c \alpha \cdot \nu \big( (\Phi_\lambda \varphi)_+ - (\Phi_\lambda \varphi)_-\big) + \frac{1}{2} (\eta I_4 + \tau \beta) \big( (\Phi_\lambda \varphi)_+ + (\Phi_\lambda \varphi)_-\big) \\
      &= (I_4 + (\eta I_4 + \tau \beta) \mathcal{C}_\lambda) \varphi,
    \end{split}$$ i.e. $-1$ is an eigenvalue of $(\eta I_4 + \tau \beta) \mathcal{C}_\lambda$.

Conversely, assume that $-1$ is an eigenvalue of $(\eta I_4 + \tau \beta) \mathcal{C}_\lambda$ with eigenfunction $\varphi \neq 0$. Then it follows first from item (i) that $\varphi \in H^{1/2}(\Sigma)^4$ and hence $f_\lambda := \Phi_\lambda \varphi \neq 0$ belongs by  to $H^1(\Omega_+)^4 \oplus H^1(\Omega_-)^4$. Using again  and and $\big( I_4 + (\eta I_4 + \tau \beta) \mathcal{C}_\lambda \big) \varphi = 0$ we obtain in the same way as in  that $f_\lambda$ fulfills the coupling condition . This shows $f_\lambda \in {\mathrm{dom}\,}A_{\eta, \tau}$. Finally, equation  yields $$(A_{\eta, \tau} - \lambda) f_\lambda = (A_{\eta, \tau} - \lambda) \Phi_\lambda \varphi = 0$$ and hence $\lambda \in \sigma_\textup{p}(A_{\eta, \tau})$.

\(iii) To show the claim it suffices to prove that $I_4 + (\eta I_4 + \tau \beta) \mathcal{C}_\lambda$ is bijective. By (ii) it is clear that this operator is injective, as $\lambda \in \mathbb{C} \setminus \mathbb{R}$ and $A_{\eta, \tau}$ is symmetric by Lemma \[lemma\_symmetric\]. Moreover, $$\label{equation_C_lambda_range}
    \begin{split}
      \operatorname{ran}\big[I_4 