[step:Show $\mathbf{e}^{(k)}$ is the $A$-orthogonal projection of $\mathbf{e}^{(0)}$ away from $K_k(A, \mathbf{r}^{(0)})$]From the [Error Projection Along Search Direction](/theorems/1398), the error update satisfies $\mathbf{e}^{(j+1)} = \mathbf{e}^{(j)} - \alpha_j \mathbf{d}^{(j)}$ for each $j = 0, \ldots, k-1$. Summing these telescopically:
\begin{align*}
\mathbf{e}^{(0)} - \mathbf{e}^{(k)} = \sum_{j=0}^{k-1} \alpha_j \mathbf{d}^{(j)} \in \operatorname{span}\{\mathbf{d}^{(0)}, \ldots, \mathbf{d}^{(k-1)}\}.
\end{align*}
By the [Properties of the Conjugate Gradient Method](/theorems/1400), $\operatorname{span}\{\mathbf{d}^{(0)}, \ldots, \mathbf{d}^{(k-1)}\} = K_k(A, \mathbf{r}^{(0)})$. Moreover, by the [Finite Termination of Conjugate Direction Methods](/theorems/1399), $\langle \mathbf{e}^{(k)}, \mathbf{d}^{(j)} \rangle_A = 0$ for all $j = 0, \ldots, k-1$, which means $\mathbf{e}^{(k)}$ is $A$-orthogonal to $K_k(A, \mathbf{r}^{(0)})$.
This is exactly the characterisation of the $A$-orthogonal projection: $\mathbf{e}^{(0)} - \mathbf{e}^{(k)} \in K_k(A, \mathbf{r}^{(0)})$ and $\mathbf{e}^{(k)} \perp_A K_k(A, \mathbf{r}^{(0)})$. By the Pythagorean theorem in the $A$-inner product, for any $\mathbf{v} \in K_k(A, \mathbf{r}^{(0)})$:
\begin{align*}
\|\mathbf{e}^{(0)} - \mathbf{v}\|_A^2 = \|\mathbf{e}^{(k)}\|_A^2 + \|(\mathbf{e}^{(0)} - \mathbf{e}^{(k)}) - \mathbf{v}\|_A^2 \geq \|\mathbf{e}^{(k)}\|_A^2,
\end{align*}
with equality iff $\mathbf{v} = \mathbf{e}^{(0)} - \mathbf{e}^{(k)}$. Therefore:
\begin{align*}
\|\mathbf{e}^{(k)}\|_A = \min_{\mathbf{v} \in K_k(A, \mathbf{r}^{(0)})} \|\mathbf{e}^{(0)} - \mathbf{v}\|_A.
\end{align*}[/step]