I am working on the following question in review for my algebra final:
Classify, up to similarity, the $3$ by $3$ matrices with coefficients in $\mathbb{Q}$ that satisfy $A^6=I$.
My work:
As $A^6 – I = 0$, we know that the minimal polynomial of $A$, $m_A(x)$ divides $x^6 – 1$. We factor
$$x^6 – 1 = (x^3 – 1)(x^3 + 1) = (x – 1)(x^2 + x + 1)(x + 1)(x^2 – x + 1).$$
Thus, the possible minimal polynomial must have degree of the matrix (which is $3$), so $m_A(x)=$
$1$. $(x-1)(x^2 + x + 1)$
$2$. $(x-1)(x^2 – x + 1)$
$3$. $(x+1)(x^2 – x + 1)$
$4$. $(x+1)(x^2 + x + 1)$
These have rational canonical forms
$$\begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & -1 \\
0 & 1 & -1
\end{pmatrix} \quad \begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & -1 \\
0 & 1 & 1
\end{pmatrix} \quad \begin{pmatrix}
-1 & 0 & 0 \\
0 & 0 & -1 \\
0 & 1 & 1
\end{pmatrix} \quad \begin{pmatrix}
-1 & 0 & 0 \\
0 & 0 & -1 \\
0 & 1 & -1
\end{pmatrix}$$
Is this correct? Edit: I forgot $I$.
I am also trying to find the number of classes if the matrices are over $\mathbb{C}$. My idea is that is is $6$ choose $3$ because over $\mathbb{C}$ there are 6 total roots, but the correct answer is apparently $56$??
Best Answer
As $\deg m(x) \leq 3$, the following are your candidates for the minimal polynomial:
$x^2 + x + 1$ is impossible because there is no polynomial in $\mathbb{Q}[x]$ of degree $1$ dividing it. (Recall that in your Smith Normal Form, your invariant factors $f_1 \mid f_2 \mid \dots \mid f_{k} $ satisfy $f_1 f_2 \dots f_k = c(x)$ and $f_k = m(x)$.) Similar for $x^2 - x + 1$.
This gives rise to the following eight invariant factors:
Since the number of conjugacy classes is in bijection with the number of invariant factors, there are eight such conjugacy classes.
For the case over $\mathbb{C}$, since $c(x)$ has distinct roots, your minimal polynomial always splits, and hence $A$ is diagonalizable. Then split by cases based on the number of distinct entries on the diagonal. You should get: