# Real $2\times2$ matrices of finite order and rotational matrices

Let $$M$$ be a real $$2\times2$$ matrix $$M = \left(\begin{matrix}a & b \\ c & d\end{matrix}\right)$$ Suppose $$M$$ has a finite order (thus there is some natural $$n$$ that $$M^n=E$$, where $$E$$ is identity matrix), is it necessary a rotational matrix $$R_\varphi$$ where $$\varphi = \frac{m}k\pi$$?

First of all I noticed, that $$\det(M)=\pm1$$. Consider now only $$+1$$, thus $$ad-bc=1$$. I thought that if there is at least one real eigenvalue $$\lambda$$, then there is at least one real eigenvector, then the whole plane stretches to that direction. It means that constant multiplication by $$M$$ would not reverse this stretching. Thus eigenvalues must be complex: $$\lambda^2 - (a+d)\lambda + ad-bc = 0$$ $$\lambda^2 - (a+d)\lambda + 1 = 0$$ $$\lambda \notin \mathbb{R} \Leftrightarrow|a+d| < 2$$ Rotational matrix $$R_\varphi$$ has $$a = d$$, but the condition above doesn't have this restriction. So it is not hard to construct a matrix with determinant 1 and with $$a\ne d$$, for example $$M = \left(\begin{matrix}\frac34 & \frac12 \\ -\frac12 & 1\end{matrix}\right)$$ Does this matrix has a finite order too? I think not. What am I missing? Is it true, that finite order $$\Rightarrow$$ rotation on the rational fraction of $$\pi$$? How to derive this fact?

• When you conclude "Thus eigenvalues must be complex.", why can't you have eigenvalues $\pm1$? Also, don't forget about reflections. – Servaes Feb 12 at 17:22
• @Servaes If only one of eigenvalue is equal to $\pm1$, then the other must be real and not equal to $\pm1$, then $\mathbb{R}^2$ stiil stretches across second eigenvector. If both eigenvalues are $1$, then it is $E$, i guess? – grentank Feb 12 at 17:31
• The determinant is equal to the product of the eigenvalues. If $\det(M)=\pm1$ and one eigenvalue is $\pm1$, then the other eigenvalue must also be $\pm1$, with an appropriate choice of sign. – amd Feb 12 at 19:20

You made a good start, but then went a bit astray when you concluded that if $$\det(M)=1$$, then its eigenvalues must be complex. This is clearly not true, though: $$E^n=E$$ for all $$n$$, and its only eigenvalue is $$1$$.
$$M$$ having finite order $$n$$ means that the polynomial $$\lambda^n-1$$ annihilates $$M$$. Since the minimal polynomial of $$M$$ divides this polynomial, this in turn implies that the only possible eigenvalues of $$M$$ are $$n$$th roots of unity. The characteristic polynomial of $$M$$ is quadratic, so its eigenvalues are either both real or both complex.
There are three cases to consider for real eigenvalues: repeated $$1$$, repeated $$-1$$ and $$\{1,-1\}$$. (Note that we can only have $$-1$$ as an eigenvalue if $$n$$ is even.) In the first case, $$M=E$$ and in the second, $$M=-E$$ (why?), which can be interpreted as a rotation through an angle of $$\pi$$. If we have both $$1$$ and $$-1$$ as eigenvalues, then $$M^2=E$$ (why?), which is one of the hallmarks of a reflection. Looking at it geometrically, $$M$$ reverses one direction and leaves another fixed. Note that this is the only case in which $$\det(M)=-1$$, so if you require that $$\det(M)=1$$, then you can exclude reflections.
Unfortunately, even this additional hypothesis doesn’t get you pure rotations only. If $$M$$ has complex eigenvalues, they must be a complex conjugate pair $$e^{\pm i\phi}$$, and their product is $$1$$. In this case, $$M$$ is similar to (but not necessarily equal to!) a matrix of the form $$R_\phi = \pmatrix{\cos\phi&-\sin\phi\\\sin\phi&\cos\phi}.$$ Why not equal? Well, if $$R_\phi^n=E$$ and $$P$$ is any invertible $$2\times2$$ matrix, then $$\left(PR_\phi P^{-1}\right)^n = PR_\phi^nP^{-1} = E$$ as well. Matrices of the form $$PR_\phi P^{-1}$$ are known as conjugate rotations. Geometrically, $$R_\phi\mathbf v$$ traces out a circle as $$\phi$$ varies (with $$\mathbf v\ne0$$, of course), while $$PR_\phi P^{-1}\mathbf v$$ traces an ellipse.