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Math Help - Norm of sum of vectors

  1. #1
    Senior Member Pinkk's Avatar
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    Norm of sum of vectors

    Given vectors x_{1}, x_{2}, ..., x_{k}\in \mathbb{R}^{n}, what is |x_{1} + x_{2} + ... + x_{k}|^{2}?

    So |x_{1} + x_{2}|^{2} = |x_{1}|^{2} + |x_{2}|^{2} + 2x_{1}\cdot x_{2} and extending it to the sum of three vectors it seems to be |x_{1}|^{2} + |x_{2}|^{2} + |x_{3}|^{2} + 2( x_{1}\cdot x_{2} + x_{1}\cdot x_{3} + x_{2}\cdot x_{3}). So it seems to be that for the norm of k vectors we have |x_{1} + x_{2} + ... + x_{k}|^{2} = |x_{1}|^{2} + |x_{2}|^{2} + ... + |x_{k}|^{2} + 2(\sum x_{i}x_{j}) for i\ne j, 1\le i, j \le k. Is this correct and is there a better way to formulate this? And how would I go about giving a formal proof for this problem? Thanks.

    Edit: I think I found a more simplified formulation: \sum_{j=1}^{k}\sum_{i=1}^{k} x_{i}\cdot x_{j}, but how would I go about giving a formal proof? Thanks.
    Last edited by Pinkk; August 30th 2010 at 06:28 PM.
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  2. #2
    A Plied Mathematician
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    The norm of a vector can always be written as follows:

    \|x\|=\sqrt{\langle x|x\rangle},

    where \langle x|y\rangle is the inner product of vector x with vector y. The inner product is just a generalization of the dot product. Therefore,

    \|x\|^{2}=\langle x|x\rangle. It follows that

    \|x+y\|^{2}=\langle (x+y)|(x+y)\rangle=\langle x|x\rangle+\langle x|y\rangle+\langle y|x\rangle+\langle y|y\rangle.

    I think this line of reasoning should enable you to see how to do your proof. How would you start?
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