subspace of $mathbb R^4$.

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$W=ab=cd.$



How do I prove that $W$ is subspace of $mathbb R^4$?



I have $(0,0,0,0)in W$, so $W$ not empty.



and $kinmathbb R$,$win W$ so $kwin W$.



so I'm not sure that if $w_1,w_2in W$ so $w_1+w_2in W$ ?




vector-spaces




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  • 2




    $(1,1,1,1)$ and $(2,3,6,1)$ are in $W$. Is $(3,4,7,2)$?
    – Paul
    Aug 21 at 13:36















up vote
1
down vote

favorite












$W=ab=cd.$



How do I prove that $W$ is subspace of $mathbb R^4$?



I have $(0,0,0,0)in W$, so $W$ not empty.



and $kinmathbb R$,$win W$ so $kwin W$.



so I'm not sure that if $w_1,w_2in W$ so $w_1+w_2in W$ ?




vector-spaces




share|cite|improve this question


















  • 2




    $(1,1,1,1)$ and $(2,3,6,1)$ are in $W$. Is $(3,4,7,2)$?
    – Paul
    Aug 21 at 13:36













up vote
1
down vote

favorite









up vote
1
down vote

favorite











$W=ab=cd.$



How do I prove that $W$ is subspace of $mathbb R^4$?



I have $(0,0,0,0)in W$, so $W$ not empty.



and $kinmathbb R$,$win W$ so $kwin W$.



so I'm not sure that if $w_1,w_2in W$ so $w_1+w_2in W$ ?




vector-spaces




share|cite|improve this question














$W=ab=cd.$



How do I prove that $W$ is subspace of $mathbb R^4$?



I have $(0,0,0,0)in W$, so $W$ not empty.



and $kinmathbb R$,$win W$ so $kwin W$.



so I'm not sure that if $w_1,w_2in W$ so $w_1+w_2in W$ ?





vector-spaces





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 21 at 13:32









Bernard

111k635103




111k635103










asked Aug 21 at 13:28









ruler maroon

85




85







  • 2




    $(1,1,1,1)$ and $(2,3,6,1)$ are in $W$. Is $(3,4,7,2)$?
    – Paul
    Aug 21 at 13:36













  • 2




    $(1,1,1,1)$ and $(2,3,6,1)$ are in $W$. Is $(3,4,7,2)$?
    – Paul
    Aug 21 at 13:36








2




2




$(1,1,1,1)$ and $(2,3,6,1)$ are in $W$. Is $(3,4,7,2)$?
– Paul
Aug 21 at 13:36





$(1,1,1,1)$ and $(2,3,6,1)$ are in $W$. Is $(3,4,7,2)$?
– Paul
Aug 21 at 13:36











2 Answers
2






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oldest

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up vote
4
down vote



accepted










It isn't, as it's not defined by linear relations.



Counter-example: set $;w_1=(1,0,0,1)$, $w_2=(0,2,1,0)$. We have:
$$w_1+w_2=(1,2,1,1)$$
and it doesn't satisfy the relation.






share|cite|improve this answer





























    up vote
    2
    down vote













    $(1,0,0,0) in W$ because $1 cdot 0 = 0 cdot 0$.



    Also



    $(0, 1,0,0) in W$ because $0 cdot 1 = 0 cdot 0$.



    $(1,0,0,0)+(0,1,0,0)=(1,1,0,0)$ but $1cdot 1 ne 0 cdot 0$.



    Hence $W$ is not closed under addition.






    share|cite|improve this answer






















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      4
      down vote



      accepted










      It isn't, as it's not defined by linear relations.



      Counter-example: set $;w_1=(1,0,0,1)$, $w_2=(0,2,1,0)$. We have:
      $$w_1+w_2=(1,2,1,1)$$
      and it doesn't satisfy the relation.






      share|cite|improve this answer


























        up vote
        4
        down vote



        accepted










        It isn't, as it's not defined by linear relations.



        Counter-example: set $;w_1=(1,0,0,1)$, $w_2=(0,2,1,0)$. We have:
        $$w_1+w_2=(1,2,1,1)$$
        and it doesn't satisfy the relation.






        share|cite|improve this answer
























          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          It isn't, as it's not defined by linear relations.



          Counter-example: set $;w_1=(1,0,0,1)$, $w_2=(0,2,1,0)$. We have:
          $$w_1+w_2=(1,2,1,1)$$
          and it doesn't satisfy the relation.






          share|cite|improve this answer














          It isn't, as it's not defined by linear relations.



          Counter-example: set $;w_1=(1,0,0,1)$, $w_2=(0,2,1,0)$. We have:
          $$w_1+w_2=(1,2,1,1)$$
          and it doesn't satisfy the relation.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 21 at 18:48

























          answered Aug 21 at 13:37









          Bernard

          111k635103




          111k635103




















              up vote
              2
              down vote













              $(1,0,0,0) in W$ because $1 cdot 0 = 0 cdot 0$.



              Also



              $(0, 1,0,0) in W$ because $0 cdot 1 = 0 cdot 0$.



              $(1,0,0,0)+(0,1,0,0)=(1,1,0,0)$ but $1cdot 1 ne 0 cdot 0$.



              Hence $W$ is not closed under addition.






              share|cite|improve this answer


























                up vote
                2
                down vote













                $(1,0,0,0) in W$ because $1 cdot 0 = 0 cdot 0$.



                Also



                $(0, 1,0,0) in W$ because $0 cdot 1 = 0 cdot 0$.



                $(1,0,0,0)+(0,1,0,0)=(1,1,0,0)$ but $1cdot 1 ne 0 cdot 0$.



                Hence $W$ is not closed under addition.






                share|cite|improve this answer
























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  $(1,0,0,0) in W$ because $1 cdot 0 = 0 cdot 0$.



                  Also



                  $(0, 1,0,0) in W$ because $0 cdot 1 = 0 cdot 0$.



                  $(1,0,0,0)+(0,1,0,0)=(1,1,0,0)$ but $1cdot 1 ne 0 cdot 0$.



                  Hence $W$ is not closed under addition.






                  share|cite|improve this answer














                  $(1,0,0,0) in W$ because $1 cdot 0 = 0 cdot 0$.



                  Also



                  $(0, 1,0,0) in W$ because $0 cdot 1 = 0 cdot 0$.



                  $(1,0,0,0)+(0,1,0,0)=(1,1,0,0)$ but $1cdot 1 ne 0 cdot 0$.



                  Hence $W$ is not closed under addition.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Aug 21 at 15:57

























                  answered Aug 21 at 14:47









                  Siong Thye Goh

                  80.2k1453100




                  80.2k1453100






















                       

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