Direct Sum Test for more than 2 subspaces

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I know that if 2 subspace $U_1,U_2$ of V ,can be written as direct sum iff

1) $U_1+U_2$=V

2) $U_1cap U_2=0$


I can prove this .But Now to extend this defination to more than 2 subspaces then there is problem even though that subspaces are mutually intersect only at 0.

Example:$U_1$=$x,y,0,$U_2$=zin R$,$U_3$=yin R$ these are subspaces of $R^3$.As any 2 subspaces has intersection 0 but still this is not direct sum

As we cannot write as 0 uniquely as 0,0,0=0,0,0+0,0,0+0,0,0=0,1,0+0,0,1+0,-1,-1 So $U_1+U_2+U_3$ is not direct sum of $R^3$.




So my question what is more condition required to make more than 2 subspace to be direct sum?





linear-algebra vector-spaces




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  • One condition is, the dimensions have to add up. In your example, $2+1+1>3$. But this is not sufficient.
    – Gerry Myerson
    2 days ago















up vote
2
down vote

favorite












I know that if 2 subspace $U_1,U_2$ of V ,can be written as direct sum iff

1) $U_1+U_2$=V

2) $U_1cap U_2=0$


I can prove this .But Now to extend this defination to more than 2 subspaces then there is problem even though that subspaces are mutually intersect only at 0.

Example:$U_1$=$x,y,0,$U_2$=zin R$,$U_3$=yin R$ these are subspaces of $R^3$.As any 2 subspaces has intersection 0 but still this is not direct sum

As we cannot write as 0 uniquely as 0,0,0=0,0,0+0,0,0+0,0,0=0,1,0+0,0,1+0,-1,-1 So $U_1+U_2+U_3$ is not direct sum of $R^3$.




So my question what is more condition required to make more than 2 subspace to be direct sum?





linear-algebra vector-spaces




share|cite|improve this question





















  • One condition is, the dimensions have to add up. In your example, $2+1+1>3$. But this is not sufficient.
    – Gerry Myerson
    2 days ago













up vote
2
down vote

favorite









up vote
2
down vote

favorite











I know that if 2 subspace $U_1,U_2$ of V ,can be written as direct sum iff

1) $U_1+U_2$=V

2) $U_1cap U_2=0$


I can prove this .But Now to extend this defination to more than 2 subspaces then there is problem even though that subspaces are mutually intersect only at 0.

Example:$U_1$=$x,y,0,$U_2$=zin R$,$U_3$=yin R$ these are subspaces of $R^3$.As any 2 subspaces has intersection 0 but still this is not direct sum

As we cannot write as 0 uniquely as 0,0,0=0,0,0+0,0,0+0,0,0=0,1,0+0,0,1+0,-1,-1 So $U_1+U_2+U_3$ is not direct sum of $R^3$.




So my question what is more condition required to make more than 2 subspace to be direct sum?





linear-algebra vector-spaces




share|cite|improve this question













I know that if 2 subspace $U_1,U_2$ of V ,can be written as direct sum iff

1) $U_1+U_2$=V

2) $U_1cap U_2=0$


I can prove this .But Now to extend this defination to more than 2 subspaces then there is problem even though that subspaces are mutually intersect only at 0.

Example:$U_1$=$x,y,0,$U_2$=zin R$,$U_3$=yin R$ these are subspaces of $R^3$.As any 2 subspaces has intersection 0 but still this is not direct sum

As we cannot write as 0 uniquely as 0,0,0=0,0,0+0,0,0+0,0,0=0,1,0+0,0,1+0,-1,-1 So $U_1+U_2+U_3$ is not direct sum of $R^3$.




So my question what is more condition required to make more than 2 subspace to be direct sum?






linear-algebra vector-spaces





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited 2 days ago
























asked 2 days ago









SRJ

994216




994216











  • One condition is, the dimensions have to add up. In your example, $2+1+1>3$. But this is not sufficient.
    – Gerry Myerson
    2 days ago

















  • One condition is, the dimensions have to add up. In your example, $2+1+1>3$. But this is not sufficient.
    – Gerry Myerson
    2 days ago
















One condition is, the dimensions have to add up. In your example, $2+1+1>3$. But this is not sufficient.
– Gerry Myerson
2 days ago





One condition is, the dimensions have to add up. In your example, $2+1+1>3$. But this is not sufficient.
– Gerry Myerson
2 days ago











1 Answer
1






active

oldest

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1
down vote



accepted










Let consider



$$U_4=U_2+U_3$$



which is a direct sum since $U_2cap U_3=emptyset$ and then $U_4$ has dimension 2.



Therefore since also $U_1$ has dimension $2$



$$U_1cap U_4neq emptyset$$



and $U_1+U_4=U_1+U_2+U_3$ is not a direct sum of $mathbbR^3$.






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    1 Answer
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    active

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    Let consider



    $$U_4=U_2+U_3$$



    which is a direct sum since $U_2cap U_3=emptyset$ and then $U_4$ has dimension 2.



    Therefore since also $U_1$ has dimension $2$



    $$U_1cap U_4neq emptyset$$



    and $U_1+U_4=U_1+U_2+U_3$ is not a direct sum of $mathbbR^3$.






    share|cite|improve this answer

























      up vote
      1
      down vote



      accepted










      Let consider



      $$U_4=U_2+U_3$$



      which is a direct sum since $U_2cap U_3=emptyset$ and then $U_4$ has dimension 2.



      Therefore since also $U_1$ has dimension $2$



      $$U_1cap U_4neq emptyset$$



      and $U_1+U_4=U_1+U_2+U_3$ is not a direct sum of $mathbbR^3$.






      share|cite|improve this answer























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        Let consider



        $$U_4=U_2+U_3$$



        which is a direct sum since $U_2cap U_3=emptyset$ and then $U_4$ has dimension 2.



        Therefore since also $U_1$ has dimension $2$



        $$U_1cap U_4neq emptyset$$



        and $U_1+U_4=U_1+U_2+U_3$ is not a direct sum of $mathbbR^3$.






        share|cite|improve this answer













        Let consider



        $$U_4=U_2+U_3$$



        which is a direct sum since $U_2cap U_3=emptyset$ and then $U_4$ has dimension 2.



        Therefore since also $U_1$ has dimension $2$



        $$U_1cap U_4neq emptyset$$



        and $U_1+U_4=U_1+U_2+U_3$ is not a direct sum of $mathbbR^3$.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered 2 days ago









        gimusi

        63.6k73480




        63.6k73480






















             

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