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

1. ## 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.

2. ## Re: For A sym. pos. def. and D pos. diagonal, when is AD pos. def?

Originally Posted by Gene
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?