Product of Row Sum Unity Matrices
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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:
\(\ds \sum_{i \mathop = 1}^n \paren {\mathbf A \mathbf B}_{i j}\) | \(=\) | \(\ds \sum_{i \mathop = 1}^n \paren {\sum_{k \mathop = 1}^n a_{i k} b_{k j} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i, k \mathop = 1}^n a_{i k} b_{k j}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \paren {b_{k j} \sum_{i \mathop = 1}^n a_{i k} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \paren {b_{k j} \cdot 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1\) |
$\blacksquare$