Suppose that u is a non-zero vector in R^n.
Prove that ||u-v|| = ||u|| - ||v||
if and only if v = ku for k is ubset of R with 0 <= k <= 1.
v and u are vectors. k is a scalar
I subbed v=ku into the right side, and I got
||u|| - k ||u|| as my last step for the right side, but I have no idea how to make it equal to the left side
I can't see how can you say that you already proved iff they're orthogonal and STILL believe the very same thing is true iff v = ku and etc.!!
If we ALREADY know that the claim is true iff v, u orthogonal, then it can NOT be true iff v is a scalar multiple of u, no matter what the scalar (different from zero is), sin u and ku are NOT orthogonal!
And indeed: take u = (1,1) and v = (1/2,1/2) in IR^2 with the standard Euclidean norm, and then u - v = (1/2, 1/2), and
||u - v|| = Sqrt(1/2) =/= Sqrt(2) - Sqrt(1/2) = ||u|| - ||v||...
I suppose u and u-v are orthogonal, we should use the condition v = ku for k subset R with 0<= k<= 1 to prove u and u-v are orthogonal, then (u-v)^2 = v^2 - u^2 which implies ||u-v|| = ||u|| - ||v||
You're right! I was confused before, and then your confussion added to mine. Let us see according to norm definition:
|| u - v ||^2 = (||u|| - ||v||)^2 <==> <u-v,u-v> = ||u||^2 - 2||u||||v|| + ||v||^2 <==> ||u||^2 + ||v||^2 - 2<u,v> <==>
||u||||v|| = <u,v> <==> cos w = <u,v>/||u||||v|| = 1 , with w = the angle between u and v <==> w = 0 <==> u,v are collinear (i.e. linearly dependent) <==> v = ku, for some scalar k , but I can't see why k must be in (0,1]...
Okay, I figured out the solution, thanks for your help, can you take a look at my other questions too, theay are in the Calculus section, are much harder, I appreciate it. http://www.mathhelpforum.com/math-he...-question.html