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.


1 Answer 1


First step, show that the determinant is not zero.

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. ...

  • $\begingroup$ can you please help with this question also, If you have some spare time. This is also related to Vandermonde matrix and linear algebra. Actually, I am not good at linear algebra. math.stackexchange.com/questions/3640746/… $\endgroup$
    – user775699
    Apr 23, 2020 at 21:59
  • $\begingroup$ @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. $\endgroup$
    – orangeskid
    Apr 24, 2020 at 1:25

You must log in to answer this question.