LINEAR-ALGEBRA · Interactive Practice | Unit 21 · Video 2

The 100th Power Problem: Why Eigenvalues Decide Everything

IKey Formulas

Formula Name Meaning
A=SΛS−1A = S \Lambda S^{-1} Diagonalization Factor AA into eigenvectors SS and eigenvalue diagonal Λ\Lambda
Ak=SΛkS−1A^k = S \Lambda^k S^{-1} Matrix power formula All the action lives in Λk\Lambda^k
Λk=diag(λ1k,λ2k,…,λnk)\Lambda^k = \text{diag}(\lambda_1^k, \lambda_2^k, \dots, \lambda_n^k) Diagonal power Just raise each eigenvalue to the kk-th power
Ak→0  ⟺  ∥λi∥<1 for all iA^k \to 0 \iff \|\lambda_i\| < 1 \text{ for all } i Stability theorem Every eigenvalue must lie strictly inside the unit circle

IIVisualization 1 — How λk\lambda^k Evolves

Each power of λ\lambda either spirals in toward 00 or flings outward to infinity — its magnitude alone decides which.

IIIVisualization 2 — The Unit Circle, Boundary of Stability

Ak→0A^k \to 0 exactly when every eigenvalue lies strictly inside the unit circle.

💡 Challenge: park λ2\lambda_2 deep inside the circle, then drag λ1\lambda_1 just past the edge — one eigenvalue outside is enough to make AkA^k diverge.

IVVisualization 3 — Watching Λk\Lambda^k Collapse or Explode

In diagonal Λk\Lambda^k each eigenvalue evolves on its own — one entry above 11 dooms the whole matrix.

💡 Challenge: set λ1=1.1\lambda_1 = 1.1, λ2=0.5\lambda_2 = 0.5 and run to k=100k = 100 — λ2100\lambda_2^{100} is already invisible (≈10−30\approx 10^{-30}) yet AkA^k still explodes.

VQuiz Questions

Question 1

Suppose a 3×33 \times 3 matrix AA is diagonalizable with eigenvalues λ1=0.9\lambda_1 = 0.9, λ2=0.5\lambda_2 = 0.5, λ3=0.1\lambda_3 = 0.1.

What happens to AkA^k as k→∞k \to \infty?

✅ Correct! All eigenvalues satisfy ∣λi∣<1|\lambda_i| < 1, so Λk→0\Lambda^k \to 0 and hence Ak→0A^k \to 0.

❌ Not quite. The stability theorem says Ak→0A^k \to 0 precisely when every eigenvalue is strictly inside the unit circle. Check what 0.9k0.9^k, 0.5k0.5^k, and 0.1k0.1^k each do as kk grows large.

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Solution:

Using the diagonalization Ak=SΛkS−1A^k = S \Lambda^k S^{-1}, all the dynamics live inside Λk\Lambda^k:

Λk=diag(0.9k,  0.5k,  0.1k)\Lambda^k = \text{diag}(0.9^k,\; 0.5^k,\; 0.1^k)

Since every eigenvalue satisfies ∣λi∣<1|\lambda_i| < 1, each diagonal entry of Λk\Lambda^k tends to 00 as k→∞k \to \infty. Therefore Λk→0\Lambda^k \to 0, and consequently Ak=SΛkS−1→S⋅0⋅S−1=0A^k = S \Lambda^k S^{-1} \to S \cdot 0 \cdot S^{-1} = 0.

This is the stability theorem in action: Ak→0  ⟺  ∣λi∣<1A^k \to 0 \iff |\lambda_i| < 1 for every ii.

Question 2

True or False: If AA has an eigenvalue λ=−0.9\lambda = -0.9, then the corresponding entry of Λk\Lambda^k grows without bound as kk increases (because −0.9-0.9 is "negative and could explode").

✅ Correct! Since ∣−0.9∣=0.9<1|-0.9| = 0.9 < 1, the entry (−0.9)k(-0.9)^k alternates in sign but decays to zero.

