Questions: Phase Portraits for Linear Systems

Question 1

A 2D linear system x′ = Ax has eigenvalues λ₁ = −2 and λ₂ = 3. Without solving the system explicitly, what can you conclude about trajectories near the origin?

• A. All trajectories spiral toward the origin because one eigenvalue is negative
• B. All trajectories converge to the origin because the negative eigenvalue eventually dominates
• C. The origin is a saddle point: trajectories along one eigendirection flow in, but almost every other trajectory eventually escapes to infinity
• D. The origin is a center because the two eigenvalues have equal magnitude

Answer: (C)

Opposite-sign eigenvalues produce a saddle point. The eigenvector for λ₁ = −2 is a stable direction: trajectories starting along it decay toward the origin. The eigenvector for λ₂ = 3 is an unstable direction: trajectories starting along it blow up. For almost every other initial condition, the unstable component eventually dominates and the trajectory escapes to infinity. A saddle is not a spiral (no complex eigenvalues), and neither eigenvalue 'wins'—they control behavior along their respective eigendirections. Option B is the key misconception: in a saddle, the negative eigenvalue does not dominate globally.

Question 2

A linear system has eigenvalues λ = −0.1 ± 3i. What does the phase portrait look like near the origin?

• A. Closed ellipses (center), because the imaginary part is much larger than the real part
• B. An unstable spiral outward, because the imaginary part is positive
• C. A stable spiral inward, because the real part is negative—trajectories rotate and gradually decay toward the origin
• D. A node, because the real part is nonzero

Answer: (C)

Complex eigenvalues λ = α ± βi produce spirals. The imaginary part β drives rotation in the phase plane; the real part α drives growth or decay. Here α = −0.1 < 0, so trajectories decay toward the origin while rotating—a stable spiral. The magnitude of β relative to α determines how 'tight' the spiral is (many rotations before reaching the origin), but not whether it is stable. Option A is the most tempting wrong answer: a center requires α = 0 exactly; even a tiny negative real part creates a spiral that eventually converges.

Question 3

For a 2D linear system x′ = Ax, whether the equilibrium at the origin is stable depends entirely on the signs of the real parts of the eigenvalues of A, not on the specific initial conditions chosen.

• T. True
• F. False

Answer: True

Correct. Stability is a property of the equilibrium itself, not of particular trajectories. If both eigenvalues have negative real parts, every trajectory converges to the origin regardless of initial conditions. If any eigenvalue has a positive real part, almost every trajectory diverges (saddles have one of each). The phase portrait makes this global structure visible: you can classify the equilibrium as stable, unstable, or mixed by reading the eigenvalue signs, without computing any individual solution.

Question 4

A phase portrait is a plot of the state variables x₁(t) and x₂(t) as separate functions of time t.

• T. True
• F. False

Answer: False

False—this confuses a phase portrait with time-domain plots. A phase portrait plots trajectories in the phase plane: the (x₁, x₂) plane, where each axis is one state variable, not time. Time is implicit—a point on a trajectory tells you the state of the system at some moment, and the arrow shows the direction of evolution. This representation reveals the global geometric structure of all solutions simultaneously: you can see whether trajectories converge, diverge, or rotate without ever computing x₁(t) or x₂(t) explicitly.

Question 5

Explain why a phase portrait conveys qualitative information about a dynamical system that a single solution curve x(t) does not.

Model answer: A single solution curve x(t) shows how one specific initial condition evolves over time, in coordinates where time is an axis. A phase portrait plots all trajectories in the state space simultaneously, without explicit time. This reveals the global structure: which directions are stable (attracting), which are unstable (repelling), whether the equilibrium is a node, saddle, or spiral, and how nearby trajectories relate to each other. From the phase portrait you can answer stability questions for all initial conditions at once—something impossible from any single solution.

The deeper point is that eigenvalues give you the phase portrait structure without solving anything. Reading λ signs tells you the qualitative behavior—stable/unstable, spiral/node/saddle—immediately. This geometric intuition built from linear systems then generalizes to nonlinear systems via linearization: the Jacobian at an equilibrium gives a local phase portrait that governs nearby behavior.