Objectives
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Define a linear combination of vectors and a span of vectors.
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Show that a set of vectors span another set of vectors.
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Determine if a set of vectors span another set.
Section 2.5 Linear Combinations and the Span of a Set of Vectors
In Section 1.3, we discussed a linear combination of rows in a matrix. We can find a linear combination of any vector (or as we will see more generally other things) as well. For example, a linear combination of the vectors \([1\;\;2]\) and \([3\;\;4]\) is
\begin{equation*}
c_1 \begin{bmatrix}
1 \\2
\end{bmatrix} + c_2 \begin{bmatrix}
3 \\4
\end{bmatrix}
\end{equation*}
and these can be any number of vectors and the vectors can be any length.
Definition 2.5.1.
Let \(\vec{s}_1, \vec{s}_2, \ldots, \vec{s}_n\) be vectors in \(\mathbb{R}^n\text{.}\) A linear combination of a set of vectors \(\{ \vec{s}_1, \vec{s}_2, \ldots, \vec{s}_n \}\) is
\begin{equation*}
c_1 \vec{s}_1+ c_2 \vec{s}_2 + \cdots + c_n \vec{s}_n
\end{equation*}
for some \(c_1, c_2, \ldots, c_n \in \mathbb{R}\text{.}\)
For example, the solutions of homogeneous systems is an excellent example of a linear combination. Example 1.4.5 shows a linear system. The associated homogenous system is
\begin{align*}
x_2 + 3x_3 -9 x_4 + 11 x_5 \amp = 0, \\
2x_3 \phantom{-9x_4} + 4x_5 \amp = 0, \\
3x_5 \amp = 0,
\end{align*}
and the solution in vector form is
\begin{equation*}
\left\{ \begin{bmatrix}
1 \\ 0 \\ 0 \\ 0 \\0
\end{bmatrix} x_1 +
\begin{bmatrix}
0 \\ 9 \\ 0 \\ 1 \\ 0
\end{bmatrix} x_4 \; | \; x_1, x_4 \in \mathbb{R} \right\}.
\end{equation*}
and thus this solution is a linear combination of the vectors
\begin{equation*}
\left\{
\begin{bmatrix}
1 \\ 0 \\ 0 \\ 0 \\0
\end{bmatrix},
\begin{bmatrix}
0 \\ 9 \\ 0 \\ 1 \\ 0
\end{bmatrix} \right\}
\end{equation*}
The solution is the set of all linear combinations and this plays an important role, so it defined as the span of a set of vectors.
Definition 2.5.2.
The span of a nonempty set subset \(S=\{\vec{s}_1,\ldots, \vec{s}_n\}\) of a vector space is the set of all linear combinations of the vectors in \(S\text{.}\) In particular,
\begin{equation*}
\text{span}(S) = \{ c_1 \vec{s}_1 + \cdots + c_n \vec{s}_n\; | \; \text{$c_1, c_2, \ldots,
c_n \in \mathbb{R}$} \}.
\end{equation*}
Understanding the span in a more general way is important. Often the span of a set of vectors is something we already know as the next example shows.
Example 2.5.3.
Show that
\begin{equation*}
\text{span}\biggl( \biggl\{\begin{bmatrix}
1 \\ 0
\end{bmatrix}, \begin{bmatrix}
1 \\ 1
\end{bmatrix} \biggr\} \biggr)
\end{equation*}
can be written as
\begin{equation*}
\left\{ \begin{bmatrix}
x_1 \\ x_2
\end{bmatrix} \; | \; x_1,x_2 \in \mathbb{R} \right\}
\end{equation*}
Solution.
In order to show this, we show that for any vector of the form
\begin{equation*}
\begin{bmatrix}
x_1 \\ x_2
\end{bmatrix}
\end{equation*}
there exist \(c_1, c_2\) such that
\begin{equation*}
c_1 \begin{bmatrix}
1 \\ 0
\end{bmatrix} + c_2 \begin{bmatrix}
1 \\ 1
\end{bmatrix} = \begin{bmatrix}
x_1 \\ x_2
\end{bmatrix}.
