In this chapter, we have been looking at bases for \(\real^p\text{,}\) sets of vectors that are linearly independent and span \(\real^p\text{.}\) Frequently, however, we focus on only a subset of \(\real^p\text{.}\) In particular, if we are given an \(m\times n\) matrix \(A\text{,}\) we have been interested in both the span of the columns of \(A\) and the solution space to the homogeneous equation \(A\xvec = \zerovec\text{.}\) In this section, we will expand the concept of basis to describe sets like these.
Explain how this parametric description produces two vectors \(\wvec_1\) and \(\wvec_2\) whose span is the solution space to the equation \(A\xvec = \zerovec\text{.}\)
Let’s denote the columns of \(A\) as \(\vvec_1\text{,}\)\(\vvec_2\text{,}\)\(\vvec_3\text{,}\) and \(\vvec_4\text{.}\) Explain why \(\vvec_3\) and \(\vvec_4\) can be written as linear combinations of \(\vvec_1\) and \(\vvec_2\text{.}\)
Our goal is to develop a common framework for describing subsets like the span of the columns of a matrix and the solution space to a homogeneous equation. That leads us to the following definition.
Since we have explored the concept of span in some detail, this definition just gives us a new word to describe something familiar. Let’s look at some examples.
In Activity 2.3.3 and the following discussion, we looked at subspaces in \(\real^3\) without explicitly using that language. Let’s recall some of those examples.
Suppose we have a single nonzero vector \(\vvec\text{.}\) The span of \(\vvec\) is a subspace, which we’ll write as \(S =
\laspan{\vvec}\text{.}\) As we have seen, the span of a single vector consists of all scalar multiples of that vector, and these form a line passing through the origin.
If instead we have two linearly independent vectors \(\vvec_1\) and \(\vvec_2\text{,}\) the subspace \(S=\laspan{\vvec_1,\vvec_2}\) is a plane passing through the origin.
Consider the three vectors \(\evec_1\text{,}\)\(\evec_2\text{,}\) and \(\evec_3\text{.}\) Since we know that every 3-dimensional vector can be written as a linear combination, we have \(S = \laspan{\evec_1, \evec_2,
\evec_3} = \real^3\text{.}\)
One more subspace worth mentioning is \(S=\laspan{\zerovec}\text{.}\) Since any linear combination of the zero vector is itself the zero vector, this subspace consists of a single vector, \(\zerovec\text{.}\)
We will look at some sets of vectors and the subspaces they form.
If \(\vvec_1, \vvec_2,\ldots,\vvec_n\) is a set of vectors in \(\real^m\text{,}\) explain why \(\zerovec\) can be expressed as a linear combination of these vectors. Use this fact to explain why the zero vector \(\zerovec\) belongs to any subspace in \(\real^m\text{.}\)
As the activity shows, it is possible to represent some subspaces as the span of more than one set of vectors. We are particularly interested in representing a subspace as the span of a linearly independent set of vectors.
A basis for a subspace \(S\) of \(\real^p\) is a set of vectors in \(S\) that are linearly independent and whose span is \(S\text{.}\) We say that the dimension of the subspace \(S\text{,}\) denoted \(\dim
S\text{,}\) is the number of vectors in any basis.
Suppose we have the 4-dimensional vectors \(\vvec_1\text{,}\)\(\vvec_2\text{,}\) and \(\vvec_3\) that define the subspace \(S = \laspan{\vvec_1,\vvec_2,\vvec_3}\) of \(\real^4\text{.}\) Suppose also that
From the reduced row echelon form of the matrix, we see that \(\vvec_2 = -\vvec\text{.}\) Therefore, any linear combination of \(\vvec_1\text{,}\)\(\vvec_2\text{,}\) and \(\vvec_3\) can be rewritten
Furthermore, the reduced row echelon form of the matrix shows that \(\vvec_1\) and \(\vvec_3\) are linearly independent. Therefore, \(\{\vvec_1,\vvec_3\}\) is a basis for \(S\text{,}\) which means that \(S\) is a two-dimensional subspace of \(\real^4\text{.}\)
Notice that the columns of \(A\) are vectors in \(\real^m\text{,}\) which means that any linear combination of the columns is also in \(\real^m\text{.}\) Since the column space is described as the span of a set of vectors, we see that \(\col(A)\) is a subspace of \(\real^m\text{.}\)
Explain why the vectors \(\vvec_1\) and \(\vvec_2\) form a basis for \(\col(A)\) and why \(\col(A)\) is a 2-dimensional subspace of \(\real^3\) and therefore a plane.
