Commutativity of Incidence Matrix with its Transpose for Symmetric Design

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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 $A^\intercal J = \paren {J A}^\intercal = \paren {k J}^\intercal = k J$, and likewise $J A^\intercal = k J$
$(2): \quad J^2 = v J$
$(3): \quad$ If a design is symmetric, then $A A^\intercal = \paren {r - \lambda} I + \lambda J = \paren {k - \lambda} I + \lambda J$



From $(3)$, we get:

\(\ds \paren {A^\intercal - \sqrt {\paren {\frac \lambda v} J} } \paren {A + \sqrt {\paren {\frac \lambda v} J} }\) \(=\) \(\ds A^\intercal A + \sqrt {\frac \lambda v} \paren {A^\intercal J - J A} - \frac \lambda v J^2\)
\(\ds \) \(=\) \(\ds A^\intercal A - \lambda J = \paren {k - \lambda} I\)


We now have that:

$\ds \frac 1 {k - \lambda} \paren {A + \sqrt {\paren {\frac \lambda v} J} }$

is the inverse of:

$A^\intercal - \sqrt {\paren {\dfrac \lambda v} J}$

which implies that they commute with each other.



Thus:

\(\ds \paren {k - \lambda} I\) \(=\) \(\ds \paren {A + \sqrt {\frac \lambda v} J} \paren {A^\intercal - \sqrt {\frac \lambda v} J}\)
\(\ds \) \(=\) \(\ds A A^\intercal + \sqrt {\frac \lambda v} \paren {J A^\intercal - A J} - \frac \lambda v J^2\)
\(\ds \) \(=\) \(\ds A A^\intercal - \lambda J\)



whence:

$A A^\intercal = \paren {k - \lambda} + \lambda J = A^\intercal A$

$\blacksquare$