Commutativity of Incidence Matrix with its Transpose for Symmetric Design

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

Then:
 * $A A^T = A^T A$

where $A^T$ is the transpose of $A$.

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

From $(3)$, we get:

We now have that $1/(k-\lambda)(A+\sqrt{(\lambda/v)J})$ is the inverse of $A^T-\sqrt{(\lambda/v)J}$, which implies that they commute with each other.

Thus:

whence:
 * $AA^T=(k-\lambda)+\lambda J=A^TA$