Questions: Stability Classification of Linear Systems

Question 1

A 2×2 linear system x' = Ax has eigenvalues λ = ±3i (purely imaginary, distinct). What is the stability classification of the origin?

• A. Asymptotically stable, because the zero real part means solutions neither grow nor decay, which satisfies the stability definition
• B. Marginally stable (center), because the eigenvalues are distinct, the matrix is non-defective, and solutions are purely oscillatory
• C. Unstable, because imaginary eigenvalues always indicate oscillatory growth in linear systems
• D. Cannot be determined without also knowing the eigenvectors

Answer: (B)

Purely imaginary eigenvalues give marginal stability if and only if the matrix is non-defective (geometric multiplicity = algebraic multiplicity for each eigenvalue). Here the two eigenvalues ±3i are distinct, so the matrix must be diagonalizable over ℂ — non-defective by definition. Solutions take the form c₁e^(3it) + c₂e^(-3it), which are bounded oscillations that neither grow nor decay. If the eigenvalues were repeated and the matrix defective, terms like te^(3it) would appear, growing without bound and making the origin unstable despite purely imaginary eigenvalues.

Question 2

For a 2×2 linear system, tr(A) = 0 and det(A) = 9. What is the stability classification of the origin, and why?

• A. Asymptotically stable, because zero trace means eigenvalues sum to zero, so they must have negative real parts
• B. Marginally stable (center), because tr = 0 and det > 0 imply purely imaginary eigenvalues with no real part
• C. Unstable, because positive determinant with zero trace indicates a saddle point
• D. Stability cannot be determined from trace and determinant alone

Answer: (B)

For a 2×2 system, the eigenvalues satisfy λ² − tr(A)λ + det(A) = 0. With tr = 0 and det = 9: λ² + 9 = 0, giving λ = ±3i. Purely imaginary eigenvalues → all Re(λ) = 0. Since the eigenvalues are distinct (non-defective), the origin is a center — marginally stable. The trace-determinant diagram codifies this: det > 0 and tr = 0 places the system exactly on the boundary between stable and unstable spirals, giving a center.

Question 3

A linear system can be unstable even if all eigenvalues have zero real part, provided the matrix is defective.

• T. True
• F. False

Answer: True

This is the subtle condition for marginal stability. When a matrix has a repeated eigenvalue λ with Re(λ) = 0 but insufficient independent eigenvectors (geometric multiplicity < algebraic multiplicity), the general solution contains terms like tⁿe^(λt). Even with Re(λ) = 0, the polynomial factor tⁿ grows without bound as t → ∞, and trajectories escape to infinity. Marginal stability requires both Re(λ) = 0 for all eigenvalues AND that the matrix is non-defective. The purely imaginary condition alone is not enough.

Question 4

For a 2×2 linear system where det(A) > 0 and tr(A) < 0, the origin is marginally stable.

• T. True
• F. False

Answer: False

With det > 0 and tr < 0, consider the eigenvalue equation λ² − tr(A)λ + det(A) = 0. Since tr < 0, −tr > 0, and since det > 0, the product of eigenvalues is positive and their sum is negative. Either both eigenvalues are real and negative, or they are complex conjugates with negative real part (stable spiral). In either case, all Re(λ) < 0, making the origin asymptotically stable — trajectories decay to the origin. Marginal stability requires tr = 0 (with det > 0) to get purely imaginary eigenvalues.

Question 5

The condition Re(λ) = 0 for all eigenvalues is necessary but not sufficient for marginal stability. What additional condition is required, and what goes wrong when it fails?

Model answer: The additional condition is that the matrix must be non-defective: for every eigenvalue, its geometric multiplicity (the number of linearly independent eigenvectors) must equal its algebraic multiplicity (its multiplicity as a root of the characteristic polynomial). When this fails — when the matrix is defective — the solution contains generalized eigenvector terms of the form tⁿe^(λt). Even when Re(λ) = 0, the polynomial factor tⁿ grows without bound as t → ∞. Trajectories that start near the origin move away from it, so the origin is unstable despite the purely imaginary eigenvalues. The canonical example is a 2×2 matrix with a single repeated eigenvalue λ = 0i and only one independent eigenvector; the solution then contains a term growing linearly in time. Marginal stability requires the system to be able to produce purely oscillatory solutions with no polynomial growth, which demands a complete set of independent eigenvectors.