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Let $V$ be a vector space over $\mathbb{F}$ and $S \subseteq V$ a nonempty set.

Theorem: $S$ is an affine subset of $V$ if and only if $$ \forall u,v \in S, \forall \lambda \in \mathbb{F}, \lambda u + (1-\lambda) v \in S$$

The only if direction is easy, but the converse is not. Let $v_0 \in S$ and define $W = \{w \in V\mid w + v_0 \in S\}$. We wish to show that $W$ is a subspace of $V$. Trivially $0 \in W$. If $u',v' \in W,\lambda \in \mathbb{F}$, there exist $u,v \in S\mid u' = u - v_0, v' = v - v_0$, and $$\begin{align*} u' + \lambda v' &= u - v_0 + \lambda v - \lambda v_0 \\ &= \lambda u - v_0 + \lambda v + (1 - \lambda) v_0 - v_0 - \lambda u + u \\ &= \underbrace{\lambda u + (1 - \lambda)v_0}_{\text{in } S} - v_0 - v_0 + \lambda v + u - \lambda u \\ &= \underbrace{\lambda u + (1 - \lambda)v_0}_{\text{in } S} - v_0 - v_0 + \underbrace{\lambda v + (1 - \lambda) u}_{\text{in } S} \end{align*}$$ How do I show that this is in the form of $w' - v_0, w' \in S$?

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We have a claim that $W$ is a subspace That is for $u',\ v'\in W$ we must show that $u'+cv'\in W$

(1) Hence $$v_0\in S, \ v_0+v'\in S $$

Hence $$ cv' +v_0=c(v_0+v') + (1-c) v_0 \in S $$ so that $$ cv'\in W $$

(2) Consider $$ u'+v_0,\ cv'+v_0 \in S$$

That is $$ (1-d)(u'+v_0) +d(cv' + v_0) \in S$$ If $d=1/2$ then $$ 1/2 (u' +cv') + v_0 \in S \Rightarrow 1/2(u'+cv')\in W $$

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  • $\begingroup$ What if $1/2 \not\in \mathbb{F}$? $\endgroup$ – Henricus V. Mar 1 '16 at 23:58
  • $\begingroup$ $(1-d)u' + dcv' \in W$ so that $u' + \frac{dc}{1-d}v'\in W$ If we redefine $c$ to be $ \frac{dc}{1-d}$, then we proved. $\endgroup$ – HK Lee Mar 27 '16 at 8:25

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