# Does this transformation preserve the order of Krylov subspace?

Let us denote two sets of matrices: \begin{align*} \mathcal S = \{A \in M_n(\mathbb R): \rho(A) < 1\}, \end{align*} where $$\rho(\cdot)$$ denotes the spectral radius of a matrix and \begin{align*} \mathcal H = \{A \in M_n(\mathbb R): \max_{i} \text{Re}(\lambda_i(A)) < 0\}, \end{align*} i.e., matrices with eigenvalues lying on the left open half plane. It is not hard to see the transformation $$f: \mathcal S \to \mathcal H$$ given by $$A \mapsto (A-I)^{-1}(A+I)$$ is well-defined and indeed a diffeomorphism.

I am trying to determine whether $$f$$ preserves the order of Krylov subspace. More precisely, let $$A \in \mathcal S$$, $$v$$ be a vector and $$\text{Krov}(A, v) = \{v, Av, \dots, A^r v\}$$, where $$\{v, Av, \dots, A^rv\}$$ is a linearly independent set. Now would $$\{v, f(A)v, \dots, [f(A)]^r v\}$$ be a linearly independent set in general?