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

David Austin, Erica Bernstein

Section 7.2 How to recognize linear dependence

Activity 7.2.1.

We would like to develop a means to detect when a set of vectors is linearly dependent. This activity will point the way.
  1. Suppose we have five vectors in \(\real^4\) that form the columns of a matrix having reduced row echelon form
    \begin{equation*} \left[\begin{array}{rrrrr} \vvec_1 \amp \vvec_2 \amp \vvec_3 \amp \vvec_4 \amp \vvec_5 \end{array}\right] \sim \left[\begin{array}{rrrrr} 1 \amp 0 \amp -1 \amp 0 \amp 2 \\ 0 \amp 1 \amp 2 \amp 0 \amp 3 \\ 0 \amp 0 \amp 0 \amp 1 \amp -1 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \\ \end{array}\right]\text{.} \end{equation*}
    Is it possible to write one of the vectors \(\vvec_1,\vvec_2,\ldots,\vvec_5\) as a linear combination of the others? If so, show explicitly how one vector appears as a linear combination of some of the other vectors. Is this set of vectors linearly dependent or independent?
  2. Suppose we have another set of three vectors in \(\real^4\) that form the columns of a matrix having reduced row echelon form
    \begin{equation*} \left[\begin{array}{rrr} \wvec_1 \amp \wvec_2 \amp \wvec_3 \\ \end{array}\right] \sim \left[\begin{array}{rrr} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 0 \\ \end{array}\right]\text{.} \end{equation*}
    Is it possible to write one of these vectors \(\wvec_1\text{,}\) \(\wvec_2\text{,}\) \(\wvec_3\) as a linear combination of the others? If so, show explicitly how one vector appears as a linear combination of some of the other vectors. Is this set of vectors linearly dependent or independent?
  3. By looking at the pivot positions, how can you determine whether the columns of a matrix are linearly dependent or independent?
  4. If one vector in a set is the zero vector \(\zerovec\text{,}\) can the set of vectors be linearly independent?
  5. Suppose a set of vectors in \(\real^{10}\) has twelve vectors. Is it possible for this set to be linearly independent?
Solution.
  1. Let’s focus on the first three vectors and view the matrix as an augmented one:
    \begin{equation*} \left[\begin{array}{rr|r} \vvec_1 \amp \vvec_2 \amp \vvec_3 \end{array}\right] \sim \left[\begin{array}{rr|r} 1 \amp 0 \amp -1 \\ 0 \amp 1 \amp 2 \\ 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \\ \end{array}\right]\text{.} \end{equation*}
    This shows that \(\vvec_3=-\vvec_1+2\vvec_2\) so it is possible to write one of the vectors as a linear combination of the others. Therefore, the set is linearly dependent.
  2. Applying the same reasoning as in the previous part, we see that we cannot write any of the vectors as a linear combination of the others. Therefore, the set is linearly independent.
  3. The columns of a matrix are linearly independent exactly when there is a pivot position in every column of the matrix.
  4. No, because we can write the zero vector \(\zerovec\) as a linear combination of the other vectors: \(\zerovec = 0\vvec_2 + \ldots + 0\vvec_n\text{.}\)
  5. No, because the matrix formed by the vectors would have 12 columns and only 10 rows. There can at most be 10 pivot positions so there are at least two columns without pivot positions.
By now, we should expect that the pivot positions play an important role in determining whether the columns of a matrix are linearly dependent. For instance, suppose we have four vectors and their associated matrix
\begin{equation*} \left[\begin{array}{rrrr} \vvec_1 \amp \vvec_2 \amp \vvec_3 \amp \vvec_4 \end{array}\right] \sim \left[\begin{array}{rrrrr} 1 \amp 0 \amp 2 \amp 0 \\ 0 \amp 1 \amp -3 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array}\right]. \end{equation*}
Since the third column does not contain a pivot position, let’s just focus on the first three columns and view them as an augmented matrix:
\begin{equation*} \left[\begin{array}{rr|r} \vvec_1 \amp \vvec_2 \amp \vvec_3 \end{array}\right] \sim \left[\begin{array}{rr|r} 1 \amp 0 \amp 2 \\ 0 \amp 1 \amp -3 \\ 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \\ \end{array}\right]. \end{equation*}
This says that
\begin{equation*} \vvec_3 = 2\vvec_1 -3\vvec_2, \end{equation*}
which tells us that the set of vectors \(\vvec_1,\vvec_2,\vvec_3,\vvec_4\) is linearly dependent. Moreover, we see that
\begin{equation*} \laspan{\vvec_1,\vvec_2,\vvec_3,\vvec_4} = \laspan{\vvec_1,\vvec_2,\vvec_4}. \end{equation*}
More generally, the same reasoning implies that a set of vectors is linearly dependent if the associated matrix has a column without a pivot position. Indeed, as illustrated here, a vector corresponding to a column without a pivot position can be expressed as a linear combination of the vectors whose columns do contain pivot positions.
Suppose instead that the matrix associated to a set of vectors has a pivot position in every column.
\begin{equation*} \left[\begin{array}{rrrr} \wvec_1 \amp \wvec_2 \amp \wvec_3 \amp \wvec_4 \\ \end{array}\right] \sim \left[\begin{array}{rrrr} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0\\ 0 \amp 0 \amp 1 \amp 0\\ 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array}\right]. \end{equation*}
Viewing this as an augmented matrix again, we see that the linear system is inconsistent since there is a pivot in the rightmost column, which means that \(\wvec_4\) cannot be expressed as a linear combination of the other vectors. Similarly, \(\wvec_3\) cannot be expressed as a linear combination of \(\wvec_1\) and \(\wvec_2\text{.}\) In fact, none of the vectors can be written as a linear combination of the others so this set of vectors is linearly independent.
The following proposition summarizes these findings.
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