1. ## Inner Product Space

Let V be a real inner product space. Show that

$\displaystyle (x , y) = \frac{1}{4}(||x + y||^2 - ||x - y||^2)$

for all $\displaystyle x , y \in V$.

Do i start by proving that (x , y) = (y , x)? which i can see because of the absolute value signs (is that what the || lines are called..?) that it is.

Then is it just showing that $\displaystyle \lambda x + \mu y , z = \lambda (x,z) + \mu (y,z)$
and also
(x , x) = 0

2. $\displaystyle \left\| {x + y} \right\|^2 = \left( {x + y,x + y} \right) = (x,x) + 2(x,y) + (y,y) = \left\| x \right\|^2 + 2(x,y) + \left\| y \right\|^2$
$\displaystyle \left\| {x - y} \right\|^2 = \left( {x - y,x - y} \right) = (x,x) - 2(x,y) + (y,y) = \left\| x \right\|^2 - 2(x,y) + \left\| y \right\|^2$
Can you finish?

3. Oh yeah i see now.
Yeah can finish, ||x||'s and ||y||'s cancel leaving 4(x , y) so need a 1/4 to make it = (x , y). Thanks. Think i was just confused with what the question was asking. Been doing a lot of inner product proofs involving what i said i thought had to be done in my first post.