Question on Linear Algebra (related to Affine Combination)












0












$begingroup$


The book I am using mentions this theorem:



$$Asubseteq Vtext{ is a flat} iff Atext{ is closed under affine combinations, that is}\ x_1,ldots,x_kin A, alpha_1,ldots,alpha_k,in F, alpha_1+cdots+ alpha_k=1Rightarrow alpha_1x_1+cdots+alpha_kx_kin A$$



The following question within the exercise then refers to this theorem (specifically the RHS of the above iff statement):




Let $ 1+1neq0$ in $F$.



Show that closure under affine combinations for $k=2$ implies closure under affine combinations for all $kgeq1$.





My approach so far has been to use Strong Induction.



Assuming the result holds for $kleq n$, we show it holds for $k=n+1$



consider $sum_{i=1}^{k+1}alpha_ix_i $



choose $lin{1,2,ldots, k+1}$ such that $sum_{i=1}^{k+1}alpha_ix_i=betasum_{i=1}^{l}frac{alpha_i}{beta}x_i+gammasum_{i={l+1}}^{k+1}frac{alpha_i}{gamma}x_i$



(Note: $beta = sum_{i=1}^{l}alpha_i neq0$ and $gamma = sum_{i=l+1}^{k+1}alpha_i neq0$)



Using induction hypothesis, $sum_{i=1}^{l}frac{alpha_i}{beta}x_iin A$ and $sum_{i={l+1}}^{k+1}frac{alpha_i}{gamma}x_iin A$. Call them $x$ and $y$ respectively.}



Now, $sum_{i=1}^{k+1}alpha_ix_i = beta x+ gamma y$



Since the result holds for $k=2$, $sum_{i=1}^{k+1}alpha_ix_iin A
$




My question is twofold. First, I never made use of the fact that $1+1neq0$ in F. What am I missing here? Second, how can I show that the condition in line 3 of my solution (when I introduce l and break the summation into two sums) can be assumed without loss of generality?



Any input would be appreciated.






linear-algebra







share|cite|improve this question











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    0












    $begingroup$


    The book I am using mentions this theorem:



    $$Asubseteq Vtext{ is a flat} iff Atext{ is closed under affine combinations, that is}\ x_1,ldots,x_kin A, alpha_1,ldots,alpha_k,in F, alpha_1+cdots+ alpha_k=1Rightarrow alpha_1x_1+cdots+alpha_kx_kin A$$



    The following question within the exercise then refers to this theorem (specifically the RHS of the above iff statement):




    Let $ 1+1neq0$ in $F$.



    Show that closure under affine combinations for $k=2$ implies closure under affine combinations for all $kgeq1$.





    My approach so far has been to use Strong Induction.



    Assuming the result holds for $kleq n$, we show it holds for $k=n+1$



    consider $sum_{i=1}^{k+1}alpha_ix_i $



    choose $lin{1,2,ldots, k+1}$ such that $sum_{i=1}^{k+1}alpha_ix_i=betasum_{i=1}^{l}frac{alpha_i}{beta}x_i+gammasum_{i={l+1}}^{k+1}frac{alpha_i}{gamma}x_i$



    (Note: $beta = sum_{i=1}^{l}alpha_i neq0$ and $gamma = sum_{i=l+1}^{k+1}alpha_i neq0$)



    Using induction hypothesis, $sum_{i=1}^{l}frac{alpha_i}{beta}x_iin A$ and $sum_{i={l+1}}^{k+1}frac{alpha_i}{gamma}x_iin A$. Call them $x$ and $y$ respectively.}



    Now, $sum_{i=1}^{k+1}alpha_ix_i = beta x+ gamma y$



    Since the result holds for $k=2$, $sum_{i=1}^{k+1}alpha_ix_iin A
    $




    My question is twofold. First, I never made use of the fact that $1+1neq0$ in F. What am I missing here? Second, how can I show that the condition in line 3 of my solution (when I introduce l and break the summation into two sums) can be assumed without loss of generality?



    Any input would be appreciated.






    linear-algebra







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      The book I am using mentions this theorem:



      $$Asubseteq Vtext{ is a flat} iff Atext{ is closed under affine combinations, that is}\ x_1,ldots,x_kin A, alpha_1,ldots,alpha_k,in F, alpha_1+cdots+ alpha_k=1Rightarrow alpha_1x_1+cdots+alpha_kx_kin A$$



      The following question within the exercise then refers to this theorem (specifically the RHS of the above iff statement):




      Let $ 1+1neq0$ in $F$.



