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Inner product of complex vectors

  1. Dec 18, 2012 #1
    I have three (N x 1) complex vectors, a, b and c.

    I know the following conditions:

    (1) a and b are orthonormal (but length of c is unknown)
    (2) c lies in the same 2D plane as a and b
    (3) aHc = x (purely real, known)
    (4) bHc = iy (purely imaginary, unknown)

    where (.)H denotes Hermitian (conjugate) transpose, i is the imaginary unit and x,y are real numbers.

    Given that I know x, can I deduce y?

    My hunch is that (without the "purely real/imaginary" statements), these conditions would define y up to an arbitrary complex phase, but the "purely real/imaginary" conditions allow the phase to be known too. However, my reasoning relies on there being some sense of "angle" between a and c and between b and c... such that these angles sum to 90° for the orthonormality condition (1). I don't know if this is valid.
     
  2. jcsd
  3. Dec 18, 2012 #2

    tiny-tim

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    hi weetabixharry! :smile:
    doesn't that mean that c must be a linear combination of a and b ?
     
  4. Dec 18, 2012 #3
    Yes, where I guess the coefficients of the linear combination are complex scalars.
     
  5. Dec 18, 2012 #4

    haruspex

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    From which it follows that y=0?
     
  6. Dec 19, 2012 #5
    Why?
     
  7. Dec 19, 2012 #6
    c is a linear combination of a and b:

    c = Aa + Bb

    for A,B complex scalars.

    Therefore, from (3) and (1), A = x
    and, from (4) and (1), B = iy

    I can't see why y=0

    I guess, from this I have:

    c = xa + iyb

    which is 1 equation in 2 unknowns (y and c)... so I'm stumped.
     
  8. Dec 19, 2012 #7

    haruspex

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    You know facts about aHc, bHc. How can you combine that with with knowing c = Aa + Bb? Actually I was wrong to suggest y=0, but you can at least make progress this way.
     
  9. Dec 19, 2012 #8
    I combined these in my previous post to write A,B as functions of x,y.
    Beyond that, I guess I can say:

    |c|2 = x2 - y2
     
  10. Dec 19, 2012 #9

    haruspex

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    Suppose you found a y and c =xa+iyb which satisfied all the conditions. Wouldn't 2y and c' =xa+i2yb also satisfy them?
     
  11. Dec 19, 2012 #10

    tiny-tim

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    hi weetabixharry! :smile:

    (just got up :zzz:)
    that's right! :smile:
    so what's the answer to :wink: … ?
     
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