Jacobi method

The splitting A=L+D+U, the choice M=D that defines Jacobi, its component-wise iterative scheme and a worked example.

Taking M = D

AA is split into its strictly lower part LL, its diagonal DD and its strictly upper part UU. Jacobi picks the simplest preconditioner, M=DM=D, with N=(L+U)N=-(L+U):

A=L+D+U,x(k+1)=D1(L+U)x(k)+D1bA=L+D+U,\qquad x^{(k+1)}=-D^{-1}(L+U)x^{(k)}+D^{-1}b

Component-wise, each unknown is solved from its equation using the previous iteration's values:

xi(k+1)=1aii(bijiaijxj(k)),aii0x_i^{(k+1)}=\frac{1}{a_{ii}}\left(b_i-\sum_{j\ne i} a_{ij}x_j^{(k)}\right),\qquad a_{ii}\ne 0

Derivation of the iteration matrix

DerivationDerivation: Jacobi iteration matrixView as its own page →
  1. In each equation ii we solve for xix_i (possible because aii0a_{ii}\ne 0):

    xi=1aii(bijiaijxj)x_i=\frac{1}{a_{ii}}\left(b_i-\sum_{j\ne i}a_{ij}x_j\right)
  2. Writing this for all rows and separating the diagonal D from the parts L and U, the matrix form appears:

    x=D1(b(L+U)x)=D1(L+U)x+D1bx=D^{-1}\bigl(b-(L+U)x\bigr)=-D^{-1}(L+U)x+D^{-1}b
  3. As an iteration, it is the Jacobi scheme with iteration matrix H_J=−D⁻¹(L+U):

    x(k+1)=D1(L+U)x(k)+D1bx^{(k+1)}=-D^{-1}(L+U)x^{(k)}+D^{-1}b