 Cyclic operator has equal minimal and characteristic polynomials

On a $d$-dimensional vector space, a cyclic vector $v$ gives a basis $v,Tv,\dots ,T^{d-1}v$. Write $T^{d}v=-{\sum }_{j<d}{c}_j{T}^{j}v$ and set $f(t)=t^d+{\sum }_{j<d}{c}_j{t}^j$. Then $f(T)v=0$. Since polynomials in $T$ commute with $T$, $f(T)$ kills every basis vector, so $f(T)=0$. No nonzero polynomial of degree less than $d$ can kill $v$, by linear independence. Hence the minimal polynomial is $f$. The matrix in this basis is $C(f)$, and the characteristic polynomial of a companion matrix is $f$ as well.