How to show that $x_{n+1} - x_* = M(x_n) (x_n - x_*)$ dynamics converges?

Question

I am trying to find conditions of convergence (or non-convergence) of a system that behaves in the following manner (quasi-linear since the matrix is not stationary):

$$ x_{n+1} - x_* = M(x_n) (x_n - x_*), \text{ where } x_{n+1}, x_n, x_* \in [0,1]^k, M(x_n) \in \mathbb{R}^{k \times k}. $$

My question is: what are the (weakest) conditions (on $M(x_n)$) under which I can claim that this system converges, i.e., $x_n \to x_*$?

There are two special structures in this problem: (a) $x_n$'s are always bounded, and (b) $M(x_n)$ has first column to be zeros always (this is because of the structure of my problem), hence one eigenvalue is always zero. I have tried various conditions, and I am looking for the weakest condition that makes it convergent.

1. A simple condition would be if the eigenvalues $\lambda_i(x_n)$ of $M(x_n)$ be distinct and $|\lambda_i(x_n)| < 1, \forall x_n$ (since the matrix is non-stationary, so will be the eigenvalues). I must mention that in this setup, $M(x_n)$ can be asymmetric. I guess this can be shown in many ways - either via the matrix norm being $<1$ or decomposing every vector $x_n - x_*$ into the linearly independent eigenvectors corresponding to the eigenvalues. I can explain more if needed, but unfortunately this condition does not always hold in the problem that I'm considering.
2. Considering an additional structure in my problem, i.e., $x_n$'s are always bounded, I suppose showing non-negative eigenvalues should also be fine -- this is similar to (in 1-dim) showing that if a function $f$ is continuous, increasing and bounded, then the fixed point iteration $x_{n+1}=f(x_n)$ converges to a fixed point. I feel that my problem tries to find a multidimensional fixed point.

Any pointers to study reading materials that consider quasi-linear systems of this kind will be appreciated. I looked up the literature that uses Lyapunov functions for stability -- is there an obvious Lyapunov function that can be constructed for this dynamics?

Answer 1

One simple sufficient condition is that each $M(x)$ is a strict contraction in the Euclidean norm. This is equivalent to all the singular values of $M(x)$ being strictly less than $1$.

EDIT: It is not sufficient to have all eigenvalues less than $1$.
An interesting example is

$$x_\star = \pmatrix{0\cr 0\cr},\ M\pmatrix{1 \cr 0\cr} = \pmatrix{0 & 0\cr 1 & 1/2\cr}, \ M\pmatrix{0 \cr 1\cr} = \pmatrix{1/2 & 1\cr 0 & 0\cr} $$

All these matrices have eigenvalues $0$ and $1/2$, but you get a 2-cycle: $$ x_0 = \pmatrix{1\cr 0\cr},\ x_1 = \pmatrix{0 & 0\cr 1 & 1/2\cr} \pmatrix{1\cr 0\cr} = \pmatrix{0\cr 1\cr},\ x_2 = \pmatrix{1/2 & 1\cr 0 & 0\cr} \pmatrix{0\cr 1\cr} = \pmatrix{1\cr 0\cr},\ \text{etc}$$