We revisit the definitions of linear independence, bases, and dimension in the context of abstract vector spaces.

### VSP-0060: Bases and Dimension for Abstract Vector Spaces

When working with and subspaces of we developed several fundamental ideas
including *span*, *linear independence*, *bases* and *dimension*. We will find that these
concepts generalize easily to abstract vector spaces and that analogous results hold in
these new settings.

#### Linear Independence

*linearly independent*if the only solution to is the

*trivial solution*.

If, in addition to the trivial solution, a *non-trivial solution* (not all are zero) exists,
then we say that the set is *linearly dependent*.

#### Bases and Dimension

Recall that our motivation for defining a basis of a subspace of was to have a collection of vectors such that every vector of the subspace can be expressed as a unique linear combination of the vectors in that collection. Definition def:basis we gave in VSP-0030 generalizes to abstract vector spaces as follows.

- Proof
- By the definition of a basis, we know that can be written as a linear
combination of . Suppose there are two such representations. Then,
But then we have:
Because are linearly independent, we have for . Consequently for .

In VSP-0035 we defined the dimension of a subspace of to be the number of elements
in a basis. (Definition def:dimension) We will adopt this definition for abstract vector spaces.
As before, to ensure that *dimension* is well-defined we need to establish
that this definition is independent of our choice of a basis. The proof of the
following theorem is identical to the proof of its counterpart in . (Theorem
th:dimwelldefined)

In our discussion to this point we have always assumed that a basis is nonempty and hence that the dimension of the space is at least . However, the zero space has no basis. To accommodate for this, we will say that the zero vector space is defined to have dimension :

Our insistence that amounts to saying that the empty set of vectors is a basis of . Thus the statement that “the dimension of a vector space is the number of vectors in any basis” holds even for the zero space.

It was shown in Example ex:centralizerofA of VSP-0050 that is a subspace for any choice of the matrix .

Let . Show that and find a basis of .

Let

The set is linearly independent. (See Practice Problem prob:CABlinind) Every element of can be written as a linear combination of elements of . Thus So is a basis of and .

Let be a subspace of consisting of all symmetric matrices. Find the dimension of .

#### Finite-Dimensional Vector Spaces

Our definition of dimension of a vector space depends on the vector space having a basis. In this section we will establish that any vector space spanned by finitely many vectors has a basis.

Given a finite-dimensional vector space we will find a basis for by starting with a linearly independent subset of and expanding it to a basis. The following results are more general versions of Lemmas lemma:atmostnlinindinrn and lemma:expandinglinindset, and Theorem th:dimwelldefined of VSP-0035. The proofs are identical and we will omit them.

#### Coordinate Vectors

Recall that in the context of (and subspaces of ) the requirement that elements of a basis be linearly independent guarantees that every element of the vector space has a unique representation in terms of the elements of the basis. (See Theorem th:linindbasis of VSP-0030) We proved the same property for abstract vector spaces in Theorem th:uniquerep.

Uniqueness of representation in terms of the elements of a basis allows us to associate
every element of a vector space with a unique *coordinate vector* with respect to a
given basis. Coordinate vectors were first introduced in VSP-0030. We now give a
formal definition.

*coordinate vector for with respect to*. We denote the coordinate vector by and write:

Next, we need to show that spans . To this end, we will consider a generic element of and attempt to express it as a linear combination of the elements of . then Setting the coefficients of like terms equal to each other gives us Solving this linear system of , and gives us (You need to verify this) This shows that every element of can be written as a linear combination of elements of . Therefore is a basis for .

To find the coordinate vector for with respect to we need to express as a linear combination of the elements of . Fortunately, we have already done all the necessary work. For , , and . This gives us the coefficients of the linear combination: , , . We now write as a linear combination The coordinate vector for with respect to is

Coordinate vectors will play a vital role in establishing one of the most fundamental results in linear algebra, that all -dimensional vector spaces have the same structure as . In Example ex:p2isor3 of LTR-0060, for instance, we will show that is essentially the same as .

### Practice Problems

### Text Source

The discussion of the zero space was adapted from Section 6.3 of Keith Nicholson’s Linear Algebra with Applications. (CC-BY-NC-SA)

W. Keith Nicholson, Linear Algebra with Applications, Lyryx 2018, Open Edition, p. 349

### Example Source

Examples ex:polyindset and ex:CAbasis were adapted from Examples 6.3.1 and 6.3.10 of Keith Nicholson’s Linear Algebra with Applications. (CC-BY-NC-SA)

W. Keith Nicholson, Linear Algebra with Applications, Lyryx 2018, Open Edition, p. 346, 350

### Exercise Source

Practice Problems prob:linindabstractvsp1, prob:linindabstractvsp2 and prob:linindabstractvsp3 are Exercises 6.3(a)(b)(c) from Keith Nicholson’s Linear Algebra with Applications. (CC-BY-NC-SA)

W. Keith Nicholson, Linear Algebra with Applications, Lyryx 2018, Open Edition, p. 351