I have a matrix whose determinant is zero:
$$\det\begin{bmatrix}A-I\lambda&B\\C &D-I\lambda \end{bmatrix} = 0$$
where $\lambda$ is a vector of complex scalars, $I$ is an identity matrix, and $A$, $B$, $C$, $D$ are square, complex, and invertible.
From block matrix identities using the Schur complement, we know that we can then write
$$\det(A-I\lambda) \cdot \det\left( D - I \lambda - C (A-I\lambda)^{-1} B \right) = 0$$
Does the above equation suggest that there are two solutions (sets of eigenvalues) $\lambda$? One being $\mbox{eig}(A)$ and the other being whatever $\det(D-I\lambda - C (A-I\lambda)^{-1} B)$ works out to be?
I've done some numerical tests, and it seems that the first determinant does not give the correct eigenvalues, which does not make sense to me.
What about the slightly modified case of
$$\det\begin{bmatrix}A-I\lambda_x&B\\C &D-I\lambda_y \end{bmatrix} = 0$$
in which we can say
$$\det(A-I\lambda_x) \cdot \det(D-I\lambda_y - C (A-I\lambda_x)^{-1} B) = \det(D-I\lambda_y) \cdot \det(A-I\lambda_x - B (D-I\lambda_y)^{-1} C) = 0$$
Can we then say that simply $\lambda_x = eig(A)$ and $\lambda_y = eig(D)$? This seems fundamentally incorrect to me, but I'm not sure as to why.
@copper.hat 's comment was the correct answer -- since det$(A-I\lambda)$ and det$(D-I\lambda)$ must be invertible, they must have a non-zero determinant.
Therefore, they cannot possibly be the solution to the eigenmode equation.