$\newcommand{\Z}{\mathbb{Z}} \newcommand{\J}{\mathcal{J}} \newcommand{\la}{\lambda} \newcommand{\1}{\mathbf{1}} \newcommand{\R}{\mathbb{R}}$

Take any $n\in[3;\infty]$. Here and in what follows, $[k;\ell]:=[k,\ell]\cap\Z$. Take then any $s\in[2;n-1]$. Let $\J:=\J_s:=\binom{[n]}s$, the set of all $s$-sets in $[n]:=[1;n]$, that is, the set of all subsets of the set $[n]$ of size (cardinality) $s$. Let $N:=N_s:=|\J_s|=\binom ns$, the cardinality of the set $\J$. Consider the $N\times N$ matrix \begin{equation} A:=A_s:=(|J\cap K|\colon J,K\in\J). \end{equation} The problem is to prove

Theorem:$\quad$ The eigenvalues of the matrix $A$ are $\la_1:=s\binom{n-1}{s-1}$ (of multiplicity $1$), $\la_2:=\binom{n-2}{s-1}$ (of multiplicity $n-1$), and $\la_3:=0$ (of multiplicity $N-n$).

**Comments:**
It is easy to see that the symmetric matrix
\begin{equation}
P:=\tfrac1{\la_1}\,A
\end{equation}
is double stochastic, which implies that $\la_1$ is indeed an eigenvalue of $A$, with a corresponding eigenvector $\1:=(1\colon J\in\J)$. That is, the vector $\pi:=\frac1N\,\1$ is the stationary distribution of the random walk/Markov chain on the set $\J_s$ of the $s$-sets with the transition probability matrix $P$. The eigenvalues of the matrix $P$ are $1$ (of multiplicity $1$), $\nu:=\la_2/\la_1:=\frac{n-s}{s(n-1)}\le\frac{n-2}{2(n-1)}\in(0,1/2)$ (of multiplicity $n-1$), and $0$ (of multiplicity $N-n$). So, $P=P_1+\nu P_2$, where $P_1$ and $P_2$ are the orthoprojectors onto the eigenspaces belonging to the respective eigenvalues $1$ and $\nu$. Take now any initial distribution $p$ on $\J$. Then for all natural $m$ we have $pP^m=pP_1+\nu^m pP_2=\pi+\nu^m pP_2$, so that we have the exponential convergence of the distribution $pP^m$ of the chain at time $m$ to the stationary distribution $\pi$:
\begin{equation}
pP^m-\pi=\nu^m\,pP_2.
\end{equation}
The difference $1-\nu$ between the two largest distinct eigenvalues of $P$ is called its spectral gap, which determines the rate of the exponential convergence.

From the spectral decomposition \begin{equation} A=\la_1 P_1+\la_2 P_2 \end{equation} it also immediately follows that for any $x=(x_J)\in\R^\J$ \begin{equation} \|Ax\|_2^2\ge\la_1^2\|P_1x\|_2^2=\la_1^2\Big(\sum_J x_J\Big)^2. \end{equation} For $s=2$, the latter inequality was proved by Fedor Petrov at Is this bound uniform in $N$? ; however, his proof seems to be easy to extend to general $s$.