# Prove the W is the orthogonal complement of its orthogonal complement

1. Feb 14, 2014

### chipotleaway

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 $(W^{\perp})^{\perp}=W$

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 $W^{\perp}$ 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 $W^{\perp} \cap (W^{\perp})^{\perp}=0$

2. Feb 14, 2014

### vela

Staff Emeritus
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?

3. Aug 25, 2016