What is the eigenvalue equation?

A non-zero vector v is an eigenvector of A with eigenvalue λ if

The eigenvalues of a 22 matrix are 3 and -1. State its trace and determinant. [2 marks]

SEAB Original-style practice: Find the eigenvalues of B = pmatrix 3 & 0 & 0 \\ 1 & 2 & 0 \\ 2 & 1 & 1 pmatrix and an eigenvector corresponding to the largest eigenvalue.

Since B is lower triangular, its eigenvalues are the diagonal entries: λ = 3, 2, 1. (For a triangular matrix (B - λI) is the product of the diagonal terms (3-λ)(2-λ)(1-λ).) The largest eigenvalue is λ = 3. Solve (B - 3I)v = 0: $pmatrix 0 & 0 & 0 \\ 1 & -1 & 0 \\ 2 & 1 & -2 pmatrixpmatrix x \\ y \\ z pmatrix = 0.$ Row 2: x - y = 0 x = y. Row 3: 2x + y - 2z = 0 2x + x - 2z = 0 z = 32x. Take x = 2: eigenvector pmatrix 2 \\ 2 \\ 3 pmatrix. Markers reward reading eigenvalues off the diagonal of the triangular matrix, setting up (B - 3I)v = 0, solving the two independent equations, and a valid eigenvector.

SingaporeFurther MathsSyllabus dot point

What are the eigenvalues and eigenvectors of a matrix, and how do we find them?

Find the eigenvalues and eigenvectors of 2x2 and 3x3 matrices using the characteristic equation

A focused answer to the H2 Further Mathematics outcome on eigenvalues and eigenvectors. The eigenvalue equation, the characteristic polynomial, finding eigenvalues from det(A - lambda I) = 0, solving for eigenvectors, and their geometric meaning as invariant directions.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-06

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

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What this dot point is asking

SEAB wants you to find the eigenvalues and eigenvectors of $2\times2$ and $3\times3$ matrices. An eigenvalue is a scalar $\lambda$ for which the matrix has a non-zero vector that it merely scales; that vector is the eigenvector. You find eigenvalues from the characteristic equation $\det(\mathbf{A} - \lambda\mathbf{I}) = 0$ and then solve a homogeneous system for each eigenvector.

The answer

The eigenvalue equation

A non-zero vector $\mathbf{v}$ is an eigenvector of $\mathbf{A}$ with eigenvalue $\lambda$ if

\mathbf{A}\mathbf{v} = \lambda\mathbf{v}.

Geometrically, $\mathbf{A}$ leaves the direction of $\mathbf{v}$ unchanged (or exactly reversed if $\lambda < 0$ ), stretching it by the factor $\lambda$ . Such directions are the invariant directions of the linear map.

The characteristic equation

Rearrange $\mathbf{A}\mathbf{v} = \lambda\mathbf{v}$ to $(\mathbf{A} - \lambda\mathbf{I})\mathbf{v} = \mathbf{0}$ . For a non-zero $\mathbf{v}$ to exist, the matrix $\mathbf{A} - \lambda\mathbf{I}$ must be singular, so

\det(\mathbf{A} - \lambda\mathbf{I}) = 0.

This is the characteristic equation, a polynomial in $\lambda$ of degree equal to the size of the matrix. Its roots are the eigenvalues.

Finding eigenvectors

For each eigenvalue $\lambda$ , substitute back and solve the homogeneous system $(\mathbf{A} - \lambda\mathbf{I})\mathbf{v} = \mathbf{0}$ . The system is always singular (by construction), so it has non-trivial solutions forming a line (or plane) of eigenvectors; quote any convenient non-zero representative.

Useful shortcuts

For a triangular matrix the eigenvalues are simply the diagonal entries. Two checks on a full matrix: the sum of the eigenvalues equals the trace (sum of the diagonal), and the product of the eigenvalues equals the determinant. These catch arithmetic slips quickly.

Worked example

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

Step 1: Form the characteristic equation

\det(\mathbf{A} - \lambda\mathbf{I}) = \det\begin{pmatrix} 4 - \lambda & 2 \\ 1 & 3 - \lambda \end{pmatrix} = (4-\lambda)(3-\lambda) - 2 = 0.

Step 2: Expand and solve

\lambda^2 - 7\lambda + 12 - 2 = \lambda^2 - 7\lambda + 10 = 0 \Rightarrow (\lambda - 2)(\lambda - 5) = 0,

so $\lambda = 2$ or $\lambda = 5$ . (Check: trace $= 7 = 2 + 5$ ; determinant $= 10 = 2 \times 5$ .)

