= Solution
The displayed <matrix> is the transpose convention for a <companion matrix>. Expanding $\det(xI-A)$ along its last row gives
$$
\det(xI-A)=x^n+\alpha_{n-1}x^{n-1}+\cdots+\alpha_1x+\alpha_0=f(x).
$$
Indeed, the last entry contributes $(x+\alpha_{n-1})x^{n-1}$ and the entry in column $j<n$ contributes $\alpha_{j-1}x^{j-1}$; the cofactor signs cancel those from the superdiagonal entries $-1$. Consequently
$$
\boxed{\chi_T(x)=(-1)^nf(x).}
$$
The <Cayley-Hamilton theorem> gives $f(A)=0$, so $m_T$ divides $f$. To exclude a smaller <minimal polynomial>, let $e_1,\ldots,e_n$ be the <standard basis>. The first $n-1$ rows of $A$ show
$$
e_1^{\mathsf T}A^j=e_{j+1}^{\mathsf T}\qquad(0\leq j<n).
$$
If $\sum_{j=0}^{n-1}c_jA^j=0$, multiplying on the left by $e_1^{\mathsf T}$ gives $\sum_jc_je_{j+1}^{\mathsf T}=0$, so every coefficient is zero. No nonzero <polynomial> of degree below $n$ vanishes at $A$. Since $f$ and $m_T$ are <monic polynomials>,
$$
\boxed{m_T=f.}
$$
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