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Learning objectives
Core idea
Key formulae
Worked examples
Explore the graph
Common mistakes
Exam tip
Guided practice
Practice questions
Challenge
Summary
Quick quiz
Same topic, different board
Past paper questions

Learn9231Paper 2 · Further Pure Mathematics 2MatricesEigenvalues and diagonalisation

Matrices · Subtopic 1 of 1

Eigenvalues, Eigenvectors and Diagonalisation

An eigenvector of $\mathbf M$ keeps its direction: $\mathbf M\mathbf v=\lambda\mathbf v$ , where $\lambda$ is the eigenvalue. Eigenvalues solve $\det(\mathbf M-\lambda\mathbf I)=0$ . Putting eigenvectors in $\mathbf P$ and eigenvalues in $\mathbf D$ gives $\mathbf M=\mathbf{PDP}^{-1}$ , so $\mathbf M^n=\mathbf{PD}^n\mathbf P^{-1}$ .

Cambridge 9231 Paper 2Edexcel YFM01 WFM03Edexcel 9FM0 OptionAQA 7367 Paper 1AQA 7367 Paper 2≈ 50 minRevision 1 · updated 2026-10-27

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Section 1 of 14

Learning objectives

Find the characteristic equation and eigenvalues of a $2\times2$ or $3\times3$ matrix.
Find eigenvectors for each eigenvalue.
Diagonalise a matrix and use $\mathbf M^n=\mathbf{PD}^n\mathbf P^{-1}$ .
Use the Cayley–Hamilton theorem to find inverses and powers.

Pause and think

A transformation stretches the plane. Are there directions that come out pointing the same way they went in?

Section 2 of 14

Core idea

Start hereThe directions that don't turn

Stretch a sheet of rubber diagonally. Most arrows drawn on it twist to new directions, but along certain lines the arrows only get longer or shorter. Those special directions are a matrix's eigenvectors, and they make huge matrix powers easy to compute.

The big idea

$\mathbf M\mathbf v=\lambda\mathbf v$ ( $\mathbf v\neq\mathbf 0$ ). Eigenvalues solve $\det(\mathbf M-\lambda\mathbf I)=0$ . With eigenvectors as columns of $\mathbf P$ : $\mathbf M=\mathbf{PDP}^{-1}$ , so $\mathbf M^n=\mathbf{PD}^n\mathbf P^{-1}$ .

1. Eigenvalues

The characteristic equation

$\mathbf M\mathbf v=\lambda\mathbf v$
Definition.
$(\mathbf M-\lambda\mathbf I)\mathbf v=\mathbf 0$
$\lambda\mathbf v=\lambda\mathbf I\mathbf v$ ; collect.
$\mathbf v\neq\mathbf 0\Rightarrow\mathbf M-\lambda\mathbf I\text{ is singular}$
If it had an inverse, $\mathbf v$ would have to be $\mathbf 0$ .
$\det(\mathbf M-\lambda\mathbf I)=0$
The characteristic equation.

For

2\times2

\lambda^2-(a+d)\lambda+(ad-bc)=0

, i.e.

\lambda^2-(\operatorname{tr})\lambda+\det=0

Worked example $\mathbf M=\begin{pmatrix}4&1\\2&3\end{pmatrix}$

Eigenvalues and eigenvectors.

Goal: $\lambda$ , $\mathbf v$ .
Strategy: Characteristic equation; then solve $(\mathbf M-\lambda\mathbf I)\mathbf v=\mathbf 0$ .

$\lambda^2-7\lambda+10=0\Rightarrow\lambda=2,5$
Trace 7, determinant 10.
$\lambda=5:\ \begin{pmatrix}-1&1\\2&-2\end{pmatrix}\mathbf v=\mathbf 0\Rightarrow y=x$
Both rows give the same equation (they must).
$\lambda=2:\ \begin{pmatrix}2&1\\2&1\end{pmatrix}\mathbf v=\mathbf 0\Rightarrow y=-2x$
Again one equation.

