 Solution

For a finite set $S=\{s_1,\dots ,s_k\}$ of vectors in a real vector space, its span is the set of all linear combinations:The span of the empty set is $\{0\}$. The set $S$ is linearly independent if ${\sum }_i{a}_i{s}_i=0$ implies $a_i=0$ for every $i$. It is a spanning set for $V$ if $\mathrm{Sp}(S)=V$. A basis is both a linearly independent set and a spanning set.

For basis extension, start with the given linearly independent set. If it does not span $V$, choose $v\notin \mathrm{Sp}(S)$ and adjoin it. The enlarged set remains linearly independent: a relation $av+{\sum }_i{a}_i{s}_i=0$ with $a\ne 0$ would express $v$ as a linear combination of $S$, a contradiction; if $a=0$, independence of $S$ forces all other coefficients to be zero.

Repeat whenever the current span is proper. Each step increases the set's size by one, while the allowed bound says its size cannot exceed $\dim V$. Thus the process terminates with a linearly independent set spanning $V$, hence a basis containing the original set. This also covers an initially empty set and the zero-dimensional vector space.