      Show that closure under affine combinations for $k=2$ implies closure under affine combinations for all $kgeq1$.





      My approach so far has been to use Strong Induction.



      Assuming the result holds for $kleq n$, we show it holds for $k=n+1$



      consider $sum_{i=1}^{k+1}alpha_ix_i $



      choose $lin{1,2,ldots, k+1}$ such that $sum_{i=1}^{k+1}alpha_ix_i=betasum_{i=1}^{l}frac{alpha_i}{beta}x_i+gammasum_{i={l+1}}^{k+1}frac{alpha_i}{gamma}x_i$



      (Note: $beta = sum_{i=1}^{l}alpha_i neq0$ and $gamma = sum_{i=l+1}^{k+1}alpha_i neq0$)



      Using induction hypothesis, $sum_{i=1}^{l}frac{alpha_i}{beta}x_iin A$ and $sum_{i={l+1}}^{k+1}frac{alpha_i}{gamma}x_iin A$. Call them $x$ and $y$ respectively.}



      Now, $sum_{i=1}^{k+1}alpha_ix_i = beta x+ gamma y$



      Since the result holds for $k=2$, $sum_{i=1}^{k+1}alpha_ix_iin A
      $




      My question is twofold. First, I never made use of the fact that $1+1neq0$ in F. What am I missing here? Second, how can I show that the condition in line 3 of my solution (when I introduce l and break the summation into two sums) can be assumed without loss of generality?



      Any input would be appreciated.






      linear-algebra







      share|cite|improve this question











      $endgroup$




      The book I am using mentions this theorem:



      $$Asubseteq Vtext{ is a flat} iff Atext{ is closed under affine combinations, that is}\ x_1,ldots,x_kin A, alpha_1,ldots,alpha_k,in F, alpha_1+cdots+ alpha_k=1Rightarrow alpha_1x_1+cdots+alpha_kx_kin A$$



      The following question within the exercise then refers to this theorem (specifically the RHS of the above iff statement):




      Let $ 1+1neq0$ in $F$.



      Show that closure under affine combinations for $k=2$ implies closure under affine combinations for all $kgeq1$.





      My approach so far has been to use Strong Induction.



      Assuming the result holds for $kleq n$, we show it holds for $k=n+1$



      consider $sum_{i=1}^{k+1}alpha_ix_i $



      choose $lin{1,2,ldots, k+1}$ such that $sum_{i=1}^{k+1}alpha_ix_i=betasum_{i=1}^{l}frac{alpha_i}{beta}x_i+gammasum_{i={l+1}}^{k+1}frac{alpha_i}{gamma}x_i$



      (Note: $beta = sum_{i=1}^{l}alpha_i neq0$ and $gamma = sum_{i=l+1}^{k+1}alpha_i neq0$)



      Using induction hypothesis, $sum_{i=1}^{l}frac{alpha_i}{beta}x_iin A$ and $sum_{i={l+1}}^{k+1}frac{alpha_i}{gamma}x_iin A$. Call them $x$ and $y$ respectively.}



      Now, $sum_{i=1}^{k+1}alpha_ix_i = beta x+ gamma y$



      Since the result holds for $k=2$, $sum_{i=1}^{k+1}alpha_ix_iin A
      $




      My question is twofold. First, I never made use of the fact that $1+1neq0$ in F. What am I missing here? Second, how can I show that the condition in line 3 of my solution (when I introduce l and break the summation into two sums) can be assumed without loss of generality?



      Any input would be appreciated.






      linear-algebra




      linear-algebra






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 7 '18 at 20:02









      Henning Makholm

      239k17304541




      239k17304541










      asked Dec 7 '18 at 19:07









      s0ulr3aper07s0ulr3aper07

      1499




      1499






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          The assumption about the characteristic of $F$ comes into play when we justify the claim you ask about, so the questions are one in the same. Here's a start...



          Firstly note that you can assume all the $alpha_i$ are nonzero.

          Then start by dividing the sum into $beta' = alpha_n$ and $gamma' = sum_{i<n} alpha_i$. We split the problem into cases.



          Case (1) $gamma' not = 0$ and there is nothing to prove.



          Case (2) $gamma' = 0$ and $alpha_j not= -alpha_n$ for some $j <n$. Then we modify the division as $beta = alpha_n + a_j$ and $gamma = sum_{i not= j,n}alpha_i$ and are done.



