# 'Vandermonde-esque' matrix has nonzero determinant.

I want to show that the matrix $$M=\begin{pmatrix} z_1^{y_1} & z_1^{y_2} & \cdots & z_1^{y_n} \\ z_2^{y_1} & z_2^{y_2} & \cdots & z_2^{y_n} \\ \vdots & \vdots & \ddots & \vdots \\ z_n^{y_1} & z_n^{y_2} & \cdots & z_n^{y_n} \end{pmatrix},$$ where $$0 and $$0 < y_1 < \cdots < y_n$$ has nonzero determinant.

This is the last 'step' of a longer combinatorics problem from class asking to prove total positivity of this type of matrix. I can argue that, for rational exponents $$y_j$$, the matrix is totally positive, so extending via continuity tells me that in the general case the determinant is at least nonnegative.

The 'nice' proofs of nonsingularity of Vandermonde matrices that I am familiar with use properties of polynomials, which don't seem to be available in this case. I suspect that an approach could be using something about growth of exponentials to show that there cannot be a nontrivial linear dependency amongst the rows (or columns), but analysis is one of my weaker spots.

Edit: This question follows from a similar one that is answered on mo: https://mathoverflow.net/questions/118225/how-to-show-a-certain-determinant-is-non-zero

Let $$V$$ denote the space of functions of the form $$f(z) = a_1z^{y_1} + \cdots + a_n z^{y_n}$$. Let $$\phi: V \to \Bbb F^n$$ denote the map $$\phi(f) = (f(z_1),\dots,f(z_n)).$$ Note that $$\phi$$ is invertible if and only if $$\phi$$ has a trivial kernel, which is to say that there are no non-zero functions $$f \in V$$ for which $$f(z_i) = 0$$ for all $$i$$.
On the other hand, note that $$M$$ is the matrix of $$\phi$$ relative to the basis $$\{z^{y_1},\dots,z^{y_n}\}$$ and the standard basis of $$\Bbb F^n$$.
In fact, as a consequence Decartes' rule of signs, we can see that $$\phi$$ indeed has a trivial kernel. The matrix $$M$$ will necessarily invertible.
• No problem. Note that the condition that $z_i > 0$ is necessary to ensure that the determinant is non-zero Commented Oct 7, 2020 at 0:45