IB Math AI HL Eigenvalues, Eigenvectors & Markov Chains — Complete Cheatsheet
Every eigenvalue, eigenvector and Markov-chain method for IB Mathematics Applications & Interpretation HL — the characteristic equation, diagonalisation and matrix powers, transition matrices, n-step states and the long-run steady state. Hand-built by an IBO-certified Singapore tutor, with a print-ready PDF to download.
Eigenvalues, eigenvectors and Markov chains are where IB Mathematics Applications & Interpretation HL turns a matrix into a time machine. AI is a modelling subject, so the point of this AHL 1.15 material is rarely the algebra for its own sake — it is predicting the long-run behaviour of a system: how a population splits between habitats, how customers churn between two providers, or where a market share settles years down the line. Because a calculator is always allowed, the marks live in setting the model up correctly and reading the long-run answer — not in grinding matrix powers by hand.
This cheatsheet condenses the whole of AHL 1.15 — eigenvalues from the characteristic equation, eigenvectors, diagonalisation and matrix powers, transition (Markov) matrices, the state after $n$ steps and the all-important steady state — onto one page, and flags the traps that quietly cost method marks. The print-ready PDF is at the bottom, free to download.
§1 — Eigenvalues & the characteristic equation AHL 1.15
An eigenvector of a square matrix $A$ points in a direction that $A$ only stretches — it does not rotate it — and the stretch factor is its eigenvalue $\lambda$. That single idea unlocks the whole topic.
Definition:$A\mathbf{v}=\lambda\mathbf{v}$ with $\mathbf{v}\neq\mathbf{0}$ — applying $A$ just scales $\mathbf{v}$ by $\lambda$
Characteristic eq.:$\det(A-\lambda I)=0$ — the equation whose roots are the eigenvalues
In words:$\lambda^2-(\operatorname{tr}A)\lambda+\det A=0$ — sum of roots $=\operatorname{tr}A$, product $=\det A$
NoteThe trace $a+d$ is the sum of the eigenvalues and $\det A=ad-bc$ is their product. Use these as an instant check once you have your two values — but neither number is itself an eigenvalue.
TrapYou subtract $\lambda$ only on the leading diagonal: $A-\lambda I=\begin{pmatrix}a-\lambda&b\\c&d-\lambda\end{pmatrix}$. Subtracting it from every entry is the classic set-up error.
§2 — Finding the eigenvalues AHL 1.15
The characteristic equation is just a quadratic in $\lambda$, so solve it on the GDC, by factorising, or with the formula. Some matrices have a shortcut worth spotting.
Symmetric shortcut:for $\begin{pmatrix}a&b\\b&a\end{pmatrix}$ the equation $(a-\lambda)^2-b^2=0$ is a difference of squares $\Rightarrow\lambda=a\pm b$
Example:$\begin{pmatrix}2&1\\1&2\end{pmatrix}\Rightarrow\lambda=2\pm1$, so $\lambda=3$ or $\lambda=1$
TrickA symmetric matrix ($b=c$) always has two real eigenvalues, because $(a-d)^2+4b^2\ge0$. When the diagonal entries are equal, factor by difference of squares and skip the formula entirely — the larger value $a+b$ is the one that controls long-run growth.
NoteIn AI you may simply enter the matrix and let the GDC return the eigenvalues. Reach for the algebra when a question says "show that" or gives the matrix in terms of a parameter.
§3 — Finding the eigenvector AHL 1.15
Each eigenvalue $\lambda$ has its own eigenvector, found from $(A-\lambda I)\mathbf{v}=\mathbf{0}$. Because that system is deliberately singular, you only ever need one of its two rows.
Set-up:write $\mathbf{v}=\begin{pmatrix}1\\k\end{pmatrix}$; the top row of $(A-\lambda I)\mathbf{v}=\mathbf 0$ gives $(a-\lambda)+bk=0$
Component:$k=\dfrac{\lambda-a}{b}$, so $\mathbf{v}=\begin{pmatrix}1\\ \frac{\lambda-a}{b}\end{pmatrix}$ (or any scalar multiple)
NoteAn eigenvector defines a direction, not a fixed length — any non-zero multiple $\begin{pmatrix}2\\2\end{pmatrix}$, $\begin{pmatrix}-1\\-1\end{pmatrix}$ is equally valid. Examiners accept any correct multiple unless a normalisation is demanded.
