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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?

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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 $t^2-t+1$.

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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

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