Powers of $2\times 2$ matrices, such that $A^n = I$

Question

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?

It's a more interesting question if you ask for the entries to be all integers (and $n$ to be minimal). — Robert Israel, Jan 11 '15 at 21:25
I do... I didn't realize, but it seems like simply (2pi/n) rotational matrix should work. — Jake1234, Jan 11 '15 at 21:27
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. — John Hughes, Jan 11 '15 at 21:28
Sorry...I was giving an answer for the same question in higher dimensions, which wasn't asked -- I've delete that part. — John Hughes, Jan 11 '15 at 21:31
I've turned Robert's "integer" version of the problem into its own question: http://math.stackexchange.com/questions/1100574/integer-2-times-2-matrices-such-that-an-i — John Hughes, Jan 11 '15 at 21:56

score 1 · Accepted Answer · answered Jan 11 '15 at 23:54

1

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.

answered Jan 11 '15 at 23:54

Ben Grossmann

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Powers of $2\times 2$ matrices, such that $A^n = I$

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