Prove that $AB = 0 \iff \mathrm{im}(B) \subseteq \ker(A)$ Problem 1. Prove that for any $\ell \times m$-matrix $A$ and any $m \times n$-matrix $B$,
$$
AB = 0 \quad\text{ if and only if }\quad \mathrm{im}(B) \subseteq \ker(A)
$$
I have no idea on how to start this... 
I'm new to proofs and this is my first proof. 

Suppose $AB = 0$. Let $x\in \operatorname{Im}(B)$. Then $x = By$ for some $y$. So $$Ax = A(By) = (AB)y = 0y = 0.$$ Thus $x\in \operatorname{Ker}(A)$. 
Conversely, assume $\operatorname{Im}(B) \subseteq \operatorname{Ker}(A)$. For every $x$, $Bx \in \operatorname{Im}(B)$ and thus $Bx \in \operatorname{Ker}(A)$, i.e., $ABx = 0$. Since $x$ was arbitrary, $AB = 0$.
