Skip to main content

Eigenvalues defined via linear transformation

Short answer

A non-zero vector X with A X = λ X is an eigenvector and λ is the eigenvalue. For the given matrix, characteristic equation λ³+λ²−21λ−45=0 gives eigenvalues λ = 5, −3, −3.

Define eigenvalues and eigenvectors in terms of a linear transformation with matrices as operator. Find the eigenvalues of the matrix [−2 2 −3; 2 1 −6; −1 −2 0].

Definition through a linear transformation

A square matrix A of order n acts as an operator (a linear transformation) T on vectors: Y = AX takes the vector X to a new vector Y, usually with a different direction and length.

A non-zero vector X is called an eigenvector (characteristic vector) of A if the transformation only rescales it, i.e. keeps its line unchanged:

AX = λX ,   X ≠ 0

The scalar λ is the corresponding eigenvalue (characteristic root). Equivalently (A − λI)X = 0 has a non-trivial solution, which happens only if

|A − λI| = 0      (characteristic equation)

Example: for A = [2 0; 0 3], the vector (1, 0)ᵀ is mapped to (2, 0)ᵀ = 2(1, 0)ᵀ, so λ = 2 with eigenvector (1, 0)ᵀ: the map stretches that direction without turning it.

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) = −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: λ = 5, −3, −3.

(Check: sum = 5 − 3 − 3 = −1 = trace ✓; product = 5 · 9 = 45 = |A| ✓.)

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Mathematics III chapter 1 questions →