.

You got Schur's complement lemma wrong, the matrix $$\left(\begin{array}{cc} 1 & 1 \newline 1 & -1 \end{array}\right)$$ satisfies you conditions, $A=1\ge 0$ and $A-BC^{-1}B^T=1-(-1)\ge 0$, but it is clearly not positive semi-definite. Replace the first condition by $C > 0$ (see also http://en.wikipedia.org/wiki/Schur_complement).

Then Chris Godsil's comment shows that the answer to your first question is yes.