$\newcommand{\vb}[1]{\boldsymbol{#1}}$ $\newcommand{\R}{\mathbb{R}}$ $\newcommand{\matspace}{\mathcal{M}_{n\times n}(\R)}$ $\newcommand{\diag}{\mathrm{diag}}$ $\newcommand{\defeq}{\stackrel{\scriptsize\text{def}}{=}}$ $\newcommand{\N}{\mathbb{N}}$ $\newcommand{\norm}[1]{\left\| #1 \right\|_{\infty}}$

I was reading in Frank Poole's Introduction to Linear Algebra, section 7.2, a passage on the convergence of the Jacobi method for diagonally dominant matrices. To use the same notation as the book, the system we are trying to solve is $$A\vb{x} = \vb{b}\,,$$ where $A\in\matspace$, $\vb{x}\in\R^n$ and $\vb{b}\in\R^n$. Also, we define $D\in\matspace$ as the diagonal matrix containing the elements in $A$'s diagonal, $U\in\matspace$ as the upper triangular matrix containing the elements above the diagonal, and $L\in\matspace$ as the lower triangular matrix containing the elements below the diagonal—such that $A = D + L + U\, $. To simplify notation, we define $G \defeq -D^{-1}(L + U)$ and $\vb{c} \defeq -D^{-1}\vb{b}$.

The equation $A\vb{x} = \vb{x}$ is equivalent to $$ \vb{x} = G\vb{x} + \vb{c}\,; $$ the iterates of the Jacobi method are thus obtained as $$ \vb{x}^{k+1} = G\vb{x}^k + \vb{c}\qquad \forall k \in\N \,. $$

It can be proven that $$\norm{\vb{x}-\vb{x}^{k+1}} \le \norm{G}\norm{\vb{x}^{k+1} - \vb{x}^k} < \norm{\vb{x}^{k+1} - \vb{x}^k}$$ since $\norm{G}<1$ (which can be proven through the fact that $A$ is diagonally dominant), implying that the sequence of iterates is convergent. The inequality above also gives rise, when used recursively, to the fact that $$ \norm{\vb{x}-\vb{x}^{k}} \le \norm{G}^k\norm{\vb{x} - \vb{x}^0}\,, $$ which can be used to bound the error of an approximation to the desired precision (by finding a lower bound for the number of iterations).

Let $\varepsilon \in \R^+$, and suppose we want to ensure $\norm{\vb{x}-\vb{x}^{k}} < \varepsilon$. It would be sufficient to ensure $$ \norm{G}^k\norm{\vb{x} - \vb{x}^0} < \varepsilon \quad\Leftrightarrow\quad k > \frac{\log\varepsilon - \log\norm{\vb{x} - \vb{x}^0}}{\log\norm{G}} = \frac{\log\norm{\vb{x} - \vb{x}^0} - \log\varepsilon}{\log\norm{G}^{-1}} $$ (the last equality is just for convenience, since $\norm{G} < 1$ implies $\log\norm{G}^{-1} > 0$).

Now, the problem is the expression $\norm{\vb{x} - \vb{x}^0}\,$: if we were to use it to estimate the amount of iterations we needed to make, we would need to know the "exact" solution, $\vb{x}$, which is impractical. So, in a practical situation, we would need to estimate that expression. Concretely, we would need to find an upper bound $M\in\R^+\,$, such that $\norm{\vb{x} - \vb{x}^0} \le M\,.$ Indeed, the inequality $$ k > \frac{\log M - \log\varepsilon}{\log\norm{G}^{-1}} $$ would necessarily imply $$ k > \frac{\log\norm{\vb{x} - \vb{x}^0} - \log\varepsilon}{\log\norm{G}^{-1}}\,, $$ and thus, through all our previous reasoning, the condition $\norm{\vb{x}-\vb{x}^{k}} < \varepsilon$ would be satisfied—so we could set the number of iterations to $$ k = \left\lceil \frac{\log M - \log\varepsilon}{\log\norm{G}^{-1}} \right\rceil\,. $$

But the author of the book simply takes $\norm{\vb{x} - \vb{x}^0} \approx \norm{\vb{x}^1 - \vb{x}^0}\,$, without making sure that $\norm{\vb{x} - \vb{x}^0} \le \norm{\vb{x}^1 - \vb{x}^0}\,$. Wouldn't that allow the possibility that $\norm{\vb{x} - \vb{x}^0} > \norm{\vb{x}^1 - \vb{x}^0}\,,$ and thus imply the possibility of the condition $\norm{\vb{x}-\vb{x}^{k}} < \varepsilon$ not being satisfied (by taking $k = \left\lceil\frac{\log \norm{\vb{x}^1 - \vb{x}^0} - \log\varepsilon}{\log\norm{G}^{-1}} \right\rceil$)?

If so, what would be a way of finding an actual upper bound approximation of $\norm{\vb{x}-\vb{x}^{0}} = \norm{\vb{x}}$ (supposing $\vb{x}^0=\vb{0}$), therefore guaranteeing the condition $\norm{\vb{x}-\vb{x}^{k}} < \varepsilon$?

  • $\begingroup$ What can you say about the decay of the residual $r_j = b - Ax_j$? Can you establish a connection between the size of the residual and the size of the error? Can you involve the dominance factor? $\endgroup$ – Carl Christian Jan 25 at 0:25
  • 1
    $\begingroup$ Probably more useful is to require that $\|x-x_k\|/\|x\|\leq \epsilon$ which is easy to obtain with $x_0=0$ knowing the norm of the iteration matrix and gives roughly an estimate on the number of valid digits in the solution (assuming the problem is well scaled so that the values of $x$ do not vary much in magnitude). $\endgroup$ – Algebraic Pavel Jan 25 at 13:35
  • $\begingroup$ Take $k=0$ in the first of your estimates and you’ll get what you want $\endgroup$ – VorKir Feb 3 at 5:19

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