IB Maths AI HLEigenvalues & EigenvectorsPaper 1 & 2Characteristic polynomial~9 min read
The Characteristic Polynomial, Eigenvalues & Eigenvectors
An eigenvector of a matrix is a special direction that the matrix stretches without rotating — the stretch factor is the eigenvalue. Finding them turns Ax = λx into a quadratic (the characteristic polynomial) plus a quick linear-equation step for each root.
📘 What you need to know
Definition: Ax = λx with x ≠ 0; x is an eigenvector, λ its eigenvalue.
Characteristic polynomial: p(λ) = det(λI − A). For 2×2 always a quadratic. Not in the formula booklet.
2×2 shortcut: p(λ) = λ2 − (trace A)λ + det A.
Eigenvalues: roots of p(λ) = 0 — real distinct, real repeated, or complex conjugate pair.
Eigenvectors: for each λ solve (λI − A)x = 0; the two rows give the same relation; set x = 1.
Trace check: leading-diagonal sum of A = sum of eigenvalues.
The characteristic polynomial
The characteristic polynomial of an n × n matrix A is p(λ) = det(λI − A). For 2×2, three steps: form λI − A, take its determinant with ad − bc, then expand. The result is always a quadratic.
Quick 2×2 shortcut: p(λ) = λ2 − (trace A)λ + det A. For A = ((a, b), (c, d)) the trace is a + d and det is ad − bc — you can read both straight off the matrix.
Eigenvalues — finding them
An eigenvector is a non-zero x that A just rescales: Ax = λx. Rearranging gives (λI − A)x = 0, which has a non-zero solution only when det(λI − A) = 0. So the eigenvalues are the roots of p(λ) = 0 — either two real distinct, one real repeated, or a complex conjugate pair, depending on the discriminant.
Eigenvectors are the directions a matrix scales without rotating; everything else gets sent off its line.
Eigenvalue equation & characteristic polynomialAx = λx, x ≠ 0p(λ) = det(λI − A) (2×2 shortcut: p(λ) = λ2 − (trace A)λ + det A)
Eigenvectors — finding them
Once an eigenvalue λ is known, the corresponding eigenvector solves (λI − A)x = 0. The two rows always reduce to one relation between x and y (because λI − A is singular at an eigenvalue). Pick a non-zero value for x (typically 1) and read off y. Any non-zero scalar multiple of your answer is also a valid eigenvector for the same λ.
🧠Recipe — finding eigenvalues and eigenvectors of a 2×2 matrix
Form λI − A: diagonal becomes λ − a, λ − d; off-diagonal entries flip sign.
Take det with ad − bc — this is p(λ).
Solve p(λ) = 0; check sum of eigenvalues equals trace.
For each λ: sub into (λI − A)x = 0 — reduce to one linear relation.
Pick a non-zero value (usually x = 1) and read off the eigenvector.
Worked examples
WE 1
Find the characteristic polynomial
Find the characteristic polynomial of M = ((2, 1), (4, 5)).
form λI − MλI − M = ((λ − 2, −1), (−4, λ − 5))take the determinant: ad − bcdet = (λ − 2)(λ − 5) − (−1)(−4)= λ² − 7λ + 10 − 4p(λ) = λ² − 7λ + 6shortcut check: trace 7, det 6 ⇒ same polynomial ✓
WE 2
Verify an eigenvector by multiplication
Verify that v = (1, 1)T is an eigenvector of A = ((3, 2), (1, 4)) and find the corresponding eigenvalue.
compute AvAv = ((3)(1) + (2)(1), (1)(1) + (4)(1))T= (5, 5)Tis Av a scalar multiple of v?(5, 5)T = 5(1, 1)T = 5vv is an eigenvector with λ = 5just one matrix-vector product needed — ratio Av:v gives λ.
WE 3
Full process — two real distinct eigenvalues
Find the eigenvalues and an eigenvector for each, for A = ((6, −1), (2, 3)).
