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

David Austin, Erica Bernstein

Section 7.3 Linear independence and homogeneous equations

Subsection 7.3.1 Homogeneous equations

If \(A\) is a matrix, we call the equation \(A\xvec = \zerovec\) a homogeneous equation. As we’ll see, the uniqueness of solutions to this equation reflects on the linear independence of the columns of \(A\text{.}\)

Activity 7.3.1. Linear independence and homogeneous equations.

  1. Explain why the homogeneous equation \(A\xvec = \zerovec\) is consistent no matter the matrix \(A\text{.}\)
  2. Consider the matrix
    \begin{equation*} A = \left[\begin{array}{rrr} 3 \amp 2 \amp 0 \\ -1 \amp 0 \amp -2 \\ 2 \amp 1 \amp 1 \end{array}\right] \end{equation*}
    whose columns we denote by \(\vvec_1\text{,}\) \(\vvec_2\text{,}\) and \(\vvec_3\text{.}\) Describe the solution space of the homogeneous equation \(A\xvec = \zerovec\) using a parametric description, if appropriate.
  3. Find a nonzero solution to the homogeneous equation and use it to find weights \(c_1\text{,}\) \(c_2\text{,}\) and \(c_3\) such that
    \begin{equation*} c_1\vvec_1 + c_2\vvec_2 + c_3\vvec_3 = \zerovec\text{.} \end{equation*}
  4. Use the equation you found in the previous part to write one of the vectors as a linear combination of the others.
  5. Are the vectors \(\vvec_1\text{,}\) \(\vvec_2\text{,}\) and \(\vvec_3\) linearly dependent or independent?
Solution.
  1. The vector \(\zerovec\) is always a solution.
  2. We have
    \begin{equation*} \left[\begin{array}{rrr} 3 \amp 2 \amp 0 \\ -1 \amp 0 \amp -2 \\ 2 \amp 1 \amp 1 \end{array}\right] \sim \left[\begin{array}{rrr} 1 \amp 0 \amp 2 \\ 0 \amp 1 \amp -3 \\ 0 \amp 0 \amp 0 \end{array}\right]. \end{equation*}
    From the reduced row echelon form, we see that \(x_3\) is a free variable and that we have
    \begin{equation*} \begin{alignedat}{2} x_1 \amp {}={} -2x_3 \\ x_2 \amp {}={} 3x_3\text{.} \\ \end{alignedat} \end{equation*}
    The solution space is then written parametrically as
    \begin{equation*} \xvec=\threevec{x_1}{x_2}{x_3} = x_3\threevec{-2}{3}{1}\text{.} \end{equation*}
  3. If we set \(x_3=1\text{,}\) then we have the solution \(\xvec=\threevec{-2}{3}{1}\text{,}\) which says that
    \begin{equation*} -2\vvec_1+3\vvec_2+\vvec_3 = \zerovec\text{.} \end{equation*}
  4. We may rewrite this expression as \(\vvec_3=2\vvec_1-3\vvec_2\text{,}\) showing that \(\vvec_3\) is a linear combination of \(\vvec_1\) and \(\vvec_2\text{.}\)
  5. The vectors \(\vvec_1\text{,}\) \(\vvec_2\text{,}\) and \(\vvec_3\) are linearly dependent, and we know this in two ways. We have seen how to express one vector as a linear combination of the others. Also, we have seen that the associated matrix has a column without a pivot position.
This activity shows how the solution space of the homogeneous equation \(A\xvec = \zerovec\) indicates whether the columns of \(A\) are linearly dependent or independent. First, we know that the equation \(A\xvec = \zerovec\) always has at least one solution, the vector \(\xvec = \zerovec\text{.}\) Any other solution is a nonzero solution.

Example 7.3.1.

