Elementary Properties of Linear Independence

Let $K$ be a \reftext{def:field-c54-2026b}{field}, let $V$ be a \reftext{def:vector-space-2026a}{vector space over $K$} with \reftext{lem:vector-space-basic-identities-2026a}{zero vector} $0_{V}$, let $n$ be a \reftext{def:natural-numbers-2026a}{natural number} with the \reftext{def:order-natural-numbers-2026a}{order relations} $<$ and $\le$, and let $v\in V^{n}$ be an \reftext{def:finite-tuple-power-2026a}{$n$-tuple} in $V$. For $j\in[n]$, with $[j]$ the \reftext{def:initial-segment-natural-numbers-2026a}{initial segment} determined by $j$, write $v|_{[j]}\in V^{j}$ for the restriction of $v$ to $[j]$. Then the following hold.

\textbf{1. (Restriction)} If $v$ is \reftext{def:linear-independence-finite-family-2026a}{linearly independent} and $j\in[n]$, then $v|_{[j]}$ is linearly independent.

\textbf{2. (Predecessors)} If $v$ is linearly independent, then $v_{1}\ne 0_{V}$, and for every $m\in\mathbb{N}$ with $m+1\in[n]$ the component $v_{m+1}$ does not lie in the \reftext{def:span-finite-family-2026b}{span} of $v|_{[m]}$.

\textbf{3. (Dependence)} Suppose $n=p+1$ for some $p\in\mathbb{N}$, and that $v$ is not linearly independent. Then there is $j\in[n]$ with

where $v^{(j)}\in V^{p}$ is obtained from $v$ by omitting the $j$-th component, as in \ref{lem:finite-sum-extraction-2026a}.