Characteristic equation and verification [1,0,2;0,2,1;…]

The characteristic equation of A is |A − λI| = 0. By the Cayley–Hamilton theorem, A satisfies it.

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

Characteristic equation

|A − λI| = | 1−λ   0    2  |
           |  0   2−λ   1  |
           |  2    0   3−λ |

Expand along the second column (only (2−λ) is non-zero):

= (2 − λ)[(1 − λ)(3 − λ) − 4]
= (2 − λ)(λ² − 4λ − 1)
= −(λ − 2)(λ² − 4λ − 1) = −(λ³ − 6λ² + 7λ + 2)

So the characteristic equation is

λ³ − 6λ² + 7λ + 2 = 0

(Check with the formula λ³ − (trace)λ² + (sum of principal minors)λ − |A| = 0: trace 6, minors 6 − 1 + 2 = 7, |A| = −2.)

Verification: A³ − 6A² + 7A + 2I = O

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

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

Every entry is zero, so A³ − 6A² + 7A + 2I = O, i.e. A satisfies its characteristic equation. ∎