In exact arithmetic, CG residuals are mutually orthogonal and the search directions are $A$-conjugate

We prove by induction on $k\ge 0$ that $r_0,\dots ,r_k$ are pairwise orthogonal and $p_0,\dots ,p_k$ are pairwise $A$-conjugate. For $k=0$ there is nothing to prove. Assume the claim through step $k$. Because $p_j=r_j+{\beta }_{j-1}{p}_{j-1}$ for $j\ge 1$ and $r_j=p_j-{\beta }_{j-1}{p}_{j-1}$, the spans $\mathrm{span}\{p_0,\dots ,p_j\}$ and $\mathrm{span}\{r_0,\dots ,r_j\}$ agree for every $j$.

For $i<k$, the induction hypothesis and step 1.1 give $⟨r_i,r_{k+1}⟩=⟨r_i,r_k⟩-{\alpha }_k⟨r_i,A{p}_k⟩=-{\alpha }_k⟨r_i,p_k{⟩}_A=0,$ using [L2], because $r_i$ lies in $\mathrm{span}\{p_0,\dots ,p_i\}$ and $p_k$ is $A$-conjugate to each of those earlier directions. Also, $⟨r_k,r_{k+1}⟩=⟨r_k,r_k⟩-{\alpha }_k⟨r_k,p_k{⟩}_A\mathrm{.}$ Since $r_k=p_k-{\beta }_{k-1}{p}_{k-1}$ for $k\ge 1$, the induction hypothesis gives $⟨r_k,p_k{⟩}_A=⟨p_k,p_k{⟩}_A$, and for $k=0$ the same identity is immediate. Using [F1], we obtain $⟨r_k,r_{k+1}⟩=⟨r_k,r_k⟩-{\alpha }_k⟨p_k,p_k{⟩}_A=0$. Thus $r_{k+1}$ is orthogonal to every earlier residual.

For $i<k$, Hermitian symmetry and [F1] give $p_i^{\ast }A{p}_{k+1}=p_i^{\ast }A{r}_{k+1}+{\beta }_k{p}_i^{\ast }A{p}_k=(A{p}_i{)}^{\ast }{r}_{k+1}\mathrm{.}$ Because $A{p}_i={\alpha }_i^{-1}(r_i-r_{i+1})$ from [F1], step 2.1 implies $(A{p}_i{)}^{\ast }{r}_{k+1}=0$, so $p_i^{\ast }A{p}_{k+1}=0$. For $i=k$, step 2.1 yields $⟨r_k,r_{k+1}⟩=0$, hence $p_k^{\ast }A{r}_{k+1}=(A{p}_k{)}^{\ast }{r}_{k+1}={\alpha }_k^{-1}(r_k-r_{k+1}{)}^{\ast }{r}_{k+1}=-{\alpha }_k^{-1}⟨r_{k+1},r_{k+1}⟩\mathrm{.}$ Therefore $p_k^{\ast }A{p}_{k+1}=p_k^{\ast }A{r}_{k+1}+{\beta }_k{p}_k^{\ast }A{p}_k=-{\alpha }_k^{-1}⟨r_{k+1},r_{k+1}⟩+{\beta }_k{p}_k^{\ast }A{p}_k=0,$ because ${\beta }_k=⟨r_{k+1},r_{k+1}⟩\mathrm{/}⟨r_k,r_k⟩$ and ${\alpha }_k=⟨r_k,r_k⟩\mathrm{/}(p_k^{\ast }A{p}_k)$ by [F1], with positivity from [L1] ensuring these divisions are valid before convergence. This closes the induction.