Eigenvalues of A — system modes, stability: Exercises

Definition The imaginary axis, and the three stability words

The complex plane has two axes. The horizontal one is the real axis (values of $σ$ ); the vertical one is the imaginary axis (values of $ω$ ) — it is the set of points where $σ = 0$ , i.e. purely up-down numbers like $2 j$ . When this page says "the axis" with no other word, it always means this imaginary axis — the vertical dividing line in Figure s01. Three words we will use constantly:

Asymptotically stable — every eigenvalue has $σ < 0$ (strictly left of the axis); every mode decays and $x \to 0$ .
Marginally stable — no eigenvalue has $σ > 0$ , and any eigenvalue sitting on the axis ( $σ = 0$ ) is simple (not repeated). The response neither grows nor dies — it stays bounded.
Unstable — some eigenvalue has $σ > 0$ , or a repeated eigenvalue sits on the axis. At least one mode grows without bound.

Every eigenvalue we find is written $λ = σ + j ω$ , where $σ$ (the real part) is the growth/decay rate and $ω$ (the imaginary part) is the oscillation frequency. Keep the complex plane picture in your head the whole way down:

Figure — Eigenvalues of A — system modes, stability

Figure s01 — the stability map (visual key). Read it once carefully, because every problem below just asks "which region is my pole in?":

The horizontal white line is the real axis; the red dashed vertical line is the imaginary axis — the border between settling down and blowing up.

Everything left of the red line (shaded cyan, $σ < 0$ ) is stable; everything right ( $σ > 0$ ) is unstable.

Each × is one eigenvalue (a "pole"). Cyan ×'s are stable: a real one at $- 2$ and a complex pair at $- 1 \pm 3 j$ . The amber × at $+ 0.2$ is unstable. The white ×'s at $\pm 2 j$ sit exactly on the imaginary axis ( $σ = 0$ ) — the marginal edge case.

L1.2 uses the amber and white points directly, and L5.4 lives entirely on a white pure-imaginary pair like these.

Level 1 — Recognition

L1.1 — Read stability straight off the plane

A system $\dot{x} = A x$ has eigenvalues ${- 2, - 5}$ . Is it asymptotically stable, marginally stable, or unstable?

Recall Solution

Both eigenvalues are real and negative, so both sit strictly to the left of the imaginary axis (look at the cyan × at $- 2$ in Figure s01). Every mode $e^{- 2 t}$ and $e^{- 5 t}$ decays to zero. $\Rightarrow ==asymptotically stable== .$ Picture: two dots on the negative real axis, both inside the left half-plane.

L1.2 — Spot the unstable one

Which of these eigenvalue sets belongs to an unstable system? $(a) {- 1 \pm 3 j} (b) {- 4, 0} (c) {- 1, + 0.2}$

Recall Solution

Check the real part of each — the horizontal position in Figure s01:

(a) $σ = - 1 < 0$ for both → stable (oscillatory but decaying); these are the cyan pair at $- 1 \pm 3 j$ .
(b) real parts $- 4$ and $0$ → the $0$ is on the imaginary axis and simple, so marginally stable (bounded, per the definition above), not unstable.
(c) $+ 0.2 > 0$ → one mode grows as $e^{0.2 t}$ → ==(c) is unstable==; this is the amber × at $+ 0.2$ . Only a strictly positive real part means unstable.

Level 2 — Application

L2.1 — Find the eigenvalues of a 2×2

$A = (- 1 0 2 - 3) .$ Compute the eigenvalues and classify stability.

Recall Solution

$A$ is upper-triangular (the entry below the diagonal is $0$ ), so its eigenvalues are just the diagonal entries — see Characteristic Polynomial & Determinants: $det (A - λ I) = (- 1 - λ) (- 3 - λ) = 0 \Rightarrow λ_{1} = - 1, λ_{2} = - 3.$ Both $σ < 0$ → asymptotically stable. The slow mode $- 1$ dominates the settling time.

L2.2 — Companion form of a 2nd-order ODE

A rotor obeys $\ddot{θ} + 3 \dot{θ} + 2 θ = 0$ . Write the state-space $A$ with $x_{1} = θ, x_{2} = \dot{θ}$ , then find the eigenvalues.

Recall Solution

With $x_{1} = θ$ , $x_{2} = \dot{θ}$ : $\overset{x}{˙}_{1} = x_{2}$ and $\overset{x}{˙}_{2} = \ddot{θ} = - 2 θ - 3 \dot{θ}$ : $A = (0 - 2 1 - 3) .$ Characteristic equation (for this companion form it reproduces the ODE exactly): $det (A - λ I) = λ^{2} + 3 λ + 2 = (λ + 1) (λ + 2) = 0 \Rightarrow λ = - 1, - 2.$ Both real, both negative → stable, overdamped (two distinct real roots, no oscillation).

