681: For Finite-Dimensional Vectors Space, Linearly Independent Subset with Dimension Number of Elements Is Basis

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description/proof of that for finite-dimensional vectors space, linearly independent subset with dimension number of elements is basis

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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 %field name% vectors space.
• The reader knows a definition of basis of module.
• The reader admits the proposition that for any finite-dimensional vectors space, any linearly independent subset can be expanded to be a basis by adding a finite number of elements.
• The reader admits the proposition that for any finite-dimensional vectors space, there is no basis that has more than the dimension number of elements.

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Target Context

• The reader will have a description and a proof of the proposition that for any finite-dimensional vectors space, any linearly independent subset with dimension number of elements is a basis.

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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:
$F$: $\in \{\text{ the fields }\}$
$V$: $\in \{\text{ the }d\text{ -dimensional }F\text{ vectors spaces }\}$
$S$: $\in \{\text{ the linearly independent subsets of }V\}$
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Statements:
$|S|=d$, where $|S|$ denotes the cardinality of $S$
$\wedge$
$S\in \{\text{ the bases of }V\}$

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2: Natural Language Description

For any field, $F$, and any $d$-dimensional $F$ vectors space, $V$, any linearly independent subset, $S\subseteq V$, with $d$ elements is a basis of $V$.

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3: Proof

Whole Strategy: Step 1: augment $S$ to be a basis; Step 2: see that $S$ is not really augmented.

Step 1:

$S$ can be augmented to become a basis, by the proposition that for any finite-dimensional vectors space, any linearly independent subset can be expanded to be a basis by adding a finite number of elements.

Step 2:

But if some positive number of elements were added to become a basis, the basis would have more than $d$ elements, a contradiction against the proposition that for any finite-dimensional vectors space, there is no basis that has more than the dimension number of elements. That means that $S$ has been already a basis with nothing added.

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References

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