Can it be proven that matrix $(AB^{-1}A)$ Is positive definite, when B is positive definite?

Question

I am trying to prove the following result:

$(AB^{-1}A)$

is a positive definite matrix when B is a positive definite matrix (that is $B^{-1}$ also positive definite). Can this be proven or is it simply not the case?

Answer 1

Firstly $A$ can not be zero matrix or singular. You have to give some condition on $A.$ $A$ is real and symmetric is enough to prove this. As $B$ is PD, So is $B^{-1}.$ Hence $x^T(AB^{-1}A)x = (Ax)^TB^{-1}(Ax) > 0$ for $0 \neq Ax \in R^{n}.$

Answer 2

No, it is not true in general, consider the case when $A$ is singular.

Answer 3

It is simply not the case. That is, there is the simple counterexample of $A$ is the zero matrix.

If we force a restriction on $A$ that it is also positive definite, then yes this product is positive definite.

Can you share the context of the problem?