# 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 ${A}^{n}=0$ |
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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 |