Eigenvalues of A — system modes, stability: Visual walkthrough

Step 1 — What "coupled" actually looks like

WHAT. We have two numbers that describe our system right now. Call them $x_{1}$ and $x_{2}$ . For a spacecraft you can picture $x_{1}$ = how far the attitude is off from where we want it, and $x_{2}$ = how fast that error is changing. Together they form a single arrow (a state vector) $x = (x_{1}, x_{2})$ living on a 2-D plane.

The rule that tells us how the arrow moves is $\dot{x} = A x .$

$x$ :: the arrow (where we are), a point on the plane.
$\dot{x}$ :: the tiny velocity arrow attached to that point — which way and how fast the state slides next instant.
$A$ :: the machine that reads your position and hands you back your velocity.

WHY it's hard. Look at the figure: at most points the velocity arrow (orange) does not line up with the position arrow (blue). The machine $A$ took your arrow and rotated it. That rotation is the coupling — $x_{1}$ 's future leans on $x_{2}$ and vice-versa, so you cannot solve them one at a time.

PICTURE.

Step 2 — The lucky directions where rotation vanishes

WHAT. Most arrows get rotated by $A$ . But there can exist special directions where the output arrow points the exact same way as the input — $A$ only stretches or shrinks it, never turns it. Such a direction $v$ satisfies $A v = λ v .$

$v$ :: the special direction, called an eigenvector ("eigen" = German for own/proper).
$λ$ :: the stretch factor, the eigenvalue. $λ = 2$ doubles the arrow, $λ = 0.5$ halves it, $λ = - 1$ flips it.

WHY we hunt for these. If the state ever points along $v$ , then its velocity $A v = λ v$ also points along $v$ . The state can only slide along its own line — it can never leave. A 2-D tangle just became a 1-D problem on a rail.

PICTURE. Green arrows are eigen-directions: input and output are parallel. Gray arrows are ordinary directions: output is twisted away.

Step 3 — Guessing the solution (the ansatz)

WHAT. We guess a solution that keeps a fixed shape $v$ and only changes its length over time: $x (t) = e^{λ t} v .$

$v$ :: the frozen direction — the state never leaves this line.
$e^{λ t}$ :: a single scalar "volume knob" that grows or shrinks the arrow as time $t$ runs.
Together :: an arrow that stays on one rail and only breathes in and out.

WHY guess this. We just proved (Step 2) that along an eigen-direction motion is 1-D. So a solution that lives entirely on that rail is the simplest thing that could possibly work — and the exponential $e^{λ t}$ is the one function whose rate of change is proportional to itself, exactly what a "grow at a fixed percentage" rail demands.

Why the exponential and nothing else? We need a knob $f (t)$ with $\dot{f} = λ f$ (velocity proportional to size — the 1-D version of our ODE). The only function satisfying that is $f (t) = e^{λ t}$ . That is the whole reason $e$ shows up in linear systems.

PICTURE. The state stays on the green rail; only its length rides the $e^{λ t}$ curve.

Step 4 — Substitute and watch the vectors cancel

WHAT. Plug the guess into $\dot{x} = A x$ . First differentiate the guess: $\dot{x} = knob’s speed λ e^{λ t} v .$ Now demand it obeys the physics: $λ e^{λ t} v = A e^{λ t} v .$

Left side :: the velocity our guess predicts.
Right side :: the velocity the machine $A$ actually hands back.
Setting them equal :: forcing the guess to be honest.

WHY. A guess is only a solution if it satisfies the equation everywhere. This line is that demand written out.

PICTURE. Two velocity arrows — predicted (blue) and machine-produced (orange) — must coincide.

Since $e^{λ t}$ is a positive number that is never zero, we can cancel it from both sides: $A v = λ v .$ The whole time-dependence disappeared. The guess works iff $(λ, v)$ is an eigenpair — exactly the special directions of Step 2. The clock was baked into $A$ all along.

Step 5 — The condition for a real (nonzero) direction

WHAT. Rearrange $A v = λ v$ by moving everything left: $(A - λ I) v = 0 .$

$I$ :: the identity matrix — the "do nothing" machine, so $λ I$ means "scale by $λ$ in every direction."
$A - λ I$ :: the machine "apply $A$ , then subtract a pure $λ$ -stretch."
$0$ :: the zero arrow (a dot at the origin).

We banned $v = 0$ (the origin is not a direction). So we need a nonzero arrow that $A - λ I$ crushes down to the origin.

