# Jacobi method

(Redirected from Jacobi iteration)

In numerical linear algebra, the Jacobi method (or Jacobi iterative method[1]) is an algorithm for determining the solutions of a diagonally dominant system of linear equations. Each diagonal element is solved for, and an approximate value is plugged in. The process is then iterated until it converges. This algorithm is a stripped-down version of the Jacobi transformation method of matrix diagonalization. The method is named after Carl Gustav Jacob Jacobi.

## Description

Given a square system of n linear equations:

$A\mathbf x = \mathbf b$

where:

$A=\begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix}, \qquad \mathbf{x} = \begin{bmatrix} x_{1} \\ x_2 \\ \vdots \\ x_n \end{bmatrix} , \qquad \mathbf{b} = \begin{bmatrix} b_{1} \\ b_2 \\ \vdots \\ b_n \end{bmatrix}.$

Then A can be decomposed into a diagonal component D, and the remainder R:

$A=D+R \qquad \text{where} \qquad D = \begin{bmatrix} a_{11} & 0 & \cdots & 0 \\ 0 & a_{22} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\0 & 0 & \cdots & a_{nn} \end{bmatrix} \text{ and } R = \begin{bmatrix} 0 & a_{12} & \cdots & a_{1n} \\ a_{21} & 0 & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & 0 \end{bmatrix}.$

The solution is then obtained iteratively via

$\mathbf{x}^{(k+1)} = D^{-1} (\mathbf{b} - R \mathbf{x}^{(k)}).$

The element-based formula is thus:

$x^{(k+1)}_i = \frac{1}{a_{ii}} \left(b_i -\sum_{j\ne i}a_{ij}x^{(k)}_j\right),\quad i=1,2,\ldots,n.$

The computation of xi(k+1) requires each element in x(k) except itself. Unlike the Gauss–Seidel method, we can't overwrite xi(k) with xi(k+1), as that value will be needed by the rest of the computation. The minimum amount of storage is two vectors of size n.

## Algorithm

Choose an initial guess $x^{(0)}$ to the solution
$k = 0$
while convergence not reached do
for i := 1 step until n do
$\sigma = 0$
for j := 1 step until n do
if j ≠ i then
$\sigma = \sigma + a_{ij} x_j^{(k)}$
end if
end (j-loop)
$x_i^{(k+1)} = {{\left( {b_i - \sigma } \right)} \over {a_{ii} }}$
end (i-loop)
check if convergence is reached
$k = k + 1$
loop (while convergence condition not reached)

## Convergence

The standard convergence condition (for any iterative method) is when the spectral radius of the iteration matrix is less than 1:

$\rho(D^{-1}R) < 1.$

The method is guaranteed to converge if the matrix A is strictly or irreducibly diagonally dominant. Strict row diagonal dominance means that for each row, the absolute value of the diagonal term is greater than the sum of absolute values of other terms:

$\left | a_{ii} \right | > \sum_{j \ne i} {\left | a_{ij} \right |}.$

The Jacobi method sometimes converges even if these conditions are not satisfied.

## Example

A linear system of the form $Ax=b$ with initial estimate $x^{(0)}$ is given by

$A= \begin{bmatrix} 2 & 1 \\ 5 & 7 \\ \end{bmatrix}, \ b= \begin{bmatrix} 11 \\ 13 \\ \end{bmatrix} \quad \text{and} \quad x^{(0)} = \begin{bmatrix} 1 \\ 1 \\ \end{bmatrix} .$

We use the equation $x^{(k+1)}=D^{-1}(b - Rx^{(k)})$, described above, to estimate $x$. First, we rewrite the equation in a more convenient form $D^{-1}(b - Rx^{(k)}) = Tx^{(k)} + C$, where $T=-D^{-1}R$ and $C = D^{-1}b$. Note that $R=L+U$ where $L$ and $U$ are the strictly lower and upper parts of $A$. From the known values

