[proofplan] We first prove that a Hurwitz matrix has exponentially decaying semigroup $e^{tA}$, using the complex Jordan form. This decay makes the Lyapunov integral converge and permits differentiation under finite truncations, producing $A^\top P + PA = -Q$ after passing the upper endpoint to infinity. Positivity and symmetry follow directly from the integrand, while uniqueness follows by applying the same semigroup conjugation to a homogeneous solution. Finally, a negative definite Lyapunov certificate forces every eigenvalue to have negative real part by testing the quadratic identity on a complex eigenvector. [/proofplan]

[step:Derive exponential decay of $e^{tA}$ from the Hurwitz condition]Assume that $A$ is a Hurwitz matrix, meaning that every complex eigenvalue of $A$ has negative real part. Choose $\alpha > 0$ such that $\operatorname{Re}(\lambda) \leq -2\alpha$ for every eigenvalue $\lambda$ of $A$. View the real matrix $A$ as the complex-[linear map](/page/Linear%20Map) $A_{\mathbb{C}}: \mathbb{C}^n \to \mathbb{C}^n$ given by the same matrix entries. By the [Canonical Form Over Algebraically Closed Fields](/theorems/428), applied over $\mathbb{C}$ to $A_{\mathbb{C}}$, let $S \in GL(n,\mathbb{C})$ be an invertible complex matrix putting $A_{\mathbb{C}}$ into Jordan form: \begin{align*} A_{\mathbb{C}} = S J S^{-1}. \end{align*} Write $J$ as the block diagonal matrix with Jordan blocks $J_k = \lambda_k I_{m_k} + N_k$, where $\lambda_k \in \mathbb{C}$ is an eigenvalue of $A$, $m_k \in \mathbb{N}$ is the block size, and $N_k \in \mathbb{C}^{m_k \times m_k}$ is nilpotent with $N_k^{m_k} = 0$. We use the Euclidean operator norm on both $\mathbb{R}^n$ and $\mathbb{C}^n$; the estimate obtained for $e^{tA_{\mathbb{C}}}$ restricts to the same estimate for $e^{tA}$ on the real subspace $\mathbb{R}^n \subset \mathbb{C}^n$. For $t \geq 0$, \begin{align*} e^{tJ_k} = e^{t\lambda_k} \sum_{j=0}^{m_k-1} \frac{t^j N_k^j}{j!}. \end{align*} We use the elementary exponential domination lemma for polynomials: for every polynomial $p_0: [0,\infty) \to [0,\infty)$ and every $\beta>0$, there is a constant $C(p_0,\beta)>0$ such that $p_0(t)\leq C(p_0,\beta)e^{\beta t}$ for all $t\geq 0$. Applying this with $\beta=\alpha$ to the finite polynomial bound for $\sum_{j=0}^{m_k-1} t^jN_k^j/j!$, there exists $C_k > 0$ such that \begin{align*} \|e^{tJ_k}\|_{\mathrm{op}} \leq C_k e^{-\alpha t} \end{align*} for all $t \geq 0$. Let $B_J \geq 1$ be the finite-dimensional block-diagonal comparison constant such that every block diagonal matrix $M=\operatorname{diag}(M_1,\dots,M_r)$ satisfies \begin{align*} \|M\|_{\mathrm{op}} \leq B_J \max_k \|M_k\|_{\mathrm{op}}. \end{align*} Defining \begin{align*} C_A := \|S\|_{\mathrm{op}} \|S^{-1}\|_{\mathrm{op}} B_J \max_k C_k, \end{align*} we obtain \begin{align*} \|e^{tA}\|_{\mathrm{op}} \leq C_A e^{-\alpha t} \end{align*} for all $t \geq 0$.[/step]

