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

David Austin, Erica Bernstein

Section 6.3 The span of vectors in \(\real^3\)

Activity 6.3.1.

In this activity, we will look at the span of sets of vectors in \(\real^3\text{.}\)
  1. Suppose \(\vvec=\threevec{1}{2}{1}\text{.}\) Give a geometric description of \(\laspan{\vvec}\) and a rough sketch of \(\vvec\) and its span in Figure 6.3.1.
    Figure 6.3.1. A three-dimensional coordinate system for sketching \(\vvec\) and its span.
  2. Now consider the two vectors
    \begin{equation*} \evec_1 = \threevec{1}{0}{0},~~~ \evec_2 = \threevec{0}{1}{0}\text{.} \end{equation*}
    Sketch the vectors below. Then give a geometric description of \(\laspan{\evec_1,\evec_2}\) and a rough sketch of the span in Figure 6.3.2.
    Figure 6.3.2. A coordinate system for sketching \(\evec_1\text{,}\) \(\evec_2\text{,}\) and \(\laspan{\evec_1,\evec_2}\text{.}\)
  3. Let’s now look at this situation algebraically by writing write \(\bvec = \threevec{b_1}{b_2}{b_3}\text{.}\) Determine the conditions on \(b_1\text{,}\) \(b_2\text{,}\) and \(b_3\) so that \(\bvec\) is in \(\laspan{\evec_1,\evec_2}\) by considering the linear system
    \begin{equation*} \left[\begin{array}{rr} \evec_1 \amp \evec_2 \\ \end{array}\right] ~\xvec = \bvec \end{equation*}
    or
    \begin{equation*} \left[\begin{array}{rr} 1 \amp 0 \\ 0 \amp 1 \\ 0 \amp 0 \\ \end{array}\right] \xvec = \threevec{b_1}{b_2}{b_3}\text{.} \end{equation*}
    Explain how this relates to your sketch of \(\laspan{\evec_1,\evec_2}\text{.}\)
  4. Consider the vectors
    \begin{equation*} \vvec_1 = \threevec{1}{1}{-1},~~ \vvec_2 = \threevec{0}{2}{1}. \end{equation*}
    1. Is the vector \(\bvec=\threevec{1}{-2}{4}\) in \(\laspan{\vvec_1,\vvec_2}\text{?}\)
    2. Is the vector \(\bvec=\threevec{-2}{0}{3}\) in \(\laspan{\vvec_1,\vvec_2}\text{?}\)
    3. Give a geometric description of \(\laspan{\vvec_1,\vvec_2}\text{.}\)
  5. Consider the vectors
    \begin{equation*} \vvec_1 = \threevec{1}{1}{-1}, \vvec_2 = \threevec{0}{2}{1}, \vvec_3 = \threevec{1}{-2}{4}\text{.} \end{equation*}
    Form the matrix \(\left[\begin{array}{rrrr} \vvec_1 \amp \vvec_2 \amp \vvec_3 \end{array}\right]\) and find its reduced row echelon form.
    What does this tell you about \(\laspan{\vvec_1,\vvec_2,\vvec_3}\text{?}\)
  6. If the span of a set of vectors \(\vvec_1,\vvec_2,\ldots,\vvec_n\) is \(\real^3\text{,}\) what can you say about the pivot positions of the matrix \(\left[\begin{array}{rrrr} \vvec_1\amp\vvec_2\amp\ldots\amp\vvec_n \end{array}\right]\text{?}\)
  7. What is the smallest number of vectors such that \(\laspan{\vvec_1,\vvec_2,\ldots,\vvec_n} = \real^3\text{?}\)
Solution.
  1. \(\laspan{\vvec}\) is the line defined by \(\vvec\text{.}\)
  2. \(\laspan{\evec_1,\evec_2}\) is the \(xy\)-plane.
  3. For the linear system to be consistent, we need \(b_3=0\text{,}\) which means that the third coordinate of the vector \(\bvec\) must be 0 for \(\bvec\) to be in \(\laspan{\evec_1,\evec_2}\text{.}\) In other words, \(\bvec\) must lie in the \(xy\)-plane.
  4. We consider the two cases.
    1. We have the augmented matrix
      \begin{equation*} \left[\begin{array}{rr|r} 1 \amp 0 \amp 1 \\ 1 \amp 2 \amp -2 \\ -1 \amp 1 \amp 4 \end{array}\right] \sim \left[\begin{array}{rr|r} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{array}\right]\text{,} \end{equation*}
      which shows that the system is inconsistent. Therefore, \(\bvec\) is not in \(\laspan{\vvec_1, \vvec_2}\text{.}\)
    2. We have the augmented matrix
      \begin{equation*} \left[\begin{array}{rr|r} 1 \amp 0 \amp -2 \\ 1 \amp 2 \amp 0 \\ -1 \amp 1 \amp 3 \end{array}\right] \sim \left[\begin{array}{rr|r} 1 \amp 0 \amp -2 \\ 0 \amp 1 \amp 1 \\ 0 \amp 0 \amp 0 \end{array}\right]\text{,} \end{equation*}
      which shows that the system is consistent. Therefore, \(\bvec\) is in \(\laspan{\vvec_1, \vvec_2}\text{.}\)
    3. The span is the plane in \(\real^3\) defined by \(\vvec_1\) and \(\vvec_2\text{.}\)
  5. We have the reduced row echelon form
    \begin{equation*} \left[\begin{array}{rrr} 1 \amp 0 \amp 1 \\ 1 \amp 2 \amp -2 \\ -1 \amp 1 \amp 4 \end{array}\right] \sim \left[\begin{array}{rrr} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{array}\right]\text{.} \end{equation*}
