Eigenvalues and eigenvectors [2,2,1;1,3,1;…]

A = [2 2 1; 1 3 1; 1 2 2]

Characteristic equation

• Trace = 2 + 3 + 2 = 7
• Sum of principal minors = (3·2 − 1·2) + (2·2 − 1·1) + (2·3 − 2·1) = 4 + 3 + 4 = 11
• |A| = 2(6 − 2) − 2(2 − 1) + 1(2 − 3) = 8 − 2 − 1 = 5

λ³ − 7λ² + 11λ − 5 = 0   ⇒   (λ − 1)²(λ − 5) = 0

Eigenvalues: λ = 1, 1, 5.

Eigenvectors

λ = 5: A − 5I = [−3 2 1; 1 −2 1; 1 2 −3]. Adding R₁ and R₂ gives −2x + 2z = 0, so x = z. Putting x = z = 1 in R₂ gives 1 − 2y + 1 = 0, so y = 1. X₃ = (1, 1, 1)ᵀ.

λ = 1 (repeated): A − I = [1 2 1; 1 2 1; 1 2 1]. All rows give the single equation x + 2y + z = 0, so rank = 1 and there are 3 − 1 = 2 independent eigenvectors.

• y = 1, z = 0 ⇒ x = −2: X₁ = (−2, 1, 0)ᵀ
• y = 0, z = 1 ⇒ x = −1: X₂ = (−1, 0, 1)ᵀ

Answer:

Since there are three independent eigenvectors, A is diagonalizable.