When is a $\Bbb Q$-homogeneous bijective function on a $\Bbb Q$-vector space $\Bbb Q$-linear?

Question

Let $V$ be a vector space over $\Bbb Q$. Let $f: V \to V$ be a bijective function such that $f(qx)=qf(x), \forall q\in \Bbb Q, \ \forall x \in V$. Then is $f$ necessarily linear? If not in general, then can we put some condition on $\dim V$, so that any such $\Bbb Q$-homogeneous bijective function $f$ on $V$ becomes $\Bbb Q$-linear?

Answer 1

It is not necessarily linear unless $\dim(V)\leq1$. You can construct a non-linear homogeneous function as follows: Pick any $v\in V\setminus0$ and define $$ f(x) = \begin{cases} 2x&\text{when $x$ is a multiple of $v$}\\ x&\text{otherwise}. \end{cases} $$ This is clearly a homogeneous bijection, but it is easy to check that linearity fails unless the space is spanned by $v$.

Answer 2

No, and it doesn't depend on $\mathrm{dim}V$ as long as it's $\geq 2$ (of course for $\mathrm{dim}V\leq 1$ the answer is yes)

Consider the following equivalence relation on $V$: $x\sim y \iff \exists \lambda \in \mathbb{Q}^*, x=\lambda y$

For each equivalence $C$ pick a nonzero rational $r_C$ and put $f(x) = r_Cx$ for $x\in C$, and then extend $f$ to the whole of $V$.

Now since $\mathrm{dim}V \geq 2$ you can pick $x,y$ two linearly independent vectors.

Choose then $r_C \neq r_{C'}$, and the other $r_D$'s don't matter, just make them nonzero to ensure that $f$ is bijective. Clearly the so constructed $f$ satisfies the condition.

Then if $f$ were linear, $f(x+y) = f(x) + f(y)= r_C x + r_{C'} y$, but also the construction yields that there is $r$ with $f(x+y) = r(x+y)$ so $rx+ry = r_Cx + r_{C'}y$. But as $r_C\neq r_{C'}$ this yields a nontrivial relation between $x,y$, contradicting their linear independence. Hence $f$ isn't linear.

So the only restriction on the dimension you can give to have a positive answer is that it be $\leq 1$