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Chapter 4: Problem 86

Prove that \(\operatorname{span}(S)\) is the intersection of all subspaces of \(V\) containing \(S\).

Short Answer

Expert verified
To prove that \(\operatorname{span}(S)\) is the intersection of all subspaces of \(V\) containing \(S\), we showed that \(\operatorname{span}(S) \subseteq I\) and \(I \subseteq \operatorname{span}(S)\), where \(I\) is the intersection of all subspaces containing \(S\). This implies that \(\operatorname{span}(S) = I\), proving that the span of \(S\) is indeed the intersection of all subspaces of \(V\) containing \(S\).

Step by step solution

01

Define given elements and terms

Let \(V\) be a vector space and let \(S\) be a subset of \(V\). We want to prove that \(\operatorname{span}(S)\) is the intersection of all subspaces of \(V\) containing S. We denote the intersection of all subspaces containing S by I.
02

Prove any element in \(\operatorname{span}(S)\) is in \(I\)

Take any vector \(v\) in \(\operatorname{span}(S)\). By definition of span, \(v\) can be written as a linear combination of elements of \(S\): \[v = \sum_{i=1}^{n} a_i s_i\] where \(a_i\) are scalars, \(s_i\) in \(S\), and \(n\) is a positive integer. Since every subspace containing \(S\) must also contain linear combinations of elements of \(S\), \(v\) is in every subspace containing \(S\). Hence, \(v\) is in \(I\). This shows that \(\operatorname{span}(S) \subseteq I\).
03

Prove any element in \(I\) is in \(\operatorname{span}(S)\)

Now take any vector \(w\) in \(I\). By definition, since \(w\) is in the intersection of all subspaces containing \(S\), it is also in \(\operatorname{span}(S)\), which is itself a subspace of \(V\) containing \(S\). Thus, \(w \in \operatorname{span}(S)\) and this shows that \(I \subseteq \operatorname{span}(S)\).
04

Conclusion

Since we have demonstrated that \(\operatorname{span}(S) \subseteq I\) and \(I \subseteq \operatorname{span}(S)\), it follows that \(\operatorname{span}(S) = I\). Therefore, the span of \(S\) is indeed the intersection of all subspaces of \(V\) containing \(S\).

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