I Conditions on negative definiteness

p4wp4w
Messages
8
Reaction score
5
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.
Thank you all in advance.
 
  • Like
Likes perplexabot
Physics news on Phys.org
Have you tried looking at Schur complements? I'm not sure it would help, but maybe.
 
  • Like
Likes p4wp4w
perplexabot said:
Have you tried looking at Schur complements? I'm not sure it would help, but maybe.
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.
 
  • Like
Likes perplexabot
p4wp4w said:
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.
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
 
  • Like
Likes p4wp4w
perplexabot said:
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
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.
 
  • Like
Likes perplexabot
p4wp4w said:
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.
Can you assume ##A## or ##D## to be positive definite?
 
Thread 'Derivation of equations of stress tensor transformation'
Hello ! I derived equations of stress tensor 2D transformation. Some details: I have plane ABCD in two cases (see top on the pic) and I know tensor components for case 1 only. Only plane ABCD rotate in two cases (top of the picture) but not coordinate system. Coordinate system rotates only on the bottom of picture. I want to obtain expression that connects tensor for case 1 and tensor for case 2. My attempt: Are these equations correct? Is there more easier expression for stress tensor...
Back
Top