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Eigenvalues and eigenvectors [3,−4,4;1,−2,4;…]

Short answer

Eigenvalues: −1, 2, 3. Egenvectors: for −1, k(1,1,0)^T; for 2, k(0,1,1)^T; for 3, k(1,1,1)^T.

Find the eigenvalues and eigenvectors of A = [3 −4 4; 1 −2 4; 1 −1 3].

A = [3 −4 4; 1 −2 4; 1 −1 3]

Characteristic equation

  • Trace = 3 − 2 + 3 = 4
  • Sum of principal minors = (−6 + 4) + (9 − 4) + (−6 + 4) = −2 + 5 − 2 = 1
  • |A| = 3(−6 + 4) + 4(3 − 4) + 4(−1 + 2) = −6 − 4 + 4 = −6
λ³ − 4λ² + λ + 6 = 0
(λ + 1)(λ − 2)(λ − 3) = 0

Eigenvalues: λ = −1, 2, 3.

Eigenvectors

λ = −1: A + I = [4 −4 4; 1 −1 4; 1 −1 4]. Equations x − y + z = 0 and x − y + 4z = 0 ⇒ z = 0, x = y. X₁ = (1, 1, 0)ᵀ.

λ = 2: A − 2I = [1 −4 4; 1 −4 4; 1 −1 1]. Equations x − 4y + 4z = 0 and x − y + z = 0. Subtract: −3y + 3z = 0 ⇒ y = z, and then x = 0. X₂ = (0, 1, 1)ᵀ.

λ = 3: A − 3I = [0 −4 4; 1 −5 4; 1 −1 0]. R₁: y = z; R₃: x = y. X₃ = (1, 1, 1)ᵀ.

Answer:

EigenvalueEigenvector
−1k(1, 1, 0)ᵀ
2k(0, 1, 1)ᵀ
3k(1, 1, 1)ᵀ

(Check: sum of eigenvalues −1 + 2 + 3 = 4 = trace; product −6 = |A|.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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