[proofplan] Solutions of the linear system are given by the matrix exponential $x(t)=e^{tA}x_0$, so stability is reduced to estimates on $e^{tA}$. When all eigenvalues lie in the open left half-plane, the complex Jordan form gives a polynomial-times-exponential bound for every Jordan block, and the negative exponential dominates the polynomial factor. This yields a uniform estimate $|e^{tA}x_0|\le Ce^{-ct}|x_0|$, which proves both [Lyapunov stability](/page/Lyapunov%20Stability) and convergence to $0$. If some eigenvalue has positive real part, an associated real invariant line or plane contains arbitrarily small initial data whose solution grows without bound, so no Lyapunov stability estimate can hold. [/proofplan]

[step:Express every solution through the matrix exponential] Define the matrix exponential map \begin{align*} \exp_A:\mathbb{R}\to \mathbb{R}^{n\times n}, \qquad t\mapsto e^{tA}:=\sum_{k=0}^{\infty}\frac{t^kA^k}{k!}. \end{align*} The series converges absolutely in any matrix norm, and termwise differentiation gives \begin{align*} \frac{d}{dt}e^{tA}=Ae^{tA}. \end{align*} Also $e^{0A}=I_n$. Hence for each initial vector $x_0\in\mathbb{R}^n$, the map \begin{align*} x:\mathbb{R}\to\mathbb{R}^n, \qquad t\mapsto e^{tA}x_0 \end{align*} satisfies $\dot{x}(t)=Ax(t)$ and $x(0)=x_0$. Conversely, if $y:\mathbb{R}\to\mathbb{R}^n$ is a differentiable solution with $y(0)=x_0$, define \begin{align*} z:\mathbb{R}\to\mathbb{R}^n, \qquad t\mapsto e^{-tA}y(t). \end{align*} Using the product rule and the fact that $A$ commutes with $e^{-tA}$, we get \begin{align*} z'(t)=-Ae^{-tA}y(t)+e^{-tA}Ay(t)=0. \end{align*} Therefore $z(t)=z(0)=x_0$, and so $y(t)=e^{tA}x_0$ for all $t\in\mathbb{R}$. [/step]

[step:Bound the matrix exponential when all eigenvalues have negative real part]Assume that every complex eigenvalue $\lambda$ of $A$ satisfies $\operatorname{Re}(\lambda)<0$. Let $\sigma(A)$ denote the set of complex eigenvalues of $A$, and define \begin{align*} \alpha:=\max\{\operatorname{Re}(\lambda):\lambda\in\sigma(A)\}. \end{align*} Then $\alpha<0$. Define \begin{align*} c:=-\frac{\alpha}{2}>0. \end{align*} By the complex [Jordan normal form theorem](/theorems/412), there exist an invertible matrix $P\in\mathbb{C}^{n\times n}$ and a block diagonal Jordan matrix $J\in\mathbb{C}^{n\times n}$ such that \begin{align*} A=PJP^{-1}. \end{align*} Each Jordan block has the form $J_\lambda=\lambda I_m+N_m$, where $m\in\{1,\dots,n\}$ is the block size and $N_m\in\mathbb{C}^{m\times m}$ is the nilpotent matrix with ones on the superdiagonal and zeros elsewhere. For $t\ge 0$, \begin{align*} e^{tJ_\lambda}=e^{t\lambda}\sum_{r=0}^{m-1}\frac{t^rN_m^r}{r!}. \end{align*} Using the Euclidean operator norm on complex coordinate spaces, $\|N_m^r\|_{\mathrm{op}}\le 1$ for $0\le r\le m-1$. Define the finite constant \begin{align*} B:=\sum_{r=0}^{n-1}\frac{1}{r!}\sup_{t\ge 0}\left(t^re^{-ct}\right). \end{align*} Since $\operatorname{Re}(\lambda)\le \alpha=-2c$, the block estimate is \begin{align*} \|e^{tJ_\lambda}\|_{\mathrm{op}}\le e^{-2ct}\sum_{r=0}^{m-1}\frac{t^r}{r!}\le Be^{-ct}. \end{align*} Because $J$ is block diagonal, the same bound holds for $e^{tJ}$ after increasing no constant: \begin{align*} \|e^{tJ}\|_{\mathrm{op}}\le Be^{-ct}. \end{align*} Define \begin{align*} C:=\|P\|_{\mathrm{op}}\|P^{-1}\|_{\mathrm{op}}B. \end{align*} Then, for all $t\ge 0$, \begin{align*} \|e^{tA}\|_{\mathrm{op}}\le Ce^{-ct}. \end{align*}[/step]

