For every integer $n > 0$, let $C\_n$ be the $4n \times 4n$ matrix having $1$'s in all positions $(i, j)$ such that $i - j$ is even, $3$'s in the two diagonals determined by $|i - j| = 2n + 1$, and $0$'s everywhere else. For example, we have
$$C\_2 = \begin{bmatrix}
1 & 0 & 1 & 0 & 1 & 3 & 1 & 0 \\
0 & 1 & 0 & 1 & 0 & 1 & 3 & 1 \\
1 & 0 & 1 & 0 & 1 & 0 & 1 & 3 \\
0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\
1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\
3 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\
1 & 3 & 1 & 0 & 1 & 0 & 1 & 0 \\
0 & 1 & 3 & 1 & 0 & 1 & 0 & 1\end{bmatrix} .$$


I'd like to prove a formula for the characteristic polynomial of $C\_n$. From some numerical experiments, I believe that it is 
$$(\lambda - 3)^{2n - 2} (\lambda + 3)^{2n - 2} (\lambda^2 - (2n-3)\lambda - 3) (\lambda^2 - (2n+3)\lambda + 3),$$
but I failed to prove that.


Any suggestion is welcome. Thanks.


**Note 1.** What makes things difficult are the $3$'s. If instead of them there were $0$'s, then we would have a [circulant matrix](https://en.wikipedia.org/wiki/Circulant_matrix), and using the theory of circulant matrices the characterist polynomial would be easily proved to be $\lambda^{4n - 2}(\lambda - 2n)^2$.


**Note 2.** Following Pat Devlin's suggestion, I checked the eigenspace of $\lambda = 3$ and it seems to be spanned by the row vectors of the following matrix $(2n-2)\times 4n$ matrix
$$\begin{bmatrix}\begin{matrix}-1 \\ -1 \\ \vdots \\ -1\end{matrix} & I\_{2n-2} & \begin{matrix}0 & 0 & -1\\ 0 & 0 & -1 \\ \vdots \\ 0 & 0 & -1\end{matrix} & I\_{2n-2}\end{bmatrix} .$$
This shouldn't be difficult to prove, and similarly for the eigenspace of $\lambda = -3$. But I have no idea how to deal with the eigenvalues related to the factor $(\lambda^2 - (2n-3)\lambda - 3) (\lambda^2 - (2n+3)\lambda + 3)$.



Characteristic polynomial of checker matrix
Ok! Your conjecture is true.


Let $W$ be the space spanned by the eigenvectors for $\lambda \in \{-3, 3\}$ as described in my comments. Let $V$ be the subspace of $\mathbb{R}^{4n}$ consisting of vectors of the form


$$V = \{(a,b,a,b,a,b, \ldots, a, x, y, b, a, b, \ldots, a,b)\},$$


where the entries corresponding to $x,y$ are in positions $2n$ and $2n+1$ of the vector. (So $V$ is the orthogonal complement of $W$.)


Let $T : V \to \mathbb{R}^4$ by $T(\vec{v}) = (a,b,x,y)$ in the obvious way (so $T$ is an isomorphism).


We can check that $V$ is invariant under the action of $C\_{n}$. And moreover, that $$T \circ C\_{n} \circ T^{-1} \begin{pmatrix}a\\b\\x\\y \end{pmatrix} = \begin{pmatrix}(2n-1)a +y + 3b\\(2n-1)b+x+3a\\(2n-1)b+x\\(2n-1)a+y \end{pmatrix}.$$


Thus, $C\_{n}$ restricted to $V$ is isomorphic to the above linear map on $\mathbb{R}^4$, namely


$$\begin{pmatrix}a\\b\\x\\y\end{pmatrix} \mapsto \begin{pmatrix}2n-1 & 3 & 0 &1\\
3 & 2n-1 & 1 &0\\
0 & 2n-1 & 1 &0\\
2n-1 & 0 & 0 &1\end{pmatrix} \begin{pmatrix}a\\b\\x\\y\end{pmatrix},$$


and this map has the desired remaining four eigenvalues as in your conjecture.


