Stability and Classification of Equilibria

We prove the theorem by explicit construction of the fundamental solutions in each case, using the real Jordan canonical form of $A$, and then reading off the asymptotics as $t\to\infty$. This is the standard technique of classification via Jordan form, and it is non-circular: it uses only linear algebra (existence of Jordan/real-canonical form for real $2\times2$ matrices) and elementary calculus of the exponential and trigonometric functions, none of which presuppose the present theorem.

Step 1: Reduction to canonical form. Every real $2\times 2$ matrix $A$ is similar, via some invertible real or complex change of basis $P$, to exactly one of the following canonical forms $J$, according to its eigenvalue type: (a) $J=\begin{pmatrix}\lambda_1&0\\0&\lambda_2\end{pmatrix}$ if $A$ has two distinct real eigenvalues $\lambda_1\ne\lambda_2$; (b) $J=\begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}$ if $A$ has a repeated real eigenvalue $\lambda$ and is not itself $\lambda I$ (non-diagonalizable case); (c) $J = \lambda I$ if $A = \lambda I$ exactly (repeated eigenvalue, already diagonal — the proper node/star case); (d) $J=\begin{pmatrix}\alpha&-\beta\\\beta&\alpha\end{pmatrix}$, $\beta>0$, if $A$ has complex eigenvalues $\alpha\pm i\beta$. This is a standard fact from linear algebra (Jordan canonical form over $\mathbb C$, converted to a real canonical form by taking real and imaginary parts of a complex eigenvector in case (d)); we take it as known, since it is prerequisite algebra, not part of the ODE theory being developed. If $\mathbf x = P\mathbf y$, then $\dot{\mathbf x}=A\mathbf x$ becomes $\dot{\mathbf y} = J\mathbf y$ (since $P$ is a constant invertible matrix, $\dot{\mathbf y}=P^{-1}\dot{\mathbf x}=P^{-1}AP\mathbf y = J\mathbf y$), and $P$ being invertible and continuous with continuous inverse means $\mathbf x(t)\to\mathbf 0$ iff $\mathbf y(t)\to\mathbf 0$, and stability of the origin for one system is equivalent to stability for the other (a linear change of coordinates is a homeomorphism fixing the origin, so it preserves every topological stability property and preserves orbit-closedness). Trace and determinant are similarity invariants ($\operatorname{tr}(P^{-1}AP)=\operatorname{tr}(A)$, $\det(P^{-1}AP)=\det(A)$), so $T,D$ computed from $J$ agree with those from $A$. It therefore suffices to prove the theorem for each canonical form $J$.

Step 2: Case (a), distinct real eigenvalues — proves items 1, 2. Here $\mathbf y(t) = (c_1 e^{\lambda_1 t}, c_2 e^{\lambda_2 t})$ for arbitrary constants $c_1,c_2$ (immediate by solving the decoupled scalar equations $\dot y_1=\lambda_1y_1$, $\dot y_2 = \lambda_2 y_2$). Here $T=\lambda_1+\lambda_2$, $D=\lambda_1\lambda_2$. If $D<0$: $\lambda_1,\lambda_2$ have opposite signs; say $\lambda_1<0<\lambda_2$. Along the $y_2$-axis ($c_1=0$), solutions grow like $e^{\lambda_2 t}\to\infty$; along the $y_1$-axis they decay to $0$. Every neighborhood of the origin contains points with $c_2\ne0$ arbitrarily small, and for these, $\|\mathbf y(t)\| \ge |c_2|e^{\lambda_2 t}\to\infty$, so such trajectories leave any fixed bounded neighborhood of the origin. Hence the origin is not stable: this is the saddle, unstable, proving item 1. If $D>0$ and $\Delta = (\lambda_1-\lambda_2)^2>0$ (automatic here since eigenvalues distinct) with $T<0$: then $\lambda_1+\lambda_2<0$ and $\lambda_1\lambda_2>0$ force both $\lambda_1,\lambda_2<0$ (if one were $\ge 0$, the product could not be positive while the sum is negative, unless both are negative — check: if both were positive the sum would be positive, contradiction; if one were zero the product would be zero, contradiction; if they had opposite sign the product would be negative, contradiction; so both negative). Then $\|\mathbf y(t)\|^2 = c_1^2e^{2\lambda_1t}+c_2^2e^{2\lambda_2t} \to 0$ as $t\to\infty$, and moreover for $t\ge0$, $\|\mathbf y(t)\| \le \|\mathbf y(0)\|\, e^{\max(\lambda_1,\lambda_2)t}\le \|\mathbf y(0)\|$, so trajectories starting close to $0$ stay close (indeed monotonically closer along each axis, and by the triangle inequality bounded by the initial norm times a decaying exponential overall once $t\ge 0$). This gives both stability and convergence to $0$: asymptotically stable node. If $D>0$, $T>0$: symmetric argument with both $\lambda_1,\lambda_2>0$ gives $\|\mathbf y(t)\|\to\infty$ for every nonzero initial condition, so the origin is unstable: unstable node. This proves item 2 in both sub-cases.

