= Solution
Since $v$ is a <cyclic vector> and $V\ne0$, $v\ne0$. Finite <dimension of a vector space> forces a first dependence in the sequence $v,Tv,T^2v,\ldots$. Let $k\geq1$ be the smallest index with $T^kv$ in the <linear span> of its predecessors. Then
$$
B=(v,Tv,\ldots,T^{k-1}v)
$$
is <linearly independent>. Its <linear span> $W$ is invariant under $T$: all but the last <basis> candidate map to the next, and the last maps to $T^kv\in W$. Induction therefore puts every $T^jv$ in $W$. Cyclicity gives $W=V$, so \b[$B$ is a <basis> and $k=n$].
Write
$$
T^nv=-a_0v-a_1Tv-\cdots-a_{n-1}T^{n-1}v.
$$
The <matrix> of $T$ in this <basis> is the <companion matrix>
$$
C=\begin{pmatrix}
0&0&\cdots&0&-a_0\\
1&0&\cdots&0&-a_1\\
0&1&\cdots&0&-a_2\\
\vdots&\vdots&\ddots&\vdots&\vdots\\
0&0&\cdots&1&-a_{n-1}
\end{pmatrix}.
$$
For $p(t)=t^n+a_{n-1}t^{n-1}+\cdots+a_0$, the defining relation says $p(T)v=0$. Since $p(T)$ commutes with $T$, it also annihilates every $T^jv$ in $B$, hence $p(T)=0$. No nonzero <polynomial> of degree less than $n$ can annihilate $v$, because that would contradict <linear independence> of $B$. Thus $m_T=p$.
Finally the <characteristic polynomial of a companion matrix> is $p$, by direct <determinant> expansion. Explicitly, expansion along the first row of $tI-C$ gives the recurrence
$$
D_n(t;a_0,\ldots,a_{n-1})
=tD_{n-1}(t;a_1,\ldots,a_{n-1})+a_0,
$$
starting from $D_1=t+a_0$. It yields $D_n=t^n+a_{n-1}t^{n-1}+\cdots+a_0$. Consequently
$$
\boxed{m_T(t)=\chi_T(t).}
$$
This proves that a <cyclic operator has equal minimal and characteristic polynomials> by the cyclic <basis> and a direct determinant calculation, with no appeal to the <Cayley-Hamilton theorem>.
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