# Thread: Inner Product Space

1. ## Inner Product Space

Could someone explain to me how to go from (1) to (2)?

$||x - \langle x,y\rangle y||^2 = \langle x - \langle x,y\rangle y,x - \langle x,y\rangle y\rangle$

$= \langle x,x\rangle - \langle x,\langle x,y\rangle y\rangle - \langle \langle x,y\rangle y,x\rangle + \langle \langle x,y\rangle y,\langle x,y\rangle y\rangle$ (1)

$||x||^2 - \overline{\langle x,y\rangle}\langle x,y\rangle - \langle x,y\rangle\langle y,x\rangle + \langle x,y\rangle\overline{\langle x,y\rangle}$ (2)

$||x||^2 - |\langle x,y\rangle|^2$.

(Just before this it states assume $||y|| = 1$)

2. Hello there!

Originally Posted by Deadstar
Could someone explain to me how to go from (1) to (2)?

$||x - \langle x,y\rangle y||^2 = \langle x - \langle x,y\rangle y,x - \langle x,y\rangle y\rangle$

$= \langle x,x\rangle - \langle x,\langle x,y\rangle y\rangle - \langle \langle x,y\rangle y,x\rangle + \langle \langle x,y\rangle y,\langle x,y\rangle y\rangle$ (1)

$||x||^2 - \overline{\langle x,y\rangle}\langle x,y\rangle - \langle x,y\rangle\langle y,x\rangle + \langle x,y\rangle\overline{\langle x,y\rangle}$ (2)

$||x||^2 - |\langle x,y\rangle|^2$.

(Just before this it states assume $||y|| = 1$)
There are some rules by definition of the dot product,

like <x,x> = ||x||^2

By definition it is

$ = \overline{\lambda}$, but

$<\lambda x, y> = \lambda,$

So let $\lambda := $

$y > = \overline{}$

And that is why $<y,x > = $

And the last one

$<y,y> = \overline{}$

$= \overline{}||y||^2$

and $||y||^2 = 1 = ||y||$ so

$= \overline{}$

Yours
Rapha