Rank of product of matrices: $\operatorname{rank}(ABA') = \operatorname{rank}(A)$ when $\operatorname{rank}(A) = \operatorname{rank}(B)$

Question

Suppose $\mathbf{A}$ is an $m \times n$ matrix, $\mathbf{B}$ is a symmetric $n \times n$ matrix, and $m < n$. If $\operatorname{Rank}(\mathbf{A}) = \operatorname{Rank}(\mathbf{B}) = m$ then what is the rank of $\mathbf{ABA}^\top$?

Is it $\operatorname{Rank}(\mathbf{ABA}^\top) = m$.

I found this result for a particular choice of $\mathbf{A}$ and $\mathbf{B}$. So I want to know whether it is a general result.

Answer 1

With $m=1$ and $n=2$, we have $$\begin{bmatrix} 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = 0$$ which has rank zero. You can construct larger counterexamples with $n=2m$ similarly: $$\begin{bmatrix} I_m & 0_{m \times (n-m)} \end{bmatrix} \begin{bmatrix} 0_{m \times m} & 0_{m \times m} \\ 0_{m \times m} & I_m \end{bmatrix} \begin{bmatrix} I_m \\ 0_{(n-m) \times m} \end{bmatrix} = 0_{m \times m}.$$