2024-04-14

533: For Finite Set of Points on Real Vectors Space, if for Point, Set of Subtractions of Point from Other Points Is Linearly Independent, It Is So for Each Point

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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


  • 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:
\(V\): \(\in \{\text{ the real vectors spaces }\}\)
\(\{p_0, ..., p_n\}\): \(\subseteq V\)
//

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
\(\implies\)
\(\forall p_l (\{p_0 - p_l, ..., \widehat{p_l - p_l}, p_n - p_l\} \text{ is linearly independent })\).
//


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.


References


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