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Given V a vector space with vectors and scalars $\mathbb{C}$, does there exists a non linear transformation $T:V\rightarrow V$ such that $T(x+y)=T(x)+T(y)$ for all $x,y\in V$?

I think such a transformation will be 'like' one that satisfies Cauchy's functional equation $f(x+y)=f(x)+f(y)$ without any other conditions, but other than that, I have no idea.

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At the end of the first paragraph, I think you mean for all $x, y \in V$. – Michael Albanese Oct 11 '12 at 14:24
edited, thanks! :) – Ivan Wangsa Oct 11 '12 at 14:25
Hint: With $\mathbb C$ regarded as a one-dimensional vector space over itself, does $T(x) = \text{Re}(x)$ define a linear transformation or a nonlinear transformation from $\mathbb C$ to $\mathbb C$? – Dilip Sarwate Oct 11 '12 at 14:28
Thanks for the hint (although it might as well be an answer) :D – Ivan Wangsa Oct 11 '12 at 14:31
Without the assumption that $T(kx) = kT(x)$ for all $k \in \mathbb{C}$, you still have $T(kx) = kT(x)$ for all $k \in \mathbb{Q}$ (fun exercise). – Michael Albanese Oct 11 '12 at 14:34

The function $T \colon V \to V$ is a homomorphism of abelian groups or equivalently $\mathbb{Z}$-modules. If we use Zorn's Lemma we can realise $V$ as a freely generated module over $\mathbb{Z}$ with generators $X$ say. Then $X$ is uncountable. Picking $X$ is like picking a basis of a complex vector space. You can define every $T$ by taking any function $X \to V$ and extending it to the whole of $V$ to be a homomorphism like you extend any map of a basis to be a linear map. Unfortunately I don't think there is any constructive way of finding $X$ so all this shows is there are lots of maps $T$.

Might these all be linear ? No as you can arrange them not to be. As $X$ is uncountable you can choose $\dim(V) + 1$ distinct generators from $X$. These must be linearly dependent so arrange $T$ to not be linear on them and then extend it to the rest of $X$ arbitrarily.

EDIT: Previous comment on automorphisms of $\mathbb{C}$ removed as not really relevant.

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