## Show that two matrices over a finite field are similar

Let $p$ be a prime. Show that the following $p \times p$ matrices are similar in $\mathsf{Mat}_p(\mathbb{F}_p)$: $A = \mathcal{C}_{x^p-1}$ (the companion matrix of $x^p-1$) and $B = J_p(1)$ (the $p \times p$ Jordan block with eigenvalue 1).

Recall that the characteristic polynomial of $\mathcal{C}_{x^p-1}$ is $x^p-1$ (as we showed here), and that over $\mathbb{F}_p$ we have $x^p-1 = (x-1)^p$ (as can be deduced from this precious exercise).

We claim that in fact $x^p-1$ is the minimal polynomial of $A$. Recall (from this previous exercise) that $\mathsf{Sym}(p)$ (the symmetric group on $p$ objects) is embedded in $\mathsf{GL}_p(\mathbb{F}_p)$ by letting a permutation $\sigma$ act index-wise on the elements of an ordered basis (i.e. $\sigma \cdot e_i = e_{\sigma(i)}$). Now $A$ is in the image of this representation, and in fact is the matrix representing the permutation whose cycle decomposition is $\sigma = (1\ 2\ 3\ \ldots\ p-1\ p)$. In particular, $A^{p-1} = A^{-1}$ is not the identity transformation, so the minimal polynomial of $A$ is indeed $x^p-1$.

So the elementary divisors of $A$ are just $(x-1)^p$, and so $A$ is similar to the Jordan block $B$.