Does the k-th approximate solution of a stationary iteration belong to the k-th Krylov subspace?

Question

For an stationary iteration method solving $Ax=b$ as follows: $$ Mx_k = Nx_{k-1}+b, $$ I have known that when $M = I$, i.e., the Richardson iteration, the k-th solution $x_k = x_{k-1}+r_{k-1}$ is in the k-th Krylov subspace $K_k(A,b)$ with $x_0=0$. So, if we use a Krylov subspace method e.g., GMRES, the k-th solution $x_k$ not only belongs to the k-th Krylov subspace $K_k(A,b)$, but also minimize the k-th residual over this subspace. From this, we know that GMRES automatically chooses the best solution from the k-th Krylov subspace. So, it will perform "better" in some sense, than the stationary Richardson iteration method (because Richardson iteration just produces a solution lying in the same subspace, not minimizes some function), right?

My question is when the matrix $M$ in the stationary iterative method $$Mx_k =Nx_{k-1}+b $$ is not taken as the identity matrix, e.g., Jacobi iteration, $M = diag(A)$, then the k-th solution are as follows: $$x_k = x_{k-1}+M^{-1}r_{k-1}.$$ Does in this case, the $x_k$ still belong to the k-th subspace $K_k(A,b)$? If so, then Jacobi like Richardson iteration just produces a solution lying in the k-th Krylov subspace and not minimizes some function in this subspace, either. So, it still performs worse than gmres. If not, which subspace does the k-th solution $x_k$ obtained (from Jacobi iteration) lie in? Is the preconditioned subspace $K_k(M^{-1}A,b)$? any suggestions are welcome.

Answer 1

Taking $x_0 = 0$, we have that $x_1 \in <M^{-1}b>$.

For the next iteration, we get $x_2 \in <M^{-1}b, M^{-1}AM^{-1}b>$.

For the next iteration, we get $x_3 \in <M^{-1}b, (M^{-1}A)^2M^{-1}b>$.

Continuing this argument, you will eventually find that $x_k \in \mathcal{K}_k(M^{-1}A,M^{-1}b)$.

This space can also be described as $Mx_k \in \mathcal{K}_k(AM^{-1},b)$. This is the space a Krylov method like GMRES will search in when using right preconditioning.