Matrix Multiplication Distributes over Matrix Addition

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Theorem

Matrix multiplication (conventional) is distributive over matrix entrywise addition.


Proof

Let $\mathbf A = \sqbrk a_{m n}, \mathbf B = \sqbrk b_{n p}, \mathbf C = \sqbrk c_{n p}$ be matrices over a ring $\struct {R, +, \circ}$.

Consider $\mathbf A \paren {\mathbf B + \mathbf C}$.

Let $\mathbf R = \sqbrk r_{n p} = \mathbf B + \mathbf C, \mathbf S = \sqbrk s_{m p} = \mathbf A \paren {\mathbf B + \mathbf C}$.

Let $\mathbf G = \sqbrk g_{m p} = \mathbf A \mathbf B, \mathbf H = \sqbrk h_{m p} = \mathbf A \mathbf C$.

Then:

\(\displaystyle s_{i j}\) \(=\) \(\displaystyle \sum_{k \mathop = 1}^n a_{i k} \circ r_{k j}\)
\(\displaystyle r_{k j}\) \(=\) \(\displaystyle b_{k j} + c_{k j}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle s_{i j}\) \(=\) \(\displaystyle \sum_{k \mathop = 1}^n a_{i k} \circ \paren {b_{k j} + c_{k j} }\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 1}^n a_{i k} \circ b_{k j} + \sum_{k \mathop = 1}^n a_{i k} \circ c_{k j}\)
\(\displaystyle \) \(=\) \(\displaystyle g_{i j} + h_{i j}\)


Thus:

$\mathbf A \paren {\mathbf B + \mathbf C} = \paren {\mathbf A \mathbf B} + \paren {\mathbf A \mathbf C}$

A similar construction shows that:

$\paren {\mathbf B + \mathbf C} \mathbf A = \paren {\mathbf B \mathbf A} + \paren {\mathbf C \mathbf A}$

$\blacksquare$


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