Scalar Product with Product
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Theorem
Let $\struct {G, +_G}$ be an abelian group.
Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Let $x \in G, \lambda \in R, n \in \Z$.
Then:
- $\lambda \circ \paren {n \cdot x} = n \cdot \paren {\lambda \circ x} = \paren {n \cdot \lambda} \circ x$
Proof
First let $n = 0$.
The assertion follows directly from Scalar Product with Identity.
Next, let $n > 0$.
The assertion follows directly from Scalar Product with Sum and Product with Sum of Scalar, by letting $m = n$ and making all the $\lambda$'s and $x$'s the same.
Finally, let $n < 0$.
The assertion follows from Scalar Product with Product for positive $n$, Scalar Product with Inverse, and from Negative Index Law for Monoids.
$\blacksquare$
Also see
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Theorem $26.2 \ (5)$