Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to solve $Ax=b$ system, but the conditioning of $A$ is bad. After some research, I found that the conditioning of $A$ can be improved by multiplying the system by a preconditioning matrix $P$ such as :

$Ax = b \Rightarrow P^{-1}Ax = P^{-1}b$

Can I calculate $P$? how?

I tried the case when $P^{-1}=A^{-1}$ and I found $x=b$, is there someone who can explain me that result?

Thank you.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

You want $P^{-1}$ to come as close as possible to being an inverse of $A$, but you also want $P^{-1} b$ to be easy to compute. These desires conflict with each other. Two extreme cases are $P = A$ and $P = I$. In the first case, the resulting system $x = P^{-1}b$ is trivial to solve, but computing $P^{-1} b$ is as difficult as solving the original system. In the other case ($P = I$), $P^{-1} b = b$ is trivial to compute, but the new system $P^{-1} A x = P^{-1} b$ is no easier to solve than the original system. What we want is something in between these two extreme cases.

Trefethen's really good book Numerical Linear Algebra has a survey of popular preconditioners at the end. They say that finding a good preconditioner is difficult, and usually depends on your specific application.

share|improve this answer
    
@Bek A good preconditioner clusters the eigenvalues around the minimal number of distinct numbers (ideally 1) as possible. This is highly dependent on the matrix $A$, but preconditioners based on incomplete factorisations (either $LU$ or cholesky) can be helpful initially. This should be how you decide on the precondition between the two extreme types of preconditioners littleO describes. –  Daryl Oct 26 '12 at 11:21

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.