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I wonder if there exists a matrix $A\neq I$ such that $A^n=1$ for any $n>1$ (positive I guess)? I think of rotation by $2\pi/n$, and the rotation matrix applied $n$ times gives the identity matrix?

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    $\begingroup$ Well, if $A \ne I$ and $A^n=I$ then $A^{n+1} = A \ne I$. $\endgroup$ – Zoran Loncarevic Dec 14 '16 at 11:51
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    $\begingroup$ It cannot work for any $n$ (just choose $n=1$). Perhaps you meant for some $n$? $\endgroup$ – Martin R Dec 14 '16 at 11:53
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    $\begingroup$ Do you mean, for any $n>1$ there is matrix $A_n \ne I$ such that $A_n^n=I$? In that case, the answer is yes, for example $$A_n = \begin{bmatrix} \cos \frac{2\pi}{n} & -\sin \frac{2\pi}{n} \\ \sin \frac{2\pi}{n} & \cos \frac{2\pi}{n} \end{bmatrix}$$ $\endgroup$ – Zoran Loncarevic Dec 14 '16 at 12:02
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For all positive $n$ simultaneously is impossible: for one $$ A^1 \neq I, $$ and also if $A^n=I$ for some $n \in \mathbb N$ then $$ A^{n+1}=A^n\cdot A=A\neq I. $$

But if you want it to hold for some $n$ then the answer is yes: $$ A= \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix} \quad \text{and} \ A^n=I, \quad \text{for all even} \ n \in \mathbb N. $$

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Yes, for every $n$, a rotation by $2\pi/n$ has the property that $A^n=I$.

In a certain sense this is the only solution: One can prove that if $A$ is a real matrix satisfying $A^n=I$, then $A$ is similar to a block diagonal matrix where each block is either a 2×2 rotation by some multiple of $2\pi/n$, or $I_{1\times 1}$, or (if $n$ is even) $-I_{1\times1}$.

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If $A$ is the rotation by $2\pi/n$, then we indeed have $A\neq I$ and $A^n = I$.

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