Math 2114

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\title{MATH 2114 Notes}

\author{Tyler Kruszewski}

\date{\today}

\begin{document}

\maketitle

\begin{flushleft}

\textcolor{red}{Disclaimer: These notes are in no way a replacement for your own, there may be typos, wrong information, difference in notation or a difference caused by a different instructor. You should always prioritize your own notes and at best use these as a review.}

\end{flushleft}

\tableofcontents

\newpage

\begin{flushleft}

\section{Section 1.1 The Geometry and Algebra of Vectors}

\subsection{Introduction}

\begin{center}

Scalars \hspace{10ex} Vectors \hspace{10ex} Matrices

\end{center}

Scalars are Real Numbers ($\mathbb{R}$)

\begin{definition}{Definition: Vectors}

An \textbf{$n$-dimensional column vector} $\vv{v}$ is an ordered list of $n$ components:

$\vv{v} = \begin{bsmallmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{bsmallmatrix}$.

\end{definition}

The set of all real $n$-dimensional column vectors is $\mathbb{R}^n=\left\{\begin{bsmallmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{bsmallmatrix}\Big| v_1, v_2, \hdots, v_n\in \mathbb{R} \right\}$ \\

With the $\Big|$ meaning ``such that'' and $\in$ meaning ``as an element of'' as well as $n$ meaning number of components and $\mathbb{R}$ being the type of components (real numbers)

\begin{example}{Example: Vectors In A Set Of Real Numbers}

\begin{itemize}

\item $\begin{bmatrix} 2 \\ -3 \end{bmatrix} \in \mathbb{R}^2$

\item $\begin{bmatrix} 1 \\ 0 \\ -3 \\ 8\end{bmatrix} \in \mathbb{R}^4$

\end{itemize}

\end{example}

\pagebreak

\subsection{Two Basic Vector Operations: Addition And Scalar Multiplication}

To add 2 vectors, they must have the same number of components.

\begin{example}{Example: Adding Two Vectors}

\begin{itemize}

\item $\begin{bmatrix}3 \\ -2 \\ 1 \\ 8 \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \\ -1 \\ 3 \end{bmatrix} = \begin{bmatrix} 4 \\ -1 \\ 0 \\ 11 \end{bmatrix}$ \\

\item $\begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix} + \begin{bmatrix} 1 \\ 8\end{bmatrix} = \text{undefined}$ \\

\item $2\begin{bmatrix} 1 \\ 4 \\ -2 \end{bmatrix} = \begin{bmatrix}2\(1\) \\ 2\(4\) \\ 2\(-2\) \end{bmatrix} = \begin{bmatrix} 2 \\ 8\\ -4\end{bmatrix}$

\end{itemize}

\end{example}

For vectors $\vv{u}, \vv{v} \in \mathbb{R}^n$ and scalars $c,d \in \mathbb{R}$ we say $c\vv{u}+d\vv{v}$ is a \textbf{linear combination} of $\vv{u}$ and $\vv{v}$.

\begin{example}{Example: Linear Combination}

$$2\begin{bmatrix}1\\1\\-1\\3\end{bmatrix}-1\begin{bmatrix}2\\1\\4\\1\end{bmatrix}=\begin{bmatrix}2\\2\\-2\\6\end{bmatrix}+\begin{bmatrix}-2\\-1\\-4\\-1\end{bmatrix}=\begin{bmatrix}0\\1\\-6\\5\end{bmatrix}$$

\begin{center}

$\begin{bsmallmatrix}0\\1\\-6\\5\end{bsmallmatrix}$ is a linear combination of $\begin{bsmallmatrix}1\\1\\-1\\3\end{bsmallmatrix}$ and $\begin{bsmallmatrix}2\\1\\4\\1\end{bsmallmatrix}$

\end{center}

\end{example}

\begin{definition}{Definition: Linear Combination}

Let $\vv{v_1},\vv{v_2},\hdots,\vv{v_k}\in\mathbb{R}^n$ then a \textbf{linear combination} of $\vv{v_1},\vv{v_2},\hdots,\vv{v_k}$ is a vector $\vv{w}$ in $\mathbb{R}^n$ such that $\vv{w}=c_1\vv{v_1}+c_2\vv{v_2}+\hdots+c_k\vv{v_k}$ for some scalars $c_1,c_2,\hdots,c_k\in\mathbb{R}$.

\end{definition}

\begin{example}{Example: Is It A Linear Combination?}

\begin{itemize}

\item {Is $\begin{bsmallmatrix}1\\2\\3\end{bsmallmatrix}$ a linear combination of $\begin{bsmallmatrix}1\\1\\0\end{bsmallmatrix},\begin{bsmallmatrix}2\\-1\\0\end{bsmallmatrix},\begin{bsmallmatrix}1\\-1\\0\end{bsmallmatrix}$?

$$c_1\begin{bmatrix}1\\1\\0\end{bmatrix}+c_2\begin{bmatrix}2\\-1\\0\end{bmatrix}+c_3\begin{bmatrix}1\\-1\\0\end{bmatrix}=\begin{bmatrix}c_1\\c_1\\0\end{bmatrix} + \begin{bmatrix}2c_2\\-c_2\\0\end{bmatrix}+\begin{bmatrix}c_3\\-c_3\\0\end{bmatrix}\ne\begin{bmatrix}1\\2\\3\end{bmatrix}$$

\begin{center} Doesn't work as row 3 is all zeros and $0 + 0 + 0 \ne 3$ thus the answer is no.\end{center}}

