= Solution
{c}
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 $\operatorname{End}(V)$; 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.
Write
$$
T^nv=-a_0v-a_1Tv-\cdots-a_{n-1}T^{n-1}v.
$$
In the cyclic basis, the <companion matrix> is
$$
C=\begin{pmatrix}
0&0&\cdots&0&-a_0\\
1&0&\cdots&0&-a_1\\
0&1&\cdots&0&-a_2\\
\vdots&&\ddots&\vdots&\vdots\\
0&0&\cdots&1&-a_{n-1}
\end{pmatrix}.
$$
For $q(t)=t^n+a_{n-1}t^{n-1}+\cdots+a_0$, the displayed vector relation gives $q(T)v=0$. Polynomial evaluation commutes with $T$, so $q(T)T^jv=T^jq(T)v=0$ 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 $v,Tv,\ldots,T^{n-1}v$. Hence $m_T=q$. Finally, direct expansion of $\det(tI-C)$, or induction along the last column, gives $\chi_T(t)=t^n+a_{n-1}t^{n-1}+\cdots+a_0=q(t)$. Therefore
$$
\boxed{m_T=\chi_T=q.}
$$
The argument computes the determinant and annihilator directly and does not invoke the <Cayley-Hamilton theorem>.
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