TrapThe two rows of $(A-\lambda I)$ are dependent (that is exactly why $\det=0$). Use one row only — trying to solve both as independent equations forces $\mathbf{v}=\mathbf 0$ and loses the eigenvector.
§4 — Diagonalisation & matrix powers AHL 1.15
Eigenvalues turn a high power of a matrix — the engine behind $n$-step Markov chains — into a single line of arithmetic. Collect the eigenvectors as the columns of $P$ and the eigenvalues on the diagonal of $D$.
Diagonalise:$A=PDP^{-1}$, where $P=\big(\mathbf{v}_1\ \ \mathbf{v}_2\big)$ and $D=\begin{pmatrix}\lambda_1&0\\0&\lambda_2\end{pmatrix}$
$n$-th power:$A^{n}=PD^{\,n}P^{-1}$ with $D^{\,n}=\begin{pmatrix}\lambda_1^{\,n}&0\\0&\lambda_2^{\,n}\end{pmatrix}$ — just raise the eigenvalues
TrickFor a plain numerical power the GDC gives $A^{n}$ directly — no diagonalisation needed. Save $A=PDP^{-1}$ for a "show that $A^{n}=\dots$" or a "behaviour as $n\to\infty$" question, where the eigenvalues carry the meaning: $|\lambda|<1$ decays to $0$, $\lambda=1$ holds steady, $|\lambda|>1$ grows.
TrapIn the triangular formula the off-diagonal entry subtracts, $q^{\,n}-p^{\,n}$ — a very common slip is to add. The bottom-left stays $0$, and you raise the eigenvalues to the power $n$, never the whole matrix entry-by-entry.
§5 — Transition (Markov) matrices AHL 1.15
A Markov chain steps a probability state forward using a transition matrix $T$. This sheet uses the column convention: each column lists where one state goes, so every column sums to $1$.
Transition matrix:$T_{ij}=P(\text{from state }j\text{ to state }i)$; every column of $T$ sums to $1$
General 2-state:$T=\begin{pmatrix}1-a&b\\a&1-b\end{pmatrix}$, where $a=P(1\!\to\!2)$ and $b=P(2\!\to\!1)$
One step:$\mathbf{s}_1=T\mathbf{s}_0$, i.e. $P(\text{X next})=P(X\!\to\!X)\,P(X)+P(Y\!\to\!X)\,P(Y)$
A two-state Markov chain: the arrows out of each state are its transition probabilities, and the options leaving a state sum to 1 — so every column of T totals 1. Here $a$ leaves state 1 and $b$ leaves state 2.
TrapWith this convention the columns sum to $1$, not the rows, and the update is $\mathbf{s}_{n+1}=T\mathbf{s}_n$ with $\mathbf{s}$ a column vector. Do not transpose $T$ — swapping its rows and columns is the single most common error on the whole topic.
NoteEvery entry of $T$ is a probability, so $0\le T_{ij}\le1$. If a column of your matrix does not add to $1$, you have mislabelled a transition — fix it before stepping forward.
§6 — The state after n steps AHL 1.15
To run the chain forward, apply $T$ once per time step. That is exactly the matrix power from §4.
$n$ steps:$\mathbf{s}_n=T^{\,n}\mathbf{s}_0$ — the distribution after $n$ time steps
Step by step:$\mathbf{s}_1=T\mathbf{s}_0$, then $\mathbf{s}_2=T\mathbf{s}_1$, and so on
TrickStore $T$ and $\mathbf{s}_0$ on the GDC and compute $T^{\,n}\mathbf{s}_0$ in one keystroke — far quicker and safer than multiplying $n$ times by hand. If a state vector holds counts (people, animals), keep the total constant at every step as a self-check.
