description/proof of that for vectors space, finite generator can be reduced to be 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: Note 1
- 4: Proof
- 5: Note 2
Starting Context
- The reader knows a definition of basis of module.
Target Context
- The reader will have a description and a proof of the proposition that for any vectors space, any finite generator can be reduced to be 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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(S\): \(\in \{\text{ the finite subsets of } V\}\)
//
Statements:
\(\forall v \in V (\exists r^j \in F (v = \sum_{v_j \in S} r^j v_j)))\)
\(\implies\)
\(\exists B \in \{\text{ the subsets of } S\} (B \in \{\text{ the bases of } V\})\)
//
2: Natural Language Description
For any field, \(F\), any \(F\) vectors space, \(V\), and any finite subset, \(S \subseteq V\), such that \(\forall v \in V (\exists r^j \in F (v = \sum_{v_j \in S} r^j v_j))\), there is a subset, \(B \subseteq S\), that is a basis of \(V\).
3: Note 1
Compare with the proposition that for any finite-dimensional vectors space, any subset that spans the space can be reduced to be a basis, which presupposes that the space is finite-dimensional but the subset may be infinite, while this proposition does not presuppose that the space is finite but the subset has to be finite.
4: Proof
Whole Strategy: Step 1: deal with the case, \(V = \{0\}\), and suppose otherwise thereafter; Step 2: remove any \(0\) from \(S\), denote the remained subset as \(S'\), and see that \(S'\) still spans \(V\); Step 3: if \(S'\) is linearly independent, it is a basis, otherwise, remove an element that is a linear combination of the other elements of \(S'\), denote the remained subset again as \(S'\), and see that \(S'\) still spans \(V\); Step 3: and so on, see that \(S'\) eventually becomes linearly independent without ceasing to span \(V\).
Step 1:
Let us suppose that \(V = \{0\}\).
Inevitably, \(S = \{0\}\).
\(B = \emptyset \subseteq S\) will do.
Let us suppose otherwise hereafter.
Step 2:
Let us remove any \(0\) from \(S\) and denote the remained subset as \(S'\).
\(S'\) is not empty, because otherwise, \(V = \{0\}\).
Let \(S' = \{v_1, ..., v_n\}\).
\(S'\) still spans \(V\), because \(0\) has played no role in spanning \(V\): \(0\) can be realized as \(0 v^1\) anyway.
Step 3:
If \(S'\) is linearly independent, it is a basis.
Otherwise, there is a not-all-zero \((r^1, ..., r^n)\) where \(r^j \in F\) such that \(r^1 v_1 + ... + r^n v_n = 0\). Supposing that \(r^j \neq 0\), \(v_j = {r^j}^{-1} (- r^1 v_1 - ... \widehat{- r^j v_j} - ... - r^n v_n)\).
Then, let us think of \(S' \setminus \{v_j\}\).
It still spans \(V\), because for each linear combination of \(S'\), \(v_j\) can be replaced with \({r^j}^{-1} (- r^1 v_1 - ... \widehat{- r^j v_j} - ... - r^n v_n)\), and it is a linear combination of \(S' \setminus \{v_j\}\).
Let us denote \(S' \setminus \{v_j\}\) again as \(S'\).
Step 4:
If \(S'\) is linearly independent, \(S'\) is a basis.
Otherwise, let us remove any element that is a linear combination of the other elements, and denote the remained subset as \(S'\), as before.
\(S'\) still spans \(V\) as before.
And so on, and eventually, \(S'\) becomes linearly independent, because at least, if \(S'\) has become to have only 1 element, it is linearly independent.
\(S'\) is a basis for \(V\).
As a result, \(V\) is finite-dimensional, although we have not presupposed so.
5: Note 2
In fact, any not-necessarily-finite generator can be reduced to be a basis, by the proposition that for any vectors space, any generator of the space, and any linearly independent subset contained in the generator, the generator can be reduced to be a basis with the linearly independent subset retained, but also this proof seems to be worth existing, because the proof of the more general proposition does not show any concrete way of getting the basis (that uses Zorn's lemma, which guarantees the existence of the basis without showing how to get it).
