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Given the system

$$\left(\begin{matrix}5&1&2\\-2&5&2\\-1&3&3\end{matrix}\right)\left(\begin{matrix}x_3\\x_1\\x_2\end{matrix}\right) = \left(\begin{matrix}0\\1\\0\end{matrix}\right)$$

I have to say something concerning the convergence of Gauss-Seidel Method.

My work: What I understand is that first of all I should look and see if it fits the convergence criteria for the method.

  1. The line criteria, i.e., the strictly or irreducibly diagonally dominant criteria for the matrix of coefficients is not true because that $3 < 1 + 3$.
  2. The symmetric positive-definite it is not true either because the coefficient matrix is not symmetric: e.g. $1\neq-2$.
  3. Also the Sassenfeld Criteria does not fit.

Could I say as an answer that the method could still converge even though these criteria doesn't hold? Does there exists other ways to prove that the method will converge? Most importantly:

Does there exists an iff statement like 'Gauss-Seidel will converge iff something is true' or at least can such a statement exist? Or could such a thing never exist?

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    $\begingroup$ Try to compute the spectal radius of the iteration matrix $G = L^{-1}U$ (compute eigenvalues and show that they are $|\lambda|<1$ - this is the case here). For more info see math.stackexchange.com/questions/270181/… $\endgroup$
    – Winther
    Feb 7, 2017 at 1:01
  • $\begingroup$ Ok. I am calculating your suggestion above now, just dont know how to find $L$, $U$ and $D$ given this $A$ but I understand that you just answer the most important question that I have made. Exist's an iff statement. $\endgroup$
    – R.W
    Feb 7, 2017 at 1:21
  • $\begingroup$ I liked your edit. My mother-tongue is not English so I commit these types of mistake. $\endgroup$
    – R.W
    Feb 7, 2017 at 1:23
  • $\begingroup$ $L$ is the lower triangular part of $A$, $L = \pmatrix{5 & 0 & 0 \\ -2 & 5 & 0 \\-1 & 3 & 3}$ and $U = A - L$ is the upper triangular part. $\endgroup$
    – Winther
    Feb 7, 2017 at 1:24
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    $\begingroup$ Yes that is correct (in my notation above $L$ is $L+D$). $\endgroup$
    – Winther
    Feb 7, 2017 at 1:30

1 Answer 1

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Using the comments above - I'll answer my own question just for the question dont remain unanswered - and the paper given in the answer we have to use the theorem:

Let $G := - (D+L)^{-1}U$ such that $D$,$L$ and $U$ are the diagonal, lower and upper parts of the matrix $A$. Let $\sigma(G)$ be the spectrum of $G$, i.e., be such that $\sigma(G)$ is the set of eigenvalues of $G$ and define $$\rho(G):= \max_{\lambda \in \sigma(G)}\vert \lambda \vert$$Then the method converges iff $$\rho(G)<1$$

So, for this A we have

$$D + L = \pmatrix{5 & 0 & 0 \\ -2 & 5 & 0 \\-1 & 3 & 3}$$

and $$U = \pmatrix{0 & 1 & 2 \\ 0 & 0 & 2 \\0 &0 &0}$$

For that then we have that $$(D+L)^{-1} = \pmatrix{1/5 & 0 & 0\\ 2/25 & 1/5 & 0 \\-1/75 &-1/5 &1/3}$$

So

$$G = \pmatrix{0 & -1/5 & -2/5 \\ 0 & -2/25 & -14/25 \\0 &1/75 &32/75}$$

Now, finding the eigenvalues of $G$ we have to solve $\det(G - \lambda I)=0$

$$\lambda\left(-\lambda^2-\frac{26}{75}\lambda - \frac{50}{1875}\right)=0$$

The solutions are

$\lambda_0 = 0 $, $\lambda_{+}=\frac{26+2\sqrt{11\cdot29}}{150} \approx 0.41$, $\lambda_{-} =\frac{26-2\sqrt{11\cdot29}}{150} \approx -0.064 $.This way we get that $\rho(G) = 0.41 < 1$ so the method converges.

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