Section 19.5 Theory
Subsection 19.5.1 Dimension as size of a basis
Since dimension is defined in terms of basis, it is important to know that we can always form 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 18.5.1 (which itself is a restatement of Proposition 17.5.5).
Theorem 19.5.1.
Every finite-dimensional vector space has a basis.
Proof.
By definition, a vector space is finite-dimensional when it has a finite spanning set. Proposition 18.5.1 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.
Lemma 19.5.2.
A basis for a finite-dimensional vector space must contain a finite number of vectors.
Proof.
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 17.5.7, 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.
Theorem 19.5.3. Uniformity of dimension.
Every basis for a finite-dimensional vector space has the same number of vectors.
Proof.
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 19.5.2. Now, \(\basisfont{B}_1\) is a basis, so it is a spanning set, and so by Lemma 17.5.7 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 19.5.2 Consequences for the theory of linear dependence/independence and spanning
Now we extend Proposition 17.5.6 to establish a “building-up” counterpart to Proposition 18.5.1.
Proposition 19.5.4. Enlarging an independent set to a basis.
In a finite-dimensional vector space, every linearly independent set of vectors can be enlarged to a basis. That is, if \(S\) is a linearly independent set of vectors in a finite-dimensional vector space, then there exists a basis for the space that contains \(S\) as a subcollection.
Proof.
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 17.5.6, 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 17.5.7). 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.
Proposition 19.5.5. Using dimension to help test basis.
Suppose \(S\) is a set of vectors in a finite-dimensional vector space, and the number of vectors in \(S\) is exactly equal to the dimension of the vector space.
- If \(S\) is linearly independent, then we can conclude that \(S\) is also a spanning set without checking.
- If \(S\) is a spanning set, then we can conclude that \(S\) is also linearly independent without checking.
Proof of Statement 1.
Assume that \(S\) is linearly independent. By Proposition 19.5.4, \(S\) can be enlarged to a basis for the vector space. But every basis for that space contains the same number of vectors (Theorem 19.5.3), 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.
Proof of Statement 2.
Assume that \(S\) is a spanning set. By Proposition 18.5.1, \(S\) can be reduced to a basis for the vector space. But every basis for that space contains the same number of vectors (Theorem 19.5.3), 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.
Corollary 19.5.6.
Suppose \(S\) is a set of vectors in a finite-dimensional vector space, and the number of vectors in \(S\) is exactly equal to the dimension of the vector space. If \(S\) is either known to be linearly independent or known to be a spanning set, then \(S\) must also have the other property, and hence must be a basis for the vector space.
Remark 19.5.7.
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 19.5.3 Dimension of subspaces
As discussed in Subsection 19.3.4, 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 19.5.4 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.
Proposition 19.5.8.
Suppose \(U\) is a subspace of a finite-dimensional vector space \(V\text{.}\) Then the following all hold true.
- Every basis for \(U\) can be enlarged to a basis for \(V\text{.}\)
- We have \(\dim U \le \dim V\text{.}\)
- It is the case that \(\dim U = \dim V\) only if \(U\) is actually the whole space \(V\text{.}\)
Proof of Statement 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 19.5.4 tells us can be enlarged to a basis for \(V\text{.}\)
Proof of Statement 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 \(V\text{.}\)
Proof of Statement 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 19.5.5, it must also be a spanning set for \(V\text{.}\) Thus, \(U = \Span\basisfont{B} = V\text{.}\)
