Quick answer

p(λ) = det(A − λI). For 2×2 matrices, p(λ) = λ² − tr(A)λ + det(A) when using the common monic form with leading coefficient 1 on λ².

Formula

  • p(λ) = det(A − λI)
  • 2×2: p(λ) = λ² − (a+d)λ + (ad−bc)
  • p(0) = det(A)

Introduction

The characteristic polynomial formula is the bridge between matrix notation and eigenvalue algebra. Every identity in this article comes from rewriting eigenvalue conditions as det(A − λI) = 0 and expanding the determinant. Use the Characteristic Polynomial Calculator while you practice the identities below.

Courses introduce the formula immediately after determinants because it reuses cofactor expansion with a parameter λ on the diagonal.

If the definition felt abstract, read the vocabulary first, then return here for compact formulas you can apply on timed exams.

Determinant formula and identity matrix relationship

The core formula is p(λ) = det(A − λI). The identity matrix I has ones on the diagonal, so subtracting λI subtracts λ from each diagonal entry of A.

Polynomial expansion means treating λ as an indeterminate while cofactor-multiplying entries of A − λI. For 3×3 matrices, expect up to λ³ after full expansion unless zeros simplify cofactors.

Matrix notation (A − λI)ij = aij − λδij makes the diagonal pattern explicit. Off-diagonal positions never pick up λ from this definition.

The definition of the characteristic polynomial explains why this determinant encodes eigenvalues; this article focuses on manipulating the formula efficiently.

Coefficient patterns: constant term is det(A). For det(A − λI) with leading term (−1)nλn, the λn−1 coefficient involves trace. These checks catch sign errors before you factor.

Formula explanation for instructors: p(λ) is not a matrix. It is a scalar polynomial whose coefficients are functions of the entries of A.

Expanded formula lines

  • p(λ) = det(A − λI)
  • det(λI − A) = (−1)^n det(A − λI)
  • 2×2 A = [[a,b],[c,d]]: p(λ) = λ² − (a+d)λ + (ad−bc)
  • Characteristic equation: p(λ) = 0

Memorize the 2×2 shortcut after you derive it once from cofactors. Derivation anchors the signs better than memorizing alone.

For 3×3, there is no universally short shortcut beyond structure (triangular, diagonal, sparse zeros). Plan cofactor tables.

When a problem asks for eigenvalues only, you may factor p(λ) after expansion. The worked characteristic polynomial examples show factoring patterns that appear on exams.

After expansion, compare coefficients with the home calculator before you solve p(λ) = 0.

Step-by-step overview

  1. Match size of I to A. λI is n×n when A is n×n. A common error is using the wrong identity size.
  2. Form A − λI entry by entry. Diagonal: aii − λ. Off-diagonal: copy aij.
  3. Choose a cofactor row or column. Pick the line with the most zeros in numeric examples to reduce arithmetic.
  4. Expand and simplify. Multiply cofactors carefully; λ terms accumulate quickly on 4×4.
  5. Verify degree and det(A). Degree n and p(0) = det(A) are quick validation rules.
  6. Hand off to solving if needed. After p(λ) is correct, solve p(λ) = 0 only when the problem asks for eigenvalues.

Formula in practice

A = [[1, 2], [0, 3]] gives A − λI = [[1−λ, 2], [0, 3−λ]] and p(λ) = (1−λ)(3−λ) = λ² − 4λ + 3.

Trace is 4 and det(A) is 3, matching the 2×2 shortcut λ² − tr(A)λ + det(A) up to the leading coefficient convention you use.

A = [[0, 1], [−1, 0]] gives p(λ) = λ² + 1. No real roots appear, which is normal for real matrices.

Repeat the expansion on a 3×3 upper triangular matrix to see how diagonal entries produce a product form immediately.