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I am trying to find all possible real $2×2$ matrices $A$ such that $A^{2019} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$

I am really stuck with this one, could someone provide some insight to help me solve it? Thanks!

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Hint Let $A= \begin{bmatrix} x & y \\ z &t \end{bmatrix}$. Then $$\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\begin{bmatrix} x & y \\ z &t \end{bmatrix}=A^{2019}A=AA^{2019}=\begin{bmatrix} x & y \\ z &t \end{bmatrix}\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$$

This gives, $$\begin{bmatrix} z & t \\ -x & -y \end{bmatrix}=\begin{bmatrix} -y & x \\ -t & z \end{bmatrix}$$ This implies $z=-y$ and $t=x$, and hence $A$ must be of the form $$\begin{bmatrix} x & y \\ -y &x \end{bmatrix}$$

Now, identify $A$ with a complex number.

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We may identify $$ \begin {bmatrix}a&b\\-b&a\end{bmatrix}$$ with the complex number $$a+bi$$

Thus the matrix $$ \begin {bmatrix}0&1\\-1&0\end{bmatrix}$$ will be identified with $i$

The problem is finding all $2019$ complex numbers $z$ such that $$z^{2019}=i=e^{i \pi /2 }$$

These are $$\cos (\theta_k ) + i \sin (\theta_k)$$ for $$\theta_k = \frac {\pi/2 +2k\pi }{2019}$$ for $k=0,1,2,...,2018$

The matrices then are $$ \begin {bmatrix}\cos \theta_k&\sin \theta_k\\-\sin \theta_k&\cos \theta_k\end{bmatrix}$$

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    $\begingroup$ Why are you assuming that $A$ has that particular form? $\endgroup$ – N. S. Oct 19 at 0:02
  • $\begingroup$ We are talking about roots of $i$ $\endgroup$ – Mohammad Riazi-Kermani Oct 19 at 0:03
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    $\begingroup$ The question is asking to find all such $A$. You are assuming that $A$ has that given form... Note that $\begin{bmatrix} 2 &1 \\ -3 &-2\end{bmatrix}^2=I_2$ but that matrix is NOT a corresponding to a complex root of unity.... The fact that $A^{2019}$ corresponds to a complex number, doesn't necessarily mean that $A$ corresponds to a complex number, if it does you need to prove it. $\endgroup$ – N. S. Oct 19 at 0:05
  • $\begingroup$ You are correct, I should have proved that the solutions have that particular form. Thanks for your informative comment. $\endgroup$ – Mohammad Riazi-Kermani Oct 19 at 0:21
  • $\begingroup$ In particular, does it also work to identify $ \begin {bmatrix}0&1\\-1&0\end{bmatrix}$ with $-i$ instead of with $i$? And if it does, do you get any additional solutions? $\endgroup$ – Robert Shore Oct 19 at 2:49

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