description/proof of that when convex set spanned by non-affine-independent set of base points on real vectors space is affine simplex, point whose original coefficients are all positive is on simplex interior of simplex, but point one of whose original coefficients is 0 is not necessarily on simplex boundary of simplex
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 convex set spanned by possibly-non-affine-independent set of base points on real vectors space.
- The reader knows a definition of affine simplex.
- The reader knows a definition of simplex interior of affine simplex.
- The reader knows a definition of simplex boundary of affine simplex.
- The reader admits the proposition that when the convex set spanned by any non-affine-independent set of base points on any real vectors space is an affine simplex, it is spanned by an affine-independent subset of the base points.
Target Context
- The reader will have a description and a proof of the proposition that when the convex set spanned by any non-affine-independent set of base points on any real vectors space is an affine simplex, any point whose original coefficients are all positive is on the simplex interior of the simplex, but a point one of whose original coefficients is 0 is not necessarily on the simplex boundary of the simplex.
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:
\(V\): \(\in \{\text{ the real vectors spaces }\}\)
\(\{p_0, ..., p_n\}\): \(\subseteq V\), \(\in \{\text{ the non-affine-independent sets of base points on } V\}\)
\(S\): \(= \{\sum_{j \in \{0, ..., n\}} t^j p_j \in V \vert t^j \in \mathbb{R}, \sum_{j \in \{0, ..., n\}} t^j = 1 \land 0 \le t^j\}\)
\(p'\): \(= \sum_{j \in \{0, ..., n\}} t'^j p_j \in S\), such that \(\forall j \in \{0, ..., n\} (0 \lt t'^j)\)
\(p''\): \(= \sum_{j \in \{0, ..., n\}} t''^j p_j \in S\), such that \(\exists j \in \{0, ..., n\} (0 = t''^j)\)
//
Statements:
\(S \in \{\text{ the affine simplexes }\}\)
\(\implies\)
(
\(\exists \{p'_0, ..., p'_m\} \subseteq \{p_0, ..., p_n\} (\{p'_0, ..., p'_m\} \in \{\text{ the affine-independent sets of base points on } V\} \land S = [p'_0, ..., p'_m])\)
\(\land\)
\(p' \in [p'_0, ..., p'_m]^\circ\)
\(\land\)
\(\text{ not necessarily } p'' \in bou [p'_0, ..., p'_m]\)
).
//
2: Natural Language Description
For any real vectors space, \(V\), any non-affine-independent set of base points, \(\{p_0, ..., p_n\} \subseteq V\), any point, \(p' = \sum_{j \in \{0, ..., n\}} t'^j p_j \in S \), such that \(\forall j \in \{0, ..., n\} (0 \lt t'^j)\), and any point, \(p'' = \sum_{j \in \{0, ..., n\}} t''^j p_j \in S\), such that \(\exists j \in \{0, ..., n\} (0 = t''^j)\), if the convex set spanned by the set of the base points, \(S := \{\sum_{j \in \{0, ..., n\}} t^j p_j \in V \vert t^j \in \mathbb{R}, \sum_{j \in \{0, ..., n\}} t^j = 1 \land 0 \le t^j\}\), is an affine simplex, \([p'_0, ..., p'_m]\), where \(\{p'_0, ..., p'_m\} \subseteq \{p_0, ..., p_n\}\) is an affine-independent subset of the base points, \(p' \in [p'_0, ..., p'_m]^\circ\) but not necessarily \(p'' \in bou [p'_0, ..., p'_m]\).
3: Proof
Let us see that the convex set spanned by any not-necessarily-affine-independent set of \((m + 1)\) base points on \(V\) can be expressed as \(\{\sum_{j \in \{1, ..., m\}} u^j (p_j - p_0) + p_0 \in V \vert \sum_{j \in \{1, ..., m\}} u^j \le 1 \land 0 \le u^j\}\) where \(p_0\) is any one of the base points. \(\sum_{j \in \{0, ..., m\}} u^j p_j = \sum_{j \in \{0, ..., m\}} u^j (p_j - p_0) + \sum_{j \in \{0, ..., m\}} u^j p_0 = \sum_{j \in \{1, ..., m\}} u^j (p_j - p_0) + p_0\). Note that any \(p_j\) can be taken instead of \(p_0\). When the set of the base points is affine-independent, \(\{p_1 - p_0, ..., p_m - p_0\}\) is linearly independent, while otherwise, \(\{p_1 - p_0, ..., p_m - p_0\}\) is not linearly independent.
Let us suppose that \(S\) is an affine simplex, which means that \(S = [p'_0, ..., p'_m]\), by the proposition that when the convex set spanned by any non-affine-independent set of base points on any real vectors space is an affine simplex, it is spanned by an affine-independent subset of the base points.
While \(p'_0 \in \{p_0, ..., p_n\}\), let us suppose that \(p'_0 = p_0\) for just the sake of simplicity of expressions: think of re-indexing \(\{p_0, ..., p_n\}\). Then, \(S = \{\sum_{j \in \{1, ..., m\}} u^j (p'_j - p_0) + p_0 \in V \vert u^j \in \mathbb{R}, \sum_{j \in \{1, ..., m\}} u^j \le 1 \land 0 \le u^j\}\).
\(\sum_{j \in \{1, ..., n\}} t^j (p_j - p_0) + p_0 = \sum_{j \in \{1, ..., m\}} u^j (p'_j - p_0) + p_0\), which means that \(\sum_{j \in \{1, ..., n\}} t^j (p_j - p_0) = \sum_{j \in \{1, ..., m\}} u^j (p'_j - p_0)\).
Taking \(t^j = 1\) (inevitably, \(t^k = 0\) for \(k \neq j\)), \(p_j - p_0 = \sum_{l \in \{1, ..., m\}} s_j^l (p'_l - p_0)\) where \(0 \le s_j^l\) and \(\sum_{l \in \{1, ..., m\}} s_j^l \le 1\). So, \(\sum_{j \in \{1, ..., n\}} t^j (p_j - p_0) = \sum_{j \in \{1, ..., n\}} t^j (\sum_{l \in \{1, ..., m\}} s_j^l (p'_l - p_0)) = \sum_{l \in \{1, ..., m\}} \sum_{j \in \{1, ..., n\}} (t^j s_j^l (p'_l - p_0))\). As \(\{p'_1 - p_0, ..., p'_m - p_0\}\) is linearly independent, \(u^l = \sum_{j \in \{1, ..., n\}} (t^j s_j^l)\), where \(s_k^l = 1\) for a \(k \in \{1, ..., n\}\), where \(s_k^l\) is the maximum of \(\{s_1^l, ..., s_n^l\}\): see Proof of the proposition that when the convex set spanned by any non-affine-independent set of base points on any real vectors space is an affine simplex, it is spanned by an affine-independent subset of the base points.
For \(p'\), as \(0 \le s_j^l\), \(s_k^l = 1\), and \(0 \lt t'^k\), \(0 \lt u^l\).
For \(p''\), a counterexample suffices. Let \(V = \mathbb{R}^2\), \(\{p_0, p_1, p_2\} = \{((-1, 0), (0, 0), (1, 0))\}\), and \(p'' = 0 p_0 + 1 p_1 + 0 p_2\). \(\{p'_0, p'_1\} = \{(-1, 0), (1, 0)\}\) such that \(S = [p'_0, p'_1]\), and \(p'' = 1 / 2 p'_0 + 1 / 2 p'_1 \notin bou [p'_0, p'_1]\).
