Krylov subspace

Remarks

Equivalently, $\mathcal{K}_r(A, b) = \set{p(A) b : p \in F[t], \ \deg p < r}$. The subspaces grow, $\mathcal{K}_1 \subseteq \mathcal{K}_2 \subseteq \cdots$, and their dimensions increase by one until they stop growing. Once $\mathcal{K}_{r+1} = \mathcal{K}_r$, the subspace is invariant under $A$ and all later ones are equal to it. This happens at the least degree $d$ of a monic polynomial $p$ with $p(A) b = 0$. That degree is at most the degree of the minimal polynomial of $A$, hence at most $n$. If $A$ is invertible, then $A^{-1} b$ lies in $\mathcal{K}_n(A, b)$, since the minimal polynomial of $A$ then has nonzero constant term. Example: for $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and $b = e_2$, one has $A b = e_1$, so $\mathcal{K}_2(A, b) = F^2$. Non-example: for $A = I$, every Krylov subspace $\mathcal{K}_r(I, b)$ is the line through $b$ or $\set{0}$. So for $n \ge 2$ the whole space $F^n$ is not a Krylov subspace of the identity.