3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

Look for instance at:

https://www.maa.org/sites/default/files/pdf/cms_upload/Square_Roots-Sullivan13884.pdf

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...

$\endgroup$
4
$\begingroup$

"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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

$\DeclareMathOperator{\Tr}{Tr}$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 $\frac 12\phi + \pi$.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.