Eigenvalues, eigenvectors and diagonalisation

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Most vectors change direction under a transformation, but a few are only stretched. These eigenvectors, and their stretch factors the eigenvalues, reveal what a matrix does and make its powers easy to compute. This lesson finds eigenvalues from the characteristic equation and eigenvectors from the null space of A-λI, diagonalises a matrix as A = PDP⁻¹, and uses the largest eigenvalue to describe the long-term behaviour of a repeated process.

Some directions are only stretched

A square matrix M moves most vectors to new directions. Take

M = (4, 1; 2, 3).

Then M (1; 0) = (4; 2), a different direction. But

M (1; 1) = (5; 5) = 5 (1; 1), M (1; -2) = (2; -4) = 2 (1; -2).

Along these two lines through the origin, M only stretches. A nonzero vector v with Mv = λv is an eigenvector of M, and the scalar λ is its eigenvalue. The zero vector is excluded: M0 = λ0 for every λ, so it would tell us nothing.

Any nonzero multiple of an eigenvector is an eigenvector for the same eigenvalue. An eigenvector names a direction; its length is your choice.

First component of Mv: 4.00
Second component of Mv: 2.00
det(v, Mv): zero means parallel: 2.00

Predict which k makes the dashed input and solid image parallel, then try k = 1 and k = −2. The fixed matrix is M = [[4, 1], [2, 3]]; only v = (1, k) changes.

The characteristic equation finds the eigenvalues

Mv = λv is the same as (M-λI)v = 0. A nonzero solution v exists exactly when M-λI is singular, that is when

det(M-λI) = 0.

This is the characteristic equation. For an n × n matrix it is a polynomial equation of degree n in λ. In this syllabus, its roots are real.

Find the eigenvalues of M = (4, 1; 2, 3) and an eigenvector for each.

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Solve the characteristic equation
Method
Subtract λ from each diagonal entry and set the determinant to zero.
Reason
Only values of λ that make M-λI singular allow a nonzero v.
Working
Answer
det(4-λ, 1; 2, 3-λ) = (4-λ)(3-λ)-2; = λ²-7λ + 10; = (λ-2)(λ-5) = 0.
The eigenvalues are λ = 5 and λ = 2.
Solve for each eigenvector
Method
Substitute each eigenvalue and solve (M-λI)v = 0.
Reason
Because the matrix is singular, its two rows give the same condition; one equation fixes the direction.
Working
Answer
For λ = 5: (-1, 1; 2, -2) (x; y) = 0 gives y = x, so (1; 1).
For λ = 2: (2, 1; 2, 1) (x; y) = 0 gives y = -2x, so (1; -2).

Quick check

The eigenvalues of a 2 × 2 matrix add to its trace and multiply to its determinant: 5 + 2 = 7 = 4 + 3 and 5 × 2 = 10 = 4 × 3-1 × 2. If both rows of M-λI do not give the same condition, the only solution is v = 0, which means the eigenvalue was found incorrectly.

Find the eigenvalues of A = (1, 2; 2, 1) and an eigenvector for each. Check your eigenvalues with the trace and determinant.

Eigenvalues, eigenvectors and your check

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(1-λ)²-4 = 0, so 1-λ = ±2 and λ = 3 or λ = -1. For λ = 3, -2x + 2y = 0 gives (1; 1). For λ = -1, 2x + 2y = 0 gives (1; -1). Check: 3 + (-1) = 2 is the trace and 3 × (-1) = -3 = 1-4 is the determinant.

For a 3 × 3 matrix the characteristic equation is a cubic. When the matrix is triangular, M-λI is also triangular, and its determinant is the product of its diagonal entries. So a triangular matrix has its diagonal entries as eigenvalues.

B = (2, 1, 1; 0, 3, 1; 0, 0, 1). State the eigenvalues and find an eigenvector for each.

Three eigenvalues and an eigenvector for each

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The eigenvalues are 2, 3 and 1.

λ = 2: the third row gives -z = 0, then y + z = 0, so y = 0. Eigenvector (1; 0; 0).
λ = 3: -2z = 0 and z = 0; the first row gives -x + y = 0. Eigenvector (1; 1; 0).
λ = 1: 2y + z = 0 and x + y + z = 0, so z = -2y and x = y. Eigenvector (1; 1; -2).

Check the last one: B (1; 1; -2) = (2 + 1-2; 3-2; -2) = (1; 1; -2).

Diagonalisation turns matrix powers into number powers

Put eigenvectors as the columns of Q and their eigenvalues, in the same order, on the diagonal of D. Then MQ = QD, column by column. If the eigenvectors are linearly independent, Q is invertible and

M = QDQ⁻¹.

Now the inner factors cancel in a power:

Mⁿ = (QDQ⁻¹)(QDQ⁻¹)…(QDQ⁻¹) = QDⁿQ⁻¹,

and Dⁿ just raises each diagonal entry to the power n. An n × n matrix with n distinct eigenvalues always has independent eigenvectors, so it can be diagonalised.

For M = (4, 1; 2, 3), find Q and D with M = QDQ⁻¹, and hence find Mⁿ for every positive integer n.

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Match columns of Q to entries of D
Method
Use the eigenpairs (5, (1; 1)) and (2, (1; -2)).
Reason
Column j of MQ is M times the jth eigenvector, which must equal its eigenvalue times that eigenvector.
Working
Answer
Q = (1, 1; 1, -2), D = (5, 0; 0, 2), Q⁻¹ = 1/(-3) (-2, -1; -1, 1) = 1/3 (2, 1; 1, -1).
Multiply out Q Dⁿ Q⁻¹
Method
Multiply QDⁿ first, then multiply by Q⁻¹.
Reason
Scaling each column of Q by its eigenvalue power keeps the arithmetic short.
Working
Answer
Mⁿ = 1/3 (5ⁿ, 2ⁿ; 5ⁿ, -2ⁿ⁺¹) (2, 1; 1, -1); = 1/3 (2 · 5ⁿ + 2ⁿ, 5ⁿ-2ⁿ; 2 · 5ⁿ-2ⁿ⁺¹, 5ⁿ + 2ⁿ⁺¹).

