in ℝ3

ℒ= (12-18-85-8-311-15-8) p(x) = det( xℐ3 -ℒ ) = 2+5⁢x+4⁢x2+x3 = (x+2)⁢(x+1)2

Eigenvalue λ1=-2

(-141880-5630-111560) → (10-1003-100000)

Eigenvalue λ2=-1

(-131880-5730-111570) → (10-2001-100000)