# Inverse of Vandermonde Matrix/Eisinberg Formula

## Theorem

Let:

 $\ds \prod_{k \mathop = 1}^n \paren {x - x_k}$ $=$ $\ds a_n x^n + \sum_{m \mathop = 0}^{n - 1} a_m x^m$ Polynomial expansion in powers of $x$ $\ds$ $=$ $\ds x^n + \sum_{m \mathop = 0}^{n - 1} \paren {-1}^{n - m} \map {e_{n - m} } {x_1, \ldots, x_n} \, x^m$ Viète's Formulas and Definition of Elementary Symmetric Function $\ds W_n$ $=$ $\ds \begin{bmatrix} 1 & x_1 & \cdots & {x_1}^{n - 1} \\ 1 & x_2 & \cdots & {x_2}^{n - 1} \\ \vdots & \vdots & \ddots & \vdots \\ 1 & x_n & \cdots & {x_n}^{n - 1} \\ \end{bmatrix}$ Definition of Vandermonde Matrix of Order $n$

Let $W_n$ have a matrix inverse $W_n^{-1} = \begin {bmatrix} d_{ij} \end {bmatrix}$.

Let $a_n = \map {e_0} {x_1, \ldots, x_n} = 1$.

Then:

 $\text {(1)}: \quad$ $\ds d_{ij}$ $=$ $\ds \dfrac {\ds \sum_{k \mathop = 0}^{n - i} a_{i + k} \, {x_j}^k} {\ds \prod_{m \mathop = 1, m \mathop \ne j }^n \paren {x_j - x_m} }$ for $i, j = 1, \ldots, n$ $\ds$ $=$ $\ds \dfrac {\ds \sum_{k \mathop = 0}^{n - i} \paren {-1}^{n - i - k} \map {e_{n - i - k} } {x_1, \ldots, x_n} \, {x_j}^k} {\ds \prod_{m \mathop = 1, m \mathop \ne j }^n \paren {x_j - x_m} }$ for $i, j = 1, \ldots, n$

## Proof

### Lemma 1

Given values $z_1, \ldots, z_{p + 1}$ and $1 \le m \le p$, then:

 $\text {(2)}: \quad$ $\ds \ds \map {e_m} {z_1, \ldots, z_p, z_{p + 1} }$ $=$ $\ds z_{p + 1} \map {e_{m - 1} } {z_1, \ldots, z_p} + \map {e_m} {z_1, \ldots, z_p}$ Recursion Property of Elementary Symmetric Function

$\Box$

### Lemma 2

Let $X = \set {x_1, \ldots, x_n}$ and $\mathbf u = x_j$ for some $j = 1, \ldots, n$.

Then:

 $\text {(3)}: \quad$ $\ds \ds \sum_{k \mathop = 0}^{n - i} \paren {-1}^k \map {e_{n - i - k} } X \mathbf u^k$ $=$ $\ds \map {e_{n - i} } {X \setminus \set {\mathbf u} }$ Eisinberg (1981)

Proof of Lemma 2:

Let $S$ denote the left hand side of $(3)$.

Let $U = X \setminus \set {\mathbf u}$.

Then:

 $\ds S$ $=$ $\ds \paren {-1}^{n - i} \mathbf u^{n - i} + \sum_{k \mathop = 0}^{n - i - 1} \paren {-1}^k \map {e_{n - i - k} } X {\mathbf u}^k$ splitting off the term for $k = n - i$ $\ds$ $=$ $\ds \paren {-1}^{n - i} \, {\mathbf u}^{n - i} + \sum_{k \mathop = 0}^{n - i - 1} \paren {-1}^k \, \paren {\mathbf u \, \map {e_{n - i - k - 1} } U + \map {e_{n - i - k} } U} \, \mathbf u^k$ by $(2)$ in Lemma 1 with $p = n - 1$ and $m = n - i - k$ $\ds$ $=$ $\ds \sum_{k \mathop = 0}^{n - i - 1} \paren {-1}^k \, \map {e_{n - i - k - 1} } U \mathbf u^{k + 1} + \sum_{k \mathop = 0}^{n - i} \paren {-1}^k \map {e_{n - i - k} } U \, \mathbf u^k$ reassembling summations $\ds$ $=$ $\ds \sum_{k \mathop = 1}^{n - i} \paren {-1}^{k - 1} \, \map {e_{n - i - k} } U \mathbf u^k + \sum_{k \mathop = 0}^{n - i} \paren {-1}^k \map {e_{n - i - k} } U \, \mathbf u^k$ reindexing the first sum $\ds$ $=$ $\ds \sum_{k \mathop = 1}^{n - i} \paren {-1}^k \paren {-\map {e_{n - i - k} } U + \map {e_{n - i - k} } U} \mathbf u^k + \paren {-1}^0 \map {e_{n - i} } U \mathbf u^0$ splitting off the term for $k = 0$ and collecting under one summation $\ds$ $=$ $\ds \map {e_{n-i} } U$

$\Box$

Proof of the Theorem

 $\ds d_{ij}$ $=$ $\ds \dfrac {\ds \paren {-1}^{n - i} \map { e_{n - i} } {X \setminus \set {x_j} } } {\ds \prod_{m \mathop = 1, m \mathop \ne j }^n \paren {x_j - x_m} }$ transposing $W_n$, then applying corollary to Inverse of Vandermonde Matrix $\ds$ $=$ $\ds \dfrac {\ds \sum_{k \mathop = 0}^{n - i} \paren {-1}^k \, \map {e_{n - i - k} } X \, x_j^k} {\ds \prod_{m \mathop = 1, m \mathop \ne j }^n \paren {x_j - x_m} }$ $(3)$ in Lemma 2 $\ds$ $=$ $\ds \dfrac {\ds \sum_{k \mathop = 0}^{n - i} a_{i + k} \, x_j^k} {\ds \prod_{m \mathop = 1, m \mathop \ne j}^n \paren {x_j - x_m} }$ Viète's Formulas

$\blacksquare$