I'm trying to solve the following problem, but cannot seem to complete it:
Let $V$ be a vector space of dimension $n \in \mathbb{N}, n>0$ over a field $\mathbb{F}$, and let $A \in M_n(\mathbb{F})$.
Every $v \in V$ has a polynomial $f_v \in \mathbb{F}[x]$ such that $\langle f_v \rangle = \{p \in \mathbb{F}[x] \mid p(A)v=0 \}$. When is it the case that $f_v=m_A(x)$? (where $m_A(x)$ is the minimal polynomial of $A$).
EDIT: the approach I took below may be wrong. It is possible that it there is a way to go about this using the fact that there is a one-to-one correspondence between $(V,T)$ (where $T$ is the linear transformation corresponding to $A$) and $(V, \cdot)$ as an $F[x]$-module, and then viewing $f_v$ and $m_A(x)$ as part of the fundamental theorem for finitely generated $F[x]$-modules (e.g. $F[x]/\langle m_A(x) \rangle$).
My approach:
If $v=0$ then every polynomial $p$ satisfies the condition $p(A)v = 0$ so $f_v = 1$ and this cannot be a minimal polynomial. So a necessary condition must be that $v \neq 0$.
Assume $v \neq 0$.
In this case, I started by trying to assume that $f_v\neq m_A(x)$ to see if this bears any consequence on the rational canonical form of $A$ (i.e. looking for some condition such as 'the rational canonical form cannot have a block of the form X'). Under this assumption, there must be a polynomial $p(x)$ such that $m_A(x) \nmid p(x)$ and for which $p(A)v = 0$.
Since $p(A) \neq 0$ there must be at least one block, $C(a_i(x))$ in the rational canonical form of $A$ such that $P(C(a_i(x)))\neq 0$.
From this point, I wasn't sure what else I could do, so I tried plugging in options for simple forms of rational canonical forms and basis vectors $e_i = (0,1,0..)$ to see what might come of it, but this didn't lead to anything fruitful.
What am I missing?