Skip to main content

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

Short answer

Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic equation. For A, the characteristic equation is λ³ − 6λ² + 7λ + 2 = 0. Computing A³ − 6A² + 7A + 2I gives the zero matrix, confirming the theorem.

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

Cayley–Hamilton theorem: every square matrix satisfies its own characteristic equation. If |A − λI| = λ³ + c₁λ² + c₂λ + c₃ = 0, then A³ + c₁A² + c₂A + c₃I = O.

A = [1 0 2; 0 2 1; 2 0 3]

Characteristic equation

  • Trace = 1 + 2 + 3 = 6
  • Sum of principal minors = (2·3 − 1·0) + (1·3 − 2·2) + (1·2 − 0) = 6 − 1 + 2 = 7
  • |A| = 1(6 − 0) − 0 + 2(0 − 4) = 6 − 8 = −2
λ³ − 6λ² + 7λ + 2 = 0

Verification: A³ − 6A² + 7A + 2I = O

A² = [5 0  8]         A³ = A²·A = [21 0 34]
     [2 4  5]                      [12 8 23]
     [8 0 13]                      [34 0 55]

Entry by entry:

Row 1: 21 − 30 + 7 + 2 = 0 ;  0 ;  34 − 48 + 14 = 0
Row 2: 12 − 12 + 0 = 0 ;  8 − 24 + 14 + 2 = 0 ;  23 − 30 + 7 = 0
Row 3: 34 − 48 + 14 = 0 ;  0 ;  55 − 78 + 21 + 2 = 0

All entries vanish, so A³ − 6A² + 7A + 2I = O. The theorem is verified. ∎

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Mathematics III chapter 1 questions →