Stability Concepts and Definitions

We prove (1), (2), and then (3), which is the substantive classification result. The technique for (3) is direct computation with the matrix exponential together with a normal-form (Jordan) decomposition, plus explicit $\varepsilon$–$\delta$ verification.

(1) Immediate from the definition: "asymptotically stable" is defined as "stable, and moreover attracting." So it logically implies stability by conjunction elimination. There is no independent content to prove here; the point of stating it is that the definition of asymptotic stability does not say merely "attracting," precisely because attraction alone can fail to give stability (see the counterexample block).

(2) Take $n=2$, $A = \begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}$, so $\dot x = y,\ \dot y = -x$. The general solution with $\mathbf{x}_0=(x_0,y_0)$ is

which one verifies by direct substitution satisfies $\dot x = y$ and $\dot y = -x$ and matches the initial condition at $t=0$. Then

so $\|\phi(t,\mathbf{x}_0)\| = \|\mathbf{x}_0\|$ for all $t$: every trajectory is a circle centered at the origin, traversed at constant speed. Given $\varepsilon>0$, choose $\delta = \varepsilon$; then $\|\mathbf{x}_0\|<\delta$ forces $\|\phi(t,\mathbf{x}_0)\| = \|\mathbf{x}_0\| < \varepsilon$ for all $t\ge 0$. This proves stability. But if $\mathbf{x}_0 \ne \mathbf{0}$, then $\|\phi(t,\mathbf{x}_0)\| = \|\mathbf{x}_0\| \not\to 0$, so $\phi(t,\mathbf{x}_0)$ does not converge to the origin. Hence the origin is stable but not asymptotically stable, proving (2).

(3) Write $\phi(t,\mathbf{x}_0) = e^{At}\mathbf{x}_0$, the unique solution of the linear IVP (existence and uniqueness of this exponential-form solution is a standard fact about constant-coefficient linear systems, proved independently of stability theory, so invoking it here is not circular). Operator norm bound via Jordan form. Let $A = PJP^{-1}$ be a Jordan decomposition over $\mathbb{C}$ (or the real Jordan form over $\mathbb{R}$; the argument is identical up to bounded real changes of basis). Then $e^{At} = Pe^{Jt}P^{-1}$, and since $P, P^{-1}$ are fixed invertible matrices there are constants $0<c_1\le c_2$ with

so it suffices to analyze $e^{Jt}$ block by block. For a single Jordan block $J_k(\lambda)$ of size $k$ with eigenvalue $\lambda = \alpha+i\beta$,

where $N(t)$ has entries that are polynomials in $t$ of degree at most $k-1$ (bounded contribution from the nilpotent part) and $|e^{i\beta t}|=1$ (rotation, norm-preserving). Thus every entry of $e^{J_k(\lambda)t}$ has magnitude bounded by $C(1+|t|)^{k-1}e^{\alpha t}$ for a constant $C$ depending only on $k$. Proof of (a). Suppose every eigenvalue has $\operatorname{Re}\lambda_i = \alpha_i < 0$. Let $\alpha = \max_i \alpha_i <0$. By the block estimate above, each block satisfies $\|e^{J_k(\lambda)t}\| \le C(1+t)^{k-1}e^{\alpha t}$, and since $(1+t)^{k-1}e^{(\alpha-\alpha')t}\to 0$ for any $\alpha < \alpha' < 0$, polynomial growth is dominated by exponential decay: for any $\alpha' \in(\alpha, 0)$ there is $M\ge 1$ with

