Given the matrix A, prove that the Gauss-Seidel method converges and the Jacobi method does not. The matrix A = $$\begin{bmatrix}2 & 1 & -1\\ -2 & 2 & -2\\ 1 & 1 & 2\\\end{bmatrix}$$. 
I was able to prove that the Jacobi method does not converge by calculating the B matrix, finding its eigenvalues, which were imaginary, and showing that $a^2 + b^2 <1$. However, I am stuck when trying to prove that Gauss-Seidel converges. Using the same method of finding the B matrix and calculating its eigenvalues, I am getting that this method diverges as well. I have tried row and column exchanges, but still cannot figure out how to prove that it converges. 
 A: For iterational methods in the canonical form
$$
\mathbf P(\mathbf x_{n+1} - \mathbf x_n) = \mathbf f - \mathbf A \mathbf x_n
$$
the nessesary and sufficient condition is given by roots of
$$
\det (\mathbf A - \mathbf P + \lambda \mathbf P) = 0 \tag{*}
$$
which all should satisfy $|\lambda| < 1$. It can be easily shown that those $\lambda$ are simply eigenvalues of $\mathbf B$, where $\mathbf B$ is iteration matrix
$$
\mathbf B = \mathbf P^{-1}(\mathbf P - \mathbf A).
$$


*

*Jacobi case. The matrix $\mathbf P$ for the Jacobi method is given by
$$
\mathbf P = \operatorname{diag}(\mathbf A).
$$
The equation $(*)$ gives
$$
\det (\mathbf A - \mathbf P + \lambda \mathbf P) = 
\begin{vmatrix}
2\lambda & 1 & -1\\
-2 & 2\lambda & -2\\
1 & 1 & 2\lambda
\end{vmatrix} = 8\lambda^3 + 10\lambda = 0.
$$
The roots are $\lambda_1 = 0, \lambda_{2,3} = \pm\sqrt\frac{5}{4}i$. Obviously, $|\lambda_{2,3}| = 5/4 > 1$. The method does not converge.

*Seidel case. The matrix $\mathbf P$ for the Seidel method is given by
$$
\mathbf P = \operatorname{diag}(\mathbf A) + \operatorname{lower}(\mathbf A).
$$
The equation $(*)$ gives
$$
\det (\mathbf A - \mathbf P + \lambda \mathbf P) = 
\begin{vmatrix}
2\lambda & 1 & -1\\
-2\lambda & 2\lambda & -2\\
\lambda & \lambda & 2\lambda
\end{vmatrix} = 2\lambda(4\lambda^2 + 6\lambda - 1) = 0.
$$
The roots are $\lambda_1 = 0, \lambda_{2,3} = \frac{-3 \pm \sqrt{13}}{4} \approx {-1.65139, 0.151388}$. The eigenvalue $\lambda_3$ violates the $|\lambda|<1$ condition, thus the method wont converge. Numerical experiment shows the same.
