Prove $\mathrm{rank}(BAC)=\mathrm{rank}(BA)=\mathrm{rank}(AC)=\mathrm{rank}(A)$

Question

Would anyone know how to prove the following? It is stated as a theorem in the textbook without further explanations.

Let $A$ be an $m \times n$ matrix, $B$ an $m\times m$ matrix, and $C$ a $n\times n$ matrix. Then if $B$ and $C$ are nonsingular matrices, it follows that:

$$ \mathrm{rank}(BAC)=\mathrm{rank}(BA)=\mathrm{rank}(AC)=\mathrm{rank}(A)$$

I have searched for other similar questions but the proofs seem to all rely on some notion of fields, and $\mathrm{dim}()$, but the textbook has yet to touch on such concepts at this point.

Your help would be greatly appreciated.

Answer 1

HINT

It is a known fact: $\text{Rank}(AB)\leq \min(\text{Rank}(A), \text{Rank}(B))\tag 1$

Using (1) you can prove $\text{Rank}(AB)= \text{Rank}(A)$ if $B$ nonsingular.

See this and this for proofs.