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Complex matrix factoring

  1. Mar 15, 2009 #1
    *The subject should have been complex matrix square root.

    I need to find one solution to the follow set of complex equations:

    a*a+b*b = 1-K-K/2
    a*b+c*d=K/2
    c*c+d*d=1-K/2

    where K is some real constant.

    This is equivalent to finding a 2x2 complex matric M such that [tex]M^\dagger M=[/tex]

    1-K-K/2, -K/2
    -K/2, 1-K/2

    I can't seem to do get started.
     
  2. jcsd
  3. Mar 15, 2009 #2

    Mark44

    Staff: Mentor

    Are you using "complex" as a synonym for "complicated?" These are not linear equations, but I don't see any evidence that they involve complex numbers.

    Is there some reason you wrote the expression on the right side of the first equation as 1 - K - K/2 instead of 1 - 3K/2?

    Also, I don't see any connection to matrices.

    Your system of three equations is in four variables, so at least one variable can be set arbitrarily. You might start by setting a to zero, and seeing what you come up with for b, c, and d.
     
  4. Mar 15, 2009 #3
    "Complex" as in complex numbers. a* is complex conjugation. I realize now that it might have been confused with multiplication.
     
  5. Mar 16, 2009 #4
    Here is as close as I can get to your equations

    [a b][a c]
    [c d][b d]

    a*a+b*b=1-k/2
    a*c+b*d=k/2
    c*a+d*b=k/2
    c*c+d*d=1-k/2

    Lets assume a lower triangle solution (AKA c=0)

    |d|=sqrt(1-K/2)
    |b|=(k/2)/sqrt(1-K/2)
    angle(d)=angle(b)
    |a|=sqrt(1-k/2-(k^2/4)/(1-K/2))

    Now I think that all solutions can be found by multiplying a particular solution by all possible orthonormal transforms.
     
  6. Mar 16, 2009 #5
    What's an "orthonormal transform"? Do you mean unitary matrices?
     
  7. Mar 17, 2009 #6
    Yes. I guess I should say unitary matrix as according to Wikipedia it is more general.
     
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