User:Tbackstr/Sandbox

Theorem
Let $V$ be Vandermonde's matrix of order $n$ given by:


 * $V = \begin{bmatrix}

1 & 1 & \cdots & 1 \\ x_1 & x_2 & \cdots & x_n \\ x_1^2 & x_2^2 & \cdots & x_n^2 \\ \vdots & \vdots & \ddots & \vdots \\ x_1^{n-1} & x_2^{n-1} & \cdots & x_n^{n-1} \end{bmatrix}$

Then its inverse $V^{-1} = B$ can be specified as:


 * $\displaystyle B_{k,h} = (-1)^{k-1} \frac{\sum_{1\leq m_1 < m_2 < \cdots < m_{N+1-k}\leq N} x_{m_1}x_{m_2}\cdots x_{N+1-k}}{\prod_{\substack{i=1\\i\neq k}}^N (x_i - x_k)}$

The inverse exists if $x_k\neq x_h$ for all $k\neq h$.

Proof
We will proceed in two steps, 1. create a matrix which diagonalizes $V$ when multiplied from the left and 2. scale the diagonal matrix to unity.

Define matrices $A_k(z)$ with coefficients $a_{k,l}$ such that they have the zeros $x_h$, for all $h$ except $h=k$:
 * $\displaystyle A_k(z) := \sum_{h=1}^N a_{k,h} z^{h-1} := \prod_{\substack{h=1\\h\neq k}}^N (x_h - z).$

where the coefficients are defined as
 * $\displaystyle a_{k,h} := (-1)^{k-1} \sum_{1\leq m_1 < m_2 < \cdots < m_{N+1-k}\leq N} x_{m_1}x_{m_2}\cdots x_{N+1-k}.$

It follows the $A_k(x_h)=0$ for all $h\neq k$ and if $A$ is the matrix with coefficients $a_{k,l}$, then
 * $AV = \begin{bmatrix}

A_1(x_1) & 0 & \cdots & 0 \\ 0 & A_2(x_2) & \ddots & \vdots \\ \vdots & \ddots & \ddots & 0 \\ 0 & \cdots & 0 & A_n(x_n) \end{bmatrix}.$ By defining a diagonal matrix $D$ as
 * $D = \begin{bmatrix}

A_1^{-1}(x_1) & 0 & \cdots & 0 \\ 0 & A_2^{-1}(x_2) & \ddots & \vdots \\ \vdots & \ddots & \ddots & 0 \\ 0 & \cdots & 0 & A_n^{-1}(x_n) \end{bmatrix},$ we can define the inverse as
 * $B:=DA.$

Note that the scalar inverses in $D$ exist if $A_k(x_k)\neq 0$, that is, if $x_k\neq x_h$ for all $k\neq h$. Since the determinant of $V$ is
 * $\det(V) = \prod_{1 \le i < j \le n} (x_j - x_i),$

then $D$ is well-defined and the inverse exists only if $\det(V)\neq 0$.

The only thing remaining is to develop explicit expressions for the coefficients of $A_k(z)$.