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^T \cdot A = \begin{bmatrix}

r & \lambda & \cdots & \lambda \\ \lambda & r & \cdots & \lambda \\ \vdots & \vdots & \ddots & \vdots \\ \lambda & \lambda & \cdots & r \\ \end{bmatrix}$

Using Necessary Condition For The Existence of a BIBD, we have that $r > \lambda$.

So we can write $r = \lambda + \mu$ and so:


 * $A^T \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 $\det(A^TA) \ne 0$.

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

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