+\section{Convergence results}
+\label{sec:04}
+
+
+We can now claim that,
+\begin{proposition}
+\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 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 $M$ 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}
+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 $r_1$ 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, $\forall m \in \{1,2,\hdots, k-1\}$, $||r_m|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||$ 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 $k \not\equiv 0 ~(\textrm{mod}\ m)$, 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,\\
+$\min_{\alpha \in \mathbb{R}^s} ||b-R\alpha ||_2 = \min_{\alpha \in \mathbb{R}^s} ||b-AS\alpha ||_2$
+
+$\begin{array}{ll}
+& = \min_{x \in span\left(S_{k-s+1}, S_{k-s+2}, \hdots, S_{k} \right)} ||b-AS\alpha ||_2\\
+& = \min_{x \in span\left(x_{k-s+1}, x_{k-s}+2, \hdots, x_{k} \right)} ||b-AS\alpha ||_2\\
+& \leqslant \min_{x \in span\left( x_{k} \right)} ||b-Ax ||_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||, \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}
