# determinant of anti-diagonal matrix

Let $A=\mathrm{adiag}({a}_{1},\mathrm{\dots},{a}_{n})$ be an anti-diagonal matrix. Using the sum over all permutations^{} formula for the determinant^{} of a matrix and since all but possibly the anti-diagonal elements are null we get directly at the result

$$\mathrm{det}A=\mathrm{sgn}(n,n-1,\mathrm{\dots},1)\prod _{i=1}^{n}{a}_{i}$$ |

so all that remains is to calculate the sign of the permutation. This can be done directly.

To bring the last element to the beginning $n-1$ permutations are needed so

$$\mathrm{sgn}(n,n-1,\mathrm{\dots},1)={(-1)}^{n-1}\mathrm{sgn}(1,n,n-1,\mathrm{\cdots},2)$$ |

Now bring the last element to the second position. To do this $n-2$ permutations are needed. Repeat this procedure $n-1$ times to get the permutation $(1,\mathrm{\dots},n)$ which has positive sign.

Summing every permutation, it takes

$$\sum _{k=1}^{n-1}k=\frac{n(n-1)}{2}$$ |

permutations to get to the desired permutation.

So we get the final result that

$$\mathrm{det}\mathrm{adiag}({a}_{1},\mathrm{\dots},{a}_{n})={(-1)}^{\frac{n(n-1)}{2}}\prod _{i=1}^{n}{a}_{i}$$ |

Notice that the sign is positive if either $n$ or $n-1$ is a multiple of $4$ and negative otherwise.

Title | determinant of anti-diagonal matrix |
---|---|

Canonical name | DeterminantOfAntidiagonalMatrix |

Date of creation | 2013-03-22 15:50:25 |

Last modified on | 2013-03-22 15:50:25 |

Owner | cvalente (11260) |

Last modified by | cvalente (11260) |

Numerical id | 6 |

Author | cvalente (11260) |

Entry type | Result |

Classification | msc 15-00 |