Symmetric Balanced Incomplete Block Designs

# Symmetric Balanced Incomplete Block Designs

Recall from The Block Number of a Balanced Incomplete Block Design page that if $(X, \mathcal A)$ is a $(v, k, \lambda)$-BIBD then the block number $b$ of this BIBD is the number of blocks contained in $\mathcal A$, i.e., $b = \mathcal A \mid$ and we derived the formula:

(1)\begin{align} \quad b = \frac{\lambda (v^2 - v)}{k^2 - k} \end{align}

It is often nice to consider BIBDs for which the number of points in the BIBD equals the number of blocks, i.e., $v = b$. Such BIBDs are given a special name which we define below.

Definition: A $(v, k, \lambda)$-BIBD $(X, \mathcal A)$ is said to be Symmetric if $v = b$. |

The following theorem tells us that every symmetric BIBD has the additional property that $r = k$.

Theorem: If $(X, \mathcal A)$ is a symmetric $(v, k, \lambda)$-BIBD then $r = k$. |

**Proof:**Suppose that $(X, \mathcal A)$ is a $(v, k, \lambda)$-BIBD and that $v = b$. From the formula for the replication number $b$ we must have that:

\begin{align} \quad v &= \frac{\lambda (v^2 - v)}{k^2 - k} \\ &= \frac{\lambda v(v - 1)}{k(k-1)} \\ \end{align}

- Dividing both sides above by $v$ (which is nonzero since $2 \leq k < v$ by the definition of a BIBD) and we get:

\begin{align} \quad 1 &= \frac{\lambda (v - 1)}{k(k-1)} \\ &= \frac{1}{k} \frac{\lambda (v - 1)}{k - 1} \\ &= \frac{r}{k} \end{align}

- Therefore $r = k$. $\blacksquare$