MathTools

Matrix Diagonalization

Write A as P D P⁻¹, where D is diagonal (the eigenvalues) and the columns of P are the eigenvectors. Exact when the eigenvalues are rational.
Diagonalizable
A = P D P⁻¹ is shown below.
Input matrix A3×3

Integers, negatives, decimals or fractions (e.g. 3, -2, 1/2). Empty cells count as 0.

Exact result
All eigenvalues are rational, so P, D and P⁻¹ are exact.

P (eigenvectors)

1
0
0
0
1
0
0
0
1

D (eigenvalues)

0
0
0

P⁻¹

1
0
0
0
1
0
0
0
1

A = P · D · P⁻¹

Step-by-step solution

  1. 1

    Characteristic polynomial

    Compute p(λ) = det(λI − A) with the Faddeev-LeVerrier algorithm. Each coefficient is built from traces of exactly-computed matrix products, so no rounding occurs.

    p(λ)=λ3p(\lambda) = \lambda^{3}
    • p(λ) = λ^3
  2. 2

    Search for rational roots

    Apply the Rational Root Theorem: every rational root p/q (in lowest terms) has p | constant term and q | leading coefficient. Roots found: 0 (×3).

    • λ = 0, algebraic multiplicity 3
  3. 3

    Eigenspace of λ = 0

    Solve (A − 0 I)v = 0 by Gaussian elimination. Algebraic multiplicity 3, geometric multiplicity 3. Basis: (1, 0, 0), (0, 1, 0), (0, 0, 1).

    (A−λI) v=0(A - \lambda I)\,v = 0
  4. 4

    Summary

    A is diagonalizable: its eigenvectors form a basis of ℝ^3, so A = P D P⁻¹.