TITLE: For $m$ distinct subsets $A_1,A_2,...,A_m$ of $\{1,2,...,n\}$ such that $|A_i \cap A_j|=k \neq 0$ for any $i \neq j$, prove $m\leq n$.
QUESTION [6 upvotes]: A lovely combinatorial problem:

For $m$ distinct subsets $A_1,A_2,...,A_m$ of $\{1,2,...,n\}$ such that $|A_i \cap A_j|=k \neq 0$ for any $i \neq j$,  prove $m\leq n$.

I think some forms of inclusion-exclusion principles may be useful here. Nevertheless, it's unclear to me how to apply. The case that $k=1$ already stumped me. So I started in the reverse order and have figured out the case that $k=n$ or $n-1$ or $n-2$. But it seems hard to argue for general $k$ (maybe backward induction?). Any idea is appreciated. Thanks in advance.

REPLY [7 votes]: Consider the "incidence matrix" $M$ of your system. This is the $m\times n$
matrix with a one in position $(i,j)$ if $j\in A_i$ and a zero otherwise. Then
$MM^t$ is an $m\times m$ matrix with diagonal entries $|A_i|>k$ and $k$s elsewhere.
This is a positive definite matrix and so of rank $m$. But if $m>n$, the rank
of $M$ is at most $n$, and so that of $MM^t$ is at most $n$, a contradiction.