Scalar Product with Sum

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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 $\sequence {x_m}$ be a sequence of elements of $G$.


$\ds \lambda \circ \paren {\sum_{k \mathop = 1}^m x_k} = \sum_{k \mathop = 1}^m \paren {\lambda \circ x_k}$


This follows by induction from Module Axiom $\text M 1$: Distributivity over Module Addition, as follows:

For all $m \in \N_{>0}$, let $\map P m$ be the proposition:

$\ds \lambda \circ \paren {\sum_{k \mathop = 1}^m x_k} = \sum_{k \mathop = 1}^m \paren {\lambda \circ x_k}$

Basis for the Induction

$\map P 1$ 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 $\map P n$ is true, where $n \ge 1$, then it logically follows that $\map P {n + 1}$ is true.

So this is our induction hypothesis:

$\ds \lambda \circ \paren {\sum_{k \mathop = 1}^n x_k} = \sum_{k \mathop = 1}^n \paren {\lambda \circ x_k}$

Then we need to show:

$\ds \lambda \circ \paren {\sum_{k \mathop = 1}^{n + 1} x_k} = \sum_{k \mathop = 1}^{n + 1} \paren {\lambda \circ x_k}$

Induction Step

This is our induction step:

\(\ds \lambda \circ \paren {\sum_{k \mathop = 1}^{n + 1} x_k}\) \(=\) \(\ds \lambda \circ \paren {\sum_{k \mathop = 1}^n x_k + x_{n + 1} }\)
\(\ds \) \(=\) \(\ds \lambda \circ \paren {\sum_{k \mathop = 1}^n x_k} + \lambda \circ x_{n + 1}\) Module Axiom $\text M 1$: Distributivity over Module Addition
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n \paren {\lambda \circ x_k} + \lambda \circ x_{n + 1}\) Induction hypothesis
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^{n + 1} \paren {\lambda \circ x_k}\)

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


$\ds \forall m \in \N_{>0}: \lambda \circ \paren {\sum_{k \mathop = 1}^m x_k} = \sum_{k \mathop = 1}^m \paren {\lambda \circ x_k}$


Also see