Euclidean space: $k$ points in $mathbb{R}^n$












1












$begingroup$



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.






linear-algebra euclidean-geometry







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$endgroup$

















    1












    $begingroup$



    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.






    linear-algebra euclidean-geometry







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    $endgroup$















      1












      1








      1





      $begingroup$



      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.






      linear-algebra euclidean-geometry







      share|cite|improve this question











      $endgroup$





      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.






      linear-algebra euclidean-geometry




      linear-algebra euclidean-geometry






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      edited Jan 11 at 17:46







      Zachary

















      asked Jan 11 at 17:32









      ZacharyZachary

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          $begingroup$

          (1) By the observation in the first line ("Since …"):
          $$p_1-p_j=(p_1-p_i)-(p_j-p_i)inoperatorname{span}(p_1-p_i,ldots,p_k-p_i),$$
          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
          $$operatorname{span}(p_1-p_j,ldots,p_k-p_j)subseteqoperatorname{span}(p_1-p_i,ldots,p_k-p_i).$$
          By reversing their roles, we can show that the opposite inclusion is also true. Therefore, the two subspaces are equal.



          (2) $p_i-p_jinoperatorname{span}(p_1-p_j,dots,p_k-p_j)$ 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 $p_i-p_jin A$. In a bit more detail:
          $$p_i-p_jin A implies (p_i-p_j)+A=A implies p_i+A=p_j+A.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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











          Your Answer





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          $begingroup$

          (1) By the observation in the first line ("Since …"):
          $$p_1-p_j=(p_1-p_i)-(p_j-p_i)inoperatorname{span}(p_1-p_i,ldots,p_k-p_i),$$
          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
          $$operatorname{span}(p_1-p_j,ldots,p_k-p_j)subseteqoperatorname{span}(p_1-p_i,ldots,p_k-p_i).$$
          By reversing their roles, we can show that the opposite inclusion is also true. Therefore, the two subspaces are equal.



          (2) $p_i-p_jinoperatorname{span}(p_1-p_j,dots,p_k-p_j)$ 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 $p_i-p_jin A$. In a bit more detail:
          $$p_i-p_jin A implies (p_i-p_j)+A=A implies p_i+A=p_j+A.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          1












          $begingroup$

          (1) By the observation in the first line ("Since …"):
          $$p_1-p_j=(p_1-p_i)-(p_j-p_i)inoperatorname{span}(p_1-p_i,ldots,p_k-p_i),$$
          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
          $$operatorname{span}(p_1-p_j,ldots,p_k-p_j)subseteqoperatorname{span}(p_1-p_i,ldots,p_k-p_i).$$
          By reversing their roles, we can show that the opposite inclusion is also true. Therefore, the two subspaces are equal.



          (2) $p_i-p_jinoperatorname{span}(p_1-p_j,dots,p_k-p_j)$ 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 $p_i-p_jin A$. In a bit more detail:
          $$p_i-p_jin A implies (p_i-p_j)+A=A implies p_i+A=p_j+A.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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














          1












          1








          1





          $begingroup$

          (1) By the observation in the first line ("Since …"):
          $$p_1-p_j=(p_1-p_i)-(p_j-p_i)inoperatorname{span}(p_1-p_i,ldots,p_k-p_i),$$
          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
          $$operatorname{span}(p_1-p_j,ldots,p_k-p_j)subseteqoperatorname{span}(p_1-p_i,ldots,p_k-p_i).$$
          By reversing their roles, we can show that the opposite inclusion is also true. Therefore, the two subspaces are equal.



          (2) $p_i-p_jinoperatorname{span}(p_1-p_j,dots,p_k-p_j)$ 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 $p_i-p_jin A$. In a bit more detail:
          $$p_i-p_jin A implies (p_i-p_j)+A=A implies p_i+A=p_j+A.$$






          share|cite|improve this answer









          $endgroup$



          (1) By the observation in the first line ("Since …"):
          $$p_1-p_j=(p_1-p_i)-(p_j-p_i)inoperatorname{span}(p_1-p_i,ldots,p_k-p_i),$$
          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
          $$operatorname{span}(p_1-p_j,ldots,p_k-p_j)subseteqoperatorname{span}(p_1-p_i,ldots,p_k-p_i).$$
          By reversing their roles, we can show that the opposite inclusion is also true. Therefore, the two subspaces are equal.



          (2) $p_i-p_jinoperatorname{span}(p_1-p_j,dots,p_k-p_j)$ 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 $p_i-p_jin A$. In a bit more detail:
          $$p_i-p_jin A implies (p_i-p_j)+A=A implies p_i+A=p_j+A.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 11 at 17:57









          zipirovichzipirovich

          11.3k11631




          11.3k11631












          • $begingroup$
            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


















          • $begingroup$
            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
















          $begingroup$
          Very clear explanation! Thank you so much!
          $endgroup$
          – Zachary
          Jan 11 at 18:23




          $begingroup$
          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




          $begingroup$
          @Zachary: You're welcome! Glad I could help. :-)
          $endgroup$
          – zipirovich
          Jan 11 at 18:54


















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