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Math Help - Sum of squared dot products, seeking proof for maximisation

  1. #1
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    Sum of squared dot products, seeking proof for maximisation

    Hi, I'm not sure if this is the right topic to post this in, so please let me know if there is another place I am better off asking. But for now, I have a set of vectors \{ \mathbf{a}_i \} indexed by i, and my problem is that I am trying to find a proof that the function
    \sum_i (\mathbf{a}_i\cdot\mathbf{b})^2 is maximised when \mathbf{b}=\sum_i \mathbf{a}_i or when \mathbf{b} \parallel \sum_i \mathbf{a}_i (because I only care about vectors with a norm of 1).


    Just to clarify, I am sure that this is true, I just don't have a clean proof of it.
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  2. #2
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    Re: Sum of squared dot products, seeking proof for maximisation

    Hey jadedscholar.

    You should take a look at the cauchy-schwartz inequality for the proof, but you may also want to show that if there if you use a decomposition of a into a component linear to b and perpendicular component then you only get the maximization when the perpendicular component is zero.
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  3. #3
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    Re: Sum of squared dot products, seeking proof for maximisation

    Hi Chiro,
    Thanks for the suggestions, but I am pretty sure that neither Cauchy-Schwarz or vector decomposition are enough on their own to sort this out. I can only get those methods to help if showing that my postulate would hold for \( \sum_i \mathbf{a}_i \cdot\mathbf{b} \)^2.

    In fact, I'm not completely sure any more that it is true. I have found that it is not true when \sum_i \mathbf{a}_i is equal to the identity. I am still pretty sure that my original postulate is true in every case except for when that happens (I've learned that it's called a spherical t-design), but even if I am right about that, it will be difficult to prove.
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