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Section 20.5 Theory

Subsection 20.5.1 Dimension as size of a basis

Since dimension is defined in terms of basis, it is important to know that we can always find a basis. The following fact is true for all vector spaces, but we will state and prove it only for finite-dimensional spaces. It is essentially just a restatement of Proposition 19.5.1 (which itself is a restatement of Proposition 18.5.5).

By definition, a vector space is finite-dimensional when it has a finite spanning set. Proposition 19.5.1 states that every finite spanning set can be reduced to a basis. So if a finite spanning set exists for a space, so does a basis.

The next two facts allow us to attach a single number to a vector space as the dimension of the space.

By definition, a finite-dimensional vector space has at least one example of a spanning set that contains a finite number of vectors. By Lemma 18.5.7, any other set of vectors from this space that contains more vectors than this example spanning set must be linearly dependent. But a basis is always linearly independent, and so cannot have more vectors than the finite number in this example spanning set.

Suppose \(\basisfont{B}_1\) and \(\basisfont{B}_2\) are two different bases for a finite-dimensional vector space \(V\text{.}\) First, both \(\basisfont{B}_1\) and \(\basisfont{B}_2\) must contain a finite number of vectors, by Lemma 20.5.2. Now, \(\basisfont{B}_1\) is a basis, so it is a spanning set, and so by Lemma 18.5.7 any set that contains more vectors than \(\basisfont{B}_1\) must be linearly dependent. But \(\basisfont{B}_2\) is also a basis, so it is linearly independent. Therefore, \(\basisfont{B}_2\) cannot contain more vectors than there are in \(\basisfont{B}_1\text{.}\)

The same reasoning works the other way: \(\basisfont{B}_1\) cannot contain more vectors than there are in the spanning set \(\basisfont{B}_2\text{,}\) otherwise it would be linearly dependent. Since neither set of vectors can contain more vectors than the other, the two sets must contain exactly the same number of vectors.

Subsection 20.5.2 Consequences for the theory of linear dependence/independence and spanning

Now we extend Proposition 18.5.6 to establish a “building-up” counterpart to Proposition 19.5.1.

Suppose \(S\) is a linearly independent set of vectors in a finite-dimensional vector space. If it is also a spanning set, then it is already a basis and does not need to be enlarged. If it is not a spanning set, then there are vectors in the space that are not in \(\Span S\text{.}\) Choose a vector \(\uvec{v}\) not in \(\Span S\text{,}\) and let \(S'\) be the set that contains all the vectors of \(S\) as well as \(\uvec{v}\text{.}\) By Proposition 18.5.6, the set \(S'\) is still linearly independent. If \(S'\) is also a spanning set, then it is a basis and we have the desired enlargement from \(S\text{.}\) Otherwise, we could again enlarge \(S'\) by some vector that is not in \(\Span S'\) and still have a linearly independent set. We can continue in this fashion, but we will have to reach a point where we will not be able to enlarge our set any further without it becoming linearly dependent, since we know that in a finite-dimensional space, once a set of vectors gets too large it can no longer be linearly independent (Lemma 18.5.7). At this point, our enlarged linearly independent set must also be a spanning set (and hence a basis), since if it weren't we would be able to enlarge it again as before, with the enlarged set remaining independent.

The concept of dimension gives us another way to know whether a set of vectors is a basis, since it is the “just-right” size for a set of vectors to be a basis.

  1. Assume that \(S\) is linearly independent. By Proposition 20.5.4, \(S\) can be enlarged to a basis for the vector space. But every basis for that space contains the same number of vectors (Theorem 20.5.3), and we have assumed that \(S\) already contains that number of vectors. So \(S\) must not need to be enlarged to become a basis — it must already be a basis itself, and so must be a spanning set.
  2. Assume that \(S\) is a spanning set. By Proposition 19.5.1, \(S\) can be reduced to a basis for the vector space. But every basis for that space contains the same number of vectors (Theorem 20.5.3), and we have assumed that \(S\) already contains that number of vectors. So \(S\) must not need to be reduced to become a basis — it must already be a basis itself, and so must be linearly independent.
Remark 20.5.7.

In a space whose dimension is known, the above corollary effectively reduces the amount of work required to check whether a set of vectors is a basis in half, since if we start with the right number of vectors in a basis-candidate set then we only need to check one of the requirements in the definition of basis. In practice, it is usually easier to carry out the Test for Linear Dependence/Independence than it is to check for spanning.

Subsection 20.5.3 Dimension of subspaces

As discussed in Subsection 20.3.4, a set of linearly independent vectors in a subspace is still linearly independent when considered as a set of vectors in the larger space. So we can use Proposition 20.5.4 to relate a basis for a subspace to a basis for the whole space, and then also the dimension of the subspace to the dimension of the whole space.

  1. Since \(U\) is a subspace of \(V\text{,}\) each vector of \(U\) is also a vector of \(V\text{.}\) So a basis for \(U\) will be a linearly independent set of vectors in \(V\text{,}\) which Proposition 20.5.4 tells us can be enlarged to a basis for \(V\text{.}\)
  2. Recall that the dimenion of a vector space (whether a subspace of another space or not) is defined to be the number of vectors in a basis for the space. Since every basis for \(U\) can be enlarged to a basis for \(V\text{,}\) the number of vectors in a basis for \(U\) cannot be larger than the number of vectors in a basis for \(V\text{.}\)
  3. Let \(\basisfont{B}\) be a basis for \(U\text{,}\) so that \(U = \Span\basisfont{B}\text{.}\) If we have \(\dim U = \dim V\text{,}\) then the number of vectors in \(\basisfont{B}\) is exactly equal to the dimension of \(V\text{.}\) But \(\basisfont{B}\) is also linearly independent in \(V\text{,}\) so by Statement 1 of Proposition 20.5.5, it must also be a spanning set for \(V\text{.}\) Thus, \(U = \Span\basisfont{B} = V\text{.}\)