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Math 211 Activities Workbook: Adapted from: Understanding Linear Algebra by David Austin

David Austin, Erica Bernstein

Section 6.2 The span of vectors in \(\real^2\)

Activity 6.2.1.

Let’s look at two examples to develop some intuition for the concept of span.
  1. First, we will consider the set of vectors
    \begin{equation*} \vvec = \twovec{1}{2}, ~~~\wvec = \twovec{-2}{-4}\text{.} \end{equation*}

    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 \(\vvec\) and \(\wvec\) are outlined while the linear combination
    \begin{equation*} c\vvec + d\wvec \end{equation*}
    is shaded in red.
    Figure 6.2.1. An interactive diagram for constructing linear combinations of the vectors \(\vvec\) and \(\wvec\text{.}\)
    1. What vector is the linear combination of \(\vvec\) and \(\wvec\) with weights:
      • \(c = 2\) and \(d=0\text{?}\)
      • \(c = 1\) and \(d=1\text{?}\)
      • \(c = 0\) and \(d=-1\text{?}\)
    2. Can the vector \(\twovec{2}{4}\) be expressed as a linear combination of \(\vvec\) and \(\wvec\text{?}\) Is the vector \(\twovec{2}{4}\) in the span of \(\vvec\) and \(\wvec\text{?}\)
    3. Can the vector \(\twovec{3}{0}\) be expressed as a linear combination of \(\vvec\) and \(\wvec\text{?}\) Is the vector \(\twovec{3}{0}\) in the span of \(\vvec\) and \(\wvec\text{?}\)
    4. Describe the set of vectors in the span of \(\vvec\) and \(\wvec\text{.}\)
    5. For what vectors \(\bvec\) does the equation
      \begin{equation*} \left[\begin{array}{rr} 1 \amp -2 \\ 2 \amp -4 \end{array}\right] \xvec = \bvec \end{equation*}
      have a solution?
  2. We will now look at an example where
    \begin{equation*} \vvec = \twovec{2}{1}, ~~~\wvec = \twovec{1}{2}\text{.} \end{equation*}

    Instructions.

    In a similar way, the diagram below can be used to construct linear combinations
    \begin{equation*} c\vvec + d\wvec. \end{equation*}
    Figure 6.2.2. An interactive diagram for constructing linear combinations of the vectors \(\vvec\) and \(\wvec\text{.}\)
    1. What vector is the linear combination of \(\vvec\) and \(\wvec\) with weights:
      • \(c = 2\) and \(d=0\text{?}\)
      • \(c = 1\) and \(d=1\text{?}\)
      • \(c = 0\) and \(d=-1\text{?}\)
    2. Can the vector \(\twovec{-2}{2}\) be expressed as a linear combination of \(\vvec\) and \(\wvec\text{?}\) Is the vector \(\twovec{-2}{2}\) in the span of \(\vvec\) and \(\wvec\text{?}\)
    3. Can the vector \(\twovec{3}{0}\) be expressed as a linear combination of \(\vvec\) and \(\wvec\text{?}\) Is the vector \(\twovec{3}{0}\) in the span of \(\vvec\) and \(\wvec\text{?}\)
    4. Describe the set of vectors in the span of \(\vvec\) and \(\wvec\text{.}\)
    5. For what vectors \(\bvec\) does the equation
      \begin{equation*} \left[\begin{array}{rr} 2 \amp 1 \\ 1 \amp 2 \end{array}\right] \xvec = \bvec \end{equation*}
      have a solution?
Solution.
  1. For the first set of vectors, we find:
    1. We can form the linear combinations:
      • When \(c = 2\) and \(d=0\text{,}\) the linear combination is \(\twovec{2}{4}\text{.}\)
      • When \(c = 1\) and \(d=1\text{,}\) the linear combination is \(\twovec{-1}{-2}\text{.}\)
      • When \(c = 0\) and \(d=-1\text{,}\) the linear combination is \(\twovec{2}{4}\text{.}\)
    2. Yes, we saw that there are at least two ways. For instance, when the weights are \(c=2\) and \(d=0\text{.}\) This means that \(\twovec24\) is in the span of \(\vvec\) and \(\wvec\text{.}\)
    3. No. No matter how we change the weights, the linear combination lies on the line through \(\vvec\) and \(\wvec\text{.}\) This means that \(\twovec{3}{0}\) is not in the span of \(\vvec\) and \(\wvec\text{.}\)
    4. The span of \(\vvec\) and \(\wvec\) is the set of all vectors on the line through \(\vvec\text{.}\)
    5. If the equation has a solution, \(\bvec\) must lie on the line defined by \(\vvec\text{.}\)
  2. For the second set of vectors, we have:
    1. We can form the linear combinations:
      • When \(c = 2\) and \(d=0\text{,}\) the linear combination is \(\twovec42\text{.}\)
      • When \(c = 1\) and \(d=1\text{,}\) the linear combination is \(\twovec{3}{3}\text{.}\)
      • When \(c = 0\) and \(d=-1\text{,}\) the linear combination is \(\twovec{-1}{-2}\text{.}\)
    2. Yes. Using the diagram, we see that \(\twovec{-2}{2}=-2\vvec + 2\wvec\text{.}\) This means that \(\twovec{-2}{2}\) is in the span of \(\vvec\) and \(\wvec\text{.}\)
    3. Yes. Using the diagram, we see that \(\twovec{3}{0}=2\vvec-1\wvec\text{.}\) This means that \(\twovec{3}{0}\) is in the span of \(\vvec\) and \(\wvec\text{.}\)
    4. Every two-dimensional vector is in the span of \(\vvec\) and \(\wvec\text{.}\)
    5. The equation has a solution for every \(\bvec\text{.}\)
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