679: For Finite-Dimensional Vectors Space, There Is No Linearly Independent Subset That Has More Than Dimension Elements

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description/proof of that for finite-dimensional vectors space, there is no linearly independent subset that has more than dimension elements

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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 dimension of vectors space.
• 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, there is no linearly independent subset that has more than the dimension number of elements.

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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:
$F$: $\in \{\text{ the fields }\}$
$V$: $\in \{\text{ the }d\text{ -dimensional }F\text{ vectors spaces }\}$
$S$: $\subseteq V$
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Statements:
$d<|S|$
$⟹$
$S\notin \{\text{ the linearly independent subsets of }V\}$
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2: Natural Language Description

For any field, $F$, and any $d$-dimensional $F$ vectors space, $V$, there is no linearly independent subset, $S\subseteq V$, that has more than $d$ elements.

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3: Proof

Whole Strategy: Step 1: suppose that $S$ was linearly independent; Step 2: augment $S$ to be a basis; Step 3: find a contradiction: there could not be such a basis.

Step 1:

Let us suppose that $S$ had more than $d$ elements.

Step 2:

Let us add some finite number of vectors to $S$ to form a basis, which would be possible 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 3:

The new basis would have more than the dimension number of 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.

So, $S$ is not linearly independent.

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References

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