Past exam of the rings and modules course of the University of Oxford / 2017 / 3 / a

Write $d=\dim V$. A cyclic vector for $S$ is a vector $v$ whose successive iterates span $V$; equivalently, when $d>0$,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 $c_0,\dots ,c_{d-1}$ with $S^{d}v=-{\sum }_{j<d}{c}_j{S}^{j}v$. Put $f(t)=t^d+{\sum }_{j<d}{c}_j{t}^j$. Then $f(S)v=0$, and commutation with powers of $S$ gives $f(S)S^{j}v=S^{j}f(S)v=0$. 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 $(-c_0,\dots ,-c_{d-1}{)}^{\mathsf{T}}$. Expanding $\det (tI-C(f))$ along its last column givesthe cofactors are $1,t,\dots ,t^{d-1}$. This proves the characteristic polynomial of a companion matrix formula here. HenceThis 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.