Linear Algebra · step 6 of 13

Eigenvalues & eigenvectors

A non-zero v is an eigenvector of A with eigenvalue λ when Av = λv — the map only stretches v, without rotating it. The eigenvalues are the roots of det(A − λI) = 0, and for each one the eigenvectors are the non-zero solutions of (A − λI)v = 0. That system is always singular by construction, so a free variable is guaranteed.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Form the characteristic polynomialCompute det(A − λI) and set it to zero. For 2×2 it is λ² − (trace)λ + det.
Solve for the eigenvaluesFactor the polynomial. Exam matrices are chosen so the roots are small integers.
Solve (A − λI)v = 0 for each λRow reduce; the system must have a free variable, and its solution space is the eigenspace.
Present eigenvectors up to scaleAny non-zero multiple is an eigenvector, so give a clean representative such as (1, 2).

Worked example

Find the eigenvalues and eigenvectors of A = [[4, 1], [2, 3]].

trace = 7, det = 12 − 2 = 10, so the characteristic polynomial is λ² − 7λ + 10.
Factor: (λ − 5)(λ − 2) = 0, giving λ = 5 and λ = 2.
For λ = 5: A − 5I = [[−1,1],[2,−2]], so −x + y = 0 and v = (1, 1).
For λ = 2: A − 2I = [[2,1],[2,1]], so 2x + y = 0 and v = (1, −2).

Answer. λ = 5 with eigenvector (1,1); λ = 2 with eigenvector (1,−2). Check: A(1,1) = (5,5) ✓.

Where marks get dropped

These are the specific errors that cost credit on eigenvalues & eigenvectors questions — QED's rubric penalises each of them separately.

Accepting v = 0 as an eigenvector. It satisfies Av = λv for every λ and is excluded by definition.
Forgetting that (A − λI)v = 0 must be singular. If your row reduction gives only the zero solution, the eigenvalue is wrong.
Reporting only one eigenvector for a repeated eigenvalue without checking the eigenspace dimension — it may be 1 or 2.

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Eigenvalues & eigenvectors — frequently asked questions

What do the trace and determinant tell me?

The eigenvalues sum to the trace and multiply to the determinant. It is the fastest sanity check on a 2×2 computation.

Can eigenvalues be complex?

Yes — a rotation matrix has no real eigenvalues, since no direction is preserved. Over ℂ every n×n matrix has n eigenvalues with multiplicity.

What is an eigenspace?

The null space of A − λI, containing all eigenvectors for λ plus the zero vector. Its dimension is the geometric multiplicity and never exceeds the algebraic one.

The rest of Linear Algebra

Matrices, systems, determinants, eigenvalues. Each subtopic below has its own method, worked example and mark-losing traps.

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