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:
\(\ds s_{i j}\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n a_{i k} \circ r_{k j}\) | ||||||||||||
\(\ds r_{k j}\) | \(=\) | \(\ds b_{k j} + c_{k j}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds s_{i j}\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n a_{i k} \circ \paren {b_{k j} + c_{k j} }\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^n a_{i k} \circ b_{k j} + \sum_{k \mathop = 1}^n a_{i k} \circ c_{k j}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds g_{i j} + h_{i j}\) |
Thus:
- $\mathbf A \paren {\mathbf B + \mathbf C} = \paren {\mathbf A \mathbf B} + \paren {\mathbf A \mathbf C}$
A similar construction shows that:
- $\paren {\mathbf B + \mathbf C} \mathbf A = \paren {\mathbf B \mathbf A} + \paren {\mathbf C \mathbf A}$
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices
- 1998: Richard Kaye and Robert Wilson: Linear Algebra ... (previous) ... (next): Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.2$ Addition and multiplication of matrices: $11$, $12$