Generalized Notion of Krylov Subspaces

Question

Let $\mathcal{X}$ be a vector space over a field $\mathbb{K}$ and let $x_0 \in \mathcal{X} \setminus \{0_{\mathcal{X}}\}$ (here, $0_{\mathcal{X}}$ denotes the zero vector in $\mathcal{X}$). We denote by $\textrm{End}(\mathcal{X}),$ the following set: $$ \textrm{End}(\mathcal{X}) = \{ T : \mathcal{X} \rightarrow \mathcal{X} \, | \, T~\textrm{is linear} \}. $$ Traditionally (and as defined by J. S. Golan in "The Linear Algebra a Beginning Graduate Student Ought to Know" (3ed) - Page 297), for any $T \in \textrm{End}(\mathcal{X})$ with $T^{0} \equiv \textrm{Id}_{\mathcal{X}}$ (the identity operator on $\mathcal{X}$), the subspace $$ \mathscr{K}(x_0, T) = \textrm{span}\{ x_0, T x_0, T^{2} x_0, \dots \} = \textrm{span}\{ T^{n} x_0 \}_{n \in \mathbb{N}} $$ is called the $\textit{Krylov subspace}$ of $\mathcal{X}$ defined by $T$ and $x_0$ (here, for any $n \in \mathbb{N} \setminus \{ 0 \},$ $T^{k} \equiv \underset{k~\textrm{times}}{\underbrace{T \circ T \circ \cdots \circ T}}$). Additionally, for any strictly positive integer $m,$ we also write

$$\mathscr{K}_{m}(x_0, T) = \textrm{span}\{ x_0, T x_0, T^{2} x_0, \dots, T^{m-1} x_0 \} \subset \mathcal{X}.$$

$\textbf{Question}$:

Is there a generalized notion of this collection? For example, if I relax the assumption of $\mathcal{X}$ being a vector space to $\mathcal{X}$ being a free module, then can I say something about the set $\mathscr{K}(x_{0}, T)$ (is it a submodule in this case)? Do a significant majority (if not all) of the expected properties that we see in the former case carry over to the latter? Any assistance in understanding this topic (whether it be a definitive answer, or references that I could read to figure these things out for myself) is deeply appreciated.

Answer 1

I don't know if this will be sufficient to you, but there are no such generalization (that I know) of Krylov spaces to modules. Why? I can think of several reasons:

• What would be the purpose of these? Are they useful for reduction? Is reduction even a thing for modules endomorphism?
• A ring $R$ is not necessarily a domain, hence you don't have factorisation of polynomials in it. Let us suppose we add this hypothesis ($ab = 0 \implies a = 0 \text{ or } b = 0$).
• Defining the determinant requires us to have a notion of dimension and a bimodule $M$ (and not just module) then we would have $$\det: \bigwedge\nolimits_R^n (M) \to R$$ Which would require a choice, but the choice would not be relevant since our ring is a domain and we only need to be able to tell when it is zero.
• Imagining that it is the case, we would have well defined characteristic function but would it be a polynomial? One should look into the $R$-algebra $\operatorname{End}(M)$, it may be possible but it is to check.
• Even so, a Krylov space is defined with $x \in V$ and $T \in \hom(V,V)$ by $$K(x,T) : \operatorname{Span}\{T^n x , \, n \in \mathbb{N}\}$$ We can choose a finite number of these spanning vectors since we have a finite dimension, but replace everything with module vocabulary, you get $K(x,T)$ as a submodule of a finite rank free module $V$, but it is not necessarily finitely generated and even if it is, it doesn't necessarily have a basis, since it is not necessarily free.

In summary, a lot of reasons, but maybe you could try to invent a new theory that takes the things into account? I can't know for sure.