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Consider $k$ points $p_1,\dots,p_k$ in $\mathbb{R}^n$. Then, $\forall i,j=1,\dots,k: p_i + \operatorname{span}(p_1-p_i,\dots,p_k-p_i) = p_j + \operatorname{span}(p_1-p_j,\dots,p_k-p_j).$ Assume that $D$ is an affine space with $p_1,\dots,p_k \in D$, then $p_1 + \operatorname{span}(p_2-p_1,\dots,p_k-p_1) \subseteq D$.

Proof: Since $p_l - p_j = (p_l - p_i)-(p_j-p_i)$ for all $l,i,j=1,\dots,k$, we have that $\underline{\operatorname{span}(p_1-p_i,\dots,p_k-p_i) = \operatorname{span}(p_1-p_j,\dots,p_k-p_j)}$.

From $\underline{p_i-p_j \in \operatorname{span}(p_1-p_j,\dots,p_k-p_j)}$ follows that $\underline{p_i + \operatorname{span}(p_1-p_i,\dots,p_k-p_i) = p_j + \operatorname{span}(p_1-p_j,\dots,p_k-p_j)}$.

Now assume $p_1,\dots,p_k \in D$, then $D = p_1 + D_0$. This implies that $p_2-p_1,\dots,p_k-p_1 \in D_0$, and therefor $\operatorname{span}(p_2-p_1,\dots,p_k-p_1) \subseteq D_0$. This completes the proof.

The underlined parts aren't clear to me. Could someone please help me understand them.

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(1) By the observation in the first line ("Since …"): $$p_1-p_j=(p_1-p_i)-(p_j-p_i)\in\operatorname{span}(p_1-p_i,\ldots,p_k-p_i),$$ and similarly for $p_2-p_j,\ldots,p_k-p_j$. Since all these vectors turn out to lie in $\operatorname{span}(p_1-p_i,\ldots,p_k-p_i)$, we have demonstrated that $$\operatorname{span}(p_1-p_j,\ldots,p_k-p_j)\subseteq\operatorname{span}(p_1-p_i,\ldots,p_k-p_i).$$ By reversing their roles, we can show that the opposite inclusion is also true. Therefore, the two subspaces are equal.

(2) $p_i-p_j\in\operatorname{span}(p_1-p_j,\dots,p_k-p_j)$ by definition, since it's one of these vectors whose span we're taking. From the previous part, we already know that $$\operatorname{span}(p_1-p_j,\ldots,p_k-p_j)=\operatorname{span}(p_1-p_i,\ldots,p_k-p_i)=A,$$ where $A$ is just a name I want to give to this subspace for brevity. Then the second property simply states that $p_i+A=p_j+A$, which is true precisely because $p_i-p_j\in A$. In a bit more detail: $$p_i-p_j\in A \implies (p_i-p_j)+A=A \implies p_i+A=p_j+A.$$

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  • $\begingroup$ Very clear explanation! Thank you so much! $\endgroup$ – Zachary Jan 11 at 18:23
  • $\begingroup$ @Zachary: You're welcome! Glad I could help. :-) $\endgroup$ – zipirovich Jan 11 at 18:54

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