# 9.4: Subspaces and Basis

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In this section we will examine the concept of subspaces introduced earlier in terms of \(\mathbb{R}^n\). Here, we will discuss these concepts in terms of abstract vector spaces.

Consider the definition of a subspace.

The span of a set of vectors as described in Definition [def:span] is an example of a subspace. The following fundamental result says that subspaces are subsets of a vector space which are themselves vector spaces.

Consider the following useful Corollary.

When determining spanning sets the following theorem proves useful.

In other words, this theorem claims that any subspace that contains a set of vectors must also contain the span of these vectors.

The following example will show that two spans, described differently, can in fact be equal.

To prove that a set is a vector space, one must verify each of the axioms given in Definition [def:vectorspaceaxiomsaddition] and [def:vectorspaceaxiomsscalarmult]. This is a cumbersome task, and therefore a shorter procedure is used to verify a subspace.

Therefore it suffices to prove these three steps to show that a set is a subspace.

Consider the following example.

The two subspaces described above are called **improper subspaces**. Any subspace of a vector space \(V\) which is not equal to \(V\) or \(\left\{ \vec{0} \right\}\) is called a **proper subspace**.

Consider another example.

Recall the definition of basis, considered now in the context of vector spaces.

Consider the following example.

The next theorem is an essential result in linear algebra and is called the exchange theorem.

The following corollary follows from the exchange theorem.

Given the result of the previous corollary, the following definition follows.

Notice that the dimension is well defined by Corollary [cor:baseslength]. It is assumed here that \(n<\infty\) and therefore such a vector space is said to be **finite dimensional**.

It is important to note that a basis for a vector space is not unique. A vector space can have many bases. Consider the following example.

Consider the following theorem.

If in fact \(W\) has \(n\) vectors, then it follows that \(W=V\).

Consider the following example.

The following theorem claims that a spanning set of a vector space \(V\) can be shrunk down to a basis of \(V\). Similarly, a linearly independent set within \(V\) can be enlarged to create a basis of \(V\).

Recall Example [exa:addinglinind] in which we added a matrix to a linearly independent set to create a larger linearly independent set. By Theorem [thm:basisfromspanninglinind] we can extend a linearly independent set to a basis.

Next we consider the case where you have a spanning set and you want a subset which is a basis. The above discussion involved adding vectors to a set. The next theorem involves removing vectors.

Consider the following example of this concept.

Consider the following example.

Now the following is a fundamental result about subspaces.

This also proves the following corollary. Let \(V\) play the role of \(W\) in the above theorem and begin with a basis for \(W\), enlarging it to form a basis for \(V\) as discussed above.

Consider the following example.