677: For Finite-Dimensional Vectors Space, Linearly Independent Subset Can Be Expanded to Be Basis by Adding Finite Elements

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description/proof of that for finite-dimensional vectors space, linearly independent subset can be expanded to be basis by adding finite 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
• 4: Note

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Starting Context

• The reader knows a definition of basis of module.
• The reader knows a definition of linearly independent subset of module.

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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 can be expanded to be a basis by adding a finite 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:
$F$: $\in \{\text{ the fields }\}$
$V$: $\in \{\text{ the }d\text{ -dimensional }F\text{ vectors spaces }\}$
$S$: $\in \{\text{ the linearly independent subsets of }V\}$
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Statements:
$\mathrm{\exists }{S}^{\prime }\subseteq V\text{ such that }{S}^{\prime }\in \{\text{ the finite sets }\}(S\cup S^{\prime }\in \{\text{ the bases of }V\})$
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2: Natural Language Description

For any field, $F$, any $d$-dimensional $F$ vectors space, $V$, and any linearly independent subset, $S\subseteq V$, there is a finite subset, $S^{\prime }\subseteq V$, such that $S\cup S^{\prime }$ is a basis of $V$.

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3: Proof

Whole Strategy: Step 1: take a $d$-cardinality basis; Step 2: iteratively add an element of the basis to $S$ to still be linearly independent; Step 3: see that eventually, the augmented $S$ (called "S''") will be a basis.

Step 1:

There is a basis, $B=\{e_1,e_2,...,e_d\}$.

Step 2:

If $S$ does not span $V$, let us add to $S$ the 1st element of the basis, $e_j$, that (the element) is independent from the elements of $S$, calling the result $S^{″}$. Such an $e_j$ exists, because otherwise, each element of $B$ would be a linear combination of $S$, which would mean that $S$ spanned $V$.

If $S^{″}$ does not span $V$, let us add to $S^{″}$ the next 1st element of the basis, $e_j$, that (the element) is independent from the elements of $S^{″}$, keeping to call the result $S^{″}$. As before, such an $e_j$ exists. And so on.

Step 3:

After all, $S^{″}$ spans $V$ with some elements of $B$ added, because $B$ spans $V$: in the worst case, all the elements of $B$ are added and $S^{″}$ spans $V$.

$S^{″}$ is linearly independent and spans $V$, so, $S^{″}$ is a basis.

$S^{\prime }\subseteq B$ is the elements of $B$ added to $S$, and is a finite set.

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4: Note

In fact, inevitably $|S|\le d$, but as that fact has not been proved as far as this proposition is concerned (it is proved using this proposition), the possibility of $d<|S|$ is not ruled out.

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References

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