From 59bde6ffdfbc419978262601f80b4a376f26d4d0 Mon Sep 17 00:00:00 2001 From: Christophe Guyeux Date: Fri, 10 Oct 2014 15:27:04 +0200 Subject: [PATCH] Typos --- paper.tex | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/paper.tex b/paper.tex index d9880d1..6ae2692 100644 --- a/paper.tex +++ b/paper.tex @@ -742,7 +742,7 @@ Suppose that $A$ is a positive real matrix with symmetric part $M$. Then the res \begin{equation} ||r_m|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_0|| , \end{equation} -where $\alpha = \lambda_min(M)^2$ and $\beta = \lambda_max(A^T A)$, which proves +where $\alpha = \lambda_{min}(M)^2$ and $\beta = \lambda_{max}(A^T A)$, which proves the convergence of GMRES($m$) for all $m$ under that assumption regarding $A$. \end{proposition} @@ -767,7 +767,7 @@ The base case is obvious, as for $k=1$, the TSIRM algorithm simply consists in a Suppose now that the claim holds for all $m=1, 2, \hdots, k-1$, that is, $\forall m \in \{1,2,\hdots, k-1\}$, $||r_m|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_0||$. We will show that the statement holds too for $r_k$. Two situations can occur: \begin{itemize} -\item If $k \mod m \neq 0$, then +\item If $k \mod m \neq 0$, then the TSIRM algorithm consists in executing GMRES once. In that case, we obtain $||r_k|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_{k-1}||\leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_0||$. \item Else, let $\operatorname{span}(S) = \left \{ {\sum_{i=1}^k \lambda_i v_i \Big| k \in \mathbb{N}, v_i \in S, \lambda _i \in \mathbb{R}} \right \}$ be the linear span of a set of real vectors $S$. So,\\ $\min_{\alpha \in \mathbb{R}^s} ||b-R\alpha ||_2 = \min_{\alpha \in \mathbb{R}^s} ||b-AS\alpha ||_2$ -- 2.39.5