description/proof of that affine subset of finite-dimensional real vectors space is spanned by finite affine-independent set of base points
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
Target Context
- The reader will have a description and a proof of the proposition that any affine subset of any finite-dimensional real vectors space is spanned by a finite affine-independent set of base points.
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:
\(V\): \(\in \{\text{ the } d \text{ -dimensional real vectors spaces }\}\)
\(S\): \(\subseteq V\), \(\in \{\text{ the affine subsets of } V\}\)
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Statements:
\(\exists \{p_0, ..., p_n\} \in \{\text{ the affine-independent sets of points on } V\} \text{ where } 0 \le n \le d (S = \{\sum_{j = 0 \sim n} t^j p_j \in V \vert t^j \in \mathbb{R}, \sum_{j = 0 \sim n} t^j = 1\})\).
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2: Natural Language Description
For any \(d\)-dimensional real vectors space, \(V\), any affine subset, \(S \subseteq V\), is the affine set spanned by an affine-independent set of some \(n\) points, \(\{p_0, ..., p_n\}\), where \(0 \le n \le d\).
3: Proof
Let us denote the affine subset spanned by \(\{p_0, ..., p_n\}\), as \(A (\{p_0, ..., p_n\})\).
When \(S\) has no point, \(S = A (\{\})\).
When \(S\) has only 1 point, \(p_0\), \(S = A (\{p_0\})\).
Let us suppose that \(S\) has more than 1 point.
Let us take any distinct points, \(p_0, p_1 \in S\).
\(\{p_0, p_1\}\) is an affine-independent set of points. \(A (\{p_0, p_1\}) \subseteq S\), because for each \(p = \sum_{j = 0 \sim 1} t^j p_j\) where \(\sum_{j = 0 \sim 1} t^j = 1\), \(p = p_1 + t^0 p_0 + (t^1 - 1) p_1 = p_1 + t^0 p_0 - t^0 p_1 = p_1 + t^0 (p_0 - p_1) \in S\).
If \(A (\{p_0, p_1\}) = S\), we are finished.
Otherwise, there is a point, \(p_2 \in S\), such that \(p_2 \notin A (\{p_0, p_1\})\).
\(\{p_0, p_1, p_2\}\) is affine-independent, because \(p_1 - p_0, p_2 - p_0\) are linearly independent, because otherwise, \(p_2 - p_0\) would be a scalar multiple of \(p_1 - p_0\), which would mean \(p_2 \in A (\{p_0, p_1\})\). \(A (\{p_0, p_1, p_2\}) \subseteq S\), because for each \(p = \sum_{j = 0 \sim 2} t^j p_j\) where \(\sum_{j = 0 \sim 2} t^j = 1\) and \(p_2 \neq 1\), which means that \(\sum_{j = 0 \sim 1} t^j \neq 0\), \(p = p_2 + \sum_{j = 0 \sim 1} t^j p_j + (t^2 - 1) p_2 = p_2 + \sum_{j = 0 \sim 1} t^j p_j - (\sum_{j = 0 \sim 1} t^j) p_2 = p_2 + \sum_{j = 0 \sim 1} t^j ((\sum_{j = 0 \sim 1} t^j p_j) / \sum_{j = 0 \sim 1} t^j - p_2) \in S\), because \((\sum_{j = 0 \sim 1} t^j p_j) / \sum_{j = 0 \sim 1} t^j \in A (\{p_0, p_1\}) \subseteq S\), because \(\sum_{j = 0 \sim 1} t^j / \sum_{j = 0 \sim 1} t^j = 1\); when \(p_2 = 1\), \(p = p_2 \in S\).
If \(A (\{p_0, p_1, p_2\}) = S\), we are finished.
Otherwise, there is a point, \(p_3 \in S\), such that \(p_3 \notin A (\{p_0, p_1, p_2\})\).
\(\{p_0, p_1, p_2, p_3\}\) is affine-independent, because \(p_1 - p_0, p_2 - p_0, p_3 - p_0\) are linearly independent, because otherwise, \(p_3 - p_0\) would be a linear combination of \(p_1 - p_0, p_2 - p_0\), which would mean \(p_3 \in A (\{p_0, p_1, p_2\})\). \(A (\{p_0, p_1, p_2, p_3\}) \subseteq S\), because for each \(p = \sum_{j = 0 \sim 3} t^j p_j\) where \(\sum_{j = 0 \sim 3} t^j = 1\) and \(p_3 \neq 1\), which means that \(\sum_{j = 0 \sim 2} t^j \neq 0\), \(p = p_3 + \sum_{j = 0 \sim 2} t^j p_j + (t^3 - 1) p_3 = p_3 + \sum_{j = 0 \sim 2} t^j p_j - (\sum_{j = 0 \sim 2} t^j) p_3 = p_3 + \sum_{j = 0 \sim 2} t^j ((\sum_{j = 0 \sim 2} t^j p_j) / \sum_{j = 0 \sim 2} t^j - p_3) \in S\), because \((\sum_{j = 0 \sim 2} t^j p_j) / \sum_{j = 0 \sim 2} t^j \in A (\{p_0, p_1, p_2\}) \subseteq S\), because \(\sum_{j = 0 \sim 2} t^j / \sum_{j = 0 \sim 2} t^j = 1\); when \(p_3 = 1\), \(p = p_3 \in S\).
And so on.
After all, \(A (\{p_0, p_1, p_2, ..., p_n\}) = S\) for an \(n \le d\), because any more-than-\(d\) vectors cannot be independent on \(V\).
So, \(S\) is the affine set spanned by \(\{p_1, ..., p_n\}\).
