1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Prove the W is the orthogonal complement of its orthogonal complement

  1. Feb 14, 2014 #1
    1. The problem statement, all variables and given/known data
    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 [itex](W^{\perp})^{\perp}=W[/itex]

    3. 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 [itex]W^{\perp}[/itex] is the orthogonal complement of [itex]W[/itex], then [itex]w.u=0, \forall w\in W, \forall u \in W^{\perp}[/itex].
    Similarly, [itex]x.u=0, \forall u\in W^{\perp}[/itex] and [itex] \forall x\in (W^{\perp})^{\perp}[/itex].
    All vectors [itex]w \in W[/itex] must also be in [itex](W^{\perp})^{\perp}[/itex] since [itex]W[/itex] consists of only vectors to perpendicular to those in [itex]W^{\perp}[/itex]. Now we need to show that [itex](W^{\perp})^{\perp}[/itex] cannot contain anything not in [itex]W[/itex]. Suppose there is a vector [itex]x' \in (W^{\perp})^{\perp}[/itex] but not in [itex]W[/itex]. Then [itex]x'.u=0[/itex], but it must also have some property that prevents it from being in [itex]W[/itex], namely [itex]x'[/itex] would also have to be orthogonal to [itex]W[/itex] (if not, then it must be in [itex]W[/itex]). But this would imply it is in [itex]W^{\perp}[/itex] and contradict the property [itex]W^{\perp} \cap (W^{\perp})^{\perp}=0[/itex]
  2. jcsd
  3. Feb 14, 2014 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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?

  4. Aug 25, 2016 #3
    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.
    <Mod note: deleted personal information -- email address>
    Last edited by a moderator: Aug 25, 2016
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted