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$\begingroup$

It suffices just to consider a linear transformation $f$ such that $f^n=id$ and require $V$ to have no proper subspace invariant under $f$. But I still don't have a picture of what's going on.

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    Hint: rotations.2012-10-23
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    Hi @nik do you mean $V$ must be less than 2 dimensional?2012-10-23
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    It depends on $n$. If $n = 1,2$, then you can easily find a representation of degree 1; otherwise, if you want to find all the representations, for some of them you will need $V$ to be of dimension at least 2 (and in fact 2 is enough).2012-10-23
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    You mean if the dimension is greater than 2 I can always decompose into direct sums of 1 or 2 dimensional subspaces? Why2012-10-23
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    How much do you know about linear algebra and representations? One possible answer is that if $f \in \mathrm{GL}(V)$ is such that $f^n = id$, then $f$ is diagonalizable over $\mathbb{C}$; its eigenvalues are roots of unity, and it's then easy to show that the eigenvalues are either real or come in pairs of conjugate numbers. Then you're left with a decomposition of $f$ as a direct sum of $\pm 1 \in \mathrm{GL}(\mathbb{R})$ and of rotations $\in \mathrm{GL}(\mathbb{R}^2)$.2012-10-23
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    @NajibIdrissi Please consider converting your comment into an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2015-11-07
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    @JulianKuelshammer Done.2015-11-08

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