Theorem 1. The intersection of any collection of subspaces is a subspace.
Proof. If we use an index \(\nu\) to tell apart the members of the collection, so that the given subspaces are \(\mathcal{M}_{\nu}\) , let us write \[\mathcal{M}=\bigcap_{\nu} \mathcal{M}_{\nu}.\] Since every \(\mathcal{M}_{\nu}\) , contains \(0\) , so does \(\mathcal{M}\) , and therefore \(\mathcal{M}\) is not empty. If \(x\) and \(y\) belong to \(\mathcal{M}\) (that is, to all \(\mathcal{M}_{\nu}\) ), then \(\alpha x+\beta y\) belongs to all \(\mathcal{M}_{\nu}\) , and therefore \(\mathcal{M}\) is a subspace. ◻
To see an application of this theorem, suppose that \(\mathcal{S}\) is an arbitrary set of vectors (not necessarily a subspace) in a vector space \(\mathcal{V}\) . There certainly exist subspaces \(\mathcal{M}\) containing every element of \(\mathcal{S}\) (that is, such that \(\mathcal{S} \subset \mathcal{M}\) ); the whole space \(\mathcal{V}\) is, for example, such a subspace. Let \(\mathcal{M}\) be the intersection of all the subspaces containing \(\mathcal{S}\) ; it is clear that \(\mathcal{M}\) itself is a subspace containing \(\mathcal{S}\) . It is clear, moreover, that \(\mathcal{M}\) is the smallest such subspace; if \(\mathcal{S}\) is also contained in the subspace \(\mathcal{N}\) , \(\mathcal{S} \subset \mathcal{N}\) , then \(\mathcal{M} \subset \mathcal{N}\) . The subspace \(\mathcal{M}\) so defined is called the subspace spanned by \(\mathcal{S}\) or the span of \(\mathcal{S}\) . The following result establishes the connection between the notion of spanning and the concepts studied in Sections 5–9.
Theorem 2. If \(\mathcal{S}\) is any set of vectors in a vector space \(\mathcal{V}\) and if \(\mathcal{M}\) is the subspace spanned by \(\mathcal{S}\) , then \(\mathcal{M}\) is the same as the set of all linear combinations of elements of \(\mathcal{S}\) .
Proof. It is clear that a linear combination of linear combinations of elements of \(\mathcal{S}\) may again be written as a linear combination of elements of \(\mathcal{S}\) . Hence the set of all linear combinations of elements of \(\mathcal{S}\) is a subspace containing \(\mathcal{S}\) ; it follows that this subspace must also contain \(\mathcal{M}\) . Now turn the argument around: \(\mathcal{M}\) contains \(\mathcal{M}\) and is a subspace; hence \(\mathcal{M}\) contains all linear combinations of elements of \(\mathcal{S}\) . ◻
We see therefore that in our new terminology we may define a linear basis as a set of linearly independent vectors that spans the whole space.
Our next result is an easy consequence of Theorem 2; its proof may be safely left to the reader.
Theorem 3. If \(\mathcal{H}\) and \(\mathcal{K}\) are any two subspaces and if \(\mathcal{M}\) is the subspace spanned by \(\mathcal{H}\) and \(\mathcal{K}\) together, then \(\mathcal{M}\) is the same as the set of all vectors of the form \(x+y\) , with \(x\) in \(\mathcal{H}\) and \(y\) in \(\mathcal{K}\) .
Prompted by this theorem, we shall use the notation \(\mathcal{H}+\mathcal{K}\) for the subspace \(\mathcal{M}\) spanned by \(\mathcal{H}\) and \(\mathcal{K}\) . We shall say that a subspace \(\mathcal{K}\) of a vector space \(\mathcal{V}\) is a complement of a subspace \(\mathcal{H}\) if \(\mathcal{H} \cap \mathcal{K}=\mathscr{O}\) and \(\mathcal{H}+\mathcal{K}=\mathcal{V}\) .