          Case (3) $gamma' = 0$ and $alpha_j = -alpha_n$ for all $j <n$. By assumption, $2alpha_j not= 0$. Hence, if we set $$gamma = gamma' + 2alpha_n = sum_{j< n-2} alpha_j$$ $$beta = beta' - 2alpha_n = sum_{igeq n-2} alpha_i = -alpha_n$$ then $beta, gamma not= 0$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            If $gamma = 0$, then $beta = alpha_n = 1$. Going to case (3) that you've mentioned, it would imply that $gamma = 0=-alpha_n-alpha_n-cdots -alpha_n=-1-1-cdots - 1$. Now how would I proceed from this point and use $1+1neq 0$?
            $endgroup$
            – s0ulr3aper07
            Dec 7 '18 at 22:18






          • 1




            $begingroup$
            i edited a better explanation for case (3), hope that helps
            $endgroup$
            – Badam Baplan
            Dec 7 '18 at 23:04











          Your Answer





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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          The assumption about the characteristic of $F$ comes into play when we justify the claim you ask about, so the questions are one in the same. Here's a start...



          Firstly note that you can assume all the $alpha_i$ are nonzero.

          Then start by dividing the sum into $beta' = alpha_n$ and $gamma' = sum_{i<n} alpha_i$. We split the problem into cases.



          Case (1) $gamma' not = 0$ and there is nothing to prove.



          Case (2) $gamma' = 0$ and $alpha_j not= -alpha_n$ for some $j <n$. Then we modify the division as $beta = alpha_n + a_j$ and $gamma = sum_{i not= j,n}alpha_i$ and are done.



          Case (3) $gamma' = 0$ and $alpha_j = -alpha_n$ for all $j <n$. By assumption, $2alpha_j not= 0$. Hence, if we set $$gamma = gamma' + 2alpha_n = sum_{j< n-2} alpha_j$$ $$beta = beta' - 2alpha_n = sum_{igeq n-2} alpha_i = -alpha_n$$ then $beta, gamma not= 0$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            If $gamma = 0$, then $beta = alpha_n = 1$. Going to case (3) that you've mentioned, it would imply that $gamma = 0=-alpha_n-alpha_n-cdots -alpha_n=-1-1-cdots - 1$. Now how would I proceed from this point and use $1+1neq 0$?
            $endgroup$
            – s0ulr3aper07
            Dec 7 '18 at 22:18






          • 1




            $begingroup$
            i edited a better explanation for case (3), hope that helps
            $endgroup$
            – Badam Baplan
            Dec 7 '18 at 23:04
















          2












          $begingroup$

          The assumption about the characteristic of $F$ comes into play when we justify the claim you ask about, so the questions are one in the same. Here's a start...



          Firstly note that you can assume all the $alpha_i$ are nonzero.

          Then start by dividing the sum into $beta' = alpha_n$ and $gamma' = sum_{i<n} alpha_i$. We split the problem into cases.



          Case (1) $gamma' not = 0$ and there is nothing to prove.



          Case (2) $gamma' = 0$ and $alpha_j not= -alpha_n$ for some $j <n$. Then we modify the division as $beta = alpha_n + a_j$ and $gamma = sum_{i not= j,n}alpha_i$ and are done.



          Case (3) $gamma' = 0$ and $alpha_j = -alpha_n$ for all $j <n$. By assumption, $2alpha_j not= 0$. Hence, if we set $$gamma = gamma' + 2alpha_n = sum_{j< n-2} alpha_j$$ $$beta = beta' - 2alpha_n = sum_{igeq n-2} alpha_i = -alpha_n$$ then $beta, gamma not= 0$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            If $gamma = 0$, then $beta = alpha_n = 1$. Going to case (3) that you've mentioned, it would imply that $gamma = 0=-alpha_n-alpha_n-cdots -alpha_n=-1-1-cdots - 1$. Now how would I proceed from this point and use $1+1neq 0$?
            $endgroup$
            – s0ulr3aper07
            Dec 7 '18 at 22:18






          • 1




            $begingroup$
            i edited a better explanation for case (3), hope that helps
            $endgroup$
            – Badam Baplan
            Dec 7 '18 at 23:04














          2












          2








          2





          $begingroup$

          The assumption about the characteristic of $F$ comes into play when we justify the claim you ask about, so the questions are one in the same. Here's a start...



