# nilpotent matrix

The square matrix^{} $A$ is said to be nilpotent if ${A}^{n}=\underset{\text{n times}}{\underset{\u23df}{AA\mathrm{\cdots}A}}=\mathrm{\U0001d7ce}$ for some positive integer $n$ (here $\mathrm{\U0001d7ce}$
denotes the matrix where every entry is 0).

###### Theorem (Characterization of nilpotent matrices).

A matrix is nilpotent iff its eigenvalues^{} are all 0.

###### Proof.

Let $A$ be a nilpotent matrix^{}. Assume ${A}^{n}=\mathrm{\U0001d7ce}$. Let $\lambda $ be an eigenvalue of $A$.
Then $A\mathbf{x}=\lambda \mathbf{x}$ for some nonzero vector $\mathbf{x}$.
By induction ${\lambda}^{n}\mathbf{x}={A}^{n}\mathbf{x}=0$, so $\lambda =0$.

Conversely, suppose that all eigenvalues of $A$ are zero. Then the chararacteristic
polynomial of $A$: $det(\lambda I-A)={\lambda}^{n}$. It now follows from the
Cayley-Hamilton theorem^{} that ${A}^{n}=\mathrm{\U0001d7ce}$.
∎

Since the determinant^{} is the product of the eigenvalues it follows that a nilpotent matrix has determinant 0. Similarly, since the trace of a square matrix is the sum of the eigenvalues, it follows that it has trace 0.

One class of nilpotent matrices are the http://planetmath.org/node/4381strictly triangular matrices^{} (lower or upper), this follows from the fact that the eigenvalues of a triangular matrix are the diagonal elements, and thus are all zero in the case of *strictly* triangular matrices.

Note for $2\times 2$ matrices $A$ the theorem implies that $A$ is nilpotent iff $A=\mathrm{\U0001d7ce}$ or ${A}^{2}=\mathrm{\U0001d7ce}$.

Also it is worth noticing that any matrix that is similar to a nilpotent matrix is nilpotent.

Title | nilpotent matrix |
---|---|

Canonical name | NilpotentMatrix |

Date of creation | 2013-03-22 13:05:56 |

Last modified on | 2013-03-22 13:05:56 |

Owner | jgade (861) |

Last modified by | jgade (861) |

Numerical id | 17 |

Author | jgade (861) |

Entry type | Definition |

Classification | msc 15-00 |