Answer

\lambda=5

\begin{pmatrix}1\\1\end{pmatrix}

;

\lambda=2

\begin{pmatrix}1\\-2\end{pmatrix}

✓ Check

\mathbf M\begin{pmatrix}1\\-2\end{pmatrix}=\begin{pmatrix}2\\-4\end{pmatrix}=2\begin{pmatrix}1\\-2\end{pmatrix}

✓.

Key takeawayTrace and determinant Sum of eigenvalues = trace; product = determinant (any size). Quick check:

2+5=7

2\times5=10

✓.

2. Diagonalisation

Why

\mathbf M=\mathbf{PDP}^{-1}

$\mathbf M\mathbf v_1=\lambda_1\mathbf v_1,\ \mathbf M\mathbf v_2=\lambda_2\mathbf v_2$
Two independent eigenvectors.
$\mathbf M[\mathbf v_1\ \mathbf v_2]=[\lambda_1\mathbf v_1\ \lambda_2\mathbf v_2]$
Multiplying column by column.
$\mathbf{MP}=\mathbf{PD},\quad\mathbf D=\begin{pmatrix}\lambda_1&0\\0&\lambda_2\end{pmatrix}$
$\mathbf{PD}$ scales each column of $\mathbf P$ .
$\mathbf M=\mathbf{PDP}^{-1}$
Multiply on the right by $\mathbf P^{-1}$ .
$\mathbf M^n=\mathbf{PD}\cancel{\mathbf P^{-1}\mathbf P}\mathbf D\cdots\mathbf P^{-1}=\mathbf{PD}^n\mathbf P^{-1}$
Inner $\mathbf P^{-1}\mathbf P$ pairs cancel.

\mathbf D^n

just raises each diagonal entry to the power

n

Worked exampleA power of $\mathbf M$

Using $\mathbf M=\begin{pmatrix}4&1\\2&3\end{pmatrix}$ , find $\mathbf M^n$ .

Goal: Formula.
Strategy: $\mathbf P=\begin{pmatrix}1&1\\1&-2\end{pmatrix}$ , $\mathbf D=\operatorname{diag}(5,2)$ .

$\mathbf P^{-1}=-\tfrac13\begin{pmatrix}-2&-1\\-1&1\end{pmatrix}=\tfrac13\begin{pmatrix}2&1\\1&-1\end{pmatrix}$
$\det\mathbf P=-3$ .
$\mathbf M^n=\tfrac13\begin{pmatrix}1&1\\1&-2\end{pmatrix}\begin{pmatrix}5^n&0\\0&2^n\end{pmatrix}\begin{pmatrix}2&1\\1&-1\end{pmatrix}$
Substitute.
$=\tfrac13\begin{pmatrix}2\cdot5^n+2^n&5^n-2^n\\2\cdot5^n-2\cdot2^n&5^n+2\cdot2^n\end{pmatrix}$
Multiply out.

AnswerAs above.

✓ Check

n=1

\frac13\begin{pmatrix}12&3\\6&9\end{pmatrix}=\begin{pmatrix}4&1\\2&3\end{pmatrix}

✓.

3. Cayley–Hamilton

Worked exampleAn inverse without the formula

Find $\mathbf M^{-1}$ for $\mathbf M=\begin{pmatrix}4&1\\2&3\end{pmatrix}$ using Cayley–Hamilton.

Goal: $\mathbf M^{-1}$ .
Strategy: Replace $\lambda$ by $\mathbf M$ in $\lambda^2-7\lambda+10=0$ .

$\mathbf M^2-7\mathbf M+10\mathbf I=\mathbf 0$
Cayley–Hamilton (constant term times $\mathbf I$ ).
$\mathbf M-7\mathbf I+10\mathbf M^{-1}=\mathbf 0$
Multiply by $\mathbf M^{-1}$ .
$\mathbf M^{-1}=\tfrac1{10}(7\mathbf I-\mathbf M)=\tfrac1{10}\begin{pmatrix}3&-1\\-2&4\end{pmatrix}$
Rearrange.

