Let V be a real inner product space and let u v sum V such t

Let V be a real inner product space, and let u, v sum V such that ||u|| = ||v|| = (u, v) = 1. Show that u = v.

Solution

Let u = (u₁,u₂,…,u_n) and v = (v₁,v₂,…,v_n). Then ||u||= (u₁² +u₂²+…+u_n²) and ||v||=   (v₁² +v₂²+…+v_n²).       Since ||u||= ||v|| = 1, we have (u₁² +u₂²+…+u_n²)=(v₁² +v₂²+…+v_n²)= 1.   Also <u, v> = u₁v₁+u₂v₂+…+u_nv_n =1. Then ||u-v|| = [(u₁-v₁)²+(u₂-v₂)²+…+(u_n-v_n)² = (u₁² +u₂²+…+u_n²)+(v₁² +v₂²+…+v_n²)-2(u₁v₁+u₂v₂+…+u_nv_n)]^1/2 = (1+1-2)^1/2 = 0. Hence u = v.