Finding eigenvalues of a block matrix I have a block matrix of size $2N \times 2N$ of the form
$$B = \begin{bmatrix} A_N & C_N \\ C_N & A_N \end{bmatrix}$$
where $A_N$ and $C_N$ are both $N \times N$ matrices. Specifically,
$$A_N = \begin{bmatrix}
0 & \cdots & 1 \\
\vdots & \ddots & \vdots \\
1 & \cdots & 0
\end{bmatrix}
\qquad
C_N =
\begin{bmatrix}
1 & \cdots & 0 \\
\vdots & \ddots & \vdots \\
0 & \cdots & 1
\end{bmatrix}
$$
That is, $A_N$ has zeroes on the diagonal, and all other entries $1$; $C_N$ has zeroes along the minor diagonal, and all other entries are $1$.
I would like to find the eigenvalues of the matrix $B$.
 A: I will answer your question just for the cases $N = 2$ and $N = 3$:
Let 
$$ B_2 = \left(
\begin{array}{cccc}
 0 & 1 & 1 & 0 \\
 1 & 0 & 0 & 1 \\
 1 & 0 & 0 & 1 \\
 0 & 1 & 1 & 0
\end{array}
\right), \quad B_3 = \left(
\begin{array}{cccccc}
 0 & 1 & 1 & 1 & 1 & 0 \\
 1 & 0 & 1 & 1 & 0 & 1 \\
 1 & 1 & 0 & 0 & 1 & 1 \\
 1 & 1 & 0 & 0 & 1 & 1 \\
 1 & 0 & 1 & 1 & 0 & 1 \\
 0 & 1 & 1 & 1 & 1 & 0
\end{array}
\right), $$
then, $\text{Spec}{(B_2)} = {(-2,2,0,0)} \, $ and $\text{Spec}{(B_3)} = (-2,-2,0,0,0,4). $
With the help of numerics, I've been able to show (at least for sufficiently large values of $N$) that the characteristic polynomial is given by:

$$\color{blue}{p(\lambda) =\det{(B_N - \lambda I_{2N})} = (\lambda - 2N +2)(\lambda+2)^{N-1} \lambda^N  }$$ 

which tells you that the only eigenvalues of this kind of matrices are $-2,0,2N-2 \ $ with the corresponding multiplicities given by $p(\lambda)$.
Here is an animation showing the spectrum of the matrices $B_N$ for $N \in (2,30)$:


Here's the same approach in the case we have the $B_N$ matrices defined as:
$$B_N = \begin{bmatrix} C_N & A_N \\ A_N & C_N \end{bmatrix},$$
then:

$$\color{blue}{p(\lambda) =\det{(B_N - \lambda I_{2N})} = \left\{ \begin{array}{ll}
\left(\lambda - 2N + 2\right) (\lambda-2)^{N/2}(\lambda+2)^{(N-2)/2} \lambda^{N} & N \text{ even} \\
\left(\lambda - 2N + 2\right) (\lambda-2)^{(N-1)/2}(\lambda+2)^{(N-1)/2} \lambda^{N} & N \text{ odd} 
\end{array}\right.}$$ 

which tells you that the only eigenvalues of this kind of matrices are $-2,0,2,2N-2 \ $ with the corresponding multiplicities given by $p(\lambda)$.
Here is another animation showing the spectrum of the matrices $B_N$ for $N \in (2,30)$:

pretty cool!
Hope somebody can shed some light on these results.
Cheers!
A: A bit late, but still:
The $2N\times 2N$ matrix $B$ is the adjacency matrix of a $2N-2$-regular
bipartite graph. Each row sums to the same value:
$2N-2$. It thus follows that $2N-2$ is an eigenvalue with eigenvector $[1,\ldots,1]^{\top}$.
Since $B$ is symmetric, it possesses an orthogonal set of eigenvectors. Since we found one eigenvector ($[1,\ldots,1]^{\top}$),
the remaining eigenvectors must be orthogonal to it, i.e., they must have entries summing to zero. One way is to take
the $2N$-length vector with $N$ ones and $N$ minus ones as the eigenvector, and this gives the eigenvalue $2$. Finally, we can see that since
there are repeated rows and columns in $B$, zero is an eigenvalue.