Quick check

At n = 1 the formula gives 1/3 (12, 3; 6, 9) = (4, 1; 2, 3), and at n = 0 it gives the identity. Every entry grows like 5ⁿ: the larger eigenvalue dominates high powers.

A student diagonalises the same M with Q = (1, 1; 1, -2) and D = (2, 0; 0, 5), and writes Mⁿ = QDⁿQ⁻¹. Without multiplying everything out, explain what is wrong and how to fix it.

Write your answer, then compare it with the solution

What is wrong, and the fix

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Compare each column with its eigenvalue
Method
The first column of Q is (1; 1), whose eigenvalue is 5, but the first entry of D is 2.
Reason
MQ = QD holds only when column j of Q belongs to the jth diagonal entry of D.
Working
Answer
QDQ⁻¹ is then a different matrix, so the student’s Mⁿ is wrong for every n ≥ 1. Either swap the entries of D or swap the columns of Q.

Common mistake

Both D and Q are individually correct lists of eigenvalues and eigenvectors; the error is the pairing. Substituting n = 1 into any power formula and comparing it with M catches this mistake, as well as writing Q⁻¹DⁿQ in the wrong order.

Not every matrix can be diagonalised. (2, 1; 0, 2) has the single eigenvalue 2, and (M-2I)v = 0 forces y = 0. Every eigenvector is a multiple of (1; 0), so no invertible Q of eigenvectors exists.

Long-term behaviour from the largest eigenvalue

Common mistakes

Solving det(A-λI) = 0 with an arithmetic slip and not checking that the eigenvalues add to the trace and multiply to the determinant.
Giving the zero vector as an eigenvector.
Putting the eigenvectors in P in a different order from their eigenvalues in D.
Writing Aⁿ = P⁻¹DⁿP instead of PDⁿP⁻¹.
Assuming every square matrix can be diagonalised, even when a repeated eigenvalue has too few independent eigenvectors.

Before you move on

Each year, 80% of a town’s residents stay in the town and 20% move to the suburbs, while 70% of suburban residents stay and 30% move to the town. Nobody else arrives or leaves. With xₙ town and yₙ suburban residents after n years,

(xₙ₊₁; yₙ₊₁) = (0.8, 0.3; 0.2, 0.7) (xₙ; yₙ).

Initially x₀ = 1000 and y₀ = 0. Find xₙ and yₙ, and describe what happens in the long run.

Write your answer, then compare it with the solution

Eigenpairs, formulas for x and y, and the long-run populations

Hint 1

One eigenvalue is 1

Each column adds to 1, so the total population is unchanged. Check that λ = 1 satisfies the characteristic equation, then find the other eigenvalue from the trace.

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Find the eigenpairs
Method
The trace is 1.5 and the determinant is 0.56-0.06 = 0.5.
Reason
λ²-1.5λ + 0.5 = (λ-1)(λ-0.5), so the eigenvalues are 1 and 0.5.
Working
Answer
For λ = 1: -0.2x + 0.3y = 0, so (3; 2). For λ = 0.5: 0.3x + 0.3y = 0, so (1; -1).
Write the start as a combination of eigenvectors
Method
Solve (1000; 0) = a (3; 2) + b (1; -1).
Reason
Each part is simply multiplied by its own eigenvalue every year, which is QDⁿQ⁻¹ applied to the start.
Working
Answer
3a + b = 1000 and 2a-b = 0 give a = 200 and b = 400. So
(xₙ; yₙ) = 200 (3; 2) + 400(0.5)ⁿ (1; -1) = (600 + 400(0.5)ⁿ; 400-400(0.5)ⁿ).
Let n grow
Method
(0.5)ⁿ → 0.
Reason
Only the eigenvalue-1 part survives.
Working
Answer
xₙ → 600 and yₙ → 400: in the long run the residents split 3:2 between the town and the suburbs, whatever the starting split of the same total.

Quick check

After one year the matrix gives (800,200), and the formula gives (600 + 200, 400-200). The total stays 1000 for every n. Compare your explanation of the long-run split with the solution; this is self-checked, not marked.

Row reduction, rank and nullity are taught in Row reduction, rank and null space, and linear transformations and linear spaces in Linear transformations and linear spaces.

Syllabus and review details

GCE A-Level H2 Further Mathematics 2027 · 2027

Last reviewed: 2026-11-06

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Guided steps

Reading settings

Some directions are only stretched

The characteristic equation finds the eigenvalues

Eigenvalues before eigenvectors

Solve the characteristic equation

Solve for each eigenvector

A symmetric matrix

Three eigenvalues at once

Diagonalisation turns matrix powers into number powers

A formula for every power

Match columns of Q to entries of D

Multiply out Q Dⁿ Q⁻¹

A diagonalisation that does not check

Compare each column with its eigenvalue

Long-term behaviour from the largest eigenvalue

Common mistakes

Before you move on

Where a population settles

Find the eigenpairs

Write the start as a combination of eigenvectors

Let n grow

Lesson done

Some directions are only stretched

The characteristic equation finds the eigenvalues

Eigenvalues before eigenvectors

A symmetric matrix

Three eigenvalues at once

Diagonalisation turns matrix powers into number powers

A formula for every power

A diagonalisation that does not check

Long-term behaviour from the largest eigenvalue

Common mistakes

Before you move on

Where a population settles

Questions