Hi All!


I was supposed to find a solution of Ax=b using Jacobi and Gauss-Seidel method.
The A is 100x100 symetric, positive-definite matrix and b is a vector filled with 1's.
I am iterating(k = 1,2,....) those methods until the norm of (x(k+1) - x(k)) < precision
which means that x is not changing and it is senseless to iterate more. 
Both methods ends with the same k(changed precision and k is still the same).
So is it possible that the convergence of Jacobi and Gauss-Seidel is the same? 


I will appreciate every resposne. 



The Convergence of Jacobi and Gauss-Seidel Iteration
There are several important cases where it is proved that $\rho(G)<\rho(J)$, with $G$ and $J$ the iteration matrices associated to the Gauss-Seidel and Jacobi methods. See for instance my book *Matrices. GTM 216, Springer-Verlag*. For instance, in the tridiagonal case, $\rho(G)=\rho(J)^2$ thus G-S is twice faster as Jacobi.


What means *twice faster* (or just $k$-times faster) ? These are order one methods, in the sense that a fixed number of exact digits are gained at each step. This number is $\tau=-\log\_{10}\rho$. A method is twice faster than an other if the ratio $\tau\_{one}/\tau\_{other}$ equals $2$. Thus you should see a significant difference between both methods. If not, there might be two reasons. Either you are in an exceptional case where $\rho(G)=\rho(J)$, or something is wrong in your code.


In general, I do not recommend Jacobi and G-S. They are good examples in a course to beginners. But a slight change of G-S yields the relaxation method. With an optimal parameter, it is much faster. This is because $\rho(G)$ is very close to $1$ when $n$ is large, and thus $\tau$ is very small. 


