Find 10 commuting $2\times 2$ matrices of the same order

Question

Prove that there exists 10 distinct real $2\times 2$ matrices which are pairwise commuting and all of the same finite order.

Here, the order of matrix A is the smallest integer $k > 0$ such that $A^k = I.$ Also by 'pairwise commuting', I mean that if $\{A_1, \cdots, A_{10}\}$ are the ten matrices, then $A_i A_j = A_j A_i$ for any $i,j = 1,2,\cdots, 10$ and $i\neq j.$

Linear algebra is not my forte yet, I haven't got a clue where to begin. The matrices commute so they all have the same eigenvectors... so letting $A_1$ be a matrix of just these eigenvectors, $A_2 = A_1 + I$ would have the same eigenvectors, thus $A_2A_1 = A_1A_2.$ However I am not sure how to deal with making them the same order.

All you need is one single matrix of finite order $n$ with $\varphi(n) \geq 10$. — MooS, Apr 25 '16 at 10:16

score 4 · Accepted Answer · answered Apr 25 '16 at 10:13

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First Hint. Define $$R(\theta) = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos\theta\end{bmatrix}$$

Then $R(\theta)$ represents the linear transform that rotates vectors counter-clockwise by an angle $\theta$.

Second Hint. $11$ is prime.

answered Apr 25 '16 at 10:13

goblin GONE

67,744

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This is a cool question, by the way. – goblin GONE Apr 25 '16 at 10:16
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Jeez I'm surprised at the simplicity of the answer. Thanks! – Merkh Apr 25 '16 at 10:18

Find 10 commuting $2\times 2$ matrices of the same order

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