[proofplan] By [citetheorem:7899] applied to the real matrix $A\in \mathbb R^{n\times n}$, [Lyapunov stability](/page/Lyapunov%20Stability) of the zero equilibrium is equivalent to uniform boundedness of the real solution semigroup $(e^{tA})_{t\geq 0}$. We pass to the complexification, put the finite-dimensional complex-[linear map](/page/Linear%20Map) $A_{\mathbb C}$ in [Jordan normal form](/theorems/864), and compute the exponential of each Jordan block. Blocks with strictly negative real part are bounded because the explicitly defined suprema $\sup_{t\geq 0}e^{\operatorname{Re}(\lambda)t}t^q$ are finite, while size-$1$ blocks on the imaginary axis have constant modulus. Conversely, a positive real part produces exponential growth, and a nontrivial Jordan block on the imaginary axis produces polynomial growth, so either obstruction contradicts uniform boundedness. [/proofplan]

[step:Reduce Lyapunov stability to boundedness of the complexified semigroup]Let $A_{\mathbb C}: \mathbb C^n\to \mathbb C^n$ be the complex-linear map defined by the same matrix entries as $A$. Let $|\cdot|_{\mathbb R^n}$ denote the Euclidean norm on $\mathbb R^n$, and let $|\cdot|_{\mathbb C^n}$ denote the Hermitian Euclidean norm on $\mathbb C^n$. For a complex matrix $B\in \mathbb C^{n\times n}$, let $\|B\|_{\mathrm{op}}$ denote the operator norm of the map $B: \mathbb C^n\to \mathbb C^n$ with respect to $|\cdot|_{\mathbb C^n}$. The real matrix $A\in \mathbb R^{n\times n}$ satisfies the hypothesis of [citetheorem:7899], whose input is a real $n\times n$ matrix defining the linear system $\frac{dx}{dt}=Ax$ on $\mathbb R^n$. Applying that theorem gives that the equilibrium $0$ of this system is Lyapunov stable if and only if there exists $M_{\mathbb R}>0$ such that \begin{align*} |e^{tA}x|_{\mathbb R^n}\leq M_{\mathbb R}|x|_{\mathbb R^n} \end{align*} for every $x\in \mathbb R^n$ and every $t\geq 0$. We claim that this is equivalent to the existence of $M_{\mathbb C}>0$ such that \begin{align*} |e^{tA_{\mathbb C}}z|_{\mathbb C^n}\leq M_{\mathbb C}|z|_{\mathbb C^n} \end{align*} for every $z\in \mathbb C^n$ and every $t\geq 0$. If the complex estimate holds, its restriction to $z=x\in \mathbb R^n\subset \mathbb C^n$ gives the real estimate with $M_{\mathbb R}=M_{\mathbb C}$. Conversely, assume the real estimate holds. For $z\in \mathbb C^n$, write $z=x+iy$ with $x,y\in \mathbb R^n$. Since $A$ has real entries, \begin{align*} e^{tA_{\mathbb C}}z=e^{tA}x+i e^{tA}y. \end{align*} Therefore, by the triangle inequality and the real bound, \begin{align*} |e^{tA_{\mathbb C}}z|_{\mathbb C^n}\leq |e^{tA}x|_{\mathbb R^n}+|e^{tA}y|_{\mathbb R^n}. \end{align*} Hence \begin{align*} |e^{tA_{\mathbb C}}z|_{\mathbb C^n}\leq M_{\mathbb R}|x|_{\mathbb R^n}+M_{\mathbb R}|y|_{\mathbb R^n}\leq 2M_{\mathbb R}|z|_{\mathbb C^n}. \end{align*} Thus the complex estimate holds with $M_{\mathbb C}=2M_{\mathbb R}$.[/step]

