Question

Question 11: Let \(S\) be a finite minimal spanning set of a vector space \(V\). That is, \(S\) has the property that if a vector is removed from \(S\), then the new set will no longer span \(V\). Prove that \(S\) must be a basis for \(V\).

Short answer

It is proved that \(S\) must be a basis for \(V\).

Step-by-step solution

State the condition for a basis

Let \(H\) be a subspaceof vector space \(V\). An indexed set of vectors \(B = \left\{ {{{\mathop{\rm b}\nolimits} _1},...,{{\mathop{\rm b}\nolimits} _p}} \right\}\) in \(V\) is a basis for \(H\) if

• \(B\)is a linearly independent set,and
• The subspace spanned by \(B\) coincides with \(H\), that is, \(H = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm b}\nolimits} _1},...,{{\mathop{\rm b}\nolimits} _p}} \right\}\).

Show that \(S\) must be a basis for \(V\)

When \(S\) is a finite spanning set for \(V\), a subset of \(S\) is a basis for \(V\). The subset of \(S\) is represented by \(S'\). \(S'\) must span \(V\) because \(S'\) is a basis for \(V\). \(S'\) cannot be a proper subset of \(S\) since \(S\) is a minimal spanning set. Therefore, \(S' = S\) and \(S\) is a basis for \(V\).

Thus, it is proved that \(S\) is a basis for \(V\).