Key Idea: Eigenvalues tell you how a matrix model behaves far into the future. For a city where people move between two areas each year, they show which part fades and which part stays.
From matrix to long run
- Eigenvalues — Solve det(A − λI) = 0.
- Eigenvectors — Find one for each λ; then A = PDP⁻¹.
- Each step — Each piece is multiplied by λⁿ: |λ| < 1 fades to zero, λ = 1 stays put.
- Steady state — Scale the λ = 1 eigenvector to the total.
| Step | Result |
|---|---|
| λ = 1 eigenvector | (3, 2) |
| Downtown | 3/5 × 50 000 = 30 000 |
| Suburbs | 2/5 × 50 000 = 20 000 |
In the long run, only the λ = 1 eigenvector is left.
One eigenvalue is 0.5. What happens to its piece over time? It shrinks to zero, because 0.5ⁿ → 0.
How can you check (30 000, 20 000) is the steady state? Multiply it by A. You get the same vector back.
Does the starting split change the long-run answer? No. Any start settles to 60% Downtown.
What share of the city ends up in the Suburbs? 2/5, which is 40%.