Suppose that \(A\) is an \(8\times 10\) matrix and that \(\col(A) = \real^8\text{.}\) If \(\bvec\) is an 8-dimensional vector, what can you say about the equation \(A\xvec = \bvec\text{?}\)
It is certainly true that \(\col(A) =
\laspan{\vvec_1,\vvec_2,\ldots,\vvec_5}\) by the definition of the column space. However, the reduced row echelon form of the matrix shows us that the vectors are not linearly independent so \(\vvec_1,\vvec_2,\ldots,\vvec_5\) do not form a basis for \(\col(A)\text{.}\)
This means that any linear combination of \(\vvec_1,\vvec_2,\ldots,\vvec_5\) can be written as a linear combination of just \(\vvec_1\) and \(\vvec_2\text{.}\) Therefore, we see that \(\col(A) = \laspan{\vvec_1,\vvec_2}\text{.}\)
Moreover, the reduced row echelon form shows that \(\vvec_1\) and \(\vvec_2\) are linearly independent, which implies that they form a basis for \(\col(A)\text{.}\) This means that \(\col(A)\) is a 2-dimensional subspace of \(\real^3\text{,}\) which is a plane in \(\real^3\text{,}\) having basis
In general, a column without a pivot position can be written as a linear combination of the columns that have pivot positions. This means that a basis for \(\col(A)\) will always be given by the columns of \(A\) having pivot positions. This leads us to the following definition and proposition.
If \(A\) is an \(m\times n\) matrix, then \(\col(A)\) is a subspace of \(\real^m\) whose dimension equals \(\rank(A)\text{.}\) The columns of \(A\) that contain pivot positions form a basis for \(\col(A)\text{.}\)
For example, the rank of the matrix \(A\) in Example 3.5.7 is two because there are two pivot positions. A basis for \(\col(A)\) is given by the first two columns of \(A\) since those columns have pivot positions.
As a note of caution, we determine the pivot positions by looking at the reduced row echelon form of \(A\text{.}\) However, we form a basis of \(\col(A)\) from the columns of \(A\) rather than the columns of the reduced row echelon matrix.
If \(A\) is an \(m\times n\) matrix, we call the subset of vectors \(\xvec\) in \(\real^n\) satisfying \(A\xvec = \zerovec\) the null space of \(A\) and denote it by \(\nul(A)\text{.}\)
Remember that a subspace is a subset that can be represented as the span of a set of vectors. The column space of \(A\text{,}\) which is simply the span of the columns of \(A\text{,}\) fits this definition. It may not be immediately clear how the null space of \(A\text{,}\) which is the solution space of the equation \(A\xvec
= \zerovec\text{,}\) does, but we will see that \(\nul(A)\) is a subspace of \(\real^n\text{.}\)
and give a parametric description of the solution space to the equation \(A\xvec = \zerovec\text{.}\) In other words, give a parametric description of \(\nul(A)\text{.}\)
This parametric description shows that the vectors satisfying the equation \(A\xvec=\zerovec\) can be written as a linear combination of a set of vectors. In other words, this description shows why \(\nul(A)\) is the span of a set of vectors and is therefore a subspace. Identify a set of vectors whose span is \(\nul(A)\text{.}\)
The parametric description gives a set of vectors that span \(\nul(B)\text{.}\) Explain why this set of vectors is linearly independent and hence forms a basis. What is the dimension of \(\nul(B)\text{?}\)
To find a parametric description of the solution space to \(A\xvec=\zerovec\text{,}\) imagine that we augment both \(A\) and its reduced row echelon form by a column of zeroes, which leads to the equations
It is straightforward to check that these vectors are linearly independent, which means that \(\vvec_1\text{,}\)\(\vvec_2\text{,}\) and \(\vvec_3\) form a basis for \(\nul(A)\text{,}\) a 3-dimensional subspace of \(\real^5\text{.}\)
As illustrated in this example, the dimension of \(\nul(A)\) is equal to the number of free variables in the equation \(A\xvec=\zerovec\text{,}\) which equals the number of columns of \(A\) without pivot positions or the number of columns of \(A\) minus the number of pivot positions.
Once again, we find ourselves revisiting our two fundamental questions concerning the existence and uniqueness of solutions to linear systems. The column space \(\col(A)\) contains all the vectors \(\bvec\) for which the equation \(A\xvec = \bvec\) is consistent. The null space \(\nul(A)\) is the solution space to the equation \(A\xvec = \zerovec\text{,}\) which reflects on the uniqueness of solutions to this and other equations.
A subspace \(S\) of \(\real^p\) is a subset of \(\real^p\) that can be represented as the span of a set of vectors. A basis of \(S\) is a linearly independent set of vectors whose span is \(S\text{.}\)
A basis for \(\nul(A)\) is found through a parametric description of the solution space of \(A\xvec =
\zerovec\text{,}\) and we have that \(\dim~\nul(A) = n -
\rank(A)\text{.}\)
The null space \(\nul(A)\) is a subspace of \(\real^p\) for what \(p\text{?}\) The column space \(\col(A)\) is a subspace of \(\real^p\) for what \(p\text{?}\)
Determine whether the following statements are true or false and provide a justification for your response. Unless otherwise stated, assume that \(A\) is an \(m\times n\) matrix.
If \(A\) is a \(127\times 341\) matrix, then \(\nul(A)\) is a subspace of \(\real^{127}\text{.}\)
In this section, we saw that the solution space to the homogeneous equation \(A\xvec = \zerovec\) is a subspace of \(\real^p\) for some \(p\text{.}\) In this exercise, we will investigate whether the solution space to another equation \(A\xvec = \bvec\) can form a subspace.
