Product with Sum of Scalar

Theorem
Let $\struct {G, +_G}$ be an abelian group whose identity is $e$.

Let $\struct {R, +_R, \times_R}$ be a ring whose zero is $0_R$.

Let $\struct {G, +_G, \circ}_R$ be an $R$-module.

Let $x \in G, \lambda \in R$.

Let $\sequence {\lambda_m}$ be a sequence of elements of $R$, that is scalars.

Then:
 * $\ds \paren {\sum_{k \mathop = 1}^m \lambda_k} \circ x = \sum_{k \mathop = 1}^m \paren {\lambda_k \circ x}$

Proof
This follows by induction from, as follows.

For all $m \in \N_{>0}$, let $\map P m$ be the proposition:
 * $\ds \paren {\sum_{k \mathop = 1}^m \lambda_k} \circ x = \sum_{k \mathop = 1}^m \paren {\lambda_k \circ x}$

Basis for the Induction
$\map P 1$ is true, as this just says:
 * $\lambda_1 \circ x = \lambda_1 \circ x$

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P n$ is true, where $n \ge 1$, then it logically follows that $\map P {n + 1}$ is true.

So this is our induction hypothesis:


 * $\ds \paren {\sum_{k \mathop = 1}^n \lambda_k} \circ x = \sum_{k \mathop = 1}^n \paren {\lambda_k \circ x}$

Then we need to show:


 * $\ds \paren {\sum_{k \mathop = 1}^{n + 1} \lambda_k} \circ x = \sum_{k \mathop = 1}^{n + 1} \paren {\lambda_k \circ x}$

Induction Step
This is our induction step:

So $\map P n \implies \map P {n + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\ds \forall m \in \N_{>0}: \paren {\sum_{k \mathop = 1}^m \lambda_k} \circ x = \sum_{k \mathop = 1}^m \paren {\lambda_k \circ x}$

Also see

 * Basic Results about Modules
 * Basic Results about Unitary Modules