# Proof of intersection and sum of vector spaces

• Jan 12th 2009, 11:38 AM
lukaszh
Proof of intersection and sum of vector spaces
Hi,
how to prove this:
$\displaystyle V^{\bot}\cap W^{\bot}=(V+W)^{\bot}$
• Jan 12th 2009, 11:58 AM
NonCommAlg
Quote:

Originally Posted by lukaszh
Hi,
how to prove this:
$\displaystyle V^{\bot}\cap W^{\bot}=(V+W)^{\bot}$

if $\displaystyle x \in V^{\bot}\cap W^{\bot},$ and $\displaystyle v \in V, \ w \in W,$ then $\displaystyle <x,v+w>=<x,v>+<x,w>=0+0=0.$ thus $\displaystyle V^{\bot}\cap W^{\bot} \subseteq (V+W)^{\bot}.$

conversely, we have $\displaystyle V \subseteq V+W,$ and hence $\displaystyle (V+W)^{\bot} \subseteq V^{\bot}.$ similarly $\displaystyle (V+W)^{\bot} \subseteq W^{\bot}.$ thus $\displaystyle (V+W)^{\bot} \subseteq V^{\bot}\cap W^{\bot}.$
• Jan 12th 2009, 01:25 PM
lukaszh
I dont understand this:
Quote:

Originally Posted by NonCommAlg
... then $\displaystyle <x,v+w>=<x,v>+<x,w>=0+0=0.$ thus $\displaystyle V^{\bot}\cap W^{\bot} \subseteq (V+W)^{\bot}....$

How this $\displaystyle V^{\bot}\cap W^{\bot} \subseteq (V+W)^{\bot}$ emerges of $\displaystyle <x,v+w>=<x,v>+<x,w>=0+0=0$
Thanks(Worried)
• Jan 12th 2009, 02:35 PM
NonCommAlg
Quote:

Originally Posted by lukaszh
I dont understand this:

How this $\displaystyle V^{\bot}\cap W^{\bot} \subseteq (V+W)^{\bot}$ emerges of $\displaystyle <x,v+w>=<x,v>+<x,w>=0+0=0$
Thanks(Worried)

in order to prove that $\displaystyle V^{\bot}\cap W^{\bot} \subseteq (V+W)^{\bot},$ you need to show that if $\displaystyle x \in V^{\bot}\cap W^{\bot},$ then $\displaystyle x \in (V+W)^{\bot}.$ to prove that $\displaystyle x \in (V+W)^{\bot},$ you need to show that $\displaystyle x$ is orthogonal to every

element of $\displaystyle V+W,$ i.e. $\displaystyle <x,z>=0, \ \forall z \in V+W.$ so you choose an element of $\displaystyle z \in V+W.$ then $\displaystyle z=v+w,$ for some $\displaystyle v \in V, w \in W.$ we have $\displaystyle <x,z>=<x,v>+<x,w>=0,$ because

$\displaystyle x \in V^{\bot}\cap W^{\bot}.$
• Jan 13th 2009, 01:14 AM
lukaszh
Thank you, it's simple :-)