517: Linear Map

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definition of linear map

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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: Note

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Starting Context

• The reader knows a definition of %ring name% module.
• The reader knows a definition of map.

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Target Context

• The reader will have a definition of linear map.

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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:
$R$: $\in \{\text{ the rings }\}$
$M_1$: $\in \{\text{ the }R\text{ modules }\}$
$M_2$: $\in \{\text{ the }R\text{ modules }\}$
$\ast f$: $:M_1\to M_2$
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Conditions:
$\mathrm{\forall }{r}_1,r_2\in R,\mathrm{\forall }{m}_1,m_2\in M_1(f(r_1{m}_1+r_2{m}_2)=r_{1}f(m_1)+r_{2}f(m_2))$.
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2: Note

The rings of $M_1$ and $M_2$ have to be the same in order for $r_{1}f(m_1)+r_{2}f(m_2)$ to make sense; there may be an argument that the ring of $M_2$ could be a superset of that of $M_1$, but we do not see any immediate necessity to include that case. For example, the complex numbers field (a ring) is not exactly any superset of the real numbers field (a ring) ($1+0i\in \mathbb{C}$ and $1\in \mathbb{R}$ are not exactly same according to the standard definitions: intuitively speaking, $(1,0)$ and $1$ are not same), although there is the canonical map from $\mathbb{R}$ into $\mathbb{C}$.

The $0$ element is inevitably mapped to the $0$ element, because $f(0)=f(0m)=0f(m)=0$ for any $m\in M_1$.

Any linear map is nothing but a module homomorphism.

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References

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