MATH1961 · Mathematics 1a (advanced)
Eigenvalues and Eigenvectors
For most vectors x, multiplying by a matrix A rotates and rescales x into a new direction. A few special non-zero vectors are only scaled — Ax lands on the same line through the origin. Those are the eigenvectors, and the scale factor is the eigenvalue λ: Av = λv. The eigenvalues are found from the characteristic polynomial det(A − λI) = 0 (the matrix A − λI must be singular for a non-zero v to exist), and each eigenspace is the null space of A − λI — report it by a basis. The chapter then draws the line between algebraic and geometric multiplicity (the trap), and uses it for diagonalisation A = PDP−1, which makes powers and Markov applications easy. A real matrix can have complex eigenvalues (a rotation has λ = ±i).
What this chapter covers
- 01The eigen-equation Av = λv — directions the matrix only stretches
- 02The characteristic polynomial det(A − λI) = 0 & the eigenspace as a null space
- 03Algebraic vs geometric multiplicity — the multiplicity trap
- 04Diagonalisation A = PDP−1, fully worked
- 05Applications: matrix powers, Markov chains; real matrices with complex eigenvalues
Worked example: eigenvalues, eigenvectors and diagonalisability
- +1(a) Characteristic polynomial. det(A − λI) = det([[2 − λ, 1], [0, 2 − λ]]) = (2 − λ)². So λ = 2 is the only eigenvalue, with algebraic multiplicity 2.
- +1(b) Eigenspace = null(A − 2I). A − 2I = [[0, 1], [0, 0]]. Solving [[0,1],[0,0]] v = 0 forces the second component to be 0, leaving v = (t, 0).
- +1Basis. A basis for the λ = 2 eigenspace is { (1, 0) }, so its geometric multiplicity is 1.
- +1(c) Compare multiplicities. Geometric (1) < algebraic (2): A does not have enough independent eigenvectors to fill ℝ².
- +1Conclude. A is not diagonalisable — a matrix is diagonalisable iff for every eigenvalue the geometric multiplicity equals the algebraic multiplicity, which fails here. ■
Key terms
- Eigenvalue / eigenvector
- A non-zero vector v with Av = λv keeps its direction under A; the scalar λ (the eigenvalue) is the stretch factor — λ > 1 stretches, 0 < λ < 1 shrinks, λ < 0 flips, λ = 1 fixes. The zero vector solves Av = λv trivially but is never called an eigenvector.
- Characteristic polynomial
- χA(λ) = det(A − λI), a degree-n polynomial whose roots are the eigenvalues. A non-zero v solves (A − λI)v = 0 only when A − λI is singular, i.e. its determinant vanishes — that single condition is the characteristic equation.
- Eigenspace
- Eλ(A) = { v : Av = λv } = null(A − λI), the set of all eigenvectors for λ together with 0. It is a subspace, so it is reported by a basis; its dimension is the geometric multiplicity of λ.
- Algebraic vs geometric multiplicity
- Algebraic multiplicity is how many times λ is a root of the characteristic polynomial; geometric multiplicity is the dimension of its eigenspace (the number of independent eigenvectors). Always geometric ≤ algebraic, and equality for every eigenvalue is exactly the diagonalisability condition.
- Diagonalisation
- Writing A = PDP−1 where D is diagonal (the eigenvalues) and P has the corresponding eigenvectors as columns. It exists iff A has a full basis of eigenvectors, and it makes Ak = PDkP−1 — the key to matrix powers and Markov-chain limits.
Eigenvalues and Eigenvectors FAQ
How do I actually find eigenvalues and eigenvectors?
Four steps. (1) Form A − λI (subtract λ down the diagonal). (2) Compute det(A − λI), a degree-n polynomial in λ. (3) Solve χA(λ) = 0 for the eigenvalues. (4) For each λ, find null(A − λI) by row reduction — that null space is the eigenspace, and you report it by a basis. The zero vector is excluded from being an eigenvector.
What is the multiplicity trap?
Confusing how often λ is a root (algebraic multiplicity) with how many independent eigenvectors it has (geometric multiplicity). They are not always equal: geometric ≤ algebraic always, and when geometric < algebraic the matrix is ‘defective’. This gap is exactly what decides diagonalisability, so the exam plants matrices where the two differ.
When is a matrix diagonalisable?
Exactly when, for every eigenvalue, the geometric multiplicity equals the algebraic multiplicity — equivalently, when A has a full basis of eigenvectors. A quick sufficient case: n distinct eigenvalues guarantees diagonalisability. A repeated eigenvalue with too few independent eigenvectors (like [[2,1],[0,2]]) is the standard non-diagonalisable example.
Can a real matrix have no real eigenvalues?
Yes. A rotation matrix has characteristic polynomial λ² + 1, whose roots are ±i — no real eigenvalues, so over ℝ it is not diagonalisable, though over ℂ it is. This is why complex numbers and linear algebra meet in MATH1961: eigenvalues naturally live in ℂ even for real matrices.
Exam move
Make the four-step recipe automatic: form A − λI, take the determinant, solve for the eigenvalues, then row-reduce A − λI for each eigenspace and write a basis. Keep the multiplicity distinction sharp — always compute both the algebraic multiplicity (root count) and the geometric multiplicity (eigenspace dimension), because their comparison is what every diagonalisability question turns on. For diagonalisation, practise assembling P (eigenvectors as columns) and D (eigenvalues on the diagonal) and using Ak = PDkP−1. Remember the two traps: 0 is never an eigenvector, and a real matrix can have complex eigenvalues.
