Suppose $p \in \mathbb{Q}[x]$ is irreducible and $\alpha \in \mathbb{C}$ is a root of $p$. Then consider the linear transformation $T : \mathbb{Q}(\alpha) \rightarrow \mathbb{Q}(\alpha)$, $x \mapsto \alpha x$.
What is the Jordan canonical form of $T$, considering $\mathbb{Q}(\alpha)$ to be a $\mathbb{Q}$ vector space?
Suppose $\deg(p) = n$ and $p = x^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$. Then $\mathbb{Q}(\alpha)$ has dimension $n$ with basis $\{1, \alpha, \alpha^2, \cdots, \alpha^{n}\}$. Moreover, obviously the transformation $T$ acts on the basis element as $\alpha^i \mapsto \alpha^{i+1}$ for $i < n-1$. But $\alpha^{n-1} \mapsto \alpha^n = -a_{n-1}\alpha^{n-1} - \cdots - a_1\alpha - a_0$ in $\mathbb{Q}(\alpha)$.
Thus $T$ is represented by the companion matrix, $C_p$
Not sure what the next steps would be- I'm confusing myself a bit here. It is well known that the companion matrix $C_p$'s characteristic and minimal polynomial is $p$ itself. Moreover, we know that $\alpha$ is thus an eigenvalue, which means that $\bar{\alpha}$ is as well. Thus $(x-\alpha)(x-\bar{\alpha}) | p$. One cannot conclude at this point that since $p$ is irreducible, $(x-\alpha)(x-\bar{\alpha}) = p$, since it is possible that $(x- \alpha)(x-\bar{\alpha}) \in \mathbb{R}[x]\setminus \mathbb{Q}[x]$.
Even if $p = (x-\alpha)(x-\bar{\alpha})$ after all, I'm still a bit confused. Now that we've expanded our scope to $\mathbb{C}$, it's pretty clear that $\alpha$ is the only eigenvalue, so would the JCF simply be $\alpha I$? I'm getting a bit confused with the JCF since we traditionally start with an algebraically closed field and $\mathbb{Q}$ isn't. Would appreciate some help - I believe the intent is to proceed as if $C_p \in \mathbb{C}^{n\times n}$.
Would appreciate some help.
Since $p$ is irreducible every root of it has multiplicity $1$. Hence the Jordan form is a diagonal matrix.