Orthogonal complement propertiss

Question

I'm required to prove that $W^{\perp} + U^{\perp} \subseteq (W \cap U)^{\perp}$.
I've already proven $U \subseteq W \to W^{\perp} \subseteq U^{\perp}$ and $(U+W)^{\perp} = W^{\perp} \cap U^{\perp}$.
Plus, I would appreciate an explaination about the difference between $u \in W+U$ and $u \in W \cup U$, I think I might mixing them up in my head :) Thanks.

Answer 1

$x\in W+U$ means there exist $w\in W$ and $u\in U$ such that $x=w+u$. (Note that the pair $(w,u)$ doesn't have to be unique here.) On the other hand, $x\in W\cup U$ means that $x\in W$ or $x\in U$ (or both).

This problem is an excellent example of one where you should do the "obvious" expansions of definitions before trying to think hard. (Tim Gowers has some excellent blog posts about this.) Of course it takes some practice to learn what the "obvious" steps are; so let's practice on this one.

Statement to be proved: $$ W^\perp + U^\perp \subseteq (W\cap U)^\perp $$ By the definition of $\subseteq$, this is the same statement as $$ \text{if } x \in W^\perp + U^\perp \text{, then } x \in (W\cap U)^\perp $$ By the definition of a sum of subspaces (as given above), this is equivalent to $$ \text{if there exist } y \in W^\perp \text{ and } z\in U^\perp \text{ such that } x=y+z \text{, then } x \in (W\cap U)^\perp $$ Finally, by the definition of (subspace)${}^\perp$, we see that an equivalent staement is $$ \text{if there exist } y \in W^\perp \text{ and } z\in U^\perp \text{ such that } x=y+z \text{, then } x \perp v \text{ for every } v \in W\cap U. $$ (Here I'm using $x\perp v$ to mean $\langle x,v\rangle=0$.)

All we've done so far is expand the definitions of the various pieces of notation. Now it's time to think: can we prove this last statement? I'll leave this to you, after noting that the variable $x$ is kind of redundant: we could just as well prove $$ \text{if } y \in W^\perp \text{ and } z\in U^\perp \text{, then } (y+z) \perp v \text{ for every } v \in W\cap U. $$

Answer 2

First note that what you've proven already: $$U \subseteq W \Longrightarrow W^{\perp} \subseteq U^{\perp}$$ Can be turned into: $$\quad\quad U\subseteq W \iff W^{\perp} \subseteq U^{\perp} \quad\text{(1)}$$ Once you prove that $$\quad\quad\quad\quad\,\,\,\,\,(L^{\perp})^{\perp} = L \quad\quad\quad\quad \text{(2)} $$ But I'll leave that to you. Now, starting with the statement you want to prove: $$W^{\perp} + U^{\perp} \subseteq (W \cap U)^{\perp}\stackrel{\text{(1)}}{\iff} (\,\,(W \cap U)^{\perp})^{\perp}\subseteq (W^{\perp} + U^{\perp})^{\perp}\stackrel{\text{(2)}}{\iff}W\cap U\subseteq W\cap U $$

which is true, so the original statement is also true.

Note: I've also used your formula for the orthogonal complement of the sum.

Answer 3

$W \cup U$ may not be a subspace, while $W + U$ is a subspace. Actually it is the smallest subspace which contain $W \cup U$.

$x \in W^{\perp} + U^{\perp} \Rightarrow x = w' + u'$, where $w' \in W^{\perp}$ and $u' \in U ^{\perp}$. This means that for every $w \in W$ and $u \in U$ we have $\langle w,w'\rangle = 0$ and $\langle u, u' \rangle = 0$. Let $y \in W \cap U$, then $\langle y, w' \rangle = 0$ and $\langle y, u' \rangle = 0$. So $\langle y, x \rangle = \langle y, w' + u' \rangle = \langle y, w' \rangle + \langle y, u' \rangle = 0$ (thanks to the distributivity law), which proves that $x \in (W \cap U)^{\perp}$.