743: Linear Injection Between Same-Finite-Dimensional Vectors Spaces Is 'Vectors Spaces - Linear Morphisms' Isomorphism

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description/proof of that linear injection between same-finite-dimensional vectors spaces is 'vectors spaces - linear morphisms' isomorphism

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Topics

About: vectors space
About: category

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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: Proof

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Starting Context

• The reader knows a definition of linear map.
• The reader knows a definition of dimension of vectors space.
• The reader knows a definition of injection.
• The reader knows a definition of %category name% isomorphism.
• The reader admits the proposition that for any finite-dimensional vectors space, there is no linearly independent subset that has more than the dimension number of elements.
• The reader admits the proposition that any linear surjection from any finite-dimensional vectors space to any same-dimensional vectors space is a 'vectors spaces - linear morphisms' isomorphism.

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Target Context

• The reader will have a description and a proof of the proposition that any linear injection between any same-finite-dimensional 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:
$F$: $\in \{\text{ the fields }\}$
$V_1$: $\in \{\text{ the }d\text{ -dimensional }F\text{ vectors spaces }\}$
$V_2$: $\in \{\text{ the }d\text{ -dimensional }F\text{ vectors spaces }\}$
$f$: $:V_1\to V_2$, $\in \{\text{ the linear injections }\}$
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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 $d$-dimensional $F$ vectors spaces, $V_1,V_2$, and any linear injection, $f:V_1\to V_2$, $f$ is a 'vectors spaces - linear morphisms' isomorphism.

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3: Proof

Whole Strategy: see that $f$ is surjective; Step 1: choose any basis of $V_1$, $\{e_1,...,e_d\}$; Step 2: see that $\{f(e_1),...,f(e_d)\}$ is linearly independent on $V_2$; Step 3: see that $\{f(e_1),...,f(e_d)\}$ spans $V_2$; Step 4: conclude the proposition.

Step 1:

Let us choose any basis of $V_1$, $\{e_1,...,e_d\}$.

Step 2:

Let us see that $\{f(e_1),...,f(e_d)\}$ is linearly independent on $V_2$.

Any element, $v\in V_1$, is $v=c^j{e}_j$ (with the Einstein convention).

If $v\ne 0$, $f(v)=c^{j}f(e_j)\ne 0$, because otherwise, $f(v^{\prime }+v)=f(v^{\prime })+f(v)=f(v^{\prime })$, a contradiction against $f$'s being injective. That means that $c^{j}f(e_j)=0$ only if $v=0$, which implies that $c^j$ s are all $0$. So, $\{f(e_1),...,f(e_d)\}$ is linearly independent on $V_2$.

Step 3:

Let us see that $\{f(e_1),...,f(e_d)\}$ spans $V_2$.

$\{f(e_1),...,f(e_d)\}$ spans $V_2$, because otherwise, there would be an element, $v^{\prime }\in V_2$, that would be linearly independent from $\{f(e_1),...,f(e_d)\}$, a contradiction against the proposition that for any finite-dimensional vectors space, there is no linearly independent subset that has more than the dimension number of elements.

So, $f$ is surjective.

Step 4:

So, $f$ is a 'vectors spaces - linear morphisms' isomorphism, by the proposition that any linear surjection from any finite-dimensional vectors space to any same-dimensional vectors space is a 'vectors spaces - linear morphisms' isomorphism.

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References

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