Let’s consider the vectors
\begin{equation*} \vvec_1 = \fourvec2{-4}10,~~~ \vvec_2 = \fourvec113{-2},~~~ \vvec_3 = \fourvec3{-3}4{-2} \end{equation*}
and their associated matrix \(A = \begin{bmatrix} \vvec_1 \amp \vvec_2 \amp \vvec_3 \end{bmatrix} \text{.}\)
The homogeneous equation \(A\xvec = \zerovec\) has the associated augmented matrix
\begin{equation*} \left[ \begin{array}{rrr|r} 2 \amp 1 \amp 3 \amp 0 \\ -4 \amp 1 \amp -3 \amp 0 \\ 1 \amp 3 \amp 4 \amp 0 \\ 0 \amp -2 \amp -2 \amp 0 \\ \end{array} \right] \sim \left[ \begin{array}{rrr|r} 1 \amp 0 \amp 1 \amp 0 \\ 0 \amp 1 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array} \right]. \end{equation*}
Therefore, \(A\) has a column without a pivot position, which tells us that the vectors \(\vvec_1\text{,}\) \(\vvec_2\text{,}\) and \(\vvec_3\) are linearly dependent. However, we can also see this fact in another way.
The reduced row echelon matrix tells us that the homogeneous equation has a free variable so that there must be infinitely many solutions. In particular, we have
\begin{equation*} \begin{aligned} x_1 \amp {}={} -x_3 \\ x_2 \amp {}={} -x_3 \\ \end{aligned} \end{equation*}
so the solutions have the form
\begin{equation*} \xvec = \threevec{x_1}{x_2}{x_3} = \threevec{-x_3}{-x_3}{x_3} = x_3\threevec{-1}{-1}1. \end{equation*}
If we choose \(x_3=1\text{,}\) then we obtain the nonzero solution to the homogeneous equation \(\xvec = \threevec{-1}{-1}1\text{,}\) which implies that
\begin{equation*} A\threevec{-1}{-1}1 = \begin{bmatrix} \vvec_1\amp\vvec_2\amp\vvec_3 \end{bmatrix} \threevec{-1}{-1}1 = -\vvec_1 -\vvec_2 + \vvec_3 = \zerovec. \end{equation*}
In other words,
\begin{equation*} \begin{aligned} -\vvec_1 -\vvec_2 + \vvec_3 \amp{}={} \zerovec \\ \vvec_3 \amp{}={} \vvec_1 + \vvec_2. \\ \end{aligned} \end{equation*}
Because \(\vvec_3\) is a linear combination of \(\vvec_1\) and \(\vvec_2\text{,}\) we know that this set of vectors is linearly dependent.
As this example demonstrates, there are many ways we can view the question of linear independence, some of which are recorded in the following proposition.

Subsection 7.3.2 Summary

This section developed the concept of linear dependence of a set of vectors. More specifically, we saw that:
  • A set of vectors is linearly dependent if one of the vectors is a linear combination of the others.
  • A set of vectors is linearly independent if and only if the vectors form a matrix that has a pivot position in every column.
  • A set of linearly independent vectors in \(\real^m\) contains no more than \(m\) vectors.
  • The columns of the matrix \(A\) are linearly dependent if the homogeneous equation \(A\xvec = \zerovec\) has a nonzero solution.
  • A set of vectors \(\vvec_1,\vvec_2,\ldots,\vvec_n\) is linearly dependent if there are weights \(c_1,c_2,\ldots,c_n\text{,}\) not all of which are zero, such that
    \begin{equation*} c_1\vvec_1 + c_2\vvec_2 + \ldots + c_n\vvec_n = \zerovec\text{.} \end{equation*}
At the beginning of the section, we said that this concept addressed the second of our [cross-reference to target(s) "fundamental-questions" missing or not unique] concerning the uniqueness of solutions to a linear system. It is worth comparing the results of this section with those of the previous one so that the parallels between them become clear.
As usual, we will write a matrix as a collection of vectors,
\begin{equation*} A=\left[\begin{array}{rrrr} \vvec_1\amp\vvec_2 \amp \ldots \amp \vvec_n \end{array}\right]. \end{equation*}
Table 7.3.3. Span and Linear Independence
Span Linear independence
A vector \(\bvec\) is in the span of a set of vectors if it is a linear combination of those vectors.
A set of vectors is linearly dependent if one of the vectors is a linear combination of the others.
A vector \(\bvec\) is in the span of \(\vvec_1, \vvec_2, \ldots, \vvec_n\) if there exists a solution to \(A\xvec = \bvec\text{.}\)
The vectors \(\vvec_1, \vvec_2, \ldots, \vvec_n\) are linearly independent if \(\xvec=\zerovec\) is the unique solution to \(A\xvec = \zerovec\text{.}\)
The columns of an \(m\times n\) matrix span \(\real^m\) if the matrix has a pivot position in every row.
The columns of a matrix are linearly independent if the matrix has a pivot position in every column.
A set of vectors that span \(\real^m\) has at least \(m\) vectors.
A set of linearly independent vectors in \(\real^m\) has at most \(m\) vectors.
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