TrapThe order is $T^{\,n}\mathbf{s}_0$, with the matrix on the left of the column vector — $\mathbf{s}_0T^{\,n}$ is undefined (or gives nonsense) in this convention.
§7 — The long-run steady state AHL 1.15
Run the chain long enough and the distribution stops changing: it reaches a steady state $\mathbf{s}$ that $T$ leaves fixed. Because $T\mathbf{s}=\mathbf{s}=1\cdot\mathbf{s}$, the steady state is precisely the eigenvector for eigenvalue $1$, scaled so its entries sum to $1$.
Steady state:solve $T\mathbf{s}=\mathbf{s}$ together with $s_1+s_2=1$
Balance idea:inflow $=$ outflow for state 1: $a\,s_1=b\,s_2$, so the share in state 1 is $s_1=\dfrac{b}{a+b}$
In words:$P(\text{state 1})=\dfrac{P(Y\!\to\!X)}{P(X\!\to\!Y)+P(Y\!\to\!X)}$
TrickThe steady state does not depend on where you start: whatever the initial split $\mathbf{s}_0$, the chain settles to the same $\mathbf{s}$. That is why the answer is simply $\tfrac{b}{a+b}$ — it is the $\lambda=1$ eigenvector normalised to sum to $1$.
TrapThe long-run split is almost never $50/50$. It tilts toward whichever state is harder to leave (smaller outflow), so a chain with $a=0.1$, $b=0.2$ settles at $\tfrac{0.2}{0.3}=\tfrac23$ in state 1, not one-half.
§8 — Application: long-run counts & expected value AHL 1.15
Once you have the steady-state proportions, real AI questions ask for a count or a value: how many animals, how many subscribers, or how much profit in the long run.
Long-run count:$N\times P(\text{state})$ — multiply the steady-state proportion by the population size $N$
Long-run value:read $\mathbf{s}$ as a distribution and attach a value $x_i$ to each state: $E(X)=\displaystyle\sum x_i\,P(x_i)$
TrickFirst solve $T\mathbf{s}=\mathbf{s}$ for the steady-state probabilities, then weight: $E=x_1\,P(\text{state 1})+x_2\,P(\text{state 2})$. Weight each value by its own probability — never take a plain average of $x_1$ and $x_2$.
NoteThis is a Markov × probability-distributions synthesis (a favourite HL crossover). The steady state supplies the probabilities; the expected-value formula $E(X)=\sum x_iP(x_i)$ then turns them into an average profit, cost or count.
§9 — Complex eigenvalues, stability & coupled systems AHL 1.15
Not every matrix has real eigenvalues. When the discriminant of the characteristic equation is negative the eigenvalues are a complex pair — and the system spirals rather than settling straight down.
Complex pair:if $(\operatorname{tr}A)^2-4\det A<0$ then $\lambda=\alpha\pm\beta i$, and trajectories spiral
Stability:the real part $\alpha=\tfrac12\operatorname{tr}A$ decides it — $\alpha<0$ spirals inward (stable), $\alpha>0$ outward (unstable)
Coupled DE link:with eigenvectors $\begin{pmatrix}1\\1\end{pmatrix},\begin{pmatrix}1\\-1\end{pmatrix}$: $\mathbf{v}=c_1e^{\lambda_1 t}\begin{pmatrix}1\\1\end{pmatrix}+c_2e^{\lambda_2 t}\begin{pmatrix}1\\-1\end{pmatrix}$
TrickYou do not need the full quadratic formula for the real part — just halve the trace, $\alpha=\tfrac12\operatorname{tr}A$. Reach for the discriminant only to confirm the roots really are complex before you talk about a spiral.
NoteThe same eigenvectors reappear in the AHL 5.17 coupled-DE method, now as growth rates $e^{\lambda t}$. Fitting the start values at $t=0$ (where every $e^{\lambda t}=1$) gives $c_1+c_2=x_0$ and $c_1-c_2=y_0$, so $c_1=\tfrac12(x_0+y_0)$ and $c_2=\tfrac12(x_0-y_0)$ — add the equations to isolate $c_1$, subtract for $c_2$.
