I have the following system for $\lambda \in \Bbb C, \lambda \neq 0$ and $\pmb{p},\pmb{q} \in \Bbb C^n$, $(\pmb{p}^T, \pmb{q}^T)\neq0$:
$$\begin{cases} F(\lambda) \pmb{p} - g(\lambda) \pmb{q} - \lambda \alpha\beta M^{-1}C \pmb{q} = 0 \ , \\ g(\lambda) \pmb{p} + F(\lambda) \pmb{q} + k \alpha\beta M^{-1}C \pmb{p} = 0 \ . \end{cases} $$ Where $F(\lambda) = \lambda^3 I + \lambda^2(\alpha I + \beta D) + \lambda \alpha \beta D$, $g(\lambda) = \lambda^2\beta + \lambda \alpha \beta$, $\alpha, \beta, k$ are positive scalars, $D,M$ are $n\times n$ diagonal matrices with positive elements on its diagonals, $C$ is $n\times n$ real positive semi-definite matrix (non-diagonal).
For which matrix $D$ phase angles of $\pmb{p}$ are opposite to $\pmb{q}$ (namely, $p_k = |p_k|e^{i\phi_k}$ and $q_k = |q_k|e^{-i\phi_k}, \ k=1,\cdots,n$)?
I suspect it is valid only if $D = dI$. Because in this case, $F(\lambda) = [\lambda^3 + \lambda^2(\alpha + \beta d) + \lambda \alpha \beta d]I = f(\lambda)I$ and $\pmb{p}$ will be an eigenvector of $M^{-1}C$. Further, $\pmb{q}= -\frac{g(\lambda) + k\alpha\beta\mu}{f(\lambda)} \pmb{p}$, where $\mu$ is the correspoding eigenvalue of $M^{-1}C$ such that difference between phase angles of $\pmb{p}$ and $\pmb{q}$ is $\phi_0 = \angle \frac{g(\lambda) + k\alpha\beta\mu}{f(\lambda)}$, and the pair $(\pmb{p} = e^{-i\frac{\phi_0}{2}}\pmb{\psi}, \pmb{q}= - e^{-i\frac{\phi_0}{2}}\frac{g(\lambda) + k\alpha\beta\mu}{f(\lambda)}\pmb{\psi})$ with $M^{-1}C$ eigenvector $\pmb{\psi} \in \Bbb R^n$ will have the opposite phase angles. I do not know how to show $D=dI$ is necessary or find conterexample with $D\neq d I$.