Understand the definition of a basis of a subspace.
Understand the basis theorem.
Recipes: basis for a column space, basis for a null space, basis of a span.
Picture: basis of a subspace of or
Theorem: basis theorem.
Essential vocabulary words:basis, dimension.
Subsection2.7.1Basis of a Subspace
As we discussed in Section 2.6, a subspace is the same as a span, except we do not have a set of spanning vectors in mind. There are infinitely many choices of spanning sets for a nonzero subspace; to avoid reduncancy, usually it is most convenient to choose a spanning set with the minimal number of vectors in it. This is the idea behind the notion of a basis.
Let be a subspace of A basis of is a set of vectors in such that:
the set is linearly independent.
Recall that a set of vectors is linearly independent if and only if, when you remove any vector from the set, the span shrinks (Theorem 2.5.12). In other words, if is a basis of a subspace then no proper subset of will span it is a minimal spanning set. Any subspace admits a basis by this theorem in Section 2.6.
A nonzero subspace has infinitely many different bases, but they all contain the same number of vectors.
We leave it as an exercise to prove that any two bases have the same number of vectors; one might want to wait until after learning the invertible matrix theorem in Section 3.5.
Let be a subspace of The number of vectors in any basis of is called the dimension of and is written
Since is a square matrix, it has a pivot in every row if and only if it has a pivot in every column. We will see in Section 3.5 that the above two conditions are equivalent to the invertibility of the matrix
Now we show how to find bases for the column space of a matrix and the null space of a matrix. In order to find a basis for a given subspace, it is usually best to rewrite the subspace as a column space or a null space first: see this important note in Section 2.6.
A basis for the column space
First we show how to compute a basis for the column space of a matrix.
The above theorem is referring to the pivot columns in the original matrix, not its reduced row echelon form. Indeed, a matrix and its reduced row echelon form generally have different column spaces. For example, in the matrix below:
the pivot columns are the first two columns, so a basis for is
The first two columns of the reduced row echelon form certainly span a different subspace, as
but contains vectors whose last coordinate is nonzero.
The dimension of is the number of pivots of
A basis of a span
Computing a basis for a span is the same as computing a basis for a column space. Indeed, the span of finitely many vectors is the column space of a matrix, namely, the matrix whose columns are
In order to compute a basis for the null space of a matrix, one has to find the parametric vector form of the solutions of the homogeneous equation
The vectors attached to the free variables in the parametric vector form of the solution set of form a basis of
The proof of the theorem has two parts. The first part is that every solution lies in the span of the given vectors. This is automatic: the vectors are exactly chosen so that every solution is a linear combination of those vectors. The second part is that the vectors are linearly independent. This part was discussed in this example in Section 2.5.
A basis for a general subspace
As mentioned at the beginning of this subsection, when given a subspace written in a different form, in order to compute a basis it is usually best to rewrite it as a column space or null space of a matrix.
Recall that forms a basis for if and only if the matrix with columns has a pivot in every row and column (see this example). Since is an matrix, these two conditions are equivalent: the vectors span if and only if they are linearly independent. The basis theorem is an abstract version of the preceding statement, that applies to any subspace.
Let be a subspace of dimension Then:
Any linearly independent vectors in form a basis for
Suppose that is a set of linearly independent vectors in In order to show that is a basis for we must prove that If not, then there exists some vector in that is not contained in By the increasing span criterion in Section 2.5, the set is also linearly independent. Continuing in this way, we keep choosing vectors until we eventually do have a linearly independent spanning set: say Then is a basis for which implies that But we were assuming that has dimension so must have already been a basis.
Now suppose that spans If is not linearly independent, then by this theorem in Section 2.5, we can remove some number of vectors from without shrinking its span. After reordering, we can assume that we removed the last vectors without shrinking the span, and that we cannot remove any more. Now and is a basis for because it is linearly independent. This implies that But we were assuming that so must have already been a basis.
In other words, if you already know that and if you have a set of vectors in then you only have to check one of:
is linearly independent, or
in order for to be a basis of If you did not already know that then you would have to check both properties.
To put it yet another way, suppose we have a set of vectors in a subspace Then if any two of the following statements is true, the third must also be true:
is linearly independent,
For example, if is a plane, then any two noncollinear vectors in form a basis.