The Orthogonal Complement

1. Nov 4, 2011

LikeMath

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. Nov 4, 2011

micromass

Staff Emeritus
So, what did you try already??

3. Nov 4, 2011

LikeMath

$x\in (M\cap N)^\bot$ means
$<x,y>=0$ for all $y\in M \text{ and } y\in N$ and hence

$x\in M^\bot\cap N^\bot.$. The other direction is obvious, so we get $(M\cap N)^\bot=M^\bot\cap N^\bot$
This is for the intersection, but I strongly think that I had a mistake.

4. Nov 4, 2011

LikeMath

$x\in (M\cap N)^\bot$ means
$<x,y>=0$ for all $y\in M \text{ and } y\in N$ and hence

$x\in M^\bot\cap N^\bot.$. The other direction is obvious, so we get $(M\cap N)^\bot=M^\bot\cap N^\bot$
This is for the intersection, but I strongly think that I had a mistake.

5. Nov 4, 2011

micromass

Staff Emeritus
I don't think the other inclusion is obvious.

In fact, I suspect $(M\cap N)^\bot = M^\bot + N^\bot$...

6. Nov 4, 2011

LikeMath

I do not know what do you mean by the sum of tow sets? and how can I prove that?

7. Nov 4, 2011

Deveno

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. Nov 4, 2011

LikeMath

How did you guess that it is the sum of?

since my goal is to reach this equality.

How can we prove it?

9. Nov 4, 2011

Deveno

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

10. Nov 4, 2011

LikeMath

Of course, but I did not manege tp solve it!
On the other hand, $(M\cap N)^\bot=M^\bot+N^\bot$ means $M^\bot+ N^\bot\subset M^\bot\cap N^\bot$, but this seems senseless, does not it?

11. Nov 4, 2011

Deveno

your definition of $(M\cap N)^\bot$ 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. Nov 4, 2011

LikeMath

Oh yes, thats the problem. But the problem now how can we prove that
$(M\cap N)^\bot=M^\bot+N^\bot$

13. Nov 5, 2011

LikeMath

I think we should have some contraintes on M and N, do not I?

Last edited: Nov 5, 2011
14. Nov 5, 2011

micromass

Staff Emeritus
Yeah, M and N should both be subspaces.