# Is there a additive non-linear non-complex map?

Let $V$ and $W$ be vector spaces over a field $\mathbb{F}\neq \mathbb{C}$. Give an example of a non-linear map $T:V\to W$ such that $$T(x+y) = T(x)+T(y), \forall x,y\in V.$$

I asked myself this question when I was resolving an excercise list os Linear Algebra. This example is pretty easy when $\mathbb{F}=\mathbb{C}$. We take $V=W=\mathbb{C}$ and $T:z\mapsto \bar{z}$, and we have that $T(\lambda z) = \bar{\lambda}T(z)$. However, I couldn't find any examples for non-complex vector spaces.

• Possible duplicate of Does a nonlinear additive function on R imply a Hamel basis of R? Commented Jan 17, 2017 at 21:48
• @FlybyNight but that's specifically asking about $\Bbb R$ Commented Jan 17, 2017 at 21:49
• @Omnomnomnom The OP said they "couldn't find any examples for non-complex vector spaces." The reals are such an example. A simple Google search of "additive non linear functions" turns up lots of hits, e.g. on stackexchange, stackoverflow and some other sites. Commented Jan 17, 2017 at 21:51
• @FlybyNight still, there is no one source that answers the question sufficiently. That seems like enough of a reason to leave it open Commented Jan 17, 2017 at 21:52

Every such map with $\Bbb F = \Bbb Q$ will be linear. The same will be true for finite fields.
With $\Bbb F = \Bbb R$, this may or may not be true, depending on your axioms (see the wiki page for more).
If $\Bbb F$ is an extension of $\Bbb Q$ with degree $1<[\Bbb F: \Bbb Q] < \infty$, then such maps exist (for example: $\Bbb Q(\sqrt{2})$). This same idea works if $\Bbb F$ satisfies $1 < [\Bbb F: \Bbb F'] < \infty$ for any subfield $\Bbb F'$, for that matter.
This is just a slight variation on your counterexample. Consider the field $\mathbb{Q}(\sqrt{2}) = \{a+b\sqrt{2}\mid a,b\in\mathbb{Q}\}$. Then $T: \mathbb{Q}(\sqrt{2}) \to \mathbb{Q}(\sqrt{2}),\ a+b\sqrt{2}\mapsto a-b\sqrt{2}$ is such a non-linear map.
A discontinuous additive function $a\colon\Bbb R\to\Bbb R$ is non-linear map of a linear space $\Bbb R$ over a field $\Bbb R$ to itself. Observe that any additive map is linear, if the field of scalars is $\Bbb Q$.