Prove $V$ is a cyclic $F[A]$ module $\iff$ it has the same characteristic and minimal polynomials

Question

Suppose $A \in M_n(F)$ where $F$ is a field. Let $V=F^n$ be the $F[A]$-module made from $A$ (that is, for any $v\in V$ and $f(x)\in F[x]$ we have $f(x)v = f(A)v \in V$).

Prove that $V$ is cyclic if and only if the characteristic and minimal polynomials are equal.

My thoughts are:

If $m(x)=c(x)$ then there's only one invariant factor. By the structure theorem for modules we have $V\cong F[x]/(m(x))$ (because $V$ is torsion only and there's no free part, I think). Thus $V$ is cyclic.

For the other direction, suppose $V$ is cyclic. Then by definition: $\exists v \in V$ such that $\{f(A)v : f \in F[x]\} = V$. How might I show $c(x)=m(x)$?

Answer 1

You've left the easiest part of the question. If $v$ is a cyclic vector and $d=\dim(V)$, then there is an injective morphism (in fact an isomorphism) of $F$-vector spaces $F[X]_{<d}\to V$ (form the polynomials of degree less than$~d$ to $V$) given by $P\mapsto P[A]\cdot v$. Therefore the degree of the minimal polynomial of$~A$ is at least$~d$; but then by the Cayley-Hamilton theorem it is exactly$~d$ and the minimal polynomial is the characteristic polynomial. You can avoid using the Cayley-Hamilton theorem at the expense of a bit of computation: after changing basis to $[\,A^i\cdot v \mid 0\leq i<d\,]$ the matrix becomes a companion matrix of some monic polynomial$~P$ of degree$~d$, and the question boils down to showing that its minimal and characteristic polynomials of it are both$~P$; this is a classic exercise, see for instance the question The characteristic and minimal polynomial of a companion matrix.