Linear Algebra for Team-Based Inquiry Learning

2023 Edition

Steven Clontz Drew Lewis
University of South Alabama

August 24, 2023

Section 2.1: Linear Combinations (EV1)

Definition 2.1.2

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 2.1.3

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*}

Remark 2.1.4

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 2.1.1 (~10 min)

Consider \(\vspan\left\{\left[\begin{array}{c}1\\2\end{array}\right]\right\}\text{.}\)

Part 1.

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.

Activity 2.1.1 (~10 min)

Consider \(\vspan\left\{\left[\begin{array}{c}1\\2\end{array}\right]\right\}\text{.}\)

Part 2.

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?
  1. A line
  2. A plane
  3. A parabola
  4. A circle

Activity 2.1.2 (~10 min)

Consider \(\vspan\left\{\left[\begin{array}{c}1\\2\end{array}\right], \left[\begin{array}{c}-1\\1\end{array}\right]\right\}\text{.}\)

Part 1.

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*}

Activity 2.1.2 (~10 min)

Consider \(\vspan\left\{\left[\begin{array}{c}1\\2\end{array}\right], \left[\begin{array}{c}-1\\1\end{array}\right]\right\}\text{.}\)

Part 2.

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?
  1. A line
  2. A plane
  3. A parabola
  4. A circle

Activity 2.1.3 (~5 min)

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?

  1. A line
  2. A plane
  3. A parabola
  4. A cube

Activity 2.1.4 (~15 min)

Consider the following questions to discover whether a Euclidean vector belongs to a span.

Part 1.

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

Activity 2.1.4 (~15 min)

Consider the following questions to discover whether a Euclidean vector belongs to a span.

Part 2.

Use technology to find \(\RREF\) of the corresponding augmented matrix, and then use that matrix to find the solution set of the vector equation.

Activity 2.1.4 (~15 min)

Consider the following questions to discover whether a Euclidean vector belongs to a span.

Part 3.

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 2.1.5

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 2.1.5 (~10 min)

Consider this claim about a vector equation:

\(\left[\begin{array}{c} -6 \\ 2 \\ -6 \end{array}\right]\)is a linear combination of the vectors \(\left[\begin{array}{c} 1 \\ 0 \\ 2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 0 \\ 6 \end{array}\right] , \left[\begin{array}{c} 2 \\ 0 \\ 4 \end{array}\right] , \text{ and } \left[\begin{array}{c} -4 \\ 1 \\ -5 \end{array}\right]\text{.}\)

Part 1.

Write a statement involving the solutions of a vector equation that’s equivalent to this claim.

Activity 2.1.5 (~10 min)

Consider this claim about a vector equation:

\(\left[\begin{array}{c} -6 \\ 2 \\ -6 \end{array}\right]\)is a linear combination of the vectors \(\left[\begin{array}{c} 1 \\ 0 \\ 2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 0 \\ 6 \end{array}\right] , \left[\begin{array}{c} 2 \\ 0 \\ 4 \end{array}\right] , \text{ and } \left[\begin{array}{c} -4 \\ 1 \\ -5 \end{array}\right]\text{.}\)

Part 2.

Explain why the statement you wrote is true.

Activity 2.1.5 (~10 min)

Consider this claim about a vector equation:

\(\left[\begin{array}{c} -6 \\ 2 \\ -6 \end{array}\right]\)is a linear combination of the vectors \(\left[\begin{array}{c} 1 \\ 0 \\ 2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 0 \\ 6 \end{array}\right] , \left[\begin{array}{c} 2 \\ 0 \\ 4 \end{array}\right] , \text{ and } \left[\begin{array}{c} -4 \\ 1 \\ -5 \end{array}\right]\text{.}\)

Part 3.

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 \(\left[\begin{array}{c} -5 \\ -1 \\ -7 \end{array}\right]\text{.}\)

Activity 2.1.6 (~10 min)

Consider this claim about a vector equation:

\(\left[\begin{array}{c} -5 \\ -1 \\ -7 \end{array}\right]\) belongs to \(\vspan\left\{\left[\begin{array}{c} 1 \\ 0 \\ 2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 0 \\ 6 \end{array}\right] , \left[\begin{array}{c} 2 \\ 0 \\ 4 \end{array}\right] , \left[\begin{array}{c} -4 \\ 1 \\ -5 \end{array}\right]\right\}\text{.}\)

Part 1.

Write a statement involving the solutions of a vector equation that’s equivalent to this claim.

Activity 2.1.6 (~10 min)

Consider this claim about a vector equation:

\(\left[\begin{array}{c} -5 \\ -1 \\ -7 \end{array}\right]\) belongs to \(\vspan\left\{\left[\begin{array}{c} 1 \\ 0 \\ 2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 0 \\ 6 \end{array}\right] , \left[\begin{array}{c} 2 \\ 0 \\ 4 \end{array}\right] , \left[\begin{array}{c} -4 \\ 1 \\ -5 \end{array}\right]\right\}\text{.}\)

Part 2.

Explain why the statement you wrote is false, to conclude that the vector does not belong to the span.