Questions: Matrix Exponential Method

Question 1

The system dx/dt = Ax has A = PDP⁻¹ where D has eigenvalues λ₁ = -2 and λ₂ = 3. As t → ∞, what determines the long-term behavior of solutions?

• A. The solution decays to zero because the average of the eigenvalues is 0.5, which is small
• B. The solution grows without bound because the eigenvalue with the largest real part (λ₂ = 3) dominates
• C. The solution oscillates because one eigenvalue is negative and one is positive
• D. The behavior depends entirely on the initial condition x₀, not on the eigenvalues

Answer: (B)

The matrix exponential e^(At) = Pe^(Dt)P⁻¹ contains terms e^(λ₁t) = e^(-2t) (decaying) and e^(λ₂t) = e^(3t) (growing). As t → ∞, the decaying mode vanishes and the growing mode dominates — the solution grows like e^(3t). The eigenvalue with the largest real part (here λ₂ = 3) always governs long-term behavior, regardless of the initial condition (unless x₀ is perfectly aligned with the decaying eigenvector). The system is therefore unstable.

Question 2

Why is diagonalization essential to making the matrix exponential computationally tractable?

• A. Diagonalization converts the matrix into a form where the power series for e^(At) terminates after finitely many terms
• B. For a diagonal matrix D, e^(Dt) is simply the diagonal matrix with entries e^(λᵢt), reducing matrix exponentiation to scalar exponentiation
• C. Diagonalization eliminates complex-valued eigenvalues, ensuring all terms in the solution are real
• D. Without diagonalization, the matrix exponential cannot be defined for matrices larger than 2×2

Answer: (B)

The power series definition of e^(At) = I + At + (At)²/2! + ⋯ requires summing infinitely many matrix powers — impractical in general. But if A = PDP⁻¹, the key identity (PDP⁻¹)ⁿ = PD^n P⁻¹ means e^(At) = Pe^(Dt)P⁻¹. And e^(Dt) for a diagonal matrix D is trivial: it's just diag(e^(λ₁t), e^(λ₂t), ...). Each eigenvalue's exponential can be computed as a scalar. Diagonalization doesn't terminate the series — it reorganizes it so only scalar exponentials remain.

Question 3

The matrix exponential e^(At) is defined by the same power series as the scalar exponential e^(at), with scalars replaced by matrices.

• T. True
• F. False

Answer: True

Yes: e^(At) = I + At + (At)²/2! + (At)³/3! + ⋯, where I plays the role of 1 and each term is a matrix power. This series always converges. The scalar analogy is not just notational — it is the precise generalization. The solution x(t) = e^(At)x₀ has exactly the same structure as the scalar solution x(t) = e^(at)x₀, which is why the matrix exponential is the natural framework for linear systems of ODEs.

Question 4

If a 2×2 matrix A has purely imaginary eigenvalues λ = ±iω, the solution x(t) = e^(At)x₀ will decay to zero as t increases.

• T. True
• F. False

Answer: False

Purely imaginary eigenvalues mean e^(λt) = e^(±iωt), which has unit magnitude for all t — it oscillates without growing or decaying. The real part of the eigenvalue determines growth/decay: positive real part → growth, negative real part → decay, zero real part → neither. Purely imaginary eigenvalues (zero real part) produce oscillatory solutions of constant amplitude. Only eigenvalues with negative real parts produce decay to zero.

Question 5

Explain in your own words why the eigenvalue structure of the matrix A completely determines the long-term behavior of solutions to dx/dt = Ax.

Model answer: The solution x(t) = e^(At)x₀ decomposes — via diagonalization — into a sum of modes, one per eigenvalue. Each mode evolves like e^(λᵢt) times the corresponding eigenvector direction. The real part of each eigenvalue determines whether that mode grows, decays, or oscillates at constant amplitude. As t → ∞, modes with positive real-part eigenvalues dominate (growing), modes with negative real-part eigenvalues vanish (decaying), and purely imaginary eigenvalues produce sustained oscillation. The eigenvalue with the largest real part controls the system's eventual fate.

The key insight is that diagonalization decouples the coupled system into independent scalar equations, each solvable separately. The matrix exponential then reassembles these into the original coordinates. This is why stability analysis focuses on eigenvalues: they are the 'natural frequencies' of the system in the eigenvector basis, and each one independently governs one mode of the solution.