TITLE: The characteristic and minimal polynomial of a companion matrix
QUESTION [32 upvotes]: The companion matrix of a monic polynomial $f \in \mathbb F\left[x\right]$ in $1$ variable $x$ over a field $\mathbb F$ plays an important role in understanding the structure of finite dimensional $\mathbb F[x]$-modules.
It is an important fact that the characteristic polynomial and the minimal polynomial of $C(f)$ are both equal to $f$. This can be seen quite easily by induction on the degree of $f$.
Does anyone know a different proof of this fact? I would love to see a graph theoretic proof or a non inductive algebraic proof, but I would be happy with anything that makes it seem like more than a coincidence!

REPLY [8 votes]: Surprisingly, the following (in my opinion) quite elegant proof is still missing:
Look at the $F$-vector space $F[x]/(f)$. The map
$$\phi : F[x]/(f)\to F[x]/(f),\quad g + (f)\mapsto x\cdot g + (f)$$
is well-defined and $F$-linear.
Let $m_\phi = \sum_{i=0}^d a_i x^i\in F[x]$ be the minimal polynomial and $\chi_\phi\in F[x]$ the characteristic polynomial of $\phi$. Then $m_\phi(\phi)$ is the zero map in $\operatorname {End}(F[x]/(f))$. Thus
$$0 + (f) = m_\phi(\phi)(1 + (f)) = \sum_{i=0}^d a_i \phi^i(1 + (f)) = \left(\sum_{i=0}^d a_i x^i\right) + (f) = m_\phi + (f).$$
So
$$f\mid m_\phi \mid \chi_{\phi},$$
where the scecond divisibility follows from Cayley-Hamilton.
Because of $m_\phi \neq 0$ and $\deg(f) = \dim_F(K[x]/(f)) = \deg(\chi_\phi)$ and because all the polynomials are monic, this forces
$$ f = m_\phi = \chi_\phi.$$
With respect to the basis $(1 + (f), x + (f),\ldots, x^{n-1} + (f))$, the transformation matrix of $\phi$ is the companion matrix $C(f)$ of $f$. So the minimal polynomial of $C(f)$ equals $m_\phi$ and the characteristic polynomial of $C(f)$ equals $\chi_\phi$.