Phase Portraits for Linear Systems

We use the eigenvalue method: solve the characteristic equation, exhibit the general solution explicitly in each regime, and read off the geometry directly from the exponential/trigonometric formulas. This is a direct proof by case analysis on the discriminant, not relying on any classification theorem stated for nonlinear systems (avoiding circularity with Stability and Classification of Equilibria, which is a consequence of this result, not a tool used here).

Step 1: the characteristic equation and its roots. The eigenvalues of $A$ satisfy

This is immediate from expanding $\det(A - \lambda I) = \lambda^2 - (\mathrm{tr} A)\lambda + \det A$ for a $2\times 2$ matrix, which follows by direct computation writing $A = \begin{pmatrix} a & b\\ c& d\end{pmatrix}$: $\det(A-\lambda I) = (a-\lambda)(d-\lambda) - bc = \lambda^2 - (a+d)\lambda + (ad-bc) = \lambda^2 - T\lambda + D$. Since we assumed $D = \lambda_1\lambda_2 \ne 0$, neither eigenvalue is zero, so $\mathbf{x}=\mathbf{0}$ is the only equilibrium (the only solution of $A\mathbf{x}=\mathbf{0}$ when $A$ is invertible).

Step 2: Case $D<0$ (saddle). Since $\Delta = T^2 - 4D > T^2 \geq 0$ (as $-4D>0$), the eigenvalues are real and distinct, say $\lambda_1 < \lambda_2$. Since $\lambda_1\lambda_2 = D<0$, they have opposite signs; write $\lambda_1<0<\lambda_2$. By the eigenvalue method (a prerequisite result), $A$ has linearly independent real eigenvectors $\mathbf{v}_1,\mathbf{v}_2$ (distinct real eigenvalues of a real matrix always admit linearly independent eigenvectors, by the standard linear-algebra fact that eigenvectors of distinct eigenvalues are linearly independent), and the general solution is

If $c_2 = 0$, $c_1\ne0$: the trajectory lies on the line through $\mathbf{v}_1$, and since $\lambda_1<0$, $\mathbf{x}(t)\to \mathbf{0}$ as $t\to\infty$ and $\|\mathbf{x}(t)\|\to\infty$ as $t\to -\infty$. This is the stable eigenline. If $c_1=0,c_2\ne0$: symmetric argument gives the unstable eigenline, with $\mathbf{x}(t)\to\mathbf 0$ as $t\to-\infty$. If $c_1,c_2\ne 0$: as $t\to+\infty$, $e^{\lambda_2 t}$ dominates $e^{\lambda_1 t}$ (since $\lambda_2>0>\lambda_1$), so

since $(\lambda_1-\lambda_2)<0$ forces the second term to vanish. Symmetrically, as $t\to-\infty$, the trajectory is asymptotic to the $\mathbf{v}_1$ direction. Thus every trajectory off the eigenlines approaches $\mathbf{v}_2$'s line forward in time and $\mathbf{v}_1$'s line backward in time, the classic hyperbolic saddle geometry. Since $\|\mathbf{x}(t)\|\to\infty$ along every trajectory except the stable line, the origin is unstable.

Step 3: Case $D>0$, $\Delta > 0$ (improper node). As in Step 2, real distinct eigenvalues $\lambda_1\ne \lambda_2$ give independent eigenvectors and the same general solution formula. But now $\lambda_1\lambda_2 = D>0$ forces $\lambda_1,\lambda_2$ to share a sign, and $\lambda_1+\lambda_2=T$ tells us that sign is the sign of $T$ (if both were negative their sum would be negative, and conversely). Assume $T<0$, so $\lambda_1<\lambda_2<0$ (relabeling so $\lambda_1$ is more negative). Then for any $c_1,c_2$ not both zero,

since both exponentials decay. To see the direction of approach, factor out the slower-decaying (less negative) exponential $e^{\lambda_2 t}$:

Since $\lambda_1 - \lambda_2 < 0$, the term $e^{(\lambda_1-\lambda_2)t}\to 0$ as $t\to\infty$ provided $c_2 \ne 0$, so the direction of $\mathbf{x}(t)$ tends to that of $\mathbf{v}_2$: all trajectories with $c_2\ne 0$ become tangent to the $\mathbf{v}_2$-eigenline (the "slow" direction, associated to the eigenvalue closer to $0$) as $t\to\infty$. Trajectories with $c_2=0$ lie exactly on the $\mathbf{v}_1$ line. As $t\to -\infty$, $\|\mathbf{x}(t)\|\to\infty$ for all nonzero solutions, so the origin is asymptotically stable ($T<0$) and this is a stable improper node. Reversing all inequalities handles $T>0$, giving an unstable node with all motion reversed in time.

