552: Affine Subset of Finite-Dimensional Real Vectors Space Is Spanned by Finite Affine-Independent Set of Base Points

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description/proof of that affine subset of finite-dimensional real vectors space is spanned by finite affine-independent set 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: Proof

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Starting Context

• The reader knows a definition of affine subset of real vectors space.
• The reader knows a definition of affine-independent set of points on real vectors space.
• The reader knows a definition of affine 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 subset of any finite-dimensional real vectors space is spanned by a finite affine-independent set of 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:
$V$: $\in \{\text{ the }d\text{ -dimensional real vectors spaces }\}$
$S$: $\subseteq V$, $\in \{\text{ the affine subsets of }V\}$
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Statements:
$\mathrm{\exists }\{p_0,...,p_n\}\in \{\text{ the affine-independent sets of points on }V\}\text{ where }0\le n\le d(S=\{\sum _{j=0\sim n}{t}^j{p}_j\in V|t^j\in \mathbb{R},\sum _{j=0\sim n}{t}^j=1\})$.
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2: Natural Language Description

For any $d$-dimensional real vectors space, $V$, any affine subset, $S\subseteq V$, is the affine set spanned by an affine-independent set of some $n$ points, $\{p_0,...,p_n\}$, where $0\le n\le d$.

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3: Proof

Let us denote the affine subset spanned by $\{p_0,...,p_n\}$, as $A(\{p_0,...,p_n\})$.

When $S$ has no point, $S=A(\{\})$.

When $S$ has only 1 point, $p_0$, $S=A(\{p_0\})$.

Let us suppose that $S$ has more than 1 point.

Let us take any distinct points, $p_0,p_1\in S$.

$\{p_0,p_1\}$ is an affine-independent set of points. $A(\{p_0,p_1\})\subseteq S$, because for each $p=\sum _{j=0\sim 1}{t}^j{p}_j$ where $\sum _{j=0\sim 1}{t}^j=1$, $p=p_1+t^0{p}_0+(t^1-1)p_1=p_1+t^0{p}_0-t^0{p}_1=p_1+t^0(p_0-p_1)\in S$.

If $A(\{p_0,p_1\})=S$, we are finished.

Otherwise, there is a point, $p_2\in S$, such that $p_2\notin A(\{p_0,p_1\})$.

$\{p_0,p_1,p_2\}$ is affine-independent, because $p_1-p_0,p_2-p_0$ are linearly independent, because otherwise, $p_2-p_0$ would be a scalar multiple of $p_1-p_0$, which would mean $p_2\in A(\{p_0,p_1\})$. $A(\{p_0,p_1,p_2\})\subseteq S$, because for each $p=\sum _{j=0\sim 2}{t}^j{p}_j$ where $\sum _{j=0\sim 2}{t}^j=1$ and $p_2\ne 1$, which means that $\sum _{j=0\sim 1}{t}^j\ne 0$, $p=p_2+\sum _{j=0\sim 1}{t}^j{p}_j+(t^2-1)p_2=p_2+\sum _{j=0\sim 1}{t}^j{p}_j-(\sum _{j=0\sim 1}{t}^j)p_2=p_2+\sum _{j=0\sim 1}{t}^j((\sum _{j=0\sim 1}{t}^j{p}_j)/\sum _{j=0\sim 1}{t}^j-p_2)\in S$, because $(\sum _{j=0\sim 1}{t}^j{p}_j)/\sum _{j=0\sim 1}{t}^j\in A(\{p_0,p_1\})\subseteq S$, because $\sum _{j=0\sim 1}{t}^j/\sum _{j=0\sim 1}{t}^j=1$; when $p_2=1$, $p=p_2\in S$.

If $A(\{p_0,p_1,p_2\})=S$, we are finished.

Otherwise, there is a point, $p_3\in S$, such that $p_3\notin A(\{p_0,p_1,p_2\})$.

$\{p_0,p_1,p_2,p_3\}$ is affine-independent, because $p_1-p_0,p_2-p_0,p_3-p_0$ are linearly independent, because otherwise, $p_3-p_0$ would be a linear combination of $p_1-p_0,p_2-p_0$, which would mean $p_3\in A(\{p_0,p_1,p_2\})$. $A(\{p_0,p_1,p_2,p_3\})\subseteq S$, because for each $p=\sum _{j=0\sim 3}{t}^j{p}_j$ where $\sum _{j=0\sim 3}{t}^j=1$ and $p_3\ne 1$, which means that $\sum _{j=0\sim 2}{t}^j\ne 0$, $p=p_3+\sum _{j=0\sim 2}{t}^j{p}_j+(t^3-1)p_3=p_3+\sum _{j=0\sim 2}{t}^j{p}_j-(\sum _{j=0\sim 2}{t}^j)p_3=p_3+\sum _{j=0\sim 2}{t}^j((\sum _{j=0\sim 2}{t}^j{p}_j)/\sum _{j=0\sim 2}{t}^j-p_3)\in S$, because $(\sum _{j=0\sim 2}{t}^j{p}_j)/\sum _{j=0\sim 2}{t}^j\in A(\{p_0,p_1,p_2\})\subseteq S$, because $\sum _{j=0\sim 2}{t}^j/\sum _{j=0\sim 2}{t}^j=1$; when $p_3=1$, $p=p_3\in S$.

And so on.

After all, $A(\{p_0,p_1,p_2,...,p_n\})=S$ for an $n\le d$, because any more-than-$d$ vectors cannot be independent on $V$.

So, $S$ is the affine set spanned by $\{p_1,...,p_n\}$.

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References

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