I have the matrix of the form $$M=\begin{pmatrix}A_N & B_N\\ I_N & 0_N \end{pmatrix},$$ I am looking for the values $\mu$ such that $$\det \begin{pmatrix}A_N-\mu I_N & -D_N^2\\ I_N & -\mu I_N \end{pmatrix}=0,$$ where $I_N$ is the $N\times N$ identity matrix, $D_N$ is a $N\times N$ diagonal matrix, and $A_N=-D_NP_N+\alpha D_N,$ where $\alpha$ is a constant and $P_N=\begin{pmatrix}A &\mathrm{e}^{ik_x'} & &\mathrm{e}^{-ik_x'} \\ \mathrm{e}^{-ik_x} & A &\ddots & \\ & \ddots & \ddots & \mathrm{e}^{ik_x'}\\ \mathrm{e}^{ik_x'}& & \mathrm{e}^{-ik_x'} & A\end{pmatrix},$ is a matrix with $\mathrm{e}^{ik_x'}$ on the upper diagonal and one in the bottom left corner, $\mathrm{e}^{-ik_x'}$ on the lower diagonal and one in the top right corner and the gaps are zeros.
Thus $$\det \begin{pmatrix}A_N-\mu I_N & -D_N^2\\ I_N & -\mu I_N \end{pmatrix}=(-1)^N\det\left(-D_NP_N\mu+\alpha D_N\mu-\mu^2 I_N-D_N^2\right).$$ If $D_N=\text{diag}(\lambda_1,...,\lambda_{N-1})$ then $$(-1)^N\det\left(-D_NP_N\mu+\alpha D_N\mu-\mu^2 I_N-D_N^2\right)=(-1)^N\det(-D_NP_N\mu+B_N-\mu^2I_N),$$ where $B_N=\text{Diag}(\alpha\lambda_0-\lambda_0^2,...,\alpha\lambda_{N-1}-\lambda_{N-1}^2).$ The only way I can see of doing this is by the brute force approach as in simplifying further and just looking at the matrix and just coming up with a possible solution. Does anybody know if there is a more elegant way of doing this in the hope of an exact form?