proof of the determinant condition for a sequence of vectors

Theorem.

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

(1)

for every $n\mathrm{\in }\mathrm{N}$. Then $\det \mathit{}\mathrm{[}{x}_{n_{\mathrm{1}}}\mathrm{,}{x}_{n_{\mathrm{2}}}\mathrm{,}\mathrm{\dots }\mathrm{,}{x}_{n_d}\mathrm{]}\mathrm{=}\mathrm{0}$ for all $\mathrm{(}{n}_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{n}_d\mathrm{)}\mathrm{\in }{\mathrm{N}}^d$.

Proof.

Introduce a linear order over the set of ordered tuples: $(n_1,n_2,\mathrm{\dots },n_d)\prec ({\widehat{n}}_1,{\widehat{n}}_2,\mathrm{\dots },{\widehat{n}}_d)$ if $({\sum }_{i=1}^d{n}_i,{\widehat{n}}_d,{\widehat{n}}_{d-1},\mathrm{\dots },{\widehat{n}}_1)$ precedes $({\sum }_{i=1}^d{\widehat{n}}_i,n_d,n_{d-1},\mathrm{\dots },n_1)$lexicographically. Let $(n_1,n_2,\mathrm{\dots },n_d)$ be the minimal (according to the above order) ordered tuple for which

$$det[x_{n_1},x_{n_2},\mathrm{\dots },x_{n_d}]\ne 0.$$(2)

Take another ordered tuple, $({\widehat{n}}_1,{\widehat{n}}_2,\mathrm{\dots },{\widehat{n}}_d)$, such that ${\sum }_{i=1}^d{n}_i={\sum }_{i=1}^d{\widehat{n}}_i$. By minimality, if$(n_d,n_{d-1},\mathrm{\dots },n_1)$ precedes $({\widehat{n}}_d,{\widehat{n}}_{d-1},\mathrm{\dots },{\widehat{n}}_1)$ lexicographically then$det[x_{{\widehat{n}}_1},x_{{\widehat{n}}_2},\mathrm{\dots },x_{{\widehat{n}}_d}]=0$. Otherwise, let $i\in \{0,1,\mathrm{\dots },d-1\}$ be the first index such that $n_{d-i}\ne {\widehat{n}}_{d-i}$ (more specifically, $n_{d-i}>{\widehat{n}}_{d-i}$). Then,${\widehat{n}}_{d-j}=n_{d-j}$ for $j=0,\mathrm{\dots },i-1$ and for $j=i,\mathrm{\dots },d-1$. Therefore,

$$det[x_{n_1},\mathrm{\dots },x_{n_{d-i-1}},x_{{\widehat{n}}_m},x_{n_{d-i+1}},\mathrm{\dots },x_{n_d}]=0$$

for all $m=1,2,\mathrm{\dots },d$ (some because of repeated columns and the othersbecause ). Since the vectors $x_{n_1},x_{n_2},\mathrm{\dots },x_{n_d}$ are linearly independent, we getthat

$$\{x_{{\widehat{n}}_1},x_{{\widehat{n}}_2},\mathrm{\dots },x_{{\widehat{n}}_d}\}\subset \mathrm{span}\left(\{x_{n_1},x_{n_2},\mathrm{\dots },x_{n_d}\}\setminus \{x_{n_{d-i}}\}\right).$$

In particular $det[x_{{\widehat{n}}_1},x_{{\widehat{n}}_2},\mathrm{\dots },x_{{\widehat{n}}_d}]=0$. Therefore, (1) reduces to $det[x_{n_1},x_{n_2},\mathrm{\dots },x_{n_d}]=0$which contradicts (2).

∎