\(\newcommand{\circledNumber}[1]{\boxed{#1}} \newcommand{\IR}{\mathbb{R}} \newcommand{\IC}{\mathbb{C}} \renewcommand{\P}{\mathcal{P}} \renewcommand{\Im}{\operatorname{Im}} \newcommand{\RREF}{\operatorname{RREF}} \newcommand{\vspan}{\operatorname{span}} \newcommand{\setList}[1]{\left\{#1\right\}} \newcommand{\setBuilder}[2]{\left\{#1\,\middle|\,#2\right\}} \newcommand{\unknown}{\,{\color{gray}?}\,} \newcommand{\drawtruss}[2][1]{ \begin{tikzpicture}[scale=#1, every node/.style={scale=#1}] \draw (0,0) node[left,magenta]{C} -- (1,1.71) node[left,magenta]{A} -- (2,0) node[above,magenta]{D} -- cycle; \draw (2,0) -- (3,1.71) node[right,magenta]{B} -- (1,1.71) -- cycle; \draw (3,1.71) -- (4,0) node[right,magenta]{E} -- (2,0) -- cycle; \draw[blue] (0,0) -- (0.25,-0.425) -- (-0.25,-0.425) -- cycle; \draw[blue] (4,0) -- (4.25,-0.425) -- (3.75,-0.425) -- cycle; \draw[thick,red,->] (2,0) -- (2,-0.75); #2 \end{tikzpicture} } \newcommand{\trussNormalForces}{ \draw [thick, blue,->] (0,0) -- (0.5,0.5); \draw [thick, blue,->] (4,0) -- (3.5,0.5); } \newcommand{\trussCompletion}{ \trussNormalForces \draw [thick, magenta,<->] (0.4,0.684) -- (0.6,1.026); \draw [thick, magenta,<->] (3.4,1.026) -- (3.6,0.684); \draw [thick, magenta,<->] (1.8,1.71) -- (2.2,1.71); \draw [thick, magenta,->] (1.6,0.684) -- (1.5,0.855); \draw [thick, magenta,<-] (1.5,0.855) -- (1.4,1.026); \draw [thick, magenta,->] (2.4,0.684) -- (2.5,0.855); \draw [thick, magenta,<-] (2.5,0.855) -- (2.6,1.026); } \newcommand{\trussCForces}{ \draw [thick, blue,->] (0,0) -- (0.5,0.5); \draw [thick, magenta,->] (0,0) -- (0.4,0.684); \draw [thick, magenta,->] (0,0) -- (0.5,0); } \newcommand{\trussStrutVariables}{ \node[above] at (2,1.71) {$x_1$}; \node[left] at (0.5,0.866) {$x_2$}; \node[left] at (1.5,0.866) {$x_3$}; \node[right] at (2.5,0.866) {$x_4$}; \node[right] at (3.5,0.866) {$x_5$}; \node[below] at (1,0) {$x_6$}; \node[below] at (3,0) {$x_7$}; } \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} \definecolor{fillinmathshade}{gray}{0.9} \newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}} \)

Linear Algebra for Team-Based Inquiry Learning

2022 Edition

Steven Clontz	Drew Lewis
University of South Alabama	University of South Alabama

August 2, 2022

Section 2.2: Linear Combinations (VS2)

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 (~10 min)

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

Part 1.

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.

Activity 2.2.3 (~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 $\vspan\setList{\left[\begin{array}{c}1\\2\end{array}\right]} = \setBuilder{a\left[\begin{array}{c}1\\2\end{array}\right]}{a\in\IR}$ in the $xy$ plane.

Activity 2.2.4 (~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 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*}

Activity 2.2.4 (~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 $\vspan\left\{\left[\begin{array}{c}1\\2\end{array}\right], \left[\begin{array}{c}-1\\1\end{array}\right]\right\}=\setBuilder{a\left[\begin{array}{c}1\\2\end{array}\right]+b\left[\begin{array}{c}-1\\1\end{array}\right]}{a, b \in \IR}$ in the $xy$ plane.

Activity 2.2.5 (~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.

Activity 2.2.6 (~15 min)

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 $x_1\left[\begin{array}{c}1\\0\\-3\end{array}\right]+ x_2\left[\begin{array}{c}-1\\-3\\2\end{array}\right] =\left[\begin{array}{c}-1\\-6\\1\end{array}\right]\text{.}$

Part 1.

Reinterpret this vector equation as a system of linear equations.

Activity 2.2.6 (~15 min)

Part 2.

Find its solution set, using technology to find $\RREF$ of its corresponding augmented matrix.

Activity 2.2.6 (~15 min)

Part 3.

Given this solution set, does $\left[\begin{array}{c}-1\\-6\\1\end{array}\right]$ belong to $\vspan\left\{\left[\begin{array}{c}1\\0\\-3\end{array}\right], \left[\begin{array}{c}-1\\-3\\2\end{array}\right]\right\}\text{?}$

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 (~10 min)

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 (~5 min)

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 (~10 min)

Does the third-degree polynomial $3y^3-2y^2+y+5$ in $\P_3$ belong to $\vspan\{y^3-3y+2,-y^3-3y^2+2y+2\}\text{?}$

Part 1.

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*}

Activity 2.2.11 (~10 min)

Does the third-degree polynomial $3y^3-2y^2+y+5$ in $\P_3$ belong to $\vspan\{y^3-3y+2,-y^3-3y^2+2y+2\}\text{?}$

Part 2.

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*}

Activity 2.2.11 (~10 min)

Does the third-degree polynomial $3y^3-2y^2+y+5$ in $\P_3$ belong to $\vspan\{y^3-3y+2,-y^3-3y^2+2y+2\}\text{?}$

Part 3.

Answer this equivalent question, and use its solution to answer the original question.

Activity 2.2.12 (~5 min)

Does the polynomial $x^2+x+1$ belong to $\vspan\{x^2-x,x+1, x^2-1\}\text{?}$

Activity 2.2.13 (~5 min)

Does the matrix $\left[\begin{array}{cc}3&-2\\1&5\end{array}\right]$ belong to $\vspan\left\{\left[\begin{array}{cc}1&0\\-3&2\end{array}\right], \left[\begin{array}{cc}-1&-3\\2&2\end{array}\right]\right\}\text{?}$

Part 1.

Reinterpret this question as a question about the solution(s) of a matrix equation.

Activity 2.2.13 (~5 min)

Part 2.

Answer this equivalent question, and use its solution to answer the original question.