Suppose $\mathbf{A}_{n\times n}$ is nonsingular. Then, for any $\mathbf{B}_{m\times n}$ it's true that $\mathrm{rank}(\mathbf{AB})=\mathrm{rank}(\mathbf{B})$ since:
\begin{eqnarray*} \mathbf{A}\ nonsingular &\Rightarrow& \mathcal{N}(\mathbf{A})=\{\mathbf{0}\}\\ &\Rightarrow& \mathcal{N}(\mathbf{A})\cap \mathcal{R}(\mathbf{B})=\{\mathbf{0}\}\\ &\Rightarrow& \mathrm{dim}(\mathcal{N}(\mathbf{A})\cap \mathcal{R}(\mathbf{B}))=0\\ &\Rightarrow& \mathrm{rank}(\mathbf{B})-\mathrm{rank}(\mathbf{AB})=0 \end{eqnarray*}
But, I am having a difficulty proving a similar result that if $\mathbf{A}$ is nonsingular then the following is also true: $$\mathrm{rank}(\mathbf{B}^T\mathbf{A})=\mathrm{rank}(\mathbf{B}^T)$$
Note that: $\text{rank}(B^{T}A) = \text{rank}(A^{T}B)$.
Suppose $v \in \ker A^{T}B$. Then $A^{T}Bv = 0$. Since $A$ is non-singular, $A^{T}$ is also non-singular. Hence $Bv = 0$. So that $\ker A^{T}B \subset \ker B$. It is trivially true that $\ker B \subset \ker A^{T}B$. It follows that $\text{rank}(B^{T}A)= \text{rank} A^{T}B = \text{rank} B = \text{rank} B^{T}$.