Learning Outcomes
- Determine if a Euclidean vector can be written as a linear combination of a given set of Euclidean vectors by solving an appropriate vector equation.
Section 2.2 Linear Combinations (VS2)
Subsection 2.2.1 Class Activities
Definition 2.2.1.
A linear combination of a set of vectors \(\{\vec v_1,\vec v_2,\dots,\vec v_m\}\) is given by \(c_1\vec v_1+c_2\vec v_2+\dots+c_m\vec v_m\) for any choice of scalar multiples \(c_1,c_2,\dots,c_m\text{.}\)
For example, we can say \(\left[\begin{array}{c}3 \\0 \\ 5\end{array}\right]\) is a linear combination of the vectors \(\left[\begin{array}{c} 1 \\ -1 \\ 2 \end{array}\right]\) and \(\left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right]\) since
\begin{equation*}
\left[\begin{array}{c} 3 \\ 0 \\ 5 \end{array}\right] =
2 \left[\begin{array}{c} 1 \\ -1 \\ 2 \end{array}\right] +
1\left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right]\text{.}
\end{equation*}
Definition 2.2.2.
The span of a set of vectors is the collection of all linear combinations of that set:
\begin{equation*}
\vspan\{\vec v_1,\vec v_2,\dots,\vec v_m\} =
\setBuilder{c_1\vec v_1+c_2\vec v_2+\dots+c_m\vec v_m}{
c_i\in\IR}\text{.}
\end{equation*}
For example:
\begin{equation*}
\vspan\setList
{
\left[\begin{array}{c} 1 \\ -1 \\ 2 \end{array}\right],
\left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right]
} = \setBuilder
{
a\left[\begin{array}{c} 1 \\ -1 \\ 2 \end{array}\right]+
b\left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right]
}{
a,b\in\IR
}\text{.}
\end{equation*}
Activity 2.2.3.
Consider \(\vspan\left\{\left[\begin{array}{c}1\\2\end{array}\right]\right\}\text{.}\)
(a)
Sketch \(1\left[\begin{array}{c}1\\2\end{array}\right]=\left[\begin{array}{c}1\\2\end{array}\right]\text{,}\) \(3\left[\begin{array}{c}1\\2\end{array}\right]=\left[\begin{array}{c}3\\6\end{array}\right]\text{,}\) \(0\left[\begin{array}{c}1\\2\end{array}\right]=\left[\begin{array}{c}0\\0\end{array}\right]\text{,}\) and \(-2\left[\begin{array}{c}1\\2\end{array}\right]=\left[\begin{array}{c}-2\\-4\end{array}\right]\) in the \(xy\) plane.
(b)
Sketch a representation of all the vectors belonging to \(\vspan\setList{\left[\begin{array}{c}1\\2\end{array}\right]}
=
\setBuilder{a\left[\begin{array}{c}1\\2\end{array}\right]}{a\in\IR}\) in the \(xy\) plane.
Activity 2.2.4.
Consider \(\vspan\left\{\left[\begin{array}{c}1\\2\end{array}\right],
\left[\begin{array}{c}-1\\1\end{array}\right]\right\}\text{.}\)
(a)
Sketch the following linear combinations in the \(xy\) plane.
\begin{equation*}
1\left[\begin{array}{c}1\\2\end{array}\right]+
0\left[\begin{array}{c}-1\\1\end{array}\right]\hspace{3em}
0\left[\begin{array}{c}1\\2\end{array}\right]+
1\left[\begin{array}{c}-1\\1\end{array}\right]\hspace{3em}
1\left[\begin{array}{c}1\\2\end{array}\right]+
1\left[\begin{array}{c}-1\\1\end{array}\right]
\end{equation*}
\begin{equation*}
-2\left[\begin{array}{c}1\\2\end{array}\right]+
1\left[\begin{array}{c}-1\\1\end{array}\right]\hspace{3em}
-1\left[\begin{array}{c}1\\2\end{array}\right]+
-2\left[\begin{array}{c}-1\\1\end{array}\right]
\end{equation*}
(b)
Sketch a representation of all the vectors belonging to \(\vspan\left\{\left[\begin{array}{c}1\\2\end{array}\right],
\left[\begin{array}{c}-1\\1\end{array}\right]\right\}=\setBuilder{a\left[\begin{array}{c}1\\2\end{array}\right]+b\left[\begin{array}{c}-1\\1\end{array}\right]}{a, b \in \IR}\) in the \(xy\) plane.
Activity 2.2.5.
Sketch a representation of all the vectors belonging to \(\vspan\left\{\left[\begin{array}{c}6\\-4\end{array}\right],
\left[\begin{array}{c}-3\\2\end{array}\right]\right\}\) in the \(xy\) plane.
Activity 2.2.6.
The vector \(\left[\begin{array}{c}-1\\-6\\1\end{array}\right]\) belongs to \(\vspan\left\{\left[\begin{array}{c}1\\0\\-3\end{array}\right],
\left[\begin{array}{c}-1\\-3\\2\end{array}\right]\right\}\) exactly when there exists a solution to the vector equation \(x_1\left[\begin{array}{c}1\\0\\-3\end{array}\right]+
x_2\left[\begin{array}{c}-1\\-3\\2\end{array}\right]
=\left[\begin{array}{c}-1\\-6\\1\end{array}\right]\text{.}\)
(a)
Reinterpret this vector equation as a system of linear equations.
(b)
Find its solution set, using technology to find \(\RREF\) of its corresponding augmented matrix.
