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

Short answer

The characteristic equation λ³−2λ²−4λ+8=0 yields λ=2 (double) and −2. For λ=−2, eigenvector (0,−1,1)ᵀ; for λ=2, the only independent vector is (2,−1,1)ᵀ, so the matrix is not diagonalizable.

Find the eigenvalues and eigenvectors of the matrix [2 1 1; −2 1 3; 2 1 −1].

A = [2 1 1; −2 1 3; 2 1 −1]

Characteristic equation

  • Trace = 2 + 1 − 1 = 2
  • Sum of principal minors = (−1 − 3) + (−2 − 2) + (2 + 2) = −4
  • |A| = 2(−1 − 3) − 1(2 − 6) + 1(−2 − 2) = −8 + 4 − 4 = −8
λ³ − 2λ² − 4λ + 8 = 0   ⇒   (λ − 2)²(λ + 2) = 0

Eigenvalues: λ = 2, 2, −2.

Eigenvectors

λ = −2: A + 2I = [4 1 1; −2 3 3; 2 1 1]. R₁ − R₃ gives 2x = 0 ⇒ x = 0, and then y + z = 0. X₁ = (0, −1, 1)ᵀ.

λ = 2: A − 2I = [0 1 1; −2 −1 3; 2 1 −3]. R₁: y = −z. R₂: −2x − y + 3z = 0 ⇒ −2x + z + 3z = 0 ⇒ x = 2z. R₃ is then satisfied. Rank = 2, so there is only one independent vector (take z = 1): X₂ = (2, −1, 1)ᵀ.

Answer:

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

The repeated eigenvalue 2 has only one independent eigenvector, so A is not diagonalizable.

✓ Answer checked by Claude Opus 5.5 (high effort)

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