Step 3: Case (b), repeated eigenvalue, non-diagonalizable — proves item 3. Here $J = \begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}$ with $T=2\lambda$, $D=\lambda^2$, so $\Delta = 4\lambda^2-4\lambda^2=0$ automatically, consistent with item 3's hypothesis. Solving $\dot y_2=\lambda y_2$ gives $y_2(t)=c_2e^{\lambda t}$; then $\dot y_1 = \lambda y_1+c_2e^{\lambda t}$ is linear first-order, solved by integrating factor $e^{-\lambda t}$:

So $\mathbf y(t) = e^{\lambda t}(c_1+c_2t,\ c_2)$. If $\lambda<0$ (i.e. $T<0$), then $e^{\lambda t}\to 0$ exponentially, and $e^{\lambda t}t\to0$ as well (exponential decay dominates any polynomial growth, a standard calculus fact: for $\lambda<0$, $\lim_{t\to\infty}te^{\lambda t}=0$ by L'Hôpital or by comparison with $e^{\lambda t/2}$ for large $t$). Hence $\mathbf y(t)\to\mathbf 0$. For stability: given $\varepsilon>0$, since $|y_1(t)|\le e^{\lambda t}(|c_1|+|c_2|t)$ is bounded on $[0,\infty)$ by a constant times $\|\mathbf y(0)\|$ (the function $t\mapsto e^{\lambda t}(1+t)$ is bounded on $[0,\infty)$ for $\lambda<0$, attaining a finite maximum $M$), we get $\|\mathbf y(t)\|\le M\|\mathbf y(0)\|$ for all $t\ge0$, giving Lyapunov stability, hence together with convergence, asymptotic stability. If $\lambda>0$ ($T>0$), taking $c_2=0,c_1\ne0$ gives $y_1(t)=c_1e^{\lambda t}\to\infty$, so unstable. This is the improper (degenerate) node, proving item 3's non-scalar sub-case; the case $A=\lambda I$ exactly gives $\mathbf y(t)=e^{\lambda t}\mathbf y(0)$, straight-line trajectories through the origin (a proper node or star point), with the identical stability dichotomy by the same exponential estimate.

Step 4: Case (d), complex eigenvalues — proves items 4 and 5. Here $J=\begin{pmatrix}\alpha&-\beta\\\beta&\alpha\end{pmatrix}$, $\beta>0$, $T=2\alpha$, $D=\alpha^2+\beta^2>0$, and $\Delta = 4\alpha^2-4(\alpha^2+\beta^2)=-4\beta^2<0$, consistent with the case-4/5 hypotheses ($D>0,\ \Delta<0$). Passing to polar coordinates $y_1=r\cos\theta$, $y_2=r\sin\theta$ in the $\mathbf y$-plane, direct differentiation of $r^2=y_1^2+y_2^2$ along solutions gives

