Solution
= Solution
At $u_0=0$, the <linear-system residual> is $r_0=b=(1,1)^T$. Here $Ar_0=r_0$, so the <steepest-descent step size> is
$$
\frac{r_0^Tr_0}{r_0^TAr_0}=\frac22=1.
$$
Consequently \b[$u_1=(1,1)^T$], and $Au_1=b$ gives residual zero. The stated exact-arithmetic stopping test with $\epsilon=0$ 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>.