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 $r = \lambda + \mu$ 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:

Now, since $k < v$ (this is obvious) and using the properties of a BIBD, we get that $r>\lambda$.

So $\det(A^\intercal A) \ne 0$.

Since $A^\intercal A$ is a $v\times v$ matrix, then the rank, $\rho$, of $A^\intercal A = v$.

Using the facts that $\rho(A^\intercal A) \le \rho(A)$, and $\rho(A) \le b$ (since A only has b cols), then $v \le \rho(A) \le b$.