Matrix to the infinite power shall be zero

Question

Hello fellow people here.

I found a problem, which started to haunt me: Defining the system: $$ e_{k+1} = A e_{k} $$ With $A \in \mathbb{R}^{n \times n}, e_k \in \mathbb{R}^n, e_0 \text{ given} $ (Hoping I do not mix here something up with the dimensions.

And now the problem: Which conditions must $A$ satisfy, such that: $\lim_{k\to \infty} ||e_k|| = 0$

I made the following considerations : $\lim_{k\to \infty} ||e_k|| = \lim_{n\to\infty} A^n = 0$ This due to the fact, that $e_{k+1} = A^k e_0$. So we have to make conditions for A. Am I missing something? But the problem is just starting here, what shall I do with that? This basically means, that the norm for A must be less then 1, and what shall I do with that? Which conditions must A fullfill, that this satisfied?

Thank you for your answer in advance.

$|A|<1$ is sufficient but not necessary. Nilpotent matrices also satisfy this property. — Kavi Rama Murthy, Dec 28 '21 at 09:12
If $|e_k| \to 0$ for all choices of $e_0$ then every eigen value $\lambda$ of $A$ necessarily has modulus less than $1$. — Kavi Rama Murthy, Dec 28 '21 at 09:40

unobservable_node · Accepted Answer · 2021-12-28T11:19:43.100

All the eigenvalues $\lambda_i$ of $A\in\mathbb{R}^{n\times n}$ must be such that $|\lambda_i|<1$, for $i=1,\dots,n$, is necessary and sufficient for $\|e_k\|\rightarrow 0$ as $k\rightarrow\infty$. That is, all the eigenvalues are inside a unit disk in the complex plane.

It is easy to see through the eigenvalue decomposition $$ A^k = V \Lambda^k V^{-1} $$ where $V$ is the matrix of eigenvectors and $\Lambda = \text{diag}(\lambda_1,\dots,\lambda_n)$. Therefore, $0\leq \|A^k\| \leq \|V\| \|\Lambda^k\| \|V^{-1}\|$ and $$ \lim_{k\rightarrow\infty} \|\Lambda^k\| = 0 $$ if and only if $|\lambda_i|<1$, for $i=1,\dots,n$. Similar arguments can be made using the Jordan normal form (see https://math.stackexchange.com/a/910658/442550).

Matrix to the infinite power shall be zero

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