Powers of A: which parts grow, shrink or stay
Slide 3 showed that powers are easy once you count in eigenvector steps: each step of A multiplies each count by its own λ. Many real systems apply the same matrix again and again: uₖ₊₁ = Auₖ, one step per year, per second, or per update. Think of money in two jars. One jar gains 5% every year (it is multiplied by 1.05), and the other loses 30% (it is multiplied by 0.7). After a few decades the first jar is all that matters. Split the starting vector into eigenvectors, and the same thing happens: parts with |λ| > 1 grow, parts with |λ| < 1 die out, and a part with λ = 1 stays. When every part dies out, the system is stable. A Markov matrix, which moves fractions of a total around, always has λ = 1, and usually settles into a steady state (a vector A leaves unchanged). (It isn't “stable” in the strict sense, because the λ = 1 part never dies.)
Lesson 7 · slide 4Stability- Split u₀ = c₁x₁ + c₂x₂ into eigenvectors. Then uₖ = Aᵏu₀ = c₁λ₁ᵏx₁ + c₂λ₂ᵏx₂.
- |λ| < 1: that part shrinks to 0. |λ| > 1: it grows. λ = 1: it stays.
- A Markov matrix has nonnegative entries and columns adding to 1. It always has λ = 1.
- Its eigenvector for λ = 1, scaled to add to 1, is the steady state. When every other eigenvalue has |λ| < 1 (always true if all entries are positive), the system settles into it.
The rule: each part is multiplied by its λ, every step
Write the start as u₀ = c₁x₁ + c₂x₂. One step of A multiplies the x₁ part by λ₁ and the x₂ part by λ₂. After k steps:
A small example: A = 2000.5 and u₀ = 11. The eigenvectors are the axes, with λ = 2 and λ = 0.5:
So the size of |λ| decides each part's future: bigger than 1 grows, smaller than 1 shrinks, exactly 1 stays the same. A negative λ also flips the sign of that part every step.
Two cities, people moving every year
Each year 10% of city 1 moves to city 2, and 20% of city 2 moves to city 1. Everyone else stays. Let uₖ = (city 1, city 2) be the shares of the people after k years:
A matrix like this, with entries that are 0 or more and columns that add to 1, is a Markov matrix. With 100 people, the head counts are just 100 × the shares. Start with all 100 people in city 2 and apply A year by year:
The eigenvalues explain it
Split the start (as shares, everyone in city 2 is u₀ = 01) into the two eigenvectors: c₁21 + c₂1−1 = 01. The two entries say 2c₁ + c₂ = 0 and c₁ − c₂ = 1, so c₁ = ⅓ and c₂ = −⅔:
The x₁ part is multiplied by 1 every year, so it stays. The x₂ part shrinks by 0.7ᵏ; after 20 years less than 0.1% of it is left. Only x₁ survives. Scaled so its entries add to 1, it is the steady state s = 2/31/3. These are shares, so they add to 1; for head counts multiply by the total, here 100·s ≈ 66.733.3.
Why a Markov matrix always has λ = 1
Each column of A adds to 1, so each column of A − I adds to 0. Then the rows of A − I add up to the zero row (entry j of the sum of the rows is the total of column j minus 1, which is 0). The rows are dependent, A − I is singular, and det(A − I) = 0. So λ = 1 is always an eigenvalue. Check here: A − I above has row 2 = −1 times row 1.
Stability
A system uₖ₊₁ = Auₖ is stable (every disturbance, meaning any small change or error added to u, dies out) exactly when every eigenvalue has |λ| < 1. A small example: A = 0.50.20.20.5 has λ = 0.7 with 11 and λ = 0.3 with 1−1. Check: A11 = 0.70.7 and A1−1 = 0.3−0.3. Both parts shrink every step, so every start goes to 0.
One eigenvalue with |λ| > 1 is enough to make some disturbance grow without limit.
Keep reading with the full course
Finish lesson 7 and you will be able to: tell whether a system settles down or blows up.
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Examples
Example 1 · Growth and decay
A = 2000.5, u₀ = 11.
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