= Solution
The $n$ spanning vectors form a <basis>. Write
$$
T^nx=-a_0x-a_1Tx-\cdots-a_{n-1}T^{n-1}x.
$$
In this cyclic basis the matrix is the <companion matrix>
$$
C=\begin{pmatrix}0&0&\cdots&0&-a_0\\1&0&\cdots&0&-a_1\\0&1&\cdots&0&-a_2\\\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&\cdots&1&-a_{n-1}\end{pmatrix}.
$$
Expanding $\det(tI-C)$ along its last column gives
$$
\chi_T(t)=t^n+a_{n-1}t^{n-1}+\cdots+a_1t+a_0.
$$
For example, the term containing $a_j$ has minor $t^j$ and the signs from the subdiagonal cancel the cofactor sign; the diagonal term supplies $t^n$. The defining relation says $\chi_T(T)x=0$. Every polynomial in $T$ commutes with $T$, so
$$
\chi_T(T)T^jx=T^j\chi_T(T)x=0\qquad(0\leq j<n).
$$
It therefore vanishes on a <basis>, proving
$$
\boxed{\chi_T(T)=0.}
$$
This direct argument for a <cyclic vector> does not assume the <Cayley-Hamilton theorem>.
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