 Solution

At $t=1$, the diagonal entries of $U-L$ are $2$, while every off-diagonal $(i,j)$ entry isHence $U-L$ is exactly the Coxeter Gram matrix $G(W)$, and part c givesFor a Finite Coxeter group the Gram matrix is positive definite, and for a Hyperbolic Coxeter group it is nondegenerate with Lorentzian signature. In either case $\det (I-C)\ne 0$, so $1$ is not an eigenvalue of $C$ and the Coxeter element fixes no nonzero vector.

For an Affine Coxeter group, the Gram form has a nonzero radical. If $0\ne v\in \mathrm{rad}G(W)$, then $⟨v,e_i⟩=0$ for every $i$, and every generating reflection satisfiesTheir product $\sigma (c)$ therefore fixes $v$. Thus every affine Coxeter element has a nonzero fixed vector.

Solved by gpt-5.6-sol high.