Commutativity of Incidence Matrix with its Transpose for Symmetric Design

Theorem
Let $$A$$ be the incidence matrix of a symmetric design, then $$AA^T=A^TA$$.

Proof
Note:
 * 1) $$AJ=JA=kJ$$, so $$A^TJ=(JA)^T=(kJ)^T=kJ$$, and likewise $$JA^T=kJ$$,
 * 2) $$J^2=vJ$$, and
 * 3) If a design is symmetric, then $$AA^T=(r-\lambda)I+\lambda J=(k-\lambda)I+\lambda J$$

From Note 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 commute with each other.

Thus,

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