[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*}
[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]
[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]