          Firstly note that you can assume all the $alpha_i$ are nonzero.

          Then start by dividing the sum into $beta' = alpha_n$ and $gamma' = sum_{i<n} alpha_i$. We split the problem into cases.



          Case (1) $gamma' not = 0$ and there is nothing to prove.



          Case (2) $gamma' = 0$ and $alpha_j not= -alpha_n$ for some $j <n$. Then we modify the division as $beta = alpha_n + a_j$ and $gamma = sum_{i not= j,n}alpha_i$ and are done.



          Case (3) $gamma' = 0$ and $alpha_j = -alpha_n$ for all $j <n$. By assumption, $2alpha_j not= 0$. Hence, if we set $$gamma = gamma' + 2alpha_n = sum_{j< n-2} alpha_j$$ $$beta = beta' - 2alpha_n = sum_{igeq n-2} alpha_i = -alpha_n$$ then $beta, gamma not= 0$






          share|cite|improve this answer











          $endgroup$



          The assumption about the characteristic of $F$ comes into play when we justify the claim you ask about, so the questions are one in the same. Here's a start...



          Firstly note that you can assume all the $alpha_i$ are nonzero.

          Then start by dividing the sum into $beta' = alpha_n$ and $gamma' = sum_{i<n} alpha_i$. We split the problem into cases.



          Case (1) $gamma' not = 0$ and there is nothing to prove.



          Case (2) $gamma' = 0$ and $alpha_j not= -alpha_n$ for some $j <n$. Then we modify the division as $beta = alpha_n + a_j$ and $gamma = sum_{i not= j,n}alpha_i$ and are done.



          Case (3) $gamma' = 0$ and $alpha_j = -alpha_n$ for all $j <n$. By assumption, $2alpha_j not= 0$. Hence, if we set $$gamma = gamma' + 2alpha_n = sum_{j< n-2} alpha_j$$ $$beta = beta' - 2alpha_n = sum_{igeq n-2} alpha_i = -alpha_n$$ then $beta, gamma not= 0$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 7 '18 at 23:02

























          answered Dec 7 '18 at 20:01









          Badam BaplanBadam Baplan

          4,501722




          4,501722












          • $begingroup$
            If $gamma = 0$, then $beta = alpha_n = 1$. Going to case (3) that you've mentioned, it would imply that $gamma = 0=-alpha_n-alpha_n-cdots -alpha_n=-1-1-cdots - 1$. Now how would I proceed from this point and use $1+1neq 0$?
            $endgroup$
            – s0ulr3aper07
            Dec 7 '18 at 22:18






          • 1




            $begingroup$
            i edited a better explanation for case (3), hope that helps
            $endgroup$
            – Badam Baplan
            Dec 7 '18 at 23:04


















          • $begingroup$
            If $gamma = 0$, then $beta = alpha_n = 1$. Going to case (3) that you've mentioned, it would imply that $gamma = 0=-alpha_n-alpha_n-cdots -alpha_n=-1-1-cdots - 1$. Now how would I proceed from this point and use $1+1neq 0$?
            $endgroup$
            – s0ulr3aper07
            Dec 7 '18 at 22:18






          • 1




            $begingroup$
            i edited a better explanation for case (3), hope that helps
            $endgroup$
            – Badam Baplan
            Dec 7 '18 at 23:04
















          $begingroup$
          If $gamma = 0$, then $beta = alpha_n = 1$. Going to case (3) that you've mentioned, it would imply that $gamma = 0=-alpha_n-alpha_n-cdots -alpha_n=-1-1-cdots - 1$. Now how would I proceed from this point and use $1+1neq 0$?
          $endgroup$
          – s0ulr3aper07
          Dec 7 '18 at 22:18




          $begingroup$
          If $gamma = 0$, then $beta = alpha_n = 1$. Going to case (3) that you've mentioned, it would imply that $gamma = 0=-alpha_n-alpha_n-cdots -alpha_n=-1-1-cdots - 1$. Now how would I proceed from this point and use $1+1neq 0$?
          $endgroup$
          – s0ulr3aper07
          Dec 7 '18 at 22:18




          1




          1




          $begingroup$
          i edited a better explanation for case (3), hope that helps
          $endgroup$
          – Badam Baplan
          Dec 7 '18 at 23:04




          $begingroup$
          i edited a better explanation for case (3), hope that helps
          $endgroup$
          – Badam Baplan
          Dec 7 '18 at 23:04


















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