= Solution
Write $d=\dim V$. A <cyclic vector> for $S$ is a vector $v$ whose successive iterates span $V$; equivalently, when $d>0$,
$$
v,Sv,\ldots,S^{d-1}v
$$
are a <basis>. The equivalence follows because the first linear dependence in the iterates makes their preceding span invariant under $S$, so no later iterate can enlarge it.
In this basis there are unique coefficients $c_0,\ldots,c_{d-1}$ with $S^dv=-\sum_{j<d}c_jS^jv$. Put $f(t)=t^d+\sum_{j<d}c_jt^j$. Then $f(S)v=0$, and commutation with powers of $S$ gives $f(S)S^jv=S^jf(S)v=0$. Therefore $f(S)$ kills an entire basis and is the zero map. No nonzero polynomial of degree less than $d$ annihilates $v$, because that would be a linear dependence of the displayed basis. Thus $m_S=f$.
The matrix of $S$ in the cyclic basis is the <companion matrix> $C(f)$ with ones on its subdiagonal and last column $(-c_0,\ldots,-c_{d-1})^{\mathsf T}$. Expanding $\det(tI-C(f))$ along its last column gives
$$
c_0+c_1t+\cdots+c_{d-2}t^{d-2}+(t+c_{d-1})t^{d-1}=f(t):
$$
the cofactors are $1,t,\ldots,t^{d-1}$. This proves the <characteristic polynomial of a companion matrix> formula here. Hence
$$
\boxed{m_S(t)=c_S(t)=f(t).}
$$
This proves that a <cyclic operator has equal minimal and characteristic polynomials>. If $V=\{0\}$, both polynomials are one by convention, so the equality remains valid.
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