so $\dot r = \alpha r$ (for $r>0$), giving $r(t)=r(0)e^{\alpha t}$. Meanwhile $\theta$ satisfies $\dot\theta = \beta$ (a similar computation with $y_1\dot y_2-y_2\dot y_1=\beta r^2 = r^2\dot\theta$), so $\theta(t)=\theta(0)+\beta t$, a steady rotation at constant angular speed $\beta>0$. If $\alpha = T/2 <0$ ($T<0$): $r(t)=r(0)e^{\alpha t}\to0$ monotonically and, since $r$ is monotonically non-increasing in $t\ge0$ (as $\alpha<0$), $r(t)\le r(0)$ for all $t\ge0$: this gives both convergence to the origin and Lyapunov stability, hence asymptotic stability. Trajectories spiral inward while rotating: stable spiral. If $\alpha>0$ ($T>0$): $r(t)\to\infty$, giving instability: unstable spiral. This proves item 4. If $\alpha=0$ ($T=0$): $r(t)=r(0)$ for all $t$ — every trajectory is a circle of constant radius in the $\mathbf y$-coordinates, traversed at constant angular speed $\beta=\sqrt D>0$, hence a closed periodic orbit of period $2\pi/\beta$. In the original $\mathbf x$-coordinates, $\mathbf x(t) = P\mathbf y(t)$ traces a closed curve (an ellipse, the linear image of a circle) since $\mathbf y$ is periodic with period $2\pi/\beta$ and $P$ is a fixed linear map, so $\mathbf x(t+2\pi/\beta)=\mathbf x(t)$ for all $t$. Since $r(t)=r(0)$ exactly, for any $\varepsilon>0$ choosing $\delta=\varepsilon/\|P\|\cdot\|P^{-1}\|^{-1}$ (more simply: the norm of $\mathbf y(t)$ never changes, and $\|\mathbf x(t)\|\le\|P\|\|\mathbf y(t)\| = \|P\|\|\mathbf y(0)\|\le \|P\|\|P^{-1}\|\|\mathbf x(0)\|$) shows every trajectory starting within $\delta = \varepsilon/(\|P\|\|P^{-1}\|)$ of the origin stays within $\varepsilon$ of it for all time. This gives Lyapunov stability. But $\mathbf x(t)$ does not converge to $\mathbf 0$ for any nonzero initial condition (it has constant, positive $\|\mathbf y(t)\|=\|\mathbf y(0)\|>0$ for all $t$), so it is not asymptotically stable. This is the center, proving item 5.

Step 5: Case $D=0$ — proves item 6. If $D=\lambda_1\lambda_2=0$, one eigenvalue (say $\lambda_2$) is zero. If $\lambda_1\ne0$ we are in case (a) with $\lambda_2=0$: $\mathbf y(t)=(c_1e^{\lambda_1t},c_2)$. The entire $y_1=0$ axis (i.e. the $\lambda_2$-eigenspace, $c_1=0$) consists of fixed points of the flow, since $\dot{\mathbf y}=\mathbf 0$ there. If $\lambda_1<0$ ($T=\lambda_1<0$), trajectories with $c_2\ne 0$ converge to the point $(0,c_2)$ on this line, not to the origin (unless $c_2=0$), yet stay within $|c_1|$ of their starting horizontal distance, bounded — giving stability of the origin (small perturbations stay small, though they do not decay to $\mathbf 0$) but not asymptotic stability, since points on the equilibrium line never move. If $\lambda_1>0$ ($T>0$), $c_1e^{\lambda_1t}\to\infty$ for any $c_1\ne0$, giving instability. If $\lambda_1=\lambda_2=0$ then $T=D=0$ and either $A=0$ (every point is a fixed point, trivially stable, not asymptotically stable) or $A$ is nilpotent nonzero (case (b) with $\lambda=0$), giving $\mathbf y(t)=(c_1+c_2t,c_2)$, unbounded whenever $c_2\ne0$: unstable. This establishes item 6.

Step 6: Hyperbolic summary. Collecting cases 1–4: asymptotic stability occurred precisely when $D>0$ and $T<0$, across the diagonalizable, repeated-eigenvalue, and complex sub-cases uniformly — and this is exactly the condition that both eigenvalues have strictly negative real part, since $\operatorname{Re}(\lambda_{1,2}) = T/2$ when complex, and both real eigenvalues are negative iff their sum is negative and product positive (shown in Step 2). Item 6, $D=0$, was excluded from "hyperbolic" by definition since a zero eigenvalue has zero real part. This completes the proof of every case, hence the theorem. $\blacksquare$