hence $\|e^{At}\| \le c_2 M e^{\alpha' t}/c_1 =: M' e^{\alpha' t}$ for all $t \ge 0$ (absorbing constants). Given $\varepsilon>0$, set $\delta = \varepsilon/M'$; then $\|\mathbf{x}_0\|<\delta$ gives $\|e^{At}\mathbf{x}_0\| \le M'\|\mathbf{x}_0\| e^{\alpha' t} < M'\delta = \varepsilon$ for all $t \ge 0$ since $e^{\alpha' t}\le 1$. This proves stability, and since $M'\|\mathbf{x}_0\|e^{\alpha' t}\to 0$ as $t\to\infty$ (as $\alpha'<0$), every trajectory converges to $\mathbf 0$; taking $\rho=\infty$ shows this is in fact global asymptotic stability. Conversely, suppose some eigenvalue $\lambda$ has $\operatorname{Re}\lambda \ge 0$. If $\operatorname{Re}\lambda>0$ we show instability directly below in the proof of (c); if $\operatorname{Re}\lambda = 0$, pick the corresponding eigenvector direction $\mathbf v$ (real or, for a complex pair, the real invariant plane it spans); the solution started on this eigenspace has $\|e^{At}\mathbf v\|$ either constant (semisimple case, purely oscillatory, as in the center example) or growing polynomially (non-semisimple case, e.g. a Jordan block with $\lambda=0$ gives solutions with a term linear in $t$). In neither case does $e^{At}\mathbf x_0 \to 0$ for $\mathbf x_0$ in that eigenspace, so asymptotic stability fails. This establishes the "only if" direction of (a). Proof of (b) and (c). Suppose every eigenvalue has $\operatorname{Re}\lambda_i \le 0$, and every eigenvalue with $\operatorname{Re}\lambda_i=0$ is semisimple, meaning its Jordan blocks all have size $k=1$. For blocks with $\operatorname{Re}\lambda<0$ the estimate above still gives $\|e^{J_k(\lambda)t}\|\to 0$. For a $1\times 1$ block with $\lambda = i\beta$ purely imaginary, $e^{J_1(i\beta)t} = e^{i\beta t}$ has modulus exactly $1$ for all $t$ — no polynomial growth factor appears since $k=1$ means there is no nilpotent part. Hence every block of $e^{Jt}$ is bounded uniformly in $t\ge 0$ by a constant depending only on $A$ (not growing polynomially), so there is $M\ge 1$ with $\|e^{At}\| \le M$ for all $t\ge0$. Given $\varepsilon>0$ take $\delta=\varepsilon/M$: then $\|\mathbf x_0\|<\delta \Rightarrow \|e^{At}\mathbf x_0\|\le M\|\mathbf x_0\|<\varepsilon$ for all $t\ge0$, proving stability. It need not be asymptotic: the center example in part (2) is exactly the case of a single semisimple purely-imaginary pair, and there $\|e^{At}\mathbf x_0\|=\|\mathbf x_0\|$ does not tend to $0$. Finally, for (c): if some eigenvalue $\lambda$ has $\operatorname{Re}\lambda = \alpha>0$, let $\mathbf v\ne 0$ be a corresponding (generalized) eigenvector, and consider $\mathbf x_0 = s\mathbf v$ for small $s>0$. Along the associated invariant subspace, $\|e^{At}(s\mathbf v)\| \ge s\, c\, e^{\alpha t}\to\infty$ for a constant $c>0$ (the $k=1$ eigenvector case gives this directly from $e^{At}\mathbf v = e^{\lambda t}\mathbf v$, and $|e^{\lambda t}| = e^{\alpha t}$; for a genuine Jordan chain the growth is even faster). No matter how small $s$ is chosen (i.e. no matter how small $\delta$ the adversary is handed), this norm eventually exceeds any fixed $\varepsilon_0>0$ at some finite time $t_1$. This exhibits, for every $\delta>0$, a point $\mathbf x_0$ with $\|\mathbf x_0\|<\delta$ whose trajectory leaves the ball of radius $\varepsilon_0$ — verifying the instability definition. The non-semisimple purely-imaginary case was already handled in the converse direction of (a): a Jordan block of size $k\ge2$ with $\lambda = i\beta$ contributes a factor growing like $t^{k-1}$ (unbounded), giving trajectories with $\|e^{At}\mathbf x_0\|\to\infty$ along that generalized eigenspace, again witnessing instability. This exhausts all cases and completes the proof of (3), hence of the theorem. $\blacksquare$