# Automorphism on a finite dimensional vector space

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!?

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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 $\mathbf{Q}$ :-)] – Jyrki Lahtonen Aug 15 '11 at 8:00
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. – Asaf Karagila Aug 15 '11 at 8:47

First, the statement should be that there isn't such an automorphism.

Second, you should review the definition of automorphisms on a vector space, and generally the definition of vector space and of operations on vectors.

Third, a hint: what would the determinant of such an automorphism be? (Recall that since determinants don't change under conjugation, the determinant is an invariant of the linear map, independently of choices of basis.)

Fourth: when you do this exercise, identify exactly where you used that the dimension is 3 and make sure you understand for that dimensions $d$ your proof works.

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What do you mean by the determinant of the automorphism? Do you mean by considering the determinant of the matrix $A$ representing $\phi$ ? In this case $A$ is a $3\times3$ matrix, so $det(A)=\frac{\pm 1}{\sqrt {2}}$ irrational! – Emma Aug 15 '11 at 9:56
@Emma The determinant of an automorphism is the determinant of any matrix representing it (I commented in the post why it is well-defined). Are you sure you followed my Fourth point in your solution? What is the determinant of $2A$? – Alex B. Aug 15 '11 at 12:15
It should be $det(A)=\frac{\pm 1}{2\sqrt{2}}$. – Emma Aug 15 '11 at 15:01

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.

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Like Alex said, there is no such automorphism. If you replace the condition $\phi^{-1} = 2\phi$ with $2\phi^2 = id$ you can have such a linear application for $\mbox{dim}\, V = 2$ or for $\mbox{dim}\,V=4$ (in general, for even dimensional vector spaces), but not for $\mbox{dim}\,V = 3$. Anyway, it won't be an automorphism.

Suppose there were such a $\phi$, the conditions on $\phi$ imply that the minimal polynomial of $\phi$ divides (and hence is) $x^2 - \frac{1}{2}$. So, your only eigenvalues are $\frac{1}{\sqrt{2}}$ and $\frac{1}{\sqrt{2}}$.

Therefore the characteristic polynomial of $\phi$ is either $$\left(x - \frac{1}{\sqrt 2}\right)^2 \left(x + \frac{1}{\sqrt 2}\right)$$ or $$\left(x - \frac{1}{\sqrt 2}\right) \left(x + \frac{1}{\sqrt 2}\right)^2$$ none of which has rational coefficients.

Edit I just realized that Alex's determinant argument is much easier. Well, consider this an alternative argument which works also if we don't require $\phi$ to be an automorphism and replace the condition $\phi^{-1} = 2 \phi$ with $2\phi^2 = id$.

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