Kernel of product Let $B \in \mathbb R^{{k}\times {m}}$ and $A \in \mathbb R^{{m}\times {n}}$. Further assume that $\operatorname{ker}(B) \cap \operatorname{ran}(A) = \{0\}.$ Show that this implies $\operatorname{ker}(A) = \operatorname{ker}(BA).$
I have no idea how to tackle this problem.
 A: $\ker(A)\subseteq\ker(BA)$ is obvious.
Suppose $v\in\ker(BA)$. Then $B(Av)=0$, so $Av\in\ker(B)\cap\operatorname{ran}(A)$. Hence $Av=0$, so $v\in\ker(A)$.
A: It is not hard, but the structure is a bit verbose, hope that helps.
$$\text{Ker}(A)=\{v \in \mathbb{R}^n ~|~Av=0\}$$
$$\text{Ker}(BA)=\{v \in \mathbb{R}^n ~|~BAv=0\}$$
If $v\in\text{Ker}(A)\Rightarrow Av=0 \Rightarrow BAv=0\Rightarrow v\in\text{Ker}(BA)$ 
or if $Av=u\neq0\Rightarrow (u\in\text{Ran}(A)$ and $B(Av)=0 \iff Bu=0)$, thus:
$$\text{Ker}(BA)= \text{Ker}(A) \cup \{v\in \mathbb{R}^n/\text{Ker}(A)~|~u=Av~\text{and}~u\in\text{Ker}(B) \}$$
But, $\text{Ker}(B)\cap\text{Ran}(A)=\{0\} \Rightarrow (u\in\text{Ran}(A)/\{0\}\Rightarrow u\notin\text{Ker}(B))$
If $v\in \mathbb{R}^n/\text{Ker}(A) \Rightarrow  Av\in\text{Ran}(A)/\{0\}$, thus (from above) $u\notin \text{Ker}(B)$, so no element of $v\in \mathbb{R}^n/\text{Ker}(A)$ satisfies $u\in \text{Ker}(B)$ and $\{v\in \mathbb{R}^n/\text{Ker}(A)~|~u=Av~\text{and}~u\in\text{Ker}(B) \}=\{\}$, finally:
$$\text{Ker}(BA)= \text{Ker}(A) \cup \{\} = \text{Ker}(A)$$
