Theorem 1   Let $x_1,x_2,...$ be a sequence of $d$ dimensional vectors. Assume that there is $C: \N^d \to \R \smallsetminus \{0\}$ such that \begin{equation}\label{Eq:assump} \sum_{\substack{n_1+\cdots+n_d=n \\ 0< n_1 < \cdots < n_d }}C(n_1,...,n_d) \det[x_{n_1},x_{n_2},...,x_{n_d}] = 0 \end{equation}for every $n \in \N$ Then $\det[x_{n_1},x_{n_2},...,x_{n_d}]\!=\!0$ for all $(n_1,...,n_d) \in \N^d$

Proof. Introduce a linear order over the set of ordered tuples: $(n_1,n_2,...,n_d) \prec (\hat{n}_1,\hat{n}_2,...,\hat{n}_d)$ if $\left(\sum_{i=1}^d n_i,\hat{n}_d,\hat{n}_{d-1},...,\hat{n}_1\right)$ precedes $\left(\sum_{i=1}^d\hat{n}_i,n_d,n_{d-1},...,n_1\right)$ lexicographically. Let $(n_1,n_2,...,n_d)$ be the minimal (according to the above order) ordered tuple for which \begin{equation}\label{Eq:det-is-not-zero} \det[x_{n_1},x_{n_2},...,x_{n_d}] \ne 0. \end{equation}Take another ordered tuple, $(\hat{n}_1,\hat{n}_2,...,\hat{n}_d)$ such that $\sum_{i=1}^d n_i = \sum_{i=1}^d \hat{n}_i$ By minimality, if $(n_d,n_{d-1},...,n_1)$ precedes $(\hat{n}_d,\hat{n}_{d-1},...,\hat{n}_1)$ lexicographically then $\det[x_{\hat{n}_1},x_{\hat{n}_2},...,x_{\hat{n}_d}]=0$ Otherwise, let $i \in \{0,1,...,d-1\}$ be the first index such that $n_{d-i} \ne \hat{n}_{d-i}$ (more specifically, $n_{d-i} > \hat{n}_{d-i}$ . Then, $\hat{n}_{d-j} = n_{d-j}$ for $j=0,...,i-1$ and $\hat{n}_{d-j} < n_{d-i}$ for $j=i,...,d-1$ Therefore, $$\det[x_{n_1},...,x_{n_{d-i-1}},x_{\hat{n}_m},x_{n_{d-i+1}},...,x_{n_d}] = 0$$ for all $m=1,2,...,d$ (some because of repeated columns and the others because $\sum_{j=1}^d n_j - n_{d-i} + \hat{n}_m < \sum_{j=1}^d n_j$ . Since the vectors $x_{n_1},x_{n_2},...,x_{n_d}$ are linearly independent, we get that $$\{x_{\hat{n}_1},x_{\hat{n}_2},...,x_{\hat{n}_d}\} \subset \spn \left(\{x_{n_1},x_{n_2},...,x_{n_d}\} \smallsetminus \{x_{n_{d-i}}\}\right).$$ In particular $\det[x_{\hat{n}_1},x_{\hat{n}_2},...,x_{\hat{n}_d}]=0$ Therefore, () reduces to $\det[x_{n_1},x_{n_2},...,x_{n_d}]=0$ which contradicts ().