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:

$\displaystyle A$ - real square symmetric positive definite.

$\displaystyle D$ - real square positive diagonal.

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

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

$\displaystyle AD + (AD)^T$ .

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

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

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.

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

Quote:

Originally Posted by

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

Expanding the Subject:

$\displaystyle A$ - real square symmetric positive definite.

$\displaystyle D$ - real square positive diagonal.

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

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

$\displaystyle AD + (AD)^T$ .

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

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

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 $\displaystyle AD$ has positive eigenvalues, and so if it's Hermitian, then you're done. But, both $\displaystyle A$ and $\displaystyle D$ are Hermitian, so $A,D$ commuting seems to be what you're looking for. Am I missing something?