Eigenfunction of a Jones Vector (System)
I am trying to find out just how to solve for the eigenfunction given a system, namely the parameters of an optical system (say a polarizer) in the form of a 2 by 2 Jones Vector. I know how to derive the eigenvalue, using the the constituent det(λI A) = 0, 'A' being the system at hand and 'λ' the eigenvalue. How do you go about solving for the eigenfunction?

Re: Eigenfunction of a Jones Vector (System)
This is a pure math question. Try the following:
http://mathworld.wolfram.com/Eigenvalue.html http://www.math.hmc.edu/calculus/tutorials/eigenstuff/ http://www.sosmath.com/matrix/eigen2/eigen2.html 
Re: Eigenfunction of a Jones Vector (System)
Once you've found λ, you can substitute its value into Av = λv. If you then multiply out the left hand side and equate components, v_{1} and v_{2}, of v on either side, you'll get two equivalent equations linking v_{1} and v_{2}. Eiter will give you the ratio v_{1}/v_{2}. This is fine: the eigenvalue equation is consistent with any multiplied constant in the eigenvector. There will be a normalisation procedure for fixing the constant.

Re: Eigenfunction of a Jones Vector (System)
so I got λ = 1. 'A' I assume is the system matrix or my Jones Vector, which is given as a 2 by 2 matrix. So that makes Av=v, thus A must be 1?? The 'v' values must be the same, but isn't 'v' the eigenfunction itself? The equation Av=v eliminates the 'v' value. What am I doing wrong here?

Re: Eigenfunction of a Jones Vector (System)
v is the vector and A is the matrix. The matrix isn't a vector, but is an operator which operates on the vector.
Try it with a matrix A representing a linear polariser at 45° to the base vectors. This matrix has all four elements equal to 1/2. This gives eigenvalue of 1, and on substituting as I explained above, shows the two components, v_{1} and v_{2}, of the vector to be equal, which is just what you'd expect. 
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