Eigenvalues and eigenvectors [−2,2,−3;2,1,−6;…]

Two properties of eigenvalues

1. The sum of the eigenvalues of a square matrix equals its trace, and their product equals its determinant: Σλ = a₁₁ + a₂₂ + …, Πλ = |A|.
2. If λ is an eigenvalue of A, then λᵏ is an eigenvalue of Aᵏ, and (for non-singular A) 1/λ is an eigenvalue of A⁻¹. A and Aᵀ have the same eigenvalues.

(Other valid properties: the eigenvalues of a triangular matrix are its diagonal entries; eigenvalues of a Hermitian matrix are real; eigenvectors of distinct eigenvalues are independent.)

Eigenvalues of A = [−2 2 −3; 2 1 −6; −1 −2 0]

• Trace = −2 + 1 + 0 = −1
• Sum of principal minors = (0 − 12) + (0 − 3) + (−2 − 4) = −12 − 3 − 6 = −21
• |A| = −2(0 − 12) − 2(0 − 6) + (−3)(−4 + 1) = 24 + 12 + 9 = 45

λ³ + λ² − 21λ − 45 = 0   ⇒   (λ − 5)(λ + 3)² = 0

Eigenvalues: 5, −3, −3.

Eigenvectors

λ = 5: A − 5I = [−7 2 −3; 2 −4 −6; −1 −2 −5]. From R₂: x = 2y + 3z; R₃: −x − 2y − 5z = 0 ⇒ −(2y + 3z) − 2y − 5z = 0 ⇒ y = −2z. Take z = −1: y = 2, x = 1. X₁ = (1, 2, −1)ᵀ. Check: AX₁ = (5, 10, −5) = 5X₁ ✓.

λ = −3: A + 3I = [1 2 −3; 2 4 −6; −1 −2 3]. All rows give x + 2y − 3z = 0 (rank 1), so there are two independent solutions:

• y = 1, z = 0: X₂ = (−2, 1, 0)ᵀ
• y = 0, z = 1: X₃ = (3, 0, 1)ᵀ

Answer: