$T\in \mbox{End}(V)$. If $p|m_T,$ then there is a vector $v$ such that the minimal polynomial of $v$ is exactly $p$. Let $T:V\rightarrow V$ be linear a linear map and let $p\in \mathcal P(\mathbb C)$ be a non-constant polynomial such that $p | m_T$, where $m_T$ is the minimal polynomial of the endomorphism $T$. Prove that exists $v\in V$ such that $m_v=p$, that is, such that the minimal polynomial for $v$ is $p$ (the minimal polynomial for $v$ is the lowest degree polynomial such that the family $(T^j(v))_{j\geq 0}$ is linear dependent, ie, such that $m_v(T)(v) = 0$)
My attempt:
Since $p|m_T$, we may write $m_T = ph$. Writing $V_p := \{v\in V: p(T)(v) =0\}$, we must have $V_p \neq \{0\},$ otherwise we have $p(T)(u)\neq 0$ for all $u\in V$ and, since for all $u\in V$ we have $0=m_T(T)(u) = p(T)(h(T)(u)),$ it must be the case that $h(T)(u) = 0 $ for all $u\in V$. But the degree of $h$ is less than the degree of $m_T$, which is a contradiction. 
Ok, now we may consider $w\in V_p, w\neq 0$. It is true that $p(T)(w) = 0$ and since $m_w$ is the lowest degree polynomial satisfying the condition $m_w(T)(w) = 0,$ it follows that $m_w|p$. No argument can help me proceed here to conclude that $p|m_w$, because it may not be the case. I'm stuck at this point. I guess that the correct path should try to construct such a vector $v$ using somehow the irreducible factors of $p$, but i'm not sure how to proceed. 
EDIT: I guess that I could solve it, I just need to formalize it properly. Since $p|m_T$, the idea is to write $m_T(t) = \prod_{i=1}^s p_i^{\ell_i}$ and $p(t)= \prod_{i=1}^s p_i^{k_i}$, where $0\leq k_i\leq \ell_i$ and each $p_i$ is a monic irreducible factor and relatively prime with one another. Using a lemma prior to the primary decomposition theorem and the notation $V_p=\{v\in V: p(T)(v) = 0\}$, we may write: $V_p = \bigoplus V_{p_i^{k_i}}$ and since $V_p\neq 0$, we may consider $0\neq v = v_1+\cdots + v_s \in V_p.$ Since $p(T)(v) = 0$ and every $p_i$ is relatively prime with each other, it must be the case that $p_i^{k_i}(T)(v_i) = 0$ for every $i=1,\cdots, s$ and $p_i^{k_i}(T)(v_j) \neq 0$ for $i\neq j$. Therefore, we may consider $m_1,\cdots, m_s\in \mathbb N$ minimal such that $p_i^{m_i}(T)(v_i) \neq 0 $ and $p_i^{m_i+1}(T)(v_i)= 0.$ Define then $0 \neq \tilde v = \sum p_i^{m_i}(T)(v_i)$. It is the case that $m_{\tilde v} = p$.
Is this correct?

Use the fact that $V^T$ has cyclic submodule $C \cong \mathbb C[x]/(m_T)$ and then
$h$ has order $p$ in it.
Here are more details:
Cyclic Modules, Characteristic Polynomial and Minimal Polynomial
