description/proof of that for finite-dimensional vectors space, there is no linearly independent subset that has more than dimension 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
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.
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.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V\): \(\in \{\text{ the } d \text{ -dimensional } F \text{ vectors spaces }\}\)
\(S\): \(\subseteq V\)
//
Statements:
\(d \lt \vert S \vert\)
\(\implies\)
\(S \notin \{\text{ the linearly independent subsets of } V\}\)
//
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.
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.
