why is $ {(0,x,z)|x,zin R}$ is a two dimensional subspace space of $R^{3}$ over $R$ but ${(0,0,z)|zin R} $...












0












$begingroup$


why is $ A={(0,x,z)|x,zin R}$ is a two dimensional subspace space of $R^{3}$ over $R$
but $B={(0,0,z)|zin R} $ $cup$ ${(0,x,0)|xin R}$ is not?



i Think both are two dimensional as A has Basis={(0,1,0),(0,0,1)} and



B is direct sum of two subspace of $ R^{3}$ with same basis .






linear-algebra matrices linear-transformations







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$endgroup$








  • 2




    $begingroup$
    "B is direct sum of two subspace of $ R^{3}$ with same basis" No, $B$ is a union of subspaces.
    $endgroup$
    – Arthur
    Jan 21 at 12:35








  • 1




    $begingroup$
    I suspect your confusion lies in the fact that every $v in B$ has a unique decomposition $v = (0,0,a) +(0,b,0)$. The issue is not in the existence or uniqueness of this decomposition, it is that the space $B$ isn't closed under addition.
    $endgroup$
    – Calvin Khor
    Jan 21 at 12:38
















0












$begingroup$


why is $ A={(0,x,z)|x,zin R}$ is a two dimensional subspace space of $R^{3}$ over $R$
but $B={(0,0,z)|zin R} $ $cup$ ${(0,x,0)|xin R}$ is not?



i Think both are two dimensional as A has Basis={(0,1,0),(0,0,1)} and



B is direct sum of two subspace of $ R^{3}$ with same basis .






linear-algebra matrices linear-transformations







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    "B is direct sum of two subspace of $ R^{3}$ with same basis" No, $B$ is a union of subspaces.
    $endgroup$
    – Arthur
    Jan 21 at 12:35








  • 1




    $begingroup$
    I suspect your confusion lies in the fact that every $v in B$ has a unique decomposition $v = (0,0,a) +(0,b,0)$. The issue is not in the existence or uniqueness of this decomposition, it is that the space $B$ isn't closed under addition.
    $endgroup$
    – Calvin Khor
    Jan 21 at 12:38














0












0








0





$begingroup$


why is $ A={(0,x,z)|x,zin R}$ is a two dimensional subspace space of $R^{3}$ over $R$
but $B={(0,0,z)|zin R} $ $cup$ ${(0,x,0)|xin R}$ is not?



i Think both are two dimensional as A has Basis={(0,1,0),(0,0,1)} and



B is direct sum of two subspace of $ R^{3}$ with same basis .






linear-algebra matrices linear-transformations







share|cite|improve this question









$endgroup$




why is $ A={(0,x,z)|x,zin R}$ is a two dimensional subspace space of $R^{3}$ over $R$
but $B={(0,0,z)|zin R} $ $cup$ ${(0,x,0)|xin R}$ is not?



i Think both are two dimensional as A has Basis={(0,1,0),(0,0,1)} and



B is direct sum of two subspace of $ R^{3}$ with same basis .






linear-algebra matrices linear-transformations




linear-algebra matrices linear-transformations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 21 at 12:32









sejysejy

1539




1539








  • 2




    $begingroup$
    "B is direct sum of two subspace of $ R^{3}$ with same basis" No, $B$ is a union of subspaces.
    $endgroup$
    – Arthur
    Jan 21 at 12:35








  • 1




    $begingroup$
    I suspect your confusion lies in the fact that every $v in B$ has a unique decomposition $v = (0,0,a) +(0,b,0)$. The issue is not in the existence or uniqueness of this decomposition, it is that the space $B$ isn't closed under addition.
    $endgroup$
    – Calvin Khor
    Jan 21 at 12:38














  • 2




    $begingroup$
    "B is direct sum of two subspace of $ R^{3}$ with same basis" No, $B$ is a union of subspaces.
    $endgroup$
    – Arthur
    Jan 21 at 12:35








  • 1




    $begingroup$
    I suspect your confusion lies in the fact that every $v in B$ has a unique decomposition $v = (0,0,a) +(0,b,0)$. The issue is not in the existence or uniqueness of this decomposition, it is that the space $B$ isn't closed under addition.
    $endgroup$
    – Calvin Khor
    Jan 21 at 12:38








