- #1

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What are orthogonal complements of following sets:

1) The union of M and N.

2) The intersection of M and N.

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- Thread starter LikeMath
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- #1

- 62

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What are orthogonal complements of following sets:

1) The union of M and N.

2) The intersection of M and N.

- #2

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So, what did you try already??

- #3

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[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,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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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

Science Advisor

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A + B = {a+b : a is in A, b is in B}.

- #8

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since my goal is to reach this equality.

How can we prove it?

- #9

Deveno

Science Advisor

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normally, to prove 2 sets are equal, you show they contain each other.

- #10

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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

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[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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[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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