For
pairwise different (i. e.
for
) matrix is invertible, as the following theorem proves:
Theorem 10.2:
Let
be the Vandermonde matrix associated to the pairwise different points
. Then the matrix
whose
-th entry is given by
![{\displaystyle \mathbf {b} _{k,m}:={\begin{cases}{\frac {\sum _{1\leq l_{1}<\cdots <l_{n-m}\leq n \atop l_{1},\ldots ,l_{n-m}\neq k}(-1)^{m-1}x_{l_{1}}\cdots x_{l_{n-m}}}{x_{k}\prod _{1\leq l\leq n \atop l\neq k}(x_{l}-x_{k})}}&m<n\\{\frac {1}{x_{k}\prod _{1\leq l\leq n \atop l\neq k}(x_{l}-x_{k})}}&m=n\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e77740203582b5868da77cbd3b4c00bdc8ec17e3)
is the inverse matrix of
.
Proof:
We prove that
, where
is the
identity matrix.
Let
. We first note that, by direct multiplication,
.
Therefore, if
is the
-th entry of the matrix
, then by the definition of matrix multiplication
.![{\displaystyle \Box }](https://wikimedia.org/api/rest_v1/media/math/render/svg/029b77f09ebeaf7528fc831fe57848be51f2240b)
Lemma 10.3:
Let
be pairwise different. The solution to the equation
![{\displaystyle {\begin{pmatrix}x_{1}&\cdots &x_{n}\\x_{1}^{2}&\cdots &x_{n}^{2}\\\vdots &\ddots &\vdots \\x_{1}^{n}&\cdots &x_{n}^{n}\end{pmatrix}}{\begin{pmatrix}y_{1}\\\vdots \\y_{n}\end{pmatrix}}={\begin{pmatrix}0\\\vdots \\0\\1\end{pmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/89e1b375d19cda5517c8eba7c6009a9085eb3a64)
is given by
,
.
Proof:
We multiply both sides of the equation by
on the left, where
is as in theorem 10.2, and since
is the inverse of
,
we end up with the equation
.
Calculating the last expression directly leads to the desired formula.