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

Eigenvalues are the roots of |A − λI| = 0; the eigenvector for λ solves (A − λI)X = 0.

A = [2 0 1; 0 2 −1; 0 0 2]

Eigenvalues

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

|A − λI| = (2 − λ)³ = 0   ⇒   λ = 2, 2, 2

So λ = 2 has algebraic multiplicity 3.

Eigenvectors for λ = 2

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

Rank of (A − 2I) = 1. The equations reduce to z = 0, with x and y free. Number of independent vectors = 3 − 1 = 2.

Taking x = 1, y = 0 and x = 0, y = 1:

X₁ = (1, 0, 0)ᵀ ,   X₂ = (0, 1, 0)ᵀ

The general eigenvector is X = (k₁, k₂, 0)ᵀ, with k₁, k₂ not both zero.

Remark

The algebraic multiplicity (3) exceeds the geometric multiplicity (2), so there are only two independent eigenvectors and A is not diagonalizable.

Answer: eigenvalue 2 (repeated three times); eigenvectors (1, 0, 0)ᵀ and (0, 1, 0)ᵀ and their non-zero combinations.