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Math Help - Inequation with unit vectors

  1. #1
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    Inequation with unit vectors

    Hi! I want to show that sum(norm(x_i-x_j), 1<=i<j<=n) <= n^2 for any vector space over R or C, where x_i are unit vectors.

    I haven't got any idea how to show this. Can somebody please help me?

    Banach
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  2. #2
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    There are {n\choose 2} ways to select two indices i<j. So
    \sum_{i<j}||x_i-x_j||\leq \sum_{i<j}||x_i||+||x_j||=2{n\choose 2}=n(n-1)<n^2
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  3. #3
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    Hi! Thanks for your answer. I realized that i forgot to square the norm, so it should be shown: \sum_{i<j}||x_i-x_j||^2 \leq n^2 and thus your idea does not work, since one has 4*{n \choose 2}? How can this be solved?
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  4. #4
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    Quote Originally Posted by Banach View Post
    Hi! I want to show that \sum_{1\leqslant i<j\leqslant n}||x_i-x_j||^2 \leq n^2 for any vector space over R or C, where x_i are unit vectors.
    This is true if you are using a euclidean or Hilbert space norm, but it need not hold for other norms. For example, in the space \mathbb{R}^2 with the norm \|(x,y)\| = |x|+|y|, the four vectors (\pm1,0),\ (0\pm1) are all distance 2 from each other, so the sum of the squares of the six distances is 24 which is greater than 16.

    Here's how to prove the result for a real vector space where the norm comes from an inner product \langle\,,\,\rangle.

    \sum_{1\leqslant i<j\leqslant n}||x_i-x_j||^2 = \tfrac12\sum_{i,j=1}^n\langle x_i-x_j,x_i-x_j\rangle

    . . . . . . . . . . . . . . = \sum_{i=1}^n\langle x_i,x_i\rangle - \sum_{i,j=1\atop i\ne j}^n\langle x_i,x_j\rangle

    . . . . . . . . . . . .. . \leqslant \sum_{i=1}^n\langle x_i,x_i\rangle + \sum_{i,j=1\atop i\ne j}^n\langle x_i,x_j\rangle

    . . . . . . . . . . . .. . = \sum_{i,j=1}^n\langle x_i,x_j\rangle = \Bigl\|\sum_{i=1}^nx_i\Bigr\|^2 \leqslant n^2

    The proof for complex spaces is more or less the same, with complex conjugates inserted where appropriate.
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  5. #5
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    Hi, thanks for your answer. You are right, i forgot the prerequisite euclidian space.

    Why does the inequation \leqslant \sum_{i=1}^n\langle x_i,x_i\rangle + \sum_{i,j=1\atop i\ne j}^n\langle x_i,x_j\rangle<br />
hold? If \sum_{i,j=1\atop i\ne j}^n\langle x_i,x_j\rangle<br />
would be positive then it would be clear. But this is not the case. Can you explain this step to me please?

    Greetings
    Banach
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  6. #6
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    \sum_{1\leqslant i<j\leqslant n}||x_i-x_j||^2 = \tfrac12\sum_{i,j=1}^n\langle x_i-x_j,x_i-x_j\rangle

    . . . . . . . . . . . . . . = \sum_{i=1}^n\langle x_i,x_i\rangle - \sum_{i,j=1\atop i\ne j}^n\langle x_i,x_j\rangle

    . . . . . . . . . . . .. . =n-2\sum_{i<j}\langle x_i,x_j\rangle

    . . . . . . . . . . . .. . \leq n+2\sum_{i<j}||x_i||\cdot ||x_j|| by Cauchy-Schwarz
    . . . . . . . . . . . .. . n+2{n\choose 2}=n^2
    Last edited by watchmath; October 9th 2008 at 06:23 AM.
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  7. #7
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    Quote Originally Posted by Banach View Post
    Hi, thanks for your answer. You are right, i forgot the prerequisite euclidian space.

    Why does the inequation \leqslant \sum_{i=1}^n\langle x_i,x_i\rangle + \sum_{i,j=1\atop i\ne j}^n\langle x_i,x_j\rangle<br />
hold? If \sum_{i,j=1\atop i\ne j}^n\langle x_i,x_j\rangle<br />
would be positive then it would be clear. But this is not the case.
    You're right, that was a careless mistake. See watchmath's correction above.
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  8. #8
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    Alright, thank you both very much. Now everything is clear.

    Another proof came to my mind:
    One can show that \sum_{1\leqslant i<j \leqslant n}||x_i - x_j||^2+||\sum_{i=1}^n x_i||^2=n*(\sum_{i=1}^n ||x_i||^2). Therefore the claim follows immediately.

    Good night!

    Banach
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