§10 — Exam attack plan All sections
Question cue
What to do
Watch for
"Find the eigenvalues"
Solve $\lambda^2-(\operatorname{tr}A)\lambda+\det A=0$ (GDC or formula)
Subtract $\lambda$ on the diagonal only; check sum $=\operatorname{tr}$, product $=\det$
"Find the eigenvector"
$(A-\lambda I)\mathbf{v}=\mathbf 0$; set $\mathbf{v}=\binom{1}{k}$, $k=\tfrac{\lambda-a}{b}$
Use one row only; any scalar multiple is valid
"Find $A^{n}$ / large power"
Diagonalise $A=PDP^{-1}$, so $A^{n}=PD^{\,n}P^{-1}$ (or GDC)
Raise the eigenvalues to $n$, not the whole matrix
"State / population after $n$ steps"
Compute $\mathbf{s}_n=T^{\,n}\mathbf{s}_0$ on the GDC
Columns of $T$ sum to 1; keep $\mathbf{s}$ a column; do not transpose
"Long-run / steady state"
Solve $T\mathbf{s}=\mathbf{s}$ with $\textstyle\sum s_i=1$ (or $\tfrac{b}{a+b}$)
Independent of the start; rarely $50/50$
"Long-run number in a state"
Steady-state proportion $\times\,N$
Round to whole items; the total is conserved
"Long-run profit / expected value"
$E=\sum x_i\,P(x_i)$ with steady-state probabilities
Weight by probability — not a plain average
"Eigenvalues are complex"
Write $\alpha\pm\beta i$; stability from $\alpha=\tfrac12\operatorname{tr}A$
Real part sets in/out spiral; sign of $\alpha$ decides stability
Worked Example — AI HL-Style Markov Chain
Question (AI HL Paper 2 style — 8 marks)
A fitness app classifies each user as Active or Lapsed every week. Of active users, $90\%$ stay active and $10\%$ lapse; of lapsed users, $20\%$ return to active and $80\%$ stay lapsed. The app currently has $500$ active and $500$ lapsed users ($1000$ in total). Let state 1 be "Active" and state 2 be "Lapsed". (a) Write down the transition matrix $T$ (columns summing to 1) and the initial state vector $\mathbf{s}_0$. (b) Find the eigenvalues of $T$. (c) Find the number of active users after 2 weeks. (d) Find the long-run (steady-state) number of active users.
Solution
(a) Each column lists where one state goes. From active: $0.9$ stay, $0.1$ lapse; from lapsed: $0.2$ return, $0.8$ stay. So $T=\begin{pmatrix}0.9&0.2\\0.1&0.8\end{pmatrix}$, $\mathbf{s}_0=\begin{pmatrix}500\\500\end{pmatrix}$. (A1)(A1)(each column of $T$ sums to 1 — a quick self-check.)
(b) $\operatorname{tr}T=1.7$ and $\det T=0.9(0.8)-0.2(0.1)=0.70$, so the characteristic equation is $\lambda^2-1.7\lambda+0.7=0$. (M1) Solving, $\lambda=\dfrac{1.7\pm\sqrt{1.7^2-4(0.7)}}{2}=\dfrac{1.7\pm0.3}{2}$, giving $\lambda=1$ and $\lambda=0.7$. (A1)(The $\lambda=1$ guarantees a steady state; $|0.7|<1$ means the chain converges to it.)
(c) After two weeks $\mathbf{s}_2=T^{2}\mathbf{s}_0$. On the GDC $T^{2}=\begin{pmatrix}0.83&0.34\\0.17&0.66\end{pmatrix}$, so $\mathbf{s}_2=\begin{pmatrix}0.83&0.34\\0.17&0.66\end{pmatrix}\begin{pmatrix}500\\500\end{pmatrix}=\begin{pmatrix}585\\415\end{pmatrix}$. (M1)(A1) There are 585 active users after 2 weeks.
(d) The steady state solves $T\mathbf{s}=\mathbf{s}$ with $s_1+s_2=1$. Balancing state 1: $0.1\,s_1=0.2\,s_2\Rightarrow s_1=\dfrac{0.2}{0.1+0.2}=\dfrac23$. (M1) So the long-run active count is $\tfrac23\times1000\approx\textbf{667 users}$ (with about $333$ lapsed). (A1)(Equivalently, the $\lambda=1$ eigenvector $\binom{2}{1}$ normalised to sum $1$.)
