LINEAR-ALGEBRA

The Left Half-Plane Rule: Why Some Systems Decay and Others Explode

IKey Formulas

Formula	Name	Description
$\lvert e^{\lambda t} \rvert = e^{\operatorname{Re}(\lambda)\,t}$	Magnitude rule	Imaginary part only rotates; real part decides growth/decay
$\frac{du}{dt} = Au$ stable $\iff \operatorname{Re}(\lambda) < 0$ for all $\lambda$	Continuous stability	All eigenvalues in the left half-plane
$\operatorname{trace}(A) < 0$ AND $\det(A) > 0$	$2\times 2$ shortcut	Necessary and sufficient for $2\times 2$ stability
$A^k \to 0 \iff \lvert \lambda \rvert < 1$ for all $\lambda$	Discrete stability	All eigenvalues inside the unit disk

IIVisualization 1 — Eigenvalue Position vs. Decay

Only the real part of $\lambda = a + bi$ decides whether $e^{\lambda t}$ decays or explodes.

💡 Challenge: drag the eigenvalue straight up and down — the decay envelope never moves, because $\operatorname{Im}(\lambda)$ only sets the spin rate.

IIIVisualization 2 — The Trace–Determinant Plane

Trace and determinant alone pin down a $2\times 2$ system's eigenvalues — and its fate.

💡 Challenge: land the point on the dashed parabola $\det = \tfrac{1}{4}\operatorname{trace}^2$ — there the eigenvalues collide into a repeated root, the border between nodes and spirals.

IVVisualization 3 — Continuous vs. Discrete Stability

The same $\lambda$ is judged against the left half-plane for $\tfrac{du}{dt}=Au$ , the unit disk for $A^k$ .

💡 The map $z = e^{\lambda}$ carries the left half-plane onto the unit disk — which is why the stability boundary bends from a straight line into a circle.

VQuiz Questions

Question 1

A linear system $\frac{du}{dt} = Au$ has eigenvalues $\lambda_1 = -4 + 10i$ and $\lambda_2 = -1 - 7i$ .

Does $u(t) \to 0$ as $t \to \infty$ ?

✅ Correct! Both real parts are negative, so every $e^{\lambda t}$ term decays.

❌ Not quite. Imaginary parts only rotate — they have magnitude 1. The real part is the only thing that grows or shrinks.

Show solution

Solution:

The magnitude rule says $\lvert e^{\lambda t}\rvert = e^{\operatorname{Re}(\lambda)\,t}$ .

For $\lambda_1 = -4 + 10i$ : $\lvert e^{\lambda_1 t}\rvert = e^{-4t} \to 0$ .
For $\lambda_2 = -1 - 7i$ : $\lvert e^{\lambda_2 t}\rvert = e^{-t} \to 0$ .

Both pieces decay, so $u(t) \to 0$ . The imaginary parts ( $10i$ and $-7i$ ) only spin the solution as it shrinks — they don't change whether it shrinks.

Question 2

Consider the matrix $A = \begin{pmatrix} -3 & 4 \\ 2 & 1 \end{pmatrix}.$

Is the system $\frac{du}{dt} = Au$ stable?

✅ Correct! A negative trace doesn't save you when the determinant is negative.

❌ Not quite. Trace $<0$ alone isn't enough. You need trace $<0$ AND det $>0$ . Compute $\det(A) = ad - bc$ .

Show solution

Solution:

Apply the $2\times 2$ shortcut. We need both trace $<0$ AND det $>0$ .

$\operatorname{trace}(A) = -3 + 1 = -2 < 0$ ✓
$\det(A) = (-3)(1) - (4)(2) = -3 - 8 = -11 < 0$ ✗

Det is negative, so the eigenvalues have opposite signs (a saddle): $\lambda_1 \lambda_2 = -11 < 0$ .

A negative trace alone only says the average eigenvalue is negative. One eigenvalue is still positive, and that one drives $e^{\lambda t}$ to infinity.

Question 3

True or False: If the matrix $A$ is stable for $\frac{du}{dt} = Au$ , then so is $-A$ .

✅ Correct! Eigenvalues flip sign with $A$ , so stability flips into full instability.

❌ Not quite. Try the eigenvalue equation: if $Ax=\lambda x$ , what does $(-A)x$ equal?

Show solution

Solution:

If $Ax = \lambda x$ , then $(-A)x = -\lambda x$ . So the eigenvalues of $-A$ are exactly $-\lambda_1, -\lambda_2, \ldots$ .

If every $\operatorname{Re}(\lambda_i) < 0$ (so $A$ is stable), then every $\operatorname{Re}(-\lambda_i) > 0$ , which means $-A$ is fully unstable — every mode blows up.

This is the "reverse time, reverse stability" idea from the video. It's a great sanity check: if you flip the sign of $A$ and the system is still stable, you've made a sign error somewhere.

Question 4

A matrix $B$ has eigenvalues $\lambda_1 = 0.6 + 0.5i$ and $\lambda_2 = -0.4$ .

Which statement is correct?

✅ Correct! $|\lambda_1| \approx 0.78 < 1$ (discrete stable), but $\operatorname{Re}(\lambda_1) = 0.6 > 0$ (continuous unstable).

❌ Not quite. For powers $B^k$ check magnitudes $|\lambda|<1$ . For $\frac{du}{dt}=Bu$ check real parts $\operatorname{Re}(\lambda)<0$ . Compute both for each eigenvalue.

Show solution

Solution:

Two different stability tests for the same eigenvalues:

Discrete ( $B^k \to 0$ ): need $|\lambda| < 1$ for all eigenvalues.

$|\lambda_1| = \sqrt{0.6^2 + 0.5^2} = \sqrt{0.36 + 0.25} = \sqrt{0.61} \approx 0.78 < 1$ ✓
$|\lambda_2| = 0.4 < 1$ ✓

So $B^k \to 0$ . Discrete stable.

Continuous ( $\frac{du}{dt}=Bu$ ): need $\operatorname{Re}(\lambda) < 0$ for all eigenvalues.

$\operatorname{Re}(\lambda_1) = 0.6 > 0$ ✗

So $u(t)$ blows up. Continuous unstable.

Same eigenvalues, opposite verdicts — that's the key takeaway from the last frame of the video. Continuous time uses left half-plane; discrete time uses unit disk.

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