Consider $$(\lambda^2 \mathbb{M} +2ic\lambda \mathbb{K}+\mu \mathbb{C})\vec{A}=0, \quad \quad (1)$$
where $$\mathbb{M}=\begin{bmatrix} \mu & 1+\mu\\ 1+\mu & \mu \\ \end{bmatrix}; \quad \quad \mathbb{K}=\begin{bmatrix} 0 & -1\\ 1 & 0 \\ \end{bmatrix};\quad \quad \mathbb{C}=\begin{bmatrix} 1 & 1\\ 1 & 1 \\ \end{bmatrix},$$
and $\nu=-4 a^2, \mu =\nu(1-2a^2), c=\sqrt{1-4a^2}$, for $a\in \mathbb{R}^+$, and $\lambda$ is an eigenvalue with associated eigenvector $\vec{A}=(A,A^*)$ for $A$ a complex number.
I can go through the algebra to find
$$\lambda^2=-\frac{4(1-6a^2)}{(1-4a^2)^2}.$$
Which implies that $\lambda$ comes in pairs. This is a simplified example of a much larger system, where I'll be solving this numerically and would gain a lot by only having to solve for 1/2 the eigenvalues.
My question is, can I rewrite/manipulate equation (1) so that the term in parentheses becomes diagonal diagonalized (possibly times other matrices which change the form of $\vec{A}$? Generally, I know how to do this in an ordinary eigenvalue problem, but do not see how to obviously proceed here.