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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 $56$??

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    $\begingroup$ Why don’t you include $I$? $\endgroup$ Commented May 7 at 4:35
  • $\begingroup$ Hmm... what minimal polynomial gives $I$? $\endgroup$ Commented May 7 at 4:36
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    $\begingroup$ That would be $X-I$ $\endgroup$ Commented May 7 at 4:42
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    $\begingroup$ The characteristic polynomial always has degree equal to the dimension of the space, and the minimal polynomial always divides the characteristic polynomial $\endgroup$ Commented May 7 at 4:44
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    $\begingroup$ The minimal polynomial has degree at most the degree of characteristic polynomial, so even $x-1$ or $x^2-1$ or $x^2-x+1$ are candidates. $\endgroup$ Commented May 7 at 4:53

3 Answers 3

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As $\deg m(x) \leq 3$, the following are your candidates for the minimal polynomial:

  • Degree 1: $x-1$, $x + 1$.
  • Degree 2: $x^2 + x + 1$, $x^2 - x + 1$, $x^2 - 1$.
  • Degree 3: $(x-1)(x^2 + x + 1)$, $(x-1)(x^2 - x + 1)$, $(x + 1)(x^2 - x + 1)$, $(x+1)(x^2 + x + 1)$.

$x^2 + x + 1$ 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:

  • Degree $1$: $\{ x - 1, x -1 , x-1\} , \{ x+1, x+1, x+1\}$,
  • Degree $2$: $\{x - 1, x^2 - 1\}, \{ x +1, x^2 - 1\}$,
  • Degree $3$: $\{1 , (x-1)(x^2 + x + 1)\}$, and I'll let you list the other three yourself.

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:

  • One distinct entry: 6 ways.
  • Two distinct entries: 30 ways.
  • Three distinct entries: 20 ways.
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  • $\begingroup$ This looks nice. How come for three distinct entries there aren’t $6\times5\times4=\color{blue}120$ ways? $\endgroup$ Commented May 7 at 15:04
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    $\begingroup$ Ordering of the roots along the diagonal doesn't matter, so 20 comes from 6 choose 3. $\endgroup$
    – koifish
    Commented May 8 at 0:50
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As mentioned in the comment, the minimal polynomial of a $3\times 3$ matrix may have degree less than $3$. And even if the minimal polynomials are the same, the matrices might still not be similar. For example, $\operatorname{diag}(1, 1, -1), \operatorname{diag}(1, -1, -1)$ both have $(x+1)(x-1)$ as their minimal polynomials.

Two rational matrices are similar over $\mathbb Q$ iff they are similar over $\mathbb C$ (cf. Similar matrices and field extensions).

Since $x^6-1$ has no repeated roots, $A$ can be diagonalized (over $\mathbb C$), hence its class up to similarity (over $\mathbb C$ hence $\mathbb Q$) only depends on its eigenvalues. And since $3$ is odd, $A$ has at least one real eigenvalue which can only be $\pm 1$. Also if $\zeta$ is an eigenvalue of $A$, so is $\bar\zeta$. Therefore either all eigenvalues of $A$ are in $\{\pm 1\}$, or $A$ has one eigenvalue in $\{\pm 1\}$ and a pair of imaginary eigenvalues.

If $A$ has only possibly $\pm 1$ as eigenvalues, it's a matter of distributing the multiplicity $3$ and there are ${3+1\choose 1}=4$ possibilies according the stars and bars. (The representatives are $I_3, -I_3, \operatorname{diag}(1, 1, -1), \operatorname{diag}(1, -1, -1)$)

If $A$ has a pair of imaginary eigenvalues, there are two pairs that can be chosen. Also the real eigenvalue has two choices as well, hence in total there are $2\times 2=4$ choices.

So in total there are $8$ possibilities.

If the matrix is allowed to be over $\mathbb C$, the problem is simply to distribute the multiplicity $3$ to $6$ possible eigenvalues, and by the stars and bars again, the answer is ${3+5 \choose 3}=56$.

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Another perspective on this question, this is asking for the number of isomorphism classes of representations of the cyclic group of order $6$ of dimension $3$. By general theory, all such representations are semi simple, and (by computation, or general cyclic group theory) we have two one dimensional irreps, and two irreps of dimension $2$. By semi simplicity, we can compute the isomorphism class in the Grothendieck ring, with this basis of simples. So it's the number of solutions to $(a,b,c,d)$ for nonnegative integers $a,b,c,d$ with $a+b+2c+2d=3$, which gives the same answer of $8$.

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