553: 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

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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

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Topics

About: vectors space

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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

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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.

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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.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$V$: $\in \{\text{ the real vectors spaces }\}$
$\{p_0,...,p_n\}$: $\subseteq V$, $\in \{\text{ the non-affine-independent sets of base points on }V\}$
$S$: $=\{\sum _{j\in \{0,...,n\}}{t}^j{p}_j\in V|t^j\in \mathbb{R},\sum _{j\in \{0,...,n\}}{t}^j=1\wedge 0\le t^j\}$
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Statements:
$S\in \{\text{ the affine simplexes }\}$
$⟹$
$\mathrm{\exists }\{p_0^{\prime },...,p_m^{\prime }\}\subseteq \{p_0,...,p_n\}(\{p_0^{\prime },...,p_m^{\prime }\}\in \{\text{ the affine-independent sets of base points on }V\}\wedge S=[p_0^{\prime },...,p_m^{\prime }]$.
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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, $S:=\{\sum _{j\in \{0,...,n\}}{t}^j{p}_j\in V|t^j\in \mathbb{R},\sum _{j\in \{0,...,n\}}{t}^j=1\wedge 0\le t^j\}$, is an affine simplex, $S$ is spanned by an affine-independent subset of the base points, $\{p_0^{\prime },...,p_m^{\prime }\}\subseteq \{p_0,...,p_n\}$, which is $S=[p_0^{\prime },...,p_m^{\prime }]$.

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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 $S$.

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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 $\{\sum _{j\in \{1,...,m\}}{u}^j(p_j-p_0)+p_0\in V|\sum _{j\in \{1,...,m\}}{u}^j\le 1\wedge 0\le u^j\}$ 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, $\{p_0^{\prime },...,p_m^{\prime }\}$, and $S=\{\sum _{j\in \{1,...,m\}}{u}^j(p_j^{\prime }-p_0^{\prime })+p_0^{\prime }|\sum _{j\in \{1,...,m\}}{u}^j\le 1\wedge 0\le u^j\}$. Note that we are not assuming that $\{p_0^{\prime },...,p_m^{\prime }\}$ is a subset of $\{p_0,...,p_n\}$.

Let us prove that $p_0^{\prime }\in \{p_0,...,p_n\}$.

Let us suppose that $p_0^{\prime }\notin \{p_0,...,p_n\}$.

$p_0^{\prime }=\sum _{j=\in \{1,...,n\}}{t}^j(p_j-p_0)+p_0$ for a $(t^1,...,t^n)$, which would not be $(0,...,0)$, because otherwise, $p_0^{\prime }=p_0$, a contradiction against the supposition. There would be a positive, $t_k$. If $\sum _{j=\in \{1,...,n\}}{t}^j<1$, 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 $p_0^{\prime }$ in its interior. If $\sum _{j\in \{1,...,n\}}{t}^j=1$, let us think of another expression, $p_0^{\prime }=\sum _{j\in \{0,...,n\}\setminus \{k\}}{t}^j(p_j-p_k)+p_k$, then, $(t^0,...,\widehat{t^k},...,t^n)$ would not be $(0,...,0)$ (otherwise, $p_0^{\prime }=p_k$) and $\sum _{j\in \{0,...,n\}\setminus k}{t}^j<1$ (because $0<t^k$). Then, $p_0^{\prime }$ would be in the interior of a line segment that was contained in $S$.

Let us prove that that cannot happen if $p_0^{\prime }$ is indeed a base point. In order to draw a line segment from $p_0^{\prime }$ in any direction, we need to take a trajectory, $\lambda \mapsto (u^1(\lambda ),...,u^m(\lambda ))$, for $\sum _{j\in \{1,...,m\}}{u}^j(p_j^{\prime }-p_0^{\prime })+p_0^{\prime }$, 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 $p_0^{\prime }\notin \{p_0,...,p_n\}$ leads to a contradiction, $p_0^{\prime }\in \{p_0,...,p_n\}$.

So, $p_0^{\prime }=p_k$ for a $k\in \{0,...,n\}$, but let us suppose that $p_0^{\prime }=p_0$ just for the sake of the convenience of expressions: think that $\{p_0,...,p_n\}$ is re-indexed.

So, $S=\{\sum _{j\in \{1,...,m\}}{u}^j(p_j^{\prime }-p_0)+p_0\in V|u^j\in \mathbb{R},\sum _{j\in \{1,...,m\}}{u}^j\le 1\wedge 0\le u^j\}$.

$\sum _{j\in \{1,...,n\}}{t}^j(p_j-p_0)+p_0=\sum _{j\in \{1,...,m\}}{u}^j(p_j^{\prime }-p_0)+p_0$, which means that $\sum _{j\in \{1,...,n\}}{t}^j(p_j-p_0)=\sum _{j\in \{1,...,m\}}{u}^j(p_j^{\prime }-p_0)$.

Taking $t^j=1$ (inevitably, $t^k=0$ for $k\ne j$), $p_j-p_0=\sum _{l\in \{1,...,m\}}{s}_j^l(p_l^{\prime }-p_0)$ where $0\le s_j^l$ and $\sum _{l\in \{1,...,m\}}{s}_j^l\le 1$. So, $\sum _{j\in \{1,...,n\}}{t}^j(p_j-p_0)=\sum _{j\in \{1,...,n\}}{t}^j(\sum _{l\in \{1,...,m\}}{s}_j^l(p_l^{\prime }-p_0))=\sum _{l\in \{1,...,m\}}\sum _{j\in \{1,...,n\}}(t^j{s}_j^l(p_l^{\prime }-p_0))$. As $\{p_1^{\prime }-p_0,...,p_m^{\prime }-p_0\}$ 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 $o\ne l$. So, $p_k-p_0=\sum _{q\in \{1,...,m\}}{s}_k^q(p_q^{\prime }-p_0)=p_l^{\prime }-p_0$.

So, for each $l\in \{1,...,m\}$, $p_l^{\prime }-p_0=p_k-p_0$ for a $k\in \{0,...,n\}$, which means that $p_l^{\prime }=p_k$. As $\{p_1^{\prime }-p_0,...,p_m^{\prime }-p_0\}$ is linearly independent, $k$ s are distinct for $l$ s. So, $\{p_0^{\prime },...,p_m^{\prime }\}\subseteq \{p_0,...,p_n\}$.

To state just in case not be misunderstood (this has been stated in Note), the choice of $\{p_0^{\prime },...,p_m^{\prime }\}$ is not arbitrary: the proof has selectively chosen a $\{p_0^{\prime },...,p_m^{\prime }\}$.

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References

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