Inverse of Vandermonde Matrix/Proof 2

Definition 1

 * $V_n = \begin{bmatrix}

x_1      & \cdots & x_n \\ x_1^2    & \cdots & x_n^2 \\ \vdots   & \ddots & \vdots \\ x_1^{n}  & \cdots & x_n^{n} \\ \end{bmatrix} \quad $


 * $V = \begin{bmatrix}

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


 * $D =

\begin{bmatrix} x_1   & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0     & \cdots & x_n \\ \end{bmatrix}, \quad P = \begin{bmatrix} \map {p_1} {x_1} & \cdots & 0 \\ \vdots          & \ddots & \vdots \\ 0               & \cdots & \map {p_n} {x_n} \\ \end{bmatrix} \quad $ Definition:Diagonal Matrix


 * $E = \begin{bmatrix}

E_{11} & \cdots & E_{1n} \\ \vdots & \ddots & \vdots \\ E_{n1} & \cdots & E_{nn} \\ \end{bmatrix} \quad$ Matrix of symmetric functions


 * where for $\mathbf {1 \mathop \le i, j \mathop \le n}$:

Lemma 2

 * $E V = P$


 * $V^{-1} = P^{-1} E$ provided $\set {x_1, \ldots, x_n}$ is a set of distinct values.


 * $V_n^{-1} = D^{-1} V^{-1}$ provided $\set {x_1, \ldots, x_n}$ is a set of distinct values, all nonzero.

Proof of Lemma 2

Matrix multiply establishes $E V = P$, provided:


 * $(1): \quad \sum_{k \mathop = 1}^n E_{i k} x_j^{k - 1} = \begin{cases} 0 & i \ne j \\ \map {p_i} {x_i} & i = j \end{cases}$

Polynomial $\map {p_i} {x}$ is zero for $x \in \set {x_1,\ldots,x_n} \setminus \set {x_i}$. Then (1) is equivalent to


 * $(2): \quad \sum_{k \mathop = 1}^n \paren {-1}^{n - k} \map {e_{n - k} } {\set {x_1, \ldots, x_n} \setminus \set {x_i} } x_j^{k - 1} = \map {p_i} {x_j}$

Apply Viète's Formulas to degree $n - 1$ monic polynomial $\map {p_i} {\mathbf u}$:


 * $(3): \quad \sum_{k \mathop = 1}^n \paren {-1}^{n - k} \map {e_{n - k} } {set {x_1, \ldots, x_n} \setminus \set {x_i} } {\mathbf u}^{k - 1} = \map {p_i} {\mathbf u}$

Substitute $\mathbf u = x_j$ into (3), proving (2) holds.

Then (2) and (1) hold, proving $E V = P$.

Assume $\set {x_1, \ldots, x_n}$ is a set of distinct values.

Then $\map \det P$ is nonzero.

By Matrix is Invertible iff Determinant has Multiplicative Inverse, $P$ has an inverse $P^{-1}$.

Multiply $E V = P$ by $P^{-1}$, then:


 * $V^{-1} = P^{-1} E\quad$ Left or Right Inverse of Matrix is Inverse

Similarly, $D^{-1}$ exists provided $\set {x_1, \ldots, x_n}$ is a set of nonzero values.

Then $V_n^{-1} = D^{-1} V^{-1}$ by Lemma 1.

Proof of the Theorem

Assume $\set {x_1, \ldots, x_n}$ is a set of distinct values.

Let $d_{i j}$ denote the entries in $V^{-1}$. Then:

Then:

Assume $\set {x_1, \ldots, x_n}$ is a set of distinct values, all nonzero.

Let $b_{i j}$ denote the entries in $V_n^{-1}$.

Then:

Factor $\paren {-1}^{n - 1}$ from the denominator of (4) to agree with Knuth (1997).