+\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_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{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 that assumption regarding $A$.
+Such 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}.
+
+Suppose now that the claim holds for all $m=1, 2, \hdots, k-1$, that is, $\forall m \in \{1,2,\hdots, k-1\}$, $||r_m|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||$.
+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.
+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,\\
+$\min_{\alpha \in \mathbb{R}^s} ||b-R\alpha ||_2 = \min_{\alpha \in \mathbb{R}^s} ||b-AS\alpha ||_2$
+
+$\begin{array}{ll}
+& = \min_{x \in span\left(S_{k-s+1}, S_{k-s+2}, \hdots, S_{k} \right)} ||b-AS\alpha ||_2\\
+& = \min_{x \in span\left(x_{k-s+1}, x_{k-s}+2, \hdots, x_{k} \right)} ||b-AS\alpha ||_2\\
+& \leqslant \min_{x \in span\left( x_{k} \right)} ||b-Ax ||_2\\
+& \leqslant \min_{\lambda \in \mathbb{R}} ||b-\lambda Ax_{k} ||_2\\
+& \leqslant ||b-Ax_{k}||_2\\
+& = ||r_k||_2\\
+& \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||,
+\end{array}$
+\end{itemize}
+which concludes the induction and the proof.
+\end{proof}
