Prove $\operatorname{rank} (BA) = \operatorname{rank} (A) = \operatorname{rank} (AC)$, for $B$ and $C$ invertible matrices This is my question:

Let $A\in F^{m \times n}$, $B\in F^{m \times m}$ and $C\in F^{n \times n}$ be invertible matrices. Prove that $$\operatorname{rank} (BA) = \operatorname{rank} (A) = \operatorname{rank} (AC).$$

How to proceed with this question? A complete answer will help my proper understanding. :)

To show that $\operatorname{rank} (A) = \operatorname{rank} (AC)$, column space of $AC$ is a subspace of column space of $A$. Now, see that $A = (AC)C^{-1}$, which says that column space of $A$ is a subspace of column space of $AC$. For the first part you may use row rank of $A$ equals the column rank of $A$.
