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.



Linear independence of element-wise powers of positive vectors
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](https://en.wikipedia.org/wiki/Rolle%27s_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.](https://mathoverflow.net/q/31699/102946).


