Conditions on a matrix for a sequence to converge

Question

Let $x_1 \in \mathbb{R}^n$ a non-zero vetor and $A \in M_n(\mathbb{R})$ a fixed $n \times n$ matrix. If we define a sequence $\{x_k\}_{k \in \mathbb{N}}$ by $x_{k+1} = A x_k$ for $k = 1, 2, ...$, then what kind of condition can I put on $|A| = \sum_j |a_j^i|$ such that the sequence converges to $x = 0$?

What I thought was something like this: If we use the max norm in $\mathbb{R}^n$, then we have

$|x_{k+1}|_{\max} = \max |x^i_{k+1}| \leq \sum_j |a^i_j| \max|x^j_k|\leq \sum_j |a^i_j|^2 \max |x^j_{k-1}| \leq \dots \leq \sum_j |a^i_j|^k \max|x^j_1|$

so, given $\epsilon > 0$ we can take $N$ such that for $k > N$, we have $\sum_j |a^i_j|^k c < \epsilon$, where $c = \max |x^j_1|$ is a positive constant. Then using the fact that $|A^k| \leq |A|^k$, then we find that $\sum_j |a^i_j| \leq (\frac{\epsilon}{c})^{1/k}$ for all $i=1,2...$

But this doesn't seem a good condition on the matrix norm. Is there a simpler condition I am missing?

Answer 1

The necessary and sufficient condition is that the spectral radius of $A$ is less than $1$. Equivalently, there exists a positive integer $n$ and a submultiplicative matrix norm $\|\cdot \|$ such that $\|A^n\|<1$.

If you are looking for an easy-to-check sufficient condition, it's reasonable to consider the induced norms for either $\ell^1$- or $\ell^\infty$- vector norms. These are $$ \|A\|_1 = \max_{j}\sum_{i}|a_{ij}|\quad\text{ and }\quad \|A\|_\infty = \max_{i}\sum_{j}|a_{ij}| $$ If either of these norms is less than $1$, then the inequality $\|Ax\|_p\le \|A\|_p\|x\|_p$ (with $p$ being $1$ or $\infty$) yields convergence to zero.