Quick answer

After finding p(λ), solve p(λ) = 0 to obtain eigenvalues (matrix roots).

Formula

  • p(λ) = det(A − λI)
  • p(λ) = 0

Introduction

Students sometimes expand det(A − λI) correctly but lose points solving p(λ) = 0. This article separates polynomial construction from root finding. Use the Characteristic Polynomial Calculator for p(λ), then focus on the equation here.

The phrase characteristic equation calculator on this site refers to that workflow: get p(λ) from the matrix tool, solve the equation by hand or with course-approved solvers.

Keep polynomial and equation language distinct in your notes.

Characteristic equation setup

Setup begins with p(λ) = det(A − λI) expanded in standard form. The characteristic equation is the statement p(λ) = 0.

For degree 2, the equation is often a quadratic. For degree 3, factoring or cubic techniques appear. For degree 4, numerical methods become common in applications courses.

Matrix roots mean eigenvalues. They need not be integers or even real numbers.

Algebraic simplification before solving helps: factor common powers of λ, divide by a gcd of coefficients when appropriate, and check for obvious rational roots on integer polynomials, as practiced in characteristic polynomial examples.

Numerical methods belong after exact methods fail or when coefficients are messy decimals from data. Your syllabus will state what is allowed.

If p(λ) itself is uncertain, return to the expansion checklist before solving anything.

Equation forms you will see

  • det(A − λI) = 0
  • p(λ) = 0 with p expanded
  • Factored form Π(λ − λ<sub>i</sub>)<sup>m<sub>i</sub></sup> = 0

Factored form displays eigenvalues directly when factorization is available.

Expanded form is better for coefficient-based root tests.

Step-by-step overview

  1. Obtain p(λ). Expansion by hand or calculator.
  2. Write p(λ) = 0. This is the characteristic equation.
  3. Simplify. Factor common terms if possible.
  4. Solve. Quadratic formula, factoring, or numerical tools as allowed.
  5. List eigenvalues with multiplicity. Include repeated roots from squared factors.

Solving after expansion

p(λ) = λ² − 5λ + 6 = 0 factors to (λ − 2)(λ − 3) = 0, so eigenvalues are 2 and 3.

p(λ) = λ² + 1 = 0 gives λ = ±i.

A cubic with a visible rational root can be reduced by synthetic division after one root is found.

Always substitute roots back into p(λ) lightly to confirm zeros before eigenvector work.