The Orthogonal Complement

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  • #1
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Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
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So, what did you try already??
 
  • #3
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[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.
 
  • #4
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[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
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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]...
 
  • #6
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I do not know what do you mean by the sum of tow sets? and how can I prove that?
 
  • #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}.
 
  • #8
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How did you guess that it is the sum of?

since my goal is to reach this equality.

How can we prove it?
 
  • #9
Deveno
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normally, to prove 2 sets are equal, you show they contain each other.
 
  • #10
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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?
 
  • #11
Deveno
Science Advisor
906
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[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.
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.
 
  • #12
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Oh yes, thats the problem. But the problem now how can we prove that
[itex](M\cap N)^\bot=M^\bot+N^\bot[/itex]
 
  • #13
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[itex](M\cap N)^\bot=M^\bot+N^\bot[/itex]
I think we should have some contraintes on M and N, do not I?
 
Last edited:
  • #14
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I think we should have some contraintes on M and N, do not I?
Yeah, M and N should both be subspaces.
 

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