Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider $M_2(\mathbb{Q})$. Two matrices $A,B$ from this domain are similar if $ A = CBC^{-1}$ for some $ C \in GL_2(\mathbb{Q})$. How would I go about finding representatives for the similarity classes of matrices $A$ of order 6?

share|improve this question

1 Answer 1

up vote 4 down vote accepted

If a matrix has order $6$, then it satisfies the polynomial $$t^6-1 = (t^3-1)(t^3+1) = (t-1)(t+1)(t^2+t+1)(t^2-t+1).$$ Hence the minimal polynomial, which must be of degree at most $2$, is either $t-1$, $t+1$, $t^2-1$, $t^2+t+1$, or $t^2-t+1$. It cannot be $t-1$, $t+1$, or $t^2-1$ (that would mean the matrix has order $1$ or $2$).

Added. And as pointed out by Jyrki Lahtonen and Robert Israel (and overlooked by me), if the minimal polynomial is $t^2+t+1$, then the minimal polynomial divides $t^3-1$, so the matrix will have order $3$.

So you are looking for matrices with minimal polynomial $t^2-t+1$; this is also their characteristic polynomial.

So you just need to figure out the rational canonical form of matrices with characteristic polynomial $t^2-t+1$.

share|improve this answer
If it was $t^2+t+1$, the matrix would have order $3$. –  Robert Israel Mar 30 '12 at 19:23
@RobertIsrael: Oops, quite so. Silly me. –  Arturo Magidin Mar 30 '12 at 19:24
@Jyrki: Quite so, silly me. –  Arturo Magidin Mar 30 '12 at 19:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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