Finding a Complex Matrix Square Root for a Set of Equations

[Dragonfall]

*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 M^\dagger M=

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

I can't seem to do get started.

 

[Mark44]

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.

 

[Dragonfall]

"Complex" as in complex numbers. a* is complex conjugation. I realize now that it might have been confused with multiplication.

 

[John Creighto]

*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

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.

 

[Dragonfall]

What's an "orthonormal transform"? Do you mean unitary matrices?

 

[John Creighto]

What's an "orthonormal transform"? Do you mean unitary matrices?

Yes. I guess I should say unitary matrix as according to Wikipedia it is more general.