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Cayley-Hamilton theorem verification [1,3,7;4,2,3;…]

Short answer

Cayley-Hamilton theorem: every square matrix satisfies its own characteristic equation. For A, the characteristic polynomial is λ³ − 4λ² − 20λ − 35, and A³ − 4A² − 20A − 35I = O, verifying the theorem.

State Cayley-Hamilton theorem and verify the theorem for the square matrix A = [1 3 7; 4 2 3; 1 2 1].

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 3 7; 4 2 3; 1 2 1]

Characteristic equation

  • Trace = 1 + 2 + 1 = 4
  • Sum of principal minors = (2·1 − 3·2) + (1·1 − 7·1) + (1·2 − 3·4) = −4 − 6 − 10 = −20
  • |A| = 1(2 − 6) − 3(4 − 3) + 7(8 − 2) = −4 − 3 + 42 = 35
λ³ − 4λ² − 20λ − 35 = 0

Verification: A³ − 4A² − 20A − 35I = O

A² = [20 23 23]        A³ = A²·A = [135 152 232]
     [15 22 37]                     [140 163 208]
     [10  9 14]                     [ 60  76 111]

Entry by entry:

Row 1: 135 − 80 − 20 − 35 = 0 ;  152 − 92 − 60 = 0 ;  232 − 92 − 140 = 0
Row 2: 140 − 60 − 80 = 0 ;  163 − 88 − 40 − 35 = 0 ;  208 − 148 − 60 = 0
Row 3:  60 − 40 − 20 = 0 ;   76 − 36 − 40 = 0 ;  111 − 56 − 20 − 35 = 0

All entries are zero, so A³ − 4A² − 20A − 35I = O. The theorem is verified. ∎

✓ Answer checked by Claude Opus 5.5 (high effort)

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