Prove a matrix in the form $A B A^T$ is not zero

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Consider a generic real matrix $A \in \mathbb{R}^{n\times m}$ and a square matrix $B \in \mathbb{R}^{m\times m}$, where both have non-zero rank.

What assumptions may I need to prove that $rank(A B A^T) \neq 0$, or equivalently that $A B A^T$ is not the zero matrix ?

Examples:

  • If $B$ is the identity matrix, it holds that $rank(A B A^T)=rank(AA^T)=rank(A)>0$
  • If $B$ is a diagonal matrix with positive diagonal elements, it holds that $rank(A B A^T)=rank(A)>0$
  • If $B$ is positive definite, it holds that $rank(A B A^T)=rank(A)>0$
  • If $nullity(A)=nullity(B)=0$, it holds that $A B A^T \neq 0$
  • If $nullity(A)<rank(BA^T)$, it holds that $A B A^T \neq 0$

Unfortunately the case I am studying does not fall in any of these examples.