Step 3: Eigenvector for lambda = 2

Solve $(\mathbf{A} - 2\mathbf{I})\mathbf{v} = \mathbf{0}$ : $\begin{pmatrix} 2 & 2 \\ 1 & 1 \end{pmatrix}\mathbf{v} = \mathbf{0}$ gives $v_1 + v_2 = 0$ , so $\mathbf{v} = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$ .

Step 4: Eigenvector for lambda = 5

Solve $(\mathbf{A} - 5\mathbf{I})\mathbf{v} = \mathbf{0}$ : $\begin{pmatrix} -1 & 2 \\ 1 & -2 \end{pmatrix}\mathbf{v} = \mathbf{0}$ gives $v_1 = 2v_2$ , so $\mathbf{v} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$ .

Step 5: State the result

The eigenvalues are $2$ and $5$ , with eigenvectors $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$ and $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$ .

Examples in context

Example 1. Long-run behaviour of a process. In a Markov-type model the eigenvalue $\lambda = 1$ and its eigenvector give the steady-state distribution, while eigenvalues with $|\lambda| < 1$ govern how fast transient effects decay. Eigenanalysis is how the long-run state is extracted.

Example 2. Principal axes of a transformation. The eigenvectors of a symmetric matrix point along the principal axes along which a stretch is pure (no shear), which is why eigenvectors describe the natural directions of deformation and underlie diagonalisation.

Try this

Q1. Write the equation that defines an eigenvalue and eigenvector of $\mathbf{A}$ . [1 mark]

Cue. $\mathbf{A}\mathbf{v} = \lambda\mathbf{v}$ with $\mathbf{v} \neq \mathbf{0}$ .

Q2. State the characteristic equation used to find eigenvalues. [1 mark]

Cue. $\det(\mathbf{A} - \lambda\mathbf{I}) = 0$ .

Q3. The eigenvalues of a $2\times2$ matrix are $3$ and $-1$ . State its trace and determinant. [2 marks]

Cue. Trace $= 3 + (-1) = 2$ ; determinant $= 3 \times (-1) = -3$ .

Exam-style practice questions

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

Original6 marksFind the eigenvalues and corresponding eigenvectors of

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

Show worked answer →

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

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

(2-\lambda)^2 = 1

, giving

2 - \lambda = \pm 1

, hence

\lambda = 1

\lambda = 3

For $\lambda = 1$ : solve $(\mathbf{A} - \mathbf{I})\mathbf{v} = \mathbf{0}$ , that is $\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\mathbf{v} = \mathbf{0}$ , giving $v_1 + v_2 = 0$ . An eigenvector is $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$ .

For $\lambda = 3$ : solve $\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}\mathbf{v} = \mathbf{0}$ , giving $v_1 = v_2$ . An eigenvector is $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ .

Markers reward forming the characteristic equation, solving for both eigenvalues, and a correct (non-zero) eigenvector for each.

Original7 marksFind the eigenvalues of

\mathbf{B} = \begin{pmatrix} 3 & 0 & 0 \\ 1 & 2 & 0 \\ 2 & 1 & 1 \end{pmatrix}

and an eigenvector corresponding to the largest eigenvalue.

Show worked answer →

Since $\mathbf{B}$ is lower triangular, its eigenvalues are the diagonal entries: $\lambda = 3, 2, 1$ . (For a triangular matrix $\det(\mathbf{B} - \lambda\mathbf{I})$ is the product of the diagonal terms $(3-\lambda)(2-\lambda)(1-\lambda)$ .)

The largest eigenvalue is $\lambda = 3$ . Solve $(\mathbf{B} - 3\mathbf{I})\mathbf{v} = \mathbf{0}$ :

\begin{pmatrix} 0 & 0 & 0 \\ 1 & -1 & 0 \\ 2 & 1 & -2 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \mathbf{0}.

Row 2:

x - y = 0 \Rightarrow x = y

. Row 3:

2x + y - 2z = 0 \Rightarrow 2x + x - 2z = 0 \Rightarrow z = \tfrac{3}{2}x

Take $x = 2$ : eigenvector $\begin{pmatrix} 2 \\ 2 \\ 3 \end{pmatrix}$ .

Markers reward reading eigenvalues off the diagonal of the triangular matrix, setting up $(\mathbf{B} - 3\mathbf{I})\mathbf{v} = \mathbf{0}$ , solving the two independent equations, and a valid eigenvector.