[guided]The Hurwitz hypothesis says that all eigenvalues lie strictly in the open left half-plane. Because there are only finitely many eigenvalues, we can choose $\alpha > 0$ so that every eigenvalue satisfies $\operatorname{Re}(\lambda) \leq -2\alpha$. The extra factor of $2$ gives room to absorb the polynomial growth coming from nontrivial Jordan blocks. We now regard the real matrix $A$ as the complex-linear map $A_{\mathbb{C}}: \mathbb{C}^n \to \mathbb{C}^n$ with the same matrix entries. This is the correct setting for Jordan form, because the theorem is stated over an algebraically closed field. By the [Canonical Form Over Algebraically Closed Fields](/theorems/428), applied over $\mathbb{C}$ to $A_{\mathbb{C}}$, choose an invertible complex matrix $S \in GL(n,\mathbb{C})$ such that \begin{align*} A_{\mathbb{C}} = S J S^{-1}, \end{align*} where $J$ is block diagonal with Jordan blocks $J_k = \lambda_k I_{m_k} + N_k$. Here $\lambda_k$ is an eigenvalue of $A$, $I_{m_k}$ is the identity matrix in $\mathbb{C}^{m_k \times m_k}$, and $N_k$ is nilpotent with $N_k^{m_k} = 0$. For $t \geq 0$, the exponential of one block is \begin{align*} e^{tJ_k} = e^{t(\lambda_k I_{m_k} + N_k)} = e^{t\lambda_k} e^{tN_k}. \end{align*} Since $N_k^{m_k} = 0$, the nilpotent exponential is a finite sum: \begin{align*} e^{tN_k} = \sum_{j=0}^{m_k-1} \frac{t^j N_k^j}{j!}. \end{align*} Thus \begin{align*} e^{tJ_k} = e^{t\lambda_k} \sum_{j=0}^{m_k-1} \frac{t^j N_k^j}{j!}. \end{align*} The scalar factor satisfies $|e^{t\lambda_k}| = e^{t\operatorname{Re}(\lambda_k)} \leq e^{-2\alpha t}$. The finite nilpotent sum is bounded in operator norm by a polynomial $p_k: [0,\infty)\to[0,\infty)$. We use the elementary exponential domination lemma for polynomials: for every polynomial $p_0: [0,\infty) \to [0,\infty)$ and every $\beta>0$, there is a constant $C(p_0,\beta)>0$ such that $p_0(t)\leq C(p_0,\beta)e^{\beta t}$ for all $t\geq 0$. Applying this with $p_0=p_k$ and $\beta=\alpha$, there exists $C_k > 0$ such that \begin{align*} \|e^{tJ_k}\|_{\mathrm{op}} \leq C_k e^{-\alpha t} \end{align*} for every $t \geq 0$. Since $J$ is block diagonal, $e^{tJ}$ is block diagonal with blocks $e^{tJ_k}$. Let $B_J \geq 1$ denote the finite-dimensional constant comparing the operator norm of a block diagonal matrix with the maximum of its block operator norms, so \begin{align*} \|e^{tJ}\|_{\mathrm{op}} \leq B_J \max_k \|e^{tJ_k}\|_{\mathrm{op}}. \end{align*} Finally, \begin{align*} e^{tA_{\mathbb{C}}} = S e^{tJ} S^{-1}. \end{align*} Taking operator norms and defining \begin{align*} C_A := \|S\|_{\mathrm{op}} \|S^{-1}\|_{\mathrm{op}} B_J \max_k C_k, \end{align*} gives \begin{align*} \|e^{tA_{\mathbb{C}}}\|_{\mathrm{op}} \leq C_A e^{-\alpha t}. \end{align*} Because $e^{tA_{\mathbb{C}}}$ restricts to $e^{tA}$ on $\mathbb{R}^n \subset \mathbb{C}^n$, the same estimate holds for the real matrix exponential. This is the exact decay estimate needed to make the Lyapunov integral converge.[/guided]

