description/proof of that for finite dimensional vectors space basis, replacing element by linear combination of elements with nonzero coefficient for element forms 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 basis of module.
Target Context
- The reader will have a description and a proof of the proposition that for any finite dimensional vectors space basis, replacing any element by any linear combination of the elements with any nonzero coefficient for the element forms 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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V\): \(\in \{\text{ the } d \text{ -dimensional } F \text{ vectors spaces }\}\)
\(B\): \(= \{e_1, ..., e_d\}\), \(\in \{\text{ the bases of } V\}\)
\(e'_k\): \(= c^j e_j\), with the Einstein convention, where \(c^k \neq 0\)
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Statements:
\(B' := (B \setminus \{e_k\}) \cup \{e'_k\} \in \{\text{ the bases of } V\}\)
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2: Natural Language Description
For any field, \(F\), any \(d\)-dimensional \(F\) vectors space, \(V\), any basis of \(V\), \(B = \{e_1, ..., e_d\}\), and any \(e'_k := c^j e_j\), with the Einstein convention, where \(c^k \neq 0\), \(B' := (B \setminus \{e_k\}) \cup \{e'_k\}\) is a basis of \(V\).
3: Proof
Whole Strategy: Step 1: show that \(B'\) is linearly independent; Step 2: show that \(B'\) spans \(V\).
Step 1:
Let us prove that \(B'\) is linearly independent.
Let us suppose that \(d^1 e_1 + d^2 e_2 + ... + d^{k - 1} e_{k - 1} + d^k e'_k + d^{k + 1} e_{k + 1} + ... + d^d e_d = 0\). \(d^1 e_1 + d^2 e_2 + ... + d^{k - 1} e_{k - 1} + d^k c^j e_j + d^{k + 1} e_{k + 1} + ... + d^d e_d = 0\). As the coefficient of \(e_k\) is \(d^k c^k\) and \(B\) is linearly independent, \(d^k c^k = 0\), and \(d^k = 0\) as \(c^k \neq 0\). Then, \(d^1 e_1 + d^2 e_2 + ... + d^{k - 1} e_{k - 1} + 0 + d^{k + 1} e_{k + 1} + ... + d^d e_d = 0\), and each \(d^j\) is \(0\) as \(B\) is linearly independent.
So, \(B'\) is linearly independent.
Step 2:
Let us prove that any vector in \(V\) is a linear combination of the elements of \(B'\).
\(e'_k / c_k = c^j e_j / c_k = e_k + \sum_{j \neq k} (c^j / c_k) e_j\), so, \(e_k = e'_k / c_k - \sum_{j \neq k} (c^j / c_k) e_j\).
For any vector, \(v = d^j e_j\), \(e_k\) can be replaced by \(e'_k / c_k - \sum_{j \neq k} (c^j / c_k) e_j\), then, it is a linear combination of \(B'\).
