= Solution
Let $B$ be the matrix whose $i$th column consists of the coordinates of $\beta_i$ in the basis $(e_j)$. Define the upper-triangular matrix $U$ and lower-triangular matrix $L$ by
$$
U_{ij}=
\begin{cases}
1,&i=j,\\
-a_{ij},&i<j,\\
0,&i>j,
\end{cases}
\qquad
L_{ij}=
\begin{cases}
1+a_{ii}=-1,&i=j,\\
a_{ij},&i>j,\\
0,&i<j.
\end{cases}
$$
The first identity in part b says $I=BU$, so $B=U^{-1}$. The second says that the matrix $C$ of $\sigma(c)$ is $C=BL=U^{-1}L$. Since $U$ has diagonal entries one, $\det U=1$, and therefore
$$
\det(tI-C)
=\det\bigl(U^{-1}(tU-L)\bigr)
=\det(tU-L).
$$
Thus $\det(tU-L)$ is the <characteristic polynomial> of the <Coxeter element> in its geometric representation.
Solved by gpt-5.6-sol high.
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