 Solution

At $u_0=0$, the linear-system residual is $r_0=b=(1,1{)}^T$. Here $A{r}_0=r_0$, so the steepest-descent step size isConsequently $u_1=(1,1{)}^T$, and $A{u}_1=b$ gives residual zero. The stated exact-arithmetic stopping test with $ϵ=0$ is met after one iteration. Stop there rather than trying to evaluate the next, undefined $0/0$ step. This is the special case where the initial residual is an eigenvector.