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Consider the linear system

$$25a+2b+c=69 \\ 2a+10b+c=63 \\ a+2b+c=43 $$

A sufficient condition for the convergence of the Jacobi or Gauss-Seidel methods is that the system matrix is strictly diagonally dominant, which is not the case here. I have tried removing this difficulty by exchanging equation three and two, but the resulting matrix still does not meet the sufficient conditions. Any suggestions on the course of action?

PierreCarre
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Hammad Tahir
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    Note to all: The answer by @José Claudinei Ferreira contains good references including one to the Sassenfeld criterion which gives a another sufficient condition for the convergence of Gauss-Seidel. While an edit is certainly required for this question to be reopened it may be worthwhile to retain it, simply because the Sassenfeld criterion is not widely known. – Carl Christian May 20 '22 at 15:55

2 Answers2

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In a general setting the Sassenfeld criterion tell us that, if $A=[a_{ij}]_{n\times n}$, and we define \begin{eqnarray*} \beta_1&=&\frac{1}{a_{11}}\sum_{j>1}|a_{1j}|\\\\ \beta_2&=&\frac{1}{a_{22}}\left(|a_{21}|\beta_1+\sum_{j>2}|a_{2j}|\right)\\\vdots&\vdots&\vdots\\ \beta_i&=&\frac{1}{a_{ii}}\left(\sum_{j<i}|a_{ij}|\beta_j+\sum_{j>i}|a_{ij}|\right). \end{eqnarray*}

If $\beta_i<1$, to any $1\leq i\leq n$, then the Gauss-Seidel method to solve the linear equation $Ax=b$ is convergent.

In the particular system of this question $$A=\begin{bmatrix}25&2&1\\2&10&1\\1&2&1\end{bmatrix}.$$ Then \begin{eqnarray*} \beta_1&=&\frac{3}{25}\\\\ \beta_2&=&\frac{1}{10}\left(2\frac{3}{25}+1\right)\\\\ \beta_3&=&\frac{1}{1}\left(\frac{3}{25}+2\frac{31}{250}\right), \end{eqnarray*} and the Gauss-Seidel method is convergent.

You can see a related discussion in How to confirm if a system can be solved by Gauss-Seidel?, and find more searching for "\(Ax=b\) Gauss-Seidel" on SearchOnMath, for instance.

José Claudinei Ferreira
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    +1 I was delighted to learn about the Sassenfeld criterion. It appears to have interesting applications to preconditioning. That said, I recommend that you add a paragraph which states the Sassenfeld criterion and verifies that it does in fact apply to this matrix. The community favors answers that rely as little as possible on outside links. Certainly, arXiv is not going away anytime soon, but it is still good to write out the key details. – Carl Christian May 20 '22 at 13:05
  • @CarlChristian Thanks for the sugestion. – José Claudinei Ferreira May 21 '22 at 01:50
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Welcome to MSE! :D

In pratice, this system could be solved by Gauss-Sidel iteration.

A = [25 2 1;2 10 1;1 2 1];
b = [69 63 43]';
L = [25 0 0;2 10 0;1 2 1]; 
U =[0 2 1;0 0 1;0 0 0];
x = randn(3,1);
for i=1:30,
x = L\(b - U*x);
disp(mean((A*x-b).^2));
end
disp(x)

Output of an example run.

  839.1641
  16.3712
   0.6568
   0.0265
   0.0011
  4.2530e-05
  1.7014e-06
  6.8056e-08
  2.7222e-09
  1.0889e-10
  4.3556e-12
  1.7422e-13
  6.9690e-15
  2.7876e-16
  1.1150e-17
  4.4601e-19
  1.7841e-20
  7.1365e-22
  2.8529e-23
  1.1377e-24
  4.5523e-26
  2.0363e-27
  6.7316e-29
  1.6829e-29
    0
    0
    0
    0
    0
    0

1.0833
2.3646



37.1875

after 25 iterations, the error decay to numerically 0. So in fact it could converge.

But indeed as you said, it's not guaranteed to converge since it's not diagonal dominant or symmetric positive definite.

Binxu Wang 王彬旭
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