1109: 'Vectors Spaces - Linear Morphisms' Isomorphism Is Map That Maps Basis onto Basis Bijectively and Expands Mapping Linearly

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description/proof of that 'vectors spaces - linear morphisms' isomorphism is map that maps basis onto basis bijectively and expands mapping linearly

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Topics

About: vectors space

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The table of contents of this article

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

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Starting Context

• The reader knows a definition of %field name% vectors space.
• The reader knows a definition of %category name% isomorphism.
• The reader knows a definition of basis of module.
• The reader admits the proposition that for any 'vectors spaces - linear morphisms' isomorphism, the image of any linearly independent subset or any basis of the domain is linearly independent or a basis on the codomain.

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Target Context

• The reader will have a description and a proof of the proposition that any 'vectors spaces - linear morphisms' isomorphism is a map that maps any basis onto a basis bijectively and expands the mapping linearly.

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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:
$F$: $\in \{\text{ the fields }\}$
$V_1$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$V_2$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$f$: $:V_1\to V_2$, $\in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}$
$J$: $\in \{\text{ the possibly uncountable index sets }\}$
$B_1$: $=\{{b_1}_j|j\in J\}$, $\in \{\text{ the bases for }{V}_1\}$
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Statements:
$B_2:=f(B_1)\in \{\text{ the bases for }{V}_2\}$
$\wedge$
$f{|}_{B_1}:B_1\to B_2\in \{\text{ the bijections }\}$
$\wedge$
$\mathrm{\forall }v\in V_1(\mathrm{\exists }S\in \{\text{ the finite subsets of }J\}(v=\sum _{j\in S}{v}^j{b_1}_j)\wedge f(v)=\sum _{j\in S}{v}^{j}f({b_1}_j))$
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2: Note

Each of $V_1$ and $V_2$ does not need to be finite-dimensional.

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3: Proof

Whole Strategy: Step 1: see that $f(B_1)$ is a basis for $V_2$, by the proposition that for any 'vectors spaces - linear morphisms' isomorphism, the image of any linearly independent subset or any basis of the domain is linearly independent or a basis on the codomain; Step 2: see that $f{|}_{B_1}:B_1\to B_2$ is a bijection; Step 3: see that for each $v\in V_1$, $v=\sum _{j\in S}{v}^j{b_1}_j$ for a finite $S\subseteq J$ and $f(v)=\sum _{j\in S}{v}^{j}f({b_1}_j)$.

Step 1:

$B_2:=f(B_1)$ is a basis for $V_2$, by the proposition that for any 'vectors spaces - linear morphisms' isomorphism, the image of any linearly independent subset or any basis of the domain is linearly independent or a basis on the codomain.

Step 2:

$f{|}_{B_1}:B_1\to B_2$ is an injection, because $f$ is an injection.

$f{|}_{B_1}:B_1\to B_2$ is a surjection, because $B_2=f(B_1)$.

So, $f{|}_{B_1}:B_1\to B_2$ is a bijection.

Step 3:

Let $v\in V_1$ be any.

$v=\sum _{j\in S}{v}^j{b_1}_j$ for a finite $S\subseteq J$, by the definition of basis of module.

$f(v)=\sum _{j\in S}{v}^{j}f({b_1}_j)$, because any 'vectors spaces - linear morphisms' isomorphism is linear.

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References

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