$A, B $ be $n \times n$ real matrices such that $A$ is singular and $B$ is non-singular, then prove or disprove $\text{Rank}(AB)=\text{Rank}(A)$

Question

Let $A, B $ be $n \times n$ real matrices such that $A$ is singular and $B$ is non-singular, then prove or disprove $\operatorname{Rank}(AB)=\operatorname{Rank}(A)$

Since $\operatorname{Rank}(AB)\leq \operatorname{Rank}(A)$, Is there any example for the strict inequality for this assumption?

Answer 1

$\operatorname{Rank}(A)=\operatorname{Rank} ((AB)B^{-1}) \leq \operatorname{Rank} (AB)$ so equality holds.

Answer 2

$B$ nonsingular implies $B(V)=V$, where $V$ is the domain of $A$ and $B$. Thus, $AB$ and $A$ have the same image, $A(V)$. So $\operatorname {rank}AB=\operatorname {rank}A$.

Answer 3

Think in terms of the associated linear maps: $B$ is the matrix of an automorphism of $\mathbf R^n$, so $\operatorname{Im}A=\operatorname{Im}(AB)$, and $$\operatorname{rank}A\stackrel{\text{def}}{=}\dim \operatorname{Im}A=\dim\operatorname{Im}(AB)=\operatorname{rank}(AB).$$