# 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 $A\in Mat_{n}(R)$ 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 $A^{n}=0$ IfAinMnRAndAIsSupertriangularThenAn0 2013-03-22 13:44:39 2013-03-22 13:44:39 Daume (40) Daume (40) 12 Daume (40) Theorem msc 15-00