Level 3 — Analysis

L3.1 — Damping ratio and natural frequency

A mode has eigenvalues $λ = - 1 \pm j 3$ . Find the natural frequency $ω_{n}$ , the damped frequency $ω_{d}$ , and the damping ratio $ζ$ . Is it under-, over-, or critically damped?

Recall Solution

Where the standard form comes from. A damped 2nd-order system is written $\overset{y}{¨} + 2 ζ ω_{n} \overset{y}{˙} + ω_{n}^{2} y = 0$ . Why these coefficients? Its characteristic polynomial (plug $y = e^{λ t}$ ) is $λ^{2} + 2 ζ ω_{n} λ + ω_{n}^{2} = 0.$ Solving with the quadratic formula:

= -\zeta\omega_n \pm \omega_n\sqrt{\zeta^2-1}.$$ For $0<\zeta<1$ the term under the root is negative, so pull out $j$: $\lambda=-\zeta\omega_n\pm j\,\omega_n\sqrt{1-\zeta^2}$. So the **constant term** of the polynomial is $\omega_n^2$ and the **$\lambda$-coefficient** is $2\zeta\omega_n$ — that is *why* matching a polynomial to this template reads off $\omega_n$ and $\zeta$ (see [[Damping Ratio and Natural Frequency]]). **Match our eigenvalues** $\lambda=-1\pm j\sqrt3$: - $\omega_n=|\lambda|=\sqrt{(-1)^2+(\sqrt3)^2}=\sqrt{1+3}=2.$ - $\omega_d=\operatorname{Im}(\lambda)=\sqrt3\approx 1.732.$ - $\sigma=-\zeta\omega_n=-1 \Rightarrow \zeta=\dfrac{1}{\omega_n}=\dfrac{1}{2}=0.5.$ Since $0<\zeta<1$ → ==underdamped== (it oscillates while decaying). **The geometric meaning of $\zeta$ — see Figure s02.** Plot the pole $\lambda=-1+j\sqrt3$ in the complex plane. Draw the straight line (of length $\omega_n$) from the origin out to the pole. Let ==$\phi$== be the angle measured *from the negative real axis, opening toward the pole*. The horizontal distance the pole sits to the left of the axis is $\omega_n\cos\phi$, and we already named that distance $|\sigma|=\zeta\omega_n$. Equate them: $$\zeta\omega_n=\omega_n\cos\phi \;\Rightarrow\; \boxed{\zeta=\cos\phi}.$$ So a pole *near the negative real axis* ($\phi$ small) is heavily damped ($\zeta\to 1$), and a pole *near the imaginary axis* ($\phi\to 90^\circ$) is barely damped ($\zeta\to 0$). For us $\cos\phi=0.5\Rightarrow\phi=60^\circ$.

Figure s02 — the damping angle $ϕ$ . The cyan line of length

ω_{n}

runs from the origin to the pole

- 1 + j 3

; the amber angle

ϕ

is measured from the negative real axis; its horizontal leg is

ζ ω_{n}

and its vertical leg is

ω_{d}

. That is why

ζ = cos ϕ

L3.2 — The slowest mode sets settling time

$A = (- 10 1 0 - 0.5) .$ Which eigenvalue governs the long-term settling, and roughly how long until the response settles?

Recall Solution

Lower-triangular → eigenvalues are the diagonal: $λ_{1} = - 10, λ_{2} = - 0.5$ . The mode $e^{- 10 t}$ vanishes almost instantly; $e^{- 0.5 t}$ lingers longest. The eigenvalue closest to the imaginary axis ( $- 0.5$ ) dominates.

Where the $4/∣ σ ∣$ rule comes from. A decaying mode has envelope $e^{- ∣ σ ∣ t}$ . We call the response "settled" once that envelope has shrunk to within a small band (the usual engineering convention is $2%$ of its start value). Solve for the time when the envelope hits that band: $e^{- ∣ σ ∣ t_{s}} = 0.02 \Rightarrow ∣ σ ∣ t_{s} = ln \frac{1}{0.02} = ln 50 \approx 3.91.$ Rounding $ln 50 \approx 4$ gives the handy rule $t_{s} \approx 4/∣ σ ∣$ — it is just "how many time-constants to fade to $2%$ " (see Damping Ratio and Natural Frequency). Applying it to the slowest pole: $t_{s} \approx \frac{4}{∣ σ _{slow} ∣} = \frac{4}{0.5} = 8 s .$ Long-term behaviour is set by the least-negative eigenvalue, not the fast one.

