Worked examples — Eigenvalues of A — system modes, stability

Before anything, three tiny bits of vocabulary the parent used but a newcomer should have nailed down.

Now the map every answer lives on:

• The horizontal axis is $\sigma =\text{Re}(\lambda )$ — the real part, the growth/decay knob. Left of the vertical line = decays = good.
• The vertical axis is $\omega =\text{Im}(\lambda )$ — the imaginary part, the wobble knob. Higher up = faster oscillation.
• The vertical line itself ($\sigma =0$) is the imaginary axis, the knife-edge between stable and unstable.

Everything below is just: "solve $\det (A-\lambda I)=0$, then find the dot on this map."

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The scenario matrix

The $2\times 2$ examples below hit every distinct eigenvalue picture a real $2\times 2$ can produce; the final $3\times 3$ shows how the same reading extends to higher order (it treats one representative $3\times 3$ — a stable companion form — rather than every $3\times 3$ sub-case).

Cell	Eigenvalue signature	Where on the map	Behaviour	Example
A	Two real, both $<0$, distinct	Both dots on left axis	Stable, no oscillation (overdamped)	Ex 1
B	Complex pair, $\sigma <0$	Mirror pair, left side	Stable, oscillates (underdamped)	Ex 2
C	Complex pair, $\sigma >0$	Mirror pair, right side	Unstable, growing oscillation	Ex 3
D	Two real, opposite signs	One left, one right	Unstable (saddle)	Ex 4
E	Pure imaginary, $\sigma =0$, simple	Pair on vertical line	Marginally stable (constant wobble)	Ex 5
F	Repeated $\lambda =0$ (defective)	Double dot at origin	Unstable via $t$-growth	Ex 6
G	Repeated real $<0$ ($\zeta =1$, critically damped)	Double dot on left axis	Stable, fastest non-oscillating	Ex 7a
H	Limiting sweep: $\zeta$ from $0\to 1\to \infty$	Dots slide toward axis / merge / split	Damping continuum	Ex 7b
I	$3\times 3$ real-world: attitude + integrator	Three dots	Design/stability read	Ex 8

We now cover every cell.

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Cell A — two real negative eigenvalues (overdamped)

Forecast: big damping ($4$) versus modest stiffness ($3$) — guess: stable, and too damped to oscillate. Two real negative roots.

Step 1 — characteristic polynomial. $\det (A-\lambda I)={\lambda }^2+\frac{c}{m}\lambda +\frac{k}{m}={\lambda }^2+4\lambda +3=0.$ Why this step? This is a companion matrix, so (by the definition above) the determinant is just the ODE's polynomial read off the coefficients — the fastest route.

Step 2 — factor. ${\lambda }^2+4\lambda +3=(\lambda +1)(\lambda +3)=0\Rightarrow {\lambda }_1=-1,{\lambda }_2=-3.$ Why this step? Real, distinct, both negative → Cell A, both dots on the far-left of the map, no imaginary part, so no oscillation.

Step 3 — read damping. ${\omega }_n=\sqrt{k/m}=\sqrt{3}$, and $2\zeta {\omega }_n=c/m=4$, so $\zeta =\frac{4}{2\sqrt{3}}=\frac{2}{\sqrt{3}}\approx 1.155>1.$ Why this step? $\zeta >1$ confirms overdamped — the mathematical signature of two real roots.

Verify: discriminant $=4^2-4\cdot 3=4>0$ → two real roots (matches). Slowest root $-1$ sets the settling time; response is $c_1{e}^{-t}+c_2{e}^{-3t}$, purely decaying. ✓

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Cell B — complex pair, left half-plane (underdamped stable)

Forecast: small damping ($2$) versus stiffness ($4$) — guess: stable but ringing (complex pair, left side).

Step 1 — polynomial. $\det (A-\lambda I)={\lambda }^2+2\lambda +4=0.$ Why this step? Same companion-form shortcut.

