Prove that $||u+v||^2 -||u-v||^2 = 4(u\cdot v)$

Question

Prove that $\|u+v\|^2 -\|u-v\|^2 = 4(u\cdot v)$
where u and v are vectors in R^n

Edit: The subtraction part basically the second half, previously the math was incorrect.

My answer: Write u = ($u_1...u_n$) and v = ($v_1...v_n$)

$\|u +v\|^2 - \|u-v\|^2= (u_1+v_1)^2+...+(u_n+v_n)^2-(u_1-v_1)^2+...(u_n-v_n)^2 =u_1^2+2u_1v_1+v_1^2+...+u_n^2+2u_nv_n+v_n^2-u_1^2-2u_1v_1+v_1^2+u_n^2-2u_nv_n+v_n^2$

How do I get the above to become $4u_1+4v_1+...+4u_n+4v_n$?

Answer 1

You can also do it without using the seperate coordinates, by using $||x||=\sqrt{x\cdot x}$. Can you that yourself?

Answer 2

Hint. $|u|^2=u\cdot u$, so $$\|u+v\|^2 -\|u-v\|^2 = (u+v)(u+v)-(u-v)(u-v)=\cdots$$