Level 4 — Synthesis

L4.1 — Stability without solving the quadratic

For $\dot{x} = A x$ with $A = (0 - 6 1 - 5),$ use the 2×2 rule (trace and determinant) to decide stability, then confirm with the eigenvalues.

Recall Solution

Why the trace–determinant test works. For any 2×2 matrix the characteristic polynomial can be written purely from its trace and determinant: $det (A - λ I) = λ^{2} - (tr A) λ + det A = 0.$ Its two roots satisfy $λ_{1} + λ_{2} = tr A$ and $λ_{1} λ_{2} = det A$ . Now ask: when do both roots lie in the left half-plane ( $Re < 0$ )?

If the roots are real, "both negative" means their sum is negative ( $tr A < 0$ ) and their product is positive ( $det A > 0$ ; two negatives multiply to a positive).
If the roots are a complex pair $σ \pm j ω$ , then sum $= 2 σ = tr A$ , so $σ < 0 ⟺ tr A < 0$ ; and product $= σ^{2} + ω^{2} = det A > 0$ automatically, which just confirms $det A > 0$ .

In every case the pair of conditions is exactly $stable ⟺ tr A < 0 AND det A > 0.$ This is the 2×2 special case of the Routh–Hurwitz Criterion. Here $tr A = 0 + (- 5) = - 5 < 0$ ✓ and $det A = (0) (- 5) - (1) (- 6) = 6 > 0$ ✓ → stable. Confirm: $det (A - λ I) = λ^{2} + 5 λ + 6 = (λ + 2) (λ + 3) = 0 \Rightarrow λ = - 2, - 3.$ ✓

L4.2 — Design a target: place the poles by choosing a parameter

A tunable system is $A = (0 - 4 1 - a),$ with gain $a > 0$ we can pick. Choose $a$ so that the closed loop is critically damped ( $ζ = 1$ ), and give the resulting (repeated) eigenvalue.

Recall Solution

Characteristic equation: $λ^{2} + aλ + 4 = 0$ . Compare to the standard form $λ^{2} + 2 ζ ω_{n} λ + ω_{n}^{2}$ derived in L3.1 (constant term $= ω_{n}^{2}$ , $λ$ -coefficient $= 2 ζ ω_{n}$ ):

$ω_{n}^{2} = 4 \Rightarrow ω_{n} = 2$ .
$2 ζ ω_{n} = a$ . For critical damping $ζ = 1$ : $a = 2 (1) (2) = 4$ .

With $a = 4$ : $λ^{2} + 4 λ + 4 = (λ + 2)^{2} = 0 \Rightarrow λ = - 2$ (repeated). This is the fastest response with no overshoot — the sweet spot control engineers aim for (see Pole Placement & LQR).

Level 5 — Mastery

L5.1 — Repeated eigenvalue on the axis: the hidden growth

$A = (0010) .$ Both eigenvalues are $0$ . Is the system marginally stable or unstable? Solve $x (t)$ from $x (0) = (x_{10}, x_{20})^{T}$ to justify.

Recall Solution

$det (A - λ I) = λ^{2} = 0 \Rightarrow λ = 0$ (repeated). Both real parts are $0$ , so the naive check says "on the axis → maybe marginal". But the two zeros share only one eigenvector (a defective/repeated eigenvalue), so a $t$ -term appears. Solve directly: $\overset{x}{˙}_{1} = x_{2}, \overset{x}{˙}_{2} = 0 \Rightarrow x_{2} (t) = x_{20}, x_{1} (t) = x_{10} + x_{20} t .$ The $x_{20} t$ term grows without bound whenever $x_{20} \neq = 0$ → ==unstable==. Marginal stability requires axis eigenvalues to be simple (non-repeated). This one fails that.

$A = (- 3 0 1 - 2), x (0) = (21) .$ Find eigenvalues, eigenvectors, the constants $c_{i}$ , the full solution, and state which mode dominates as $t \to \infty$ .

Recall Solution

Eigenvalues (triangular): $λ_{1} = - 3, λ_{2} = - 2$ . Eigenvectors (see Diagonalization and Modal Decomposition):

$λ_{1} = - 3$ : $(A + 3 I) v = (0011) v = 0 \Rightarrow v_{1} = (1, 0)^{T}$ .
$λ_{2} = - 2$ : $(A + 2 I) v = (- 1 0 10) v = 0 \Rightarrow v_{2} = (1, 1)^{T}$ .

