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
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) = λ^3
- 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
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).
- 4
Summary
A is diagonalizable: its eigenvectors form a basis of ℝ^3, so A = P D P⁻¹.