Two conjugate-gradient steps close an SPD two-by-two residual

Let \[ A=\begin{pmatrix}4&1\\1&3\end{pmatrix},\qquad b=\begin{pmatrix}1\\2\end{pmatrix}, \] and consider the linear system $Ax=b$ in exact arithmetic. The conjugate-gradient iteration from $x_0=0$ is \begin{align*} r_0&=b-Ax_0,\qquad p_0=r_0,\\ \alpha_k&=\frac{r_k^\top r_k}{p_k^\top Ap_k},\qquad x_{k+1}=x_k+\alpha_k p_k,\qquad r_{k+1}=r_k-\alpha_k Ap_k,\\ \beta_{k+1}&=\frac{r_{k+1}^\top r_{k+1}}{r_k^\top r_k},\qquad p_{k+1}=r_{k+1}+\beta_{k+1}p_k, \end{align*} stopping when $r_{k}=0$. 1. Prove that $A$ is symmetric positive definite by leading minors (or by eigenvalues). Compute those eigenvalues as $(7\pm\sqrt{5})/2$. 2. Starting from $x_0=0$, compute $r_0$, $p_0$, $\alpha_0$, $x_1$, $r_1$, $\beta_1$, $p_1$, $\alpha_1$, $x_2$, and $r_2$ as exact rationals. Verify $Ax_2=b$. 3. Verify the two identities $r_1^\top r_0=0$ and $p_0^\top Ap_1=0$. Then use the stated exact-arithmetic CG orthogonality theorem to explain why a general SPD $n\times n$ problem terminates with $r_n=0$ after at most $n$ steps. 4. Audit both sentences: (i) "CG is just steepest descent, so the next search direction is $p_1=r_1$ and $\beta_1$ may be ignored"; (ii) "the residuals are $A$-orthogonal, while the search directions are Euclidean-orthogonal." Compute the rejected steepest-descent vector $r_1$ and show that it is not $A$-orthogonal to $p_0$. Do not factor $A$ by Gaussian elimination, and do not replace the iteration by a nonlinear Newton update.

The leading minors of $A$ are $4$ and $\det A=11$. The residual $r_0$ is $b$ itself because $x_0=0$.