iterations phase. In the remainder, the $i$-th column vector of $S$ will be
denoted by $S_i$.
-$\|r_n\| \leq \left( 1-\frac{\lambda_{\mathrm{min}}^2(1/2(A^T + A))}{ \lambda_{\mathrm{max}}(A^T A)} \right)^{n/2} \|r_0\|,$
-In the general case, where A is not positive definite, we have
-
-$\|r_n\| \le \inf_{p \in P_n} \|p(A)\| \le \kappa_2(V) \inf_{p \in P_n} \max_{\lambda \in \sigma(A)} |p(\lambda)| \|r_0\|, \,$
-
-
At each $s$ iterations, another kind of minimization step is applied in order to
compute a new solution $x$. For that, the previous residuals of $Ax=b$ are computed by
the inner iterations with $(b-AS)$. The minimization of the residuals is obtained by
\section{Convergence results}
\label{sec:04}
-Let us recall the following result, see~\cite{Saad86} for further readings.
-\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|| ,
-\end{equation}
-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}
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. Furthermore,
-let $r_k$ be the
+\label{prop:saad}
+If $A$ is either a definite positive or a positive matrix and GMRES($m$) is used as solver, then the TSIRM algorithm is convergent.
+
+Furthermore, let $r_k$ be the
$k$-th residue of TSIRM, then
-we still have:
+we have the following boundaries:
+\begin{itemize}
+\item when $A$ is positive:
\begin{equation}
||r_k|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0|| ,
\end{equation}
-where $\alpha$ and $\beta$ are defined as in Proposition~\ref{prop:saad}.
+where $M$ is the symmetric part of $A$, $\alpha = \lambda_{min}(M)^2$ and $\beta = \lambda_{max}(A^T A)$;
+\item when $A$ is positive definite:
+\begin{equation}
+\|r_k\| \leq \left( 1-\frac{\lambda_{\mathrm{min}}^2(1/2(A^T + A))}{ \lambda_{\mathrm{max}}(A^T A)} \right)^{km/2} \|r_0\|.
+\end{equation}
+\end{itemize}
+%In the general case, where A is not positive definite, we have
+%$\|r_n\| \le \inf_{p \in P_n} \|p(A)\| \le \kappa_2(V) \inf_{p \in P_n} \max_{\lambda \in \sigma(A)} |p(\lambda)| \|r_0\|, .$
\end{proposition}
\begin{proof}
-We will prove by a mathematical induction that, for each $k \in \mathbb{N}^\ast$,
-$||r_k|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{mk}{2}} ||r_0||.$
+Let us first recall that the residue is under control when considering the GMRES algorithm on a positive definite matrix, and it is bounded as follows:
+\begin{equation*}
+\|r_k\| \leq \left( 1-\frac{\lambda_{\mathrm{min}}^2(1/2(A^T + A))}{ \lambda_{\mathrm{max}}(A^T A)} \right)^{k/2} \|r_0\| .
+\end{equation*}
+Additionally, when $A$ 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|| ,
+\end{equation*}
+where $\alpha$ and $\beta$ are defined as in Proposition~\ref{prop:saad}, which proves
+the convergence of GMRES($m$) for all $m$ under such assumptions regarding $A$.
+These well-known results can be found, \emph{e.g.}, in~\cite{Saad86}.
+
+We will now prove by a mathematical induction that, for each $k \in \mathbb{N}^\ast$,
+$||r_k|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{mk}{2}} ||r_0||$ when $A$ is positive, and $\|r_k\| \leq \left( 1-\frac{\lambda_{\mathrm{min}}^2(1/2(A^T + A))}{ \lambda_{\mathrm{max}}(A^T A)} \right)^{km/2} \|r_0\|$ when $A$ is positive definite.
-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}.
+The base case is obvious, as for $k=1$, the TSIRM algorithm simply consists in applying GMRES($m$) once, leading to a new residual $r_1$ that follows the inductive hypothesis due, to the results recalled above.
-Suppose now that the claim holds for all $m=1, 2, \hdots, k-1$, that is, $\forall m \in \{1,2,\hdots, k-1\}$, $||r_m|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||$.
+Suppose now that the claim holds for all $m=1, 2, \hdots, k-1$, that is, $\forall m \in \{1,2,\hdots, k-1\}$, $||r_m|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||$ in the positive case, and $\|r_k\| \leq \left( 1-\frac{\lambda_{\mathrm{min}}^2(1/2(A^T + A))}{ \lambda_{\mathrm{max}}(A^T A)} \right)^{km/2} \|r_0\|$ in the definite positive one.
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 the TSIRM algorithm consists in executing GMRES once. In that case, we obtain $||r_k|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_{k-1}||\leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||$ by the inductive hypothesis.
-\item Else, the TSIRM algorithm consists in two stages: a first GMRES($m$) execution leads to a temporary $x_k$ whose residue satisfies $||r_k|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_{k-1}||\leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||$, and a least squares resolution.
+\item If $k \not\equiv 0 ~(\textrm{mod}\ m)$, then the TSIRM algorithm consists in executing GMRES once. In that case and by using the inductive hypothesis, we obtain either $||r_k|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_{k-1}||\leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||$ if $A$ is positive, or $\|r_k\| \leqslant \left( 1-\frac{\lambda_{\mathrm{min}}^2(1/2(A^T + A))}{ \lambda_{\mathrm{max}}(A^T A)} \right)^{m/2} \|r_{k-1}\|$ $\leqslant$ $\left( 1-\frac{\lambda_{\mathrm{min}}^2(1/2(A^T + A))}{ \lambda_{\mathrm{max}}(A^T A)} \right)^{km/2} \|r_{0}\|$ in the positive definite case.
+\item Else, the TSIRM algorithm consists in two stages: a first GMRES($m$) execution leads to a temporary $x_k$ whose residue satisfies:
+\begin{itemize}
+\item $||r_k|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_{k-1}||\leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||$ in the positive case,
+\item $\|r_k\| \leqslant \left( 1-\frac{\lambda_{\mathrm{min}}^2(1/2(A^T + A))}{ \lambda_{\mathrm{max}}(A^T A)} \right)^{m/2} \|r_{k-1}\|$ $\leqslant$ $\left( 1-\frac{\lambda_{\mathrm{min}}^2(1/2(A^T + A))}{ \lambda_{\mathrm{max}}(A^T A)} \right)^{km/2} \|r_{0}\|$ in the positive definite one,
+\end{itemize}
+and a least squares resolution.
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,\\
$\min_{\alpha \in \mathbb{R}^s} ||b-R\alpha ||_2 = \min_{\alpha \in \mathbb{R}^s} ||b-AS\alpha ||_2$
& \leqslant \min_{\lambda \in \mathbb{R}} ||b-\lambda Ax_{k} ||_2\\
& \leqslant ||b-Ax_{k}||_2\\
& = ||r_k||_2\\
-& \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||,
+& \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||, \textrm{ if $A$ is positive,}\\
+& \leqslant \left( 1-\frac{\lambda_{\mathrm{min}}^2(1/2(A^T + A))}{ \lambda_{\mathrm{max}}(A^T A)} \right)^{km/2} \|r_{0}\|, \textrm{ if $A$ is}\\
+& \textrm{positive definite,}
\end{array}$
\end{itemize}
which concludes the induction and the proof.
\end{proof}
-We can remark that, at each iterate, the residue of the TSIRM algorithm is lower
-than the one of the GMRES method.
+%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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