description/proof of that for finite-dimensional vectors space, linearly independent subset can be expanded to be basis by adding finite elements
Topics
About: vectors space
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
Starting Context
- The reader knows a definition of basis of module.
- The reader knows a definition of linearly independent subset of module.
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.
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.
Main Body
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:
\(\exists S' \subseteq V \text{ such that } S' \in \{\text{ the finite sets }\} (S \cup S' \in \{\text{ the bases of } V\})\)
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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, \(S' \subseteq V\), such that \(S \cup S'\) is a basis of \(V\).
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.
\(S' \subseteq B\) is the elements of \(B\) added to \(S\), and is a finite set.
4: Note
In fact, inevitably \(\vert S \vert \le d\), but as that fact has not been proved as far as this proposition is concerned (it is proved using this proposition), the possibility of \(d \lt \vert S \vert\) is not ruled out.
