Chapter 37 Quiz — The Matrix Exponential and Systems of ODEs
Twelve conceptual checks. Try each before opening the answer. These test understanding — whether you see that the eigenvalues of $A$ are the destiny of $\mathbf{x}'=A\mathbf{x}$, and that the matrix exponential is just those eigenvalues exponentiated.
1. What single formula solves the linear system $\mathbf{x}'=A\mathbf{x}$ with initial state $\mathbf{x}(0)$, and what scalar equation is it the matrix version of?
Answer
$\mathbf{x}(t)=e^{At}\mathbf{x}(0)$. It is the matrix version of $x'=ax$, whose solution is $x(t)=e^{at}x(0)$ — the same object, with the matrix exponential $e^{At}$ in place of the scalar $e^{at}$. The proof rests on the single fact $\frac{d}{dt}e^{At}=Ae^{At}$.2. Why does the series $e^{At}=\sum_k (At)^k/k!$ converge for every square matrix, while diagonalization $A=PDP^{-1}$ does not always exist?
Answer
The series is dominated term-by-term by the convergent scalar series for $e^{\lVert A\rVert|t|}$ (because $\lVert A^k\rVert\le\lVert A\rVert^k$ for a submultiplicative norm), so it converges absolutely with no conditions. Diagonalization requires a *full set of independent eigenvectors*, which defective matrices lack. So $e^{At}$ always exists even when you cannot diagonalize $A$ — you just need a different computation route (Jordan form).3. A system has eigenvalues $\lambda_1=-3$ and $\lambda_2=-0.1$. Is it stable? Which mode dominates the long-run approach to equilibrium, and why?
Answer
Stable — both real parts are negative, so every trajectory decays to $\mathbf{0}$. The *slow* mode $e^{-0.1t}$ (eigenvalue closest to zero) dominates the long run: the fast mode $e^{-3t}$ dies almost immediately, after which the trajectory creeps into the origin along the slow eigenvector. The least-negative eigenvalue sets the system's effective relaxation time.4. You are told an eigenvalue is $\lambda = 0.5 - 4i$. What does each part tell you about the trajectory?
Answer
The real part $\alpha=0.5>0$ means the amplitude *grows* (unstable). The imaginary part $\beta=-4$ means it *oscillates* at angular frequency $4$ (period $2\pi/4=\pi/2$). Together: an outward (growing) spiral. Only the real part decides stability; the imaginary part is the frequency.5. Why can a system with only real eigenvalues never oscillate?
Answer
Oscillation requires the periodic functions $\cos\beta t,\sin\beta t$, which enter solutions *only* through Euler's formula $e^{i\beta t}=\cos\beta t+i\sin\beta t$ — i.e. only when an eigenvalue has a nonzero imaginary part. A real eigenvalue gives a monotone mode $e^{\lambda t}$ that moves straight along its eigenvector without turning around. Oscillation is the fingerprint of complex eigenvalues.6. Classify the phase portrait when the eigenvalues are real with opposite signs, and state whether it is stable.
Answer
A **saddle**, and it is **always unstable**. Trajectories come in along the stable (negative-eigenvalue) eigenvector and go out along the unstable (positive-eigenvalue) one. Only the two trajectories lying exactly on the stable eigenvector reach the origin; every other trajectory eventually escapes along the unstable direction.7. On the trace-determinant plane, how do you tell a node from a spiral, and a stable system from an unstable one?
Answer
The discriminant $\Delta=(\operatorname{tr} A)^2-4\det A$ separates them: $\Delta>0$ ⇒ real eigenvalues ⇒ **node**; $\Delta<0$ ⇒ complex eigenvalues ⇒ **spiral**. Stability is read from the trace: $\operatorname{tr} A<0$ (and $\det A>0$) ⇒ stable; $\operatorname{tr} A>0$ ⇒ unstable. The determinant's sign also flags saddles: $\det A<0$ ⇒ saddle (always). Slogan: trace controls stability, determinant and discriminant control type.8. Why is $e^{A+B}=e^Ae^B$ false in general, and exactly when is it true?
Answer
Because matrix multiplication does not commute. Expanding $e^{A+B}$ produces the term $\frac12(A^2+AB+BA+B^2)$, while $e^Ae^B$ produces $\frac12 A^2+AB+\frac12 B^2$; these match only if $AB+BA=2AB$, i.e. only if $AB=BA$. So the law holds **if and only if $A$ and $B$ commute**. (It always holds for $A$ and $-A$, giving $e^{At}e^{-At}=I$.)9. What is $e^{A\cdot 0}$, and why must this value hold for the solution formula to work?
Answer
$e^{A\cdot 0}=I$ (every term of the series but the constant $I$ vanishes at $t=0$). This is essential because $\mathbf{x}(0)=e^{A\cdot0}\mathbf{x}(0)=I\mathbf{x}(0)=\mathbf{x}(0)$ — the formula must reproduce the given initial state at $t=0$. An exponential that did not start at $I$ could not satisfy the initial condition.10. A defective matrix produces a solution term like $t\,e^{\lambda t}$. Does this destabilize a system whose eigenvalue has $\operatorname{Re}\lambda<0$?
Answer
No. Exponentials beat polynomials: $t\,e^{\lambda t}\to 0$ as $t\to\infty$ whenever $\operatorname{Re}\lambda<0$, because the decaying exponential overwhelms the growing $t$. So a defective eigenvalue with negative real part is still stable — the $t$-factor only adds a transient "bump" before the decay wins. (The *one* exception: a defective eigenvalue with $\operatorname{Re}\lambda=0$ *is* unstable, because then there is no exponential decay to crush the growing $t$.)11. What does it mean geometrically that the columns of $e^{At}$ are trajectories, and why is $e^{At}$ called the "flow" or "state-transition matrix"?
Answer
Applying $e^{At}$ to a starting vector transports it forward by time $t$ along its trajectory; the $j$-th column is the trajectory starting from the standard basis vector $\mathbf{e}_j$. Because the system is linear, every trajectory is a linear combination of these columns. So $e^{At}$ is the single transformation that *is* the passage of time from $0$ to $t$ — it maps the entire state space forward, hence "flow" / "state-transition matrix."12. When should you compute $e^{At}$ by hand via $Pe^{Dt}P^{-1}$, and when should you call scipy.linalg.expm instead?