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Understanding Linear Algebra

David Austin

Section 2.3 The span of a set of vectors

Matrix multiplication allows us to rewrite a linear system in the form Ax=b. Besides being a more compact way of expressing a linear system, this form allows us to think about linear systems geometrically since matrix multiplication is defined in terms of linear combinations of vectors.
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We now return to our two fundamental questions, rephrased here in terms of matrix multiplication.
  • Existence: Is there a solution to the equation Ax=b?
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  • Uniqueness: If there is a solution to the equation Ax=b, is it unique?
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In this section, we focus on the existence question and see how it leads to the concept of the span of a set of vectors.
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Preview Activity 2.3.1. The existence of solutions.

(a)

If the equation Ax=b is inconsistent, what can we say about the pivot positions of the augmented matrix [Ab]?
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(b)

Consider the matrix A
A=[10−2−22211−3].
If b=[225], is the equation Ax=b consistent? If so, find a solution.
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(c)

If b=[226], is the equation Ax=b consistent? If so, find a solution.
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(d)

Identify the pivot positions of A.
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(e)

For our two choices of the vector b, one equation Ax=b has a solution and the other does not. What feature of the pivot positions of the matrix A tells us to expect this?
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Subsection 2.3.1 The span of a set of vectors

In the preview activity, we considered a 3×3 matrix A and found that the equation Ax=b has a solution for some vectors b in R3 and has no solution for others. We will introduce a concept called span that describes the vectors b for which there is a solution.
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We can write an m×n matrix A in terms of its columns
A=[v1v2⋯vn].
Remember that Proposition 2.2.4 says that the equation Ax=b is consistent if and only if we can express b as a linear combination of v1,v2,…,vn.
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Definition 2.3.1.

The span of a set of vectors v1,v2,…,vn is the set of all linear combinations that can be formed from the vectors.
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Alternatively, if A=[v1v2⋯vn], then the span of the vectors consists of all vectors b for which the equation Ax=b is consistent.
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Example 2.3.2.

Considering the set of vectors v=[−21] and w=[8−4], we see that the vector
b=3v+w=[2−1]
is one vector in the span of the vectors v and w because it is a linear combination of v and w.
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To determine whether the vector b=[52] is in the span of v and w, we form the matrix
A=[vw]=[−281−4]
and consider the equation Ax=b. We have
[−2851−42]∼[1−40001],
which shows that the equation Ax=b is inconsistent. Therefore, b=[52] is one vector that is not in the span of v and w.
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Activity 2.3.2.

Let’s look at two examples to develop some intuition for the concept of span.
  1. First, we will consider the set of vectors
    v=[12],   w=[−2−4].
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    Instructions.
    The diagram below can be used to construct linear combinations whose weights c and d may be varied using the sliders at the top. The vectors v and w are outlined while the linear combination
    cv+dw
    is shaded in red.
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    Figure 2.3.3. An interactive diagram for constructing linear combinations of the vectors v and w.
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    1. What vector is the linear combination of v and w with weights:
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    2. Can the vector [24] be expressed as a linear combination of v and w? Is the vector [24] in the span of v and w?
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    3. Can the vector [30] be expressed as a linear combination of v and w? Is the vector [30] in the span of v and w?
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    4. Describe the set of vectors in the span of v and w.
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    5. For what vectors b does the equation
      [1−22−4]x=b
      have a solution?
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  2. We will now look at an example where
    v=[21],   w=[12].
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    Instructions.
    In a similar way, the diagram below can be used to construct linear combinations
    cv+dw.
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    Figure 2.3.4. An interactive diagram for constructing linear combinations of the vectors v and w.
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    1. What vector is the linear combination of v and w with weights:
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    2. Can the vector [−22] be expressed as a linear combination of v and w? Is the vector [−22] in the span of v and w?
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    3. Can the vector [30] be expressed as a linear combination of v and w? Is the vector [30] in the span of v and w?
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    4. Describe the set of vectors in the span of v and w.
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    5. For what vectors b does the equation
      [2112]x=b
      have a solution?
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This activity aims to convey the geometric meaning of span. Remember that we can think of a linear combination of the two vectors v and w as a recipe for walking in the plane R2. We first move a prescribed amount in the direction of v and then a prescribed amount in the direction of w. The span consists of all the places we can walk to.
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Example 2.3.5.

