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The Orthogonal Complement

  1. Nov 4, 2011 #1
    Let M and N be two subsets of a hilbert space H.
    What are orthogonal complements of following sets:
    1) The union of M and N.
    2) The intersection of M and N.
     
  2. jcsd
  3. Nov 4, 2011 #2

    micromass

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    So, what did you try already??
     
  4. Nov 4, 2011 #3
    [itex] x\in (M\cap N)^\bot [/itex] means
    [itex] <x,y>=0 [/itex] for all [itex] y\in M \text{ and } y\in N [/itex] and hence

    [itex] x\in M^\bot\cap N^\bot. [/itex]. The other direction is obvious, so we get [itex] (M\cap N)^\bot=M^\bot\cap N^\bot [/itex]
    This is for the intersection, but I strongly think that I had a mistake.
     
  5. Nov 4, 2011 #4
    [itex] x\in (M\cap N)^\bot [/itex] means
    [itex] <x,y>=0 [/itex] for all [itex] y\in M \text{ and } y\in N [/itex] and hence

    [itex] x\in M^\bot\cap N^\bot. [/itex]. The other direction is obvious, so we get [itex] (M\cap N)^\bot=M^\bot\cap N^\bot [/itex]
    This is for the intersection, but I strongly think that I had a mistake.
     
  6. Nov 4, 2011 #5

    micromass

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    I don't think the other inclusion is obvious.

    In fact, I suspect [itex](M\cap N)^\bot = M^\bot + N^\bot[/itex]...
     
  7. Nov 4, 2011 #6
    I do not know what do you mean by the sum of tow sets? and how can I prove that?
     
  8. Nov 4, 2011 #7

    Deveno

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    one can't, in general, sum two sets, but if we already have an addition defined, then:

    A + B = {a+b : a is in A, b is in B}.
     
  9. Nov 4, 2011 #8
    How did you guess that it is the sum of?

    since my goal is to reach this equality.

    How can we prove it?
     
  10. Nov 4, 2011 #9

    Deveno

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    normally, to prove 2 sets are equal, you show they contain each other.
     
  11. Nov 4, 2011 #10
    Of course, but I did not manege tp solve it!
    On the other hand, [itex](M\cap N)^\bot=M^\bot+N^\bot[/itex] means [itex]M^\bot+ N^\bot\subset M^\bot\cap N^\bot[/itex], but this seems senseless, does not it?
     
  12. Nov 4, 2011 #11

    Deveno

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    your definition of [itex] (M\cap N)^\bot [/itex] isn't correct here.

    we don't know that <x,y> = 0 for all y in M and y in N, just those y that are in the intersection. if y is in M-N, all bets are off.
     
  13. Nov 4, 2011 #12
    Oh yes, thats the problem. But the problem now how can we prove that
    [itex](M\cap N)^\bot=M^\bot+N^\bot[/itex]
     
  14. Nov 5, 2011 #13
    I think we should have some contraintes on M and N, do not I?
     
    Last edited: Nov 5, 2011
  15. Nov 5, 2011 #14

    micromass

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    Yeah, M and N should both be subspaces.
     
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