User:Hjilderda
Jump to navigation
Jump to search
Theorem
Let $V_n$ be the Vandermonde matrix of order $n$ given by:
- $V_n = \begin{bmatrix}
x_1 & x_2 & \cdots & x_n \\ x_1^2 & x_2^2 & \cdots & x_n^2 \\
\vdots & \vdots & \ddots & \vdots \\
x_1^n & x_2^n & \cdots & x_n^n
\end{bmatrix}$
Then its inverse $V_n^{-1} = \sqbrk b_n$ can be specified as:
- $b_{ij} = \dfrac {\ds \sum_{\stackrel {1 \le k_1 < \ldots < k_{n - j} \le n} {k_1, \ldots, k_{n - j} \ne i} } \paren {-1}^{j - 1} x_{k_1} \ldots x_{k_{n - j} } } {\ds x_i \prod_{\stackrel {1 \le k \le n} {k \ne i} } \paren {x_k - x_i} }$