Proving that the union of a linear code with its left coset is also linear

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Suppose $C$ is a linear code, with elements $textbfc$ and $textbfv$ is a code word not in $C$, but it is an element of We form the coset of $C$, given by
$$
C + textbfv = textbfc + textbfv : textbfc in C, textbfv not in X.
$$
I want to prove that $C cup C + textbfv$. is also a linear code, but am struggling a bit.



I considered breaking this into cases by considering a partition $C cup (C + textbfv)$, and then establishing closure for each of the partitions. But, there seems to be a lapse in information about $textbfv$, save for the fact that it's a field element and thus clearly is closed under the group operation.



Any comments or insights would be greatly appreciated.




coding-theory




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    Suppose $C$ is a linear code, with elements $textbfc$ and $textbfv$ is a code word not in $C$, but it is an element of We form the coset of $C$, given by
    $$
    C + textbfv = textbfc + textbfv : textbfc in C, textbfv not in X.
    $$
    I want to prove that $C cup C + textbfv$. is also a linear code, but am struggling a bit.



    I considered breaking this into cases by considering a partition $C cup (C + textbfv)$, and then establishing closure for each of the partitions. But, there seems to be a lapse in information about $textbfv$, save for the fact that it's a field element and thus clearly is closed under the group operation.



    Any comments or insights would be greatly appreciated.




    coding-theory




    share|cite|improve this question





















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Suppose $C$ is a linear code, with elements $textbfc$ and $textbfv$ is a code word not in $C$, but it is an element of We form the coset of $C$, given by
      $$
      C + textbfv = textbfc + textbfv : textbfc in C, textbfv not in X.
      $$
      I want to prove that $C cup C + textbfv$. is also a linear code, but am struggling a bit.



      I considered breaking this into cases by considering a partition $C cup (C + textbfv)$, and then establishing closure for each of the partitions. But, there seems to be a lapse in information about $textbfv$, save for the fact that it's a field element and thus clearly is closed under the group operation.



      Any comments or insights would be greatly appreciated.




      coding-theory




      share|cite|improve this question











      Suppose $C$ is a linear code, with elements $textbfc$ and $textbfv$ is a code word not in $C$, but it is an element of We form the coset of $C$, given by
      $$
      C + textbfv = textbfc + textbfv : textbfc in C, textbfv not in X.
      $$
      I want to prove that $C cup C + textbfv$. is also a linear code, but am struggling a bit.



      I considered breaking this into cases by considering a partition $C cup (C + textbfv)$, and then establishing closure for each of the partitions. But, there seems to be a lapse in information about $textbfv$, save for the fact that it's a field element and thus clearly is closed under the group operation.



      Any comments or insights would be greatly appreciated.





      coding-theory





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      share|cite|improve this question




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      asked Aug 5 at 23:21









      Matt.P

      768313




      768313




















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          That's true iff $C$ is a binary linear code, that is it's linear
          over $Bbb F_2$, the field of two elements.



          The difficult case is showing the sum of two elements of $C+v$
          lies in $Ccup(C+v)$. If those elements are $c_1+v$ and $c_2+v$ then
          they add to
          $$(c_1+v)+(c_2+v)=c_1+c_2+2v=c_1+c_2in C$$
          as $2=0$ within $Bbb F_2$.






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            That's true iff $C$ is a binary linear code, that is it's linear
            over $Bbb F_2$, the field of two elements.



            The difficult case is showing the sum of two elements of $C+v$
            lies in $Ccup(C+v)$. If those elements are $c_1+v$ and $c_2+v$ then
            they add to
            $$(c_1+v)+(c_2+v)=c_1+c_2+2v=c_1+c_2in C$$
            as $2=0$ within $Bbb F_2$.






            share|cite|improve this answer

























              up vote
              1
              down vote













              That's true iff $C$ is a binary linear code, that is it's linear
              over $Bbb F_2$, the field of two elements.



              The difficult case is showing the sum of two elements of $C+v$
              lies in $Ccup(C+v)$. If those elements are $c_1+v$ and $c_2+v$ then
              they add to
              $$(c_1+v)+(c_2+v)=c_1+c_2+2v=c_1+c_2in C$$
              as $2=0$ within $Bbb F_2$.






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                That's true iff $C$ is a binary linear code, that is it's linear
                over $Bbb F_2$, the field of two elements.



                The difficult case is showing the sum of two elements of $C+v$
                lies in $Ccup(C+v)$. If those elements are $c_1+v$ and $c_2+v$ then
                they add to
                $$(c_1+v)+(c_2+v)=c_1+c_2+2v=c_1+c_2in C$$
                as $2=0$ within $Bbb F_2$.






                share|cite|improve this answer













                That's true iff $C$ is a binary linear code, that is it's linear
                over $Bbb F_2$, the field of two elements.



                The difficult case is showing the sum of two elements of $C+v$
                lies in $Ccup(C+v)$. If those elements are $c_1+v$ and $c_2+v$ then
                they add to
                $$(c_1+v)+(c_2+v)=c_1+c_2+2v=c_1+c_2in C$$
                as $2=0$ within $Bbb F_2$.







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Aug 6 at 3:11









                Lord Shark the Unknown

                86k951112




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