\end{equation*}
\begin{align*}
c_1 + c_2 \amp = x_1 \\
c_2 \amp = x_2
\end{align*}
Since we found by \(c_1\) and \(c_2\text{,}\) a linear combination of these two vectors can produce any vector in \(\mathbb{R}^2\text{.}\)
We saw that the span of the two vectors in the example above was \(\mathbb{R}^2\text{.}\) What if we add a third vector? The following example answers this question.
Example 2.5.4.
Find
\begin{equation*}
\text{span}\biggl( \biggl\{ \begin{bmatrix}
1 \\ 0
\end{bmatrix}, \begin{bmatrix}
1 \\ 1
\end{bmatrix}, \begin{bmatrix}
0 \\1
\end{bmatrix} \biggr\} \biggr)
\end{equation*}
Solution.
In the case, it is the set formed by the linear combination or specifically
\begin{equation*}
\left\{ c_1 \begin{bmatrix}
1 \\ 0
\end{bmatrix} + c_2 \begin{bmatrix}
1 \\ 1
\end{bmatrix} + c_3 \begin{bmatrix}
0 \\ 1
\end{bmatrix} \; | \; c_1, c_2, c_3 \in \mathbb{R} \right\}
\end{equation*}
and it seems that this set is also \(\mathbb{R}^2\text{.}\) In order to show this, we’ll take \(x_1,
x_2 \in \mathbb{R}\text{,}\) we will solve
\begin{equation*}
c_1 \begin{bmatrix}
1 \\ 0
\end{bmatrix} + c_2 \begin{bmatrix}
1 \\ 1
\end{bmatrix} + c_3 \begin{bmatrix}
0 \\ 1
\end{bmatrix} = \begin{bmatrix}
x_1 \\ x_2
\end{bmatrix}
\end{equation*}
and this can be solved with the linear system
\begin{align*}
c_1 + c_2 \amp = x_1\\
c_2 + c_3 \amp = x_2.
\end{align*}
If we take \(c_3\) as the free variable, we get
\begin{align*}
c_2 \amp = x_2 - c_3 \\
c_1 \amp = x_1 -c_2 = x_1 - (x_2-c_3) = x_1 - x_2 + c_3
\end{align*}
and since there is a solution, these vectors span \(\mathbb{R}^2\text{.}\) It appears that adding another vector does not change the span. This actually does not depend on this vector, but we’ll need information in the next section to show this.
And if we continue to investigate this, what if we have two vectors in \(\mathbb{R}^3\text{.}\) Is the span equal to \(\mathbb{R}^3\text{?}\) The next example looks as this.
Example 2.5.5.
Is
\begin{equation*}
\text{span}\biggl(\biggl\{ \begin{bmatrix}
1 \\ 0 \\ 1
\end{bmatrix}, \begin{bmatrix}
1 \\ 1 \\ 0
\end{bmatrix} \biggr\} \biggr)
\end{equation*}
equal to \(\mathbb{R}^3\text{?}\)
Solution.
\begin{equation*}
c_1 \begin{bmatrix}
1 \\ 0 \\ 1
\end{bmatrix} + c_2 \begin{bmatrix}
1 \\ 1 \\ 0
\end{bmatrix} = \begin{bmatrix}
x_1 \\ x_2 \\ x_3
\end{bmatrix}
\end{equation*}
or the linear system
\begin{align*}
c_1 + c_2 \amp = x_1\\
c_2 \amp = x_2 \\
c_1 \amp = x_3
\end{align*}
however from the latter two equations, there is a contradiction with the first equation which would say \(x_2+x_3 = x_1\) therefore there is no solution. This means that the two given vectors do not span \(\mathbb{R}^3\text{.}\)
If we consider the above example, one of the reasons that this happened is that we don’t have enough vectors. The span of two vectors in \(\mathbb{R}^3\) form a plane as we saw in section Subsection 1.6.5. One way to look at \(\mathbb{R}^3\) is that is the set of all points in \(\mathbb{R}^3\text{,}\) so there is no way a plane can equal all points.
