Inverse of Vandermonde Matrix

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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:

$b_{i j} = \begin{cases} \paren {-1}^{j - 1} \paren {\dfrac {\displaystyle \sum_{\substack {1 \mathop \le m_1 \mathop < \ldots \mathop < m_{n - j} \mathop \le n \\ m_1, \ldots, m_{n - j} \mathop \ne i} } x_{m_1} \cdots x_{m_{n - j} } } {x_i \displaystyle \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne i} } \paren {x_m - x_i} } } & : 1 \le j < n \\ \qquad \qquad \qquad \dfrac 1 {x_i \displaystyle \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne i} } \paren {x_i - x_m} } & : j = n \end{cases}$


Corollary

Define for variables $\set {y_1,\ldots, y_k}$ elementary symmetric functions:

\(\displaystyle \map {e_m} {\set {y_1, \ldots, y_k} }\) \(=\) \(\displaystyle \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \mathop \cdots \mathop < j_m \mathop \le k } y_{j_1} y_{j_2} \cdots y_{j_m}\) for $m = 0, 1, \ldots, k$

Let $\set {x_1, \ldots, x_n}$ be a set of distinct values.

Let $W_n$ and $V_n$ be Vandermonde matrices of order $n$:

$W_n = \begin{bmatrix} 1 & x_1 & \cdots & x_1^{n-1} \\ 1 & x_2 & \cdots & x_2^{n-1} \\ \vdots & \vdots & \ddots & \vdots \\ 1 & x_1^{n-1} & \cdots & x_n^{n-1} \\ \end{bmatrix}, \quad V_n = \begin{bmatrix} x_1 & x_2 & \cdots & x_n \\ x_1^2 & x_2^2 & \cdots & x_n^2 \\ \vdots & \vdots & \ddots & \vdots \\ x_1^n & x_2^n & \cdots & x_n^n \\ \end{bmatrix}$

Let their matrix inverses be written as $W_n^{-1} = \begin{bmatrix} b_{ij} \end{bmatrix}$ $V_n^{-1} = \begin{bmatrix} c_{ij} \end{bmatrix}$.


Then:

\(\displaystyle b_{ij}\) \(=\) \(\displaystyle \dfrac {\paren {-1}^{n - i} \map {e_{n - i} } {\set {x_1, \ldots, x_n} \setminus \set {x_j} } } {\prod_{m \mathop = 1, m \mathop \ne j }^n \paren {x_j - x_m} }\) for $i, j = 1, \ldots, n$
\(\displaystyle c_{ij}\) \(=\) \(\displaystyle \dfrac 1 {x_i} \, b_{j i}\) for $i, j = 1, \ldots, n$


Proof 1

First consider the classical form of the Vandermonde matrix:

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

By Vandermonde Determinant, the determinant of $W_n$ is:

$\displaystyle \det \left({W_n}\right) = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \left({x_i - x_j}\right) \ne 0$

Since this is non-zero, by Matrix is Invertible iff Determinant has Multiplicative Inverse, the inverse matrix, denoted $B = \left[{b_{ij}}\right]$, is guaranteed to exist.

Using the definition of the matrix product and the inverse:

$\displaystyle \sum_{k \mathop = 1}^n b_{kj} x_i^{k-1} = \delta_{ij}$

That is, if $P_j \left({x}\right)$ is the polynomial:

$\displaystyle P_j \left({x}\right) := \sum_{k \mathop = 1}^n b_{kj}x^{k-1}$

then:

$P_j \left({x_1}\right) = 0, \ldots, P_j \left({x_{j-1}}\right) = 0, P_j \left({x_j}\right) = 1, P_j \left({x_{j+1}}\right) = 0, \ldots, P_j \left({x_n}\right) = 0$


By the Lagrange Interpolation Formula, the $j$th row of $B$ is composed of the coefficients of the $j$th Lagrange basis polynomial:

$\displaystyle P_j \left({x}\right) = \sum_{k \mathop = 1}^n b_{kj} x^{k-1} = \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne j}} \frac {x - x_m} {x_j - x_m}$


Identifying the $k$th order coefficient in these two polynomials yields:

