Questions: Matrix Exponential and Differential Equations

Question 1

A system x' = Ax has eigenvalues −2 and 3. Starting from any nonzero initial condition, what happens to solutions as t → ∞?

• A. All solutions decay to zero, since one eigenvalue is negative
• B. All solutions grow without bound (for generic initial conditions), since the positive eigenvalue component dominates
• C. Solutions oscillate between growing and decaying phases
• D. Only initial conditions aligned with the first eigenvector will cause growth

Answer: (B)

The solution is x(t) = e^{At}x₀ = c₁e^{−2t}v₁ + c₂e^{3t}v₂, where v₁, v₂ are eigenvectors. As t → ∞, the e^{−2t} term vanishes but the e^{3t} term grows without bound. For any initial condition with a nonzero component along v₂ (i.e., generic x₀), solutions grow. The negative eigenvalue does not save the system — it only controls one mode. The positive eigenvalue governs long-term behavior. A student who says 'one eigenvalue is negative so it might decay' is missing that growth always wins over decay as t → ∞.

Question 2

Why does the diagonalization A = PDP⁻¹ make computing e^{At} tractable?

• A. It reduces A to upper triangular form, where matrix exponentials always simplify
• B. Because Aⁿ = PDⁿP⁻¹, the power series for e^{At} collapses to scalar exponentials e^{λᵢt} on the diagonal of e^{Dt}
• C. Diagonalization ensures all eigenvalues are real, eliminating oscillatory behavior
• D. It converts the matrix ODE into a system of decoupled polynomial equations

Answer: (B)

The key algebraic fact is Aⁿ = PDⁿP⁻¹, which follows by induction since (PDP⁻¹)ⁿ = PDⁿP⁻¹. Substituting into the series e^{At} = Σ (At)ⁿ/n!, you get P(Σ(Dt)ⁿ/n!)P⁻¹ = Pe^{Dt}P⁻¹. Since D is diagonal, e^{Dt} is simply diagonal with entries e^{λᵢt} — reducing the entire computation to scalar exponentials applied to eigenvalues. Diagonalization decouples the system into independent one-dimensional ODEs along eigenvector directions.

Question 3

If a 2×2 matrix A has purely imaginary eigenvalues ±iω, then e^{At} produces oscillating solutions without growth or decay.

• T. True
• F. False

Answer: True

Eigenvalues with zero real part give e^{(±iω)t} = cos(ωt) ± i·sin(ωt). The magnitude |e^{λt}| = e^{Re(λ)t} = e^0 = 1 for all t, so solutions neither grow nor shrink. The system oscillates forever at frequency ω. This is the undamped harmonic oscillator — purely imaginary eigenvalues are the hallmark of sustained periodic motion.

Question 4

The matrix exponential satisfies e^{A+B} = e^A · e^B for any square matrices A and B.

• T. True
• F. False

Answer: False

This identity holds for scalars but fails for matrices unless A and B commute (AB = BA). The proof for scalars uses the binomial theorem, which requires commutativity. For noncommuting matrices, cross terms in the series expansion do not cancel properly. A simple counterexample: A = [[0,1],[0,0]], B = [[0,0],[1,0]] gives AB ≠ BA and e^{A+B} ≠ e^A e^B. The correct generalization is the Baker–Campbell–Hausdorff formula.

Question 5

A linear system x' = Ax has all eigenvalues with strictly negative real parts. What can you conclude about solutions for any initial condition, and why does this follow from the matrix exponential?

Model answer: All solutions x(t) = e^{At}x₀ decay to zero as t → ∞. This follows because each mode in the solution is proportional to e^{λᵢt}, and |e^{λᵢt}| = e^{Re(λᵢ)t} → 0 as t → ∞ when Re(λᵢ) < 0. The matrix exponential decomposes the solution into independent modes — one per eigenvalue — and each mode decays exponentially. The system is asymptotically stable: regardless of where you start, the trajectory converges to the origin.

The matrix exponential makes long-term behavior a purely spectral question: look at the eigenvalues of A. Negative real parts → decay; positive real parts → growth; zero real parts → sustained oscillation or polynomial growth (if Jordan blocks exist). Engineers designing stable control systems, physicists modeling dissipative oscillators, and mathematicians analyzing dynamical systems all use this eigenvalue criterion as their first diagnostic.