We have a set system $A_1,A_2,\dots A_m$ where $A_i \subseteq [n]$. Given that $|A_i| = k$ and $|A_i \cap A_j| = \lambda \;\;\;\forall i \neq j \;\;\; ( k \text{ and } \lambda \text{ are some positive integers } )\;\;$ . Prove that $m \leq n $.
We represent each set $A_i$ with a vector $V_i$ of size $n$ of zeros and ones (1 in $x^{th}$ place in $V_i$ if $x \in A_i$ otherwise zero) and $B = \begin{bmatrix} V_1 \\ V_2 \\ V_3 \\ \dots \\ \dots \\ V_m \\ \end{bmatrix}_{m \times n} $
then $BB^{T} \text{ becomes } \begin{bmatrix} k&\lambda&\lambda&\dots&\lambda \\ \lambda&k&\lambda&\dots&\lambda \\ \lambda&\lambda&k&\dots&\lambda \\ \dots&\dots&\dots&\dots&\dots \\ \lambda&\lambda&k&\dots&k \\ \end{bmatrix}_{m \times m} $
How to prove that this $BB^{T}$ is of full rank (using the idea of positive definite matrices )? Then we can apply $n \geq \text{rank}(B) \geq \text{rank}(BB^{T})= m$ to prove that $m \leq n$.
Thank You !
You can assume without loss of generality that the $A_{i}$ are distinct, and then $\lambda < k$.
Note that $$ B B^{T} = (k - \lambda) I + \lambda S, $$ where $$ S = \begin{bmatrix} 1 & 1 & \dots & 1\\ 1 & 1 & \dots & 1\\ & & \ddots & \\ 1 & 1 & \dots & 1\\ \end{bmatrix} $$ So if $B B^{T} v = 0$, for some $v = [v_{1}, \dots, v_{m}]^{T} \ne 0$, we have for each $i$ $$ (k - \lambda) v_{i} = - \lambda \sum_{j=1}^{m} v_{j} $$ so that all $v_{i}$ are equal, $v_{i} = u \ne 0$ for all $u$, and thus $(k - \lambda) u = - \lambda m u$, and $k = \lambda (1 - m) \le 0$, a contradiction.