IB Maths AI HLTransition Matrices & Markov ChainsPaper 1 & 2~9 min read
Steady State & Long-term Probabilities
Keep stepping a regular chain forward and the probabilities settle down: Tn tends to a matrix with identical columns, and sn tends to a fixed steady-state vectors that satisfies Ts = s. You can find it two ways: a large power on the GDC, or exactly as the eigenvector for eigenvalue 1, scaled to sum to 1.
📘 What you need to know
Steady state: a vector s unchanged by T, i.e. Ts = s. Regular chains always have one.
Regular chain: some power Tk has no zero entries. All AI HL chains are regular.
Eigenvalues: a regular T has exactly one eigenvalue of 1, the rest < 1 in size.
Long-term: as n → ∞, Tn has identical columns, each equal to s; sn → s regardless of the start.
GDC method: take a large power; if the columns agree to the required accuracy, that column is s.
Exact method: solve Ts = s (the eigenvector for eigenvalue 1), then scale so the entries sum to 1.
What the steady state is
The defining property is simply that another step changes nothing — the long-run probabilities have stopped moving.
Steady-state conditionTs = sSet up and solve it yourself — not a booklet formula ✗
Tⁿ settles to identical columns
As the power grows the two columns converge to the same vector — that vector is the steady state s.
🤔 Why does it forget where it started?
Writing Tn = PDnP−1, every eigenvalue except the one equal to 1 has size < 1, so its power → 0. Only the eigenvalue-1 part survives, leaving a matrix whose columns are all the same steady-state vector. Multiplying any start s0 by that gives s — so the long-run behaviour is independent of the starting state.
Two ways to find it
approximatebig power, GDCCompute Tn for large n; when the columns match to the needed accuracy, read off s.
exacteigenvector, λ=1Solve Ts = s for the eigenvalue-1 eigenvector, then scale to sum to 1.
🧭 Recipe — exact steady state
Write s with unknown entries x1, x2, … and set up Ts = s.
Form the equations; they’re dependent, so drop one and fix a value (e.g. let x1 = a convenient number).
Solve for a simple integer eigenvector (any scalar multiple works).
Scale to sum 1: divide each entry by the total — these are the steady-state probabilities.
Diagonalisation route: if asked to show the result, write T = PDP−1, let n → ∞ so Dn keeps only the entry 1, and compute T∞ = PD∞P−1. Its identical columns are the steady state.
🧠 “Stays the same when stepped”
Steady = Stays. The whole idea is one vector that T leaves alone. And the quick check on a hand-computed T∞: the columns should be identical.
Worked examples
All five use the cat sanctuary again, with T =
0.2
0.9
0.8
0.1
(states B = brushed, B′ = not brushed; columns = current, rows = next).
WE 1
Set up Tv = v
Let v =
x1
x2
be an eigenvector of T with eigenvalue 1. Write out Tv = v.
v is an eigenvector with eigenvalue 1 if Tv = vTv = (0.2x₁ + 0.9x₂ , 0.8x₁ + 0.1x₂)set each component equal to the matching entry of v.
WE 2
Form the equation linking x₁ and x₂
Use the first component to find a relationship between x1 and x2.
0.2x₁ + 0.9x₂ = x₁0.9x₂ = 0.8x₁ → 9x₂ = 8x₁the second equation gives the same relationship (they’re dependent).
WE 3
Find an eigenvector
State a simple integer eigenvector for eigenvalue 1.
9x₂ = 8x₁ → pick x₁ = 9, x₂ = 8
v =
9
8
(9, 8) — or any scalar multiple
WE 4
Hence find the steady-state vector
Scale the eigenvector so its entries sum to 1.
sum = 9 + 8 = 17 → divide each entry by 17
s =
917
817
s = (9/17, 8/17) ≈ (0.529, 0.471)
WE 5
Check it with a large power
Confirm the answer by considering Tn for large n.
compute a large power on the GDC
Tn →
0.529
0.529
0.471
0.471
identical columns = (0.529, 0.471) ✓matches 9/17 and 8/17 — in the long run ~53% of cats are brushed on any day.
💡 Top tips
Ts = s is the whole idea — set it up, solve, then scale so the entries sum to 1.
Equations are dependent: drop one and fix a value to escape the infinite solutions.
Any scalar multiple of the eigenvector is fine until you scale to sum 1.
GDC check: a large power should have identical columns equal to s.
Start doesn’t matter for a regular chain — the long-run vector is the same from any s0.
Population version: multiply s by N for the long-run expected numbers at each state.
⚠ Common mistakes
Forgetting to scale — leaving the answer as (9, 8) instead of (9/17, 8/17).
Entries not summing to 1 — a steady-state vector is a probability distribution.
Using all the equations — they’re dependent; one must be dropped, with a value fixed.
Too small a power — if columns aren’t yet identical, the power isn’t large enough.
Confusing s with a column of T — the steady state comes from the limitT∞, not from T itself.
That completes Transition Matrices & Markov Chains. The unit has one clean storyline: a chain hops between states with memoryless transition probabilities → pack them into a matrix T (columns = current, rows = next, columns sum to 1) → step forward with sn = Tns0, reading powers as multi-step probabilities → and in the long run everything settles to the steady states with Ts = s, found as the eigenvalue-1 eigenvector scaled to sum to 1. Set up T carefully, let the GDC handle the arithmetic, and always state your answer as probabilities (or × N for expected numbers).
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