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Dimension of intersection of subspaces proof

  1. Dec 19, 2011 #1
    1. The problem statement, all variables and given/known data
    [itex]V[/itex] is a vector space with dimension n, [itex]U[/itex] and [itex]W[/itex] are two subspaces with dimension k and l.
    prove that if k+l > n then [itex]U \cap W[/itex] has dimension > 0



    2. Relevant equations
    Grassmann's formula

    [itex]dim(U+W) = dim(U) + dim(W) - dim(U \cap W)[/itex]


    3. The attempt at a solution
    Suppose k+l >n.
    Suppose that [itex]dim(U \cap W) \leq 0[/itex]

    since the dimension can't be negative [itex]dim(U \cap W) = 0[/itex]

    then Grassman formula reduces to

    [itex]dim(U+W) = dim(U) + dim(W)[/itex]
    [itex]dim(U+W) = k +l > n[/itex]

    this is a contraddiction because [itex]U+W[/itex] has dimension grater than the dimension of his enclosing space.




    is this a valid proof?
     
    Last edited: Dec 19, 2011
  2. jcsd
  3. Dec 19, 2011 #2

    HallsofIvy

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    What, exactly, are you trying to prove?
     
  4. Dec 19, 2011 #3
    that [itex] dim(U \cap W) > 0 [/itex]



    since you are asking me that question do i have to assume that my proof is wrong?

    EDIT: ops

    I have to prove that if k+l > n then [itex] dim(U \cap W) > 0 [/itex]
     
    Last edited: Dec 19, 2011
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