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Thread: Nondegenerate inner product proof

  1. #1
    Sep 2008

    Nondegenerate inner product proof

    Take n-tuples from the field of 4 elements. Define a Hermitian inner product (u,v)H by setting (u,v)H equal to the sum of all u * v^2 in the same place (its basically the same as a regular inner product, but the second term has its entries squared). Prove that this is nondegenerate; that is, no nonzero n-tuple u has (u,v)H = 0 for all v.

    I'm not really sure how to go about this. Can I just show it is nonzero for one v? For example, let u have a nonzero element in its ith place. Then take v such that the ith entry in v is 1 and the rest are 0's. Then (u,v)H would be equal to whatever the ith entry of u is. Does that prove it? Seems too simple to be that.
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  2. #2
    MHF Contributor

    Mar 2011

    Re: Nondegenerate inner product proof

    it will suffice to exhibit a basis for V, and then show that for at least one element v of the basis (u,v)H ≠ 0.

    the natural basis that comes to mind is: {(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)}.

    let u = (u1,u2,u3,u4).

    then (u,(1,0,0,0))H = u1(1^2) + u2(0^2) + u3(0^2) + u4(0^2) = u1

    similarly, (u,(0,1,0,0))H = u2, (u,(0,0,1,0))H = u3, (u,(0,0,0,1))H = u4.

    since by assumption, u is non-zero, one of u1,u2,u3 or u4 is non-zero, so one of those 4 inner products with basis vectors for V is non-zero.

    in my opinion, it's far more interesting to show (u,v)H is conjugate-linear in v, and conjugate symmetric (which then implies (u,v)H is positive-definite), linearity in u is obvious.
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