Prove that if $\operatorname{rank}A=n$, then $\operatorname{rank}AB=\operatorname{rank}B$ 
Let $A \in M_{m\times n}(\mathbb{R})$ and $B \in M_{n\times p}(\mathbb{R})$.
Prove that if $\operatorname{rank}(A)=n$ then $\operatorname{rank}(AB)=\operatorname{rank}(B)$.

I tried to start with definitions finding that $n \le m$, but didn't know what to do with $AB$.
Please help, thank you!

Remember, if:
(a) $rk(A + B)  \leq rk(A) + rk(B)$ for any two $mxn$ matrices $A,B$;
(b) $rk(AB)  \leq \min (rk(A),\ rk(B))$ for any $k\times l$ matrix $A$ and $l\times m$ matrix $B$;
(c) if an $n\times n$ matrix $M$ is positive definite, then $rk(M) = n$.
So that, for your question: Prove that if $rk(A)=n$ then $rk(AB)=rk(B)$
since $A \in Mat_{m\times n}(\mathbb{R})$ which is matrix $A$ is positive definite, and let for  $n \le m$  the maximum number of linearly independent columns is $n$, hence $rk(A) = n$. similar for $rk(B)$, For  $p \le n$  the maximum number of linearly independent columns is $p$, hence $rk(B) = p$.
Further since we have $A \in Mat_{m\times n}(\mathbb{R})$ and $B \in Mat_{n\times p}(\mathbb{R})$. we take $A,B$ such that $A  $ is $mxn$ and $B$ is $nxp$ and $AB=I_{mp}$
So, $rk(AB)\le rk(A) \le n \lt p $ hence $rk(I_{mp})=mp$ and $rk(AB)  \leq min (rk(A); rk(B))$
And as $rk(I_{mp})=mp$, then $rk(AB)\ne rk(I_{mp}) \Rightarrow AB \ne I_{mp}$, thus
    If $A$ is an $m × n$ as matrix of $rk(A) = n$ and , then $rk(B)$, For  $p \le n$  the maximum number of linearly independent columns is $p$, and $rk(B) = p$, therefore $rk(AB)=rk(B)$.
