Product_of_the_Incidence_Matrix_of_a_BIBD_with_its_Transpose

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 $A^\intercal \cdot A = \begin{bmatrix} r & \lambda & \cdots & \lambda \\ \lambda & r & \cdots & \lambda \\ \vdots & \vdots & \ddots & \vdots \\ \lambda & \lambda & \cdots & r \\ \end{bmatrix}$