[proofplan] We build the quadratic Lyapunov function $V(x)=x^\top Px$ and compare it to the Euclidean norm using positive definiteness of $P$. The strict negativity of $A^\top P+PA$ gives a negative upper bound for the derivative of $V$ along every solution. This produces a scalar exponential decay estimate for $V(x(t))$, and the norm comparison converts that estimate into exponential decay of $|x(t)|$. [/proofplan] [step:Choose explicit positive constants from the two quadratic forms] Define the symmetric positive definite matrix $S \in \mathbb{R}^{n \times n}$ by \begin{align*} S := -(A^\top P + PA). \end{align*} Let $Q := \{y \in \mathbb{R}^n : |y|=1\}$ denote the Euclidean unit sphere. The functions $q_P: Q \to \mathbb{R}$ and $q_S: Q \to \mathbb{R}$ defined by $q_P(y) := y^\top P y$ and $q_S(y) := y^\top S y$ are continuous. Since $Q$ is compact, the extreme value theorem for continuous real-valued functions on compact sets implies that $q_P$ and $q_S$ attain their minima and maxima on $Q$. Define \begin{align*} p_{\min} := \min_{y \in Q} y^\top P y, \qquad p_{\max} := \max_{y \in Q} y^\top P y, \qquad s_{\min} := \min_{y \in Q} y^\top S y. \end{align*} Since $P$ and $S$ are positive definite, these constants satisfy \begin{align*} 0 < p_{\min} \leq p_{\max}, \qquad 0 < s_{\min}. \end{align*} For every $x \in \mathbb{R}^n$, the following bounds hold: \begin{align*} p_{\min}|x|^2 \leq x^\top P x \leq p_{\max}|x|^2 \end{align*} and \begin{align*} x^\top Sx \geq s_{\min}|x|^2. \end{align*} Indeed, the inequalities are immediate for $x=0$, while for $x \neq 0$ they follow by applying the definitions to $y=x/|x| \in Q$. [guided] The point of this step is to replace matrix inequalities by numerical constants. Since $P$ is symmetric positive definite, the scalar quadratic form $x \mapsto x^\top Px$ is positive for every nonzero $x \in \mathbb{R}^n$. Likewise, because $A^\top P+PA$ is negative definite, the matrix \begin{align*} S := -(A^\top P + PA) \end{align*} is symmetric positive definite. Let \begin{align*} Q := \{y \in \mathbb{R}^n : |y|=1\} \end{align*} be the Euclidean unit sphere. The functions $y \mapsto y^\top Py$ and $y \mapsto y^\top Sy$ are continuous real-valued functions on $Q$. Since $Q$ is compact, the extreme value theorem for continuous real-valued functions on compact sets says that each function attains its minimum and maximum on $Q$. We define \begin{align*} p_{\min} := \min_{y \in Q} y^\top P y, \qquad p_{\max} := \max_{y \in Q} y^\top P y, \qquad s_{\min} := \min_{y \in Q} y^\top S y. \end{align*} Positive definiteness gives $p_{\min}>0$ and $s_{\min}>0$, because every $y \in Q$ is nonzero. Now fix $x \in \mathbb{R}^n$. If $x=0$, all the desired inequalities reduce to $0 \leq 0$. If $x \neq 0$, set $y=x/|x|$. Then $y \in Q$, and \begin{align*} x^\top Px = |x|^2 y^\top Py. \end{align*} Using the definitions of $p_{\min}$ and $p_{\max}$ gives \begin{align*} p_{\min}|x|^2 \leq x^\top P x \leq p_{\max}|x|^2. \end{align*} The same normalization gives \begin{align*} x^\top Sx = |x|^2 y^\top Sy \geq s_{\min}|x|^2. \end{align*} These are the constants that will convert decay