Pythagorean Law for Inner Product Spaces

Question

The Pythagorean Law: if u and v are orthogonal vectors in an inner product space $V$, then

$||$u$+$v$||^2 = ||$u$||^2+||$v$||^2$

Proof:

1. $||$u$+$v$||^2 = <$u$+$v$, $u$+$v$>$

2. $||$u$+$v$||^2 = <$u$, $u$> + 2<$u$,$v$>+<$v$, $v$>$

3. $||$u$+$v$||^2 = ||$u$||^2+||$v$||^2$

I believe part 1 is obtained by: $||$u$+$v$||^2 = (\sqrt{<u+v, u+v>})^2 = {<u+v, u+v>}$

What happens from part 1 to part 2?

Answer 1

Linearity happens!

Remember from the definition of an inner product that $\langle u+v,w \rangle = \langle u,w \rangle + \langle v,w \rangle$ and $\langle u,v+w \rangle = \langle u,v \rangle + \langle u,w \rangle$. Applying this fact gives you the expression required.

Answer 2

You are just using the fact that the internal product is bilinear.

In other words:

$<a+b,c> = <a,c> + <b,c>$

Etc.