Step 4: Case $D>0$, $\Delta=0$, repeated eigenvalue $\lambda = T/2$, diagonalizable ($A=\lambda I$, proper/star node). If $A = \lambda I$ then every nonzero vector is an eigenvector and the general solution is $\mathbf{x}(t) = e^{\lambda t}\mathbf{x}(0)$: every trajectory is a ray from the origin (or through it), traversed radially outward if $\lambda>0$ or inward if $\lambda<0$, with all directions equally "fast" — hence "star" or "proper" node.

Step 5: Case $D>0$, $\Delta = 0$, repeated eigenvalue not diagonalizable (degenerate node). By the theory of Complex and Repeated Eigenvalues, when $A$ has a repeated eigenvalue $\lambda=T/2$ with only a one-dimensional eigenspace spanned by $\mathbf v$, there is a generalized eigenvector $\mathbf{w}$ with $(A-\lambda I)\mathbf w = \mathbf v$, and the general solution is

If $c_2 = 0$: trajectory lies on the eigenline through $\mathbf v$. If $c_2\ne 0$, write

As $t\to \pm\infty$ (whichever direction makes $e^{\lambda t}\to 0$, i.e. $t\to +\infty$ if $\lambda<0$), the coefficient $(c_1+c_2 t)$ dominates the fixed vector $c_2\mathbf w$ in the bracket, so the direction of $\mathbf x(t)$ tends to $\pm\mathbf v$: every trajectory becomes tangent to the single eigenline as it approaches the origin, and $\|\mathbf x(t)\|\to 0$ because $t e^{\lambda t}\to 0$ for $\lambda <0$ (standard fact: exponential decay beats polynomial growth, provable by L'Hôpital or by the ratio test applied to $t^n e^{\lambda t}$). This produces the characteristic "combed" degenerate node, distinct in shape from the star node of Step 4 because trajectories are curved, tangent to one line, rather than straight rays.

Step 6: Case $\Delta<0$ (complex eigenvalues), $T\ne0$ (spiral). Since $\Delta<0$, the eigenvalues are complex conjugates $\lambda = \alpha\pm i\beta$ with $\alpha = T/2$, $\beta = \sqrt{-\Delta}/2>0$ (nonzero because $\Delta<0$). By the method for complex eigenvalues (prerequisite result), with complex eigenvector $\mathbf{v} = \mathbf{p}+i\mathbf{q}$ for eigenvalue $\alpha+i\beta$, two real, linearly independent solutions are

and the general real solution is $\mathbf x(t) = c_1\mathbf u(t) + c_2 \mathbf w(t)$. In the basis $\{\mathbf p, \mathbf q\}$ (linearly independent because $\mathbf v$ is a genuine complex, not real, eigenvector — otherwise the eigenvalue would be real), write $\mathbf{x}(t) = e^{\alpha t} R(t) \binom{\cos(\beta t + \phi)}{\ast}$-type expression; more directly, compute

because the bracket is a fixed quadratic form in $\cos\beta t,\sin\beta t$ evaluated along a circle, hence bounded above and below by positive constants when $\mathbf p,\mathbf q$ are linearly independent. Therefore $\|\mathbf x(t)\| \to 0$ as $t\to\infty$ if $\alpha = T/2<0$, and $\|\mathbf x(t)\|\to\infty$ if $\alpha>0$; in either case the direction rotates monotonically (the angular part $\beta t$ advances at constant rate, since $\beta\ne0$), producing a spiral: inward (stable) for $T<0$, outward (unstable) for $T>0$.

Step 7: Case $\Delta<0$, $T=0$ (center). Repeat Step 6 with $\alpha=0$: then $\|\mathbf x(t)\|^2$ is exactly periodic and bounded above and below by positive constants (no $e^{2\alpha t}$ factor), so trajectories are closed curves, retraced with period $2\pi/\beta$ exactly (since $\cos,\sin$ have period $2\pi/\beta$ in argument $\beta t$). The origin is Lyapunov stable, since nearby trajectories stay on nearby bounded closed curves and never approach or leave the origin, but it is not asymptotically stable, since $\|\mathbf x(t)\|$ does not tend to $0$.

Step 8: exhaustiveness and mutual exclusivity. Given $D\ne 0$, exactly one of $D<0$ or $D>0$ holds. If $D>0$, exactly one of $\Delta>0$, $\Delta=0$, $\Delta<0$ holds, and if $\Delta=0$, exactly one of $A=\lambda I$ (diagonalizable) or not holds. If $\Delta<0$, exactly one of $T=0$ or $T\ne0$ holds. These conditions partition $\{(T,D):D\ne0\}$ into exactly the cases enumerated, each shown above to produce the claimed geometry, and no matrix can satisfy two of these mutually exclusive numerical conditions simultaneously. $\blacksquare$