\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
+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}
\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 $k \rightarrow +\infty$.
+We will prove by a mathematical induction that, for each $k \in \mathbb{N}^\ast$,
+$||r_m|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_0||.$
-Each step of the TSIRM algorithm \\
+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}.
-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,\\
+Suppose now that the claim holds for all $m=1, 2, \hdots, k-1$, that is, $\forall m \in \{1,2,\hdots, k-1\}$, $||r_m|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_0||$.
+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 $||r_k|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_{k-1}||\leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_0||$.
+
+\item Else, 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,\\
$\min_{\alpha \in \mathbb{R}^s} ||b-R\alpha ||_2 = \min_{\alpha \in \mathbb{R}^s} ||b-AS\alpha ||_2$
$\begin{array}{ll}
& \leqslant \min_{\lambda \in \mathbb{R}} ||b-\lambda Ax_{k-1} ||_2\\
& \leqslant ||b-Ax_{k-1}||_2 .
\end{array}$
+\end{itemize}
\end{proof}
We can remark that, at each iterate, the residue of the TSIRM algorithm is lower
