Commutativity of Incidence Matrix with its Transpose for Symmetric Design

Theorem
Let $A$ be the incidence matrix of a symmetric design.

Then:
 * $A A^\intercal = A^\intercal A$

where $A^\intercal$ is the transpose of $A$.

Proof
First note, we have:
 * $(1): \quad A J = J A = k J$, so $A^\intercal J = \left({J A}\right)^\intercal = \left({k J}\right)^\intercal = k J$, and likewise $J A^\intercal = k J$
 * $(2): \quad J^2 = v J$
 * $(3): \quad$ If a design is symmetric, then $A A^\intercal = \left({r - \lambda}\right) I + \lambda J = \left({k - \lambda}\right) I + \lambda J$

From $(3)$, we get:

We now have that:
 * $\displaystyle \frac 1 {k - \lambda} \left({A + \sqrt{\left({\frac \lambda v}\right) J}}\right)$

is the inverse of:
 * $A^\intercal - \sqrt{\left({\dfrac \lambda v}\right) J}$

which implies that they commute with each other.

Thus:

whence:
 * $A A^\intercal = \left({k - \lambda}\right) + \lambda J = A^\intercal A$