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 Incoming links: Heavy-ball method

Since

\nabla F (x^{(k)}) = - r^{(k)}

x^{(k + 1)} = x^{(k)} + α r^{(k)} + β (x^{(k)} - x^{(k - 1)}) .

A

and use of

A x^{(j)} = b - r^{(j)}

r^{(k + 1)} = ((1 + β) I - α A) r^{(k)} - β r^{(k - 1)} .

With the usual initialization

x^{(- 1)} = x^{(0)} = 0

, both initial residuals equal

b

. Induction in the recurrence shows that

r^{(k)} = p_{k} (A) b

p_{k}

of degree at most

k

. Therefore

r^{(k)} \in span {b, A b, \dots, A^{k} b} .

This is

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

K_{k + 1} (A, b)