Eigenvalues of A — system modes, stability

WHY do we care about eigenvalues at all?

WHAT we have: a state vector $\mathbf{x}(t)\in\mathbb{R}^n$ (e.g. spacecraft attitude error & its rate) obeying $\dot{\mathbf{x}} = A\mathbf{x}$.

WHY it's hard: the equations are coupled — $\dot{x}_1$ depends on $x_2$, $\dot{x}_2$ on $x_1$, etc. You can't solve them one at a time.

HOW eigenvalues rescue us: find directions where $A$ acts like pure scaling. Along those directions the whole tangle collapses into $n$ separate scalar ODEs, each trivially solvable.

Derivation from scratch: guess the solution

Step 1 — Ansatz. Try a solution that keeps a fixed shape $\mathbf{v}$ and only changes size in time: $\mathbf{x}(t) = e^{\lambda t}\,\mathbf{v}$. Why this step? If any solution stays pointed in one direction, the vector ODE becomes scalar — the simplest thing that could possibly work, so we test it.

Step 2 — Substitute. Differentiate: $\dot{\mathbf{x}} = \lambda e^{\lambda t}\mathbf{v}$. Put into $\dot{\mathbf{x}}=A\mathbf{x}$: $\lambda e^{\lambda t}\mathbf{v} = A e^{\lambda t}\mathbf{v}.$ Why this step? We force the ansatz to actually obey the physics.

Step 3 — Cancel $e^{\lambda t}$ (never zero): $A\mathbf{v}=\lambda\mathbf{v}\quad\Rightarrow\quad (A-\lambda I)\mathbf{v}=\mathbf{0}.$ Why this step? We discover the ansatz works only if $(\lambda,\mathbf{v})$ is an eigenpair. The exponentials are baked into $A$.

Step 4 — Non-trivial solution condition. $(A-\lambda I)\mathbf{v}=0$ has a nonzero $\mathbf{v}$ iff the matrix is singular: $\boxed{\det(A-\lambda I)=0}$ This polynomial in $\lambda$ is the characteristic equation; its roots are the eigenvalues. Why this step? An invertible matrix would force $\mathbf{v}=0$, which we banned.

What each eigenvalue does — reading the complex plane

Write a complex eigenvalue as $\lambda=\sigma+j\omega$. Then, using Euler, $e^{\lambda t}=e^{\sigma t}\big(\cos\omega t + j\sin\omega t\big).$

• $\sigma=\text{Re}(\lambda)$ → growth/decay rate of the envelope $e^{\sigma t}$.
• $\omega=\text{Im}(\lambda)$ → oscillation frequency of that mode.

WHY only the real part matters: $|e^{\lambda t}| = e^{\sigma t}$. The oscillation $|{\cos+j\sin}|=1$ never grows; only the real-part exponential decides whether magnitude blows up or shrinks.

For a physical 2nd-order mode $\lambda=-\zeta\omega_n\pm j\omega_n\sqrt{1-\zeta^2}$: $\sigma=-\zeta\omega_n$ gives damping, $\omega_d=\omega_n\sqrt{1-\zeta^2}$ the damped frequency.

Worked Example 1 — a stable oscillator (spring–damper as state space)

$m\ddot q + c\dot q + kq=0$. Let $x_1=q,\ x_2=\dot q$. Then $A=\begin{pmatrix}0&1\\-k/m&-c/m\end{pmatrix}.$ Take $k/m=4,\ c/m=2$.

Step 1 — characteristic eqn: $\det(A-\lambda I)=\lambda^2 + \tfrac{c}{m}\lambda + \tfrac{k}{m} = \lambda^2+2\lambda+4=0.$ Why? For the companion form of a scalar 2nd-order ODE, $\det(A-\lambda I)$ reproduces the ODE's own characteristic polynomial — a great sanity check.

Step 2 — roots: $\lambda=\dfrac{-2\pm\sqrt{4-16}}{2}=-1\pm j\sqrt3.$ Why? Quadratic formula.

Step 3 — read it: $\sigma=-1<0$ → stable; $\omega_d=\sqrt3$ → it oscillates while decaying. $\omega_n=\sqrt{k/m}=2$, $\zeta=\sigma/(-\omega_n)=0.5$ (underdamped). Response: $e^{-t}\cos(\sqrt3\,t+\phi)$.

Worked Example 2 — decoupled by eigenvectors

$A=\begin{pmatrix}-3&1\\0&-2\end{pmatrix}.$ Step 1 — triangular ⇒ eigenvalues on diagonal: $\lambda_1=-3,\ \lambda_2=-2$. Why? $\det(A-\lambda I)=(-3-\lambda)(-2-\lambda)$ for triangular matrices.

Step 2 — eigenvectors. For $\lambda_1=-3$: $(A+3I)\mathbf v=0\Rightarrow\begin{pmatrix}0&1\\0&1\end{pmatrix}\mathbf v=0\Rightarrow \mathbf v_1=(1,0)^T.$ For $\lambda_2=-2$: $(A+2I)\mathbf v=0\Rightarrow\begin{pmatrix}-1&1\\0&0\end{pmatrix}\mathbf v=0\Rightarrow \mathbf v_2=(1,1)^T.$

Step 3 — solution $\mathbf x(t)=c_1 e^{-3t}(1,0)^T+c_2 e^{-2t}(1,1)^T$. The $-3$ mode dies faster; long-term behaviour dominated by the slowest (least-negative) eigenvalue $-2$. Why this matters: the eigenvalue closest to the imaginary axis sets the settling time.

Worked Example 3 — Forecast-then-Verify

$A=\begin{pmatrix}0&1\\2&-1\end{pmatrix}.$ Forecast: the $+2$ off-diagonal looks like positive feedback → probably unstable. Verify: $\det(A-\lambda I)=\lambda^2+\lambda-2=(\lambda+2)(\lambda-1)$ ⇒ $\lambda=-2,\ +1$. Since $+1>0$ → unstable. Forecast confirmed; the growing mode is $e^{+t}\mathbf v$.

Flashcards

Connections

• State-Space Representation — where $\dot{\mathbf x}=A\mathbf x+B\mathbf u$ comes from.
• Characteristic Polynomial & Determinants — how $\det(A-\lambda I)$ is computed.
• Diagonalization and Modal Decomposition — $A=V\Lambda V^{-1}$ makes modes explicit.
• Routh–Hurwitz Criterion — stability without finding roots for $n>2$.
• Pole Placement & LQR — GNC control design moves eigenvalues into the left half-plane.
• Damping Ratio and Natural Frequency — physical reading of complex eigenvalues.
• Transfer Functions and Poles — poles = eigenvalues of $A$ (of a minimal realization).

Concept Map

Hinglish (regional understanding)