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I Conditions on negative definiteness

  1. Mar 24, 2016 #1
    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.
     
  2. jcsd
  3. Mar 24, 2016 #2

    perplexabot

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    Gold Member

    Have you tried looking at Schur complements? I'm not sure it would help, but maybe.
     
  4. Mar 25, 2016 #3
    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.
     
  5. Mar 25, 2016 #4

    perplexabot

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    Gold Member

    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 [itex]R[/itex] is not square but in your original post you say it is positive definite?! Which one is it?
    If [itex]R[/itex] is positive definite and if you assume the first argument (1) is true, then one condition on [itex]R[/itex] that results in argument (2) being satisfied is if [itex]R[/itex] is simply the Identity matrix : P
     
  6. Mar 25, 2016 #5
    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.
     
  7. Mar 25, 2016 #6

    perplexabot

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    Gold Member

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