WHY the determinant. A matrix crushes a nonzero arrow to zero only when it is singular — when it flattens the plane onto a line (or a point), collapsing all area. The number that measures "how much area survives" is the determinant. Area collapsing to zero means: $det (A - λ I) = 0.$ This is the characteristic equation. See Characteristic Polynomial & Determinants for why the determinant is the exact area-scaling factor.

PICTURE. Left: a healthy $A - λ I$ keeps area, only the origin maps to the origin — no eigenvector. Right: the right $λ$ makes it collapse the unit square onto a line, so a whole green rail maps to the origin — eigenvector found.

Step 5b — When rails run out: defective matrices

WHAT. The sum-of-modes formula quietly assumed we have $n$ independent rails. Sometimes we don't: two roots coincide yet share only one eigenvector. The plane cannot be built from rails alone — one direction is missing.

diagonalizable :: $n$ independent eigenvectors exist → clean sum $\sum c_{i} e^{λ_{i} t} v_{i}$ .
defective :: fewer eigenvectors than $n$ → a rail is missing → an extra $t e^{λ t}$ term appears.

WHY it matters for stability. A missing rail forces a shear into the flow, and the solution picks up a factor of $t$ : $x (t) = (c_{1} v + c_{2} (t v + w)) e^{λ t} .$ That extra $t$ can turn an otherwise-borderline mode into a growing one — this is the whole reason "repeated eigenvalue on the axis" needs a second look in Step 7.

Crucial distinction. A repeated eigenvalue is not automatically defective. For example $A = (λ 0 0 λ)$ has $λ$ twice but every direction is an eigenvector — it is diagonalizable, no $t$ term, perfectly well-behaved. Defectiveness is a missing-eigenvector problem, not a repeated-value problem.

PICTURE. Left: repeated value but two rails (diagonalizable, pure scaling). Right: repeated value, one rail — the flow shears and drags a $t$ along.

Step 6 — Reading a complex eigenvalue: $σ$ and $ω$

WHAT. By the fundamental theorem of algebra some roots may be complex. Write one as $λ = σ + j ω .$

$σ$ :: the real part — the growth/decay rate.
$ω$ :: the imaginary part — how fast the mode spins.
$j$ :: the imaginary unit, $j^{2} = - 1$ (control engineers use $j$ so $i$ stays free for current).

Feeding this into the knob: $e^{λ t} = e^{σ t} e^{j ω t} = e^{σ t} (cos ω t + j sin ω t) .$

$e^{σ t}$ :: the envelope — a pure grow/shrink curve.
$e^{j ω t} = cos ω t + j sin ω t$ :: a point spinning on a unit circle — length always exactly $1$ .

WHY only $σ$ decides stability. The magnitude is $∣ e^{λ t} ∣ = e^{σ t} \cdot = 1 ∣ cos ω t + j sin ω t ∣ = e^{σ t} .$ The spinning part never changes length — it can wiggle the arrow around forever but can't make it big or small. Only the envelope $e^{σ t}$ controls magnitude. So the sign of $σ$ alone says grow or die. This links to Damping Ratio and Natural Frequency where $σ = - ζ ω_{n}$ .

PICTURE. Envelope $e^{σ t}$ (dashed) squeezing the oscillation $cos ω t$ inside it.

Step 6b — Complex modes are secretly real

WHAT. Our matrix $A$ has real entries, and the physical state $x (t)$ is a real arrow — no imaginary spacecraft. Yet an eigenvalue came out complex, $λ = σ + j ω$ . How can complex numbers describe a real motion?

WHY it works out. For a real matrix, complex eigenvalues arrive in conjugate pairs: if $λ = σ + j ω$ is a root, so is $\overset{ˉ}{λ} = σ - j ω$ , with conjugate eigenvector $\overset{ˉ}{v}$ . In the mode sum these two partners appear together. If the state starts real, their coefficients are conjugate too, $c_{2} = \overset{c}{ˉ}_{1}$ , and the two complex modes add up to a real thing:

