Uniqueness of solutions and linear independence

Section 7.1 Uniqueness of solutions and linear independence

Exploration 7.1.1.

Let’s begin by looking at some sets of vectors in \(\real^3\text{.}\) As we saw in the previous section, the span of a set of vectors in \(\real^3\) will be either a line, a plane, or \(\real^3\) itself.

Consider the following vectors in \(\real^3\text{:}\)

\begin{equation*} \vvec_1=\threevec{0}{-1}{2}, \vvec_2=\threevec{3}{1}{-1}, \vvec_3=\threevec{2}{0}{1}\text{.} \end{equation*}
Describe the span of these vectors, \(\laspan{\vvec_1,\vvec_2,\vvec_3}\text{,}\) as a line, a plane, or \(\real^3\text{.}\)
Now consider the set of vectors:

\begin{equation*} \wvec_1=\threevec{0}{-1}{2}, \wvec_2=\threevec{3}{1}{-1}, \wvec_3=\threevec{3}{0}{1}\text{.} \end{equation*}
Describe the span of these vectors, \(\laspan{\wvec_1,\wvec_2,\wvec_3}\text{,}\) as a line, a plane, or \(\real^3\text{.}\)
Show that the vector \(\wvec_3\) is a linear combination of \(\wvec_1\) and \(\wvec_2\) by finding weights such that

\begin{equation*} \wvec_3 = c\wvec_1 + d\wvec_2\text{.} \end{equation*}
Explain why any linear combination of \(\wvec_1\text{,}\) \(\wvec_2\text{,}\) and \(\wvec_3\text{,}\)

\begin{equation*} c_1\wvec_1 + c_2\wvec_2 + c_3\wvec_3 \end{equation*}

can be written as a linear combination of \(\wvec_1\) and \(\wvec_2\text{.}\)
Explain why

\begin{equation*} \laspan{\wvec_1,\wvec_2,\wvec_3} = \laspan{\wvec_1,\wvec_2}\text{.} \end{equation*}

Solution.

We will construct the matrix whose columns are \(\vvec_1\text{,}\) \(\vvec_2\text{,}\) and \(\vvec_3\text{:}\)

\begin{equation*} \begin{bmatrix} 0 \amp 3 \amp 2 \\ -1 \amp 1 \amp 0 \\ 2 \amp -1 \amp 1 \\ \end{bmatrix} \sim \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \\ \end{bmatrix}. \end{equation*}

Because there is a pivot in every row, Proposition 6.3.5 tells us that \(\laspan{\vvec_1,\vvec_2,\vvec_3} = \real^3\text{.}\)
Similarly,

\begin{equation*} \begin{bmatrix} 0 \amp 3 \amp 3 \\ -1 \amp 1 \amp 0 \\ 2 \amp -1 \amp 1 \\ \end{bmatrix} \sim \begin{bmatrix} 1 \amp 0 \amp 1 \\ 0 \amp 1 \amp 1 \\ 0 \amp 0 \amp 0 \\ \end{bmatrix}. \end{equation*}

As there are two pivot positions, we see that \(\laspan{\wvec_1, \wvec_2, \wvec_3}\) is a plane in \(\real^3\text{.}\)
We see that

\begin{equation*} \left[ \begin{array}{rr|r} 0 \amp 3 \amp 3 \\ -1 \amp 1 \amp 0 \\ 2 \amp -1 \amp 1 \\ \end{array} \right] \sim \left[ \begin{array}{rr|r} 1 \amp 0 \amp 1 \\ 0 \amp 1 \amp 1 \\ 0 \amp 0 \amp 0 \\ \end{array} \right], \end{equation*}

which tells us that \(\wvec_3 = \wvec_1 + \wvec_2\text{.}\)
We have

\begin{equation*} c_1\wvec_1 + c_2\wvec_2 + c_3\wvec_3 = c_1\wvec_1 + c_2\wvec_2 + c_3(\wvec_1+\wvec_2)= (c_1+c_3)\wvec_1 + (c_2+c_3)\wvec_2. \end{equation*}
Any linear combination of \(\wvec_1\text{,}\) \(\wvec_2\text{,}\) and \(\wvec_3\) is itself a linear combination of \(\wvec_1\) and \(\wvec_2\text{.}\)

Subsection 7.1.1 Linear dependence

We have seen examples where the span of a set of three vectors in \(\real^3\) is \(\real^3\) and other examples where the span of three vectors is a plane. We would like to understand the difference between these two situations.

Example 7.1.1.

