I'd be interested to see a solution to this problem! I tried...

the result is not trivial: i'll take \(\displaystyle F\) to be the ground field. let \(\displaystyle p(x) \in F[x]\) be the minimal polynomials of \(\displaystyle A.\) we have \(\displaystyle \deg p(x)=n\) because the minimal and characteristic polynomials of \(\displaystyle A\) are equal.

now for every \(\displaystyle v \in V\) define \(\displaystyle I_v= \{f(x) \in F[x]: \ f(A)(v) = 0 \}.\) see that \(\displaystyle I_v\) is an ideal of \(\displaystyle F[x]\) and thus, since \(\displaystyle F[x]\) is a PID, \(\displaystyle I_v\) is principal, i.e. \(\displaystyle I_v= \langle f_v(x) \rangle,\) for some \(\displaystyle f_v(x) \in F[x].\)

on the other hand, clearly \(\displaystyle p(x) \in I_v\) and so \(\displaystyle f_v(x) \mid p(x).\) but the number of divisors of \(\displaystyle p(x)\) is finite, which means the number of distinct elements of the set \(\displaystyle \{I_v: \ v \in V \}\) is finite. let \(\displaystyle I_{v_1}, \cdots , I_{v_m}\)

be those distinct elements and put \(\displaystyle V_i = \{v \in V: \ f_{v_i}(A)(v)=0 \}.\) it is clear that each \(\displaystyle V_i\) is a subspace of \(\displaystyle V\) and \(\displaystyle V=\bigcup_{i=1}^m V_i.\) therefore \(\displaystyle V=V_k,\) for some \(\displaystyle 1 \leq k \leq m.\) thus \(\displaystyle f_{v_k}(A)(v)=0,\) for all

\(\displaystyle v \in V,\) i.e. \(\displaystyle f_{v_k}(A)=0\). that implies \(\displaystyle p(x) \mid f_{v_k}(x),\) by the minimality of \(\displaystyle p(x),\) and so \(\displaystyle f_{v_k}(x)=p(x)\) because we already know that \(\displaystyle f_{v_k}(x) \mid p(x)\) and \(\displaystyle p(x)\) is monic.

finally, suppose that \(\displaystyle c_0v_k + c_1A(v_k) + \cdots + c_{n-1}A^{n-1}(v_k)=0,\) for some \(\displaystyle c_j \in F.\) let \(\displaystyle g(x)=c_0 + c_1x + \cdots + c_{n-1}x^{n-1} \in F[x].\) then \(\displaystyle g(A)(v_k)=0\), i.e. \(\displaystyle g(x) \in I_{v_k} = \langle f_{v_k}(x) \rangle = \langle p(x) \rangle.\) therefore

\(\displaystyle p(x) \mid g(x),\) which is not possible unless \(\displaystyle g = 0,\) because \(\displaystyle \deg p(x)=n > \deg g(x).\) so we must have \(\displaystyle c_j = 0,\) for all \(\displaystyle j. \ \Box\)

**Remark 1**.

__second solution__: very briefly, give \(\displaystyle V\) the structure of an \(\displaystyle F[x]\) module by defining the multiplication by \(\displaystyle f(x)v = f(A)(v).\) then see that \(\displaystyle V\) is a finitely generated torsion \(\displaystyle F[x]\) module and

thus, since \(\displaystyle F[x]\) is a PID, we can apply the fundamental theorem for finitely generated modules over PIDs to get some \(\displaystyle v \in V\) such that \(\displaystyle \text{ann}_{F[x]}(v)=\text{ann}_{F[x]}(V).\) now it's easy to show that the set

\(\displaystyle \{v,A(v), \cdots , A^{n-1}(v) \}\) is linearly independent over \(\displaystyle F.\)

**Remark 2.** the statement in the problem is actually an "if and only if" statement.