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:
 * $A^\intercal \cdot A = \left({a_{ij}}\right) = \left({r - \lambda}\right) I_v + \lambda J_v$

where:


 * $A$ is $v \times b$
 * $A^\intercal$ is the transpose of $A$
 * $J_v$ is the all $v\times{v}$ $ 1$'s matrix
 * $I_v$ is the $v\times{v}$ identity matrix.

That is:
 * $A^\intercal \cdot A = \begin{bmatrix}

r & \lambda & \cdots & \lambda \\ \lambda & r & \cdots & \lambda \\ \vdots & \vdots & \ddots & \vdots \\ \lambda & \lambda & \cdots & r \\ \end{bmatrix}$

Proof
Let row $i$ of $A$ be multiplied by column $i$ of $A^\intercal$.

This is the same as multiplying row $i$ of $A$ by row $i$ of $A$.

Each row of $A$ has $r$ entries (since any point must be in $r$ blocks).

Then:
 * $ \left({a_{ii}}\right) = r = \sum $ of the all the $1$'s in row $i$

This completes the main diagonal.

Let row $i$ of $A$ be multiplied by column $j$ of $A^\intercal$.

This is the same as multiplying row $i$ of $A$ by the row $j$ row of $A$.

This will give the number of times point $i$ point is the same block as point $j$.

Therefore:
 * $i \ne j \implies \left({a_{ij}}\right) = \lambda$

So:
 * $A^\intercal \cdot A = \left({a_{ij}}\right) = \left({r - \lambda}\right) I_v + \lambda{J_v}$