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!
Solving $\det(A - \lambda I) = 0 \implies \lambda_{1} = 6, \lambda_{2} = 5$ with the associated eigenvectors $v_{1} = \begin{pmatrix}1 \\1 \end{pmatrix}$ and $v_{2} = \begin{pmatrix}1 \\2 \end{pmatrix}$.
Let $P = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$. Then $P^{-1} = \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix}$.
We have $P^{-1}AP = \text{diag}(6, 5) \implies P^{-1}A^{n}P = \text{diag}(6^{n}, 5^{n})$.
Best Answer
Let us fix some notation first.
Define $$ M = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 1 & 1 & 1& 0\\ 2& 3&1&0\\ 3&4&5&1 \end{bmatrix}, \ \ M_1 = I, \ \ M_2 = \begin{bmatrix} 0 & 2 & 3 & 4 \\ 1 & 0 & 1& 0\\ 2& 3&0&0\\ 3&4&5&0\end{bmatrix}$$
Clearly, $$ M = M_1 + M_2 $$
Let $(\lambda, \mathbf{x})$ be any eigenpair for $M_2$.
Then it follows that $$ M_2 \mathbf{x} = \lambda \mathbf{x} \tag{1} $$ Also, $$ I \mathbf{x} = \mathbf{x} \tag{2} $$
Adding (1) and (2), we get $$ (I + M_2) \mathbf{x} = (\lambda + 1) \mathbf{x} $$ or equivalently, $$ M \mathbf{x} = (\lambda + 1) \mathbf{x} $$
If $(\lambda, \mathbf{x})$ is an eigenpair for $M_2$, then $(\lambda + 1, \mathbf{x})$ is an eigenpair for $M$.
Thus, the eigenvalues of $M$ are given by $$ \lambda_1 + 1, \lambda_2 + 1, \lambda_3 + 1, \lambda_4 + 1 $$ where $\lambda_1, \lambda_2, \lambda_3, \lambda_4$ are the eigenvalues of $M_2$. Moreover, $M$ and $M_2$ have the same eigenvectors.