Dear Colleague,

Good Day,

write the equation in terns of kronecker products, which replace the

matrix structure with a matrix vector structure, where

vec( CXD ) = transpose(D) X C vec(X)

=> { I X A + transpose(B) X I + Transpose(D) X C } vec(X) = vec(E)

invert/solve and map result to a matrix. Your done.

You can see this article:

Golub, G.H., Nash, S. and Van Loan, C.F., "A Hessenberg-Schur method for the problem AX + XB = C," IEEE Trans. Auto. Contr., AC-24, pp. 909–913, 1979.

You can use the efficient methods introduced in the following paper:

1. Applied Mathematical Modelling, DOI:10.1016/j.apm.2015.01.026.

2. IET Control Theory & Applications, 9 (2015) 74-81.

3.Transactions of the Institute of Measurement and Control, 36 (2014) 963-970.

4. Applied Mathematics Letters, 26 (2013) 1013–1017.

Dear Prof ,

   Do you try to use Maple to solve it ? 

In my opinion, to solve

(XA+BTX+DTXC)vec(X) = vec(E)

is more complicated than the original problem AX+XB+CXD = E.

To compare the two formulations, a simple way to write down the equivalent equation as

Fy = g,

where

F = I $\otimes$ A + transpose(B) $\otimes$ I + Transpose(D) $\otimes$ C,

y = vec(X),

g = vec(E).

Note that the operator $\otimes$ represents the Kronecker product of the operand matrices (see in linear algebra text books, or in wikipedia). This shows that the original problem est more complicated than this new formulation.

You may use a basic, iterative scheme, like Richardson, Arnoldi, BiCGstab... but formulate it in a matrix valued fashion. This has been done successfully for this kind of Sylvester equations, especially if the matrices are very large. The basic ideas of this approach can be found, e.g., here: DOI:10.1137/100799010.