Let’s consider the set of three vectors in \(\real^3\text{:}\)

\begin{equation*} \vvec_1 = \threevec220,~~~ \vvec_2 = \threevec11{-1},~~~ \vvec_3 = \threevec{-1}01. \end{equation*}

Forming the associated matrix gives

\begin{equation*} \begin{bmatrix}\vvec_1\amp\vvec_2\amp\vvec_3\end{bmatrix} = \begin{bmatrix} 2 \amp 1 \amp -1 \\ 2 \amp 1 \amp 0 \\ 0 \amp -1 \amp 1 \\ \end{bmatrix} \sim \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \\ \end{bmatrix}. \end{equation*}

Because there is a pivot position in every row, Proposition 6.3.5 tells us that \(\laspan{\vvec_1, \vvec_2, \vvec_3} = \real^3\text{.}\)

Example 7.1.2.

Now let’s consider the set of three vectors:

\begin{equation*} \wvec_1 = \threevec220,~~~ \wvec_2 = \threevec11{-1},~~~ \wvec_3 = \threevec{-5}{-5}{1}. \end{equation*}

Forming the associated matrix gives

\begin{equation*} \begin{bmatrix}\wvec_1\amp\wvec_2\amp\wvec_3\end{bmatrix} = \begin{bmatrix} 2 \amp 1 \amp -5 \\ 2 \amp 1 \amp -5 \\ 0 \amp -1 \amp 1 \\ \end{bmatrix} \sim \begin{bmatrix} 1 \amp 0 \amp -2 \\ 0 \amp 1 \amp -1 \\ 0 \amp 0 \amp 0 \\ \end{bmatrix}. \end{equation*}

Since the last row does not have a pivot position, we know that the span of these vectors is not \(\real^3\) but is instead a plane.

In fact, we can say more if we shift our perspective slightly and view this as an augmented matrix:

\begin{equation*} \left[ \begin{array}{rr|r}\wvec_1\amp\wvec_2\amp\wvec_3 \end{array} \right] = \left[ \begin{array}{rr|r} 2 \amp 1 \amp -5 \\ 2 \amp 1 \amp -5 \\ 0 \amp -1 \amp 1 \\ \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]. \end{equation*}

In this way, we see that \(\wvec_3 = -2\wvec_1 - \wvec_2\text{,}\) which enables us to rewrite any linear combination of these three vectors:

\begin{equation*} \begin{aligned} c_1\wvec_1 + c_2\wvec_2 + c_3\wvec_3 \amp {}={} c_1\wvec_1 + c_2\wvec_2 + c_3(-2\wvec_1-\wvec_2) \\ \amp {}={} (c_1-2c_3)\wvec_1 + (c_2-c_3)\wvec_2. \\ \end{aligned} \end{equation*}

In other words, any linear combination of \(\wvec_1\text{,}\) \(\wvec_2\text{,}\) and \(\wvec_3\) may be written as a linear combination using only the vectors \(\wvec_1\) and \(\wvec_2\text{.}\) Since the span of a set of vectors is simply the set of their linear combinations, this shows that

\begin{equation*} \laspan{\wvec_1,\wvec_2,\wvec_3} = \laspan{\wvec_1,\wvec_2}. \end{equation*}

As a result, adding the vector \(\wvec_3\) to the set of vectors \(\wvec_1\) and \(\wvec_2\) does not change the span.

Before exploring this type of behavior more generally, let’s think about it from a geometric point of view. Suppose that we begin with the two vectors \(\vvec_1\) and \(\vvec_2\) in Example 7.1.1. The span of these two vectors is a plane in \(\real^3\text{,}\) as seen on the left of Figure 7.1.3.

Figure 7.1.3. The span of the vectors \(\vvec_1\text{,}\) \(\vvec_2\text{,}\) and \(\vvec_3\text{.}\)

Because the vector \(\vvec_3\) is not a linear combination of \(\vvec_1\) and \(\vvec_2\text{,}\) it provides a direction to move that is independent of \(\vvec_1\) and \(\vvec_2\text{.}\) Adding this third vector \(\vvec_3\) therefore forms a set whose span is \(\real^3\text{,}\) as seen on the right of Figure 7.1.3.

Similarly, the span of the vectors \(\wvec_1\) and \(\wvec_2\) in Example 7.1.2 is also a plane. However, the third vector \(\wvec_3\) is a linear combination of \(\wvec_1\) and \(\wvec_2\text{,}\) which means that it already lies in the plane formed by \(\wvec_1\) and \(\wvec_2\text{,}\) as seen in Figure 7.1.4. Since we can already move in this direction using just \(\wvec_1\) and \(\wvec_2\text{,}\) adding \(\wvec_3\) to the set does not change the span. As a result, it remains a plane.

Figure 7.1.4. The span of the vectors \(\wvec_1\text{,}\) \(\wvec_2\text{,}\) and \(\wvec_3\text{.}\)

What distinguishes these two examples is whether one of the vectors is a linear combination of the others, an observation that leads to the following definition.

Definition 7.1.5.

A set of vectors is called linearly dependent if one of the vectors is a linear combination of the others. Otherwise, the set of vectors is called linearly independent.

For the sake of completeness, we say that a set of vectors containing only one nonzero vector is linearly independent.

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