Modal matrix of a matrix [2,1,1;−2,1,3;…]

A = [2 1 1; −2 1 3; 2 1 −1]

Eigenvalues

• 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

λ = −2, 2, 2.

Eigenvectors

λ = −2: A + 2I = [4 1 1; −2 3 3; 2 1 1]. From R₁ and R₃: 2x = 0 ⇒ x = 0, and 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 ⇒ x = 2z. Rank = 2, so only one independent vector: X₂ = (2, −1, 1)ᵀ.

The modal matrix

The repeated eigenvalue 2 has only one independent eigenvector, so A is defective and cannot be reduced to a diagonal matrix. The modal matrix is then built with a generalised eigenvector X₃ satisfying (A − 2I)X₃ = (an eigenvector for λ = 2).

Use the eigenvector 2X₂ = (4, −2, 2)ᵀ. Solve (A − 2I)X₃ = (4, −2, 2)ᵀ: y + z = 4 and −2x − y + 3z = −2. Taking z = 1 gives y = 3, x = 1:

X₃ = (1, 3, 1)ᵀ. Check: (A − 2I)X₃ = (4, −2, 2)ᵀ ✓.

P = [ 0  4  1]
    [−1 −2  3]        |P| = 16 ≠ 0
    [ 1  2  1]

(columns: X₁, 2X₂, X₃). Then

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

Answer: the modal matrix is P above; it gives the Jordan form J, not a purely diagonal form, because A has only two independent eigenvectors.