[guided]The point of passing to Jordan form is that diagonalization may fail, but Jordan blocks still have an explicit exponential. We work over $\mathbb{C}$ because the eigenvalues of the real matrix $A$ may be nonreal. Let \begin{align*} \alpha:=\max\{\operatorname{Re}(\lambda):\lambda\in\sigma(A)\}. \end{align*} The spectrum $\sigma(A)$ is finite, and the hypothesis gives $\alpha<0$. Set \begin{align*} c:=-\frac{\alpha}{2}>0. \end{align*} Thus every eigenvalue satisfies $\operatorname{Re}(\lambda)\le -2c$. By the complex [Jordan normal form](/theorems/864) theorem, there are an invertible matrix $P\in\mathbb{C}^{n\times n}$ and a block diagonal Jordan matrix $J\in\mathbb{C}^{n\times n}$ such that \begin{align*} A=PJP^{-1}. \end{align*} A Jordan block corresponding to an eigenvalue $\lambda$ has the form \begin{align*} J_\lambda=\lambda I_m+N_m, \end{align*} where $m$ is the size of the block and $N_m$ is nilpotent with ones on the superdiagonal. Since $\lambda I_m$ commutes with $N_m$, the exponential separates: \begin{align*} e^{tJ_\lambda}=e^{t\lambda}e^{tN_m}. \end{align*} Because $N_m^m=0$, the exponential of $tN_m$ is a finite sum: \begin{align*} e^{tN_m}=\sum_{r=0}^{m-1}\frac{t^rN_m^r}{r!}. \end{align*} Therefore \begin{align*} e^{tJ_\lambda}=e^{t\lambda}\sum_{r=0}^{m-1}\frac{t^rN_m^r}{r!}. \end{align*} Now we estimate the norm. In the Euclidean operator norm, the shift matrices $N_m^r$ have norm at most $1$. Hence, for $t\ge 0$, \begin{align*} \|e^{tJ_\lambda}\|_{\mathrm{op}}\le e^{t\operatorname{Re}(\lambda)}\sum_{r=0}^{m-1}\frac{t^r}{r!}. \end{align*} Since $\operatorname{Re}(\lambda)\le -2c$, this gives \begin{align*} \|e^{tJ_\lambda}\|_{\mathrm{op}}\le e^{-2ct}\sum_{r=0}^{m-1}\frac{t^r}{r!}. \end{align*} The polynomial factor is harmless because exponential decay dominates every polynomial. To make this explicit, define \begin{align*} B:=\sum_{r=0}^{n-1}\frac{1}{r!}\sup_{t\ge 0}\left(t^re^{-ct}\right). \end{align*} Each supremum is finite, so $B<\infty$. Since $m\le n$, we obtain \begin{align*} \|e^{tJ_\lambda}\|_{\mathrm{op}}\le Be^{-ct}. \end{align*} Taking the maximum over the finitely many Jordan blocks gives \begin{align*} \|e^{tJ}\|_{\mathrm{op}}\le Be^{-ct}. \end{align*} Finally, $e^{tA}=Pe^{tJ}P^{-1}$, so [submultiplicativity of the operator norm](/theorems/1054) gives \begin{align*} \|e^{tA}\|_{\mathrm{op}}\le \|P\|_{\mathrm{op}}\|e^{tJ}\|_{\mathrm{op}}\|P^{-1}\|_{\mathrm{op}}. \end{align*} With \begin{align*} C:=\|P\|_{\mathrm{op}}\|P^{-1}\|_{\mathrm{op}}B, \end{align*} we conclude that \begin{align*} \|e^{tA}\|_{\mathrm{op}}\le Ce^{-ct} \end{align*} for every $t\ge 0$.[/guided]

[step:Use the exponential estimate to prove asymptotic stability] Assume again that all eigenvalues of $A$ have negative real part, and let $C>0$ and $c>0$ be the constants obtained above. For every $x_0\in\mathbb{R}^n$ and every $t\ge 0$, \begin{align*} |e^{tA}x_0|\le \|e^{tA}\|_{\mathrm{op}}|x_0|\le Ce^{-ct}|x_0|\le C|x_0|. \end{align*} We prove Lyapunov stability. Let $\varepsilon>0$ be given, and define \begin{align*} \delta:=\frac{\varepsilon}{C}. \end{align*} If $|x_0|<\delta$, then the solution $x(t)=e^{tA}x_0$ satisfies \begin{align*} |x(t)|<\varepsilon \end{align*} for every $t\ge 0$. Hence $0$ is Lyapunov stable. We prove attraction. For every $x_0\in\mathbb{R}^n$, \begin{align*} |x(t)|=|e^{tA}x_0|\le Ce^{-ct}|x_0|\to 0 \end{align*} as $t\to\infty$. Thus all initial data sufficiently close to $0$ produce trajectories converging to $0$. Combining Lyapunov stability with attraction proves that $0$ is asymptotically stable. [/step]

