# Unable to understand the proof of positivity of generalized Vandermonde matrix

I am self-studying a research paper in analytic number theory and I am unable to understand the following proof.

Lemma : $$t\geq 1$$ natural number , let $$0 < x_1 < \ldots < x_t$$ and $$\alpha_1< \ldots<\alpha_t$$ . Then the generalized Vandermonde matrix $$[x_j^{\alpha_i} ],\;1\leq i,\;j\leq t$$ has positive determinant.

Proof as given in paper :

## Analytical proof of Lemma 4

By induction on $$t$$ one proves the following claim. A non-zero function $$f(x)=\sum_{i=1}^{t} c_{i} x^{\alpha_{i}}$$ with $$c_{i}, \alpha_{i} \in \mathbb{R},$$ has at most $$t-1$$ positive zeros. Indeed, if $$f$$ has $$t$$ positive zeros then Rolle's theorem provides $$t-1$$ positive zeros of the derivative $$(\mathrm{d} / \mathrm{d} x)\left(x^{-\alpha 1} f(x)\right)$$. The non-vanishing of the determinant in Lemma 4 is an immediate consequence of this claim. since the determinant depends continuously on the parameters $$\alpha_{i},$$ we deduce the required positivity from the positivity of the Vandermonde determinant.

Unfortunately, I am unable to understand the proof from the $$2^{\text{nd}}$$ line itself.

Question 1: why can't $$(d/dx) ( x^{-\alpha_1} f(x)$$ have $$t-1$$ positive zeroes?

Question 2: How is non-vanishing of determinant in Lemma $$4$$ an immediate consequence of this claim?

Please explain the proof. I shall be really thankful.

Assume it is. Then there exists a linear combination of the rows that gives the zero row vector. That means, there exists $$c_1$$, $$\ldots$$, $$c_t$$, not all $$0$$ such that $$\sum_{i=1}^t c_i x_j^{a_i}$$ for all $$j=1,\ldots, t$$. This means that the function $$\sum c_i x^{a_i}$$ has at least $$t$$ zeroes $$x_1$$, $$\ldots$$, $$x_t$$.
Second step. Show that such a linear combination cannot have $$t$$ distinct positive zeroes. (positive is key here). By induction. The case $$t=1$$ needs to be checked. Then, if $$f$$ (a sum of $$t$$ terms) has $$t$$ positive roots, then $$x^{-a_1} f(x)$$ also has $$t$$ roots. Note that the last term of this is a constant. Now take the derivative . By Rolle, it has at least $$t-1$$ roots. But if consists of $$t-1$$ terms. Apply induction.
Third step. Knowing that the determinant is not $$0$$, for any value of $$a_1<\ldots < a_t$$, deform them into $$(0,\ldots, t-1)$$. All along the determinant is $$\ne 0$$. Since the deformation is continuous, the determinant keeps a constant sign Now, for $$(0, 1, \ldots, t-1)$$ we have the Vandermonde, so positive. ...
• @Yannic Muller: I recorrected the statement of the question, the hypothesis is $x_i$ positive, $\alpha_i$ real numbers and $x_i$, $\alpha_i$ are ordered in the same way. I suggest for the second step, just consider the case $t=1$, $t=2$, and try to see for yourself why the statement is true. For the third step, consider the $(\alpha'_1, \ldots, \alpha'_t)$ on the segment between $(\alpha_1, \ldots,\alpha_t)$ and $(0, \ldots, t-1)$, so they are still in increasing order. Apr 24, 2020 at 1:25