 Solution

The displayed matrix is the transpose convention for a companion matrix. Expanding $\det (xI-A)$ along its last row givesIndeed, 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 $-1$. ConsequentlyThe Cayley-Hamilton theorem gives $f(A)=0$, so $m_T$ divides $f$. To exclude a smaller minimal polynomial, let $e_1,\dots ,e_n$ be the standard basis. The first $n-1$ rows of $A$ showIf ${\sum }_{j=0}^{n-1}{c}_j{A}^j=0$, multiplying on the left by $e_1^{\mathsf{T}}$ gives ${\sum }_j{c}_j{e}_{j+1}^{\mathsf{T}}=0$, so every coefficient is zero. No nonzero polynomial of degree below $n$ vanishes at $A$. Since $f$ and $m_T$ are monic polynomials,