rank(AB) = rank(A) if B is invertible 
If $B$ is invertible, show that rank($AB$) = rank($A$). 

I've seen this question asked elsewhere but all had answers I didn't understand. I know how to solve the following problem

If $A$ is invertible, then rank($AB$) = rank($B$)

Because if $Bx=0$, then $ABx = A0 = 0$, and when $ABx=0$ then $Bx=0$ because $A$ is invertible, so null($AB$)=null($A$), and by the rank-nullity theorem, rank($A$) = rank($AB$). 
However when $B$ is invertible, as in the problem we have to tackle, I don't know how to use that fact. $ABx = 0$, but $B$ is in the middle so we can't simply get rid of it to get a meaningful expression.
Does someone know how to tackle this?
 A: The rank is the dimension of the column space.
The column space of $AB$ is the same as the column space of $A$.
A: For any two matrices such that $AB$ makes sense,
$$\DeclareMathOperator{\rk}{rk}
\rk(AB)\le\rk(A)
$$
If $B$ is invertible, then
$$
\rk(A)=\rk(ABB^{-1})\le\rk(AB)\le\rk(A)
$$
A: You already have: 
If $A$ is invertible, then rank($AB$) = rank($B$)
But I am going to write it out anyway: 
if $\mathbf u$, is in the kernel of $B \implies \mathbf u$ is in the kernel of $AB$  that is $AB\mathbf u = A\mathbf 0 = \mathbf 0$
if $\mathbf v$ not in the kernel of $AB, B\mathbf v = \mathbf x,\mathbf x \ne 
\mathbf 0$ and since $A$ is non-singular $x \ne \mathbf 0 \implies A\mathbf x \ne \mathbf 0$
Now a similar argument can be used for rank($BA$) = rank($B$) but it is a little bit more work.
For every $\mathbf u$ in the kernel of $B$ there exists an $\mathbf x = A^{-1} \mathbf u$ such that $BA  \mathbf x = \mathbf 0$
For every $\mathbf v$ not in the kernel of $B$ there exists an $\mathbf y = A^{-1} \mathbf v$ such that $BA  \mathbf y = B\mathbf v
$
