950: From Module with Basis into Module, Linear Map Can Be Defined by Mapping Basis and Linearly Expanding Mapping

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description/proof of that from module with basis into module, linear map can be defined by mapping basis and linearly expanding mapping

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Topics

About: module

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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Starting Context

• The reader knows a definition of basis of module.
• The reader knows a definition of linear map.
• The reader admits the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.

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Target Context

• The reader will have a description and a proof of the proposition that from any module with any basis into any module, a linear map can be defined by mapping the basis and linearly expanding the mapping.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$R$: $\in \{\text{ the rings }\}$
$M_1$: $\in \{\text{ the }R\text{ modules }\}$
$J$: $\in \{\text{ the possibly uncountable index sets }\}$
$B$: $=\{b_j|j\in J\}$, $\in \{\text{ the bases for }{M}_1\}$
$M_2$: $\in \{\text{ the }R\text{ modules }\}$
$f$: $:B\to M_2$
$g$: $:M_1\to M_2,\sum _{j\in S}{m}^j{b}_j\mapsto \sum _{j\in S}{m}^{j}f(b_j)$
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Statements:
$g\in \{\text{ the linear maps }\}$
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$M_2$ does not need to have any basis.

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2: Proof

Whole Strategy: Step 1: see that $g$ is well-defined; Step 2: see that $g$ is linear.

Step 1:

Let us see that $g$ is well-defined.

Each $m\in M_1$ is expressed as $m=\sum _{j\in S}{m}^j{b}_j$ by the definition of basis of module, which means that each element of $M_1$ is mapped.

The expression is unique when each $m^j$ is required to be nonzero, by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.

That means that when $m=\sum _{j\in S}{m}^j{b}_j=\sum _{j\in S^{\prime }}{\tilde{m}}^j{b}_j$, $S\subseteq S^{\prime }$ and for each $j\in S$, ${\tilde{m}}^j=m^j$ and for each $j\in S^{\prime }\setminus S$, ${\tilde{m}}^j=0$.

So, $\sum _{j\in S^{\prime }}{\tilde{m}}^{j}f(b_j)=\sum _{j\in S}{\tilde{m}}^{j}f(b_j)+\sum _{j\in S^{\prime }\setminus S}{\tilde{m}}^{j}f(b_j)=\sum _{j\in S}{m}^{j}f(b_j)+\sum _{j\in S^{\prime }\setminus S}0f(b_j)=\sum _{j\in S}{m}^{j}f(b_j)$, which means that the result does not depend on the expression.

$\sum _{j\in S}{m}^{j}f(b_j)\in M_2$.

So, each element of $M_1$ is mapped into $M_2$ and there is no ambiguity in the mapping.

Step 2:

Let us see that $g$ is indeed linear.

Let $m,m^{\prime }\in M_1$ be any. Let $r,r^{\prime }\in R$ be any.

$m=\sum _{j\in S}{m}^j{b}_j$ and $m^{\prime }=\sum _{j\in S^{\prime }}{m}^{\prime j}{b}_j$.

Let us take $S\cup S^{\prime }$. For each $j\in S\cup S^{\prime }$, let ${\tilde{m}}^j=m^j$ for $j\in S$ and ${\tilde{m}}^j=0$ for $j\notin S$. For each $j\in S\cup S^{\prime }$, let ${\widetilde{m^{\prime }}}^j=m^{\prime j}$ for $j\in S^{\prime }$ and ${\widetilde{m^{\prime }}}^j=0$ for $j\notin S^{\prime }$.

Then, $m=\sum _{j\in S\cup S^{\prime }}{\tilde{m}}^j{b}_j$ and $m^{\prime }=\sum _{j\in S\cup S^{\prime }}{\widetilde{m^{\prime }}}^j{b}_j$.

$g(rm+r^{\prime }{m}^{\prime })=g(r\sum _{j\in S\cup S^{\prime }}{\tilde{m}}^j{b}_j+r^{\prime }\sum _{j\in S\cup S^{\prime }}{\widetilde{m^{\prime }}}^j{b}_j)=g(\sum _{j\in S\cup S^{\prime }}(r{\tilde{m}}^j+r^{\prime }{\widetilde{m^{\prime }}}^j)b_j)=\sum _{j\in S\cup S^{\prime }}(r{\tilde{m}}^j+r^{\prime }{\widetilde{m^{\prime }}}^j)f(b_j)=r\sum _{j\in S\cup S^{\prime }}{\tilde{m}}^{j}f(b_j)+r^{\prime }\sum _{j\in S\cup S^{\prime }}{\widetilde{m^{\prime }}}^{j}f(b_j)=rg(\sum _{j\in S\cup S^{\prime }}{\tilde{m}}^j{b}_j)+r^{\prime }g(\sum _{j\in S\cup S^{\prime }}{\widetilde{m^{\prime }}}^j{b}_j)=rg(m)+r^{\prime }g(m^{\prime })$.

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References

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