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Math Help - ||x-y||+||y-z||=||x-z|| iff y= ax + (1-a)z

  1. #1
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    Question ||x-y||+||y-z||=||x-z|| iff y= ax + (1-a)z

    Dear mathematician,

    I need to show that:

    ||x-y||+||y-z||=||x-z||

    if and only if

    y= ax + (1-a)z with 0<=a<=1

    In other words, if x,y and z are on the same line, that relation between the norms must be satisfied and the other way around.

    I didn't come really far. Writing out the inproducts didn't give me any insight.

    Can someone help me please? Would be great!

    Cheers, Eva
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by Eva BSc View Post
    Dear mathematician,

    I need to show that:

    ||x-y||+||y-z||=||x-z||

    if and only if

    y= ax + (1-a)z with 0<=a<=1

    In other words, if x,y and z are on the same line, that relation between the norms must be satisfied and the other way around.

    I didn't come really far. Writing out the inproducts didn't give me any insight.

    Can someone help me please? Would be great!

    Cheers, Eva
    You have
    \parallel x-y\parallel+\parallel y-z\parallel = \parallel x-z\parallel
    Now square both sides:
    (\parallel x-y\parallel+\parallel y-z\parallel)^2 = \parallel x-z\parallel^2
    Then bring x-y and y-z into play on the right side, like this
    \parallel x-y\parallel^2+2\parallel x-y\parallel \parallel y-z\parallel +\parallel y-z\parallel^2=\parallel (x-y)+(y-z)\parallel^2
    Then multiply out (using the inner product on the right)
    \parallel x-y\parallel^2+2\parallel x-y\parallel \parallel y-z\parallel = \parallel x-y\parallel^2+2<x-y,y-z>+\parallel y-z\parallel^2
    Thus it follows:
    \parallel x-y\parallel \parallel y-z\parallel = <x-y,y-z><br />
    (I am assuming a vector space over \mathbb{R} here.) Now consider this last equation: it should remind you of the Cauchy-Schwarz inequality. So you are asked to prove under what condition equality holds in the Cauchy-Schwarz inequality. Quite possibly, you can simply refer to your lecture notes on the Cauchy-Schwarz inequality.
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  3. #3
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    Hi Failure,

    Thanks, that really helped me!

    From Cauchy-Schwartz it follows that (x-y) and (y-z) are lineary dependent.

    So (x-y)= b(y-z) with a real number b.

    Now why does this imply that

    y= ax + (1-a)z with 0<=a<=1 ???

    In R2 it is obvious, but how to show this for Rn ??

    Cheers, Eva
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  4. #4
    Super Member Failure's Avatar
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    Quote Originally Posted by Eva BSc View Post
    Hi Failure,

    Thanks, that really helped me!

    From Cauchy-Schwartz it follows that (x-y) and (y-z) are lineary dependent.

    So (x-y)= b(y-z) with a real number b.

    Now why does this imply that

    y= ax + (1-a)z with 0<=a<=1 ???

    In R2 it is obvious, but how to show this for Rn ??

    Cheers, Eva
    Linear dependence of x-y and y-z is not equivalent to y= ax + (1-a)z with 0<=a<=1. But you know more than that x-y and y-z are linearly dependent: in addition, you now that <x-y,y-z> \geq 0, because the right side of my last equation was not the absolute value of that inner product but the value of the inner product itself.
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  5. #5
    Super Member Failure's Avatar
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    Quote Originally Posted by Eva BSc View Post
    Hi Failure,

    Thanks, that really helped me!

    From Cauchy-Schwartz it follows that (x-y) and (y-z) are lineary dependent.

    So (x-y)= b(y-z) with a real number b.
    Actually, that is not quite true. Linear dependence of x-y and y-z means that there exists a b such that x-y=b(y-z) or that b(x-y)=y-z.

    Now why does this imply that

    y= ax + (1-a)z with 0<=a<=1 ???

    In R2 it is obvious, but how to show this for Rn ??

    Cheers, Eva
    Let me add to my first reply to this question: from <x-y,y-z>\geq 0, together with the two linear dependence cases, you can further infer that b\geq 0.
    Then you solve the two cases, x-y=b(y-z) or b(x-y)=y-z, for y, which gives y=\tfrac{1}{b+1}x+\tfrac{b}{b+1}z or y=\tfrac{b}{b+1}x+\tfrac{1}{b+1}z. Finally you set a := \tfrac{1}{b+1} and a := \tfrac{b}{b+1}, respectively.

    And these two cases are indeed equivalent to there existing an a, 0 <= a <= 1, such that y=ax+(1-a)z.
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  6. #6
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    Thnx your awesome!!

    Looking at it I could have figured this out myself.. Anyhow, thnx for helping me out!
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