description/proof of that for finite-dimensional vectors space and basis, linearly independent set of elements can be augmented with some elements of basis 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
- 4: 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 with dimension number of elements is a basis.
Target Context
- The reader will have a description and a proof of the proposition that for any finite-dimensional vectors space and its any basis, any linearly independent set of elements can be augmented with some elements of the basis to become 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:
\(V\): \(\in \{\text{ the } d \text{ -dimensional vectors spaces }\}\)
\(B\): \(= \{b_1, ..., b_d\}\), \(\in \{\text{ the bases of } V\}\)
\(S\): \(= \{v_1, ..., v_k\}\), \(\in \{\text{ the linearly independent subsets of } V\}\), where \(0 \le k \le d\)
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Statements:
\(\exists \{b_{j_1}, ..., b_{j_{d - k}}\} (\{v_1, ..., v_k, b_{j_1}, ..., b_{j_{d - k}}\} \in \{\text{ the bases of } V\})\)
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2: Natural Language Description
For any finite \(d\)-dimensional vectors space, \(V\), any basis of \(V\), \(B := \{b_1, ..., b_d\}\), and any linearly independent subset of \(k\) elements, \(S := \{v_1, ..., v_k\} \subseteq V\), where \(0 \le k \le d\), there are some elements of the basis, \(b_{j_1}, ..., b_{j_{d - k}}\), such that \(\{v_1, ..., v_k, b_{j_1}, ..., b_{j_{d - k}}\}\) is a basis of \(V\).
3: Note
This proposition is not about just augmenting the linearly independent set of \(k\) elements to be a basis, but about augmenting the linearly independent set of \(k\) elements to become a basis with some elements of any fixed basis.
4: Proof
Whole Strategy: Step 1: if \(k \lt d\), add into \(S\) an element of \(B\) to form a linearly independent set, called \(S'\); Step 2: if \(k + 1 \lt d\), add into \(S'\) an element of \(B\) to form a linearly independent set, called again \(S'\); Step 3: and so on until \(k + l = d\); Step 4: conclude that \(S'\) is a basis.
Let be \(I := \{1, ..., d\}\) and \(I' := \{1, ..., k\}\).
Step 1:
If \(k = d\), let us go to the last step.
If \(k \lt d\), let us prove that there is an element of the basis, \(b_{j_1}\), such that \(S \cup \{b_{j_1}\}\) is linearly independent. Let us suppose otherwise. There would be a set of not-all-\(0\) coefficients, \(c_j^1, ..., c_j^{k + 1}\), such that \(\sum_{l \in I'} c_j^l v_l + c_j^{k + 1} b_j = 0\) for each \(j \in I\). \(c_j^{k + 1} \neq 0\), because otherwise, \(\sum_{l \in I'} c_j^l v_l = 0\), which would imply that all the coefficients were \(0\), because \(S\) was linearly independent, a contradiction. So, each \(b_j\) would be a linear combination of \(S\), which would imply that \(S\) was a basis, a contradiction against that \(V\) is \(d\)-dimensional. So, Let us choose such a \(b_{j_1}\) and call \(S \cup \{b_{j_1}\}\) "\(S'\)".
Step 2:
If \(k + 1 = d\), let us go to the last step.
If \(k + 1 \lt d\), let us prove that there is an element of the basis, \(b_{j_2}\), such that \(S' \cup \{b_{j_2}\}\) is linearly independent. Let us suppose otherwise. There would be a set of not-all-\(0\) coefficients, \(c_j^1, ..., c_j^{k + 2}\), such that \(\sum_{l \in I'} c_j^l v_l + c_j^{k + 1} b_{j_1} + c_j^{k + 2} b_j = 0\) for each \(j \in I\). \(c_j^{k + 2} \neq 0\), because otherwise, \(\sum_{l \in I'} c_j^l v_l + c_j^{k + 1} b_{j_1} = 0\), which would imply that all the coefficients were \(0\), because \(S'\) was linearly independent, a contradiction. So, each \(b_j\) would be a linear combination of \(S'\), which would imply that \((v_1, ..., v_k, b_{j_1})\) was a basis, a contradiction against that \(V\) is \(d\)-dimensional. Obviously, \(b_{j_1} \neq b_{j_2}\), because if \(b_{j_1} = b_{j_2}\), \(S' \cup b_{j_2}\) would not be linearly independent. So, Let us choose such a \(b_{j_2}\) and call \(S' \cup \{b_{j_1}\}\) again "\(S'\)".
Step 3:
And so on, which means that as far as \(k + l \lt d\), there is a linearly independent \(S' \cup \{b_{j_{l + 1}}\}\), called again \(S'\), and after all, it becomes that \(k + l = d\).
Step 4:
Then, \(S'\) is a basis of \(V\), by the proposition that for any finite-dimensional vectors space, any linearly independent subset with dimension number of elements is a basis.
