"J " <philNOSPAMsonj@spammenot.hotmailDcom> wrote in message <hu9d3m$fqq$1@fred.mathworks.com>...
> I'm well aware of eigs() for standard eigenvalue problems, but is there a function (or a way to use eigs() maybe?) to solve the following eigenvalue problem:
>
> A*x = c*conj(x)
>
> ?
>
> Here, A is a 2x2 matrix with complex entires, x is a 2x1 complex "pseudo eigen vector", and c is a constant eigenvalue. Note the only difference from the standard eigenvalue problem is that conj(x) appears on the RHS instead of x.
        
I'm afraid your original question in this thread five days ago was never completely answered. Bruno showed how to find certain of the possible solutions. I would now like to complete that discussion.
To begin with, it must be conceded that not all 2 x 2 complex matrices possess solutions of the kind you seek. Of those that do, there is an infinite continuum of possible pseudo eigenvalues, and for each one there is a corresponding eigenvector.
Let the original complex matrix be:
M = [a11+i*b11,a12+i*b12;a21+i*b21,a22+i*b22];
The solutions you seek are [x1+i*y1;x2+i*y2] which satisfy
M*[x1+i*y1;x2+i*y2] = (d+i*e)*[x1i*y1;x2i*y2]
where d+i*e is a complexvalued pseudo eigenvalue. It is to be understood that the a's, b's, x's, y's, d's, and e's here are all real values.
If the real and imaginary parts are separated, this gives rise to the four homogeneous linear equations in the four unknowns x1, x2, y1, y2:
(a11d)*x1+a12*x2+(b11e)*y1b12*y2 = 0
a21*x1+(a22d)*x2b21*y1+(b22e)*y2 = 0
(b11+e)*x1b12*x2+(a11d)*y1a12*y2 = 0
b21*x1+(b22+e)*x2a21*y1+(a22d)*y2 = 0
If the determinant of these coefficients were to be nonzero, there could be no solutions except all zeros for the x's and y's. Setting that determinant equal to zero can be shown to yield the quadratic equation:
(d^2+e^2)^2  K1*(d^2+e^2) + K2 = 0
where
K1 = a11^2+2*a12*a21+a22^2+b11^2+2*b12*b21+b22^2
K2 = det([ a11, a12,b11,b12;
a21, a22,b21,b22;
b11,b12,a11,a12;
b21,b22,a21,=a22])
Thus we have a quadratic equation in the unknown d^2+e^2 where we have specified earlier that d and e are understood to be real numbers. If the quadratic does not have real, positive roots, then no root could be equal to d^2+e^2, and clearly the original problem can have no solution.
Bruno has shown how these roots can be found if applicable by using the eig function. He does so by assuming the complex part e is zero. Then by their form, d is necessarily an eigenvalue of the matrix of the four linear equations above, and the corresponding eigenvector, [x1;x2;y1;y2], is a solution [x1+i*y1,x2+i*y2] to the pseudo problem.
What neither one of us showed you earlier was how to obtain other solutions, provided that one was found as above. It can be done as follows. Multiply the pseudo eigenvalue by an arbitrary factor, exp(i*k), and multiply the corresponding eigenvector by exp(i*k/2). These will also be solutions. This can easily be shown by observing that multiplying [x1+i*y1,x2+i*y2] by exp(i*k/2) is equivalent to multiplying [x1i*y1,x2i*y2] by exp(i*k/2). These two changes, multiplication by exp(i*k/2) on the left and exp(i*k/2) on the right, are compensated for in the required equality by multiplying the pseudo eigenvalue by exp(i*k). Moreover the eigenvectors remain complex conjugates of one another in this process.
Since exp(i*k) has an infinite continuum of possible values for k, this gives rise to an infinitude of possible eigenvalue/eigenvector combinations. Of course we could also multiply by factors that have other than magnitude one, but that would spoil the usual convention that eigenvector solutions should be normalized.
The property that is lost in the pseudo problem is that the eigenvectors alone can no longer be multiplied by arbitrary complex values, as it true in ordinary eigenvalue/eigenvector problems.
This discussion has involved only the 2 x 2 cases. However, though I have not proven it rigorously, I believe the general properties described here extend to any size square matrix, M.
Roger Stafford