Answer

\frac1{10}\begin{pmatrix}3&-1\\-2&4\end{pmatrix}

✓ CheckMatches the usual formula with

\det=10

✓.

Common misconceptionMismatched order The

k

th column of

\mathbf P

must be the eigenvector for the

k

th entry of

\mathbf D

Edge caseSymmetric matrices Symmetric matrices have perpendicular eigenvectors. Normalise them and

\mathbf P

is orthogonal:

\mathbf P^{-1}=\mathbf P^{\mathrm T}

Pause and think

What are the eigenvalues of a rotation by $90^\circ$ ?

Summary

$\det(\mathbf M-\lambda\mathbf I)=0$ .
Eigenvectors from $(\mathbf M-\lambda\mathbf I)\mathbf v=\mathbf 0$ .
$\mathbf M^n=\mathbf{PD}^n\mathbf P^{-1}$ .
Cayley–Hamilton for inverses and powers.

Eigenvalue: Scale factor along an eigenvector.
Characteristic equation: $\det(\mathbf M-\lambda\mathbf I)=0$ .
Diagonalise: Write $\mathbf M=\mathbf{PDP}^{-1}$ .

HistoryEigen = own. "Eigen" is German for "own" or "characteristic". David Hilbert popularised the term in 1904 for the "own values" of a transformation.

Fun factGoogle's first algorithm. PageRank ranks web pages using the eigenvector with eigenvalue 1 of a gigantic matrix describing links between pages.

Section 3 of 14

Key formulae

Definition

\mathbf M\mathbf v=\lambda\mathbf v,\quad\mathbf v\neq\mathbf 0

Characteristic equation

\det(\mathbf M-\lambda\mathbf I)=0;\quad2\times2:\ \lambda^2-(\operatorname{tr}\mathbf M)\lambda+\det\mathbf M=0

Diagonalisation

\mathbf M=\mathbf{PDP}^{-1},\qquad\mathbf M^n=\mathbf{PD}^n\mathbf P^{-1}

Cayley–Hamilton

\text{every matrix satisfies its characteristic equation}

TrickTrace and determinant check. For any matrix, the eigenvalues add up to the trace (sum of the leading diagonal) and multiply to the determinant. Check your eigenvalues with this before finding eigenvectors.

Section 4 of 14

Worked examples

Example 1 · 2×20 / 3 steps

Find the eigenvalues and eigenvectors of

\mathbf M=\begin{pmatrix}4&1\\2&3\end{pmatrix}

Example 2 · Cayley–Hamilton0 / 3 steps

Use Cayley–Hamilton to express

\mathbf M^{-1}

for

\mathbf M=\begin{pmatrix}4&1\\2&3\end{pmatrix}

Section 5 of 14

Explore the graph

Eigenvector directions

Loading the interactive graph…

Try this

For $\mathbf M=\begin{pmatrix}4&1\\2&3\end{pmatrix}$ , the point $(t,t)$ maps to $(5t,5t)$ : same line, stretched by 5.
$(t,-2t)$ maps to $(2t,-4t)$ : stretched by 2 along $y=-2x$ .

Along eigenvector directions the matrix just stretches.

In the real worldVibrations. The natural frequencies of a bridge or a guitar body come from eigenvalues of the structure's stiffness matrix; the eigenvectors are the shapes of the vibration modes.

Section 6 of 14

Common mistakes

Incorrect

\mathbf v=\mathbf 0

as an eigenvector

CorrectEigenvectors are non-zero

\mathbf M\mathbf 0=\lambda\mathbf 0

for every

\lambda

, so it tells you nothing.

Incorrect

\mathbf P

columns and

\mathbf D

entries in different orders

CorrectSame order

The

k

th column of

\mathbf P

must match the

k

th diagonal entry of

\mathbf D

Incorrect

\det(\mathbf M)-\lambda=0

Correct

\det(\mathbf M-\lambda\mathbf I)=0

Subtract

\lambda

from every diagonal entry first.

