$S∘T=0$ if and only if Im$T \subset$ ker$S$ Let $T:\mathbb{R}^n \to \mathbb{R}^m$ and $S:\mathbb{R}^m \to \mathbb{R}^l$ be linear maps. I have to show that:
$S∘T=0$ if and only if Im$T \subset$ ker$S$
Can someone talk me through this? I actually have no idea where to start.
 A: $$S\circ T=0\iff \forall\,v\in\Bbb R^n\;,\;\;S\circ T(v):=S(Tv)=0\iff $$
$$\iff Tv\in\ker S\;\;\forall\,v\in\Bbb R^n\iff \text{Im}\,(T)\le\ker S$$
A: I would start by drawing a picture. Represent the $\mathbb{R}^i$ by 'blobs'.
Now 
$$\mathbb{R}^n\overset{T}{\longrightarrow}\mathbb{R}^m\overset{S}\longrightarrow \mathbb{R}^\ell.$$
Now if $\text{im }T\subset\ker S$ then all of the things that are mapped into $\mathbb{R}^m$ by $T$ are in the kernel of $S$ and so will be 'killed' by $S$ so we have $S\circ T=0$.
Now on the other hand, if $S\circ T=0$, the composition of $T$ and $S$ 'kills' everything. In particular everything that $T$ sends to $\mathbb{R}^m$ aka $\text{im }T$ is 'killed' by $S$ so is in the kernel of $S$. 
This is the why? The other answers I expect will answer in a nice, technical voice. Do you know what I mean by 'kill'?
A: Hint: $\Leftarrow$ is trivial, isn't it? For the other direction; let $v\in\mathrm{Im}(T)$, so there's a vector $w$ s.t. $Tw=v$.
A: By definition, $\operatorname{Ker}(S)=\{w\in\mathbb R^m~:~ S(m)=0\}$ and
$\operatorname{Im}(T)=\{w\in\mathbb R^m~:~ \exists n\in\mathbb R^n~:~ T(n)=w\}$.
Then  $S\circ T=0\Leftrightarrow \forall v\in\mathbb R^n~ S(T(v))=0\Leftrightarrow \forall v\in\mathbb R^n~~ T(v)\in\operatorname{Ker}(S)\;$ . Can you finish the proof?
