 Solution

The minimal polynomial $m_T$ is the unique monic polynomial of least degree with $m_T(T)=0$. Existence follows from a linear dependence among powers of $T$ in the finite-dimensional vector space $\mathrm{End}(V)$; polynomial division proves uniqueness and that it divides every annihilating polynomial. The characteristic polynomial is ${\chi }_T(t)=\det (tI-[T{]}_{\mathcal{B}})$, independent of basis because a change of basis conjugates the matrix.

WriteIn the cyclic basis, the companion matrix isFor $q(t)=t^n+a_{n-1}{t}^{n-1}+\cdots +a_0$, the displayed vector relation gives $q(T)v=0$. Polynomial evaluation commutes with $T$, so $q(T)T^{j}v=T^{j}q(T)v=0$ for every basis vector. Thus $q(T)=0$.

No nonzero polynomial of degree below $n$ can annihilate $T$, since applying it to the cyclic vector $v$ would contradict independence of $v,Tv,\dots ,T^{n-1}v$. Hence $m_T=q$. Finally, direct expansion of $\det (tI-C)$, or induction along the last column, gives ${\chi }_T(t)=t^n+a_{n-1}{t}^{n-1}+\cdots +a_0=q(t)$. ThereforeThe argument computes the determinant and annihilator directly and does not invoke the Cayley-Hamilton theorem.