Eigenvalues, eigenvectors, modal matrix [2,0,1;0,2,−1;…]

A = [2 0 1; 0 2 −1; 0 0 2]

Eigenvalues

A is upper triangular, so |A − λI| = (2 − λ)³ = 0 and λ = 2, 2, 2.

Eigenvectors

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

Rank = 1, so there are 3 − 1 = 2 independent eigenvectors. The equation is z = 0, with x, y free:

X₁ = (1, 0, 0)ᵀ ,   X₂ = (0, 1, 0)ᵀ

(Any k₁X₁ + k₂X₂ ≠ 0 is an eigenvector.)

Modal matrix

A 3 × 3 matrix needs three independent eigenvectors to be diagonalized. Here the algebraic multiplicity of λ = 2 is 3 but the geometric multiplicity is 2, so A is defective: no modal matrix P exists with P⁻¹AP diagonal.

To complete a modal matrix we add a generalised eigenvector X₃ with (A − 2I)X₃ = u, where u is an eigenvector. Take X₃ = (0, 0, 1)ᵀ. Then

(A − 2I)X₃ = (1, −1, 0)ᵀ = u

u = (1, −1, 0)ᵀ is a genuine eigenvector, since it is the combination X₁ − X₂ of the two eigenvectors found above. Take the columns u, X₁ = (1, 0, 0)ᵀ and X₃. The modal matrix is

P = [ 1  1  0]
    [−1  0  0]        |P| = 1 ≠ 0
    [ 0  0  1]

For this P (Au = 2u, AX₁ = 2X₁, AX₃ = 2X₃ + u):

P⁻¹AP = J = [2 0 1]
            [0 2 0]
            [0 0 2]

which is a Jordan form, the nearest to diagonal possible.

Answer: eigenvalue 2 (three times) with eigenvectors (1, 0, 0)ᵀ and (0, 1, 0)ᵀ; the matrix is not diagonalizable, and the modal matrix (with a generalised eigenvector) is P above, giving the Jordan form J.