# Method to solve a coupled system of matrix equation

1. Aug 17, 2015

### duc

Hello everyone,

I'm struggling with a coupled of matrix equations of the general form:

AX + CY = cX
BY + DX = cY

where A, B, C and D are hermitics square matrices. X, Y and c are the eigenvector and eigenvalue to be found. I'm looking for a method or an algorithm to solve this system by using Fortran. Could you suggest a reference or paper which treats this kind of equation ?

Thanks a lot!
duc

2. Aug 17, 2015

### DEvens

Last edited by a moderator: May 7, 2017
3. Aug 17, 2015

### duc

Hi DEvens,

Maybe i should be more specific. The system of matrix equations I've mentioned is not linear system of equations that can be solved by using matrix technique. A, B, C and D here are not number but square matrices of dimension N x N (where N is integer and typically of order 1000). And X, Y and c are respectively the column vectors of N elements and the unknown value which have to be found.

Physically, X and Y are two components of a spinor associated to a spin-1/2 particle. c is the energy associated with this spinor. When studying a physical system whose Hamiltonian contains an interaction which involves spin operators in an intricate manner, I've been led to the above kind of equations.

In other word, instead of the usual eigenvalue problem: A X = e X (A: matrix, X - a vector of given dimension, e - eigenvalue), the system of matrix equations above describes the eigenvalue problem for a spinor X of 2-components and A will become a matrix of 2x2 elements (and each one is an operator) in spin space. When using a discrete truncated basis in position space for example, each "element" of A will be represented by a matrix.

Hope that would make my question clearer. I'll consult the reference you've cited any way.

duc

Last edited by a moderator: May 7, 2017
4. Aug 17, 2015

### DEvens

Oh, it's a different part of it you are having difficulty with. I sort of assumed this part was trivial.

AX + CY = cX
BY + DX = cY

Presuming X and Y are column vectors. Rearrange that as Z = {X,Y}. I'm too lazy to get into the TeX-like things here to make that line up properly. But that is a column vector with the first "n" components being the components of X, and the second "n" being the components of Y.

Then line up your matrices like so:

A C
D B

That is make a 2n x 2n matrix out of the four matrices A, C, D, and B. Call this new matrix M.

So you then have M Z = c Z. And you round up the usual suspects to solve the matrix equation.

5. Aug 24, 2015

### duc

Thank you very much DEvens. I've thought of it but had some doubts and didn't go further :D. It is indeed "trivial" as you said :">. Thanks again. :)