❌ Not quite. Remember the stability condition uses ∣λ∣|\lambda|, not λ\lambda itself. Compute (−0.9)2,(−0.9)3,(−0.9)10(-0.9)^2, (-0.9)^3, (-0.9)^{10} and watch the magnitudes.

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Solution:

The statement is False. What matters for stability is the magnitude ∣λ∣|\lambda|, not the sign.

For λ=−0.9\lambda = -0.9:

∣λ∣=0.9<1|\lambda| = 0.9 < 1

so ∣λk∣=0.9k→0|\lambda^k| = 0.9^k \to 0. The values λk\lambda^k alternate in sign (−0.9,0.81,−0.729,…-0.9, 0.81, -0.729, \dots), but their magnitudes shrink to zero.

This is why the stability theorem uses the absolute value: ∣λi∣<1|\lambda_i| < 1, not λi<1\lambda_i < 1. Geometrically, we ask whether each eigenvalue lies strictly inside the unit circle in the complex plane — distance from the origin is what counts.

Question 3

A diagonalizable matrix AA has eigenvalues λ1=1.1\lambda_1 = 1.1, λ2=0.3\lambda_2 = 0.3, λ3=0.05\lambda_3 = 0.05.

What is the long-term behavior of AkA^k as k→∞k \to \infty?

✅ Correct! One eigenvalue with ∣λ∣>1|\lambda| > 1 is enough to make AkA^k diverge, no matter how small the others are.

❌ Not quite. The stability condition requires every eigenvalue to satisfy ∣λ∣<1|\lambda| < 1. What does 1.11001.1^{100} equal? Can the other shrinking eigenvalues cancel that out?

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Solution:

The stability theorem requires every eigenvalue to satisfy ∣λi∣<1|\lambda_i| < 1. Here we have:

  • λ1=1.1\lambda_1 = 1.1: ∣λ1∣=1.1>1|\lambda_1| = 1.1 > 1 — this entry blows up: 1.1100≈13,7801.1^{100} \approx 13{,}780.
  • λ2=0.3\lambda_2 = 0.3: decays to 00.
  • λ3=0.05\lambda_3 = 0.05: decays to 00 even faster.

But Λk=diag(1.1k,0.3k,0.05k)\Lambda^k = \text{diag}(1.1^k, 0.3^k, 0.05^k) has one entry going to infinity, so Λk\Lambda^k diverges, and so does

Ak=SΛkS−1.A^k = S \Lambda^k S^{-1}.

A single rebellious eigenvalue hijacks everything. The well-behaved companions cannot save you, because SS and S−1S^{-1} are fixed, finite matrices that cannot rescue a divergent Λk\Lambda^k.

Question 4

A matrix AA has eigenvalue λ=2i\lambda = 2i (purely imaginary).

Which statement correctly describes the behavior of λk\lambda^k as k→∞k \to \infty?

✅ Correct! ∣2i∣=2|2i| = 2, which is outside the unit circle, so ∣λk∣=2k→∞|\lambda^k| = 2^k \to \infty.

❌ Not quite. Compute ∣2i∣|2i| first. What is the distance from 2i2i to the origin in the complex plane? The stability theorem cares about that distance, not the real or imaginary parts alone.

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Solution:

Stability depends on the magnitude (distance from origin in the complex plane), not the real or imaginary parts individually.

For λ=2i\lambda = 2i:

∣λ∣=∣2i∣=02+22=2|\lambda| = |2i| = \sqrt{0^2 + 2^2} = 2

Since ∣λ∣=2>1|\lambda| = 2 > 1, the eigenvalue lies outside the unit circle, so:

∣λk∣=∣2i∣k=2k→∞.|\lambda^k| = |2i|^k = 2^k \to \infty.

The values λk\lambda^k rotate around the complex plane (multiplying by ii rotates by 90°90°), but their magnitudes grow like 2k2^k — they spiral outward to infinity.

A common misconception is that complex eigenvalues are "neutral" or bounded. They aren't — only eigenvalues on the unit circle (∣λ∣=1|\lambda| = 1) are bounded but non-decaying. Anything strictly outside explodes.

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