Linear independence of element-wise powers of positive vectors

Question

Consider a vector $x$ with $0 < x_1 < \cdots < x_n < \infty$, and let $0 < \gamma_1 < \cdots < \gamma_n < \infty$.

I would like to show that $x^{\gamma_1}, \ldots, x^{\gamma_n}$ are linearly independent, where $x^{\gamma_i}$ is defined as the vector $(x_1^{\gamma_i}, \ldots, x_n^{\gamma_i})$.

It is clear that $x^{\gamma_1}$ and $x^{\gamma_2}$ are linearly independent, and I might have an overly complicated argument for the case $x^{\gamma_1}$, $x^{\gamma_2}$, and $x^{\gamma_3}$, but I'm really at a loss about how to tackle the general case.

Answer 1

The following proposition shows that $x^{\gamma_1}, \dots x^{\gamma_n}$ are indeed always linearly independent if $x \in \mathbb{R}^n$ has $n$ mutually distinct strictly positive entries.

Proposition. For all real numbers $\gamma_1 < \dots < \gamma_n$ (be they positive or not) and each tuple $0 \not= (\alpha_1, \dots, \alpha_n) \in \mathbb{R}^n$ the function $$ f_n: (0,\infty) \ni t \mapsto \alpha_1 t^{\gamma_1} + \dots + \alpha_n t^{\gamma_n} \in \mathbb{R} $$ has at most $n-1$ distinct zeros (by "distinct" I mean that we do not count multiplicities of zeros).

Proof. We show the proposition by induction. The assertion is obvious for $n = 1$, so assume that the claim has been proved for some $n \in \mathbb{N}$ and consider $n+1$ terms now.

Assume that $f_{n+1}$ has at least $n+1$ distinct zeros in $(0,\infty)$. Then all the coefficients $\alpha_1, \dots, \alpha_{n+1}$ are non-zero (otherwise $f_{n+1}$ would be a non-trivial linear combination of $n$ terms with at least $n+1$ zeros).

The function $t^{-\gamma_1}f_{n+1}(t)$ also has at least $n+1$ distinct zeros, and hence, its derivative $(t^{-\gamma_1}f_{n+1}(t))'$ has at least $n$ distinct zeros (as a consequence of Rolle's theorem). This is a contradiction since $(t^{-\gamma_1}f_{n+1}(t))'$ is a non-trivial linear combination of the $n$ terms $t^{\gamma_2-\gamma_1-1}, \dots, t^{\gamma_{n+1}-\gamma_1 - 1}$.

Remark. It is important in the above proof that we show the result for all real numbers $\gamma_k$ (and not only for positive ones), since even if all $\gamma_k$ are positive we need the induction hypotheses for negative exponents, too. So this is a nice example where strengthening the hypothesis is needed to make induction work..

Answer 2

In the case of integer $\gamma_i$, the determinant is nonzero (in fact, strictly positive). This is proved in Gantmacher, The Theory of Matrices, Vol 2, p99. Gantmacher calls it a "generalized Vandermonde matrix", though that name is also used for a different class of matrices.

Gantmacher's proof uses the properties of polynomials like Rolle's Theorem, so I don't think it will work for non-integer $\gamma_i$. But I bet that case is out there somewhere.

Answer 3

When $\gamma_i$ are rational, let $L$ be the LCM of their denominators. Then a linear dependency would imply that $x_1^{1/L},\dots,x_n^{1/L}$ are zeros of a certain polynomial with at most $n$ nonzero coefficients. However, this contradicts Descartes' rule of signs.