# Inverse of Vandermonde's Matrix

## Theorem

Let $V_n$ be the Vandermonde's 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 $V_n^{-1} = \left[{b}\right]_n$ can be specified as:

$b_{ij} = {\dfrac{\displaystyle \sum_{\substack {0 \mathop \le m_0 \mathop < \ldots \mathop < m_{n-i} \mathop \le n \\ m_0, \ldots, m_{n-i} \mathop \ne j} } (-1)^{i-1}x_{m_0} \cdots x_{m_{n-i}} } {\displaystyle x_j \prod_{\substack {0 \mathop \le m \mathop \le n \\ m \mathop \ne j} } \left({x_m - x_j}\right)}}$

## Proof

First let us consider the classical form of the Vandermonde matrix:

$W_n = \begin{bmatrix} 1& x_0 & \cdots & x_0^n \\ 1& x_1 & \cdots & x_1^n \\ \vdots & \vdots & \ddots & \vdots \\ 1& x_n & \cdots & x_n^n \\ \end{bmatrix}$

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

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

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, we see that:

$\displaystyle \sum_{k \mathop = 0}^n b_{kj} x_i^k = \delta_{ij}$

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

$\displaystyle P_j \left({x}\right) := \sum_{k \mathop = 0}^n b_{kj}x^k$

then

$P_j \left({x_0}\right) = 0, \ldots, P_j \left({x_{j-1}}\right) = 0, P_j \left({x_j}\right) = 1, P_j \left({x_{j+1}}\right) = 0, \ldots, P_j \left({x_n}\right) = 0$

Now by the Lagrange Interpolation Formula, we deduce that the $j$th row of $B$ is composed of the coefficients of the $j^\text{th}$ Lagrange basis polynomial:

$\displaystyle P_j \left({x}\right) = \sum_{k \mathop = 0}^n b_{kj} x^k = \prod_{\substack {0 \mathop \le m \mathop \le n \\ m \mathop \ne j}} \frac {x - x_m} {x_j - x_m}$

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

$b_{kj} = (-1)^{n-k-1} \left({\dfrac{\displaystyle \sum_{\substack{0 \mathop \le m_0 \mathop < \ldots \mathop < m_{n-k} \mathop \le n \\ m_0, \ldots, m_{n-k} \mathop \ne j} } x_{m_0} \cdots x_{m_{n-k}} } {\displaystyle \prod_{\substack {0 \mathop \le m \mathop \le n \\ m \mathop \ne j} } \left({x_j - x_m}\right)}}\right)$

which gives:

$b_{kj} = (-1)^{k-1} \left({\dfrac{\displaystyle \sum_{\substack{0 \mathop \le m_0 \mathop < \ldots \mathop < m_{n-k} \mathop \le n \\ m_0, \ldots, m_{n-k} \mathop \ne j} } x_{m_0} \cdots x_{m_{n-k}} } {\displaystyle \prod_{\substack {0 \mathop \le m \mathop \le n \\ m \mathop \ne j} } \left({x_m - x_j}\right)}}\right)$

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

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

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

Since the first matrix in the product expression for $V_n^{-1}$ above is diagonal, we find simply that:

$c_{kj} = \dfrac 1 {x_j} b_{kj}$

which we see establishes the result.

$\blacksquare$