$b_{k j} = \begin{cases} \left({-1}\right)^{n - k} \left({\dfrac{\displaystyle \sum_{\substack{1 \mathop \le m_1 \mathop < \ldots \mathop < m_{n-k} \mathop \le n \\ m_1, \ldots, m_{n-k} \mathop \ne j} } x_{m_1} \cdots x_{m_{n-k}} } {\displaystyle \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne j} } \left({x_j - x_m}\right)}}\right) & : 1 \le k < n \\ \qquad \qquad \qquad \dfrac 1 {\displaystyle \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne j} } \left({x_j - x_m}\right)} & : k = n \end{cases}$

which gives:

$b_{kj} = \begin{cases} \left({-1}\right)^{k - 1} \left({\dfrac{\displaystyle \sum_{\substack{1 \mathop \le m_1 \mathop < \ldots \mathop < m_{n-k} \mathop \le n \\ m_1, \ldots, m_{n-k} \mathop \ne j} } x_{m_1} \cdots x_{m_{n-k}} } {\displaystyle \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne j} } \left({x_m - x_j}\right)}}\right) & : 1 \le k < n \\ \qquad \qquad \qquad \dfrac 1 {\displaystyle \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne j} } \left({x_j - x_m}\right)} & : k = n \end{cases}$


For the general case, we observe that by simple multiplication:

$\displaystyle V_n = \begin{pmatrix} \begin{bmatrix} x_1 & 0 & \cdots & 0 \\ 0 & x_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & x_n \end{bmatrix} \cdot W_n \end{pmatrix}^\intercal$

So by Inverse of Matrix Product and Inverse of Diagonal Matrix:

$\displaystyle V_n^{-1} = \begin{bmatrix} x_1^{-1} & 0 & \cdots & 0 \\ 0 & x_2^{-1} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & x_n^{-1} \end{bmatrix} \cdot \left({W_n^{-1} }\right)^\intercal$


Let $c_{k j}$ denote the $\left({k, j}\right)$th coefficient of $V_n^{-1}$.

Since the first matrix in the product expression for $V_n^{-1}$ above is diagonal:

$c_{kj} = \dfrac 1 {x_k} b_{j k}$

which establishes the result.

$\blacksquare$


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 $ Vandermonde matrices
$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 $\mathbf {1 \mathop \le i,j \mathop \le n}$:
\(\displaystyle E_{ij}\) \(=\) \(\displaystyle \paren { -1 }^{n-j} \, e_{n-j} \paren { \set {x_1,\ldots,x_n} \setminus \set {x_i} }\) Definition:Elementary Symmetric Function $e_m \paren {U}$
\(\displaystyle \displaystyle \map {p_j} x\) \(=\) \(\displaystyle \prod_{ k \mathop = 1,\, k \mathop \neq j }^n \paren { x - x_k }\)

Lemma 1

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

$\Box$

Lemma 2

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

Proof of Lemma 2

Matrix multiply establishes $EV=P$, provided:

$(1)\quad \sum_{k = 1}^n E_{ik}\, x_j^{k-1} = \begin{cases} 0 & i \neq 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 = 1}^n \paren { -1 }^{n-k} e_{n-k} \paren { \set {x_1,\ldots,x_n} \setminus \set {x_i} } \, x_j^{k-1} = \map {p_i} {x_j} $

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

$(3)\quad \sum_{k = 1}^{n} \paren {-1}^{n-k} e_{n-k} \paren { \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 $EV=P$.

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

Then $\det \paren {P}$ is nonzero.

The Invertible Matrix Theorem implies $P$ has an inverse $P^{-1}$.

Multiply $EV=P$ 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 $d_{ij}$ denote the entries in $V^{-1}$. Then:

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

Then:

\((4):\quad\) \(\displaystyle d_{ij}\) \(=\) \(\displaystyle \paren { -1 }^{n-j} \, \dfrac { \displaystyle \sum_{ \substack { 1 \mathop \le m_1 \mathop \lt \cdots \mathop \lt m_{n-j} \mathop \le n \\ m_1,\ldots,m_{n-j} \mathop \neq i } } x_{m_1} \cdots x_{m_{n-j} } } { \displaystyle \prod_{ \substack { 1 \mathop \le m \mathop \le n \\ m \mathop \neq i } } \paren { x_i - x_m } }\) Definition: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 $b_{ij}$ denote the entries in $V_n^{-1}$. Then:

\(\displaystyle V_n^{-1}\) \(=\) \(\displaystyle D^{-1} V^{-1}\) Lemma 1
\(\displaystyle b_{ij}\) \(=\) \(\displaystyle \dfrac{ 1 }{ x_i} \, d_{ij}\) Matrix multiply

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

$\blacksquare$


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