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 $A A^\intercal = \paren {r - \lambda} I + \lambda J = \paren {k - \lambda} I + \lambda J$
This article, or a section of it, needs explaining. In particular: 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. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
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.
This article, or a section of it, needs explaining. In particular: ... why? You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
Thus:
\(\ds \paren {k - \lambda} I\) | \(=\) | \(\ds \paren {A + \sqrt {\frac \lambda v} J} \paren {A^\intercal - \sqrt {\frac \lambda v} J}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds A A^\intercal + \sqrt {\frac \lambda v} \paren {J A^\intercal - A J} - \frac \lambda v J^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds A A^\intercal - \lambda J\) |
This article, or a section of it, needs explaining. In particular: The above steps need justification You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
whence:
- $A A^\intercal = \paren {k - \lambda} + \lambda J = A^\intercal A$
$\blacksquare$