(c)
Given this solution set, does \(\left[\begin{array}{c}-1\\-6\\1\end{array}\right]\) belong to \(\vspan\left\{\left[\begin{array}{c}1\\0\\-3\end{array}\right],
\left[\begin{array}{c}-1\\-3\\2\end{array}\right]\right\}\text{?}\)
Fact 2.2.7.
A vector \(\vec b\) belongs to \(\vspan\{\vec v_1,\dots,\vec v_n\}\) if and only if the vector equation \(x_1 \vec{v}_1+\cdots+x_n \vec{v}_n=\vec{b}\) is consistent.
Observation 2.2.8.
The following are all equivalent statements:
- The vector \(\vec{b}\) belongs to \(\vspan\{\vec v_1,\dots,\vec v_n\}\text{.}\)
- The vector equation \(x_1 \vec{v}_1+\cdots+x_n \vec{v}_n=\vec{b}\) is consistent.
- The linear system corresponding to \(\left[\vec v_1\,\dots\,\vec v_n \,|\, \vec b\right]\) is consistent.
- \(\RREF\left[\vec v_1\,\dots\,\vec v_n \,|\, \vec b\right]\) doesn't have a row \([0\,\cdots\,0\,|\,1]\) representing the contradiction \(0=1\text{.}\)
Activity 2.2.9.
Determine if \(\left[\begin{array}{c}3\\-2\\1 \\ 5\end{array}\right]\) belongs to \(\vspan\left\{\left[\begin{array}{c}1\\0\\-3 \\ 2\end{array}\right],
\left[\begin{array}{c}-1\\-3\\2 \\ 2\end{array}\right]\right\}\) by solving an appropriate vector equation.
Activity 2.2.10.
Determine if \(\left[\begin{array}{c}-1\\-9\\0\end{array}\right]\) belongs to \(\vspan\left\{\left[\begin{array}{c}1\\0\\-3\end{array}\right],
\left[\begin{array}{c}-1\\-3\\2\end{array}\right]\right\}\) by solving an appropriate vector equation.
Activity 2.2.11.
Does the third-degree polynomial \(3y^3-2y^2+y+5\) in \(\P_3\) belong to \(\vspan\{y^3-3y+2,-y^3-3y^2+2y+2\}\text{?}\)
(a)
Reinterpret this question as a question about the solution(s) of a polynomial equation:
\begin{equation*}
x_1(\cdots\unknown\cdots)+x_2(\cdots\unknown\cdots)= (\cdots\unknown\cdots)
\end{equation*}
(b)
Write a Euclidean vector equation that has the same solution set:
\begin{equation*}
x_1\left[\begin{array}{c}\unknown\\\unknown\\\unknown\\\unknown\end{array}\right]+
x_2\left[\begin{array}{c}\unknown\\\unknown\\\unknown\\\unknown\end{array}\right]=
\left[\begin{array}{c}\unknown\\\unknown\\\unknown\\\unknown\end{array}\right]
\end{equation*}
(c)
Answer this equivalent question, and use its solution to answer the original question.
Activity 2.2.12.
Does the polynomial \(x^2+x+1\) belong to \(\vspan\{x^2-x,x+1, x^2-1\}\text{?}\)
Activity 2.2.13.
Does the matrix \(\left[\begin{array}{cc}3&-2\\1&5\end{array}\right]\) belong to \(\vspan\left\{\left[\begin{array}{cc}1&0\\-3&2\end{array}\right],
\left[\begin{array}{cc}-1&-3\\2&2\end{array}\right]\right\}\text{?}\)
(a)
Reinterpret this question as a question about the solution(s) of a matrix equation.
(b)
Answer this equivalent question, and use its solution to answer the original question.
Subsection 2.2.2 Videos
Subsection 2.2.3 Slideshow
Slideshow of activities available at
https://teambasedinquirylearning.github.io/linear-algebra/2023e/VS2.slides.html.Exercises 2.2.4 Exercises
Exercises available at
https://teambasedinquirylearning.github.io/linear-algebra/2023e/exercises/#/bank/VS2/.Subsection 2.2.5 Mathematical Writing Explorations
Exploration 2.2.14.
Suppose \(S = \{\vec{v_1},\ldots, \vec{v_n}\}\) is a set of vectors. Show that \(\vec{v_0}\) is a linear combination of members of \(S\text{,}\) if an only if there are a set of scalars \(\{c_0,c_1,\ldots, c_n\}\) such that \(\vec{z} = c_0\vec{v_0} + \cdots + c_n\vec{v_n}.\) We can do this in a few parts. I've used bullets here to indicate all that needs to be done. This is an "if and only if" proof, so it needs two parts.
- First, assume that \(\vec{0} = c_0\vec{v_0} + \cdots + c_n\vec{v_n}\) has a solution, with \(c_0 \neq 0\text{.}\) Show that \(\vec{v_0}\) is a linear combination of elements of \(S\text{.}\)
- Next, assume that \(\vec{v_0}\) is a linear combination of elements of \(S\text{.}\) Can you find the appropriate \(\{c_0,c_1,\ldots, c_n\}\) to make the equation \(\vec{z} = c_0\vec{v_0} + \cdots + c_n\vec{v_n}\) true?
- In either of your proofs above, does the case when \(\vec{v_0} = \vec{z}\) change your thinking? Explain why or why not.
Subsection 2.2.6 Sample Problem and Solution
Sample problem Example B.1.6.
