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Antisymmetric Eigenvalues

  1. May 16, 2006 #1
    Why do they have to purely imaginary?

    I got a proof that looks like Ax=ax
    where a = eigenvalue

    therefore Ax.x = ax.x = a|x|^2

    Ax.x = x.(A^t)x
    where A^t = transpose = -A
    x.(-A)x = -b|x|^2

    therefore a=-b, where b = conjugate of a

    Now is this as far as i need to go?
     
  2. jcsd
  3. May 16, 2006 #2

    matt grime

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    Write it as a=x+iy, b=x-iy, now you're saying what....
     
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