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I've seen the notion of an affine subspace defined differently as follows:

  • $S \subset \mathbb R^3$, non-empty, is an affine subspace if $(1-t)u + tv \in S$ whenever $u,v \in S$.

  • $S$ is an affine subspace if $S=V+x_0$, where $V\subset \mathbb R^3$ is a subspace and $x_0 \in \mathbb R^3$.

Are these two definitions equivalent ?

I see that if $S$ is an affine subspace with respect to the second definition, then it is also with respect to the first definition. However, I cannot prove the other way around ?

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  • $\begingroup$ Write the first as $u+t(v-u)$, then let $u$ be $x_0$. $\endgroup$ – KittyL Apr 18 '15 at 18:34
  • $\begingroup$ Why is $t(v-x_0)$ then a subspace ? $\endgroup$ – Shuzheng Apr 18 '15 at 19:17
  • $\begingroup$ A span of any set of vectors is a subspace. You can easily prove it by definition of a subspace. $\endgroup$ – KittyL Apr 18 '15 at 21:54
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Suppose S satisfies the first definition, and let $u\in S$. $\;\;$If we define $V=\{s-u:s\in S\}$, then

1) $0\in V$ since $u\in S$.

2) If $v\in V$, then $tv\in V$ since $\;\;\;\;v\in V\implies v+u\in S\implies t(v+u)+(1-t)u\in S\implies tv+u\in S\implies tv\in V.$

3) If $v\in V$ and $w\in V$, then $v+w\in V$ since $\;\;\;v, w\in V\Rightarrow v+u\in S \text{ and } w+u\in S\Rightarrow\frac{1}{2}(v+u)+\frac{1}{2}(w+u)\in S\Rightarrow\frac{1}{2}(v+w)+u\in S$;

$\;\;\;$so $\frac{1}{2}(v+w)\in V$ and therefore $v+w\in V$ by 2).

Thus V is a subspace of $\mathbb{R^3}$, and by definition $S=V+u$.

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