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
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
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.
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:
\(V_1\): \(\in \{\text{ the real vectors spaces }\}\)
\(V_2\): \(\in \{\text{ the real vectors spaces }\}\)
\(\{p_0, ..., p_n\}\): \(\subseteq V_1\), \(\in \{\text{ the possibly-non-affine-independent sets of base points on } V_1\}\)
\(S\): \(= \text{ the affine or convex set spanned by } \{p_0, ..., p_n\}\)
\(f\): \(: S \to V_2\), \(\in \{\text{ the affine maps }\}\)
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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 \(\sum_{j = 0 \sim n} t^j = 1 \land 0 \le t^j\).
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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 \(\sum_{j = 0 \sim n} t^j = 1 \land 0 \le t^j\).
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 \(r_1 s_1 + ... + r_m s_m \in S\).
4: Proof
\(f\) is defined based on an affine-independent subset of the base points, \(\{p'_0, ..., p'_k\} \subseteq \{p_0, ..., p_n\}\), that spans \(S\).
Let \(s_1, ..., s_m \in S\) be any points. \(s_l = \sum_{j = 0 \sim k} t'^j_l p'_j\). Let \(r^1, ..., r^m \in \mathbb{R}\) be any such that \(r^1 s_1 + ... + r^m s_m \in S\).
\(f (r^1 s_1 + ... + r^m s_m) = f (r^1 \sum_{j = 0 \sim k} t'^j_1 p'_j + ... + r^m \sum_{j = 0 \sim k} t'^j_m p'_j) = f (\sum_{j = 0 \sim k} ((r^1 t'^j_1 + ... + r^m t'^j_m) p'_j)) = \sum_{j = 0 \sim k} ((r^1 t'^j_1 + ... + r^m t'^j_m) f (p'_j)) = \sum_{j = 0 \sim k} (r^1 t'^j_1 f (p'_j)) + ... + \sum_{j = 0 \sim k} (r^m t'^j_m f (p'_j)) = r^1 f (\sum_{j = 0 \sim k} (t'^j_1 p'_j)) + ... + r^m f (\sum_{j = 0 \sim k} (t'^j_m p'_j)) = r^1 f (s_1) + ... + r^m f (s_m)\).
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 \(\sum_{j = 0 \sim n} t^j = 1 \land 0 \le t^j\), \(f (\sum_{j = 0 \sim n} t^j p_j) = \sum_{j = 0 \sim n} t^j f (p_j)\). So, although \(f\) is defined based on \(\{p'_0, ..., p'_k\}\), the expected expansion holds.
