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Math Help - For A sym. pos. def. and D pos. diagonal, when is AD pos. def?

  1. #1
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    For A sym. pos. def. and D pos. diagonal, when is AD pos. def?

    If this is better placed in the general analysis forum, please tell me. I'm new here.

    Expanding the Subject:

    A - real square symmetric positive definite.
    D - real square positive diagonal.

    Under what conditions (sufficient nice; necessary excellent) is AD also positive definite?

    It is easy to show AD has positive eigenvalues, so the "failure" case is when a negative one pops up in

    AD + (AD)^T .

    I have not been able to get traction on this. All help appreciated.

    Background: The need is to explore convergence of a very large, very nonlinear system of differential equations. My best shot at a Lyapunov function has derivative of the form

    \frac{dE}{dt} = - \vec{f}(\vec{u})^T [AD(u)] \vec{f}(\vec{u}) .

    The system is bounded, so if AD is always positive definite, it will converge. Alas, it starts meandering in experiments exactly when AD becomes indefinite. I'm searching for a fix.

    Many thanks for your thoughts.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: For A sym. pos. def. and D pos. diagonal, when is AD pos. def?

    Quote Originally Posted by Gene View Post
    If this is better placed in the general analysis forum, please tell me. I'm new here.

    Expanding the Subject:

    A - real square symmetric positive definite.
    D - real square positive diagonal.

    Under what conditions (sufficient nice; necessary excellent) is AD also positive definite?

    It is easy to show AD has positive eigenvalues, so the "failure" case is when a negative one pops up in

    AD + (AD)^T .

    I have not been able to get traction on this. All help appreciated.

    Background: The need is to explore convergence of a very large, very nonlinear system of differential equations. My best shot at a Lyapunov function has derivative of the form

    \frac{dE}{dt} = - \vec{f}(\vec{u})^T [AD(u)] \vec{f}(\vec{u}) .

    The system is bounded, so if AD is always positive definite, it will converge. Alas, it starts meandering in experiments exactly when AD becomes indefinite. I'm searching for a fix.

    Many thanks for your thoughts.
    Ok, so you know that AD has positive eigenvalues, and so if it's Hermitian, then you're done. But, both A and D are Hermitian, so $A,D$ commuting seems to be what you're looking for. Am I missing something?
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