Characteristic polynomial of checker matrix

Question

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, 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 $(\lambda^2 - (2n-3)\lambda - 3) (\lambda^2 - (2n+3)\lambda + 3)$.

Answer 1

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.

Answer 2

The methods in the paper by Junod:

Junod, Alexandre, Hankel determinants and orthogonal polynomials., Expo. Math. 21, No. 1, 63-74 (2003). ZBL1153.15304.

Answer your question (and much more general ones, too).