> 1. In general the answer is no, as there is no solution. But, there are some situations where an acceptable solution is found using appropriate pre-conditioning. It all depends on the exact circumstance you are in. (For example, if the singularity is due to a zero row in K, then using the Jacobi pre-conditioner will make it an identity row. This may or may not be acceptable depending on your application.)
> 2. Solving the linear system is almost always preferred as finding the inverse to K explicitly boils down to repeatedly solving Kx=u where u are individual columns of I. Even if you are interested in solving for many u's it is preferable to save K's factorization rather than K's inverse. Read Matrix Computations by Gene Golub, or an introductory Scientific Computing book. Many of these questions are addressed there. -Andrew On Wed, Jun 25, 2008 at 4:29 PM, <enjoywm at cs.wm.edu> wrote: > Hi, > I have a linear equations of system such as Kx=u. > 1. If K is singular, can I use KSPSolve to solve it? > 2. If K is non-singular I also can use x= inverse(K)u to do it. > Which one is better? > > Best, > Yixun > >
