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 2.1 Linear Combinations (EV1)
Subsection 2.1.1 Warm Up
Activity 2.1.1.
Discuss which of the vectors \(\vec{u}=\left[\begin{array}{c} 1 \\ -1 \\ 2 \end{array}\right]\) and \(\vec{v}=\left[\begin{array}{c} 0 \\ 3 \\ -1 \end{array}\right]\) is a solution to the given vector equation:
\begin{equation*}
x_1\left[\begin{array}{c} -1 \\ 2 \\ 3 \end{array}\right]+
x_2\left[\begin{array}{c} 2 \\ -1 \\ 0 \end{array}\right]+
x_3\left[\begin{array}{c} 1 \\ -1 \\ 1 \end{array}\right]=
\left[\begin{array}{c} -1 \\ 1 \\ 5 \end{array}\right]
\end{equation*}
Subsection 2.1.2 Class Activities
Note 2.1.2.
We’ve been working with Euclidean vector spaces of the form
\begin{equation*}
\IR^n=\setBuilder{\left[\begin{array}{c}x_1\\x_2\\\vdots\\x_n\end{array}\right]}{x_1,x_2,\dots,x_n\in\IR}\text{.}
\end{equation*}
There are other kinds of vector spaces as well (e.g. polynomials, matrices), which we will investigate in Section 3.5. But understanding the structure of Euclidean vectors on their own will be beneficial, even when we turn our attention to other kinds of vectors.
Likewise, when we multiply a vector by a real number, as in \(-3 \left[\begin{array}{c} 1 \\ -1 \\ 2 \end{array}\right]=\left[\begin{array}{c} -3 \\ 3 \\ -6 \end{array}\right]\text{,}\) we refer to this real number as a scalar.
Definition 2.1.3.
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 2.1.4.
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 2.1.5.
Consider \(\vspan\left\{\left[\begin{array}{c}1\\2\end{array}\right]\right\}\text{.}\)
(a)
Sketch the four Euclidean vectors
\begin{equation*}
1\left[\begin{array}{c}1\\2\end{array}\right]=\left[\begin{array}{c}1\\2\end{array}\right],\hspace{1em}
3\left[\begin{array}{c}1\\2\end{array}\right]=\left[\begin{array}{c}3\\6\end{array}\right],\hspace{1em}
0\left[\begin{array}{c}1\\2\end{array}\right]=\left[\begin{array}{c}0\\0\end{array}\right],\hspace{1em}
-2\left[\begin{array}{c}1\\2\end{array}\right]=\left[\begin{array}{c}-2\\-4\end{array}\right]
\end{equation*}
in the \(xy\) plane by placing a dot at the \((x,y)\) coordinate associated with each vector.
(b)
Sketch a representation of all the vectors belonging to
\begin{equation*}
\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}
\end{equation*}
in the \(xy\) plane by plotting their \((x,y)\) coordinates as dots. What best describes this sketch?
- A line
- A plane
- A parabola
- A circle
Remark 2.1.6.
It is important to remember that
\begin{equation*}
\{\vec v_1,\vec v_2,\dots,\vec v_m\}\not=\vspan\{\vec v_1,\vec v_2,\dots,\vec v_m\}\text{.}
\end{equation*}
For example,
\begin{equation*}
\setList
{
\left[\begin{array}{c} 1 \\ -1 \\ 2 \end{array}\right],
\left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right]
}
\end{equation*}
is a set containing exactly two vectors, while
\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
}
\end{equation*}
is a set containing infinitely-many vectors.
Activity 2.1.7.
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 five Euclidean vectors 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]=\unknown\hspace{2em}
0\left[\begin{array}{c}1\\2\end{array}\right]+
1\left[\begin{array}{c}-1\\1\end{array}\right]=\unknown\hspace{2em}
1\left[\begin{array}{c}1\\2\end{array}\right]+
1\left[\begin{array}{c}-1\\1\end{array}\right]=\unknown
\end{equation*}
\begin{equation*}
-2\left[\begin{array}{c}1\\2\end{array}\right]+
1\left[\begin{array}{c}-1\\1\end{array}\right]=\unknown\hspace{2em}
-1\left[\begin{array}{c}1\\2\end{array}\right]+
-2\left[\begin{array}{c}-1\\1\end{array}\right]=\unknown
\end{equation*}
(b)
Sketch a representation of all the vectors belonging to
\begin{equation*}
\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}
\end{equation*}
in the \(xy\) plane. What best describes this sketch?
- A line
- A plane
- A parabola
- A circle
Activity 2.1.8.
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. What best describes this sketch?
- A line
- A plane
- A parabola
- A cube
Activity 2.1.9.
Consider the following questions to discover whether a Euclidean vector belongs to a span.
(a)
The Euclidean 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 which of these vector equations?
