# Inverse of Vandermonde Matrix

## 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:

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

### Corollary

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

 $\ds \map {e_m} {\set {y_1, \ldots, y_k} }$ $=$ $\ds \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \mathop \cdots \mathop < j_m \mathop \le k } y_{j_1} y_{j_2} \cdots y_{j_m}$ 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:

 $\ds b_{ij}$ $=$ $\ds \dfrac {\paren {-1}^{n - i} \map {e_{n - i} } {\set {x_1, \ldots, x_n} \setminus \set {x_j} } } {\prod_{m \mathop = 1, m \mathop \ne j }^n \paren {x_j - x_m} }$ for $i, j = 1, \ldots, n$ $\ds c_{ij}$ $=$ $\ds \dfrac 1 {x_i} \, b_{j i}$ 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 Value of Vandermonde Determinant, the determinant of $W_n$ is:

$\ds \map \det {W_n} = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_i - x_j} \ne 0$

Since this is non-zero, by Matrix is Invertible iff Determinant has Multiplicative Inverse, the inverse matrix, denoted $B = \sqbrk {b_{i j} }$, is guaranteed to exist.

Using the definition of the matrix product and the inverse:

$\ds \sum_{k \mathop = 1}^n b_{k j} x_i^{k - 1} = \delta_{i j}$

That is, if $\map {P_j} x$ is the polynomial:

$\ds \map {P_j} x := \sum_{k \mathop = 1}^n b_{k j}x^{k - 1}$

then:

$\map {P_j} {x_1} = 0, \ldots, \map {P_j} {x_{j - 1} } = 0, \map {P_j} {x_j} = 1, \map {P_j} {x_{j + 1} } = 0, \ldots, \map {P_j} {x_n} = 0$

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

$\ds \map {P_j} x = \sum_{k \mathop = 1}^n b_{k j} x^{k - 1} = \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne j} } \frac {x - x_m} {x_j - x_m}$

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

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

which gives:

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

For the general case, we observe that by multiplication:

$\ds V_n = \begin {pmatrix} \begin {bmatrix} x_1 & 0 & \cdots & 0 \\ 0 & x_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & x_n \end {bmatrix} \cdot W_n \end {pmatrix}^\intercal$
$\ds V_n^{-1} = \begin {bmatrix} x_1^{-1} & 0 & \cdots & 0 \\ 0 & x_2^{-1} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & x_n^{-1} \end {bmatrix} \cdot \paren {W_n^{-1} }^\intercal$

Let $c_{k j}$ denote the $\tuple {k, j}$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

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

 $\text {(4)}: \quad$ $\ds d_{i j}$ $=$ $\ds \paren {-1}^{n - j} \dfrac {\ds \sum_{\substack {1 \mathop \le m_1 \mathop < \cdots \mathop < m_{n - j} \mathop \le n \\ m_1, \ldots, m_{n - j} \mathop \ne i} } x_{m_1} \cdots x_{m_{n - j} } } {\ds \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne i} } \paren {x_i - x_m} }$ 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).

$\blacksquare$