Rank of a Vandermonde matrix

Question

Let the $m \times n$ Vandermonde matrix be $V$, where $m\geq n$. Let $d$ such that $n \geq d$. I want a proof that $\operatorname{rank}(V)=d$ if and only if $d$ number of the elements are distinct. Could you help me out? A reference such as book, paper or a website link would be appreciated.

Answer 1

The $m\times n$ Vandermonde matrix formed from $\alpha_1,\alpha_2,\ldots,\alpha_m$ is $$V=\begin{pmatrix} 1 & \alpha_1 & \cdots &\alpha_1^{n-1}\\ 1 & \alpha_2 & \cdots &\alpha_2^{n-1}\\ \vdots & \vdots & \ddots &\vdots\\ 1 & \alpha_m & \cdots &\alpha_m^{n-1}\\ \end{pmatrix}.$$ Suppose that $\#\{\alpha_1,\alpha_2,\ldots,\alpha_m\}=d\le n\le m$. Then, without loss of generality, we may assume that $\alpha_1,\alpha_2,\ldots,\alpha_d$ are distinct. Clearly, $\operatorname{rank} V \le d$. By a well-known formula, $$\det\begin{pmatrix} 1 & \alpha_1 & \cdots &\alpha_1^{d-1}\\ 1 & \alpha_2 & \cdots &\alpha_2^{d-1}\\ \vdots & \vdots & \ddots &\vdots\\ 1 & \alpha_d & \cdots &\alpha_d^{d-1}\\ \end{pmatrix}=\prod_{1\le i< j\le d}(\alpha_i-\alpha_j)\ne 0.$$ Hence, $\operatorname{rank} V=d$. For proofs of said formula, see this or that.