Skip to main content
Logo image

Linear Algebra for Team-Based Inquiry Learning 2025 Edition

Steven Clontz, Drew Lewis

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 \(\{\vec v_1,\vec v_2,\dots,\vec v_n\}\) is given by \(c_1\vec v_1+c_2\vec v_2+\dots+c_n\vec v_n\) for any choice of scalar multiples \(c_1,c_2,\dots,c_n\text{.}\)
🔗
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_n\} = \setBuilder{c_1\vec v_1+c_2\vec v_2+\dots+c_n\vec v_n}{ 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 same \(xy\) plane by drawing an arrow to 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. Which of the following geometrical objects best describes this sketch?
  1. A line
    🔗
  2. A plane
    🔗
  3. A parabola
    🔗
  4. A circle
    🔗
🔗
🔗
🔗

Activity 2.1.6.

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 same \(xy\) 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)
Correct the SageMath code cell below to generate an illustration of several 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.
🔗
Based on this illustration, which of these geometrical objects best describes the span of these two vectors?
  1. A line
    🔗
  2. A plane
    🔗
  3. A parabola
    🔗
  4. A circle
    🔗
🔗
🔗
🔗

Activity 2.1.7.

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, or adapt the code in the previous activity to illustrate this span.
🔗
Which of these geometrical objects best describes the span of these two vectors?
  1. A line
    🔗
  2. A plane
    🔗
  3. A parabola
    🔗
  4. A cube
    🔗
🔗
🔗

Activity 2.1.8.

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?
🔗
  1. \(\displaystyle x_1\left[\begin{array}{c}-1\\-6\\1\end{array}\right]+ x_2\left[\begin{array}{c}1\\0\\-3\end{array}\right] =\left[\begin{array}{c}-1\\-3\\2\end{array}\right]\)
    🔗
  2. \(\displaystyle x_1\left[\begin{array}{c}1\\0\\-3\end{array}\right]+ x_2\left[\begin{array}{c}-1\\-3\\2\end{array}\right] =\left[\begin{array}{c}-1\\-6\\1\end{array}\right]\)
    🔗
  3. \(\displaystyle x_1\left[\begin{array}{c}-1\\-3\\2\end{array}\right]+ x_2\left[\begin{array}{c}-1\\-6\\1\end{array}\right]+ x_3\left[\begin{array}{c}1\\0\\-3\end{array}\right]=0\)
    🔗
🔗
🔗
(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 2.1.9.

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 2.1.10.

Consider the following claim:
🔗
\(\left[\begin{array}{c} -6 \\ 2 \\ -6 \end{array}\right]\)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 2.1.11.

Consider the following claim:
🔗
\(\left[\begin{array}{c} -5 \\ -1 \\ -7 \end{array}\right]\) 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.
🔗
🔗
🔗

Activity 2.1.12.

Are the sets
\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*}
and
\begin{equation*} \vspan\setList { \left[\begin{array}{c} 1 \\ -1 \\ 2 \end{array}\right], \left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right] } \end{equation*}
equal or nonequal to each other?
  1. Equal
    🔗
    🔗
  2. Non-equal
    🔗
    🔗
🔗
🔗

Remark 2.1.13.

It is important to remember that
\begin{equation*} \{\vec v_1,\vec v_2,\dots,\vec v_n\}\not=\vspan\{\vec v_1,\vec v_2,\dots,\vec v_n\}\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. See the below Sage cell for an illustration.
🔗
🔗
🔗

Subsection 2.1.3 Individual Practice

Activity 2.1.14.

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

Figure 5. Video: Linear combinations
🔗
🔗

Subsection 2.1.5 Exercises

🔗

Subsection 2.1.6 Mathematical Writing Explorations

Exploration 2.1.15.

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.
🔗
🔗
🔗
PrevTopNext