Write down the characteristic equation for pmatrix 5 & 0 \\ 0 & 2 pmatrix. [1 mark]

Find an eigenvector of pmatrix 3 & 0 \\ 0 & 7 pmatrix for λ = 7. [1 mark]

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What are eigenvalues and eigenvectors, and how do you find them for a 2x2 matrix?

Eigenvalues and eigenvectors of a 2x2 matrix, the characteristic equation, invariant lines and lines of invariant points of a transformation, and using eigenvectors to describe the geometry of a matrix.

A CCEA A2 Further Maths answer on eigenvalues and eigenvectors of a 2x2 matrix, the characteristic equation, finding eigenvectors, invariant lines and lines of invariant points, and what they reveal about a transformation.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-16

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
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What this dot point is asking

CCEA wants you to find the eigenvalues and eigenvectors of a $2 \times 2$ matrix via the characteristic equation, and to interpret them geometrically as invariant lines (and lines of invariant points) of the transformation the matrix represents.

The answer

Eigenvalues and eigenvectors

The characteristic equation

Finding the eigenvectors

Invariant lines

Eigenvalues, eigenvectors and invariant lines

Find the eigenvalues and eigenvectors of $\mathbf{A} = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$ .

step Form the characteristic equation

\det\begin{pmatrix} 4 - \lambda & 1 \\ 2 & 3 - \lambda \end{pmatrix} = (4 - \lambda)(3 - \lambda) - 2 = \lambda^2 - 7\lambda + 10 = 0.

step Solve for the eigenvalues

\lambda^2 - 7\lambda + 10 = (\lambda - 2)(\lambda - 5) = 0 \Rightarrow \lambda = 2 \text{ or } \lambda = 5.

step Find each eigenvector

For $\lambda = 2$ : $\begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix}\mathbf{v} = \mathbf{0}$ gives $2x + y = 0$ , so $\begin{pmatrix} 1 \\ -2 \end{pmatrix}$ .

For $\lambda = 5$ : $\begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix}\mathbf{v} = \mathbf{0}$ gives $x = y$ , so $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ .

The lines $y = -2x$ and $y = x$ are invariant under $\mathbf{A}$ .

Examples in context

Example 1. Principal axes of a stretch. A material stretched unevenly has two perpendicular directions, its principal axes, along which it simply lengthens or shortens. These are the eigenvectors of the deformation matrix, and the stretch factors are the eigenvalues, which is how engineers analyse strain.

Example 2. Long-term behaviour of a process. When a system evolves by repeated multiplication by a matrix (such as a population model), the dominant eigenvalue controls the eventual growth rate and its eigenvector the eventual proportions. Eigen-analysis predicts the long-run state.

Try this

Q1. Write down the characteristic equation for $\begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix}$ . [1 mark]

Cue. $(5 - \lambda)(2 - \lambda) = 0$ , so $\lambda = 5$ or $\lambda = 2$ .

Q2. An eigenvalue equals $1$ . What is special about its invariant line? [1 mark]

Cue. It is a line of invariant points (each point maps to itself).

Q3. Find an eigenvector of $\begin{pmatrix} 3 & 0 \\ 0 & 7 \end{pmatrix}$ for $\lambda = 7$ . [1 mark]

Cue. $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ .

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA A2 20207 marksFind the eigenvalues and corresponding eigenvectors of the matrix

\mathbf{A} = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}

Show worked answer →

The eigenvalues satisfy the characteristic equation $\det(\mathbf{A} - \lambda\mathbf{I}) = 0$ :

$\det\begin{pmatrix} 2 - \lambda & 1 \\ 1 & 2 - \lambda \end{pmatrix} = (2 - \lambda)^2 - 1 = 0.$

So $(2 - \lambda)^2 = 1$ , giving $2 - \lambda = \pm 1$ , hence $\lambda = 1$ or $\lambda = 3$ .

For $\lambda = 1$ : $(\mathbf{A} - \mathbf{I})\mathbf{v} = \mathbf{0}$ gives $\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\mathbf{v} = \mathbf{0}$ , so $x + y = 0$ , eigenvector $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$ .

For $\lambda = 3$ : $\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}\mathbf{v} = \mathbf{0}$ , so $x = y$ , eigenvector $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ .

Markers reward the characteristic equation, both eigenvalues, and a correct eigenvector for each.

CCEA A2 20185 marksThe matrix

\mathbf{M} = \begin{pmatrix} 3 & 0 \\ 0 & -2 \end{pmatrix}

represents a transformation. State its eigenvalues and describe the invariant lines through the origin.

Show worked answer →

The matrix is diagonal, so the eigenvalues are the diagonal entries: $\lambda = 3$ and $\lambda = -2$ .

For $\lambda = 3$ , the eigenvector is $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ , so the $x$ -axis ( $y = 0$ ) is an invariant line: points on it map to points on it (stretched by factor $3$ ).

For $\lambda = -2$ , the eigenvector is $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ , so the $y$ -axis ( $x = 0$ ) is an invariant line: points on it map onto it (scaled by $-2$ , so reflected and stretched).

Markers reward both eigenvalues, identifying the axes as invariant lines, and noting the scale factor along each.

Related dot points

Sources & how we know this

CCEA GCE Further Mathematics specification — CCEA (2018)