6
$\begingroup$

An earlier question today motivates this slight variant:

For what natural numbers $n$ does there exist a non-identity integer $2\times 2$ matrix $A$, such that $A^n = I$? (And let's say $A^k \ne I$ for $|k| < |n|$, too.)

Clearly there are solutions for $n=2,3$, because $(-I)^2 = I$, and for $n=3$ we have $\left( \begin{smallmatrix} -2 & 1\\ -3 & 1 \\\end{smallmatrix} \right)$

The solution for $n = 3$ suggests to me that something involving the euclidean algorithm might come into play...and the fact the the determinant is $\pm 1$ suggests that this is really an $SL(2, \mathbb Z)$ (or $PSL(2, \mathbb Z)$)problem...but that's a group I'm woefully ignorant about.

For $n = 4$, there's $\left( \begin{smallmatrix} 0 & -1\\ 1 & 0 \\\end{smallmatrix} \right)$. And right about there I run out of ideas.

$\endgroup$
10
$\begingroup$

For $n=6$ there's $\left( \begin{smallmatrix} 2 & -1\\ 3 & -1 \\\end{smallmatrix} \right)$, not surprisingly. And these are the only such $n$, even if you allow rational entries instead of just integers.

Suppose that $A^n=I$ but $A^k\ne I$ for $k<n$. Then the minimal polynomial of $A$ divides $x^n-1$ but doesn't divide $x^k-1$ for $k<n$. It follows that the minimal polynomial of $A$ divides the $n$th cyclotomic polynomial $\Phi_n(x)$. But that polynomial is irreducible over the rationals, and so the minimal polynomial of $A$ equals $\Phi_n(x)$. Finally, the minimal polynomial of $A$ has degree at most $2$, since it divides the characteristic polynomial of $A$ by the Cayley-Hamilton theorem. Therefore the degre of $\Phi_n$ is at most $2$, which means that $\phi(n)$ is at most $2$, and so $n\in\{1,2,3,4,6\}$.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ Nice answer! And it used only things that I already (sort of) knew, showing that I should have been able to get there myself. $\endgroup$ – John Hughes Jan 11 '15 at 22:24
  • 2
    $\begingroup$ Everything's easier in hindsight :) I do find it a good instinct, whenever I see something about a poynomial expression a matrix, to translate it into a statement about the matrix's minimal polynomial - that tends to trigger the stuff-I-know part of my brain better than the original formulation. $\endgroup$ – Greg Martin Jan 12 '15 at 1:58
3
$\begingroup$

$n=6$ is another possibility: let $A$ be any matrix with characteristic polynomial $X^2 - X + 1$.

There are no other possibilities, because the characteristic polynomial $p$ of $A$ has degree $2$. If $A$ has multiplicative order $n$, then $p$ must be a cyclotomic polynomial, which has degree $2$ only for $n\in \{1,2,3,4,6\}$.

| cite | improve this answer | |
$\endgroup$
-1
$\begingroup$

$$R_{\theta}=\begin{bmatrix} cos \theta & -sin \theta \\ sin \theta& cos \theta \end{bmatrix}\\$$if $$R_{\theta+2k\pi}=R_{\theta}\\(R_{\theta})^n=R_{n\theta}\\\theta=\frac{2k\pi}{n} \rightarrow (R_{\theta})^n=R_{n\theta}=R_{n\frac{2k\pi}{n}}=R_{2k\pi}=I $$

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Perhaps you missed the part about the entries being integers. This is a good answer, however, to the question I linked, which was about real matrices rather than integer ones. $\endgroup$ – John Hughes Jan 11 '15 at 22:23
  • $\begingroup$ Matrices with integer ,similar to R(theta) $$A^n=I\\A=B^{-1}R_{\theta}B $$ $\endgroup$ – Khosrotash Jan 11 '15 at 22:28
  • 1
    $\begingroup$ Sure...*if* you can find such a $B$, then this works. Let's take $n = 8$, so $\theta = \pi/4$. Can you find a matrix $B$ such that $B^{-1} R_\theta B$ actually has all integer entries? The other two answers claim that you cannot, and I think I believe them. $\endgroup$ – John Hughes Jan 12 '15 at 1:50

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.