# proof of determinant of the Vandermonde matrix

To begin, note that the determinant^{} of the $n\times n$ Vandermonde
matrix^{} (which we shall denote as ‘$\mathrm{\Delta}$’) is a homogeneous
polynomial^{} of order $n(n-1)/2$ because every term in the determinant
is, up to sign, the product of a zeroth power of some variable times the first
power of some other variable , $\mathrm{\dots}$, the $n-1$-st power of some
variable and $0+1+\mathrm{\cdots}+(n-1)=n(n-1)/2$.

Next, note that if ${a}_{i}={a}_{j}$ with $i\ne j$, then $\mathrm{\Delta}=0$
because two columns of the matrix would be equal. Since $\mathrm{\Delta}$ is a
polynomial^{}, this implies that ${a}_{i}-{a}_{j}$ is a factor of $\mathrm{\Delta}$.
Hence,

$$ |

where C is some polynomial. However, since both $\mathrm{\Delta}$ and the product on the right hand side have the same degree, $C$ must have degree zero, i.e. $C$ must be a constant. So all that remains is the determine the value of this constant.

One way to determine this constant is to look at the coefficient of the leading diagonal, ${\prod}_{n}{({a}_{n})}^{n-1}$. Since it equals 1 in both the determinant and the product, we conclude that $C=1$, hence

$$ |

Title | proof of determinant of the Vandermonde matrix |
---|---|

Canonical name | ProofOfDeterminantOfTheVandermondeMatrix |

Date of creation | 2013-03-22 15:44:50 |

Last modified on | 2013-03-22 15:44:50 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 10 |

Author | rspuzio (6075) |

Entry type | Proof |

Classification | msc 15A57 |

Classification | msc 65F99 |

Classification | msc 65T50 |