Inverse of Vandermonde Matrix/Proof 2

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Proof

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 $ Definition of Vandermonde Matrix
$V = \begin{bmatrix} 1 & \cdots & 1 \\ x_1 & \cdots & x_n \\ \vdots & \ddots & \vdots \\ x_1^{n-1} & \cdots & x_n^{n-1} \\ \end{bmatrix} \quad$ Definition of Vandermonde Matrix
$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 $\mathbf {1 \mathop \le i,j \mathop \le n}$:
\(\ds E_{ij}\) \(=\) \(\ds \paren {-1}^{n - j} \map {e_{n - j} } {\set {x_1, \ldots, x_n} \setminus \set {x_i} }\) Definition of Elementary Symmetric Function $\map {e_m} U$
\(\ds \map {p_j} x\) \(=\) \(\ds \prod_{k \mathop = 1, k \mathop \ne j}^n \paren {x - x_k}\)


Lemma 1

\(\ds V_n\) \(=\) \(\ds V D\) Definition of Matrix Product (Conventional)
\(\ds V_n^{-1}\) \(=\) \(\ds D^{-1} V^{-1}\) provided the inverses exist: Inverse of Matrix Product

$\Box$


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.

$\Box$


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:

\(\ds V^{-1}\) \(=\) \(\ds P^{-1} E\) Lemma 2
\(\ds d_{i j}\) \(=\) \(\ds \dfrac {E_{i j} } {\map {p_i} {x_i} }\) Definition of Matrix Product (Conventional)
\(\ds \) \(=\) \(\ds \dfrac {\paren { -1 }^{n - j} e_{n - j} \paren {\set {x_1, \ldots, x_n} \setminus \set {x_i} } } {\map {p_i} {x_i} }\) Definition 1

Then:

\(\text {(4)}: \quad\) \(\ds d_{i j}\) \(=\) \(\ds \paren {-1}^{n - j} \dfrac {\displaystyle \sum_{\substack {1 \mathop \le m_1 \mathop < \cdots \mathop < m_{n - j} \mathop \le n \\ m_1, \ldots, m_{n - j} \mathop \ne i} } x_{m_1} \cdots x_{m_{n - j} } } {\displaystyle \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne i} } \paren {x_i - x_m} }\) Definition of Elementary Symmetric Function $e_m \paren U$ Definition 1, equation for $\map {p_i} x$

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:

\(\ds V_n^{-1}\) \(=\) \(\ds D^{-1} V^{-1}\) Lemma 1
\(\ds b_{ij}\) \(=\) \(\ds \dfrac 1 {x_i} d_{i j}\) Definition of Matrix Product (Conventional)

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

$\blacksquare$


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