Let $B_n$ denote the Boolean lattice of a set with $n \geq 2$ elements and $C_n$ the matrix with entries $c_{i,j}=1$ if $i \leq j$ and $c_{i,j}=0$ else, where $i,j\in B_n$.

Let $M_n:=C_n+C_n^T$ (this is also the Cartan matrix of a certain Frobenius algebra associated to $B_n$), which is a symmetric matrix. Thus geometric and algebraic multiplicity should coincide. Let $I$ denote the identity matrix.

Question: Is there a bijective proof that for $n$ even (odd) we have that the basis of the kernel of $M_n - 2 I$ ($M_n-3 I$) is enumerated by the Catalan numbers?

This is true for $n \leq 15$.

(see also Factorisation of a polynomial from the Boolean algebra )

Here a bijective proof asks for a bijection of a basis of the kernel to known combinatorial objects that are enumerated by the Catalan numbers.

For example for $n=6$, the kernel has dimension 5 and basis vectors are given by

(0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, -1, 1, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, -1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0),

(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0),

(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0),

(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, -1, 0, 0, -1, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0, 1, 0, 0, 1, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0),

(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)

I was not able to see a pattern so far for a nice basis of the kernel as the vectors are so big for larger n.

For example for $n=2$, $C_2$ is given by $\begin{bmatrix} 1 & 1 & 1 &1 \\ 0 & 1 & 0 &1\\ 0 & 0 & 1 &1\\ 0 & 0 & 0 &1\\ \end{bmatrix}$ and $M_2$ is given by $\begin{bmatrix} 2 & 1 & 1 &1 \\ 1 & 2 & 0 &1\\ 1 & 0 & 2 &1\\ 1 & 1 & 1 &2\\ \end{bmatrix}$.

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