Can $\operatorname{Tr}[A^{-1} BAB^\top]$ be shown to be always positive if $A$ is real and positive definite?

Question

Let $A$ be a real symmetric positive definite matrix and $B$ is a real matrix with all eigenvalues zero. Can we prove or disprove that $\operatorname{Tr}[A^{-1} BAB^\top]$ is a positive number?

Answer 1

It can be proven in the case where $B\neq 0$.

Let $A^{-1/2}$ be the unique real symmetric positive definite matrix such that $A^{-1}=\left(A^{-1/2}\right)^2$. This is posible since $A^{-1}$ is also positive definite, then \begin{align*} \operatorname{Tr}\left[ A^{-1} B A B^T \right] &= \operatorname{Tr}\left[ A^{-1/2} A^{-1/2} B A B^T \right]\\ &= \operatorname{Tr}\left[ A^{-1/2} B A B^T A^{-1/2} \right] \end{align*}

But it is clear that $A^{-1/2} B A B^T A^{-1/2}$ is PSD, indeed for any vector $x$ \begin{align*} x^TA^{-1/2} B A B^T A^{-1/2} x = \left( B^T A^{-1/2} x \right) ^T A \left( B^T A^{-1/2} x \right) \geq 0 \end{align*}

Therefore $\operatorname{Tr}\left[ A^{-1} B A B^T \right]$ is non-negative. Note that it is zero if and only if $A^{-1/2} B A B^T A^{-1/2}$ is zero. This can happen in your setup, for instance obviously if $B=0$. If $B\neq 0$, then there is a vector $y$ such that $B^T y \neq 0$ and there is $x$ such that $y=A^{-1/2}x$, therefore $B^T A^{-1/2} x\neq 0$. Because $A$ is positive definite, we get that $\left( B^T A^{-1/2} x \right)^T A \left(B^T A^{-1/2} x\right) > 0$ and so $\operatorname{Tr}\left[ A^{-1} B A B^T \right] > 0$.