[guided]The theorem is stated for a real system, but the spectral condition is naturally expressed over $\mathbb C$. We therefore need to justify that boundedness over $\mathbb R^n$ and boundedness over $\mathbb C^n$ are the same condition up to a harmless constant. Let $A_{\mathbb C}: \mathbb C^n\to \mathbb C^n$ be the complex-linear map represented by the same real matrix $A$. The real system is stable exactly when the real semigroup is uniformly bounded. The theorem [citetheorem:7899] applies to a real $n\times n$ matrix defining a linear system on $\mathbb R^n$. Since the present matrix satisfies $A\in \mathbb R^{n\times n}$ and the system is exactly $\frac{dx}{dt}=Ax$, its hypotheses are met. The theorem gives that Lyapunov stability of $0$ is equivalent to the existence of $M_{\mathbb R}>0$ such that \begin{align*} |e^{tA}x|_{\mathbb R^n}\leq M_{\mathbb R}|x|_{\mathbb R^n} \end{align*} for every $x\in \mathbb R^n$ and every $t\geq 0$. Now suppose first that the complexified semigroup is bounded: there is $M_{\mathbb C}>0$ such that \begin{align*} |e^{tA_{\mathbb C}}z|_{\mathbb C^n}\leq M_{\mathbb C}|z|_{\mathbb C^n} \end{align*} for every $z\in \mathbb C^n$ and every $t\geq 0$. Taking $z=x\in \mathbb R^n$ gives the real estimate immediately. For the converse, assume the real estimate holds. Every $z\in \mathbb C^n$ has a unique decomposition $z=x+iy$ with $x,y\in \mathbb R^n$. Because $A$ has real entries, the exponential preserves real and imaginary parts separately: \begin{align*} e^{tA_{\mathbb C}}z=e^{tA}x+i e^{tA}y. \end{align*} Applying the triangle inequality in $\mathbb C^n$ gives \begin{align*} |e^{tA_{\mathbb C}}z|_{\mathbb C^n}\leq |e^{tA}x|_{\mathbb R^n}+|e^{tA}y|_{\mathbb R^n}. \end{align*} The real boundedness estimate applies to both $x$ and $y$, so \begin{align*} |e^{tA_{\mathbb C}}z|_{\mathbb C^n}\leq M_{\mathbb R}|x|_{\mathbb R^n}+M_{\mathbb R}|y|_{\mathbb R^n}. \end{align*} Since $|x|_{\mathbb R^n}\leq |z|_{\mathbb C^n}$ and $|y|_{\mathbb R^n}\leq |z|_{\mathbb C^n}$, we obtain \begin{align*} |e^{tA_{\mathbb C}}z|_{\mathbb C^n}\leq 2M_{\mathbb R}|z|_{\mathbb C^n}. \end{align*} Thus boundedness of the real semigroup is equivalent to boundedness of the complexified semigroup. This permits us to work in Jordan coordinates over $\mathbb C$ without losing the original real Lyapunov stability question.[/guided]

[step:Put the complexified matrix into Jordan normal form] Because $A_{\mathbb C}:\mathbb C^n\to\mathbb C^n$ is a linear map on a finite-dimensional [vector space](/page/Vector%20Space) over the algebraically closed field $\mathbb C$, the [Jordan normal form theorem](/theorems/412) over $\mathbb C$ applies. It gives an invertible complex matrix $P\in \mathbb C^{n\times n}$ and a block diagonal complex matrix $J\in \mathbb C^{n\times n}$ such that \begin{align*} A_{\mathbb C}=PJP^{-1}. \end{align*} The diagonal blocks of $J$ are Jordan blocks \begin{align*} J_{\lambda,m}=\lambda I_m+N_m, \end{align*} where $\lambda\in \mathbb C$ is an eigenvalue of $A_{\mathbb C}$, $m\in \mathbb N$ is the block size, $I_m\in \mathbb C^{m\times m}$ is the identity matrix, and $N_m\in \mathbb C^{m\times m}$ is the nilpotent matrix whose $(i,j)$ entry equals $1$ when $j=i+1$ and equals $0$ when $j\neq i+1$. The matrix $N_m$ satisfies $N_m^m=0$. Since \begin{align*} e^{tA_{\mathbb C}}=P e^{tJ}P^{-1} \end{align*} for all $t\geq 0$, boundedness of $(e^{tA_{\mathbb C}})_{t\geq 0}$ is equivalent to boundedness of $(e^{tJ})_{t\geq 0}$. Indeed, if $\|e^{tJ}\|_{\mathrm{op}}\leq C$ for all $t\geq 0$, then \begin{align*} \|e^{tA_{\mathbb C}}\|_{\mathrm{op}}\leq \|P\|_{\mathrm{op}} C \|P^{-1}\|_{\mathrm{op}}. \end{align*} Conversely, since $e^{tJ}=P^{-1}e^{tA_{\mathbb C}}P$, boundedness of $(e^{tA_{\mathbb C}})_{t\geq 0}$ implies boundedness of $(e^{tJ})_{t\geq 0}$. [/step]