[step:Construct the Lyapunov solution by the convergent integral]Let $Q \in \mathbb{R}^{n \times n}$ be symmetric positive definite. Define the matrix-valued map $F: [0,\infty) \to \mathbb{R}^{n \times n}$ by sending each $t \in [0,\infty)$ to \begin{align*} e^{tA^\top}Qe^{tA}. \end{align*} For $t \geq 0$, \begin{align*} \|F(t)\|_{\mathrm{op}} \leq \|e^{tA^\top}\|_{\mathrm{op}}\|Q\|_{\mathrm{op}}\|e^{tA}\|_{\mathrm{op}} \leq C_A^2 \|Q\|_{\mathrm{op}} e^{-2\alpha t}. \end{align*} Since $t \mapsto C_A^2\|Q\|_{\mathrm{op}}e^{-2\alpha t}$ is integrable over $[0,\infty)$ with respect to $\mathcal{L}^1$, the componentwise improper integral \begin{align*} P := \int_0^\infty e^{tA^\top}Qe^{tA}\,d\mathcal{L}^1(t) \end{align*} exists. For every $t \geq 0$, $F(t)$ is symmetric because \begin{align*} F(t)^\top = e^{tA^\top}Q^\top e^{tA} = e^{tA^\top}Qe^{tA}. \end{align*} Therefore $P$ is symmetric. If $x \in \mathbb{R}^n$ and $x \neq 0$, then \begin{align*} x^\top P x = \int_0^\infty (e^{tA}x)^\top Q(e^{tA}x)\,d\mathcal{L}^1(t). \end{align*} The integrand is continuous and nonnegative. At $t=0$ it equals $x^\top Qx > 0$, so it is positive on some interval $[0,\varepsilon)$ with $\varepsilon > 0$. Hence $x^\top Px > 0$, and $P$ is positive definite.[/step]

[guided]The decay estimate from the previous step is now used to define the candidate solution. Let $Q \in \mathbb{R}^{n \times n}$ be symmetric positive definite, and define the matrix-valued map $F: [0,\infty) \to \mathbb{R}^{n \times n}$ by \begin{align*} F(t)=e^{tA^\top}Qe^{tA}. \end{align*} The matrix exponential is continuous in $t$, so $F$ is continuous. For $t \geq 0$, [submultiplicativity of the operator norm](/theorems/1054) and the decay estimate give \begin{align*} \|F(t)\|_{\mathrm{op}} \leq \|e^{tA^\top}\|_{\mathrm{op}}\|Q\|_{\mathrm{op}}\|e^{tA}\|_{\mathrm{op}} \leq C_A^2 \|Q\|_{\mathrm{op}} e^{-2\alpha t}. \end{align*} The scalar function $t \mapsto C_A^2\|Q\|_{\mathrm{op}}e^{-2\alpha t}$ is integrable on $[0,\infty)$ with respect to $\mathcal{L}^1$. Therefore each matrix entry of $F$ is dominated by an integrable function, and the componentwise improper integral \begin{align*} P := \int_0^\infty e^{tA^\top}Qe^{tA}\,d\mathcal{L}^1(t) \end{align*} exists. We next check that this candidate has the required definiteness properties. For every $t \geq 0$, symmetry of $Q$ gives \begin{align*} F(t)^\top = e^{tA^\top}Q^\top e^{tA} = e^{tA^\top}Qe^{tA}. \end{align*} Hence the integral $P$ is symmetric. If $x \in \mathbb{R}^n$ and $x \neq 0$, then \begin{align*} x^\top P x = \int_0^\infty (e^{tA}x)^\top Q(e^{tA}x)\,d\mathcal{L}^1(t). \end{align*} The integrand is continuous and nonnegative because $Q$ is positive definite. At $t=0$ it equals $x^\top Qx$, which is strictly positive. By continuity it remains positive on some interval $[0,\varepsilon)$ with $\varepsilon > 0$, and the integral over that interval is strictly positive. Therefore $x^\top Px>0$ for every nonzero $x$, so $P$ is positive definite.[/guided]

