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

Eigenvalues are the roots of |A − λI| = 0, and each eigenvector solves (A − λI)X = 0.

A = [2 −2 2; 1 1 1; 1 3 −1]

Characteristic equation

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

λ³ − 2λ² − 4λ + 8 = 0
(λ − 2)²(λ + 2) = 0

Eigenvalues: λ = 2, 2, −2.

Eigenvector for λ = −2

A + 2I = [4 −2  2]
         [1  3  1]
         [1  3  1]

Equations: x + 3y + z = 0 and 4x − 2y + 2z = 0 (i.e. z = y − 2x). Substituting: x + 3y + y − 2x = 0 ⇒ x = 4y, then z = −7y.

X₁ = (4, 1, −7)ᵀ.

Eigenvector for λ = 2

A − 2I = [0 −2  2]
         [1 −1  1]
         [1  3 −3]

From R₁: y = z. From R₂: x − y + z = 0 ⇒ x = 0. R₃ is then satisfied (3y − 3z = 0). The rank of A − 2I is 2, so only one independent vector exists:

X₂ = (0, 1, 1)ᵀ.

Answer:

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