Solving A*X=B

[only1892]

Hey guys,
I have a linear equation to solve,where A is a 20000*20000 symmetric matrix (Not a sparse matrix). I am using pdgesv from scalapack. But still it takes too long. (Actually matrix A changes and I need to solve this equation over and over again,around 10000 times). So it looks to me that this calculation is not practical.
I am looking for another function or other way to solve this equation. Or can I find a way to solve this equation once and then use some kind of perturbation theory to find the new X when A changes? Thanks a lot.....

Good luck to all,
Wayne

[Julien Langou]

If A is symmetric and positive definite, then you should use: PDPOSV and not PDGESV, PDPOSV exploits the symmetry of the matrix and reduces the number of operations by 2, it also works by tile and so in general the algorithm is much more efficient that PDGESV. If A is symmetric indefinite, then the only solver from ScaLAPACK that you can use is PDGESV, you can not exploit the symmetry. (There is no LDL^T solver in ScaLAPACK.)

If the change is a small rank perturbution (with possible big norm): A_new = A_old + uv^T where u and v are n-by-k where k << n, then you can use the Sherman–Morrison–Woodbury formula (http://en.wikipedia.org/wiki/Woodbury_matrix_identity).

If the change is a small perturbation (with possible full rank): A_new = A_old + E where || E || << || A ||, then you can try to use the previous LU factorization
of A_old as a preconditioner for A_new in the context of an iterivative solve (FGMRES for example). If moreover the right-hand side also has slightly changed, then
you can start your iterative method with the previous solution.

What will work is really problem dependent in general.

Best wishes,
--julien.