% use a multiple column layout for up to two different
% affiliations
-\author{\IEEEauthorblockN{Rapha\"el Couturier\IEEEauthorrefmark{1}, Lilia Ziane Khodja \IEEEauthorrefmark{2}, and Christophe Guyeux\IEEEauthorrefmark{1}}
+\author{\IEEEauthorblockN{Rapha\"el Couturier\IEEEauthorrefmark{1}, Lilia Ziane Khodja\IEEEauthorrefmark{2}, and Christophe Guyeux\IEEEauthorrefmark{1}}
\IEEEauthorblockA{\IEEEauthorrefmark{1} Femto-ST Institute, University of Franche Comte, France\\
Email: \{raphael.couturier,christophe.guyeux\}@univ-fcomte.fr}
\IEEEauthorblockA{\IEEEauthorrefmark{2} INRIA Bordeaux Sud-Ouest, France\\
\label{sec:04}
Let us recall the following result, see~\cite{Saad86}.
\begin{proposition}
+\label{prop:saad}
Suppose that $A$ is a positive real matrix with symmetric part $M$. Then the residual norm provided at the $m$-th step of GMRES satisfies:
\begin{equation}
||r_m|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_0|| ,
We can now claim that,
\begin{proposition}
-If $A$ is a positive real matrix and GMRES($m$) is used as solver, then the TSIRM algorithm is convergent.
+If $A$ is a positive real matrix and GMRES($m$) is used as solver, then the TSIRM algorithm is convergent. Furthermore, we still have
+\begin{equation}
+||r_m|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_0|| ,
+\end{equation}
+where $\alpha$ and $\beta$ are defined as in Proposition~\ref{prop:saad}.
\end{proposition}
\begin{proof}
Let $r_k = b-Ax_k$, where $x_k$ is the approximation of the solution after the
$k$-th iterate of TSIRM.
-We will prove that $r_k \rightarrow 0$ when $k \rightarrow +\infty$.
+We will prove by a mathematical induction that, for each $k \in \mathbb{N}^\ast$,
+$||r_m|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_0||.$
+
+The base case is obvious, as for $k=1$, the TSIRM algorithm simply consists in applying GMRES($m$) once, leading to a new residual $r_1$ which follows the inductive hypothesis due to Proposition~\ref{prop:saad}.
-Each step of the TSIRM algorithm \\
+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 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,\\
$\min_{\alpha \in \mathbb{R}^s} ||b-R\alpha ||_2 = \min_{\alpha \in \mathbb{R}^s} ||b-AS\alpha ||_2$
$\begin{array}{ll}
-& = \min_{x \in Vect\left(x_0, x_1, \hdots, x_{k-1} \right)} ||b-AS\alpha ||_2\\
-& \leqslant \min_{x \in Vect\left( S_{k-1} \right)} ||b-Ax ||_2\\
-& \leqslant ||b-Ax_{k-1}||
+& = \min_{x \in span\left(S_{k-s}, S_{k-s+1}, \hdots, S_{k-1} \right)} ||b-AS\alpha ||_2\\
+& = \min_{x \in span\left(x_{k-s}, x_{k-s}+1, \hdots, x_{k-1} \right)} ||b-AS\alpha ||_2\\
+& \leqslant \min_{x \in span\left( x_{k-1} \right)} ||b-Ax ||_2\\
+& \leqslant \min_{\lambda \in \mathbb{R}} ||b-\lambda Ax_{k-1} ||_2\\
+& \leqslant ||b-Ax_{k-1}||_2 .
\end{array}$
+\end{itemize}
\end{proof}
+We can remark that, at each iterate, the residue of the TSIRM algorithm is lower
+than the one of the GMRES method.
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