of the Lyapunov function into decay of the Euclidean norm. [/guided] [/step] [step:Differentiate the quadratic Lyapunov function along a solution] Define the quadratic function $V: \mathbb{R}^n \to \mathbb{R}$ by \begin{align*} V(x) := x^\top Px. \end{align*} Let $x: [0,\infty) \to \mathbb{R}^n$ be a classical solution of $x'(t)=Ax(t)$ with $x(0)=x_0$. Differentiating the coordinate expression $V(x(t))=\sum_{i,j=1}^n P_{ij}x_i(t)x_j(t)$ and using the product rule gives \begin{align*} \frac{d}{dt}V(x(t)) = x'(t)^\top P x(t) + x(t)^\top P x'(t). \end{align*} Using $x'(t)=Ax(t)$, \begin{align*} \frac{d}{dt}V(x(t)) = x(t)^\top A^\top P x(t) + x(t)^\top PAx(t). \end{align*} Hence \begin{align*} \frac{d}{dt}V(x(t)) = x(t)^\top(A^\top P+PA)x(t) = -x(t)^\top Sx(t). \end{align*} By the bound for $S$ from the previous step, \begin{align*} \frac{d}{dt}V(x(t)) \leq -s_{\min}|x(t)|^2. \end{align*} [/step] [step:Convert the differential inequality into exponential decay of $V$] From the upper bound $V(x) \leq p_{\max}|x|^2$, we have \begin{align*} |x|^2 \geq \frac{1}{p_{\max}}V(x) \end{align*} for every $x \in \mathbb{R}^n$. Therefore, for every $t \geq 0$, \begin{align*} \frac{d}{dt}V(x(t)) \leq -\frac{s_{\min}}{p_{\max}}V(x(t)). \end{align*} Define $\lambda \in \mathbb{R}$ by \begin{align*} \lambda := \frac{s_{\min}}{p_{\max}}. \end{align*} Since $s_{\min}>0$ and $p_{\max}>0$, we have $\lambda>0$. Define $W: [0,\infty) \to \mathbb{R}$ by \begin{align*} W(t) := e^{\lambda t}V(x(t)). \end{align*} Differentiating $W$ gives \begin{align*} W'(t) = e^{\lambda t}\left(\lambda V(x(t)) + \frac{d}{dt}V(x(t))\right) \leq 0. \end{align*} The map $W$ is differentiable on $[0,\infty)$ because $x$ is a classical solution, $V$ is a polynomial function, and $t \mapsto e^{\lambda t}$ is differentiable. For any $0 \leq a < b$, applying the [one-variable mean value theorem](/theorems/186) to $W$ on the compact interval $[a,b]$ gives a point $c \in (a,b)$ such that $W(b)-W(a)=W'(c)(b-a) \leq 0$. Hence $W$ is nonincreasing on $[0,\infty)$. Thus $W(t) \leq W(0)$ for every $t \geq 0$, which is \begin{align*} V(x(t)) \leq e^{-\lambda t}V(x_0). \end{align*} [/step] [step:Translate Lyapunov decay into exponential stability of the system] Using the lower bound for $V$ at $x(t)$ and the upper bound for $V$ at $x_0$, we obtain \begin{align*} p_{\min}|x(t)|^2 \leq V(x(t)) \leq e^{-\lambda t}V(x_0) \leq p_{\max}e^{-\lambda t}|x_0|^2. \end{align*} Hence \begin{align*} |x(t)| \leq \left(\frac{p_{\max}}{p_{\min}}\right)^{1/2} e^{-\lambda t/2}|x_0|. \end{align*} Set \begin{align*} M := \left(\frac{p_{\max}}{p_{\min}}\right)^{1/2}, \qquad \alpha := \frac{s_{\min}}{2p_{\max}}. \end{align*} Then $M \geq 1$ and $\alpha>0$, and every solution satisfies \begin{align*} |x(t)| \leq M e^{-\alpha t}|x_0| \end{align*} for every $t \geq 0$. By the definition of exponential stability, exponential stability of the equilibrium $0 \in \mathbb{R}^n$ means that there exist constants $M \geq 1$ and $\alpha>0$ such that this estimate holds for every initial value $x_0 \in \mathbb{R}^n$ and every $t \geq 0$. Hence $x'=Ax$ is exponentially stable. [/step]