**Proof plan.** The eigenvalues of $A$ are determined by $\tau$ and $\Delta$ via the characteristic polynomial. The qualitative dynamics of $\dot{y} = Ay$ follow from the [Jordan normal form](/theorems/864), which depends on the sign of the discriminant $\tau^2 - 4\Delta$ and the signs of $\tau$ and $\Delta$. **Step 1 (Characteristic polynomial).** The eigenvalues of $A \in \mathbb{R}^{2 \times 2}$ satisfy $\lambda^2 - \tau\lambda + \Delta = 0$, giving \begin{align*} \lambda_\pm = \frac{\tau \pm \sqrt{\tau^2 - 4\Delta}}{2}. \end{align*} **Step 2 (Real distinct eigenvalues: $\tau^2 - 4\Delta > 0$).** Both eigenvalues are real. In a basis of eigenvectors, the general solution is $y(t) = c_1 e^{\lambda_+ t} v_+ + c_2 e^{\lambda_- t} v_-$. [claim:Saddle Case] If $\Delta < 0$, then $\lambda_+ > 0 > \lambda_-$ and the origin is a saddle. [/claim] [proof] The product $\lambda_+\lambda_- = \Delta < 0$ forces opposite signs. The solution component along $v_+$ grows while the component along $v_-$ decays, producing a saddle. [/proof] [claim:Node Case] If $\Delta > 0$ and $\tau^2 - 4\Delta > 0$, then $\lambda_\pm$ have the same sign as $\tau$. When $\tau < 0$, both eigenvalues are negative and the origin is a stable node. When $\tau > 0$, both are positive and the origin is an unstable node. [/claim] [proof] Since $\Delta = \lambda_+\lambda_- > 0$, the eigenvalues share a sign. Their sum $\lambda_+ + \lambda_- = \tau$ determines which sign. When $\tau < 0$, both are negative: the general solution $y(t) = c_1 e^{\lambda_+ t}v_+ + c_2 e^{\lambda_- t}v_-$ decays to zero as $t \to \infty$ along both eigendirections. When $\tau > 0$, both are positive and the solution diverges. [/proof] **Step 3 (Complex eigenvalues: $\tau^2 - 4\Delta < 0$).** The eigenvalues are $\lambda_\pm = \alpha \pm i\beta$ where $\alpha = \tau/2$ and $\beta = \sqrt{4\Delta - \tau^2}/2 > 0$. In appropriate real coordinates $(y_1, y_2)$, the general solution is \begin{align*} y_1(t) &= e^{\alpha t}(c_1 \cos\beta t + c_2 \sin\beta t), \\ y_2(t) &= e^{\alpha t}(-c_1 \sin\beta t + c_2 \cos\beta t). \end{align*} When $\alpha < 0$ ($\tau < 0$), the oscillatory solution decays: stable spiral. When $\alpha > 0$ ($\tau > 0$): unstable spiral. When $\alpha = 0$ ($\tau = 0$), the solution is purely periodic with period $2\pi/\beta$: centre. **Step 4 (Repeated eigenvalues: $\tau^2 - 4\Delta = 0$).** The repeated eigenvalue is $\lambda = \tau/2$. If $A$ is diagonalisable, the origin is a star node. Otherwise, the Jordan block produces the solution $y(t) = e^{\lambda t}(c_1 v + c_2(tv + w))$ where $w$ is the generalised eigenvector, giving a degenerate node. In both cases, stability is determined by the sign of $\tau$.