description/proof of that for finite-dimensional vectors space, linearly independent subset with dimension number of elements is basis
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
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.
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.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V\): \(\in \{\text{ the } d \text{ -dimensional } F \text{ vectors spaces }\}\)
\(S\): \(\in \{\text{ the linearly independent subsets of } V\}\)
//
Statements:
\(\vert S \vert = d\), where \(\vert S \vert\) denotes the cardinality of \(S\)
\(\land\)
\(S \in \{\text{ the bases of } V\}\)
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 \(V\).
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.
