Fisher's Inequality
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Theorem
For any BIBD $\left({v, k, \lambda}\right)$, the number of blocks $b$ must be greater then or equal to the number of points $v$:
- $ b \ge v$
Proof
Let $A$ be the incidence matrix.
By Product of the Incidence Matrix of a BIBD with its Transpose, we have that:
- $A^\intercal \cdot A = \begin{bmatrix} r & \lambda & \cdots & \lambda \\ \lambda & r & \cdots & \lambda \\ \vdots & \vdots & \ddots & \vdots \\ \lambda & \lambda & \cdots & r \\ \end{bmatrix}$
From Necessary Condition for Existence of BIBD:
- $r > \lambda$
So we can write $r = \lambda + \mu$ for some $\mu > 0$ and so:
- $A^\intercal \cdot A = \begin{bmatrix} \lambda + \mu & \lambda & \cdots & \lambda \\ \lambda & \lambda + \mu & \cdots & \lambda \\ \vdots & \vdots & \ddots & \vdots \\ \lambda & \lambda & \cdots & \lambda + \mu \\ \end{bmatrix}$
This is a combinatorial matrix of order $v$.
So:
| \(\displaystyle \det \left({A^\intercal \cdot A}\right)\) | \(=\) | \(\displaystyle \mu^{v-1} \left ({\mu + v \lambda}\right)\) | $\quad$ from Determinant of Combinatorial Matrix | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \left ({r + \left({v-1}\right) \lambda}\right) \left({r - \lambda}\right)^{v-1}\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle r k \left({r - \lambda}\right)^{v-1}\) | $\quad$ by Necessary Condition for Existence of BIBD | $\quad$ |
This article contains statements that are justified by handwavery, in particular:
obvious
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Now, since $k < v$ (this is obvious) and using the properties of a BIBD, we get that $r > \lambda$.
So $\det \left({A^\intercal A}\right) \ne 0$.
Since $A^\intercal A$ is a $v \times v$ matrix, then the rank, $\rho$, of $A^\intercal A = v$.
Using the facts that $\rho \left({A^\intercal A}\right) \le \rho \left({A}\right)$, and $\rho \left({A}\right) \le b$ (since $A$ only has $b$ cols), then $v \le \rho \left({A}\right) \le b$.
$\blacksquare$
Some "basic properties" here need to be linked, in order that the general thrust of this can be understood.
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We can probably lose most of the fiddly stuff at the end by noting that as $\lambda$ and $\mu$ are both $>0$ then so is the determinant, and go straight to the last-line result about rank.
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Source of Name
This entry was named for Ronald Aylmer Fisher.
