728: For Linear Surjection Between Finite-Dimensional Vectors Spaces, Dimension of Codomain Is Equal to or Smaller than That of Domain

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description/proof of that for linear surjection between finite-dimensional vectors spaces, dimension of codomain is equal to or smaller than that of domain

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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: 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 admits the proposition that for any vectors space, any finite generator can be reduced to be a basis.

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Target Context

• The reader will have a description and a proof of the proposition that for any linear surjection between any finite-dimensional vectors spaces, the dimension of the codomain is equal to or smaller than that of the domain.

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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:
$F$: $\in \{\text{ the fields }\}$
$V_1$: $\in \{\text{ the }{d}_1\text{ -dimensional }F\text{ vectors spaces }\}$
$V_2$: $\in \{\text{ the }{d}_2\text{ -dimensional }F\text{ vectors spaces }\}$
$f$: $:V_1\to V_2$, $\in \{\text{ the linear surjections }\}$
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Statements:
$dim{V}_2\le dim{V}_1$
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2: Natural Language Description

For any field, $F$, any $d_1$-dimensional $F$ vectors space, $V_1$, any $d_2$-dimensional $F$ vectors space, $V_2$, and any linear surjection, $f:V_1\to V_2$, $dim{V}_2\le dim{V}_1$.

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3: Proof

Whole Strategy: Step 1: choose any basis for $V_1$; Step 2: express the range of $f$ with the image of the basis and see that $V_2$ is spanned by the image of the basis; Step 3: conclude the proposition.

Step 1:

Let us choose any basis for $V_1$, $\{e_1,...,e_{d_1}\}$.

Step 2:

For each $v\in V_1$, $v=\sum _{j\in \{1,...,d_1\}}{v}^j{e}_j$, where $v^j\in F$, $f(v)=\sum _{j\in \{1,...,d_1\}}{v}^{j}f(e_j)$, and $f(V_1)=V_2=\{\sum _{j\in \{1,...,d_1\}}{v}^{j}f(e_j)|v^j\in F\}$, which means that $V_2$ is spanned by $\{f(e_1),...,f(e_{d_1})\}$.

Step 3:

A basis of $V_2$ is a subset of $\{f(e_1),...,f(e_{d_1})\}$, by the proposition that for any vectors space, any finite generator can be reduced to be a basis, which implies that $dim{V}_2\le dim{V}_1$.

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References

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