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Mathematics-Online course: Linear Algebra - Matrices - Determinants

Vandermonde Determinant

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The determinant of monomes $ 1,x,\ldots x^{n-1}$ evaluated at $ n$ points $ x_i$ equals

$\displaystyle \left\vert\begin{array}{cccc} 1 & x_1 & \ldots & x_1^{n-1} \\ ... ... & \ldots & x_n^{n-1} \end{array}\right\vert = \prod_{i>j} (x_i-x_j)\, . $

(Authors: Burkhardt/Höllig/Hörner)
For $ n =2$ the determinant of the Vandermonde matrix is

$\displaystyle \operatorname{det}\left(\begin{array}{rr}1 & x_1\\ 1 & x_2\end{array} \right) = x_2-x_1\,. $

For arbitrary $ n$ subtract the first row from the other rows:

$\displaystyle \left\vert\begin{array}{cccc} 1 & x_1 & \ldots & x_1^{n-1} \\ ... ...ts \\ 0 & x_n-x_1 & \ldots & x_n^{n-1}-x_1^{n-1} \end{array}\right\vert\,. $

Expansion along the first column yields:

$\displaystyle \left\vert\begin{array}{ccc} x_2-x_1 & \ldots & x_2^{n-1}-x_1^{n... ...\vdots \\ x_n-x_1 & \ldots & x_n^{n-1}-x_1^{n-1} \end{array}\right\vert\,. $

Now we can factor out $ x_k-x_1$ from the $ k$-th row:

$\displaystyle \left(\prod\limits_{k=2}^n (x_k-x_1)\right) \left\vert\begin{arr... ... &\ldots & \sum\limits_{i=0}^{n-2}x_n^{n-2-i}x_1^i \end{array}\right\vert\,. $

Starting at the second column we subtract the $ x_1$-fold of the previous column and repeat this operation on the following columns. So we obtain

$\displaystyle \left(\prod\limits_{k=2}^n (x_k-x_1)\right) \left\vert\begin{arr... ... & \vdots & & \vdots \\ 1 & x_n &\ldots & x_n^{n-2} \end{array}\right\vert $

and we can prove the formula by induction.

(Authors: Burkhardt/Höllig/Hörner)
  automatically generated 4/21/2005