If AMn(R) and A is supertriangular then An=0

theorem: Let R be commutative ring with identityPlanetmathPlanetmath. If an n-square matrix AMatn(R) is supertriangular then An=0.

proof: Find the characteristic polynomialMathworldPlanetmathPlanetmath of A by computing the determinantMathworldPlanetmath of A-tI. The square matrixMathworldPlanetmath A-tI is a triangular matrixMathworldPlanetmath. The determinant of a triangular matrix is the product of the diagonal element of the matrix. Therefore the characteristic polynomial is p(t)=tn and by the Cayley-Hamilton theoremMathworldPlanetmath the matrix A satisfies the polynomialMathworldPlanetmathPlanetmathPlanetmath. That is An=0.

Title If AMn(R) and A is supertriangular then An=0
Canonical name IfAinMnRAndAIsSupertriangularThenAn0
Date of creation 2013-03-22 13:44:39
Last modified on 2013-03-22 13:44:39
Owner Daume (40)
Last modified by Daume (40)
Numerical id 12
Author Daume (40)
Entry type Theorem
Classification msc 15-00