How do I prove that, if for a $2 \times 2$ matrix $A$ and a fixed integer $n> 0$ we have that $$A^n= \begin{bmatrix} \cos{x}& -\sin{x}\\ \sin{x} & \cos{x}\end{bmatrix} ,$$ for some real $x$, then $A$ is also a rotation matrix? I know that it is very obvious, especially if you think about the isomorphism with complex numbers, but I can't seem to come up with a simple and rigorous proof.


Even for the case $A^2=R$ there are many possible roots.

Look for instance at:


With the additional constraint $\det(R)=1$ here you get $$\operatorname{tr}(A)A=R\pm I$$

and have to discuss according values of $x$, whether the trace is zero or not and so on.

  • e.g. $R=I$ and null trace, gives $\begin{pmatrix}a&b\\c&-a\end{pmatrix}$ with $a^2+bc=1$

[ $a=0,b=1,c=1$ is the case presented by David C. Ulrich in his answer ]

For $A^n=R$ you get even more solutions.

Starting with $n=2p$ even, you have to solve $A^p=B$ for every $B^2=R$ with $B$ not necessarily a rotation already...


"Very obvious" or not, it's not true. If $A=\begin{bmatrix}0&1\\1&0\end{bmatrix}$ then $A^2=I$ but $A$ is not a rotation.

  • $\begingroup$ I see. So this means that the nth roots of a 2x2 rotation matrix are not necessarily rotation matrices? How would you then determine the nth roots of a rotation matrix? $\endgroup$ – Tanny Sieben Dec 16 '18 at 15:31
  • 2
    $\begingroup$ I doubt there's a simple algorithm to find all the $n$th roots. For example $I$ has infinitely many square roots, only two of which are rotations. Otoh if you just want $n$ $n$th roots that's easy. Say $A_\theta$ is the rotation through the angle $\theta$; then $B^n=A_\theta$ if $B=A_{(\theta+2\pi k)/n}$, $k=1,\dots,n$. $\endgroup$ – David C. Ullrich Dec 16 '18 at 15:41
  • $\begingroup$ I figured that too; What about if $A^n = R$ where $R$ is a 2x2 rotation matrix different from the identity? Then would A necessarily be a rotation matrix? $\endgroup$ – Tanny Sieben Dec 16 '18 at 15:52

$\DeclareMathOperator{\Tr}{Tr}$For the case $A^2=R$ where $R$ is a 2x2 rotation different from identity, we can apply Sullivan's article quoted by zwim, and find:

If $R=\begin{pmatrix}-1\\&-1\end{pmatrix}$, then $A=\begin{pmatrix}\alpha&\beta\\\gamma&-\alpha\end{pmatrix}$ with $\alpha^2+\beta\gamma=-1$.

If $R$ is any other rotation over an angle $\phi$, then $\displaystyle A=\pm\frac{R+I}{\sqrt{2\cos\phi+2}}$. Working through the possibilities, it means that $A$ is a rotation matrix over either $\frac 12\phi$ or $\frac 12\phi + \pi$.


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