Suppose to have two real vector spaces $V$ and $W$ and an injective map $T:V\rightarrow W$ such that $T(\alpha v)=\alpha T(v)$ for all $v\in V$ and $\alpha \in\mathbb{R}$. Do there exist some conditions on the vector spaces or the map which guarantee that $T$ is also additive, that is $T(v+u)=T(v)+T(u)$?



When a homogeneous map between vector spaces is also additive?
If $V$ is one dimensional, every homogeneous map $T:V\to W$ is necessarily linear and thus additive.
If $W$ is one dimensional, it follows from the injectivity assumption on $T$ that $\dim(V)\leq1$ and so every homogeneous map is additive.


Let us then assume that both spaces have dimension two or higher but finite.
If you are willing to accept the axiom of choice, there is always a homogeneous bijection $T:V\to W$ which is nonadditive.
To see this, let $H\_V$ and $H\_W$ be unit hemispheres (with respect to any norm) on the two spaces.
That is, for any point $x$ of unit norm on $V$ exactly one of $x$ and $-x$ is in $H\_V$, and similarly in $W$.
These sets have the same cardinality, so there is a bijection $f:H\_V\to H\_W$.
Now if we define $T(rv)=rf(v)$ for all $v\in S\_V$ and $r\in\mathbb R$, we get a homogeneous bijection.
If it happens that $T$ is additive, we can always modify $f$ so that $T$ becomes nonadditive, for example by swapping the images of two points of $H\_V$.


If $2\leq\dim(V)\leq\dim(W)$, the same construction can be done without choice so that $T$ is continuous.
To do this, it suffices to produce such a map $T:\mathbb C\to\mathbb C$; we can write the spaces as $V=\mathbb R^2\times\mathbb R^n$ and $W=\mathbb R^2\times\mathbb R^n\times\mathbb R^m$ and then let $(x,y)\mapsto(Tx,y,0)$.
For $0\leq\theta<\pi$ and $r\in\mathbb R$ let $T(re^{i\theta})=re^{i\sqrt{\pi\theta}}$.
This mapping is homogeneous and bijective but not additive.


