Question

Let S be a maximal linearly independent subset of a vector space V. In other words, S has the property that if a vector not in S is adjoined to S, the new set will no longer be linearly independent. Prove that S must be a basis of V. [Hint: What if S were linearly independent but not a basis of V?]

Short answer

S must be a basis of V.

Step-by-step solution

Find a vector v in set V

Let \(S = \left\{ {{{\bf{v}}_1},\,{{\bf{v}}_2},.....,{{\bf{v}}_p}} \right\}\).

Assume that S does not span V.

Therefore, \(V \ne {\rm{Span}}\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},....,{{\bf{v}}_p}} \right\}\).

Then, there must be a vector v in set V that is not in the \({\rm{Span}}\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},....{{\bf{v}}_p}} \right\}\).

Check for linear independency

Let \(S'\) be a set formed by

\(\begin{aligned} S' &= S \cup \left\{ {\bf{v}} \right\}\\ &= \left\{ {{{\bf{v}}_1},{{\bf{v}}_2},.....,{{\bf{v}}_p},{\bf{v}}} \right\}\end{aligned}\).

The set \(S'\) is also a linearly independent set, but the number of elements in the set \(S'\) is more than \(S\)(contradiction).

So, S must be a basis of V.