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Subspaces and perpendiculuar subspaces

  1. Mar 28, 2010 #1
    1. The problem statement, all variables and given/known data
    How do you show that M double perp is a subset of M?

    2. Relevant equations



    3. The attempt at a solution
    My prof told me to try proving that M is a subset of M perp perp, then to use the facts that if M is a subspace of Rn then T(X) = projU(X) for all X in Rn.

    I'm not sure how to go about that. I know logically that it's a subset, but I don't know how to prove it.

    I'm thinking that once I prove it, maybe I can show that the dimension of U and U perp perp are equal, so the spaces are equal too?

    Can anyone help get me started? Thanks :)
     
    Last edited: Mar 28, 2010
  2. jcsd
  3. Mar 28, 2010 #2

    Office_Shredder

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    It follows pretty much from the definition of perpendicularity. Try it out and see what happens
     
    Last edited: Mar 28, 2010
  4. Mar 28, 2010 #3
    Okay well I tried this:
    Say X is a vector in M.

    Then Y is a vector in M perp if Y ● X = 0.

    Z is a vector in M perp perp if Z ● Y = 0.

    But I can show that every vector X in M is also in M perp perp:
    X ● Y = 0 because Y ● X = 0.

    Therefore M is a subset of M perp perp. Is that part right?
     
  5. Mar 28, 2010 #4

    Office_Shredder

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    Yeah, that's all there is to it for that part
     
  6. Mar 28, 2010 #5
    Okay now for the next part.
    T(X) = projM(X) = Y
    dim(M) + dim (M perp) = n

    S(Y) = projM perp(Y) = Z
    dim(M perp) + dim(M perp perp) = n

    Then n - dim(M) = n - dim(M perp perp)
    dim (M) = dim(M perp perp)

    Since M is a subset of M perp perp then they are equal?
     
  7. Mar 28, 2010 #6

    Office_Shredder

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    I'm not sure precisely what your projection notation is supposed to say but the argument basically goes like how you posted:

    For all subspaces U, dim(U)+dim(U perp)=n

    So dim(M)+dim(M perp)=n

    M perp is a subspace also, so dim(M perp)+dim(M perp perp)= n

    And then subtract like you did to finish it off
     
  8. Mar 28, 2010 #7
    Hmm okay, so the projection thing isn't actually necessary?

    Thanks for your help, by the way!
     
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