# Scalar Product with Sum

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## Theorem

Let $\left({G, +_G}\right)$ be an abelian group whose identity is $e$.

Let $\left({R, +_R, \times_R}\right)$ be a ring whose zero is $0_R$.

Let $\left({G, +_G, \circ}\right)_R$ be an $R$-module.

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

Let $\left \langle {x_m} \right \rangle$ be a sequence of elements of $G$.

Then:

$\displaystyle \lambda \circ \left({\sum_{k \mathop = 1}^m x_k}\right) = \sum_{k \mathop = 1}^m \left({\lambda \circ x_k}\right)$

## Proof

This follows by induction from Module Axiom $(1)$, as follows:

For all $m \in \N_{>0}$, let $P \left({m}\right)$ be the proposition:

$\displaystyle \lambda \circ \left({\sum_{k \mathop = 1}^m x_k}\right) = \sum_{k \mathop = 1}^m \left({\lambda \circ x_k}\right)$

### Basis for the Induction

$P(1)$ is true, as this just says:

$\lambda \circ x_1 = \lambda \circ x_1$

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $P \left({n}\right)$ is true, where $n \ge 1$, then it logically follows that $P \left({n+1}\right)$ is true.

So this is our induction hypothesis:

$\displaystyle \lambda \circ \left({\sum_{k \mathop = 1}^n x_k}\right) = \sum_{k \mathop = 1}^n \left({\lambda \circ x_k}\right)$

Then we need to show:

$\displaystyle \lambda \circ \left({\sum_{k \mathop = 1}^{n+1} x_k}\right) = \sum_{k \mathop = 1}^{n+1} \left({\lambda \circ x_k}\right)$

### Induction Step

This is our induction step:

 $\displaystyle \lambda \circ \left({\sum_{k \mathop = 1}^{n+1} x_k}\right)$ $=$ $\displaystyle \lambda \circ \left({\sum_{k \mathop = 1}^{n} x_k + x_{n+1} }\right)$ $\displaystyle$ $=$ $\displaystyle \lambda \circ \left({\sum_{k \mathop = 1}^{n} x_k}\right) + \lambda \circ x_{n+1}$ Module Axiom $(1)$ $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 1}^n \left({\lambda \circ x_k}\right) + \lambda \circ x_{n+1}$ Induction hypothesis $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 1}^{n+1} \left({\lambda \circ x_k}\right)$

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

Therefore:

$\displaystyle \forall m \in \N^*: \lambda \circ \left({\sum_{k \mathop = 1}^m x_k}\right) = \sum_{k \mathop = 1}^m \left({\lambda \circ x_k}\right)$

$\blacksquare$