[step:Produce a real growing solution from an eigenvalue with positive real part] Assume that $A$ has an eigenvalue $\lambda\in\mathbb{C}$ with $\operatorname{Re}(\lambda)>0$. Let $v\in\mathbb{C}^n\setminus\{0\}$ be an eigenvector satisfying \begin{align*} Av=\lambda v. \end{align*} Write \begin{align*} \lambda=a+ib \end{align*} with $a,b\in\mathbb{R}$ and $a>0$, and write \begin{align*} v=p+iq \end{align*} with $p,q\in\mathbb{R}^n$. The complex-valued curve \begin{align*} u:\mathbb{R}\to\mathbb{C}^n, \qquad t\mapsto e^{\lambda t}v \end{align*} satisfies $\dot{u}(t)=Au(t)$. Its real part \begin{align*} w:\mathbb{R}\to\mathbb{R}^n, \qquad t\mapsto \operatorname{Re}(e^{\lambda t}v) \end{align*} is therefore a real solution of $\dot{x}=Ax$. Explicitly, \begin{align*} w(t)=e^{at}\bigl(p\cos(bt)-q\sin(bt)\bigr). \end{align*} We now choose a nonzero real growing solution. If $w(0)=p\ne 0$, use $w$. If $p=0$, then $q\ne 0$, and the real-valued curve \begin{align*} \widetilde{w}:\mathbb{R}\to\mathbb{R}^n, \qquad t\mapsto \operatorname{Im}(e^{\lambda t}v) \end{align*} satisfies $\widetilde{w}(0)=q\ne 0$ and is also a real solution. Denote by \begin{align*} r:\mathbb{R}\to\mathbb{R}^n \end{align*} one of these two real solutions with $r(0)\ne 0$. It remains to record that $r$ is unbounded as $t\to\infty$. If $\lambda$ is real, then $b=0$ and $r(t)=e^{at}r(0)$, so $|r(t)|=e^{at}|r(0)|\to\infty$. If $b\ne 0$, then the real plane spanned by $p$ and $q$ is invariant, and on that plane the solution is multiplication by $e^{at}$ followed by a rotation with respect to a fixed linear coordinate representation. Since the rotating factor is periodic and not identically zero, there exists a time $\tau\ge 0$ such that $r(\tau)\ne 0$. For integers $k\ge 0$, \begin{align*} r\left(\tau+\frac{2\pi k}{|b|}\right)=e^{a2\pi k/|b|}r(\tau). \end{align*} Hence $|r(t)|$ is unbounded along this sequence of times. [/step]

[step:Use the growing solution to contradict Lyapunov stability] Let $r:\mathbb{R}\to\mathbb{R}^n$ be the nonzero real solution constructed above whose norm is unbounded for $t\ge 0$. Define \begin{align*} \varepsilon_0:=\frac{|r(0)|}{2}>0. \end{align*} For any $\delta>0$, choose a scalar $\gamma>0$ such that \begin{align*} \gamma |r(0)|<\delta. \end{align*} The curve \begin{align*} x_\gamma:\mathbb{R}\to\mathbb{R}^n, \qquad t\mapsto \gamma r(t) \end{align*} is a solution of $\dot{x}=Ax$ with initial data $x_\gamma(0)=\gamma r(0)$. Since $|r(t)|$ is unbounded on $[0,\infty)$, there exists $T\ge 0$ such that \begin{align*} \gamma |r(T)|\ge \varepsilon_0. \end{align*} Thus \begin{align*} |x_\gamma(0)|<\delta \end{align*} but \begin{align*} |x_\gamma(T)|\ge \varepsilon_0. \end{align*} This shows that no matter how small the initial neighborhood radius $\delta$ is chosen, there is an initial condition inside it whose forward trajectory leaves the ball of radius $\varepsilon_0$ about $0$. Therefore the equilibrium $0$ is not Lyapunov stable. [/step]