Every other vector: gets rotated and squished

Eigenvector: just gets longer. Living its best life.

Section 7 of 14

Exam tip

Exam tipNormalise only if asked. Any non-zero multiple of an eigenvector is fine unless the question asks for unit vectors (needed for orthogonal diagonalisation of symmetric matrices).

Section 8 of 14

Guided practice

SA-FP-EIG-G012 marksL2 StandardNon-calculatorEigenvalues

Find the eigenvalues of

\begin{pmatrix}4&1\\2&3\end{pmatrix}

Salam Academy Original Practice · Proposed mark scheme

Section 9 of 14

Practice questions

SA-FP-EIG-0011 markL1 FoundationNon-calculatorEigenvalues

3\times3

matrix has trace

9

and two eigenvalues

2

and

3

. Find the third eigenvalue.

Salam Academy Original Practice · Proposed mark scheme

SA-FP-EIG-0022 marksL2 StandardNon-calculatorEigenvectors

\begin{pmatrix}1\\k\end{pmatrix}

is an eigenvector of

\begin{pmatrix}4&1\\2&3\end{pmatrix}

with eigenvalue

2

. Find

k

Salam Academy Original Practice · Proposed mark scheme

SA-FP-EIG-0033 marksL3 ExamNon-calculatorDiagonalisation

\mathbf M=\mathbf{PDP}^{-1}

with

\mathbf D=\begin{pmatrix}2&0\\0&-1\end{pmatrix}

. Find the trace of

\mathbf M^5

Salam Academy Original Practice · Proposed mark scheme

SA-FP-EIG-0043 marksL3 ExamNon-calculatorEigenvalues

Find the eigenvalues of

\begin{pmatrix}2&0&0\\1&3&0\\4&5&-1\end{pmatrix}

Salam Academy Original Practice · Proposed mark scheme

More questions on every skill in this subtopic are in the question bank under Practise.

Section 10 of 14

Challenge

SA-FP-EIG-0053 marksL4 ChallengingNon-calculatorDiagonalisation

\mathbf M

satisfies

\mathbf M^2=5\mathbf M-6\mathbf I

. Express

\mathbf M^3=p\mathbf M+q\mathbf I

. Find

p

Salam Academy Original Practice · Proposed mark scheme

Section 11 of 14

Summary

$\mathbf M\mathbf v=\lambda\mathbf v$ .
$\det(\mathbf M-\lambda\mathbf I)=0$ .
$\mathbf M=\mathbf{PDP}^{-1}$ ; $\mathbf M^n=\mathbf{PD}^n\mathbf P^{-1}$ .
Cayley–Hamilton: $\mathbf M$ satisfies its own characteristic equation.

Section 12 of 14

Quick quiz

Q1 · SA-FP-EIG-Q011 markL1 FoundationNon-calculatorEigenvectors

\mathbf v

is an eigenvector of

\mathbf M

with eigenvalue

3

, then

\mathbf M(2\mathbf v)=

Salam Academy Original Practice · Proposed mark scheme

Section 13 of 14

Same topic, different board

Specification: 9231 AS & A Level Further Mathematics
Where: Paper 2 · 2.2 Matrices
Expected: Eigenvalues and eigenvectors of 2×2 and 3×3 matrices; diagonalisation M = PDP⁻¹ (and symmetric case); Cayley–Hamilton to find inverses and powers.
Calculator: Permitted on all papers; exact answers are expected where the question asks.
Typical style: Typical structure: find the eigenvalues, then eigenvectors, then write down P and D with M = PDP⁻¹ and use it for Mⁿ.
Open in: 9231 · Paper 2

Section 14 of 14

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Salam Academy original content · Content status: reviewed and published. Syllabus mapping: Cambridge International 9231, Current syllabus (official specification).