Basic Props of Incidence Mats. of Balanced Incomplete Block Designs

# Basic Properties of Incidence Matrices of Balanced Incomplete Block Designs

Recall from The Incidence Matrix of a Balanced Incomplete Block Design page that if $(X, \mathcal A)$ is a $(v, b, r, k, \lambda)$-BIBD then the incidence matrix of this BIBD is the $v \times b$ matrix $M$ whose entries $m_{ij}$ are defined as:

(1)\begin{align} \quad m_{ij} = \left\{\begin{matrix} 0 \: \mathrm{if} \: x_i \not \in A_j \\ 1 \: \mathrm{if} \: x_i \in A_j \end{matrix}\right. \end{align}

We will now state some very basic properties regarding these incidence matrices.

Proposition 1: Let Let $(X, \mathcal A)$ be a $(v, b, r, k, \lambda)$-BIBD and let $M$ be a corresponding incidence matrix. Then every column of $M$ contains exactly $k$ many $1$s. |

**Proof:**The number $k$ represents the size of each block. Each column in $M$ represents a block in the BIBD and along each column there is a $1$ if the corresponding point is contained in that block and a $0$ otherwise. Hence each column of $M$ contains exactly $k$ many $1$s. $\blacksquare$

Proposition 2: Let $(X, \mathcal A)$ be a $(v, b, r, k, \lambda)$-BIBD and let $M$ be a corresponding incidence matrix. Then every row of $M$ contains exactly $r$ many $1$s. |

**Proof:**The number $r$ represents how many blocks contain any arbitrary point in $X$. Each row in $M$ represents a point in the BIBD and along each row there is a $1$ if the corresponding point is contained in that block and a $0$ otherwise. Hence each row of $M$ contains exactly $r$ many $1$s. $\blacksquare$

Proposition 3: Let $(X, \mathcal A)$ be a $(v, b, r, k, \lambda)$-BIBD and let $M$ be a corresponding incidence matrix. Then between any two distinct rows there are exactly $\lambda$ many $1$s in the same corresponding columns. |

**Proof:**The number $\lambda$ means that between any two distinct pair of points $x, y \in X$ with $x \neq y$ there is exactly $\lambda$ blocks containing both $x$ and $y$. So between any two distinct rows (representing points) $]] there are exactly $\lambda$ many $1$s in the same corresponding columns. $\blacksquare$