1. Orthogonal complement subspaces

Prove that:
($\displaystyle \bot$= orthogonal complement)

$\displaystyle (U \cap W)^\bot = U^\bot + W^\bot$

Attempt at solution:
$\displaystyle \rightarrow$
Let $\displaystyle x \in (U \cap W)^\bot$and $\displaystyle y \in U + W$
Then $\displaystyle <x,y> = <x,u> + <x,w> = 0 + 0$
Therefore $\displaystyle x \in U^\bot + W^\bot$

$\displaystyle \leftarrow$
Let $\displaystyle x \in U^\bot + W^\bot$ and $\displaystyle y \in U \cap W \Rightarrow y \in U$and $\displaystyle y \in W$
Then $\displaystyle <x,y> = <x,u> \cap <x,w> = 0 \cap 0 = 0$
Therfore, $\displaystyle x \in (U \cap W)^\bot$

Prove that:
($\displaystyle \bot$= orthogonal complement)

$\displaystyle (U \cap W)^\bot = U^\bot + W^\bot$

Attempt at solution:
$\displaystyle \rightarrow$
Let $\displaystyle x \in (U \cap W)^\bot$and $\displaystyle y \in U + W$
Then $\displaystyle <x,y> = <x,u> + <x,w> = 0 + 0$

Make sure here to write $\displaystyle y=u+w\,,\,u\in U\,,\,w\in W$ , and remark you're using additivity of the inner product

Therefore $\displaystyle x \in U^\bot + W^\bot$

As above, write $\displaystyle x=u'+w'\,,\,u'\in U^\perp\,,\,w'\in W^\perp$ , and don't write u,w

below: it's enough you already remarked that $\displaystyle y\in U\cap W$

Tonio

$\displaystyle \leftarrow$
Let $\displaystyle x \in U^\bot + W^\bot$ and $\displaystyle y \in U \cap W \Rightarrow y \in U$and $\displaystyle y \in W$
Then $\displaystyle <x,y> = <x,u> \cap <x,w> = 0 \cap 0 = 0$
Therfore, $\displaystyle x \in (U \cap W)^\bot$

.

Prove that:
($\displaystyle \bot$= orthogonal complement)

$\displaystyle (U \cap W)^\bot = U^\bot + W^\bot$

Attempt at solution:
$\displaystyle \rightarrow$
Let $\displaystyle x \in (U \cap W)^\bot$and $\displaystyle y \in U + W$
Then $\displaystyle <x,y> = <x,u> + <x,w> = 0 + 0$
Therefore $\displaystyle x \in U^\bot + W^\bot$
Not valid. The fact that two numbers add to 0 doesn't mean the two numbers must each be 0.

$\displaystyle \leftarrow$
Let $\displaystyle x \in U^\bot + W^\bot$ and $\displaystyle y \in U \cap W \Rightarrow y \in U$and $\displaystyle y \in W$
Then $\displaystyle <x,y> = <x,u> \cap <x,w> = 0 \cap 0 = 0$
Therfore, $\displaystyle x \in (U \cap W)^\bot$

4. Originally Posted by HallsofIvy
Not valid. The fact that two numbers add to 0 doesn't mean the two numbers must each be 0.
So if i said what tonio suggested, $\displaystyle \Rightarrow y = u + w, u \in U, w \in W$, will it be okay?

Prove that:
(= orthogonal complement)

Attempt at solution:

Let and $\displaystyle \Rightarrow y = u + w, u \in U, w \in W$
Then using addivity of inner product

Therefore

Let $\displaystyle \Rightarrow x = u' + w', u' \in U^\perp, w \in W^\perp$ and $\displaystyle y \in U \cap W$
Then $\displaystyle <x,y> = <u',y> \cap <w',y> = 0$
Therfore,