description/proof of that for finite set of points on real vectors space, if for point, set of subtractions of point from other points are linearly independent, it is so for each point
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 linearly independent subset of module.
Target Context
- The reader will have a description and a proof of the proposition that for any finite set of points on any real vectors space, if for one of the points, the set of the subtractions of the point from the other points is linearly independent, it is so for each point.
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:
\(V\): \(\in \{\text{ the real vectors spaces }\}\)
\(\{p_0, ..., p_n\}\): \(\subseteq V\)
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Statements:
\(\exists p_k (\{p_0 - p_k, ..., \widehat{p_k - p_k}, p_n - p_k\} \text{ is linearly independent })\), where the hat mark denotes that the component is missing
\(\implies\)
\(\forall p_l (\{p_0 - p_l, ..., \widehat{p_l - p_l}, p_n - p_l\} \text{ is linearly independent })\).
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2: Natural Language Description
For any real vectors space, \(V\), and any finite set of points, \(\{p_0, ..., p_n\} \subseteq V\), if for a \(p_k\), \(\{p_0 - p_k, ..., \widehat{p_j - p_k}, p_n - p_k\}\) is linearly independent, where the hat mark denotes that the component is missing, for each \(p_l\), \(\{p_0 - p_l, ..., \widehat{p_l - p_l}, p_n - p_l\}\) is linearly independent.
3: Proof
Let us suppose that \(\{p_0 - p_k, ..., \widehat{p_k - p_k}, p_n - p_k\}\) is linearly independent.
Let \(l\) be any \(l \in \{0, ..., n\}\) and \(l \neq k\) and \(J_l := \{0, ..., n\} \setminus \{l\}\).
Let \(\sum_{j \in J_l} (t^j (p_j - p_l)) = 0\). \(= \sum_{j \in J_l} (t^j (p_j - p_k + p_k - p_l)) = \sum_{j \in J_l} (t^j (p_j - p_k)) - \sum_{j \in J_l} (t^j (p_l - p_k)) = \sum_{j \in J_l} (t^j (p_j - p_k)) - (\sum_{j \in J_l} t^j) (p_l - p_k)\). Then, \(t^j = 0\) for each \(j \neq k\) and \(\sum_{j \in J_l} t^j = 0\), but \(\sum_{j \in J_l} t^j = t^k = 0\). So, each \(t^j\) is \(0\), which means that \(\{p_0 - p_l, ..., \widehat{p_l - p_l}, p_n - p_l\}\) is linearly independent.
