# 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

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

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

Yeah, M and N should both be subspaces.