[step:Differentiate the truncated integral to obtain the Lyapunov equation]For $T > 0$, define \begin{align*} P_T := \int_0^{\!T}e^{tA^\top}Qe^{tA}\,d\mathcal{L}^1(t). \end{align*} The map $F$ is continuously differentiable and satisfies \begin{align*} F'(t) = A^\top e^{tA^\top}Qe^{tA} + e^{tA^\top}Qe^{tA}A. \end{align*} Therefore \begin{align*} A^\top P_T + P_TA = \int_0^{\!T}F'(t)\,d\mathcal{L}^1(t). \end{align*} By the [Fundamental Theorem of Calculus](/theorems/632) applied componentwise to the continuously differentiable matrix-valued map $F$, \begin{align*} A^\top P_T + P_TA = F(T) - F(0). \end{align*} Since $F(0)=Q$ and $\|F(T)\|_{\mathrm{op}} \leq C_A^2\|Q\|_{\mathrm{op}}e^{-2\alpha T}$, we have $F(T) \to 0$ in operator norm as $T \to \infty$. By the definition of the improper matrix integral, $P_T \to P$ componentwise. Since $\mathbb{R}^{n\times n}$ is finite-dimensional, all norms induce the same topology; hence componentwise convergence is equivalent to convergence in operator norm. Therefore $A^\top P_T + P_TA \to A^\top P + PA$, and passing to the limit gives \begin{align*} A^\top P + PA = -Q. \end{align*}[/step]

[guided]The point of truncating the integral is that the finite-endpoint integral can be differentiated without any improper-limit issue. For $T>0$, define \begin{align*} P_T := \int_0^{\!T}e^{tA^\top}Qe^{tA}\,d\mathcal{L}^1(t). \end{align*} The standard differentiability formula for the matrix exponential gives $\frac{d}{dt}e^{tA}=Ae^{tA}=e^{tA}A$ and $\frac{d}{dt}e^{tA^\top}=A^\top e^{tA^\top}=e^{tA^\top}A^\top$. Hence the map $F: [0,\infty) \to \mathbb{R}^{n\times n}$ defined by $F(t)=e^{tA^\top}Qe^{tA}$ is continuously differentiable. Applying the product rule for finite-dimensional matrix-valued maps gives \begin{align*} F'(t) = A^\top e^{tA^\top}Qe^{tA} + e^{tA^\top}Qe^{tA}A. \end{align*} Linearity of the finite matrix integral therefore gives \begin{align*} A^\top P_T + P_TA = \int_0^{\!T}F'(t)\,d\mathcal{L}^1(t). \end{align*} We now apply the [Fundamental Theorem of Calculus](/theorems/632) componentwise. Its hypotheses are satisfied because each entry of $F$ is continuously differentiable on the compact interval $[0,T]$. Thus \begin{align*} A^\top P_T + P_TA = F(T)-F(0). \end{align*} Here $F(0)=Q$. Also, the decay estimate gives \begin{align*} \|F(T)\|_{\mathrm{op}} \leq C_A^2\|Q\|_{\mathrm{op}}e^{-2\alpha T}, \end{align*} so $F(T)\to 0$ in operator norm as $T\to\infty$. By the definition of the improper integral, $P_T\to P$ componentwise. Since $\mathbb{R}^{n\times n}$ is finite-dimensional, all norms induce the same topology; hence componentwise convergence is equivalent to operator-norm convergence. Therefore $A^\top P_T+P_TA\to A^\top P+PA$, and passing to the limit gives \begin{align*} A^\top P + PA = -Q. \end{align*}[/guided]

