# Product of Row Sum Unity Matrices

## Theorem

Let $\mathbf A = \sqbrk a_{m n}$ be an $m \times n$ matrix.

Let $\mathbf B = \sqbrk b_{n p}$ be an $n \times p$ matrix.

Let the row sum of $\mathbf A$ and $\mathbf B$ be equal to $1$.

Then the row sum of their (conventional) product is also $1$.

## Proof

We have that:

$\ds \sum_{i \mathop = 1}^n a_{i j} = \sum_{i \mathop = 1}^n b_{i j} = 1$

Then:

 $\ds \sum_{i \mathop = 1}^n \paren {\mathbf A \mathbf B}_{i j}$ $=$ $\ds \sum_{i \mathop = 1}^n \paren {\sum_{k \mathop = 1}^n a_{i k} b_{k j} }$ $\ds$ $=$ $\ds \sum_{i, k \mathop = 1}^n a_{i k} b_{k j}$ $\ds$ $=$ $\ds \sum_{k \mathop = 1}^n \paren {b_{k j} \sum_{i \mathop = 1}^n a_{i k} }$ $\ds$ $=$ $\ds \sum_{k \mathop = 1}^n \paren {b_{k j} \cdot 1}$ $\ds$ $=$ $\ds 1$

$\blacksquare$