775: Bijective Linear Map Between Vectors Spaces Is 'Vectors Spaces - Linear Morphisms' Isomorphism

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description/proof of that bijective linear map between vectors spaces is 'vectors spaces - linear morphisms' isomorphism

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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: Natural Language Description
• 3: Note
• 4: Proof

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Starting Context

• The reader knows a definition of %field name% vectors space.
• The reader knows a definition of bijection.
• The reader knows a definition of linear map.
The reader knows a definition of %category name% isomorphism.

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Target Context

• The reader will have a description and a proof of the proposition that any bijective linear map between any vectors spaces is a 'vectors spaces - linear morphisms' isomorphism.

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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:
$F$: $\in \{\text{ the fields }\}$
$V_1$: $\in \{\text{ the not necessarily finite-dimensional }F\text{ vectors spaces }\}$
$V_2$: $\in \{\text{ the not necessarily finite-dimensional }F\text{ vectors spaces }\}$
$f$: $:V_1\to V_2$, $\in \{\text{ the linear maps }\}\cap \{\text{ the bijections }\}$
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Statements:
$f\in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}$
//

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2: Natural Language Description

For any field, $F$, any not necessarily finite-dimensional vectors spaces, $V_1,V_2$, and any bijective linear map, $f:V_1\to V_2$, $f$ is a 'vectors spaces - linear morphisms' isomorphism.

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3: Note

In general, a bijective morphism is not necessarily an isomorphism: for example, a bijective continuous map is not necessarily a 'topological spaces - continuous maps' isomorphism, because the inverse is not necessarily continuous, which is the reason why we need to specifically prove this proposition.

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4: Proof

Whole Strategy: Step 1: define the inverse, $f^{-1}:V_2\to V_1$; Step 2: see that $f^{-1}$ is linear; Step 3: conclude the proposition.

Step 1:

As $f$ is bijective, the inverse, $f^{-1}:V_2\to V_1$, is well-defined.

Step 2:

Let us see that $f^{-1}$ is linear.

Let $v_1,v_2\in V_2$ and $r_1,r_2\in F$ be any. $f^{-1}(r_1{v}_1+r_2{v}_2)=r_1{f}^{-1}(v_1)+r_2{f}^{-1}(v_2)$? $f(r_1{f}^{-1}(v_1)+r_2{f}^{-1}(v_2))=r_{1}f(f^{-1}(v_1))+r_{2}f(f^{-1}(v_2))=r_1{v}_1+r_2{v}_2$, which means that $f^{-1}(r_1{v}_1+r_2{v}_2)=r_1{f}^{-1}(v_1)+r_2{f}^{-1}(v_2)$. So, $f^{-1}$ is linear.

Step 3:

So, $f$ is a 'vectors spaces - linear morphisms' isomorphism.

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References

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