math.stackexchange.com/questions/113957 Theorem: Let $T$ be a linear operator on $n$-dimensional vector space $V$. There exists a cyclic vector for $T$ if and only if minimal polynomial and characteristic polynomial are the same.
Proof: Suppose there exists a cyclic vector $v$ for $T$, that is, we have $v\in V$ such that $\{v, Tv,...,T^{n-1}v\}$ span $V$. Then matrix representation of $T$ will be some companion matrix, whose minimal and characteristic polynomial are the same.
Now conversely, if minimal and characteristic polynomial are the same, then we have a minimal polynomial which is of degree $n$. Take $v\neq 0 $, let minimal polynomial $p(x)= a_0+a_1x+\dots+a_{n-1}x^{n-1}$. Degree of $p(x)$ is equal to the cyclic subspace generate by $p_v(x)$. Consider $\{v, Tv, T^2v,...,T^{n-1}v\}$, where $T$ is annihilator linear operator for $p_v(x)$ (following Hoffman-Kunze). This is a cyclic base. Read section 7.1 in Linear Algebra by Hoffman-Kunze.