I need some help with this problem:
Let
$$A=\begin{bmatrix} 1 & p \\ -p & 1 \\ \end{bmatrix}$$
Using SOR method, for which values of $\omega$ does the system converge?
I know, there is a theorem for symmetric matrices ($0<\omega<2$), but in this case tha matrix is antisymmetric.
I guess, I need to see the expression $H_\omega = (D-\omega L)^{-1}((1-\omega)D + \omega U)$ and the fact tha $H_\omega$ should has $\lambda<1$ but when I developed the characteristic polynomial, I get lost with very great results and that make no sense
The characteristic polynomial of $H_\omega$ is
$$ r(x) = x^2 - (\omega^2p^2 + 2(1-\omega))x + (1-\omega)^2 $$
and the eigenvalues of $H_\omega$ are given by
$$ \frac{\omega^2p^2 + 2(1-\omega) \pm \omega|p|\sqrt{\omega^2p^2 + 4(1-\omega)}}{2} $$
We may assume that $\omega\in (0,2)$, otherwise the SOR method does not converge. I further assume that $p$ is real. Now suppose that $|p| \geq 1$. Then it follows that both eigenvalues are real. Moreover
$$ \left(\omega|p|\sqrt{\omega^2p^2 + 4(1-\omega)}\,\right)^2 - (w^2p^2 - 2\omega)^2 = 4\omega^2(p^2-1) \geq 0, $$
which implies that
$$ |w^2p^2 - 2\omega| \leq \omega|p|\sqrt{\omega^2p^2 + 4(1-\omega)} $$
From this result we observe that
$$ 2 \leq w^2p^2 + 2(1-\omega) + \omega|p|\sqrt{\omega^2p^2 + 4(1-\omega)} \implies \rho(H_\omega) \geq 1 $$
Thus, the SOR method does not converge for all $\omega\in (0,2)$ if $|p| \geq 1$.
Now suppose that $|p| < 1$. Then the eigenvalues are real if $0 < \omega \leq (2/p^2)(1-\sqrt{1-p^2})$ and are imaginary otherwise. In the case of imaginary eigenvalues you can show that $\rho(H_\omega) = |1-\omega| < 1$ by arguments similar to those above. In the case of real eigenvalues I will leave it up to you to show that
$$ \left|\omega^2p^2 + 2(1-\omega) \pm \omega|p|\sqrt{\omega^2p^2 + 4(1-\omega)}\,\right| < 2 $$
and hence that $\rho(H_\omega) < 1$ for all $\omega\in(0,2)$ and all $|p| < 1$.