A phase portrait draws the trajectories of $\mathbf{x}' = A\mathbf{x}$ in the $(x, y)$ plane, without time on an axis. You can predict its shape from the eigenvalues alone, which is often all you need: will the system settle, blow up, or circle forever? This builds on eigenvalue solutions.
Key idea
For $2\times 2$ systems with $\det A \ne 0$: real eigenvalues of opposite sign, saddle; both negative, stable node (sink); both positive, unstable node (source). Complex $\alpha \pm \beta i$: spiral sink if $\alpha < 0$, spiral source if $\alpha > 0$, center if $\alpha = 0$. With $\tau = \operatorname{tr}A$, $\Delta = \det A$: saddle iff $\Delta < 0$; otherwise stable iff $\tau < 0$; spirals iff $\tau^{2} < 4\Delta$.
Why it works
Every solution is a combination of $e^{\lambda_1t}\mathbf{v}_1$ and $e^{\lambda_2t}\mathbf{v}_2$. Negative eigenvalues make both parts shrink, so everything flows to the origin; a positive one makes its part grow. In a saddle, the component along the stable eigenvector dies out while the other grows, so trajectories come in along one line and leave along the other. Complex eigenvalues bring $\cos\beta t$ and $\sin\beta t$, so the motion rotates, while $e^{\alpha t}$ decides whether it spirals in or out. The eigenvalues are the roots of $\lambda^{2} - \tau\lambda + \Delta = 0$, which is why $\tau$ and $\Delta$ are enough.
Worked example: From the eigenvalues
The eigenvalues are $-1$ and $-4$. Classify the origin.
- Both are real and negative: every solution decays.
Answer: stable node (sink)
Worked example: From the matrix
Classify the origin for $A = \begin{bmatrix} 1 & -5 \\ 1 & -3 \end{bmatrix}$.
- $\tau = -2$ and $\Delta = -3 + 5 = 2 > 0$, so not a saddle; $\tau < 0$, so stable.
- $\tau^{2} - 4\Delta = 4 - 8 = -4 < 0$: complex eigenvalues $-1 \pm i$.
Answer: spiral sink
Worked example: A parameter
For $A = \begin{bmatrix} 0 & 1 \\ -k & -2 \end{bmatrix}$ (a damped spring), when is the origin a spiral sink?
- $\tau = -2$, $\Delta = k$, $\tau^{2} - 4\Delta = 4 - 4k$.
- Saddle if $k < 0$. For $k > 0$ it is stable; spirals need $4 - 4k < 0$.
Answer: $k > 1$ (and $0 < k < 1$ gives a stable node: the overdamped spring)
Common mistake
Deciding stability from the eigenvectors. Eigenvectors only set the directions of the straight-line solutions; whether motion along them grows or shrinks is decided by the signs of the eigenvalues (or of their real part). Two eigenvectors pointing anywhere can belong to a sink, a source or a saddle.
| Eigenvalues | Type | Stable? |
|---|---|---|
| real, opposite signs | saddle | no |
| real, both negative | stable node | yes |
| real, both positive | unstable node | no |
| $\alpha \pm \beta i$, $\alpha < 0$ | spiral sink | yes |
| $\alpha \pm \beta i$, $\alpha > 0$ | spiral source | no |
| $\pm\beta i$ | center | neutrally |