The composition of a positive definite operator and a self-adjoint operator is positive definite?

Question

Let $A: \mathbb{R}^n \to \mathbb{R}^n$, $n>1$ be a positive definite operator that is invertible, and it is not the identity map. And let $B$ be a self-adjoint operator.

My question is, $A^{-1}\circ B$ is positive definite?

I have the feeling that this can be true, but I don't know how to prove it.

Thank you very much.

Answer 1

It's not true in general for any $\ n\ $. Unless $\ B\ $ is positive definite it must have at least one non-positive eigenvalue $\ \lambda\ $. If $\ u\ $ is a corresponding eigenvector, then \begin{align} u^\top A^{-1}Bu&=\lambda u^\top A^{-1}u\\ &=\lambda(A^{-1}u)^\top A(A^{-1}u)\\ &\le0\ . \end{align}

Answer 2

Counter example

Take $n=1$, so any linear map is self-adjoint, $A(x) = 2 x$, $B(x) = - x$.

Then $A^{-1} \circ B (x) = - \dfrac{1}{2} x$, which is clearly negative definite.

What you could say is that $B \circ A^{-1} \circ B$ is positive semi-definite (depending on the kernel of $B$).