# 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:

$\displaystyle \paren {\sum_{k \mathop = 1}^m \lambda_k} \circ x = \sum_{k \mathop = 1}^m \paren {\lambda_k \circ x}$

## Proof

This follows by induction from Module: $(2)$, as follows.

For all $m \in \N^*$, let $\map P m$ be the proposition:

$\displaystyle \paren {\sum_{k \mathop = 1}^m \lambda_k} \circ x = \sum_{k \mathop = 1}^m \paren {\lambda_k \circ x}$

### 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:

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

Then we need to show:

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

### Induction Step

This is our induction step:

 $\displaystyle \paren {\sum_{k \mathop = 1}^{n + 1} \lambda_k} \circ x$ $=$ $\displaystyle \paren {\sum_{k \mathop = 1}^n \lambda_k + \lambda_{n + 1} } \circ x$ $\displaystyle$ $=$ $\displaystyle \paren {\sum_{k \mathop = 1}^n \lambda_k \circ x} + \lambda_{n + 1} \circ x$ Module: $(2)$ $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 1}^n \paren {\lambda_k \circ x} + \lambda_{n + 1} \circ x$ Induction hypothesis $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 1}^{n + 1} \paren {\lambda_k \circ x}$

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

Therefore:

$\displaystyle \forall m \in \N^*: \paren {\sum_{k \mathop = 1}^m \lambda_k} \circ x = \sum_{k \mathop = 1}^m \paren {\lambda_k \circ x}$

$\blacksquare$