# Subspaces.

• April 7th 2011, 12:11 PM
lisaalg
Subspaces.
Hey again!
Sorry, I don't mean to fill up the entire forum, but as I said in my other post, I have an exam and I need all the help I can get. I figured if anyone can answer these for me while I study, bonus! Here's a few questions I could use the answers for.
Attachment 21397

I put them as pictures because they're much more clear that way than if I tried to type them.

Again, thank you to whoever is willing to help! :)
• April 8th 2011, 12:13 AM
zoek
For 1:

$x \in U \Rightarrow = 0 \Rightarrow x \in \left( U^\perp\right)^\perp$.

So we have $U\subseteq \left( U^\perp\right)^\perp^{(1)}$.

$dim_\mathbb R U = dim_\mathbb R U^\perp, dim_\mathbb R U^\perp= dim_\mathbb R \left( U^\perp\right)^\perp \Rightarrow dim_ \mathbb R U= dim_\mathbb R \left( U^\perp\right)^\perp ^{(2)}$.

$\overset{(1),(2)}{\Longrightarrow} U = \left( U {^\perp}\right)^{\perp}$.
• April 8th 2011, 12:24 AM
FernandoRevilla
For 2 :

(i)

$x\in (U+V)^{\perp}\Rightarrow \ldots$

Next step?

(ii)

$x\in U^{\perp}\cap V^{\perp}\Rightarrow \ldots$

Next step?
• April 8th 2011, 12:45 AM
Deveno
two more hints:

if <x,u> = 0, and <x,v> = 0 then certainly <x,u+v> = 0.

now can we say that if u is in U, and v is in V, that u,v are in U+V?