Conditions for $I-x A$ to be a convergent matrix for some $x\in \mathbb{R}$

Question

I'm looking for interpretable necessary/sufficient conditions on $A$ which guarantee that $(I-x A)$ is a convergent matrix for some $x\in \mathbb{R}$

For instance, $A$ normal with non-zero eigenvalues having real parts of same sign seems sufficient. What about non-normal $A$?

For instance, let $A=\left(\begin{matrix}1&2\\0&1\end{matrix}\right)$, we can visualize trajectories of $\left(I-x \left(\begin{matrix}1&2\\0&1\end{matrix}\right)\right)^k$ below.

$A$ satisfies this condition because of spectral radius condition, $\rho(I-xA)<1$ for any $0<x<2$

Following this it seems the example above is provably convergent because this map is contractive under coordinate change $y=Px$ with

$$P=\left( \begin{array}{cc} \frac{2 \sqrt{\frac{29}{3}}}{3} & -\frac{4}{\sqrt{87}} \\ 0 & \frac{10}{\sqrt{87}} \\ \end{array} \right)$$

Coordinate transformation $P$ under which $x=Ax$ becomes contractive for any stable $A$ can be obtained by solving the following (comes from discrete Lyapunov equation)

$$P^T P = T\left(\frac{T^{-1}A^{-T}U}{\lambda(A)-\lambda(A^{-1})^T}\right)U^{-1}$$

with $$\begin{align} T&=\text{columns of right eigenvectors of } A\\ U&=\text{columns of left eigenvectors of } A^{-1}\\ \lambda&=\text{column vector of eigenvalues}\\ a-b&=\text{subtraction with numpy broadcasting rules}\\ a/b&=\text{componentwise (Hadamard) division} \end{align} $$

notebook

Answer 1

Let $\{v_k\}$ be some orthonormal eigen-vectors and $\{\sigma_k\}$ the corresponding eigen-values of $A$, then for $n\geq 1$, $(I-xA)^n v_k=(I-xA)^{n-1}(v_k-x\sigma_k v_k)=(1-x\sigma_k )(I-xA)^{n-1}v_k=(1-x\sigma_k)^n v_k$. Now one can argue that since any vector $y$ is a combination of the $v_k$, then $(I-xA)^n y\to 0$ for all $y$ is true when it is true for the eigen-vectors, indeed if $y=\sum_{k} y_k v_k$, then $(I-xA)^n y=\sum_k y_k(1-x\sigma_k)^n v_k$.

In order for it to be true, you need to find a $x$ such that $|1-x\sigma_k|<1$ for all $k$. In particular $\min_x\max_k |1-x\sigma_k|<1$ iff there exists $x$ such that $I-xA$ is convergent.

Now this is equivalent to $\min_x\max_k |1-x\sigma_k|^2<1$ which in turn is equivalent to $\min_x\max_k (1-x2\mathrm{Re}\{\sigma_k\} +x^2|\sigma_k^2|)<1$ which can be rewritten as \begin{align*} 0>\min_x\max_k x(|\sigma_k|^2x-2\mathrm{Re}\{\sigma_k\}) \end{align*} In particular this will hold if and only if all of the (unordered) intervals $\left]0,2\frac{\mathrm{Re}\{\sigma_k\}}{|\sigma_k|^2} \right[$ intersect, meaning that $\mathrm{Re}\{ \sigma_k \}$ should have the same sign for all $k$ (and be non $0$).