[step:Compute the exponential of each Jordan block] Fix one Jordan block $J_{\lambda,m}=\lambda I_m+N_m$. Since $\lambda I_m$ commutes with $N_m$, the [binomial theorem](/theorems/750) gives, for every integer $r\geq 0$, \begin{align*} (\lambda I_m+N_m)^r=\sum_{q=0}^{r}\binom{r}{q}\lambda^{r-q}N_m^q. \end{align*} The matrix exponential series converges absolutely in the operator norm $\|\cdot\|_{\mathrm{op}}$ on $\mathbb C^{m\times m}$. Since $N_m^q=0$ for every $q\geq m$, the nilpotent part contributes only finitely many powers. Hence the commuting product rule for exponentials and the finite truncation of $e^{tN_m}$ give, for every $t\geq 0$, \begin{align*} e^{tJ_{\lambda,m}}=e^{t\lambda I_m}e^{tN_m}=e^{\lambda t}\sum_{q=0}^{m-1}\frac{t^q}{q!}N_m^q. \end{align*} If $\operatorname{Re}(\lambda)<0$, then for each $q\in\{0,\dots,m-1\}$ define \begin{align*} S_q=\sup_{t\geq 0} e^{\operatorname{Re}(\lambda)t}t^q. \end{align*} For $q=0$, this supremum is $S_0=1$. For $q\geq 1$, the function $t\mapsto e^{\operatorname{Re}(\lambda)t}t^q$ is continuous on $[0,\infty)$, tends to $0$ as $t\to\infty$, and therefore attains a finite maximum on $[0,\infty)$. Thus each $S_q$ is finite. Define the block constant \begin{align*} C_{\lambda,m}=\sum_{q=0}^{m-1}\frac{S_q}{q!}\|N_m^q\|_{\mathrm{op}}. \end{align*} Taking operator norms in the block exponential formula and using the triangle inequality gives \begin{align*} \|e^{tJ_{\lambda,m}}\|_{\mathrm{op}}\leq C_{\lambda,m} \end{align*} for every $t\geq 0$. If $\operatorname{Re}(\lambda)=0$ and $m=1$, then $J_{\lambda,1}=(\lambda)$ and \begin{align*} \|e^{tJ_{\lambda,1}}\|_{\mathrm{op}}=|e^{\lambda t}|=e^{\operatorname{Re}(\lambda)t}=1 \end{align*} for every $t\geq 0$. [/step]

[step:Prove boundedness under the spectral condition] Assume that every eigenvalue $\lambda$ of $A_{\mathbb C}$ satisfies $\operatorname{Re}(\lambda)\leq 0$, and that every Jordan block associated to an eigenvalue with $\operatorname{Re}(\lambda)=0$ has size $1$. Let $J_{\lambda_1,m_1},\dots,J_{\lambda_r,m_r}$ be the Jordan blocks of $J$, where $r\in \mathbb N$ is the number of blocks. For each block, the previous step gives a constant $C_k>0$ such that \begin{align*} \|e^{tJ_{\lambda_k,m_k}}\|_{\mathrm{op}}\leq C_k \end{align*} for every $t\geq 0$. Define \begin{align*} C_J=\max_{1\leq k\leq r} C_k. \end{align*} Because $e^{tJ}$ is block diagonal with blocks $e^{tJ_{\lambda_k,m_k}}$, it follows that \begin{align*} \|e^{tJ}\|_{\mathrm{op}}\leq C_J \end{align*} for every $t\geq 0$. Therefore \begin{align*} \|e^{tA_{\mathbb C}}\|_{\mathrm{op}}\leq \|P\|_{\mathrm{op}} C_J \|P^{-1}\|_{\mathrm{op}} \end{align*} for every $t\geq 0$. Thus the complexified semigroup is uniformly bounded. By the reduction step, the real semigroup $(e^{tA})_{t\geq 0}$ is uniformly bounded on $\mathbb R^n$. Since $A\in \mathbb R^{n\times n}$ and the system under consideration is $\frac{dx}{dt}=Ax$, the hypotheses of [citetheorem:7899] are satisfied again. Applying the converse direction of that theorem implies that the equilibrium $0$ is Lyapunov stable. [/step]

