# Thread: Prove ||u-v|| = ||u|| - ||v||

1. ## Prove ||u-v|| = ||u|| - ||v||

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

2. Originally Posted by 450081592
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

The above isn't true: ||u - v|| = ||u|| - ||v|| iff u is orthogonal to v.

Tonio

3. Originally Posted by tonio
The above isn't true: ||u - v|| = ||u|| - ||v|| iff u is orthogonal to v.

Tonio
Yes, I have already proven ||u - v|| = ||u|| - ||v|| iff u is orthogonal to v.
But this question is asking to prove if and only if v = ku for k is subset to R with 0<= k <= 1.

4. Originally Posted by 450081592
Yes, I have already proven ||u - v|| = ||u|| - ||v|| iff u is orthogonal to v.
But this question is asking to prove if and only if v = ku for k is subset to R with 0<= k <= 1.

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||...

Tonio

5. Originally Posted by tonio
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||...

Tonio
sorry, I made a mistake, it's kind of confused now, I have proven ||u+v|| = ||u-v|| iff u and v are orthogonal, not ||u-v|| = ||u|| - ||v||, and I do not think ||u-v|| = ||u|| - ||v|| is true when u and u-v are orthogonal, not u and v are orthogonal. Also, iff v = ku for k subset R with 0<= k<= 1 must be true.

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||

6. Originally Posted by 450081592
sorry, I made a mistake, it's kind of confused now, I have proven ||u+v|| = ||u-v|| iff u and v are orthogonal, not ||u-v|| = ||u|| - ||v||, and I do not think ||u-v|| = ||u|| - ||v|| is true when u and v are orthogonal. Also, iff v = ku for k subset R with 0<= k<= 1 must be true.

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]...

Tonio

7. Originally Posted by tonio
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]...

Tonio
what does <u-v,u-v> mean? And how do you know || u - v ||^2 = (||u|| - ||v||)^2?

8. Originally Posted by 450081592
what does <u-v,u-v> mean? And how do you know || u - v ||^2 = (||u|| - ||v||)^2?

<u-v,u-v> is the inner (scalar) product of u-v with itself. It sometimes is denoted by (u-v).(u-v), as well.

Next, a = b <==> a^2 = b^2 if we know both a,b are non-negative, and we can always assume ||u| >= ||v||...

Tonio

9. Originally Posted by tonio
<u-v,u-v> is the inner (scalar) product of u-v with itself. It sometimes is denoted by (u-v).(u-v), as well.

Next, a = b <==> a^2 = b^2 if we know both a,b are non-negative, and we can always assume ||u| >= ||v||...

Tonio

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

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