# Preconditioning matrix

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.

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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.
@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