Eigenvalues and eigenvectors [1,2,−1;0,2,1;…]

A = [1 2 −1; 0 2 1; 0 0 3]

Eigenvalues

A is upper triangular, so the eigenvalues are the diagonal entries:

|A − λI| = (1 − λ)(2 − λ)(3 − λ) = 0

λ = 1, 2, 3.

Eigenvectors

λ = 1: A − I = [0 2 −1; 0 1 1; 0 0 2]. This gives z = 0, y = 0, x free. X₁ = (1, 0, 0)ᵀ.

λ = 2: A − 2I = [−1 2 −1; 0 0 1; 0 0 1]. This gives z = 0, x = 2y. X₂ = (2, 1, 0)ᵀ.

λ = 3: A − 3I = [−2 2 −1; 0 −1 1; 0 0 0]. This gives y = z and −2x + 2y − z = 0 ⇒ x = y/2. Take y = 2: X₃ = (1, 2, 2)ᵀ.

Answer:

The three eigenvalues are distinct, so the eigenvectors are linearly independent.