Commutativity of an Incidence Matrix with its Transpose for a 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 $A A^T = (r-\lambda) I + \lambda J = (k-\lambda)I + \lambda J$
We need to establish these three results. We need to confirm that $J$ is the ones matrix (I think it is, from the context) and then make sure of those above results.
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From $(3)$, we get:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \left({A^T - \sqrt{\left({\frac{\lambda} v}\right) J} }\right) \left({A + \sqrt{\left({\frac{\lambda}v}\right)J} }\right)\) | \(=\) | \(\displaystyle A^T A + \sqrt{\frac{\lambda} v} \left({A^T J - J A}\right) - \frac{\lambda} v J^2\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle A^T A - \lambda J = \left({k - \lambda}\right) I\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
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.
... why?
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Thus:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle (k-\lambda)I\) | \(=\) | \(\displaystyle (A+\sqrt{\frac{\lambda} v}J)(A^T-\sqrt{\frac{\lambda} v} J)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle AA^T+\sqrt{\frac{\lambda} v}(JA^T-AJ)-\frac{\lambda} v J^2\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle AA^T-\lambda J\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
whence:
- $AA^T=(k-\lambda)+\lambda J=A^TA$
$\blacksquare$
