# proof of the determinant condition for a sequence of vectors

###### Theorem.

Let $x_{1},x_{2},...$ be a sequence of $d$ dimensional vectors. Assume that there is $C:{\mathbb{N}}^{d}\to{\mathbb{R}}\smallsetminus\{0\}$ such that

 $\sum_{\begin{subarray}{c}n_{1}+\cdots+n_{d}=n\\ 0 (1)

for every $n\in{\mathbb{N}}$. Then $\det[x_{n_{1}},x_{n_{2}},...,x_{n_{d}}]\!=\!0$ for all $(n_{1},...,n_{d})\in{\mathbb{N}}^{d}$.

###### 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

 $\det[x_{n_{1}},x_{n_{2}},...,x_{n_{d}}]\neq 0.$ (2)

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 $n_{d-i}\neq\hat{n}_{d-i}$ (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} 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

 $\{x_{\hat{n}_{1}},x_{\hat{n}_{2}},...,x_{\hat{n}_{d}}\}\subset\operatorname{% span}\left(\{x_{n_{1}},x_{n_{2}},...,x_{n_{d}}\}\smallsetminus\{x_{n_{d-i}}\}% \right).$

In particular $\det[x_{\hat{n}_{1}},x_{\hat{n}_{2}},...,x_{\hat{n}_{d}}]=0$. Therefore, (1) reduces to $\det[x_{n_{1}},x_{n_{2}},...,x_{n_{d}}]=0$ which contradicts (2).

Title proof of the determinant condition for a sequence of vectors ProofOfTheDeterminantConditionForASequenceOfVectors 2013-03-22 14:33:46 2013-03-22 14:33:46 GeraW (6138) GeraW (6138) 5 GeraW (6138) Proof msc 15A15