0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ When you conclude "Thus eigenvalues must be complex.", why can't you have eigenvalues $\pm1$? Also, don't forget about reflections. $\endgroup$ – Servaes Feb 12 at 17:22
  • $\begingroup$ @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? $\endgroup$ – grentank Feb 12 at 17:31
  • $\begingroup$ 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. $\endgroup$ – amd Feb 12 at 19:20
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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