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Let \(S = \left\{ {{{\mathop{\rm v}\nolimits} _1},...,{{\mathop{\rm v}\nolimits} _p}} \right\}\) be a set in \(V\), and \(H = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _1},...,{{\mathop{\rm v}\nolimits} _p}} \right\}\).

1. If one of the vectors in \(S\), say \({{\mathop{\rm v}\nolimits} _k}\), is alinear combination of the remaining vectors in \(S\), then the set formed from \(S\) by removing \({{\mathop{\rm v}\nolimits} _k}\) still spans \(H\).
2. If \(H \ne \left\{ 0 \right\}\), some subset of \(S\) is abasisfor \(H\).

Theorem 10 states that if a vector space \(V\) has a basis of \(n\) vectors, then every basis of \(V\) must consist ofexactly \(n\) vectors.

It should be noted that \(n \ge 1\) since \(S\) cannot have fewer than one vector. \(V \ne 0\) because \(n \ge 1\).

Consider \(S\) spans \(V\) and \(S\) has fewer than \(n\) vectors. According to the spanning set theorem, each subset \(S'\) of \(S\) is a basis for \(V\).

If \(S'\) is a subset of \(S\), then it contains fewer than \(n\) vectors because \(S\) also contains fewer than \(n\) vectors. Therefore, there is a basis \(S'\)for \(V\) contains fewer than \(n\) vectors but this is impossible according to theorem 10since \(\dim V = n\).

Thus, subset \(S\) cannot span \(V\).