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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!?

  • 3
    Some confusion here? $V$ has no product, so I don't know what $uv$ means? [Edit: I suggested scalar multiplication, but only then realized that the field is $\mathbf{Q}$ :-)]2011-08-15
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    I kept thinking the question was about a finite vector space in the sense of a finite set. I hope you do not mind I have added a word to the title.2011-08-15

3 Answers 3