- \(\displaystyle x_1\left[\begin{array}{c}-1\\-6\\1\end{array}\right]+ x_2\left[\begin{array}{c}1\\0\\-3\end{array}\right] =\left[\begin{array}{c}-1\\-3\\2\end{array}\right]\)
- \(\displaystyle x_1\left[\begin{array}{c}1\\0\\-3\end{array}\right]+ x_2\left[\begin{array}{c}-1\\-3\\2\end{array}\right] =\left[\begin{array}{c}-1\\-6\\1\end{array}\right]\)
- \(\displaystyle x_1\left[\begin{array}{c}-1\\-3\\2\end{array}\right]+ x_2\left[\begin{array}{c}-1\\-6\\1\end{array}\right]+ x_3\left[\begin{array}{c}1\\0\\-3\end{array}\right]=0\)
(b)
Use technology to find \(\RREF\) of the corresponding augmented matrix, and then use that matrix to find the solution set of the vector equation.
(c)
Given this solution set, does \(\left[\begin{array}{c}-1\\-6\\1\end{array}\right]\) belong to \(\vspan\left\{\left[\begin{array}{c}1\\0\\-3\end{array}\right],
\left[\begin{array}{c}-1\\-3\\2\end{array}\right]\right\}\text{?}\)
Observation 2.1.10.
The following are all equivalent statements:
- The vector \(\vec{b}\) belongs to \(\vspan\{\vec v_1,\dots,\vec v_n\}\text{.}\)
- The vector \(\vec{b}\) is a linear combination of the vectors \(\vec v_1,\dots,\vec v_n\text{.}\)
- 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 2.1.11.
Consider this claim about a vector equation:
\(\left[\begin{array}{c} -6 \\ 2 \\ -6 \end{array}\right]\)is a linear combination of the vectors \(\left[\begin{array}{c} 1 \\ 0 \\ 2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 0 \\ 6 \end{array}\right] , \left[\begin{array}{c} 2 \\ 0 \\ 4 \end{array}\right] , \text{ and } \left[\begin{array}{c} -4 \\ 1 \\ -5 \end{array}\right]\text{.}\)
(a)
Write a statement involving the solutions of a vector equation that’s equivalent to this claim.
(b)
Explain why the statement you wrote is true.
(c)
Since your statement was true, use the solution set to describe a linear combination of \(\left[\begin{array}{c} 1 \\ 0 \\ 2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 0 \\ 6 \end{array}\right] , \left[\begin{array}{c} 2 \\ 0 \\ 4 \end{array}\right] , \text{ and } \left[\begin{array}{c} -4 \\ 1 \\ -5 \end{array}\right]\) that equals \(\left[\begin{array}{c} -6 \\ 2 \\ -6 \end{array}\right]\text{.}\)
Activity 2.1.12.
Consider this claim about a vector equation:
\(\left[\begin{array}{c} -5 \\ -1 \\ -7 \end{array}\right]\) belongs to \(\vspan\left\{\left[\begin{array}{c} 1 \\ 0 \\ 2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 0 \\ 6 \end{array}\right] ,
\left[\begin{array}{c} 2 \\ 0 \\ 4 \end{array}\right] , \left[\begin{array}{c} -4 \\ 1 \\ -5 \end{array}\right]\right\}\text{.}\)
(a)
Write a statement involving the solutions of a vector equation that’s equivalent to this claim.
(b)
Explain why the statement you wrote is false, to conclude that the vector does not belong to the span.
Subsection 2.1.3 Cool Down
Activity 2.1.13.
Before next class, find some time to do the following:
(a)
Without referring to your activity book, write down the definition of a linear combination of vectors.
(b)
Let \(\vec{u}=\left[\begin{array}{c} 1 \\ 2 \\0 \end{array}\right]\) and \(\vec{v}=\left[\begin{array}{c} -1 \\ 3 \\ 0\end{array}\right]\text{.}\) Write down an example \(\vec{w_1}=\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown\end{array}\right]\) of a linear combination of \(\vec{u},\vec{v}\text{.}\) Then write down an example \(\vec{w_2}=\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown\end{array}\right]\) that is not a linear combination of \(\vec{u},\vec{v}\text{.}\)
(c)
Draw a rough sketch of the vectors \(\vec{u}=\left[\begin{array}{c} 1 \\ 2 \\0 \end{array}\right]\text{,}\) \(\vec{v}=\left[\begin{array}{c} -1 \\ 3 \\ 0\end{array}\right]\text{,}\) \(\vec{w_1}=\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown\end{array}\right]\text{,}\) and \(\vec{w_2}=\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown\end{array}\right]\) in \(\IR^3\text{.}\)
Subsection 2.1.4 Videos
Exercises 2.1.5 Exercises
Exercises available at
https://teambasedinquirylearning.github.io/linear-algebra/2024/exercises/#/bank/EV1/.Subsection 2.1.6 Mathematical Writing Explorations
Exploration 2.1.14.
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 2.1.7 Sample Problem and Solution
Sample problem Example B.1.5.