Let’s consider the vectors v=[20] and w=[−11] as shown in Figure 2.3.6.
Figure 2.3.6. The vectors v and w and some linear combinations they create.
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The figure shows us that b=v+2w=[02] is a linear combination of v and w. Indeed, we can verify this algebraically by constructing the linear system
[vw] x=[02],
whose corresponding augmented matrix has the reduced row echelon form
[2−10012]∼[101012].
Because this system is consistent, we know that b=[02] is in the span of v and w.
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In fact, we can say more. Notice that the coefficient matrix
[2−101]∼[1001]
has a pivot position in every row. This means that for any other vector b, the augmented matrix corresponding to the equation [vw] x=b cannot have a pivot position in its rightmost column:
[2−1∗01∗]∼[10∗01∗].
Therefore, the equation [vw] x=b is consistent for every two-dimensional vector b, which tells us that every two-dimensional vector is in the span of v and w. In this case, we say that the span of v and w is R2.
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The intuitive meaning is that we can walk to any point in the plane by moving an appropriate distance in the v and w directions.
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Example 2.3.7.

Now let’s consider the vectors v=[−11] and w=[2−2] as shown in Figure 2.3.8.
Figure 2.3.8. The vectors v and w and some linear combinations they create.
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From the figure, we expect that b=[02] is not a linear combination of v and w. Once again, we can verify this algebraically by constructing the linear system
[vw] x=[02].
The augmented matrix has the reduced row echelon form
[−1201−22]∼[1−20001],
from which we see that the system is inconsistent. Therefore, b=[02] is not in the span of v and w.
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We should expect this behavior from the coefficient matrix
[−121−2]∼[1−200].
Because the second row of the coefficient matrix does not have a pivot position, it is possible for a linear system [vw] x=b to have a pivot position in its rightmost column:
[−12∗1−2∗]∼[1−20001].
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If we notice that w=−2v, we see that any linear combination of v and w,
cv+dw=cv−2dv=(c−2d)v,
is actually a scalar multiple of v. Therefore, the span of v and w is the line defined by the vector v. Intuitively, this means that we can only walk to points on this line using these two vectors.
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Notation 2.3.9.

We will denote the span of the set of vectors v1,v2,…,vn by Span{v1,v2,…,vn}.
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In Example 2.3.5, we saw that Span{v,w}=R2. However, for the vectors in Example 2.3.7, we saw that Span{v,w} is simply a line.
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Subsection 2.3.2 Pivot positions and span

A set of vectors v1,v2,…,vn naturally defines a matrix A=[v1v2⋯vn] whose columns are the given vectors. As we’ve seen, a vector b is in Span{v1,v2,…,vn} precisely when the linear system Ax=b is consistent.
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The previous examples point to the fact that the span is related to the pivot positions of A. While Section 2.4 and Section 3.5 develop this idea more fully, we will now examine the possibilities in R3.
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Activity 2.3.3.

