# Question about rotation $2\times 2$ rotation matrices

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.

• 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? – Tanny Sieben Dec 16 '18 at 15:31
• 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 $B=A_{(\theta+2\pi k)/n}$, $k=1,\dots,n$. – David C. Ullrich Dec 16 '18 at 15:41
• 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? – 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 $$\frac 12\phi + \pi$$.