2




2




$begingroup$
"B is direct sum of two subspace of $ R^{3}$ with same basis" No, $B$ is a union of subspaces.
$endgroup$
– Arthur
Jan 21 at 12:35






$begingroup$
"B is direct sum of two subspace of $ R^{3}$ with same basis" No, $B$ is a union of subspaces.
$endgroup$
– Arthur
Jan 21 at 12:35






1




1




$begingroup$
I suspect your confusion lies in the fact that every $v in B$ has a unique decomposition $v = (0,0,a) +(0,b,0)$. The issue is not in the existence or uniqueness of this decomposition, it is that the space $B$ isn't closed under addition.
$endgroup$
– Calvin Khor
Jan 21 at 12:38




$begingroup$
I suspect your confusion lies in the fact that every $v in B$ has a unique decomposition $v = (0,0,a) +(0,b,0)$. The issue is not in the existence or uniqueness of this decomposition, it is that the space $B$ isn't closed under addition.
$endgroup$
– Calvin Khor
Jan 21 at 12:38










2 Answers
2






active

oldest

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8












$begingroup$

$B$ is defined as the union of two subsets of $mathbb R^{3}$, not as a sum of subspaces. $(0,0,1)+(0,1,0)=(0,1,1)$ is not in $B$ so it is not a subspace of $mathbb R^{3}$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    $B$ is a union of two subspaces and is missing all of the vectors that are “in between” the subspaces. More formally, the union isn’t closed under vector addition: $(0,0,z)+(0,x,0)notin B$ unless at least one of those two vectors is zero. On the other hand, $A$ contains all of those sums and then some: if you add any two vectors of the form $(0,x,z)$ to each other, you get another vector of the same form, so $A$ is closed under vector addition. It’s easy to verify that it also contains the zero vector and is closed under scalar multiplication, so it is a subspace of $mathbb R^3$.



    To illustrate this geometrically, $B$ is the union of the $y$-and $z$-axes. This just gets you a pair of lines, which you probably know is not a vector space. On the other hand, $A$ describes the entire $y$-$z$ plane, which contains those two lines, but also all of the points “in between” them on that plane. This plane is isomorphic to $mathbb R^2$, which you know to be a vector space.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      U cleared my confusion insightful !!
      $endgroup$
      – sejy
      Jan 23 at 0:38











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    2 Answers
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    2 Answers
    2






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    active

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    8












    $begingroup$

    $B$ is defined as the union of two subsets of $mathbb R^{3}$, not as a sum of subspaces. $(0,0,1)+(0,1,0)=(0,1,1)$ is not in $B$ so it is not a subspace of $mathbb R^{3}$.






    share|cite|improve this answer









    $endgroup$


















      8












      $begingroup$

      $B$ is defined as the union of two subsets of $mathbb R^{3}$, not as a sum of subspaces. $(0,0,1)+(0,1,0)=(0,1,1)$ is not in $B$ so it is not a subspace of $mathbb R^{3}$.






      share|cite|improve this answer









      $endgroup$
















        8












        8








        8





        $begingroup$

        $B$ is defined as the union of two subsets of $mathbb R^{3}$, not as a sum of subspaces. $(0,0,1)+(0,1,0)=(0,1,1)$ is not in $B$ so it is not a subspace of $mathbb R^{3}$.






        share|cite|improve this answer









        $endgroup$



        $B$ is defined as the union of two subsets of $mathbb R^{3}$, not as a sum of subspaces. $(0,0,1)+(0,1,0)=(0,1,1)$ is not in $B$ so it is not a subspace of $mathbb R^{3}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 21 at 12:33









        Kavi Rama MurthyKavi Rama Murthy

        62.3k42262




        62.3k42262























            2












            $begingroup$

            $B$ is a union of two subspaces and is missing all of the vectors that are “in between” the subspaces. More formally, the union isn’t closed under vector addition: $(0,0,z)+(0,x,0)notin B$ unless at least one of those two vectors is zero. On the other hand, $A$ contains all of those sums and then some: if you add any two vectors of the form $(0,x,z)$ to each other, you get another vector of the same form, so $A$ is closed under vector addition. It’s easy to verify that it also contains the zero vector and is closed under scalar multiplication, so it is a subspace of $mathbb R^3$.



