Matrices: eigenvalues, eigenvectors and diagonalisation

CIE A-Level Further Mathematics· 9231 U2 Further Pure 2, Section 7.1-7.4· 45 min read

1. Deriving the Characteristic Equation★★★☆☆⏱ 10 min

The characteristic equation is the core starting point for all eigenvalue work. For any n x n matrix M, we rearrange the defining eigenvector relation $M v = λ v$ to get $(M - λ I) v = 0$ , where I is the n x n identity matrix.

\det(M - \lambda I) = 0

📘 Definition

Characteristic Polynomial

\det(M - \lambda I)

The nth degree polynomial in λ obtained by expanding the determinant of M minus λ times the identity matrix.

📐 Worked Example

Find the characteristic equation for the 3x3 matrix $M = 210120003$

1
First subtract λ times the identity matrix from M:
2
$M - \lambda I = \begin{pmatrix} 2-\lambda & 1 & 0 \\ 1 & 2-\lambda & 0 \\ 0 & 0 & 3-\lambda \end{pmatrix}$
3
Expand the determinant along the third row to simplify calculation:
4
$\det(M-\lambda I) = (3-\lambda) \times \det\begin{pmatrix}2-\lambda & 1 \\ 1 & 2-\lambda\end{pmatrix}$
5
Simplify the 2x2 determinant and rearrange terms:
6
$\det(M-\lambda I) = (3-\lambda)\left((2-\lambda)^2 -1\right) = (3-\lambda)(\lambda^2 -4\lambda +3) = -\lambda^3 +7\lambda^2 -15\lambda +9$
7
Final characteristic equation (multiply both sides by -1 for standard positive leading coefficient):
8
$\lambda^3 -7\lambda^2 +15\lambda -9 = 0$

✓ Quick check

What is the degree of the characteristic polynomial for a 4x4 matrix?
Reveal answer
4 —
The degree of the characteristic polynomial always equals the size of the square matrix.

Exam tip:

Always expand the determinant along a row or column with zero entries to reduce arithmetic errors, especially for 3x3 matrices.

2. Calculating Eigenvalues and Eigenvectors★★★★☆⏱ 12 min

Once you have the characteristic equation, factorise it to find all roots, which are the eigenvalues. For each eigenvalue λ, substitute back into $(M - λ I) v = 0$ and solve the homogeneous system to find the corresponding non-zero eigenvector.

📐 Worked Example

Find all eigenvalues and corresponding eigenvectors for the matrix M from the previous example, $M = 210120003$

1
Factorise the characteristic equation we derived earlier:
2
$\lambda^3 -7\lambda^2 +15\lambda -9 = (\lambda-3)^2(\lambda-1) = 0$
3
Eigenvalues are λ=1 and λ=3 (repeated twice)
4
For λ=1, substitute into $(M - I) v = 0$ :
5
$M-I = \begin{pmatrix}1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 2\end{pmatrix} \implies v_1 + v_2 = 0, v_3=0 \implies \mathbf{v} = k\begin{pmatrix}1 \\ -1 \\ 0\end{pmatrix}, k\neq 0$
6
For λ=3, substitute into $(M - 3 I) v = 0$ :
7
$M-3I = \begin{pmatrix}-1 & 1 & 0 \\ 1 & -1 & 0 \\ 0 & 0 & 0\end{pmatrix} \implies -v_1 + v_2 = 0 \implies \mathbf{v} = k\begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix} + m\begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}, k,m \neq 0$

3. Diagonalisability and PDP⁻¹ Decomposition★★★★☆⏱ 10 min

An n x n matrix is diagonalisable if and only if it has n linearly independent eigenvectors. If this condition holds, you can write $M = P D P^{- 1}$ , where P is the modal matrix of eigenvectors and D is the diagonal matrix of corresponding eigenvalues.

📐 Worked Example

Construct the PDP⁻¹ decomposition for the 3x3 matrix M from earlier

1
We have 3 linearly independent eigenvectors, so M is diagonalisable. Arrange eigenvectors as columns of P, matching order of eigenvalues in D:
2
$P = \begin{pmatrix} 1 & 1 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix}$
3
Calculate the inverse of P using standard 3x3 inverse method:
4
$P^{-1} = \frac{1}{2}\begin{pmatrix} 1 & -1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}$
5
Verify the decomposition by multiplying PDP⁻¹ to confirm you recover M.

4. Calculating High Matrix Powers Using Diagonalisation★★★☆☆⏱ 8 min

A key application of diagonalisation is computing $M^{n}$ for very large n, which would be impossible via repeated matrix multiplication. The identity $M^{n} = P D^{n} P^{- 1}$ holds, and $D^{n}$ is trivial to compute as you just raise each diagonal entry to the power n.

