It seems that one can write the eigenvectors of the hypercube $\{ \pm 1\}^n$ as the functions, $\{ \chi_S \}_{S \subseteq [n] }$. And these functions $\chi_S$ are defined on the vertices $x \in \{ \pm 1\}^n$ as $\chi_S (x) = \prod_{i \in S} x_i$. And the eigenvalue of $\chi_S$ is $\vert S \vert/n$
- Even on $n=2$ I am not able to check this to be true. What is the justification for this above formulation?
Your eigenvalue of $L$ is slightly wrong if $L:=I-\frac{1}{n}A$ where $A$ is the adjacency matrix of the hypercube graph $Q_n$. The correct eigenvalue for the given eigenvector is $2\frac{|S|}{n}$ as I show below.
First we show $\chi_S$ is an eigenvector for $A$ (with an eigenvalue $\lambda_S$ that will be pop out from the proof of this). To do this, you need to check that for every vertex $v$, the sum of all $\chi_S(x)$ over all vertices $x$ that are adjacent to $v$ is equal to $\lambda_S\chi_S(v)$.
Well, the vertices adjacent to $v$ are those in which $v$ and $x$ differ in one position, say $i$. When $i\not\in S$, we have $\chi_S(v)=\chi_S(x)$. If $i\in S$, then $\chi_S(x)=-\chi_S(v)$. Hence, $$\sum_{x\sim v}\chi_S(x) = (n-|S|)\chi_S(v)+|S|(-\chi_S(v))=(n-2|S|)\chi_S(v).$$
So $\chi_S$ is an eigenvector for $A$ with eigenvalue $n-2|S|$. It follows that $$1-\frac{1}{n}(n-2|S|) = \frac{2|S|}{n}$$ is an eigenvalue of $L$ with eigenvector $\chi_S$.