Let A:V->V be Linear Transformation. dim V =n. Prove that if Characteristic polynomial of A = min. Polynomial A then, there is a v \in V such that,
$\displaystyle v+ Av+ \cdots A^{n-1}v$ spans V.
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.