 Solution

One restarted step chooses $x_j=x_{j-1}+{\alpha }_j{r}_{j-1}$ to minimize the new residual norm. Since $r_{j-1}=A(x-x_{j-1})$,and induction givesThese polynomials have degree at most one; a zero step gives the constant polynomial one.

In the specified example the first search space is spanned by $b=(1,1{)}^T$. Writing $x_1=\alpha b$, its squared residual norm is $(1-\alpha {)}^2+(1+\alpha {)}^2=2+2{\alpha }^2$, uniquely minimized at $\alpha =0$. Thus restarted GMRES gives $\boxed{x_1=x_2=(0,0{)}^T}$ and stagnates forever, repeatedly facing the same minimization problem.

Without restarting, the first iterate is still zero, but $K_2(A,b)=\mathrm{span}\{(1,1{)}^T,(1,-1{)}^T\}={\mathbb{R}}^2$. The second minimization finds the exact solution $\boxed{x_2=(1,-1{)}^T}$. This one-step restarted GMRES stagnation example shows how discarding the accumulated search space can prevent convergence even for a nonsingular two-dimensional system.