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

David Austin, Erica Bernstein

Section 1.2 linear combinations

Activity 1.2.1.

In this activity, we will look at linear combinations of a pair of vectors, \(\vvec = \left[\begin{array}{r} 2 \\ 1 \end{array}\right]\) and \(\wvec = \left[\begin{array}{r} 1 \\ 2 \end{array}\right] \text{.}\)

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 red.
Figure 1.2.1. Linear combinations of vectors \(\vvec\) and \(\wvec\text{.}\)
  1. The weight \(d\) is initially set to 0. Explain what happens as you vary \(c\) while keeping \(d=0\text{.}\) How is this related to scalar multiplication?
  2. What is the linear combination of \(\vvec\) and \(\wvec\) when \(c = 1\) and \(d=-2\text{?}\) You may find this result using the diagram, but you should also verify it by computing the linear combination.
  3. Describe the vectors that arise when the weight \(d\) is set to 1 and \(c\) is varied.
  4. Can the vector \(\left[\begin{array}{r} 0 \\ 0 \end{array} \right]\) be expressed as a linear combination of \(\vvec\) and \(\wvec\text{?}\) If so, what are the weights \(c\) and \(d\text{?}\)
  5. Can the vector \(\left[\begin{array}{r} 3 \\ 0 \end{array} \right]\) be expressed as a linear combination of \(\vvec\) and \(\wvec\text{?}\) If so, what are the weights \(c\) and \(d\text{?}\)
  6. Verify the result from the previous part by algebraically computing the linear combination \(c\vvec+d\wvec \) to make sure you get the vector \(\left[\begin{array}{r} 3 \\ 0 \end{array} \right]\text{.}\)
  7. Can the vector \(\left[\begin{array}{r} -3 \\ 0 \end{array} \right]\) be expressed as a linear combination of \(\vvec\) and \(\wvec\text{?}\) If so, what are the weights \(c\) and \(d\text{?}\)
  8. Verify the result from the previous part by algebraically computing the linear combination \(c\vvec+d\wvec \) to make sure you get the vector \(\left[\begin{array}{r} -3 \\ 0 \end{array} \right]\text{.}\)
  9. Can the vector \(\left[\begin{array}{r} 1.3 \\ -1.7 \end{array} \right]\) be expressed as a linear combination of \(\vvec\) and \(\wvec\text{?}\) What about the vector \(\left[\begin{array}{r} 15.2 \\ 7.1 \end{array} \right]\text{?}\)
  10. Are there any two-dimensional vectors that cannot be expressed as linear combinations of \(\vvec\) and \(\wvec\text{?}\)
Answer.
  1. The linear combinations lie on the line defined by \(\vvec\text{.}\)
  2. \(\twovec{0}{-3}\text{.}\)
  3. They lie on the line parallel to \(\vvec\) going through through the tip of \(\wvec\text{.}\)
  4. Yes, with weights \(c=d=0\text{.}\)
  5. Yes, with weights \(c=2\) and \(d=-1\text{.}\)
  6. This can be done by writing the appropriate linear system for the weights.
  7. No, any two-dimensional vector can be expressed as a linear combination of \(\vvec\) and \(\wvec\text{.}\)
Solution.
  1. When we vary \(c\) with \(d=0\text{,}\) the linear combination moves along the line defined by \(\vvec\text{.}\)
  2. When \(c=1\) and \(d=-2\text{,}\) we find
    \begin{equation*} 1\vvec -2 \wvec = \twovec{0}{-3}\text{.} \end{equation*}
  3. When \(d=1\) and \(c\) is allowed to vary, the linear combinations lie on the line parallel to \(\vvec\) going through the tip of \(\wvec\) .
  4. If the weights \(c=0\) and \(d=0\text{,}\) then the linear combination is the vector \(\twovec{0}{0}\text{.}\)
  5. If the weights \(c=2\) and \(d=-1\text{,}\) then the linear combination is the vector \(\twovec{3}{0}\text{.}\)
  6. We find the linear system for the weights:
    \begin{equation*} \begin{alignedat}{3} 2c \amp {}+{} \amp 1d \amp {}={} \amp 3 \\ 1c \amp {}+{} \amp 2d \amp {}={} \amp 0 \\ \end{alignedat} \end{equation*}
    If we construct the corresponding augmented matrix and determine its reduced row echelon matrix, we find the weights \(c=2\) and \(d=-1\text{.}\)
  7. In the same way, we construct a linear system for the weights whose augmented matrix is
    \begin{equation*} \left[\begin{array}{rr|r} 2 \amp 1 \amp 1.3 \\ 1 \amp 2 \amp -1.7 \\ \end{array}\right] \sim \left[\begin{array}{rr|r} 1 \amp 0 \amp 1.43 \\ 0 \amp 1 \amp -1.57 \\ \end{array}\right] \text{,} \end{equation*}
    which shows that there are weights that produce the desired linear combination. The same will happen for any vector \(\bvec\) that we ask to write as a linear combination of \(\vvec\) and \(\wvec\text{.}\)
  8. Every two-dimensional vector can be written as a linear combination of \(\vvec\) and \(\wvec\) because the coefficient matrix of the linear system remains the same. Since that coefficient matrix has a pivot position in every row, the augmented matrix can never have a pivot position in the rightmost column.
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