Constants from $x (0) = c_{1} v_{1} + c_{2} v_{2}$ : $c_{1} + c_{2} = 2, c_{2} = 1 \Rightarrow c_{2} = 1, c_{1} = 1.$ Solution:

=\begin{pmatrix} e^{-3t}+e^{-2t} \\ e^{-2t}\end{pmatrix}.$$ As $t\to\infty$ the $e^{-3t}$ mode dies first; the **slower** $e^{-2t}$ mode (nearest the axis) dominates the tail. Both decay → asymptotically stable.

L5.3 — Third-order stability with Routh–Hurwitz

The eigenvalues of a system are the roots of its characteristic polynomial. Writing that polynomial in the eigenvalue variable $λ$ , suppose $det (A - λ I) = λ^{3} + 2 λ^{2} + 2 λ + K = 0.$ Find the range of gain $K$ for which the system is asymptotically stable.

Recall Solution

Each root $λ$ of this cubic is an eigenvalue, so "asymptotically stable" means every root has $Re (λ) < 0$ . The Routh–Hurwitz Criterion lets us test that without finding the roots. For a cubic $λ^{3} + a_{2} λ^{2} + a_{1} λ + a_{0}$ all four conditions must hold simultaneously: $a_{2} > 0, a_{1} > 0, a_{0} > 0, and a_{2} a_{1} > a_{0} .$ (The first three demand every coefficient be positive; the last is the extra "inner" determinant condition that a cubic needs.) Here $a_{2} = 2, a_{1} = 2, a_{0} = K$ :

$a_{2} = 2 > 0$ ✓ always.
$a_{1} = 2 > 0$ ✓ always.
$a_{0} = K > 0 \Rightarrow K > 0$ .
$a_{2} a_{1} = 2 \cdot 2 = 4 > K \Rightarrow K < 4$ .

Combining: $0 < K < 4$ . At $K = 4$ a pair of eigenvalues sits exactly on the imaginary axis (marginal); at $K = 0$ an eigenvalue sits at the origin.

L5.4 — Pure-imaginary poles: the undamped edge case

A frictionless flywheel gives $A = (0 - 9 10) .$ Find the eigenvalues, classify stability, and describe the motion $x (t)$ .

Recall Solution

$det (A - λ I) = λ^{2} + 9 = 0 \Rightarrow λ^{2} = - 9 \Rightarrow λ = \pm 3 j$ (using $j = - 1$ ). These are pure imaginary: $σ = 0, ω = \pm 3$ — they sit exactly on the imaginary axis in Figure s01 (like the white × marks, but at $\pm 3 j$ ).

Classification. Both eigenvalues have real part $σ = 0$ , so no mode grows exponentially — the system is not unstable. And the two roots are distinct ( $+ 3 j \neq = - 3 j$ ), i.e. the axis eigenvalues are simple, so no dangerous $t$ -term appears (contrast L5.1, where the repeated $0$ broke marginal stability). Both requirements of the definition are met, so the flywheel is ==marginally stable==.

The motion. The envelope is $e^{σ t} = e^{0 \cdot t} = 1$ , a constant, so the amplitude neither grows nor decays. Using Euler's formula $e^{\pm 3 j t} = cos 3 t \pm j sin 3 t$ , the real solution is a pure sustained oscillation at frequency $ω = 3 rad/s$ : $x (t) = (c_{1} cos 3 t + c_{2} sin 3 t - 3 c_{1} sin 3 t + 3 c_{2} cos 3 t),$ where $c_{1} = x_{1} (0)$ and $c_{2} = x_{2} (0) /3$ come from the initial state. The state traces a closed loop (an ellipse) in the $(x_{1}, x_{2})$ plane forever — spinning at constant amplitude, never settling and never blowing up. In damping language this is $ζ = 0$ (no damping) with $ω_{n} = 9 = 3$ , so $ω_{d} = ω_{n} 1 - ζ^{2} = 3 = ω_{n}$ . This is the knife-edge: nudge either pole a hair left of the imaginary axis → a decaying spiral (stable); a hair right → a growing spiral (unstable).

Recall One-line ladder recap

L1 read the plane ::: sign of the real part decides stability. L2 compute eigenvalues ::: solve $det (A - λ I) = 0$ , not the diagonal. L3 interpret ::: $ζ, ω_{n}, ω_{d}$ from pole location; slowest pole sets $t_{s}$ . L4 design ::: trace/det rule + pole placement pick behaviour. L5 master ::: repeated axis roots grow, simple imaginary pairs merely oscillate; Routh–Hurwitz for higher order.