Fisher's Inequality

Theorem
For any BIBD$$(v,k,\lambda)$$, the number of blocks, $$b$$, must be greater then or equal to the number of points, $$v$$
 * i.e. $$b\geq v$$

Proof
Let $$A$$ be the incidence matrix. Consider the determinant:

$$

Now, since $$k\lambda$$.

$$\therefore$$ det$$(A^TA)\neq 0$$.

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 $$v\leq\rho(A)\leq b$$.