TITLE: Automorphism on a finite dimensional vector space
QUESTION [7 upvotes]: If $V$ is a vector space over $\mathbb Q$ with $\operatorname{dim}(V)=3$. How we can prove that there is NO automorphism $\phi: V \rightarrow V$ such that $\phi^{-1}=2\phi$.
I tried: Let $\phi: V \rightarrow V$ is an automorphism such that $\phi^{-1}=2\phi$, then 
let $v,u\in V$, then since $\phi^{-1}$ is also a homomorphism we have: 
$$\phi^{-1}(uv)=\phi^{-1}(u)\phi^{-1}(v)=2\phi(u).2\phi(v)=4\phi(u)\phi(v)$$
 On the other hand, $\phi^{-1}(uv)=2\phi(uv)=2\phi(u)\phi(v)$, so
$$4\phi(u)\phi(v)=2\phi(u)\phi(v)$$
implies
$\phi(u)\phi(v)=0$, for all $u,v \in V$, which means $\phi=0$.
I think there is something missing in my argument! I didn't need the dimension of $V$ !!
Any suggestion!?

REPLY [8 votes]: If there were such an automorphism, its minimal polynomial would be $X^2-2$. This polynomial being irreducible over $\mathbb Q$, the dimension of $V$ would be even.