$D^{-1}= \begin{bmatrix} 1/2 & 0 \\ 0 & 1/7 \\ \end{bmatrix}, \ L= \begin{bmatrix} 0 & 0 \\ 5 & 0 \\ \end{bmatrix} \quad \text{and} \quad U = \begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix} .$

we determine $T=-D^{-1}(L+U)$ as

$T= \begin{bmatrix} 1/2 & 0 \\ 0 & 1/7 \\ \end{bmatrix} \left\{ \begin{bmatrix} 0 & 0 \\ -5 & 0 \\ \end{bmatrix} + \begin{bmatrix} 0 & -1 \\ 0 & 0 \\ \end{bmatrix}\right\} = \begin{bmatrix} 0 & -1/2 \\ -5/7 & 0 \\ \end{bmatrix} .$

Further, C is found as

$C = \begin{bmatrix} 1/2 & 0 \\ 0 & 1/7 \\ \end{bmatrix} \begin{bmatrix} 11 \\ 13 \\ \end{bmatrix} = \begin{bmatrix} 11/2 \\ 13/7 \\ \end{bmatrix}.$

With T and C calculated, we estimate $x$ as $x^{(1)}= Tx^{(0)}+C$:

$x^{(1)}= \begin{bmatrix} 0 & -1/2 \\ -5/7 & 0 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ \end{bmatrix} + \begin{bmatrix} 11/2 \\ 13/7 \\ \end{bmatrix} = \begin{bmatrix} 5.0 \\ 8/7 \\ \end{bmatrix} \approx \begin{bmatrix} 5 \\ 1.143 \\ \end{bmatrix} .$

The next iteration yields

$x^{(2)}= \begin{bmatrix} 0 & -1/2 \\ -5/7 & 0 \\ \end{bmatrix} \begin{bmatrix} 5.0 \\ 8/7 \\ \end{bmatrix} + \begin{bmatrix} 11/2 \\ 13/7 \\ \end{bmatrix} = \begin{bmatrix} 69/14 \\ -12/7 \\ \end{bmatrix} \approx \begin{bmatrix} 4.929 \\ -1.713 \\ \end{bmatrix} .$

This process is repeated until convergence (i.e., until $\|Ax^{(n)} - b\|$ is small). The solution after 25 iterations is

$x=\begin{bmatrix} 7.111\\ -3.222 \end{bmatrix} .$

### Another example

Suppose we are given the following linear system:

\begin{align} 10x_1 - x_2 + 2x_3 & = 6, \\ -x_1 + 11x_2 - x_3 + 3x_4 & = 25, \\ 2x_1- x_2+ 10x_3 - x_4 & = -11, \\ 3x_2 - x_3 + 8x_4 & = 15. \end{align}

Suppose we choose (0, 0, 0, 0) as the initial approximation, then the first approximate solution is given by

\begin{align} x_1 & = (6 - 0 - 0) / 10 = 0.6, \\ x_2 & = (25 - 0 - 0) / 11 = 25/11 = 2.2727 \\ x_3 & = (-11 - 0 - 0) / 10 = -1.1,\\ x_4 & = (15 - 0 - 0) / 8 = 1.875. \end{align}

Using the approximations obtained, the iterative procedure is repeated until the desired accuracy has been reached. The following are the approximated solutions after five iterations.

$x_1$ $x_2$ $x_3$ $x_4$
$0.6$ $2.27272$ $-1.1$ $1.875$
$1.04727$ $1.7159$ $-0.80522$ $0.88522$
$0.93263$ $2.05330$ $-1.0493$ $1.13088$
$1.01519$ $1.95369$ $-0.9681$ $0.97384$
$0.98899$ $2.0114$ $-1.0102$ $1.02135$

The exact solution of the system is (1, 2, −1, 1).

### An example using Python 3 and Numpy

The following numerical procedure simply iterates to produce the solution vector.

import numpy as np

ITERATION_LIMIT = 1000

# initialize the matrix
A = np.array([[10., -1., 2., 0.],
[-1., 11., -1., 3.],
[2., -1., 10., -1.],
[0.0, 3., -1., 8.]])
# initialize the RHS vector
b = np.array([6., 25., -11., 15.])

