$u+v$ is orthogonal to $u-v$ if and only if $\|u\| = \|v\|$ I'm doing an introductory linear algebra course and I'm stuck on this question.
Show that with respect to any inner product, $u+v$ is orthogonal to $u-v$ if and only if $\|u\| = \|v\|$.
I'm trying to prove the forward implication and I don't know where to go from
$\langle u+v,u-v \rangle=0$
I tried working with the cosine formula or with the fact that $\langle u+v,u-v \rangle = \langle u-v,u+v \rangle$ but I don't really know where I'm going...
Could someone show me how to prove both forward and backward implications?
Thanks in advance!
 A: Expand the inner product using the linearity rules (twice): $\langle a + b, c \rangle = \langle a,c\rangle + \langle b,c\rangle$. Can you take it from here?
A: You can start like this:
$\langle u+v,u-v\rangle =0$ $\Leftrightarrow$ 
$\langle u+v,u\rangle - \langle u+v,v\rangle=0$ $\Leftrightarrow$
Can you take it from there? (At some place you will have to use that $\langle u,v \rangle = \langle v,u\rangle$ and that $\langle u,u\rangle = \|u\|^2$.) 
A: If u+v is orthogonal to u-v,
(u+v).(u-v)=u.u - v.v = 0 by dot product and by definition of orthogonal,
Hence, u.u=v.v.
But u.u=||u||^2 and v.v=||v||^2
Hence, ||u||^2=||v||^2
It follows that ||u||=||v|| or ||u||=-||v|| (But it cannot be negative so we reject the latter answer).
Therefore, we have ||u||=||v||.
Now because this is if and only if, so we have to show that if p, then q. AND if q, then p.
Hence now we show that if ||u||=||v||, then u+v is orthogonal to u-v,
Applying the same concept because we are lazy people :),
||u||^2=||v||^2
||u||^2-||v||^2=0 which is the same as u.u - v.v = 0 and factorizing it will produce
(u + v)(u - v) = 0. By definition of orthogonal, u+v is orthogonal to u-v.
Proven
