Fisher's Inequality

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:

\(\displaystyle \det \left({A^\intercal \cdot A}\right)\)

\(=\)

\(\displaystyle \mu^{v-1} \left ({\mu + v \lambda}\right)\)

from Determinant of Combinatorial Matrix

\(\displaystyle \)

\(=\)

\(\displaystyle \left ({r + \left({v-1}\right) \lambda}\right) \left({r - \lambda}\right)^{v-1}\)

\(\displaystyle \)

\(=\)

\(\displaystyle r k \left({r - \lambda}\right)^{v-1}\)

by Necessary Condition for Existence of BIBD

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 $v \le \rho \left({A}\right) \le b$.

$\blacksquare$

Source of Name

This entry was named for Ronald Aylmer Fisher.