The Sequences $\{m^ke^{2\pi i \alpha_j m}\}_{m=1}^\infty$ are Linearly Independent over $\mathbb{C}$

Question

I am studying the book "Galois' Dream: Group Theory and Differential Equations" by Michio Kuga.

I have some trouble in proving the Lemma 18.3:

Let $\alpha_1, \ldots, \alpha_n$ be complex numbers such that $\alpha_t - \alpha_s$ is not an integer for $t \neq s$. Then for every positive integer $n$, the following $(N+1)n$ sequences are linearly independent over $\mathbb{C}$:

$$ \{m^ke^{2\pi i \alpha_j m}\}_{m=1}^\infty, $$ where $j = 1, \ldots, n$ and $k = 0, \ldots, N$. In other words, if the equation $$ \sum_{j=1}^n\sum_{k=0}^NC_{j,k} m^ke^{2\pi i \alpha_j m} = 0 $$ ($m = 1,2,\ldots$) holds for complex numbers $C_{j,k}$, then $C_{j,k} = 0$ for all $j = 1, \ldots, n$ and $k = 0, \ldots, N$.

Here is my process of thought:

First, to have a simpler notation, let $\beta_j := e^{2\pi i \alpha_j} \neq 0 $. Then, consider the linear dependence relation:

$$ \sum_{j=1}^n\sum_{k=0}^NC_{j,k} m^k \beta_j^m = 0, \hspace{10pt} m = 1, 2, \ldots $$

But I couldn't figure out how should I proceed next. I tried to do induction on one of the indicies, eg. on $j$ as: For $j = 1$ the relation becomes $$ \sum_{k = 0} ^N C_{0,k}m^k \beta_0^m = 0 $$

But even here, I couldn't conclude that all $C_{0,k}$s are zero. Since this $C$ has two indicies, I am not sure if finding all $C_{0,k}$s zero would imply that all $C_{j,k}$s are zero. So I am stuck. The book says this proof was easy, but I couldn't catch that "easy". I will appreciate any help. Thank you in advance.

Answer 1

We may replace $m^k$ by $\binom{m}{k}$ since, for $k>0$, $\binom{m}{k}$ is a linear combination of $\{m^j\mid 1\leqslant j\leqslant k\}$.

Suppose $z_1,\ldots,z_n\in\mathbb{C}$ are pairwise distinct, and $k_1,\ldots,k_n\in\mathbb{Z}_{>0}$. Then the set $$\left\{m\mapsto\binom{m}{k}z_j^{m-k}\ \middle|\ 1\leqslant j\leqslant n,0\leqslant k<k_j\right\}$$ of functions of $m\in\mathbb{Z}_{\geqslant 0}$ (i.e., sequences) is linearly independent over $\mathbb{C}$.

An "analytic" proof can be built as follows. Suppose $$\sum_{j=1}^n\sum_{k=0}^{k_j-1}C_{j,k}\binom{m}{k}z_j^{m-k}=0,$$ then, multiplying by $t^m/m!$ and summing over $m\geqslant 0$, we have $$\sum_{j=1}^n\sum_{k=0}^{k_j-1}C_{j,k}\frac{t^k}{k!}e^{z_j t}=0.$$ The rest is done by analysing $t\to\infty$ (roughly speaking). Namely, if $C_{J,K}\neq 0$ for some $(J,K)$, take (any!) $J$ with the greatest possible $|z_J|$, and take the greatest possible $K$ (with $C_{J,K}\neq 0$). Let $z_J=re^{i\theta}$. Now divide the above equality by $(t^K/K!)e^{z_J t}$, put $t=\rho e^{-i\theta}$, and take $\rho\to\infty$. Then the LHS tends to $C_{J,K}$ (all the terms with $(j,k)\neq(J,K)$ tend to zero; here we use the condition that $z_j$ are distinct), giving a contradiction.

There's also an "algebraic" proof (see this answer). Take $r=\sum_{j=1}^n k_j$ and let $V(\begin{smallmatrix}z_1&\dots&z_n\\k_1&\dots&k_n\end{smallmatrix})$ be the $r\times r$ matrix whose rows are the initial segments (of length $r$) of the sequences considered above. Namely, let $0\leqslant m<r$, $1\leqslant j\leqslant n$, and $0\leqslant k<k_j$; put $\ell=k+\sum_{p=1}^{j-1}k_p$. The $(\ell+1,m+1)$-th element of the matrix $V(\begin{smallmatrix}z_1&\dots&z_n\\k_1&\dots&k_n\end{smallmatrix})$ is taken to be $\binom{m}{k}z_j^{m-k}$. (Thus, if $k_1=\dots=k_n=1$, we get a Vandermonde matrix.) It turns out that $$\det V(\begin{smallmatrix}z_1&\dots&z_n\\k_1&\dots&k_n\end{smallmatrix})=\prod_{1\leqslant q<p\leqslant n}(z_p-z_q)^{k_p k_q}.$$ A proof by induction may be built using $\det V(\begin{smallmatrix}\dots&z_j&\dots\\\dots&k_j+1&\dots\end{smallmatrix})=\frac1{k_j!}\left.\frac{\partial^{k_j}}{\partial z^{k_j}}\det V(\begin{smallmatrix}\dots&z_j&z&\dots\\\dots&k_j&1&\dots\end{smallmatrix})\right|_{z=z_j}$.