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This is a homework problem, so please give hints or tips instead of full answers.

The problem is as follows:

Let $G$ be the iteration matrix of the Gauss-Seidel method; i.e. $$G=I-(D-L)^{-1} A$$ where $D-L$ is the lower triangular part of $A$, and $A$ is strictly row diagonally dominant. Suppose $x$ is any vector with $||x||_\infty=1$, and let $y=Gx$. Show that $||y||_\infty<1$.

The problem is that I have absolutely no idea where to even start. I've tried expanding $y=Gx$ to $y=x-((D-L)^{-1}A)x$, but that's obviously useless.

ithisa
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1 Answers1

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From $y=Gx$, we have $(D-L)y=(D-L-A)x=Ux$. Select $i$ so that $|y_i|=\|y\|_\infty$. Write down the $i$th equation

\begin{align}a_{ii}y_i+\sum\limits_{j=1}^{i-1}a_{ij}y_j=-\sum\limits_{j=i+1}^na_{ij}x_j\end{align}

and make an upper bound on $|y_i|$

\begin{align}|y_i|\leq|a_{ii}|^{-1}\left(\sum\limits_{j=1}^{i-1}|a_{ij}||y_j|+\sum\limits_{j=i+1}^n|a_{ij}||x_j|\right).\end{align}

Then use $|y_j|\leq|y_i|$, $|x_j|\leq\|x\|_\infty\leq 1$, and the strict row diagonal dominance to get an upper bound on $|y_i|$ in terms of the absolute values of the entries of $A$:

\begin{align}|y_i|\leq\frac{\sum\limits_{j=i+1}^n|a_{ij}|}{|a_{ii}|-\sum\limits_{j=1}^{i-1}|a_{ij}|}.\end{align}

Then use the diagonal dominance again to conclude that $|y_i|<1$.

Algebraic Pavel
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