From S.L linear algebra:
Let $L:V \rightarrow W$ be a linear map from one vector space to another. Then show that $L(O)=O$.
($O$ is a null vector).
There are two axioms for a linear map:
$\forall u, v \in V,\,L(u+v)=L(u)+L(v)$
$\forall c \in K,\, \forall v \in V,\,L(cv)=cL(v)$
($V$ is a vector space $K^n$; $L$ is a linear map $V \rightarrow W$; $K$ is an arbitrary field).
The proof that I've constructed seems fairly simple, but I'm not certain of how convincing it might be:
If $O \in V$ and $u=v=O$, then $L(u+v)=L(O)=L(v)+L(u)$, where $L(v)=L(O)=L(O)-L(u)=O$ and $L(u)=L(O)=L(O)-L(v)=O$, thus every function in the equation has an output of the null vector $O$.
As seen above, I've used the first vector space axiom to show that $L(O)=O$ if the mapping is linear.
Is this proof sufficient? or shall I present deeper analysis (by associating each vector $v\in V$ its coordinate vector $X$ with respect of finite basis $\{v_{1},...,v_n\}$)?
Your proof is correct from the logical point of view, but it can certainly be simplified. First of all, there is no need the put the hypothesis “If $0\in V$”; of course that $V$ has a $0$ vector. It's part of the definition of vector space.
And there is no need to give two (?) new names to $0$. You can just write that\begin{align}0&=L(0)-L(0)\\&=L(0+0)-L(0)\\&=\bigl(L(0)+L(0)\bigr)-L(0)\\&=L(0)+\bigl(L(0)-L(0)\bigr)\\&=L(0)+0\\&=L(0).\end{align}