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Math Help - Orthogonal complements

  1. #1
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    Question Please Help!!!

    Let V be an inner product space, and let U and W be subspaces of V. Show that

    (U \cap W)* = U*+W*

    and

    (U+W)=U* \cap W*


    here, U* denotes the orthogonal complement of U.

    having trouble even finding where to start this one.
    Last edited by mathisthebestpuzzle; April 26th 2008 at 06:49 PM. Reason: need help
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  2. #2
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    i know i need to prove subspaces both ways.

    x \in the perp of the intersection of U and W what does that tell me?

    i'm stuck.
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  3. #3
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    hey

    hey have you been able to figure out this problem. I also have the same problem just wondering if u found a way to do it or maybe we can but our ideas together to solve it.
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  4. #4
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    Quote Originally Posted by mathisthebestpuzzle View Post
    (U \cap W)* = U*+W*

    and

    (U+W)=U* \cap W* ====> shouldn't it be (U+W)*=U* \cap W*
    well this is basically De Morgans theorem, the way i would prove it is through drawing a Venn diagram and shade out the required region.
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  5. #5
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    Quote Originally Posted by mathisthebestpuzzle View Post
    Let V be an inner product space, and let U and W be subspaces of V. Show that

    (U \cap W)* = U*+W*

    and

    (U+W)=U* \cap W*


    here, U* denotes the orthogonal complement of U.

    having trouble even finding where to start this one.
    See here.
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  6. #6
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    the link sends me to a login for live journal? thank you though
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  7. #7
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    Quote Originally Posted by mathisthebestpuzzle View Post
    the link sends me to a login for live journal? thank you though
    Sorry, I hadn't noticed that LiveJournal page was "friends only". Here's a copy of the relevant comment. The "perp" symbol ⊥ (denoting an orthogonal complement) has come out on the line instead of as a superscript, which makes it a bit hard to read.

    You need to know that U⊥⊥ = U. Also, if G and H are subspaces with G⊆H, then H⊥⊆G⊥.

    If x = y+z with y∈U⊥ and z∈W⊥ then it is easy to see that y and z both lie in (U∩W)⊥, hence so does x. Therefore U⊥+W⊥⊆(U∩W)⊥.

    For the converse inclusion, if x∈(U⊥+W⊥)⊥ then x∈U⊥⊥ = U, and similarly x∈W⊥⊥ = W. Thus (U⊥+W⊥)⊥⊆U∩W. Take the perp of each side to see that (U∩W)⊥⊆U⊥+W⊥.

    Thus (U∩W)⊥ = U⊥+W⊥. You get the other identity by taking the perp of both sides and replacing U with U⊥ and W with W⊥.
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