If V is an inner product space, prove that for all x,y in V

Question

1/4 { $||x + y||^2$ - $||x - y||^2$} = $<x,y>$

Ok at first i really had no idea how to start this but i spend about 5 minutes trying to do this and this is what i came up with..

1) i multiplied both sides by 4 to get rid of the fraction..

$||x + y||^2$ - $||x - y||^2$ = 4$<x,y>$

2) I know that $<x,y>$ is just the dot product and will give a scalar, which we define as $\lambda$

$||x + y||^2$ - $||x - y||^2$ = 4 $\lambda$

3) the magnitude of x+y and x-y will give the squareroot of a scalar, the squareroot will cancel itself with the exponent, we will represent the result of this with n and m

n - m = 4 $\lambda$

this is as far as i can go by myself. If i tried going any further, i will only stare at my paper for hours

any help will be appreciated

Answer 1

Use this and FOIL.

$$\|x + y\|^2 = \langle x + y, x + y\rangle.$$

Answer 2

HINT

$$||x + y||^2=<x+y,x+y>=||x||^2+2<x,y>+||y||^2$$

$$||x - y||^2=<x-y,x-y>=|x||^2-2<x,y>+||y||^2$$