Step 2 — roots. $\lambda =\frac{-2\pm \sqrt{4-16}}{2}=\frac{-2\pm \sqrt{-12}}{2}=-1\pm j\sqrt{3}.$ Why this step? Negative discriminant $=-12<0$ → the $\sqrt{-12}=2\sqrt{3}j$ brings in $j$, marking a complex pair; $\sigma =-1<0$ → Cell B, mirror dots on the left half of the map.

Step 3 — physical numbers. ${\omega }_n=\sqrt{4}=2$; $\zeta {\omega }_n=1\Rightarrow \zeta =0.5$; damped frequency ${\omega }_d={\omega }_n\sqrt{1-{\zeta }^2}=2\sqrt{1-0.25}=\sqrt{3}\approx \mathrm{1.732.}$ Why this step? ${\omega }_d$ is exactly the imaginary part — the actual wobble rate you'd measure.

Verify: $\sigma =-\zeta {\omega }_n=-0.5\cdot 2=-1$ ✓; $|\lambda |=\sqrt{1+3}=2={\omega }_n$ ✓. Stable, underdamped, rings at $\sqrt{3}$ rad/s. See the parent's phase-plane picture for the inward spiral.

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Cell C — complex pair, right half-plane (growing oscillation)

Forecast: positive "damping" injects energy — guess: still oscillates (stiffness unchanged) but now grows. Complex pair on the right.

Step 1 — polynomial. $\det (A-\lambda I)={\lambda }^2-2\lambda +4=0.$ Why this step? The $+2$ makes the $\lambda$ coefficient negative — first warning sign.

Step 2 — roots. $\lambda =\frac{2\pm \sqrt{4-16}}{2}=+1\pm j\sqrt{3}.$ Why this step? Discriminant still $-12<0$ → complex pair (again the $j$), but now $\sigma =+1>0$ → Cell C, mirror dots on the right of the map.

Step 3 — verdict. Envelope $e^{+t}$ multiplies the oscillation → magnitude blows up while ringing. Unstable. Why this step? Only $\sigma$ decides stability; the wobble is irrelevant to blow-up.

Verify: compare with the Routh–Hurwitz Criterion: for ${\lambda }^2+a_1\lambda +a_0$ stability needs $a_1>0$ and $a_0>0$. Here $a_1=-2<0$ → fails → unstable ✓.

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Cell D — real, opposite signs (saddle)

Forecast: the $+2$ is positive feedback → guess one growing, one decaying mode. A saddle: Cell D.

Step 1 — polynomial. $\det (A-\lambda I)={\lambda }^2+\lambda -2=0.$ Why this step? Note the negative constant term $-2$ — for a $2\times 2$, constant term $=\det A={\lambda }_1{\lambda }_2<0$ forces opposite signs immediately.

Step 2 — factor. $(\lambda +2)(\lambda -1)=0\Rightarrow \lambda =-2,+1.$ Why this step? Real, one negative one positive → confirms Cell D.

Step 3 — verdict. The $e^{+t}$ mode grows along its eigenvector → unstable saddle.

Verify: $\det A=(0)(-1)-(1)(2)=-2={\lambda }_1{\lambda }_2=(-2)(1)=-2$ ✓. $\text{tr}A=-1={\lambda }_1+{\lambda }_2=-2+1=-1$ ✓.

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Cell E — pure imaginary, simple (marginal, constant wobble)

Forecast: a frictionless spring — energy conserved, so guess: oscillates forever, never grows, never dies. Dots on the imaginary axis. Cell E.

Step 1 — polynomial. $\det (A-\lambda I)={\lambda }^2+9=0.$ Why this step? No first-order term because damping is zero.

Step 2 — roots. ${\lambda }^2=-9\Rightarrow \lambda =\pm 3j.$ Why this step? $\sigma =0$ exactly, $\omega =3$. The dots sit on the vertical line of the map.

Step 3 — verdict. Roots are simple (distinct) on the axis → marginally stable: $\mathbf{x}(t)$ stays a bounded circle in phase space, oscillating at $3$ rad/s forever. Why this step? Marginal stability is allowed only because the axis roots are non-repeated (contrast Cell F).

