If $A\in M_n(R)$ and $A$ is supertriangular then $A^n=0$

theorem: Let $R$ be commutative ring with identity. If an n-square matrix $A\in Ma{t}_n(R)$ is supertriangular then $A^n=0$.

proof: Find the characteristic polynomial of $A$ by computing the determinant of $A-tI$. The square matrix $A-tI$ is a triangular matrix. The determinant of a triangular matrix is the product of the diagonal element of the matrix. Therefore the characteristic polynomial is $p(t)=t^n$ and by the Cayley-Hamilton theorem the matrix $A$ satisfies the polynomial. That is $A^n=0$.
QED

Title	If $A\in M_n(R)$ and $A$ is supertriangular then $A^n=0$
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