Derivations as abstract vector space

Question

I'm trying to prove that the drivations on a manifold are a vector space. For now I've proved that $$(\delta+\lambda \eta) \in \text{Der}(C^{\infty} (M)) \text{ } \delta ,\eta \in \text{Der}(C^{\infty} (M)) \text{ } \lambda \in \mathbb{R}$$ but i'm stuck at proving that there is a zero.

Answer 1

So, a derivation is defined as a linear functional on $C^{\infty}(M)$ satisfying a Leibniz rule, right? So when is a functional equal to zero? Well, it is zero if and only if it maps each function to zero.

Clearly, the functional $z$ which maps everything to zero is a derivation (you may check this yourself) and hence this is exactly your zero vector: it satisfies $\lambda z = 0$ for all $\lambda$ and $v+z = v$ for all $v$.