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. 
 A: 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, $AB = 0$.
A: suppose $AB = 0$.  you need to show $im B \subset ker A$  so pick an element $y$ in $im B$ that means $y = Bx$ for some $x.$ look at $Ay = ABx = 0$ because $AB = 0.$ now we have shown that for any $y \in im B $ implies $Ay = 0$ so $y \in kerA$ we are done showing if $AB = 0$, then $im B \subset ker A$
suppose $im B \subset ker A,$ we need to show $AB = 0$ we will show it by contradiction. suppose $AB \neq 0.$ that means there is an $x$ such that $ABx \neq 0$ since $Bx \in im B$ and $im B \subset ker A$. that means $Bx \in kerA$  and $ABx = 0$ the contradiction we sought.  
A: Basically you need to show these two facts:
1)If $im(B) \subseteq ker(A)$ then $AB=0$. Hint for this: use the fact that a vector $y$ is in the image of $B$ if there exists a vector $x$ such that $Bx=y$.
2) If $AB=0$ then $im(B) \subseteq ker(A)$. Hint: use the above characterization of $im(B)$ and the definition of $ker(A)$.
