Vandermonde matrix

In linear algebra, a Vandermonde matrix is a matrix with a geometric progression in each column, i.e;

$V={\begin{bmatrix}1&1&1&\dots &1\\\alpha _{1}&\alpha _{2}&\alpha _{3}&\dots &\alpha _{n}\\\alpha _{1}^{2}&\alpha _{2}^{2}&\alpha _{3}^{2}&\dots &\alpha _{n}^{2}\\\alpha _{1}^{3}&\alpha _{2}^{3}&\alpha _{3}^{3}&\dots &\alpha _{n}^{3}\\\vdots &\vdots &\vdots &&\vdots \\\alpha _{1}^{m-1}&\alpha _{2}^{m-1}&\alpha _{3}^{m-1}&\dots &\alpha _{n}^{m-1}\\\end{bmatrix}}$

Vandermonde matrices are named after Alexandre-Théophile Vandermonde.

In mathematical terms:

$V_{i,j}=\alpha _{j}^{i-1}$

These matrices are useful in polynomial interpolation, since solving an equation $Va=b$ for $a$, is equivalent to finding the coefficients $a_{i}$ of a polynomial that has values $b_{i}$ at $\alpha _{i}$.

The determinant of an $n\times n$ Vandermonde matrix can be expressed as follows:

$\det(V)=\prod _{1\leq i<j\leq n}(\alpha _{j}-\alpha _{i})$

If two or more exponents $\alpha _{i}$ are equal, the rank of the matrix drops (if all $\alpha _{i}$ are distinct, then $V$ is of full rank). This problem can alleviated by using a generalisation called confluent Vandermonde matrices, where the k-multiple columns are replaced by:

$V_{i,j}={\frac {i!}{(i-j-1)!}}\alpha _{j}^{i-j-1}$ where $0\leq j<k$