Examiner's note: state the convention first — here columns sum to 1, so $\mathbf{s}_{n+1}=T\mathbf{s}_n$ — and do not transpose $T$. The presence of an eigenvalue equal to $1$ is the fingerprint of a valid steady state, and the total must stay at $1000$ at every step. Give whole people in the final answer and keep exact GDC values along the way.
Common Student Questions
What actually is an eigenvector and eigenvalue, in plain words?
An eigenvector is a special direction that a matrix leaves pointing the same way — it only stretches or shrinks it, never rotates it — and the eigenvalue is that stretch factor. Most vectors get knocked off their line when you multiply by $A$; eigenvectors are the ones that survive along their own line. For a Markov chain, the eigenvector for $\lambda=1$ is the direction that does not change at all, which is exactly the long-run steady state.
Do the rows or the columns of a transition matrix add up to 1?
It depends on the convention. This sheet uses columns summing to 1, with state vectors written as columns and updated by $\mathbf{s}_{n+1}=T\mathbf{s}_n$ — because each column lists where everything in one state goes. Some textbooks use row vectors with $\mathbf{s}_{n+1}=\mathbf{s}_nT$, and then the rows sum to 1 (that $T$ is the transpose of ours). Decide which you are using at the very start and never transpose $T$ halfway through.
Why is the steady state just the eigenvector for eigenvalue 1?
Because "steady" means the distribution stops changing: $T\mathbf{s}=\mathbf{s}$. That is the eigenvalue equation $T\mathbf{s}=\lambda\mathbf{s}$ with $\lambda=1$, so the steady state is the $\lambda=1$ eigenvector, rescaled so its entries add to 1. A column transition matrix always has $\lambda=1$ as an eigenvalue, which is why a steady state always exists. In practice you rarely need eigenvectors explicitly — just solve $T\mathbf{s}=\mathbf{s}$ with $s_1+s_2=1$, or quote $\tfrac{b}{a+b}$.
My eigenvalues came out complex — have I made a mistake?
Not necessarily. When the discriminant $(\operatorname{tr}A)^2-4\det A$ is negative, the eigenvalues are a genuine complex conjugate pair $\alpha\pm\beta i$, and it means the system spirals rather than moving straight to equilibrium. The real part $\alpha=\tfrac12\operatorname{tr}A$ tells you whether it spirals inward ($\alpha<0$, stable) or outward ($\alpha>0$, unstable). Transition (Markov) matrices, though, always have real eigenvalues with $\lambda=1$ as the largest — so complex values there really would signal an arithmetic slip.
How do I raise a matrix to a high power without multiplying it out many times?
For a plain number like $T^{10}$, enter $T$ on the GDC and use the power key — one step. For an algebraic "show that $A^{n}=\dots$" question, diagonalise: write $A=PDP^{-1}$ with the eigenvectors in $P$ and eigenvalues in $D$, then $A^{n}=PD^{n}P^{-1}$, where $D^{n}$ just raises each eigenvalue to the power $n$. This is also why long-run behaviour is easy to read off: eigenvalues with $|\lambda|<1$ die away, $\lambda=1$ gives the steady state, and $|\lambda|>1$ blows up.
What's NOT in this cheatsheet
This page gives you the formulas and traps for free. The full Photon Academy IB Math AI HL library (enrolled students, or the lifetime resource library) adds:
Notes PDF — every eigenvalue, eigenvector, diagonalisation and Markov-chain method in full, with worked steady-state examples and the exact GDC keystrokes.
Tutorial booklet — 25+ AI HL-style questions from basic eigenvalues to full transition-matrix and long-run steady-state applications.
Tutorial Solutions — mark-scheme-style solutions with M1/A1 annotations for every eigenvalue and Markov method.
Predicted-paper questions — the exact eigenvalue and Markov-chain question types most likely in the next session.
