We keep all the notations from Parts I and II and assume hypotheses (H1)–(H5). Let $\mathcal{B} = (z_1, \ldots, z_\ell)$, where $\ell = 2m-2$, be a basis of $G$. For any element $u$ of $G$, we denote by $U$ (capital letter) the column vector containing the coordinates of $u$ with respect to the basis $\mathcal{B}$. We denote by $A = [a_{i,j}]_{1 \leq i,j \leq \ell}$ and $B = [b_{i,j}]_{1 \leq i,j \leq \ell}$ the two square matrices whose coefficients are defined by $$\forall 1 \leq i,j \leq \ell, \quad a_{i,j} = (z_i \mid z_j), \quad b_{i,j} = (T(z_i) \mid T(z_j))$$
(a) Let $u, v \in G$. Show that $$(u \mid v) = {}^t U A V, \quad (T(u) \mid T(v)) = {}^t U B V$$ and deduce that $A$ and $B$ are invertible.
(b) Let $\lambda \in \mathbb{R}$. Show that an element $u \in G$ is a solution of $(\mathcal{P}_\lambda)$ if and only if $$(A - \lambda B) U = 0$$ Deduce that $(\mathcal{P}_\lambda)$ admits a non-zero solution if and only if $\operatorname{det}(A - \lambda B) = 0$.
(c) We define the function $\psi$ on $\mathbb{R}$ by $$\forall t \in \mathbb{R}, \psi(t) = \frac{\operatorname{det}(A - tB)}{\operatorname{det}(B)}$$ Show that this function $\psi$ is independent of the choice of basis $\mathcal{B}$.
(d) Justify why we can choose the basis $\mathcal{B}$ so that $B = I_\ell$. Deduce that $\psi$ is a polynomial function and specify its degree.
(e) Show that the polynomial $\psi$ is split over $\mathbb{R}[X]$ and that its roots are either simple or double.
(f) Show that $$\psi(X) = \frac{1}{S(w_1, T^{2m-1}(w_1)) S(w_2, T^{2m-1}(w_2))} Q_1(X) Q_2(X)$$ (justify why necessarily the denominator is non-zero). Deduce that $Q_1$ and $Q_2$ are split over $\mathbb{R}[X]$ and have simple roots.