    Since there is a pivot position in every row, this says that every equation \(A\xvec=\bvec\) is consistent. The \(\laspan{\vvec_1, \vvec_2, \vvec_3}\) is therefore \(\real^3\text{.}\)
  6. There must be a pivot position in every row.
  7. If a set of vectors spans \(\real^3\text{,}\) its corresponding matrix must have a pivot position in every row. Because there can be at most one pivot position in a column, there must be at least three columns. Therefore, the smallest number of vectors that span \(\real^3\) is three.
The types of sets that appear as the span of a set of vectors in \(\real^3\) 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 6.3.3.
    Figure 6.3.3. The span of a single nonzero vector is a line.
    Notice that the matrix formed by this vector has one pivot position. For example,
    \begin{equation*} \threevec{-2}{3}{1} \sim \threevec{1}{0}{0}\text{.} \end{equation*}
  • The span of two vectors in \(\real^3\) that do not lie on the same line will be a plane, as seen in Figure 6.3.4.
    Figure 6.3.4. The span of these two vectors in \(\real^3\) is a plane.
    For example, the vectors
    \begin{equation*} \vvec_1=\threevec{-2}{3}{1},~~~ \vvec_2=\threevec{1}{-1}{3} \end{equation*}
    lead to the matrix
    \begin{equation*} \left[\begin{array}{rr} -2 \amp 1 \\ 3 \amp -1 \\ 1 \amp 3 \\ \end{array}\right] \sim \left[\begin{array}{rr} 1 \amp 0 \\ 0 \amp 1 \\ 0 \amp 0 \\ \end{array}\right] \end{equation*}
    with two pivot positions.
  • Finally, a set of three vectors, such as
    \begin{equation*} \vvec_1=\threevec12{-1},~~~ \vvec_2=\threevec201,~~~ \vvec_3=\threevec{-2}20 \end{equation*}
    may form a matrix having three pivot positions
    \begin{equation*} \left[\begin{array}{rrr} \vvec_1 \amp \vvec_2 \amp \vvec_3 \end{array}\right] = \left[\begin{array}{rrr} 1 \amp 2 \amp -2 \\ 2 \amp 0 \amp 2 \\ -1 \amp 1 \amp 0 \\ \end{array}\right] \sim \left[\begin{array}{rrr} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \\ \end{array}\right], \end{equation*}
    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:
    \begin{equation*} \left[\begin{array}{rrr|r} 1 \amp 2 \amp -2 \amp *\\ 2 \amp 0 \amp 2 \amp * \\ -1 \amp 1 \amp 0 \amp * \\ \end{array}\right] \sim \left[\begin{array}{rrr|r} 1 \amp 0 \amp 0 \amp *\\ 0 \amp 1 \amp 0 \amp * \\ 0 \amp 0 \amp 1 \amp * \\ \end{array}\right]. \end{equation*}
    Therefore, any linear system \(\begin{bmatrix}\vvec_1\amp\vvec_2\amp\vvec_3\end{bmatrix} ~\xvec = \bvec\) is consistent, which tells us that \(\laspan{\vvec_1,\vvec_2,\vvec_3} = \real^3\text{.}\)
To summarize, we looked at the pivot positions in a matrix whose columns are the three-dimensional vectors \(\vvec_1,\vvec_2,\ldots,\vvec_n\text{.}\) We found that with
  • one pivot position, the span was a line.
  • two pivot positions, the span was a plane.
  • three pivot positions, the span was \(\real^3\text{.}\)
Though we will return to these ideas later, for now take note of the fact that the span of a set of vectors in \(\real^3\) is a relatively simple, familiar geometric object.
The reasoning we led us to conclude that the span of a set of vectors is \(\real^3\) when the associated matrix has a pivot position in every row applies more generally.
This tells us something important about the number of vectors needed to span \(\real^m\text{.}\) Suppose we have \(n\) vectors \(\vvec_1,\vvec_2,\ldots,\vvec_n\) that span \(\real^m\text{.}\) The proposition tells us that the matrix \(A = \left[\begin{array}{rrrr} \vvec_1\amp\vvec_2\amp\ldots\amp\vvec_n \end{array}\right]\) has a pivot position in every row, such as in this reduced row echelon matrix.
\begin{equation*} \left[\begin{array}{rrrrrr} 1 \amp 0 \amp * \amp 0 \amp * \amp 0 \\ 0 \amp 1 \amp * \amp 0 \amp * \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \amp * \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 1 \\ \end{array}\right]. \end{equation*}
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 \(n\geq m\text{.}\) For instance, if we have a set of vectors that span \(\real^{632}\text{,}\) there must be at least 632 vectors in the set.
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