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For what $n$ natural number does there exist a real $2\times 2$ matrix $A$, such that $A^n = I$?

$n=2,3$ clearly works, because $(-I)^2 = I$, and for $n=3$ we have $\left( \begin{smallmatrix} -2 & 1\\ -3 & 1 \\\end{smallmatrix} \right)$

However, I only found these by luck, and I really don't know how to go about this question. By what method should I try to find the answer to this question?

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    $\begingroup$ Let $A=I_2{{}}$? $\endgroup$ – Git Gud Jan 11 '15 at 21:23
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    $\begingroup$ Do you know about rotation matrices? $\endgroup$ – Johannes Kloos Jan 11 '15 at 21:24
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    $\begingroup$ It's a more interesting question if you ask for the entries to be all integers (and $n$ to be minimal). $\endgroup$ – Robert Israel Jan 11 '15 at 21:25
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    $\begingroup$ I agree with Robert here. Just build $\begin{bmatrix} c & -s \\ s & c\end{bmatrix}$, where $c = \cos (2\pi/n)$ and $s = \sin(2\pi/n)$. But for INTEGER matrices...that's more interesting. $\endgroup$ – John Hughes Jan 11 '15 at 21:28
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    $\begingroup$ I've turned Robert's "integer" version of the problem into its own question: math.stackexchange.com/questions/1100574/… $\endgroup$ – John Hughes Jan 11 '15 at 21:56
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A canonical answer, per the comments: the matrix $$ A = \pmatrix{\cos(2 \pi /n) & -\sin(2 \pi /n)\\ \sin(2 \pi /n) & \cos(2 \pi /n)} $$ will do.

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