I am reading a lecture note and it says:
Suppose we have $$\dot{x} = Ax$$ and suppose $\tilde{x}$ is a fixed point. Then we have the following theorem:
$\tilde{x}$ is stable if and only if all eigenvalues of $A$ satisfy $\text{Re}(\lambda)\leq 0$ and whenever $\text{Re}(\lambda) = 0$ we have $\dim(\mathcal{N}(\lambda I-A)) = 1$.
I am confused of the second condition. $\dim(\mathcal{N}(\lambda I-A)) = 1$ just implies that
- the eigenspace corresponding to $\lambda$ is of dimension $1$.
- it can happen that the geometric multiplicity $\leq$ algebraic multiplicity. (if there is repeated roots of $\lambda$ )
If $\text{Re}(\lambda) = 0$, trajectory of $x(t)$ is an orbit (unstable in linear system and possibly stable in nonlinear system, for example a stable limit cycle.)
I have no idea what the second condition tells related to the stability. I mean if "$\dim(\mathcal{N}(\lambda I-A)) \geq 1$ ", what happen?
We can assume that $A$ consists of a single Jordan block $\lambda I + N$ with $N$ nilpotent. Then $\delta x(t)$ satisfies $$\delta x(t) = e^{(\lambda I + N)t}\delta x(0) = e^{\lambda t} e^{Nt}\delta x(0),$$ where $e^{Nt}$ has entries polynomial in $t$ since $N$ is nilpotent. Suppose column $i$ of $e^{Nt}$ contains a nonzero power of $t$; choose $\delta x(0) = e_i$ and now $\|\delta x(t)\| = \|e^{Nt}\delta x(0)\|$ grows unbounded as $t\to \infty$.