+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, 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 $k \rightarrow +\infty$.
+
+Each step of the TSIRM algorithm \\
+
+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 vectors $S$. So,\\
+$\min_{\alpha \in \mathbb{R}^s} ||b-R\alpha ||_2 = \min_{\alpha \in \mathbb{R}^s} ||b-AS\alpha ||_2$
+
+$\begin{array}{ll}
+& = \min_{x \in span\left(S_{k-s}, S_{k-s+1}, \hdots, S_{k-1} \right)} ||b-AS\alpha ||_2\\
+& = \min_{x \in span\left(x_{k-s}, x_{k-s}+1, \hdots, x_{k-1} \right)} ||b-AS\alpha ||_2\\
+& \leqslant \min_{x \in span\left( x_{k-1} \right)} ||b-Ax ||_2\\
+& \leqslant \min_{\lambda \in \mathbb{R}} ||b-\lambda Ax_{k-1} ||_2\\
+& \leqslant ||b-Ax_{k-1}||_2 .
+\end{array}$
+\end{proof}
+
+We can remark that, at each iterate, the residue of the TSIRM algorithm is lower
+than the one of the GMRES method.
