how to check for closed addition and closed multiplication

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When checking if a set is in the subspace, I'm checking for the $0$ vector, closed addition and closed multiplication. I've verified all set have the zero vector since $x,y,z$ and $s,t$ are all in $R$ and can be set to $0$. But how do i check closed addition/multiplication?






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    addition: you take two arbitrary elements from the set, add them and check whether the result is in the set. (scalar) mulitplication: you take a scalar and an element $x$ of the set, multiply $x$ by the scalar and check whether the result is in the set.
    – Thomas
    Sep 1 at 5:54














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When checking if a set is in the subspace, I'm checking for the $0$ vector, closed addition and closed multiplication. I've verified all set have the zero vector since $x,y,z$ and $s,t$ are all in $R$ and can be set to $0$. But how do i check closed addition/multiplication?






linear-algebra







share|cite|improve this question

















  • 1




    addition: you take two arbitrary elements from the set, add them and check whether the result is in the set. (scalar) mulitplication: you take a scalar and an element $x$ of the set, multiply $x$ by the scalar and check whether the result is in the set.
    – Thomas
    Sep 1 at 5:54












up vote
0
down vote

favorite









up vote
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down vote

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enter image description here



When checking if a set is in the subspace, I'm checking for the $0$ vector, closed addition and closed multiplication. I've verified all set have the zero vector since $x,y,z$ and $s,t$ are all in $R$ and can be set to $0$. But how do i check closed addition/multiplication?






linear-algebra







share|cite|improve this question













enter image description here



When checking if a set is in the subspace, I'm checking for the $0$ vector, closed addition and closed multiplication. I've verified all set have the zero vector since $x,y,z$ and $s,t$ are all in $R$ and can be set to $0$. But how do i check closed addition/multiplication?






linear-algebra




linear-algebra






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asked Sep 1 at 5:46









lohboys

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617







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    addition: you take two arbitrary elements from the set, add them and check whether the result is in the set. (scalar) mulitplication: you take a scalar and an element $x$ of the set, multiply $x$ by the scalar and check whether the result is in the set.
    – Thomas
    Sep 1 at 5:54












  • 1




    addition: you take two arbitrary elements from the set, add them and check whether the result is in the set. (scalar) mulitplication: you take a scalar and an element $x$ of the set, multiply $x$ by the scalar and check whether the result is in the set.
    – Thomas
    Sep 1 at 5:54







1




1




addition: you take two arbitrary elements from the set, add them and check whether the result is in the set. (scalar) mulitplication: you take a scalar and an element $x$ of the set, multiply $x$ by the scalar and check whether the result is in the set.
– Thomas
Sep 1 at 5:54




addition: you take two arbitrary elements from the set, add them and check whether the result is in the set. (scalar) mulitplication: you take a scalar and an element $x$ of the set, multiply $x$ by the scalar and check whether the result is in the set.
– Thomas
Sep 1 at 5:54










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  1. $U_1$ is not a subspace: See that $x=(1,2,3)in U_1$ but $ -1.x notin U_1$


  2. $U_2$ is not a subspace: $(1,1,1) ,(0,1,2)in U_2$ but their sum doesn't belongs to $U_2$.


  3. $U_3$ is the Linear span of $(1,0,0),(0,0,1) $, so a subspace.


  4. $U_4$ is not a subspace. you can check it is not closed under addition.


  5. $(0,0,0)$ is a subspace. since it is closed under addition and scalar multiplication.


  6. $U_6= textNull space of ~A$, where
    $A=beginbmatrix
    1 & 1 & -1\
    2 & 1 & 1 \
    endbmatrix$.
    Hence $U_6$ is also a subspace.






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    1. $U_1$ is not a subspace: See that $x=(1,2,3)in U_1$ but $ -1.x notin U_1$


    2. $U_2$ is not a subspace: $(1,1,1) ,(0,1,2)in U_2$ but their sum doesn't belongs to $U_2$.


    3. $U_3$ is the Linear span of $(1,0,0),(0,0,1) $, so a subspace.


    4. $U_4$ is not a subspace. you can check it is not closed under addition.


    5. $(0,0,0)$ is a subspace. since it is closed under addition and scalar multiplication.


    6. $U_6= textNull space of ~A$, where
      $A=beginbmatrix
      1 & 1 & -1\
      2 & 1 & 1 \
      endbmatrix$.
      Hence $U_6$ is also a subspace.






    share|cite|improve this answer
























      up vote
      1
      down vote













      1. $U_1$ is not a subspace: See that $x=(1,2,3)in U_1$ but $ -1.x notin U_1$


      2. $U_2$ is not a subspace: $(1,1,1) ,(0,1,2)in U_2$ but their sum doesn't belongs to $U_2$.


      3. $U_3$ is the Linear span of $(1,0,0),(0,0,1) $, so a subspace.


      4. $U_4$ is not a subspace. you can check it is not closed under addition.


      5. $(0,0,0)$ is a subspace. since it is closed under addition and scalar multiplication.


      6. $U_6= textNull space of ~A$, where
        $A=beginbmatrix
        1 & 1 & -1\
        2 & 1 & 1 \
        endbmatrix$.
        Hence $U_6$ is also a subspace.






      share|cite|improve this answer






















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        1
        down vote










        up vote
        1
        down vote









        1. $U_1$ is not a subspace: See that $x=(1,2,3)in U_1$ but $ -1.x notin U_1$


        2. $U_2$ is not a subspace: $(1,1,1) ,(0,1,2)in U_2$ but their sum doesn't belongs to $U_2$.


        3. $U_3$ is the Linear span of $(1,0,0),(0,0,1) $, so a subspace.


        4. $U_4$ is not a subspace. you can check it is not closed under addition.


        5. $(0,0,0)$ is a subspace. since it is closed under addition and scalar multiplication.


        6. $U_6= textNull space of ~A$, where
          $A=beginbmatrix
          1 & 1 & -1\
          2 & 1 & 1 \
          endbmatrix$.
          Hence $U_6$ is also a subspace.






        share|cite|improve this answer












        1. $U_1$ is not a subspace: See that $x=(1,2,3)in U_1$ but $ -1.x notin U_1$


        2. $U_2$ is not a subspace: $(1,1,1) ,(0,1,2)in U_2$ but their sum doesn't belongs to $U_2$.


        3. $U_3$ is the Linear span of $(1,0,0),(0,0,1) $, so a subspace.


        4. $U_4$ is not a subspace. you can check it is not closed under addition.


        5. $(0,0,0)$ is a subspace. since it is closed under addition and scalar multiplication.


        6. $U_6= textNull space of ~A$, where
          $A=beginbmatrix
          1 & 1 & -1\
          2 & 1 & 1 \
          endbmatrix$.
          Hence $U_6$ is also a subspace.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 1 at 6:06









        Indrajit Ghosh

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