= 2\,e^{\sigma t}\big(\mathbf a\cos\omega t - \mathbf b\sin\omega t\big),$$ where $\mathbf a,\mathbf b$ are the real and imaginary parts of $c_1\mathbf v$. - The imaginary halves :: cancel exactly (a number plus its conjugate is real, twice its real part). - What survives :: a real envelope $e^{\sigma t}$ times real sine/cosine — a genuine decaying (or growing) oscillation. **PICTURE.** Two conjugate spirals (mirror images across the real axis) summing to one flat real oscillation. --- ## Step 7 — The complex plane verdict (all cases) **WHAT.** Plot each eigenvalue as a dot on the complex plane: horizontal = $\sigma$, vertical = $\omega$. The **vertical axis is the finish line.** - $\sigma<0$ (left of the axis) :: envelope $e^{\sigma t}$ shrinks → mode dies → **stable**. - $\sigma>0$ (right of the axis) :: envelope grows → mode explodes → **unstable**. - $\sigma=0$ (on the axis) :: envelope is a flat $1$ → mode neither dies nor grows → **marginal**, *provided* that root brings a full set of eigenvectors (Step 5b). **WHY every eigenvalue must pass.** The full solution is a *sum* of modes. One growing mode drags the whole sum to infinity, no matter how nicely the others behave. So stability = **all** dots in the open left half-plane. Compare with the algebraic shortcut in [[Routh–Hurwitz Criterion]] and the pole language of [[Transfer Functions and Poles]]. **Degenerate case — repeated eigenvalue on the axis.** Here the diagonalizable-vs-defective split of Step 5b decides the verdict: - **Defective** repeat on the axis (missing a rail): the solution carries a $t\,e^{0t}=t$ ramp that grows without bound → **unstable**. Example $\begin{pmatrix}0&1\\0&0\end{pmatrix}$ has a double $\lambda=0$ with only one eigenvector. - **Diagonalizable** repeat on the axis (full set of rails): no $t$ term, solutions stay bounded → **marginally stable**. Example $A=\mathbf 0$ has a double $\lambda=0$ but every direction is an eigenvector, so $\mathbf x(t)=\mathbf x(0)$ sits still forever. So a repeated axis eigenvalue is *unstable only when defective* — never assume the ramp appears. **PICTURE.** Every quadrant/case with its behaviour sketch, plus both flavours of the axis-repeat. > [!mnemonic] LEFT is RIGHT > Dots in the **LEFT** half-plane ($\sigma<0$) make the system do the **RIGHT** thing — settle to > zero. Any dot on the right side, or a *defective* repeated dot sitting on the axis, spells trouble. --- ## The one-picture summary Everything on one canvas: the coupled field → a special green rail where $A$ only scales → the knob $e^{\lambda t}$ on that rail → its speed decided by $\sigma$ → the left/right verdict. > [!recall]- Feynman: the whole walkthrough in plain words > Picture a windy meadow where every blade of grass shows which way the wind blows *there* — that's > $A$ turning position into velocity, and it's a swirly mess because directions push on each other. > But there are a few magic lines where the wind blows straight along the line itself. Drop a leaf on > one of those lines and it can only slide up or down the line, never wander off. How fast it slides > is one number, $\lambda$. If that number pulls the leaf toward the center, the line is "safe"; if > it pushes the leaf out to infinity, the line is "dangerous." Sometimes two magic lines collapse into > one and a line goes missing — then the leaf gets *dragged sideways by time itself* (the extra $t$), > which is why a borderline case can secretly be dangerous. And when a line makes the leaf spiral, > that spiral is the imaginary part $\omega$; because the wind is real, spirals always come in mirror > pairs that add up to an honest real wobble. Real systems mix all these lines, so we check every one: > if *all* pull inward, the meadow calms down (stable); if even one pushes outward, everything > eventually blows away (unstable). > [!recall]- Quick self-test > Why can we cancel $e^{\lambda t}$ in Step 4? ::: Because $e^{\lambda t}$ is strictly positive and never zero, so dividing both sides by it is legal, leaving $A\mathbf v=\lambda\mathbf v$. > Why does $\det(A-\lambda I)=0$ (and not some other condition)? ::: A nonzero arrow can be crushed to $\mathbf 0$ only if the matrix flattens area, i.e. is singular, i.e. has zero determinant. > Two eigenvalues are $-1\pm 100j$. Stable or not? ::: Stable — the real part is $-1<0$; the large $\omega=100$ only makes it oscillate fast, it does not affect stability. > A matrix has a repeated eigenvalue exactly at $\lambda=0$. Verdict? ::: It depends — if defective (one eigenvector) a $t$ ramp grows → unstable; if diagonalizable (full rails, e.g. $A=\mathbf 0$) solutions stay put → marginally stable. > Why does a complex eigenvalue still give a real motion? ::: Because real $A$ has conjugate eigenpairs; the two partner modes add so their imaginary parts cancel, leaving a real $e^{\sigma t}\cos/\sin$ oscillation. > What guarantees $\det(A-\lambda I)=0$ even has roots? ::: The fundamental theorem of algebra: a degree-$n$ polynomial has exactly $n$ roots over the complex numbers, counting multiplicity.