TITLE: Kernel of a matrix and the Catalan numbers
QUESTION [11 upvotes]: 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
$\begin{bmatrix}
2 & 1 & 1 &1 \\
1 & 2 & 0 &1\\
1 & 0 & 2 &1\\
1 & 1 & 1 &2\\
\end{bmatrix}$.

REPLY [6 votes]: Another suggestion: let $A_{2n}=M_{2n}-2I$ ($I$=identity matrix), so we are
interested in $\ker A_{2n}$. Let $V$ be the vector space on which $A_{2n}$
acts, so we can regard $V$ as having a basis consisting of all subsets
of $[2n]=\{1,2,\dots,2n\}$. Thus $V$ has a grading $V_0\oplus
V_1\oplus\cdots \oplus V_{2n}$, where $V_i$ has a basis consisting of all
$i$-element subsets of $[2n]$. The symmetric group $\mathfrak{S}_{2n}$
acts on $V$ and preserves the grading. Write $M_\lambda$ for the
irreducible $\mathfrak{S}_{2n}$-module indexed by the partition $\lambda$
of $2n$. Then (as is well-known) $V_i$ decomposes as $M_{(2n)}\oplus
M_{(2n-1,1)} \oplus \cdots\oplus M_{(2n-i,i)}$ for $0\leq i\leq n$,
and $V_i\cong V_{2n-i}$ (as $\mathfrak{S}_{2n}$-modules).
Assuming that $\dim\ker A_{2n}=C_n$, it is natural to conjecture that
as an $\mathfrak{S}_{2n}$-module, $\ker A_{2n}$ is isomorphic to the
irreducible module $M_{(n,n)}$. The operator
$\frac{1}{(2n)!}\sum_{w\in\mathfrak{S}_{2n}}\chi^{(n,n)}(w)w$ will
then project $V_n$ to $\ker A_{2n}$. Here $\chi^{(n,n)}$ is the
irreducible character of $\mathfrak{S}_{2n}$ indexed by the partition
$(n,n)$.  There is something similar for $M_{2n+1}$.