A basis for a vector space is a sequence of vectors that form a set that is linearly independent and that spans the space.
We denote a basis with angle brackets to signify that this collection is a sequence — the order of the elements is significant. (The requirement that a basis be ordered will be needed, for instance, in Definition 1.13.)
This is a basis for .
It is linearly independent
and it spans .
This basis for
differs from the prior one because the vectors are in a different order. The verification that it is a basis is just as in the prior example.
The space has many bases. Another one is this.
The verification is easy.
For any ,
is the standard (or natural) basis. We denote these vectors by .
(Calculus books refer to 's standard basis vectors and instead of and , and they refer to 's standard basis vectors , , and instead of , , and .) Note that the symbol "" means something different in a discussion of than it means in a discussion of .
Consider the space of functions of the real variable .
Another basis is . Verification that these two are bases is Problem 7.
A natural for the vector space of cubic polynomials is . Two other bases for this space are and . Checking that these are linearly independent and span the space is easy.
The trivial space has only one basis, the empty one .
The space of finite-degree polynomials has a basis with infinitely many elements .
We have seen bases before. In the first chapter we described the solution set of homogeneous systems such as this one
That is, we described the vector space of solutions as the span of a two-element set. We can easily check that this two-vector set is also linearly independent. Thus the solution set is a subspace of with a two-element basis.
Parameterization helps find bases for other vector spaces, not just for solution sets of homogeneous systems. To find a basis for this subspace of
we rewrite the condition as .
Thus, this is a good candidate for a basis.
The above work shows that it spans the space. To show that it is linearly independent is routine.
Consider again Example 1.2. It involves two verifications.
In the first, to check that the set is linearly independent we looked at linear combinations of the set's members that total to the zero vector . The resulting calculation shows that such a combination is unique, that must be and must be .
The second verification, that the set spans the space, looks at linear combinations that total to any member of the space . In Example 1.2 we noted only that the resulting calculation shows that such a combination exists, that for each there is a . However, in fact the calculation also shows that the combination is unique: must be and must be .
That is, the first calculation is a special case of the second. The next result says that this holds in general for a spanning set: the combination totaling to the zero vector is unique if and only if the combination totaling to any vector is unique.
In any vector space, a subset is a basis if and only if each vector in the space can be expressed as a linear combination of elements of the subset in a unique way.
We consider combinations to be the same if they differ only in the order of summands or in the addition or deletion of terms of the form "".
By definition, a sequence is a basis if and only if its vectors form both a spanning set and a linearly independent set. A subset is a spanning set if and only if each vector in the space is a linear combination of elements of that subset in at least one way.
Thus, to finish we need only show that a subset is linearly independent if and only if every vector in the space is a linear combination of elements from the subset in at most one way. Consider two expressions of a vector as a linear combination of the members of the basis. We can rearrange the two sums, and if necessary add some terms, so that the two sums combine the same 's in the same order: and . Now
holds if and only if
holds, and so asserting that each coefficient in the lower equation is zero is the same thing as asserting that for each .
In a vector space with basis the
representation of with respect to
is the column vector of the coefficients used to express as a
linear combination of the basis vectors:
where and . The 's are the
coordinates of with respect to
We will later do representations in contexts that involve more than one basis. To help with the bookkeeping, we shall often attach a subscript to the column vector.
In , with respect to the basis , the representation of is
(note that the coordinates are scalars, not vectors). With respect to a different basis , the representation
This use of column notation and the term "coordinates" has both a down side and an up side.
The down side is that representations look like vectors from , which can be confusing when the vector space we are working with is , especially since we sometimes omit the subscript base. We must then infer the intent from the context. For example, the phrase "in , where " refers to the plane vector that, when in canonical position, ends at . To find the coordinates of that vector with respect to the basis
to get that and .
Then we have this.
Here, although we've ommited the subscript from the column, the fact that the right side is a representation is clear from the context.
The up side of the notation and the term "coordinates" is that they generalize the use that we are familiar with:~in and with respect to the standard basis , the vector starting at the origin and ending at has this representation.
Our main use of representations will come in the third chapter. The definition appears here because the fact that every vector is a linear combination of basis vectors in a unique way is a crucial property of bases, and also to help make two points. First, we fix an order for the elements of a basis so that coordinates can be stated in that order. Second, for calculation of coordinates, among other things, we shall restrict our attention to spaces with bases having only finitely many elements. We will see that in the next subsection.