In this activity, we will look at the span of sets of vectors in R3.
  1. Suppose v=[121]. Give a geometric description of Span{v} and a rough sketch of v and its span in Figure 2.3.10.
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    Figure 2.3.10. A three-dimensional coordinate system for sketching v and its span.
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  2. Now consider the two vectors
    e1=[100],   e2=[010].
    Sketch the vectors below. Then give a geometric description of Span{e1,e2} and a rough sketch of the span in Figure 2.3.11.
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    Figure 2.3.11. A coordinate system for sketching e1, e2, and Span{e1,e2}.
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  3. Let’s now look at this situation algebraically by writing write b=[b1b2b3]. Determine the conditions on b1, b2, and b3 so that b is in Span{e1,e2} by considering the linear system
    [e1e2] x=b
    or
    [100100]x=[b1b2b3].
    Explain how this relates to your sketch of Span{e1,e2}.
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  4. Consider the vectors
    v1=[11−1],  v2=[021].
    1. Is the vector b=[1−24] in Span{v1,v2}?
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    2. Is the vector b=[−203] in Span{v1,v2}?
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    3. Give a geometric description of Span{v1,v2}.
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  5. Consider the vectors
    v1=[11−1],v2=[021],v3=[1−24].
    Form the matrix [v1v2v3] and find its reduced row echelon form.
    What does this tell you about Span{v1,v2,v3}?
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  6. If the span of a set of vectors v1,v2,…,vn is R3, what can you say about the pivot positions of the matrix [v1v2…vn]?
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  7. What is the smallest number of vectors such that Span{v1,v2,…,vn}=R3?
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The types of sets that appear as the span of a set of vectors in R3 are relatively simple.
  • First, with a single nonzero vector, all linear combinations are simply scalar multiples of that vector so that the span of this vector is a line, as shown in Figure 2.3.12.
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    Figure 2.3.12. The span of a single nonzero vector is a line.
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    Notice that the matrix formed by this vector has one pivot position. For example,
    [−231]∼[100].
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  • The span of two vectors in R3 that do not lie on the same line will be a plane, as seen in Figure 2.3.13.
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    Figure 2.3.13. The span of these two vectors in R3 is a plane.
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    For example, the vectors
    v1=[−231],   v2=[1−13]
    lead to the matrix
    [−213−113]∼[100100]
    with two pivot positions.
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  • Finally, a set of three vectors, such as
    v1=[12−1],   v2=[201],   v3=[−220]
    may form a matrix having three pivot positions
    [v1v2v3]=[12−2202−110]∼[100010001],
    one in every row. When this happens, no matter how we augment this matrix, it is impossible to obtain a pivot position in the rightmost column:
    [12−2∗202∗−110∗]∼[100∗010∗001∗].
    Therefore, any linear system [v1v2v3] x=b is consistent, which tells us that Span{v1,v2,v3}=R3.
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To summarize, we looked at the pivot positions in a matrix whose columns are the three-dimensional vectors v1,v2,…,vn. We found that with
  • one pivot position, the span was a line.
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  • two pivot positions, the span was a plane.
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  • three pivot positions, the span was R3.
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Though we will return to these ideas later, for now take note of the fact that the span of a set of vectors in R3 is a relatively simple, familiar geometric object.
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The reasoning that led us to conclude that the span of a set of vectors is R3 when the associated matrix has a pivot position in every row applies more generally.
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This tells us something important about the number of vectors needed to span Rm. Suppose we have n vectors v1,v2,…,vn that span Rm. The proposition tells us that the matrix A=[v1v2…vn] has a pivot position in every row, such as in this reduced row echelon matrix.
[10∗0∗001∗0∗00001∗0000001].
Since a matrix can have at most one pivot position in a column, there must be at least as many columns as there are rows, which implies that n≥m. For instance, if we have a set of vectors that span R632, there must be at least 632 vectors in the set.
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We have thought about a linear combination of a set of vectors v1,v2,…,vn as the result of walking a certain distance in the direction of v1, followed by walking a certain distance in the direction of v2, and so on. If Span{v1,v2,…,vn}=Rm, this means that we can walk to every point in Rm using the directions v1,v2,…,vn. Intuitively, this proposition is telling us that we need at least m directions to have the flexibility needed to reach every point in Rm.
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Terminology.

Because span is a concept that is connected to a set of vectors, we say, “The span of the set of vectors v1,v2,…,vn is ....” While it may be tempting to say, “The span of the matrix A is ...,” we should instead say “The span of the columns of the matrix A is ....”
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Subsection 2.3.3 Summary

We defined the span of a set of vectors and developed some intuition for this concept through a series of examples.
  • The span of a set of vectors v1,v2,…,vn is the set of linear combinations of the vectors. We denote the span by Span{v1,v2,…,vn}.
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  • A vector b is in Span{v1,v2,…,vn} if and only if the linear system
    [v1v2…vn] x=b
    is consistent.
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  • If the m×n matrix
    [v1v2…vn]
    has a pivot position in every row, then the span of these vectors is Rm; that is,
    Span{v1,v2,…,vn}=Rm.
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  • Any set of vectors that spans Rm must have at least m vectors.
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Exercises 2.3.4 Exercises

1.

In this exercise, we will consider the span of some sets of two- and three-dimensional vectors.
  1. Consider the vectors
    v1=[1−2],v2=[43].
    1. Is b=[21] in Span{v1,v2}?
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    2. Give a geometric description of Span{v1,v2}.
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  2. Consider the vectors
    v1=[213],v2=[−202],v3=[61−1].
    1. Is the vector b=[−10−15] in Span{v1,v2,v3}?
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    2. Is the vector v3 in Span{v1,v2,v3}?
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    3. Is the vector b=[33−1] in Span{v1,v2,v3}?
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    4. Give a geometric description of Span{v1,v2,v3}.
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2.

Provide a justification for your response to the following questions.
  1. Suppose you have a set of vectors v1,v2,…,vn. Can you guarantee that 0 is in Span{v1v2,…,vn}?
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  2. Suppose that A is an m×n matrix. Can you guarantee that the equation Ax=0 is consistent?
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  3. What is Span{0,0,…,0}?
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3.

