Prove the W is the orthogonal complement of its orthogonal complement

[chipotleaway]

Homework Statement

Let W be a subspace of R^n. Show that the orthogonal complement of the orthogonal complement of W is W.

i.e. Show that (W^{\perp})^{\perp}=W

The Attempt at a Solution

This is one of those 'obvious' properties that probably has a really simple proof but which continues to elude me. Here's my latest, rather ugly attempt. I still feel like I've overlooked something, if not, I think it definitely could be refined.

If W^{\perp} is the orthogonal complement of W, then w.u=0, \forall w\in W, \forall u \in W^{\perp}.
Similarly, x.u=0, \forall u\in W^{\perp} and \forall x\in (W^{\perp})^{\perp}.
All vectors w \in W must also be in (W^{\perp})^{\perp} since W consists of only vectors to perpendicular to those in W^{\perp}. Now we need to show that (W^{\perp})^{\perp} cannot contain anything not in W. Suppose there is a vector x' \in (W^{\perp})^{\perp} but not in W. Then x'.u=0, but it must also have some property that prevents it from being in W, namely x' would also have to be orthogonal to W (if not, then it must be in W). But this would imply it is in W^{\perp} and contradict the property W^{\perp} \cap (W^{\perp})^{\perp}=0

 

[vela]

Homework Statement

Let W be a subspace of R^n. Show that the orthogonal complement of the orthogonal complement of W is W.

i.e. Show that (W^{\perp})^{\perp}=W

The Attempt at a Solution

This is one of those 'obvious' properties that probably has a really simple proof but which continues to elude me. Here's my latest, rather ugly attempt. I still feel like I've overlooked something, if not, I think it definitely could be refined.

If W^{\perp} is the orthogonal complement of W, then w.u=0, \forall w\in W, \forall u \in W^{\perp}.
Similarly, x.u=0, \forall u\in W^{\perp} and \forall x\in (W^{\perp})^{\perp}.
All vectors w \in W must also be in (W^{\perp})^{\perp} since W consists of only vectors to perpendicular to those in W^{\perp}.

Is it okay to simply assert that or should you explicitly show it? One way to show equality of two sets A and B is to show that A is a subset of B and B is a subset of A, which is essentially what you seem to be doing. But if you can simply make the claim you did, can't you do the same thing with the two sets swapped and be finished?

Now we need to show that (W^{\perp})^{\perp} cannot contain anything not in W. Suppose there is a vector x' \in (W^{\perp})^{\perp} but not in W. Then x'.u=0, but it must also have some property that prevents it from being in W, namely x' would also have to be orthogonal to W (if not, then it must be in W). But this would imply it is in W^{\perp} and contradict the property W^{\perp} \cap (W^{\perp})^{\perp}=0

 

[Ogunkoya Oluwadara]

You have neglected the fact that your claim is true only when W is closed. You can check Erwin Kreyszig - Introduction to Functional analysis.. pg 149, lemma 3.3-6 for a standard proof.
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