From: Christophe Guyeux <guyeux@gmail.com>
Date: Fri, 10 Oct 2014 13:27:04 +0000 (+0200)
Subject: Typos
X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/GMRES2stage.git/commitdiff_plain/59bde6ffdfbc419978262601f80b4a376f26d4d0?ds=inline

Typos
---

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,\\
 $\min_{\alpha \in \mathbb{R}^s} ||b-R\alpha ||_2 = \min_{\alpha \in \mathbb{R}^s} ||b-AS\alpha ||_2$