Verify: ${\omega }_n=\sqrt{9}=3$, $\zeta =0$ → ${\omega }_d={\omega }_n\sqrt{1-0}=3$ matches $\text{Im}(\lambda )$ ✓. $|e^{3jt}|=1$ for all $t$ → bounded, doesn't decay ✓.

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Cell F — repeated zero, defective (the sneaky unstable one)

Forecast: eigenvalues on the axis, so tempting to call it marginal. But watch the repeated root.

Step 1 — polynomial. $\det (A-\lambda I)={\lambda }^2=0\Rightarrow \lambda =0,0.$ Why this step? A repeated eigenvalue at the origin — the danger flag.

Step 2 — count eigenvectors. Solve $A\mathbf{v}=0$: $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\mathbf{v}=0$ gives only $\mathbf{v}=(1,0{)}^T$ — one eigenvector for a double root. The matrix is defective. Why this step? Repeated-on-axis is only marginal if it has a full set of eigenvectors (semisimple). Here it doesn't.

Step 3 — actual solution. With one missing eigenvector the solution carries a $t$ term: $\mathbf{x}(t)=\left(\begin{array}{c}{x}_1(0)+t{x}_2(0)\\ x_2(0)\end{array}\right).$ The $t{x}_2(0)$ term grows without bound → unstable. Why this step? This is mistake #3 from the parent, made concrete: $t{e}^{0t}=t\to \infty$.

Verify: integrate $\ddot{q}=0$ directly: $q(t)=q(0)+\dot{q}(0)t$ — a straight line, unbounded whenever $\dot{q}(0)\ne 0$ ✓. Repeated axis root + missing eigenvector = unstable ✓.

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Cell G — repeated real negative eigenvalue (critically damped, $\zeta =1$)

Forecast: $\zeta =\frac{c/m}{2{\omega }_n}=\frac{4}{4}=1$ exactly — the tipping point between Cell A (two real, oscillation-free but sluggish) and Cell B (complex, ringing). Guess: a single repeated real negative root, no oscillation, fastest possible non-ringing decay.

Step 1 — polynomial. $\det (A-\lambda I)={\lambda }^2+4\lambda +4=0.$ Why this step? Companion-form shortcut again.

Step 2 — roots. ${\lambda }^2+4\lambda +4=(\lambda +2{)}^2=0\Rightarrow \lambda =-2,-2.$ Why this step? Discriminant $=16-16=0$ exactly → a repeated real root. This is the knife-edge $\zeta =1$: Cell G, a double dot at $-2$ on the left axis.

Step 3 — verdict. Both roots have $\sigma =-2<0$ → stable, and no imaginary part → no oscillation. Unlike Cell F, the axis root here is strictly negative, so even a $t{e}^{-2t}$ term still decays (the exponential beats the linear growth). This is critically damped: the fastest decay you can get without ringing. Why this step? Contrast with Cell F carefully — a repeated root is only dangerous when it sits on the imaginary axis; a repeated root in the left half-plane is perfectly stable.

Verify: discriminant $=4^2-4\cdot 4=0$ ✓; $\zeta =\frac{4}{2\cdot 2}=1$ ✓; $t{e}^{-2t}\to 0$ as $t\to \infty$ (exponential dominates) → stable ✓.

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Cell H — limiting sweep: watch the dots move as $\zeta$ goes $0\to 1\to \infty$

Forecast: as we add damping the oscillation should fade; guess the two dots slide from the imaginary axis inward, collide on the real axis at $\zeta =1$, then split apart along the real axis.

Step 1 — general roots. ${\lambda }^2+4\zeta \lambda +4=0\Rightarrow \lambda =-2\zeta \pm 2\sqrt{{\zeta }^2-1}.$ Why this step? One formula covers the whole sweep.

Step 2 — the regimes, each tagged to its cell.

• $\zeta =0$: $\lambda =\pm 2j$ → Cell E, pure wobble on the axis.
• $\zeta =0.5$: $\lambda =-1\pm j\sqrt{3}$ → Cell B, spiralling in (this is exactly Ex 2).
• $\zeta =1$: $\lambda =-2,-2$ → Cell G, critically damped, dots collide on the real axis (this is exactly Ex 7a).
• $\zeta =2$: $\lambda =-4\pm 2\sqrt{3}=-0.54,-7.46$ → Cell A, two real, overdamped.