[step:Prove uniqueness of the symmetric solution]Let $P_1,P_2 \in \mathbb{R}^{n \times n}$ be symmetric matrices satisfying \begin{align*} A^\top P_1 + P_1A = -Q \end{align*} and \begin{align*} A^\top P_2 + P_2A = -Q. \end{align*} Define $H := P_1 - P_2$. Then $H$ is symmetric and \begin{align*} A^\top H + HA = 0. \end{align*} Define the matrix-valued map $G: [0,\infty) \to \mathbb{R}^{n \times n}$ by sending each $t \in [0,\infty)$ to \begin{align*} e^{tA^\top}He^{tA}. \end{align*} Using the differentiability formula for the matrix exponential and the product rule for finite-dimensional matrix-valued maps, $G$ is continuously differentiable and \begin{align*} G'(t) = e^{tA^\top}(A^\top H + HA)e^{tA} = 0. \end{align*} Thus $G(t)=G(0)=H$ for all $t \geq 0$. On the other hand, \begin{align*} \|G(t)\|_{\mathrm{op}} \leq C_A^2\|H\|_{\mathrm{op}}e^{-2\alpha t}, \end{align*} so $G(t)\to 0$ as $t\to\infty$. Hence $H=0$, and $P_1=P_2$. This proves statement 2 from statement 1, including the asserted integral formula.[/step]

[guided]Uniqueness is proved by subtracting two possible solutions and showing that the difference must vanish. Let $P_1,P_2 \in \mathbb{R}^{n\times n}$ be symmetric matrices satisfying \begin{align*} A^\top P_1+P_1A=-Q \end{align*} and \begin{align*} A^\top P_2+P_2A=-Q. \end{align*} Define $H:=P_1-P_2$. Then $H$ is symmetric and subtraction gives the homogeneous Lyapunov equation \begin{align*} A^\top H+HA=0. \end{align*} To use the exponential decay of the semigroup, define $G: [0,\infty)\to\mathbb{R}^{n\times n}$ by \begin{align*} G(t)=e^{tA^\top}He^{tA}. \end{align*} The differentiability formula for the matrix exponential and the product rule for finite-dimensional matrix-valued maps show that $G$ is continuously differentiable and \begin{align*} G'(t)=e^{tA^\top}(A^\top H+HA)e^{tA}=0. \end{align*} Thus $G$ is constant on $[0,\infty)$, so $G(t)=G(0)=H$ for all $t\geq 0$. On the other hand, the exponential decay estimate gives \begin{align*} \|G(t)\|_{\mathrm{op}}\leq C_A^2\|H\|_{\mathrm{op}}e^{-2\alpha t}, \end{align*} which tends to $0$ as $t\to\infty$. The same matrix is both constantly equal to $H$ and convergent to $0$, so $H=0$. Therefore $P_1=P_2$, proving uniqueness.[/guided]

[step:Extract a quadratic Lyapunov certificate from the equation]Assume statement 2. Let $I_n \in \mathbb{R}^{n \times n}$ denote the identity matrix. Choose $Q := I_n$, which is symmetric positive definite. By statement 2, there exists a symmetric positive definite matrix $P \in \mathbb{R}^{n \times n}$ such that \begin{align*} A^\top P + PA = -I_n. \end{align*} Since $-I_n$ is negative definite, statement 3 follows.[/step]

[guided]Statement 2 is stronger than what statement 3 asks for, so we choose a convenient positive definite right-hand side. Let $I_n \in \mathbb{R}^{n\times n}$ be the identity matrix and set $Q:=I_n$. This matrix is symmetric positive definite. By statement 2, there exists a symmetric positive definite matrix $P\in\mathbb{R}^{n\times n}$ satisfying \begin{align*} A^\top P+PA=-I_n. \end{align*} The matrix $-I_n$ is negative definite, so this same $P$ is a quadratic Lyapunov certificate. Hence statement 3 follows.[/guided]