For both parts of this exercise, give a geometric description of sets of the vectors b and include a sketch.
  1. For which vectors b in R2 is the equation
    [3−6−24]x=b
    consistent?
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  2. For which vectors b in R2 is the equation
    [3−6−22]x=b
    consistent?
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4.

Consider the following matrices:
A=[30−111−1373−215−1223],   B=[30−141−13−13−213−1221].
Do the columns of A span R4? Do the columns of B span R4?
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5.

Determine whether the following statements are true or false and provide a justification for your response. Throughout, we will assume that the matrix A has columns v1,v2,…,vn; that is,
A=[v1v2…vn].
  1. If the equation Ax=b is consistent, then b is in Span{v1,v2,…,vn}.
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  2. The equation Ax=v1 is consistent.
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  3. If v1, v2, v3, and v4 are vectors in R3, then their span is R3.
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  4. If b is a linear combination of v1,v2,…,vn, then b is in Span{v1,v2,…,vn}.
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  5. If A is an 8032×427 matrix, then the span of the columns of A is a set of vectors in R427.
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6.

This exercise asks you to construct some matrices whose columns span a given set.
  1. Construct a 3×3 matrix whose columns span R3.
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  2. Construct a 3×3 matrix whose columns span a plane in R3.
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  3. Construct a 3×3 matrix whose columns span a line in R3.
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7.

Provide a justification for your response to the following questions.
  1. Suppose that we have vectors in R8, v1,v2,…,v10, whose span is R8. Can every vector b in R8 be written as a linear combination of v1,v2,…,v10?
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  2. Suppose that we have vectors in R8, v1,v2,…,v10, whose span is R8. Can every vector b in R8 be written uniquely as a linear combination of v1,v2,…,v10?
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  3. Do the vectors
    e1=[100],e2=[010],e3=[001]
    span R3?
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  4. Suppose that v1,v2,…,vn span R438. What can you guarantee about the value of n?
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  5. Can 17 vectors in R20 span R20?
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8.

The following observation will be helpful in this exercise. If we want to find a solution to the equation ABx=b, we could first find a solution to the equation Ay=b and then find a solution to the equation Bx=y.
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Suppose that A is a 3×4 matrix whose columns span R3 and B is a 4×5 matrix. In this case, we can form the product AB.
  1. What is the shape of the product AB?
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  2. Can you guarantee that the columns of AB span R3?
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  3. If you know additionally that the span of the columns of B is R4, can you guarantee that the columns of AB span R3?
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9.

Suppose that A is a 12×12 matrix and that, for some vector b, the equation Ax=b has a unique solution.
  1. What can you say about the pivot positions of A?
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  2. What can you say about the span of the columns of A?
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  3. If c is some other vector in R12, what can you conclude about the equation Ax=c?
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  4. What can you about the solution space to the equation Ax=0?
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10.

Suppose that
v1=[313−1],v2=[0−1−22],v3=[−3−3−75].
  1. Is v3 a linear combination of v1 and v2? If so, find weights such that v3=av1+bv2.
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  2. Show that a linear combination
    av1+bv2+cv3
    can be rewritten as a linear combination of v1 and v2.
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  3. Explain why Span{v1,v2,v3}=Span{v1,v2}.
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11.

As defined in this section, the span of a set of vectors is generated by taking all possible linear combinations of those vectors. This exercise will demonstrate the fact that the span can also be realized as the solution space to a linear system.
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We will consider the vectors
v1=[10−2],v2=[210],v3=[112]
  1. Is every vector in R3 in Span{v1,v2,v3}? If not, describe the span.
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  2. To describe Span{v1,v2,v3} as the solution space of a linear system, we will write
    b=[abc].
    If b is in Span{v1,v2,v3}, then the linear system corresponding to the augmented matrix
    [121a011b−202c]
    must be consistent. This means that a pivot cannot occur in the rightmost column. Perform row operations to put this augmented matrix into a triangular form. Now identify an equation in a, b, and c that tells us when there is no pivot in the rightmost column. The solution space to this equation describes Span{v1,v2,v3}.
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  3. In this example, the matrix formed by the vectors [v1v2v2] has two pivot positions. Suppose we were to consider another example in which this matrix had had only one pivot position. How would this have changed the linear system describing Span{v1,v2,v3}?
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