# I Conditions on negative definiteness

1. Mar 24, 2016

### p4wp4w

Hello, I want to know if there exist any result in literature that answers my question:
Under which conditions on the real valued matrix $R$ (symmetric positive definite), the first argument results in and guarantees the second one:
1) for real valued matrices $A, B, C,$ and $D$ with appropriate dimensions and $A$ and $D$ being symmetric:
$X= \begin{pmatrix} A & B+RC\\ B^T+C^TR & D\\ \end{pmatrix} < 0$
2)
$Y = \begin{pmatrix} A & B+C\\ B^T+C^T & D\\ \end{pmatrix} < 0$
Congruence transformation doesn't help since it will affect the diagonal elements as well.

2. Mar 24, 2016

### perplexabot

Have you tried looking at Schur complements? I'm not sure it would help, but maybe.

3. Mar 25, 2016

### p4wp4w

No, it won't help really. The matrix inequality is already linear (LMI) and Schur complement in that sense will just cause a second problem that the off-diagonal terms will cause nonlinearty (later $R$ should be found by interior point method). I am expecting the answer to be in the form of a second LMI on $R$ but the problem is that a lot of simplifications or even assumptions can not be done since $R$ is not square.

4. Mar 25, 2016

### perplexabot

Hmmm. Maybe you are right, Schur may not be of use. I was just throwing things out there.

Wait, you say in your last post that $R$ is not square but in your original post you say it is positive definite?! Which one is it?
If $R$ is positive definite and if you assume the first argument (1) is true, then one condition on $R$ that results in argument (2) being satisfied is if $R$ is simply the Identity matrix : P

5. Mar 25, 2016

### p4wp4w

That's a nasty mistake that I made; I think deep down, I was looking for something like SPDness of $R$ to simplify things but $R$ is not square in general.

6. Mar 25, 2016

### perplexabot

Can you assume $A$ or $D$ to be positive definite?