[step:Use a quadratic Lyapunov certificate to force all eigenvalues into the left half-plane]Assume statement 3. Choose a symmetric positive definite matrix $P \in \mathbb{R}^{n \times n}$ such that \begin{align*} A^\top P + PA \end{align*} is negative definite. Define \begin{align*} R := -(A^\top P + PA). \end{align*} Then $R \in \mathbb{R}^{n \times n}$ is symmetric positive definite. Let $\lambda \in \mathbb{C}$ be an eigenvalue of $A$, and choose a nonzero vector $v \in \mathbb{C}^n$ such that \begin{align*} Av = \lambda v. \end{align*} We extend $P$ and $R$ complex-linearly to matrices acting on $\mathbb{C}^n$. Since $P$ and $R$ are real symmetric positive definite, for every nonzero $w \in \mathbb{C}^n$, \begin{align*} w^*Pw > 0 \end{align*} and \begin{align*} w^*Rw > 0. \end{align*} Indeed, writing $w = x + iy$ with $x,y \in \mathbb{R}^n$, we have \begin{align*} w^*Pw = x^\top Px + y^\top Py \end{align*} and \begin{align*} w^*Rw = x^\top Rx + y^\top Ry. \end{align*} At least one of $x$ and $y$ is nonzero, so positive definiteness over $\mathbb{R}^n$ gives the claimed strict positivity. Testing the identity $A^\top P + PA = -R$ against $v$ gives \begin{align*} v^*(A^\top P + PA)v = -v^*Rv. \end{align*} Because $A^\top$ is the real transpose of $A$, its complex adjoint relation gives $v^*A^\top = (Av)^* = \bar{\lambda}v^*$. Therefore \begin{align*} v^*(A^\top P + PA)v = \bar{\lambda}v^*Pv + \lambda v^*Pv. \end{align*} Thus \begin{align*} 2\operatorname{Re}(\lambda)v^*Pv = -v^*Rv. \end{align*} Since $v^*Pv > 0$ and $v^*Rv > 0$, division by $2v^*Pv$ gives \begin{align*} \operatorname{Re}(\lambda) = -\frac{v^*Rv}{2v^*Pv} < 0. \end{align*} Every complex eigenvalue of $A$ has negative real part, so $A$ is Hurwitz. This proves statement 1 from statement 3 and completes the equivalence.[/step]

[guided]We now prove that a negative definite Lyapunov certificate forces all eigenvalues into the open left half-plane. Assume statement 3, and choose a symmetric positive definite matrix $P\in\mathbb{R}^{n\times n}$ such that $A^\top P+PA$ is negative definite. Define \begin{align*} R:=-(A^\top P+PA). \end{align*} Then $R$ is symmetric positive definite. Let $\lambda\in\mathbb{C}$ be an eigenvalue of $A$, and choose a nonzero vector $v\in\mathbb{C}^n$ such that \begin{align*} Av=\lambda v. \end{align*} We extend $P$ and $R$ complex-linearly to matrices on $\mathbb{C}^n$. Since $P$ and $R$ are real symmetric positive definite, they remain positive on nonzero complex vectors in the Hermitian quadratic form sense. Indeed, if $w=x+iy$ with $x,y\in\mathbb{R}^n$, then \begin{align*} w^*Pw=x^\top Px+y^\top Py \end{align*} and \begin{align*} w^*Rw=x^\top Rx+y^\top Ry. \end{align*} At least one of $x$ and $y$ is nonzero, so both quantities are strictly positive for $w\neq 0$. Testing the identity $A^\top P+PA=-R$ against the eigenvector $v$ gives \begin{align*} v^*(A^\top P+PA)v=-v^*Rv. \end{align*} Because $A$ has real entries, $A^\top$ is the complex adjoint of $A$ with respect to the standard Hermitian product, so $v^*A^\top=(Av)^*=\bar{\lambda}v^*$. Therefore \begin{align*} v^*(A^\top P+PA)v=\bar{\lambda}v^*Pv+\lambda v^*Pv. \end{align*} Combining the last two identities yields \begin{align*} 2\operatorname{Re}(\lambda)v^*Pv=-v^*Rv. \end{align*} Since $v^*Pv>0$ and $v^*Rv>0$, division by $2v^*Pv$ gives \begin{align*} \operatorname{Re}(\lambda)=-\frac{v^*Rv}{2v^*Pv}<0. \end{align*} Thus every complex eigenvalue of $A$ has negative real part, which is exactly the Hurwitz condition. This proves statement 1 from statement 3 and completes the equivalence.[/guided]