Rational Canonical Form: Example #1

we discussed the rational canonical form (RCF) of a linear transformation, and we mentioned that any two similar linear transformations have the same RCF. It's this fact which allows us to classify distinct linear transformations on a given $F$-vector space $V$ for some field $F$. Today, to illustrate this, we'll work through a concrete example:
 
  

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Find representatives for the distinct conjugacy classes of matrices of finite order in GL$_2(\mathbb{Q})$.

(Notice we're asked to classify matrices instead of linear transformations. This makes little difference since linear transformations on $V$ are in one-to-one correspondence with $n\times n$ matrices with entries in $F$. And one can show (see Dummit and Foote's Abstract Algebra section 12.2 Theorem 17) that any two similar matrices also share the same RCF.)

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Set-Up

Let $g\in \text{GL}_2(\mathbb{Q})$ be a matrix satisfying the polynomial $x^n-1$ over $\mathbb{Q}$ for some $n\geq 1$. It suffices to list all the possible rational canonical forms (RCFs) of $g$ since any two matrices which are similar (i.e. which live in the same conjugacy class) have the same RCF. Finding the RCF amounts to finding the possible invariant factors $a_1(x),\ldots,a_d(x) \in \mathbb{Q}[x]$ such that our $\mathbb{Q}$-vector space $V$, on which the linear transformation represented by $g$ acts, is isomorphic to
 
 $$\mathbb{Q}[x]/(a_1(x))\oplus\cdots\oplus \mathbb{Q}[x]/(a_d(x))$$
 
 where $a_1(x)\mid\cdots\mid a_d(x)$. (For more on this, see

Let $m_g(x)$ and $c_g(x)$ denote the minimal and characteristic polynomials of $g$, respectively. We can find the invariant factors by keeping in mind that the following must hold*:
 
 

• $a_1(x)\mid\cdots\mid a_d(x)$
• $m_g(x)=a_d(x)$ is the largest (with respect to divisibility) invariant factor
• $c_g(x)=a_1(x)\cdots a_d(x)$ is the product of all the invariant factors
• both $m_g(x)$ and $c_g(x)$ have the same roots and $m_g(x)\mid c_g(x)$
• $\deg(c_g(x))=2$ since $g$ is a $2\times 2$ matrix

These bullets allow us to narrow down the possible RCFs for $g$.

Solution

In what follows, let $\mathscr{C}_{a(x)}$ denote the companion matrix of the polynomial $a(x)\in \mathbb{Q}(x)$. We proceed by considering the various values of $n$.

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Case 1: $n=1$.

  In this case $g=I$ is the identity matrix with $m_g(x)=x-1$. The characteristic polynomial must be $c_g(x)=(x-1)^2$ which means we have two invariant factors: $a_1(x)=a_2(x)=x-1$. The corresponding RCF is $$\begin{pmatrix}  \mathscr{C}_{x-1} & 0\\ 0 & \mathscr{C}_{x-1} \end{pmatrix} =
\begin{pmatrix}  1 & 0\\ 0 & 1 \end{pmatrix} .$$

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Case 2: $n\geq 2$.

 If $n$ is prime, we need only consider the cases when $n=2$ and $n=3$. In fact, this holds in more generality**: If $g\in \text{GL}_n(\mathbb{Q})$ has prime order $p$, then $p\leq n+1$. (For the proof, see the second footnote at the bottom.)

 If $n=2$, then $x^2-1=(x+1)(x-1)$ and we have two cases: either $m_g(x)=c_g(x)=x^2-1$, or $m_g(x)=x+1$ and $c_g(x)=x^2+2x+1$. In the first case, the RCF of $g$ is $$\mathscr{C}_{x^2-1}= \begin{pmatrix}  0 & 1\\ 1 & 0 \end{pmatrix}.$$ In the second case, the invariant factors are $x+1$ and $x+1$ and the RCF is $$\begin{pmatrix}  \mathscr{C}_{x+1} & 0\\ 0 & \mathscr{C}_{x+1} \end{pmatrix} =
\begin{pmatrix}  -1 & 0\\ 0 & -1 \end{pmatrix}.$$

 If $n=3$, we have $x^3-1=(x-1)(x^2+x+1)$ and thus $m_g(x)=c_g(x)=x^2+x+1$ so that $g$ has RCF $$\mathscr{C}_{x^2+x+1}=\begin{pmatrix}  0 & -1\\ 1 & -1 \end{pmatrix} .$$ (Notice this is the only possibility since $\Phi_3(x)=x^2+x+1$ is irreducible over $\mathbb{Q}$.)

 Now if $n>3$ and $n$ is not prime, we factor $x^n-1$ as
$$x^n-1=\prod_{d\mid n}\Phi_d(x)$$
to see that finding the possible RCFs for $g$ reduces to finding all
cyclotomic polynomials $\Phi_d(x)$ where $d\mid n$. For $n>3$, there are only two such possibilities: $\Phi_4(x)=x^2+1$ and $\Phi_6(x)=x^2-x+1$. This yields two more possible RCFs (since both $\Phi_4(x)$ and $\Phi_6(x)$ are irreducible). If $m_g(x)=c_g(x)=\Phi_4(x)$ then $$\mathscr{C}_{\Phi_4(x)}= \begin{pmatrix}  0 & -1\\ 1 & 0 \end{pmatrix}.$$ And if $m_g(x)=c_g(x)=\Phi_6(x)$ then $$\mathscr{C}_{\Phi_6(x)}=  \begin{pmatrix}  0 & -1\\ 1 & 1 \end{pmatrix}.$$

  So in sum, there are six possible distinct conjugacy classes for matrices of finite order in GL$_2(\mathbb{Q})$:

 
     

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Footnotes:

* See Dummit and Foote, section 12.2 Proposition 20.
 

**  Recall that the characteristic polynomial of an $n\times n$ matrix is of degree $n$ and is divisible by the minimal polynomial. By assumption $g^p=I$ which implies that the minimal polynomial of $g$ must divide $x^p-1=(x-1)(x^{p-1}+x^{p-2}+\cdots+x+1)$. Since  $g$ is not the identity we know $m_g(x)\neq x-1$ (i.e. $g-I\neq 0$) and we must have $m_g(x)=x^{p-1}+x^{p-2}+\cdots+x+1$, the irreducible cyclotomic $p$th polynomial. Since deg$(m_g(x))\leq n$, we conclude $p-1\leq n $, i.e. $p\leq n+1$.