[step:Show that a positive real part forces unbounded growth] Assume that $0$ is Lyapunov stable. By the first step and [citetheorem:7899], the complexified semigroup $(e^{tA_{\mathbb C}})_{t\geq 0}$ is uniformly bounded. Suppose, toward a contradiction, that $A_{\mathbb C}$ has an eigenvalue $\lambda\in \mathbb C$ with $\operatorname{Re}(\lambda)>0$. Let $v\in \mathbb C^n\setminus\{0\}$ be an eigenvector, so $A_{\mathbb C}v=\lambda v$. Then \begin{align*} e^{tA_{\mathbb C}}v=e^{\lambda t}v \end{align*} for every $t\geq 0$. Taking norms gives \begin{align*} |e^{tA_{\mathbb C}}v|_{\mathbb C^n}=e^{\operatorname{Re}(\lambda)t}|v|_{\mathbb C^n}. \end{align*} Since $\operatorname{Re}(\lambda)>0$, the right-hand side is unbounded as $t\to\infty$, contradicting uniform boundedness of $(e^{tA_{\mathbb C}})_{t\geq 0}$. Therefore every eigenvalue satisfies $\operatorname{Re}(\lambda)\leq 0$. [/step]

[step:Show that a nontrivial Jordan block on the imaginary axis forces unbounded growth] It remains to exclude nontrivial Jordan blocks for eigenvalues on the imaginary axis. Suppose, toward a contradiction, that $\lambda\in \mathbb C$ satisfies $\operatorname{Re}(\lambda)=0$ and that $A_{\mathbb C}$ has a Jordan block $J_{\lambda,m}$ with $m\geq 2$. Work in the Jordan basis from the second step. Let $e_m\in \mathbb C^m$ denote the last standard basis vector in the coordinate space of this block, and let $w\in \mathbb C^n$ denote the vector whose component in this block is $e_m$ and whose components in all other Jordan blocks are $0$. For the nilpotent matrix $N_m$ defined above, \begin{align*} N_m^{m-1}e_m=e_1. \end{align*} Using the block exponential formula, \begin{align*} e^{tJ_{\lambda,m}}e_m=e^{\lambda t}\sum_{q=0}^{m-1}\frac{t^q}{q!}N_m^q e_m. \end{align*} The first coordinate of this vector is \begin{align*} e^{\lambda t}\frac{t^{m-1}}{(m-1)!}. \end{align*} Since $|e^{\lambda t}|=1$, we obtain \begin{align*} |e^{tJ_{\lambda,m}}e_m|_{\mathbb C^m}\geq \frac{t^{m-1}}{(m-1)!}. \end{align*} Thus $(e^{tJ})_{t\geq 0}$ is unbounded, because it is unbounded on the vector $w$. By the equivalence under conjugation established above, $(e^{tA_{\mathbb C}})_{t\geq 0}$ is unbounded. This contradicts the boundedness obtained from Lyapunov stability. Therefore every Jordan block associated to an eigenvalue with real part $0$ has size $1$. Combining this with the previous step proves the necessity of the spectral condition, and the sufficiency was proved above. Hence the stated spectral condition is equivalent to Lyapunov stability of the equilibrium $0$. [/step]