# prints the system
print("System:")
for i in range(A.shape[0]):
row = ["{}*x{}".format(A[i, j], j + 1) for j in range(A.shape[1])]
print(" + ".join(row), "=", b[i])
print()

x = np.zeros_like(b)
for it_count in range(ITERATION_LIMIT):
print("Current solution:", x)
x_new = np.zeros_like(x)

for i in range(A.shape[0]):
s1 = np.dot(A[i, :i], x[:i])
s2 = np.dot(A[i, i + 1:], x[i + 1:])
x_new[i] = (b[i] - s1 - s2) / A[i, i]

if np.allclose(x, x_new, rtol=1e-10):
break

x = x_new

print("Solution:")
print(x)
error = np.dot(A, x) - b
print("Error:")
print(error)


Produces the output:

System:
10.0*x1 + -1.0*x2 + 2.0*x3 + 0.0*x4 = 6.0
-1.0*x1 + 11.0*x2 + -1.0*x3 + 3.0*x4 = 25.0
2.0*x1 + -1.0*x2 + 10.0*x3 + -1.0*x4 = -11.0
0.0*x1 + 3.0*x2 + -1.0*x3 + 8.0*x4 = 15.0

Current solution: [ 0.  0.  0.  0.]
Current solution: [ 0.6         2.27272727 -1.1         1.875     ]
Current solution: [ 1.04727273  1.71590909 -0.80522727  0.88522727]
Current solution: [ 0.93263636  2.05330579 -1.04934091  1.13088068]
Current solution: [ 1.01519876  1.95369576 -0.96810863  0.97384272]
Current solution: [ 0.9889913   2.01141473 -1.0102859   1.02135051]
Current solution: [ 1.00319865  1.99224126 -0.99452174  0.99443374]
Current solution: [ 0.99812847  2.00230688 -1.00197223  1.00359431]
Current solution: [ 1.00062513  1.9986703  -0.99903558  0.99888839]
Current solution: [ 0.99967415  2.00044767 -1.00036916  1.00061919]
Current solution: [ 1.0001186   1.99976795 -0.99982814  0.99978598]
Current solution: [ 0.99994242  2.00008477 -1.00006833  1.0001085 ]
Current solution: [ 1.00002214  1.99995896 -0.99996916  0.99995967]
Current solution: [ 0.99998973  2.00001582 -1.00001257  1.00001924]
Current solution: [ 1.00000409  1.99999268 -0.99999444  0.9999925 ]
Current solution: [ 0.99999816  2.00000292 -1.0000023   1.00000344]
Current solution: [ 1.00000075  1.99999868 -0.99999899  0.99999862]
Current solution: [ 0.99999967  2.00000054 -1.00000042  1.00000062]
Current solution: [ 1.00000014  1.99999976 -0.99999982  0.99999975]
Current solution: [ 0.99999994  2.0000001  -1.00000008  1.00000011]
Current solution: [ 1.00000003  1.99999996 -0.99999997  0.99999995]
Current solution: [ 0.99999999  2.00000002 -1.00000001  1.00000002]
Current solution: [ 1.          1.99999999 -0.99999999  0.99999999]
Current solution: [ 1.  2. -1.  1.]
Solution:
[ 1.  2. -1.  1.]
Error:
[ -2.81440107e-08   5.15706873e-08  -3.63466359e-08   4.17092547e-08]


## Weighted Jacobi method

The weighted Jacobi iteration uses a parameter $\omega$ to compute the iteration as

$\mathbf{x}^{(k+1)} = \omega D^{-1} (\mathbf{b} - R \mathbf{x}^{(k)}) + \left(1-\omega\right)\mathbf{x}^{(k)}$

with $\omega = 2/3$ being the usual choice.[2]

## Recent developments

In 2014, a refinement of the algorithm, called scheduled relaxation Jacobi method, was published.[1][3] The new method is alleged to provide a two-hundred fold performance improvement.