# Fisher's Inequality

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## Theorem

For any BIBD $\struct {v, k, \lambda}$, 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:

\(\ds \map \det {A^\intercal \cdot A}\) | \(=\) | \(\ds \mu^{v - 1} \paren {\mu + v \lambda}\) | Determinant of Combinatorial Matrix | |||||||||||

\(\ds \) | \(=\) | \(\ds \paren {r + \paren {v - 1} \lambda} \paren {r - \lambda}^{v - 1}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds r k \paren {r - \lambda}^{v - 1}\) | Necessary Condition for Existence of BIBD |

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

So $\map \det {A^\intercal A} \ne 0$.

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

Using the facts that $\map \rho {A^\intercal A} \le \map \rho A$, and $\map \rho A \le b$ (since $A$ only has $b$ cols), we have that:

- $v \le \map \rho A \le b$

$\blacksquare$

This article, or a section of it, needs explaining.Some "basic properties" here need to be linked, in order that the general thrust of this can be understood.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Explain}}` from the code. |

This article, or a section of it, needs explaining.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.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Explain}}` from the code. |

This article needs proofreading.Please check it for mathematical errors.If you believe there are none, please remove `{{Proofread}}` from the code.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Proofread}}` from the code. |

## Source of Name

This entry was named for Ronald Aylmer Fisher.