description/proof of that when convex set spanned by non-affine-independent set of base points on real vectors space is affine simplex, it is spanned by affine-independent subset 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: Note
- 4: 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.
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, it is spanned by an affine-independent subset of the 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:
\(V\): \(\in \{\text{ the real vectors spaces }\}\)
\(\{p_0, ..., p_n\}\): \(\subseteq V\), \(\in \{\text{ the non-affine-independent sets of base points on } V\}\)
\(S\): \(= \{\sum_{j \in \{0, ..., n\}} t^j p_j \in V \vert t^j \in \mathbb{R}, \sum_{j \in \{0, ..., n\}} t^j = 1 \land 0 \le t^j\}\)
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Statements:
\(S \in \{\text{ the affine simplexes }\}\)
\(\implies\)
\(\exists \{p'_0, ..., p'_m\} \subseteq \{p_0, ..., p_n\} (\{p'_0, ..., p'_m\} \in \{\text{ the affine-independent sets of base points on } V\} \land S = [p'_0, ..., p'_m]\).
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2: Natural Language Description
For any real vectors space, \(V\), and any non-affine-independent set of base points, \(\{p_0, ..., p_n\} \subseteq V\), 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, \(S\) is spanned by an affine-independent subset of the base points, \(\{p'_0, ..., p'_m\} \subseteq \{p_0, ..., p_n\}\), which is \(S = [p'_0, ..., p'_m]\).
3: Note
This proposition is stating that although a convex set spanned by a non-affine-independent set of base points may not be any affine simplex (see the proposition that the convex set spanned by a non-affine-independent set of base points on a real vectors space is not necessarily any affine simplex spanned by an affine-independent subset of the base points), if it is so, it is the affine simplex spanned by an affine-independent subset of the base points.
This proposition is not stating that each affine-independent subset of the base points spans the affine simplex: an appropriate subset has to be chosen. For example, when \(V = \mathbb{R}^2\) and \(\{p_0, p_1, p_2\} = \{(0, 0), (1, 0), (-1, 0)\}\), \(S\) is an affine simplex, and \(\{p_1, p_2\}\) spans \(S\), but \(\{p_0, p_1\}\) does not span \(S\).
4: 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 \(S\) is spanned by an affine-independent set of some base points, \(\{p'_0, ..., p'_m\}\), and \(S = \{\sum_{j \in \{1, ..., m\}} u^j (p'_j - p'_0) + p'_0 \vert \sum_{j \in \{1, ..., m\}} u^j \le 1 \land 0 \le u^j\}\). Note that we are not assuming that \(\{p'_0, ..., p'_m\}\) is a subset of \(\{p_0, ..., p_n\}\).
Let us prove that \(p'_0 \in \{p_0, ..., p_n\}\).
Let us suppose that \(p'_0 \notin \{p_0, ..., p_n\}\).
\(p'_0 = \sum_{j = \in \{1, ..., n\}} t^j (p_j - p_0) + p_0\) for a \((t^1, ..., t^n)\), which would not be \((0, ..., 0)\), because otherwise, \(p'_0 = p_0\), a contradiction against the supposition. There would be a positive, \(t_k\). If \(\sum_{j = \in \{1, ..., n\}} t^j \lt 1\), we could take an interval, \([t_k - \delta, t_k + \delta]\), where \(0 \le t_k - \delta\) and \(\sum_{j \in \{1, ..., n\}} t^j + \delta \le 1\), which would mean that when we moved \(t^k\) in \([t_k - \delta, t_k + \delta]\) with any other \(t^j\) fixed, the corresponding points would be on \(S\), while the points would constitute a line segment that would contain \(p'_0\) in its interior. If \(\sum_{j \in \{1, ..., n\}} t^j = 1\), let us think of another expression, \(p'_0 = \sum_{j \in \{0, ..., n\} \setminus \{k\}} t^j (p_j - p_k) + p_k\), then, \((t^0, ..., \hat{t^k}, ..., t^n)\) would not be \((0, ..., 0)\) (otherwise, \(p'_0 = p_k\)) and \(\sum_{j \in \{0, ..., n\} \setminus {k}} t^j \lt 1\) (because \(0 \lt t^k\)). Then, \(p'_0\) would be in the interior of a line segment that was contained in \(S\).
Let us prove that that cannot happen if \(p'_0\) is indeed a base point. In order to draw a line segment from \(p'_0\) in any direction, we need to take a trajectory, \(\lambda \mapsto (u^1 (\lambda), ..., u^m (\lambda))\), for \(\sum_{j \in \{1, ..., m\}} u^j (p'_j - p'_0) + p'_0\), where \(u^j (\lambda)\) has to be non-negative and at least one \(u^k (\lambda)\) has to go positive (otherwise, each \(u^j (\lambda)\) would stay \(0\) and we would not be drawing any line segment), but for the other direction, we need to take a trajectory, \(\lambda \mapsto (u^1 (\lambda), ..., u^m (\lambda))\), where \(u^k (\lambda)\) needs to go negative, which is impossible.
So, as supposing that \(p'_0 \notin \{p_0, ..., p_n\}\) leads to a contradiction, \(p'_0 \in \{p_0, ..., p_n\}\).
So, \(p'_0 = p_k\) for a \(k \in \{0, ..., n\}\), but let us suppose that \(p'_0 = p_0\) just for the sake of the convenience of expressions: think that \(\{p_0, ..., p_n\}\) is re-indexed.
So, \(S = \{\sum_{j \in \{1, ..., m\}} u^j (p'_j - p_0) + p_0 \in V \vert u^j \in \mathbb{R}, \sum_{j \in \{1, ..., m\}} u^j \le 1 \land 0 \le u^j\}\).
\(\sum_{j \in \{1, ..., n\}} t^j (p_j - p_0) + p_0 = \sum_{j \in \{1, ..., m\}} u^j (p'_j - p_0) + p_0\), 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, \(u^l = \sum_{j \in \{1, ..., n\}} (t^j s_j^l)\). Let us suppose that \(s_k^l\) is one of the maximums of \(\{s_1^l, ..., s_n^l\}\). When \((t^1, ..., t^n)\) moves around, the maximum of \(\sum_{j \in \{1, ..., n\}} (t^j s_j^l)\) is \(s_k^l\) realized at \(t^k = 1\). The maximum, \(s_k^l\), has to be \(1\), and that means that \(s_k^o = 0\) for \(o \neq l\). So, \(p_k - p_0 = \sum_{q \in \{1, ..., m\}} s_k^q (p'_q - p_0) = p'_l - p_0\).
So, for each \(l \in \{1, ..., m\}\), \(p'_l - p_0 = p_k - p_0\) for a \(k \in \{0, ..., n\}\), which means that \(p'_l = p_k\). As \(\{p'_1 - p_0, ..., p'_m - p_0\}\) is linearly independent, \(k\) s are distinct for \(l\) s. So, \(\{p'_0, ..., p'_m\} \subseteq \{p_0, ..., p_n\}\).
To state just in case not be misunderstood (this has been stated in Note), the choice of \(\{p'_0, ..., p'_m\}\) is not arbitrary: the proof has selectively chosen a \(\{p'_0, ..., p'_m\}\).
