Proof related with kernel of a function

Question

If $f : V → V ′$ and $g : V ′ → V ′′$ are linear applications and $v ∈ V$

Proof that: $v ∈ Ker f ⇒ \langle v\rangle ⊂ Ker (g∘f)$

I don't know how to do it. I know that $Kerf=\{v ∈ V/f(v)=0\}$.

Answer 1

If $v\in \ker f$ then $f(v)=0$ and $g(f(v)=g(0)=0$.

If $w\in\langle v\rangle$ then $w=kv$ for some scalar $k$, so $g(f(w))=g(f(kv))=kg(f(v))=0$.

That is $w\in\ker(g\circ f)$ and $\langle v\rangle\subset\ker(g\circ f)$.