Results 1 to 9 of 9

Math Help - Prove ||u-v|| = ||u|| - ||v||

  1. #1
    Member
    Joined
    Oct 2009
    From
    Canada
    Posts
    128

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by 450081592 View Post
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2009
    From
    Canada
    Posts
    128
    Quote Originally Posted by tonio View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by 450081592 View Post
    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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2009
    From
    Canada
    Posts
    128
    Quote Originally Posted by tonio View Post
    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||
    Last edited by 450081592; October 18th 2009 at 11:18 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by 450081592 View Post
    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
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Oct 2009
    From
    Canada
    Posts
    128
    Quote Originally Posted by tonio View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by 450081592 View Post
    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
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Oct 2009
    From
    Canada
    Posts
    128
    Quote Originally Posted by tonio View Post
    <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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove a/b and a/c then a/ (3b-7c)
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: March 23rd 2010, 05:20 PM
  2. prove,,,
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: March 1st 2010, 09:02 AM
  3. Prove |w + z| <= |w| +|z|
    Posted in the Algebra Forum
    Replies: 3
    Last Post: February 28th 2010, 05:44 AM
  4. Replies: 2
    Last Post: August 28th 2009, 02:59 AM
  5. How to prove that n^2 + n + 2 is even??
    Posted in the Algebra Forum
    Replies: 3
    Last Post: November 30th 2008, 01:24 PM

Search Tags


/mathhelpforum @mathhelpforum