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Math Help - Show Hilbert Space has a pre inner product

  1. #1
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    Show V has pre inner product

    Let V = C^n and A be in Mn(C).
    Let

    <x,y>=sum i, j, k =1 to n Akj . conjugate Aki . x(j) . conjugate y(i)

    Show this is a pre inner product on V.

    Do I start by letting say B = Akj . conjugate Aki
    Define <x, y> = sum i, j, k =1 to n Bx . conjugate y
    & find linearity of V by the linearity of B somehow?

    Thanks.
    Last edited by dazed; October 14th 2010 at 09:38 AM.
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  2. #2
    Member HappyJoe's Avatar
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    First of all, I was confused as to why the thread is called "Show Hilbert Space has a pre inner product". Certainly any Hilbert space has a pre-inner product, namely the inner product making the underlying vector space a Hilbert space in the first place.

    Second of all, your notation is not really user-friendly, but I think I've deciphered it.

    Let me show you how to prove additivity in the first variable. I want to show that \langle x+z,y\rangle = \langle x,y\rangle + \langle z,y\rangle.

    The left hand side is just

    \langle x+z,y\rangle = \displaystyle\sum_{i,j,k=1}^n A_{kj}\overline{A_{ki}}\left(x(j)+z(j)\right)\over  line{y(i)}.

    On the right hand side of this equality, for each term in this (huge) sum, you just multiply into the parenthesis, giving

    \displaystyle\sum_{i,j,k=1}^n A_{kj}\overline{A_{ki}}\left(x(j)+z(j)\right)\over  line{y(i)}=\displaystyle\sum_{i,j,k=1}^n A_{kj}\overline{A_{ki}}x(j)\overline{y(i)}+\displa  ystyle\sum_{i,j,k=1}^n A_{kj}\overline{A_{ki}}z(j)\overline{y(i)},

    and hey, this is exactly the right hand side of the equality I was trying to prove.
    Last edited by HappyJoe; October 14th 2010 at 12:23 PM. Reason: "linearity" corrected to "additivity"
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  3. #3
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    Sorry. I realised this mistake after, but did not know how to edit the title of the whole thread and I apologise for the poor notation and thank you for deciphering it!

    So to continue:

    \langle y,x\rangle = \displaystyle\sum_{i,j,k=1}^n A_{kj}\overline{A_{ki}}\left(y(i))\right)\overline  {x(j)}

    = y.transpose Akj Aki conjugate x
    = conjugate(conjugate y. conjugate (Akj Aki conjugate x))
    =conjugate <x,y>

    Sorry and thanks!
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  4. #4
    Member HappyJoe's Avatar
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    Yeah. Try comparing this to what you want to end up with, i.e. \overline{\langle x,y\rangle} . Does this suggest which steps to make to come from one to the other?
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