Product of the Incidence Matrix of a BIBD with its Transpose

Theorem
For any BIBD with parameters $$v,k,\lambda$$, let $$A$$ be its block incidence matrix. Then $$A\cdot A^T =(a_{ij})=(r-\lambda)I_v+\lambda{J_v}$$ where:
 * $$A$$ is $$v\times b$$,
 * $$A^T$$ is the transpose of $$A$$,
 * $$J_v$$ is the all $$v\times{v}$$ $$ 1$$'s matrix, and
 * $$I_v$$ is the $$v\times{v}$$ identity matrix.

i.e. $$A\cdot A^T= \begin{bmatrix} r & \lambda & \cdots & \lambda \\ \lambda & r & \cdots & \lambda \\ \vdots & \vdots & \ddots & \vdots \\ \lambda & \lambda & \cdots & r \\ \end{bmatrix} $$

Proof

 * Consider multiplying $$i^{th}$$ row of $$A$$ by the $$i^{th}$$ column of $$A^T$$. This is the same as multiplying the $$i^{th}$$ row of $$A$$ by the $$i^{th}$$ row of $$A$$. We know that each row of $$A$$ has $$r$$ enteries (since any point must be in $$r$$ blocks), then $$ (a_{ii})=r=\sum $$ of the all the $$1$$'s in row $$i$$. This completes the main diagonal.


 * Consider multiplying $$i^{th}$$ row of $$A$$ by the $$j^{th}$$ column of $$A^T$$. This is the same as multiplying the $$i^{th}$$ row of $$A$$ by the $$j^{th}$$ row of $$A$$. This will give the number of times the $$i^{th}$$ point is the same block as the $$j^{th}$$ point. Therefore when $$i\neq j$$, $$(a_{ij}) = \lambda $$

$$\therefore A\cdot A^T =(a_{ij})=(r-\lambda)I_v+\lambda{J_v}$$