Consider the map $x_{i+1} = f(x_{i})$ and let $x^{*}$ be a fixed point then the fixed point is said to be Lyapunov stable if $\forall \epsilon > 0 $ there exist $\delta > 0$ such that $x_{i} \in B_{\epsilon}(x^{*}) \forall x_{0} \in B_{\delta}(x^{*}) $ and $\forall i \geq 0$ and

the fixed point is asymptotically stable if it is Lyapunov stable and there exist $\delta >0$ such that $x_{i} \rightarrow x^{*}$ as $i \rightarrow \infty \forall x_{0} \in B_{\delta}(x^{*})$.

Now we take the case of linear maps $x_{i+1} = Ax_{i}$ where $A$ is realvalued $n \times n$ matrix and for any $x_{0} \in \Bbb{R}^n$ we have $x_{i} = A^{i}x_{0}$

Then I was thinking of proving rigorously the following -

$1. $ If $|\lambda| <1$, for every eigenvalue of $A$ then $0$ is an asymptotically stable fixed point.

$2. $ If $|\lambda|>1$ for some eigenvalue $\lambda$ of $A$ then $0$ is not a Lyapunov stable fixed point $x_{i+1} = Ax_{i}$.

I think that if $x$ is an eigenvector of which $\lambda$ is a corresponding eigenvalue then due to factor coming from $\lambda^n$ which as $n \rightarrow \infty$ tends to zero so seems to be asymptotically stable but here $x_{0}$ is any real $n$ dimensional vector.

  • $\begingroup$ Consider $x_0$ in an eigenbasis. Or use if for the spectral radius $\rho(A)<1$ that then there is a vector norm so that $\|A\|<1$ in the associated operator norm. $\endgroup$ – LutzL Sep 13 '18 at 11:42

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