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Consider

Let the Gauss-Seidel method be applied to the equations $A \boldsymbol{x}=\boldsymbol{b}$ when $A$ is the nonsymmetric $2 \times 2$ matrix $$ A=\left[\begin{array}{cc} 10 & -3 \\ 3 & 1 \end{array}\right] . $$ Find the spectral radius of the iteration matrix. Then show that the relaxation method, described in Lecture 17, can reduce the spectral radius by a factor of 2.9. Further, show that iterating twice with Gauss-Seidel with this relaxation decreases the error $\left\|\boldsymbol{x}^{(k)}-\boldsymbol{x}^{(\infty)}\right\|$ by more than a factor of ten. Estimate the number of iterations of the original Gauss-Seidel method that would be required to achieve this decrease in the error.

I have done the first two parts with the spectral radius of the relaxed scheme being $\frac{9}{29}$.

My issue is that I do not understand what is meant by

Further, show that iterating twice with Gauss-Seidel with this relaxation decreases the error $\left\|\boldsymbol{x}^{(k)}-\boldsymbol{x}^{(\infty)}\right\|$ by more than a factor of ten.

In the course so far the notation $x^{(k)}$ is used for the $k$-th itterate and $x^{\infty}$ to what it converges to. However, in the above I do not understand with respect to which scheme this is. Is this wrt to the original, relaexed, relaxed with two iterations? Moreover, what does iterating twice mean? Does it mean that I define a new sequence $$ y^{k}=x^{2k}\,? $$

I am still confused. I have tried to work with a number of these but I do not get the required end result.

Question: Can someone explain what is meant by the full sentance

Further, show that iterating twice with Gauss-Seidel with this relaxation decreases the error $\left\|\boldsymbol{x}^{(k)}-\boldsymbol{x}^{(\infty)}\right\|$ by more than a factor of ten.

Maths Wizzard
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  • Denoting $x^{(k)}$ the $k$-iterate using Gauss-Seidel, and $\tilde{x}^{(k)}$ the $k$-iterate using the relaxed Gauss-Seidel, I understand that you must show that $\left|\tilde{x}^{(k)}-\tilde{x}^{(\infty)}\right|$ is at least 10 times smaller than $\left|x^{(k)}-x^{(\infty)}\right|$. – Florian Ingels Sep 26 '23 at 10:20
  • Not entirely. The interpretation that the relaxed method converges faster is $$\left|\tilde x^{(k+2)}-x^{(\infty)}\right|\le\frac1{10},\left|\tilde x^{(k)}-x^{(\infty)}\right|.$$ The solution of the linear system and thus the limit of the sequences, if it exists, is the same independent of the method it was obtained with. – Lutz Lehmann Sep 26 '23 at 10:37

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