How to confirm if a system can be solved by Gauss-Seidel? 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. 


*

*The line criteria, i.e., the strictly or irreducibly diagonally dominant criteria for the matrix of coefficients is not true because that $3 < 1 + 3$.

*The symmetric positive-definite it is not true either because the coefficient matrix is not symmetric: e.g. $1\neq-2$. 

*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?

 A: 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.