📐 Worked Example

Compute $M^{4}$ for our example 3x3 matrix M

1
Raise each diagonal entry of D to the 4th power:
2
$D^4 = \begin{pmatrix} 1^4 & 0 & 0 \\ 0 & 3^4 & 0 \\ 0 & 0 & 3^4 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 81 & 0 \\ 0 & 0 & 81 \end{pmatrix}$
3
Multiply P, D⁴ and P⁻¹ in sequence:
4
$M^4 = \begin{pmatrix} 1 & 1 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 81 & 0 \\ 0 & 0 & 81 \end{pmatrix} \frac{1}{2}\begin{pmatrix} 1 & -1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix} = \begin{pmatrix}41 & 40 & 0 \\ 40 & 41 & 0 \\ 0 & 0 & 81\end{pmatrix}$

5. Common Pitfalls

Wrong move:

Normalising eigenvectors to unit length unnecessarily

Why:

Wastes time and introduces arithmetic errors that lose marks, as CIE does not require unit eigenvectors

Correct move:

Leave eigenvectors in simplest integer component form with no fractions or radicals

Wrong move:

Mixing up the order of eigenvectors in P and eigenvalues in D

Why:

The columns of P must exactly match the order of eigenvalues on the diagonal of D, otherwise the decomposition is invalid

Correct move:

Label each eigenvector with its corresponding eigenvalue before building P and D to confirm alignment

Wrong move:

Using the zero vector as an eigenvector

Why:

The zero vector is explicitly excluded from the definition of eigenvectors, and will not satisfy the non-trivial solution condition

Correct move:

Always specify that the scalar multiple constant k is non-zero when writing general eigenvector solutions

Wrong move:

Assuming all matrices with repeated eigenvalues are not diagonalisable

Why:

Repeated eigenvalues can still have a full set of linearly independent eigenvectors, making the matrix diagonalisable

Correct move:

Count the number of linearly independent eigenvectors before concluding a matrix is non-diagonalisable

Wrong move:

Forgetting to multiply by P⁻¹ at the end when calculating Mⁿ

Why:

This is the single most common mark-losing error in CIE diagonalisation questions, worth 2-3 marks

Correct move:

Write the full formula $M^{n} = P D^{n} P^{- 1}$ at the top of your working to remind yourself of all three terms

6. Quick Reference Cheatsheet

Step	2x2 Matrix	3x3 Matrix	CIE Mark Scheme Check
Characteristic Equation	Expand 2x2 determinant, solve quadratic	Expand along zero row/column, solve cubic	Confirm sum of eigenvalues equals trace of M
Eigenvectors	Solve 1 linear equation for non-zero solution	Solve homogeneous system for each λ	No zero vectors, any non-zero multiple accepted
Diagonalisation	2 independent eigenvectors required	3 independent eigenvectors required	P and D order must match exactly
Mⁿ Calculation	Raise D entries to n, multiply PDⁿP⁻¹	Raise D entries to n, multiply PDⁿP⁻¹	Verify with small n (e.g. n=2) to catch errors

7. Frequently Asked

Can eigenvectors have zero components?

Yes, individual components can be zero, but the eigenvector itself cannot be the zero vector, as this is not accepted by the definition Mv=λv for any λ.

Do eigenvalues have to be distinct for a matrix to be diagonalisable?

No, a matrix can be diagonalisable even with repeated eigenvalues if there are enough linearly independent eigenvectors to form a full basis for the space.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2024 · Paper 1
3x3 diagonalisation and matrix power
2023 · Paper 2
Repeated eigenvalues and diagonalisability check
2022 · Paper 1
Characteristic equation derivation and eigenvectors

Going deeper

official_syllabusCIE 9231 Syllabus 2025-2027Section 7 reference

What's Next

Mastering eigenvalues, eigenvectors and diagonalisation is the foundation for the rest of the Further Pure 2 matrices unit. You will next extend these concepts to symmetric matrices, which have the unique property that their eigenvectors are mutually orthogonal, allowing you to construct orthogonal diagonalisation. This is a heavily weighted 15-20 mark question type in recent CIE exam papers. You will also go on to learn the Cayley-Hamilton theorem, which lets you compute matrix powers and inverses without full diagonalisation, and explore applications of matrix transformations to conic sections. These concepts build directly on the skills you have practiced here, so ensure you can complete all 3x3 diagonalisation steps without arithmetic errors before moving on.