Why this step? It shows the continuous geometry: increasing $\zeta$ drags $\sigma =-2\zeta$ leftward while the imaginary part $2\sqrt{1-{\zeta }^2}$ shrinks to zero at $\zeta =1$ — and each stop lands in a cell we already worked.

Step 3 — read the picture. For $\zeta <1$ the dots ride a circle of radius ${\omega }_n=2$ (since $|\lambda |={\omega }_n$); at $\zeta =1$ they meet at $-2$; for $\zeta >1$ they walk along the real axis, one toward $0$, one toward $-\infty$. Why this step? The circle is the key limiting-behaviour insight: underdamped eigenvalues always sit at distance ${\omega }_n$ from the origin.

Verify: at $\zeta =0.5$, $|\lambda |=\sqrt{1+3}=2={\omega }_n$ ✓. At $\zeta =1$ discriminant $=4({\zeta }^2-1)=0$ → repeated $-2$ ✓. At $\zeta =2$, product of roots $=(-0.536)(-7.464)=4=k/m$ ✓.

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Cell I — a $3\times 3$ real-world design read

Forecast: three states, so three eigenvalues. A well-designed loop should put them all in the left half-plane. Guess: stable.

Step 1 — characteristic polynomial. Because the bottom row of a $3\times 3$ companion matrix lists $-a_0,-a_1,-a_2$ (here $-6,-11,-6$), we read off $\det (A-\lambda I)={\lambda }^3+6{\lambda }^2+11\lambda +6=0.$ Why this step? Same companion-matrix property as the $2\times 2$ cases, one dimension up — no need to expand a general determinant.

Step 2 — factor. ${\lambda }^3+6{\lambda }^2+11\lambda +6=(\lambda +1)(\lambda +2)(\lambda +3)=0\Rightarrow \lambda =-1,-2,-3.$ Why this step? Try small integer roots ($\lambda =-1$ works), then factor down. All three real and negative → all dots on the left. Stable.

Step 3 — cross-check with Routh–Hurwitz (so we don't have to factor next time). For ${\lambda }^3+a_2{\lambda }^2+a_1\lambda +a_0$ with $a_2=6,a_1=11,a_0=6$, the Routh–Hurwitz Criterion needs all coefficients positive and $a_2{a}_1>a_0$: $6\cdot 11=66>6.✓$ Why this step? For higher order, factoring is often impossible; Routh–Hurwitz gives the same stable/unstable verdict from the coefficients alone. This connects to Characteristic Polynomial & Determinants and Pole Placement & LQR (where you choose these coefficients).

Verify: slowest (least-negative) eigenvalue is $-1$ → sets settling time $\approx 4/1=4$ s. Sum of roots $=-6=-a_2$ ✓; product $=(-1)(-2)(-3)=-6$, and for a cubic ${\lambda }^3+\cdots +a_0$ the product of roots $=-a_0=-6$ ✓.

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Recall Which cell is which — quick self-test

Complex roots with $\sigma <0$ live in which cell? ::: Cell B — stable, underdamped (oscillating decay). A $2\times 2$ with $\det A<0$ is always which cell? ::: Cell D — real roots of opposite sign (saddle), always unstable. Repeated $\lambda =0$ with only one eigenvector is stable or unstable? ::: Unstable (Cell F) — the missing eigenvector produces a $t$-term that grows. Repeated real root at $-2$ (both copies) — stable or unstable? ::: Stable (Cell G) — a repeated root in the left half-plane still decays, since $t{e}^{-2t}\to 0$. As $\zeta \to 0$ with ${\omega }_n$ fixed, where do the eigenvalues go? ::: Onto the imaginary axis at $\pm j{\omega }_n$ (Cell E), marginally stable. For underdamped roots, how far from the origin do they sit? ::: Distance ${\omega }_n$ exactly, since $|\lambda |={\omega }_n$.

Related: State-Space Representation · Diagonalization and Modal Decomposition · Damping Ratio and Natural Frequency · Transfer Functions and Poles.