Eigenvalues of a 2x2 block matrix with invertible diagonal blocks There is something I do not understand for eigenvalues of block matrices.
If we consider a matrix \begin{equation} M = \begin{pmatrix} A & B \\ C & D\end{pmatrix}. \end{equation} Here each block is a $N\times N$ matrix. We know that, provided $A$ is not singular i.e. $\det(A) \neq 0$, the determinant of $M$ is obtained using the Schur's complement of block $A$ of the matrix $M$ via
\begin{equation} \det(M) = \det(A)\cdot \det(D - CA^{-1}B) \, . \end{equation}
Thus from this equation we know that the eigenvalues of block $A$ are also eigenvalues of $M$ because they cancel the caracteristic polynomial of $M$ as well, right ?
By symmetry, if $D$ is not singular, i.e. $\det(D) \neq 0$, the determinant of $M$ is also
\begin{equation} \det(M) = \det(D)\cdot\det(A - BD^{-1}C) \, .\end{equation}
Thus we also learn from this that eigenvalues of $D$ are also eigenvalues of $M$ for the same reason explained previsouly. Therefore the spectrum of $M$ should be composed of the $N$ eigenvalues of $A$ and the $N$ eigenvalues of $D$. In practice, this is not correct. Could someone explain me what I am missing here please ? It should be something obvious and trivial I suppose, but I am totally missing it.
Anyway, can we extract some interesting information about the eigenvalues of $M$ in the case of 4 squared blocks of the same size, which are all invertible but different ?
Thank you !
 A: No, it is not necessarily the case that the eigenvalues of $A$ (or $D$) are eigenvalues of $M$.
Note that $\lambda$ is an eigenvalue of $A$ iff $A - \lambda I$ fails to be invertible.
If $\lambda$ is not an eigenvalue of $A$, then the characteristic polynomial of $M$ can be written in the form
$$
\det(M - \lambda I) = \det(A - \lambda I) \cdot \det([D - C(A - \lambda I)^{-1}B] - \lambda I).
$$
Note that the expression on the right is only defined if $(A - \lambda I)$ is invertible.
That said, the expression on the right certainly gives us some idea about the characteristic polynomial of $M$: because the expression on the right is equal to the characteristic polynomial except at the finitely many values of $\lambda$ where it is undefined and because polynomials are continuous, we can say the following. If $\mu$ is an eigenvalue of $A$, then
$$
\det(M - \mu I) = \lim_{\lambda \to \mu} \det(M - \lambda I) = \lim_{\lambda \to \mu}\det(A - \lambda I) \det(D - C(A - \lambda I)^{-1}B).
$$
However, note that the fact that $\det(A - \mu I) = 0$ does not necessarily imply that the limit on the right is equal to zero. In particular, if $\lim_{\lambda \to \mu}\det(D - C(A - \lambda I)^{-1}B)$ is infinite, then "plugging in $\mu$" gives us the indeterminate expression $0 \cdot \infty$.

An illustrative example: suppose that $N = 1$ and $A = B = C = D = 1$. Note that the determinant of a $1 \times 1$ matrix is simply the entry of this matrix. We find that
$$
M = \pmatrix{1&1\\1&1}, \quad \det(M - \lambda I) = \lambda^2 - 2\lambda = \lambda(\lambda 
- 2).
$$
On the other hand, we compute
$$
\det(A - \lambda I) = 1 - \lambda, \quad \det(D - C(A - \lambda I)^{-1}C) = 1 - \frac{1}{1 - \lambda} = \frac{-\lambda}{1-\lambda}.
$$
We find that
$$
\lim_{\lambda \to 1} \det(A - \lambda I) \det(D - C(A - \lambda I)^{-1}C) = 
\lim_{\lambda \to 1} \frac{-\lambda(1 - \lambda)}{1-\lambda}.
$$
Although it is indeed the case that $\det(A - 1 \cdot I) = 1 - 1 = 0$, it is not the case that the above limit is equal to zero. The limit of the above expression is $-1$, which coincides with the result of $\det(M - 1 \cdot I)$.