            To illustrate this geometrically, $B$ is the union of the $y$-and $z$-axes. This just gets you a pair of lines, which you probably know is not a vector space. On the other hand, $A$ describes the entire $y$-$z$ plane, which contains those two lines, but also all of the points “in between” them on that plane. This plane is isomorphic to $mathbb R^2$, which you know to be a vector space.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              U cleared my confusion insightful !!
              $endgroup$
              – sejy
              Jan 23 at 0:38
















            2












            $begingroup$

            $B$ is a union of two subspaces and is missing all of the vectors that are “in between” the subspaces. More formally, the union isn’t closed under vector addition: $(0,0,z)+(0,x,0)notin B$ unless at least one of those two vectors is zero. On the other hand, $A$ contains all of those sums and then some: if you add any two vectors of the form $(0,x,z)$ to each other, you get another vector of the same form, so $A$ is closed under vector addition. It’s easy to verify that it also contains the zero vector and is closed under scalar multiplication, so it is a subspace of $mathbb R^3$.



            To illustrate this geometrically, $B$ is the union of the $y$-and $z$-axes. This just gets you a pair of lines, which you probably know is not a vector space. On the other hand, $A$ describes the entire $y$-$z$ plane, which contains those two lines, but also all of the points “in between” them on that plane. This plane is isomorphic to $mathbb R^2$, which you know to be a vector space.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              U cleared my confusion insightful !!
              $endgroup$
              – sejy
              Jan 23 at 0:38














            2












            2








            2





            $begingroup$

            $B$ is a union of two subspaces and is missing all of the vectors that are “in between” the subspaces. More formally, the union isn’t closed under vector addition: $(0,0,z)+(0,x,0)notin B$ unless at least one of those two vectors is zero. On the other hand, $A$ contains all of those sums and then some: if you add any two vectors of the form $(0,x,z)$ to each other, you get another vector of the same form, so $A$ is closed under vector addition. It’s easy to verify that it also contains the zero vector and is closed under scalar multiplication, so it is a subspace of $mathbb R^3$.



            To illustrate this geometrically, $B$ is the union of the $y$-and $z$-axes. This just gets you a pair of lines, which you probably know is not a vector space. On the other hand, $A$ describes the entire $y$-$z$ plane, which contains those two lines, but also all of the points “in between” them on that plane. This plane is isomorphic to $mathbb R^2$, which you know to be a vector space.






            share|cite|improve this answer











            $endgroup$



            $B$ is a union of two subspaces and is missing all of the vectors that are “in between” the subspaces. More formally, the union isn’t closed under vector addition: $(0,0,z)+(0,x,0)notin B$ unless at least one of those two vectors is zero. On the other hand, $A$ contains all of those sums and then some: if you add any two vectors of the form $(0,x,z)$ to each other, you get another vector of the same form, so $A$ is closed under vector addition. It’s easy to verify that it also contains the zero vector and is closed under scalar multiplication, so it is a subspace of $mathbb R^3$.



            To illustrate this geometrically, $B$ is the union of the $y$-and $z$-axes. This just gets you a pair of lines, which you probably know is not a vector space. On the other hand, $A$ describes the entire $y$-$z$ plane, which contains those two lines, but also all of the points “in between” them on that plane. This plane is isomorphic to $mathbb R^2$, which you know to be a vector space.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 23 at 0:52

























            answered Jan 21 at 21:35









            amdamd

            30.6k21050




            30.6k21050












            • $begingroup$
              U cleared my confusion insightful !!
              $endgroup$
              – sejy
              Jan 23 at 0:38


















            • $begingroup$
              U cleared my confusion insightful !!
              $endgroup$
              – sejy
              Jan 23 at 0:38
















            $begingroup$
            U cleared my confusion insightful !!
            $endgroup$
            – sejy
            Jan 23 at 0:38




            $begingroup$
            U cleared my confusion insightful !!
            $endgroup$
            – sejy
            Jan 23 at 0:38


















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