 Solution

Since $\nabla F(x^{(k)})=-r^{(k)}$, the Heavy-ball method isMultiplication by $A$ and use of $A{x}^{(j)}=b-r^{(j)}$ give the heavy-ball residual recurrenceWith the usual initialization $x^{(-1)}=x^{(0)}=0$, both initial residuals equal $b$. Induction in the recurrence shows thatfor a polynomial $p_k$ of degree at most $k$. ThereforeThis is ${\mathcal{K}}_k(A,b)$ under the zero-indexed Krylov subspace convention in the question; under the convention whose order-$j$ space ends at $A^{j-1}b$, it is ${\mathcal{K}}_{k+1}(A,b)$.