Jacobi method

In numerical linear algebra, the Jacobi method is an algorithm for determining the solutions of a system of linear equations with largest absolute values in each row and column dominated by the diagonal element. Each diagonal element is solved for, and an approximate value 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 German mathematician Carl Gustav Jakob Jacobi.

Contents

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%2BR \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}, \qquad 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} ,\qquad \mathbf{x}^{(k%2B1)} = D^{-1} (\mathbf{b} - R \mathbf{x}^{(k)}).

The element-based formula is thus:

x^{(k%2B1)}_i = \frac{1}{a_{ii}} \left(b_i -\sum_{j\ne i}a_{ij}x^{(k)}_j\right),\quad i=1,2,\ldots,n.

Note that 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

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 %2B a_{ij} x_j^{(k-1)}
end if
end (j-loop)
x_i^{(k)} = {{\left( {b_i - \sigma } \right)} \over {a_{ii} }}
end (i-loop)
check if convergence is reached
end (while convergence condition not reached loop)

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%2B1)}=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)} %2B C, where T=-D^{-1}R and C = D^{-1}b. Note that R=L%2BU 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%2BU) as

T= \begin{bmatrix} 1/2 & 0 \\ 0 & 1/7 \\ \end{bmatrix} \left\{ \begin{bmatrix} 0 & 0 \\ -5 & 0 \\ \end{bmatrix} %2B \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)}%2BC :

x^{(1)}= \begin{bmatrix} 0 & -1/2 \\ -5/7 & 0 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ \end{bmatrix} %2B \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} %2B \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} .

See also

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