+\r
+\r
+\subsection{$\mathfrak{R}(A)$ is positive definite}\r
+\r
+\begin{proposition}\r
+\label{prop2}\r
+Convergence of the TSIRM algorithm is at least linear when $\mathfrak{R}(A)$ is \r
+positive definite. Furthermore, the rate of convergence is lower \r
+than $$\min\left( \left(1- \dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{ \lambda_{min}^{\mathfrak{R}(A)} \lambda_{max}^{\mathfrak{R}(A)} + {\lambda_{max}^{\mathfrak{I}(A)}}^2}\right)^{\frac{m}{2}}; \r
+\left(1-\dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{||A||^2}\right)^{\frac{m}{2}}\right) ,$$\r
+where ${\lambda_{min}^{X}}$ (resp. ${\lambda_{max}^{X}}$) is the lowest (resp. largest) eigenvalue of matrix $X$.\r
+\end{proposition}\r
+\r
+\r
+\begin{proof}\r
+If $\mathfrak{R}(A)$ is positive definite, then it is positive, and so the TSIRM algorithm\r
+is convergent due to Proposition~\ref{positiveConvergent}.\r
+\r
+Furthermore, as stated in the proof of Proposition~\ref{positiveConvergent}, the GMRES residue is under control \r
+when $\mathfrak{R}(A)$ is positive. More precisely, it has been proven in the literature that the residual norm \r
+provided at the $m$-th step of GMRES satisfies:\r
+\begin{enumerate}\r
+\item $||r_m|| \leqslant \left(1- \dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{ \lambda_{min}^{\mathfrak{R}(A)} \lambda_{max}^{\mathfrak{R}(A)} + {\lambda_{max}^{\mathfrak{I}(A)}}^2}\right)^{\frac{mk}{2}} ||r_0||$, see, \emph{e.g.},~\cite{citeulike:2951999}, \r
+\item $||r_m|| \leqslant \left(1-\dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{||A||^2}\right)^{\frac{mk}{2}} ||r_0||$, see~\cite{ANU:137201},\r
+\end{enumerate}\r
+which proves the convergence of GMRES($m$) for all $m$ under such assumptions regarding $A$.\r
+\r
+We will now prove by a mathematical induction, and following the same canvas than in the proof of Prop.~\ref{positiveConvergent}, that: for each $k \in \mathbb{N}^\ast$, the TSIRM-residual norm satisfies\r
+\begin{equation}\r
+\label{induc}\r
+\begin{array}{ll}\r
+||r_k|| \leqslant & \min\left( \left(1- \dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{ \lambda_{min}^{\mathfrak{R}(A)} \lambda_{max}^{\mathfrak{R}(A)} + {\lambda_{max}^{\mathfrak{I}(A)}}^2}\right)^{\frac{m}{2}}; \right. \\\r
+& \left. \left(1-\dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{||A||^2}\right)^{\frac{m}{2}}\right) ||r_0||\r
+\end{array}\r
+\end{equation}\r
+when $A$ is positive definite.\r
+\r
+\r
+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 in the items listed above.\r
+\r
+Suppose now that the claim holds for all $u=1, 2, \hdots, k-1$, that is, $\forall u \in \{1,2,\hdots, k-1\}$, $||r_u|| \leqslant \left(1-\dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{||A||^2}\right)^{\frac{mu}{2}} ||r_0||$.\r
+We will show that the statement holds too for $r_k$. Two situations can occur:\r
+\begin{itemize}\r
+\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 \r
+$||r_k|| \leqslant \left(1- \dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{ \lambda_{min}^{\mathfrak{R}(A)} \lambda_{max}^{\mathfrak{R}(A)} + {\lambda_{max}^{\mathfrak{I}(A)}}^2}\right)^{\frac{m}{2}} \leqslant \left(1- \dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{ \lambda_{min}^{\mathfrak{R}(A)} \lambda_{max}^{\mathfrak{R}(A)} + {\lambda_{max}^{\mathfrak{I}(A)}}^2}\right)^{\frac{mk}{2}} ||r_0||$, due to~\cite{citeulike:2951999}. Furthermore, we have too that: $||r_k|| \leqslant \left(1-\dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{||A||^2}\right)^{\frac{m}{2}} ||r_{k-1}|| \leqslant \left(1-\dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{||A||^2}\right)^{\frac{mk}{2}} ||r_0||$, as proven in~\cite{ANU:137201} and by using the inductive hypothesis. So we can conclude that \r
