Let $Q = [0,1]\times [0,1]$. Let $1\leq k < \infty$ and $\{(x_l,\xi_l)\}_{l=1}^{k}\subseteq Q$ be such that $(x_i,\xi_i)\ne (x_j, \xi_j)$ for $i\ne j$. Additionally, for $1\leq l \leq k$, let $n_l\in \mathbf{N}$. Let $m = (m_1,m_2) \in \mathbf{Z}^2$ and define $ E_m(x,\xi) = e^{2\pi i m_1 x}e^{2\pi i m_2 \xi}. $ Moreover, let $$ E_m^{(\alpha, \beta)}(x,\xi) = \frac{\partial^{\alpha+\beta}}{\partial x^{\alpha}\partial \xi^{\beta}}E_m(x,\xi). $$ Furthermore, for $1\leq l \leq k$ define $$ v_{ml} = \begin{bmatrix} E_{m}(x_l,\xi_l)\\ E_m^{(1,0)}(x_l,\xi_l) \\ E_m^{(0,1)}(x_l,\xi_l) \\ \vdots \\ E_{m}^{(\alpha,\beta)}(x_l,\xi_l) \end{bmatrix} $$ where $0\leq\alpha+\beta \leq n_l-1$. In other words, we are considering all partial derivatives up to order $n_l-1$ of $E_m(x,\xi)$ evaluated at $(x_l,\xi_l)$. Now, define $$ v_m = \begin{bmatrix} v_{m1} \\ v_{m2} \\ \vdots \\ v_{mk} \end{bmatrix}. $$ Here, we are concatenating all of the $v_{ml}$ ($1\leq l \leq k$) vectors into one vector.
$\textbf{Claim.}$ $\{v_m\}_{m\in \mathbf{Z}^2}$ spans $\mathbf{C}^N$ for $$ N = \sum_{l=1}^k \frac{n_l(n_l+1)}{2}. $$ In other words, there exists some sequence $M = \{m_i\}_{i=1}^N\subseteq \mathbf{Z}^2$ such that $\{v_m\}_{m\in M}$ forms a linearly independent set. In other words, it is a basis.
$\textbf{Ideas.}$ In the 1-D case I am able to show that this holds; however in the 2-D case the partial derivatives make things much more complicated. I'm able to show that this is indeed true for specific cases, but not in the general case. Any advice would be appreciated or a counter-example.
