- #1

chipotleaway

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## 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 [itex](W^{\perp})^{\perp}=W[/itex]

## 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]