+$$\begin{array}{ll}||r_k|| \leqslant & \min\left( \left(1- \dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{ \lambda_{min}^{\mathfrak{R}(A)} \lambda_{max}^{\mathfrak{R}(A)} + {\lambda_{max}^{\mathfrak{I}(A)}}^2}\right)^{\frac{mk}{2}}; \right. \\\r
+& \left. \left(1-\dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{||A||^2}\right)^{\frac{mk}{2}}\right) \times ||r_0|| \r
+\end{array}.$$\r
+\r
+\item Else, the TSIRM algorithm consists in two stages: a first GMRES($m$) execution leads to a temporary $x_k$ whose residue satisfies, following the previous item:\r
+$$\begin{array}{ll}\r
+||r_k|| & \leqslant \min\left( \left(1- \dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{ \lambda_{min}^{\mathfrak{R}(A)} \lambda_{max}^{\mathfrak{R}(A)} + {\lambda_{max}^{\mathfrak{I}(A)}}^2}\right)^{\frac{m}{2}}; \right. \\\r
+& \left. \left(1-\dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{||A||^2}\right)^{\frac{m}{2}}\right) \times ||r_{k-1}||\\\r
+ & \leqslant \min\left( \left(1- \dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{ \lambda_{min}^{\mathfrak{R}(A)} \lambda_{max}^{\mathfrak{R}(A)} + {\lambda_{max}^{\mathfrak{I}(A)}}^2}\right)^{\frac{mk}{2}}; \right. \\\r
+& \left. \left(1-\dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{||A||^2}\right)^{\frac{mk}{2}}\right) \times ||r_0|| \r
+\end{array}$$\r
+and the least squares resolution of $\min_{\alpha \in \mathbb{R}^s} ||b-R\alpha ||_2$.\r
+\r
+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 $S$, as defined previously. So,\\\r
+$\min_{\alpha \in \mathbb{R}^s} ||b-R\alpha ||_2 = \min_{\alpha \in \mathbb{R}^s} ||b-AS\alpha ||_2$\r
+\r
+$\begin{array}{ll}\r
+& = \min_{x \in span\left(S_{k-s+1}, S_{k-s+2}, \hdots, S_{k} \right)} ||b-AS\alpha ||_2\\\r
+& = \min_{x \in span\left(x_{k-s+1}, x_{k-s}+2, \hdots, x_{k} \right)} ||b-AS\alpha ||_2\\\r
+& \leqslant \min_{x \in span\left( x_{k} \right)} ||b-Ax ||_2\\\r
+& \leqslant \min_{\lambda \in \mathbb{R}} ||b-\lambda Ax_{k} ||_2\\\r
+& \leqslant ||b-Ax_{k}||_2\\\r
+& = ||r_k||_2\\\r
+& \leqslant \min\left( \left(1- \dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{ \lambda_{min}^{\mathfrak{R}(A)} \lambda_{max}^{\mathfrak{R}(A)} + {\lambda_{max}^{\mathfrak{I}(A)}}^2}\right)^{\frac{mk}{2}}; \right. \\\r
+& \left. \left(1-\dfrac{{\lambda_{min}^{\mathfrak{R}(A)}}^2}{||A||^2}\right)^{\frac{mk}{2}}\right) \times ||r_0|| \r
+\end{array} .$\r
+\end{itemize}\r
+due to the inductive hypothesis. \r
+So the statement of Equation~\eqref{induc} holds too for the $k$-th iterate, which concludes the induction and the proof.\r
+\end{proof}\r
+\r
+\subsection{A last linear convergence}\r
+\r
+\r
+\begin{proposition}\r
+Let us define the field of values of $A$ by \r
+$$\mathfrak{F}(A) = \left\{ \dfrac{x^\ast A x}{x^\ast x}, x \in \mathds{C}^n\setminus \{0\} \right\} .$$\r
+\r
+Then if $\mathfrak{F}(A)$ is included into a closed ball of radius $r$ and center $c$,\r
+which does not contain the origin, then the convergence of the TSIRM algorithm is at least linear. \r
+\r
+More precisely, the rate of convergence is lower \r
+than $2 \dfrac{r}{|c|}$.\r
+\end{proposition}\r
+\r
+\begin{proof}\r
+This inequality comes from the fact that, in the conditions of the proposition, the GMRES residue \r
+satisfies the inequality: $|r_k| \leqslant 2 \dfrac{r}{|c|}^k |r_0|$. An induction inspired by\r
+the proofs of Propositions~\ref{prop:saad} and~\ref{prop2} can transfer this inequality to the\r
+TSIRM residue.\r
+\end{proof}\r
+\r
+\r
+\r