Product with Sum of Scalar

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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 Module Axiom $\text M 2$: Distributivity over Scalar Addition, 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

\(\ds \paren {\sum_{k \mathop = 1}^1 \lambda_1} \circ x\) \(=\) \(\ds \lambda_1 \circ x\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^1 \paren {\lambda_1 \circ x}\)

Hence $\map P 1$ is true.

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 $\map P {n + 1}$ 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:

\(\ds \paren {\sum_{k \mathop = 1}^{n + 1} \lambda_k} \circ x\) \(=\) \(\ds \paren {\sum_{k \mathop = 1}^n \lambda_k + \lambda_{n + 1} } \circ x\)
\(\ds \) \(=\) \(\ds \paren {\sum_{k \mathop = 1}^n \lambda_k \circ x} + \lambda_{n + 1} \circ x\) Module Axiom $\text M 2$: Distributivity over Scalar Addition
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n \paren {\lambda_k \circ x} + \lambda_{n + 1} \circ x\) Induction hypothesis
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^{n + 1} \paren {\lambda_k \circ x}\)

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


Therefore:

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

$\blacksquare$


Also see


Sources

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