108: Linear Surjection from Finite-Dimensional Vectors Space to Same-Dimensional Vectors Space Is 'Vectors Spaces - Linear Morphisms' Isomorphism

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description/proof of that linear surjection from finite-dimensional vectors space to same-dimensional vectors space 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 %field name% vectors space.
• The reader knows a definition of surjection.
• 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 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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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:
$F$: $\in \{\text{ the fields }\}$
$V_1$: $\in \{\text{ the d-dimensional }F\text{ vectors spaces }\}$
$V_2$: $\in \{\text{ the d-dimensional }F\text{ vectors spaces }\}$
$f$: $V_1\to V_2$, $\in \{\text{ the linear surjections }\}$
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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 space, $V_1$, and any d-dimensional $F$ vectors space, $V_2$, any linear surjection, $f:V_1\to V_2$, is a 'vectors spaces - linear morphisms' isomorphism.

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3: Proof

Whole Strategy: Step 1: take any basis for $V_1$, $\{b_1,b_2,...,b_d\}$; Step 2: see that $\{f(b_1),f(b_2),...,f(b_d)\}$ is a basis for $V_2$; Step 3: see that $f$ is injective; Step 4: see that $f^{-1}$ is linear.

Step 1:

Let us take any basis for $V_1$, $\{b_1,b_2,...,b_d\}$.

Step 2:

Let us see that $\{f(b_1),f(b_2),...,f(b_d)\}$ is a basis for $V_2$.

$\{f(b_1),f(b_2),...,f(b_d)\}$ is linearly independent on $V_2$, because otherwise, $V_2=f(V_1)=\{f(\sum _{j\in \{1,...,d\}}{c}^j{b}_j)\}=\{\sum _{j\in \{1,...,d\}}{c}^{j}f(b_j)\}$ would not be d-dimensional.

$\{f(b_1),f(b_2),...,f(b_d)\}$ spans $V_2$, because $V_2=\{\sum _{j\in \{1,...,d\}}{c}^{j}f(b_j)\}$.

Step 3:

Let us see that $f$ is injective.

For each $p=\sum _{j\in \{1,...,d\}}{c}^j{b}_j,p^{\prime }=\sum _{j\in \{1,...,d\}}{c}^{\prime j}{b}_j\in V_1$ such that $p\ne p^{\prime }$, $c^j\ne c^{\prime j}$ for a $j$, so, $f(\sum _{j\in \{1,...,d\}}{c}^j{b}_j)=\sum _{j\in \{1,...,d\}}{c}^{j}f(b_j)\ne \sum _{j\in \{1,...,d\}}{c}^{\prime j}f(b_j)=f(\sum _{j\in \{1,...,d\}}{c}^{\prime j}{b}_j)$, because $\{f(b_1),f(b_2),...,f(b_d)\}$ is a basis.

So, $f$ is bijective, and the inverse, $f^{-1}$, exists.

Step 4:

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

$f^{-1}(\sum _{j\in \{1,...,d\}}{c}^{j}f(b_j))=\sum _{j\in \{1,...,d\}}{c}^j{b}_j$, because $f(\sum _{j\in \{1,...,d\}}{c}^j{b}_j)=\sum _{j\in \{1,...,d\}}{c}^{j}f(b_j)$, $=\sum _{j\in \{1,...,d\}}{c}^j{f}^{-1}(f(b_j))$.

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References

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