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Cayley-Hamilton theorem verification [1,0,−1;1,2,1;…]

Short answer

Cayley–Hamilton theorem: every square matrix satisfies its own characteristic equation. For A, the characteristic equation is λ³−6λ²+11λ−6=0, and A³−6A²+11A−6I = O, verifying the theorem.

State Cayley-Hamilton theorem and verify it for the matrix A = [1 0 −1; 1 2 1; 2 2 3].

Cayley–Hamilton theorem: every square matrix satisfies its own characteristic equation. If |A − λI| = λⁿ + c₁λⁿ⁻¹ + … + cₙ = 0, then Aⁿ + c₁Aⁿ⁻¹ + … + cₙI = O.

Characteristic equation

A = [1 0 −1; 1 2 1; 2 2 3]

  • Sum of diagonal elements (trace) = 1 + 2 + 3 = 6
  • Sum of principal 2 × 2 minors = (2·3 − 1·2) + (1·3 − (−1)·2) + (1·2 − 0·1) = 4 + 5 + 2 = 11
  • |A| = 1(6 − 2) − 0 + (−1)(2 − 4) = 4 + 2 = 6
λ³ − 6λ² + 11λ − 6 = 0

Verification: show A³ − 6A² + 11A − 6I = O

A² = [−1 −2 −4]       A³ = A²·A = [−11 −12 −13]
     [ 5  6  4]                    [ 19  20  13]
     [10 10  9]                    [ 38  38  27]

Now compute:

A³ − 6A² + 11A − 6I
 Row 1: (−11+6+11−6, −12+12+0−0, −13+24−11−0) = (0, 0, 0)
 Row 2: (19−30+11−0, 20−36+22−6, 13−24+11−0)   = (0, 0, 0)
 Row 3: (38−60+22−0, 38−60+22−0, 27−54+33−6)   = (0, 0, 0)

Every entry is zero, so A³ − 6A² + 11A − 6I = O.

Hence the Cayley–Hamilton theorem is verified for A. (The eigenvalues are 1, 2, 3.) ∎

✓ Answer checked by Claude Opus 5.5 (high effort)

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