How to show $rank(BA)=rank(A)$

Let $\mathbb F$ be a finite field and let $A_{m\times n}$ $B_{m\times m}$ given that $rank(B)=m$ show that $$rank(BA)=rank(A)$$

I guess the idea is to use the equality $$dim(N(A))=n-rank(A)$$

so I understand from the given that $dim(N(B))=0$ and I dont understand how to procced from here, how does this given is helping me? I assume i need to show some how that $dim(N(BA))=dim(N(A))$

Solution 1:

$A_{m\times n}$ And $B_{m\times m}$ and $rank(B) =m$

Then, $N(A) =N(BA) $

$v\in N(B) \implies Bv=0$

Now, $(BA)v=B(Av) =B\cdot0=0$

$\implies v\in N(BA) $

Hence, $N(B) \subseteq N(BA) $

Again,$v\in N(BA) \implies (BA) v=0$

As $B_{m\times m}$ is Invertible.

$Av=B^{-1}\cdot 0=0$

$\implies v\in N(A) $

$\implies N(BA) \subseteq N(A) $

Hence $N(BA) =N(B) $

Now, using rank-nullity theorem,

$Null(BA) + Rank(BA) =m=Null(A) +Rank(A) $