549: Affine Map from Affine or Convex Set Spanned by Possibly-Non-Affine-Independent Set of Base Points on Real Vectors Space Is Linear

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description/proof of that affine map from affine or convex set spanned by possibly-non-affine-independent set of base points on real vectors space is linear

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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 affine map from affine set spanned by possibly-non-affine-independent set of base points on real vectors space.
• The reader knows a definition of affine map from convex set spanned by possibly-non-affine-independent set of base points on real vectors space.

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Target Context

• The reader will have a description and a proof of the proposition that any affine map from the affine or convex set spanned by any possibly-non-affine-independent set of base points on any real vectors space is linear.

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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:
$V_1$: $\in \{\text{ the real vectors spaces }\}$
$V_2$: $\in \{\text{ the real vectors spaces }\}$
$\{p_0,...,p_n\}$: $\subseteq V_1$, $\in \{\text{ the possibly-non-affine-independent sets of base points on }{V}_1\}$
$S$: $=\text{ the affine or convex set spanned by }\{p_0,...,p_n\}$
$f$: $:S\to V_2$, $\in \{\text{ the affine maps }\}$
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Statements:
$f\in \{\text{ the linear maps }\}$; especially, $f(\sum _{j=0\sim n}{t}^j{p}_j)=\sum _{j=0\sim n}{t}^{j}f(p_j)$ when $\sum _{j=0\sim n}{t}^j=1$ or $\sum _{j=0\sim n}{t}^j=1\wedge 0\le t^j$.
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2: Natural Language Description

For any real vectors spaces, $V_1,V_2$, any possibly non-affine-independent set of base points, $\{p_0,...,p_n\}\subseteq V_1$, and the affine or convex set spanned by the set of the base points, $S\subseteq V_1$, any affine map, $f:S\to V_2$, is linear. Especially, $f(\sum _{j=0\sim n}{t}^j{p}_j)=\sum _{j=0\sim n}{t}^{j}f(p_j)$ when $\sum _{j=0\sim n}{t}^j=1$ or $\sum _{j=0\sim n}{t}^j=1\wedge 0\le t^j$.

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3: Note

Saying about "linear", $S$ is not really any vectors space in general. This proposition is talking about being $f(r_1{s}_1+...+r_m{s}_m)=r_{1}f(s_1)+...+r_{m}f(s_m)$ whenever $s_1,...,s_m\in S$ and $r_1{s}_1+...+r_m{s}_m\in S$.

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4: Proof

$f$ is defined based on an affine-independent subset of the base points, $\{p_0^{\prime },...,p_k^{\prime }\}\subseteq \{p_0,...,p_n\}$, that spans $S$.

Let $s_1,...,s_m\in S$ be any points. $s_l=\sum _{j=0\sim k}{t}_l^{\prime j}{p}_j^{\prime }$. Let $r^1,...,r^m\in \mathbb{R}$ be any such that $r^1{s}_1+...+r^m{s}_m\in S$.

$f(r^1{s}_1+...+r^m{s}_m)=f(r^1\sum _{j=0\sim k}{t}_1^{\prime j}{p}_j^{\prime }+...+r^m\sum _{j=0\sim k}{t}_m^{\prime j}{p}_j^{\prime })=f(\sum _{j=0\sim k}((r^1{t}_1^{\prime j}+...+r^m{t}_m^{\prime j})p_j^{\prime }))=\sum _{j=0\sim k}((r^1{t}_1^{\prime j}+...+r^m{t}_m^{\prime j})f(p_j^{\prime }))=\sum _{j=0\sim k}(r^1{t}_1^{\prime j}f(p_j^{\prime }))+...+\sum _{j=0\sim k}(r^m{t}_m^{\prime j}f(p_j^{\prime }))=r^{1}f(\sum _{j=0\sim k}(t_1^{\prime j}{p}_j^{\prime }))+...+r^{m}f(\sum _{j=0\sim k}(t_m^{\prime j}{p}_j^{\prime }))=r^{1}f(s_1)+...+r^{m}f(s_m)$.

As $p_j\in S$ and $\sum _{j=0\sim n}{t}^j{p}_j\in S$ when $\sum _{j=0\sim n}{t}^j=1$ or $\sum _{j=0\sim n}{t}^j=1\wedge 0\le t^j$, $f(\sum _{j=0\sim n}{t}^j{p}_j)=\sum _{j=0\sim n}{t}^{j}f(p_j)$. So, although $f$ is defined based on $\{p_0^{\prime },...,p_k^{\prime }\}$, the expected expansion holds.

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References

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