Q:

If $L(\lambda x+y)=\lambda L(x)+L(y)$ for all $\lambda\in\mathbb{F}$ and $x,y\in V$, then $L$ linear?

If $V$ and $W$ are vector spaces over a field $\mathbb{F}$ and $L\colon V\to W$ is a map that satisfies $$L(\lambda x+y)=\lambda L(x)+L(y)\qquad(*)$$ for all $\lambda\in\mathbb{F}$ and $x,y\in V$. Is it then true that $L$ is linear? I dont think it is. Clearly, if we take $\lambda=1$, we see that $L$ is additive. But $$L(\lambda x)=L(\lambda x+0_{V})=\lambda L(x)+L(0_{V})$$ and (maybe?) $L(0_{V})$ can be non-zero in $W$. So I think linearity is not equivalent to $(*)$. Are there any counterexamples that show that $L(0_{V})\neq0_{W}$ under the assumption $(\ast)$?
Clearly, linearity is equivalent to $L(\lambda x+\mu y)=\lambda L(x)+\mu L(y)$ for all $\lambda,\mu\in\mathbb{F}$ and $x,y\in V$.

A:

Put $x=y=0$ to get $L(0)=\lambda L(0)+L(0)$. Since this is true for all $\lambda $ we get $L(0)=0$. 

A:

Note that $L(0) = L(0+0) = L(0) + L(0)$. Canceling $L(0)$ from both sides gives $L(0) =0$.

