Jacobi method

Let

${\displaystyle A\mathbf {x} =\mathbf {b} }$

be a square system of n linear equations, where:

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

Then A can be decomposed into a diagonal component D, a lower triangular part L and an upper triangular part U:

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

The solution is then obtained iteratively via

${\displaystyle \mathbf {x} ^{(k+1)}=D^{-1}(\mathbf {b} -(L+U)\mathbf {x} ^{(k)}),}$

where ${\displaystyle \mathbf {x} ^{(k)}}$ is the kth approximation or iteration of ${\displaystyle \mathbf {x} }$ and ${\displaystyle \mathbf {x} ^{(k+1)}}$ is the next or k + 1 iteration of ${\displaystyle \mathbf {x} }$. The element-based formula is thus:

${\displaystyle x_{i}^{(k+1)}={\frac {1}{a_{ii}}}\left(b_{i}-\sum _{j\neq i}a_{ij}x_{j}^{(k)}\right),\quad i=1,2,\ldots ,n.}$

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

Input: initial guess ${\displaystyle x^{(0)}}$ to the solution, (diagonal dominant) matrix ${\displaystyle A}$, right-hand side vector ${\displaystyle b}$, convergence criterion
Output: solution when convergence is reached
Comments: pseudocode based on the element-based formula above

${\displaystyle k=0}$
while convergence not reached do
    for i := 1 step until n do
        ${\displaystyle \sigma =0}$
        for j := 1 step until n do
            if j ≠ i then
              ${\displaystyle \sigma =\sigma +a_{ij}x_{j}^{(k)}}$
            end
        end
        ${\displaystyle x_{i}^{(k+1)}={{\frac {1}{a_{ii}}}\left({b_{i}-\sigma }\right)}}$
    end
    ${\displaystyle k=k+1}$
end

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

${\displaystyle \rho (D^{-1}(L+U))<1.}$

A sufficient (but not necessary) condition for the method to converge is that 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:

${\displaystyle \left|a_{ii}\right|>\sum _{j\neq i}{\left|a_{ij}\right|}.}$

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

Note that the Jacobi method does not converge for every symmetric positive-definite matrix. For example

${\displaystyle A={\begin{pmatrix}29&2&1\\2&6&1\\1&1&{\frac {1}{5}}\end{pmatrix}}\quad \Rightarrow \quad D^{-1}(L+U)={\begin{pmatrix}0&{\frac {2}{29}}&{\frac {1}{29}}\\{\frac {1}{3}}&0&{\frac {1}{6}}\\5&5&0\end{pmatrix}}\quad \Rightarrow \quad \rho (D^{-1}(L+U))\approx 1.0661\,.}$

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

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

We use the equation ${\displaystyle x^{(k+1)}=D^{-1}(b-(L+U)x^{(k)})}$, described above, to estimate ${\displaystyle x}$. First, we rewrite the equation in a more convenient form ${\displaystyle D^{-1}(b-(L+U)x^{(k)})=Tx^{(k)}+C}$, where ${\displaystyle T=-D^{-1}(L+U)}$ and ${\displaystyle C=D^{-1}b}$. From the known values

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

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

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

Further, ${\displaystyle C}$ is found as

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

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

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

The next iteration yields

${\displaystyle x^{(2)}={\begin{bmatrix}0&-1/2\\-5/7&0\\\end{bmatrix}}{\begin{bmatrix}5.0\\8/7\\\end{bmatrix}}+{\begin{bmatrix}11/2\\13/7\\\end{bmatrix}}={\begin{bmatrix}69/14\\-12/7\\\end{bmatrix}}\approx {\begin{bmatrix}4.929\\-1.714\\\end{bmatrix}}.}$

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

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

Another example

Suppose we are given the following linear system:

${\displaystyle {\begin{aligned}10x_{1}-x_{2}+2x_{3}&=6,\\-x_{1}+11x_{2}-x_{3}+3x_{4}&=25,\\2x_{1}-x_{2}+10x_{3}-x_{4}&=-11,\\3x_{2}-x_{3}+8x_{4}&=15.\end{aligned}}}$

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

${\displaystyle {\begin{aligned}x_{1}&=(6+0-(2*0))/10=0.6,\\x_{2}&=(25+0+0-(3*0))/11=25/11=2.2727,\\x_{3}&=(-11-(2*0)+0+0)/10=-1.1,\\x_{4}&=(15-(3*0)+0)/8=1.875.\end{aligned}}}$

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.

${\displaystyle x_{1}}$	${\displaystyle x_{2}}$	${\displaystyle x_{3}}$	${\displaystyle x_{4}}$
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).

def jacobi(A, b, x_init, epsilon=1e-10, max_iterations=500):
    D = np.diag(np.diag(A))
    LU = A - D
    x = x_init
    for i in range(max_iterations):
        D_inv = np.diag(1 / np.diag(D))
        x_new = np.dot(D_inv, b - np.dot(LU, x))
        if np.linalg.norm(x_new - x) < epsilon:
            return x_new
        x = x_new
    return x

# problem data
A = np.array([
    [5, 2, 1, 1],
    [2, 6, 2, 1],
    [1, 2, 7, 1],
    [1, 1, 2, 8]
])
b = np.array([29, 31, 26, 19])

# you can choose any starting vector
x_init = np.zeros(len(b))
x = jacobi(A, b, x_init)

print('x:', x)
print('computed b:', np.dot(A, x))
print('real b:', b)

x: [3.99275362 2.95410628 2.16183575 0.96618357]
computed b: [29. 31. 26. 19.]
real b: [29 31 26 19]

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

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

with ${\displaystyle \omega =2/3}$ being the usual choice.[1]

• Gauss–Seidel method
• Successive over-relaxation
• Iterative method § Linear systems
• Gaussian Belief Propagation
• Matrix splitting

• This article incorporates text from the article Jacobi_method on CFD-Wiki that is under the GFDL license.

• Black, Noel; Moore, Shirley; and Weisstein, Eric W. "Jacobi method". MathWorld.CS1 maint: multiple names: authors list (link)
• Jacobi Method from www.math-linux.com