\item {Is $\begin{bsmallmatrix}1\\2\\3\end{bsmallmatrix}$ a linear combination of $\begin{bsmallmatrix}1\\1\\1\end{bsmallmatrix},\begin{bsmallmatrix}1\\1\\0\end{bsmallmatrix},\begin{bsmallmatrix}-1\\1\\0\end{bsmallmatrix}$?

\begin{align*}

c_1\begin{bmatrix}1\\1\\1\end{bmatrix}+c_2\begin{bmatrix}1\\1\\0\end{bmatrix}+c_3\begin{bmatrix}-1\\1\\0\end{bmatrix}&=\begin{bmatrix}1\\2\\3\end{bmatrix} \\

\begin{bmatrix}c_1\\c_1\\c_1\end{bmatrix}+\begin{bmatrix}c_2\\c_2\\0\end{bmatrix}+\begin{bmatrix}-c_3\\c_3\\0\end{bmatrix}&=\begin{bmatrix}1\\2\\3\end{bmatrix}

\end{align*}

\begin{align*}

c_1+c_2-c_3&=1 & c_1+c_2+c_3 &= 2 & c_1 = 3 \\

3 + c_2 - c_3 &= 1 & 3+ c_2+c_3&=2 \\

c_2-c_3&=-2 & c_2+c_3 &=-1 \\

c_2 &= -2 + c_3 & -2 + c_3 + c_3 &=-1 \\

c_2 &=-2+\frac{1}{2} & 2c_3&=1 \\

c_2 &=-1.5 & c_3 &=\frac{1}{2}

\end{align*}

$$\begin{bmatrix}1\\2\\3\end{bmatrix}=3\begin{bmatrix}1\\1\\1\end{bmatrix}-\frac{3}{2}\begin{bmatrix}1\\1\\0\end{bmatrix}+\frac{1}{2}\begin{bmatrix}-1\\1\\0\end{bmatrix}\text{ thus yes}$$}

\end{itemize}

\end{example}

\begin{definition}{Definition: Zero Vector}

A \textbf{zero vector} $\vv{0}$ is a vector whose components are all zero.

\begin{example}{Example: Zero Vector}

\begin{itemize}

\item $\begin{bmatrix}0\\0\end{bmatrix}\in\mathbb{R}^2$

\item $\begin{bmatrix}0\\0\\0\\0\\0\end{bmatrix}\in\mathbb{R}^5$

\end{itemize}

\end{example}

\end{definition}

\pagebreak

\subsection{Algebraic Properties of Addition and Scalar Multiplication}

For $\vv{x}, \vv{y}, \vv{z} \in \mathbb{R}^n$ and $s, t \in \mathbb{R}$,

\begin{tasks}[label=\arabic*.,ref=\

arabic*](2)

\task $\vv{x}+\vv{y}=\vv{y}+\vv{x}$ (Commutativity)

\task $\(\vv{x}+\vv{y}\)+\vv{z}=\vv{x}+\(\vv{y}+\vv{z}\)$ (Associativity)

\task $\vv{x}+0=\vv{x}$

\task $\vv{x}+\(-\vv{x}\)=0$

\task $\(s+t\)\vv{x}=s\vv{x}+t\vv{x}$ (Distributivity)

\task $s\(\vv{x}+\vv{y}\)=s\vv{x}+s\vv{y}$ (Distributivity)

\task $\(st\)\vv{x} = s\(t\vv{x}\)$

\task $1\vv{x} = \vv{x}$

\end{tasks}

\begin{proof}

Let $\vv{u}=\begin{bmatrix}u_1,u_2,\hdots,u_n\end{bmatrix},\vv{v}=\begin{bmatrix}v_1,v_2,\hdots,v_n\end{bmatrix}\text{, and }\vv{w}=\begin{bmatrix}w_1,w_2,\hdots,w_n\end{bmatrix}.$

\begin{itemize}

\item Commutativity

\begin{align*}

\vv{u}+\vv{v}&=\begin{bmatrix}u_1,u_2,\hdots,u_n\end{bmatrix}+\begin{bmatrix}v_1,v_2,\hdots,v_n\end{bmatrix}\\

&=\begin{bmatrix}u_1+v_1,u_2+v_2,\hdots,u_n+v_n\end{bmatrix} \text{ (Definition of Vector Addition)}\\

&=\begin{bmatrix}v_1+u_1,v_2+u_2,\hdots,v_n+u_n\end{bmatrix} \text{ (Commutativity of Addition of Real Numbers)}\\

&=\begin{bmatrix}v_1,v_2,\hdots,v_n\end{bmatrix}+\begin{bmatrix}u_1,u_2,\hdots,u_n\end{bmatrix} \text{ (Definition of Vector Addition)} \\

&=\vv{v}+\vv{u}

\end{align*}

\item Associativity

\begin{align*}

\(\vv{u}+\vv{v}\)+\vv{w}&=\(\begin{bmatrix}u_1,u_2,\hdots,u_n\end{bmatrix}+\begin{bmatrix}v_1,v_2,\hdots,v_n\end{bmatrix}\)+\begin{bmatrix}w_1,w_2,\hdots,w_n\end{bmatrix} \\

&=\begin{bmatrix}u_1+v_1,u_2+v_2,\hdots,u_n+v_n\end{bmatrix}+\begin{bmatrix}w_1,w_2,\hdots,w_n\end{bmatrix} \\

&=\begin{bmatrix}\(u_1+v_1\)+w_1,\(u_2+v_2\)+w_2,\hdots,\(u_n+v_n\)+w_n\end{bmatrix} \\

&=\begin{bmatrix}u_1+\(v_1+w_1\),\hdots,u_n+\(v_n+w_n\)\end{bmatrix} \text{ (Associativity of Addition of Real Numbers)} \\

&=\begin{bmatrix}u_1,u_2,\hdots,u_n\end{bmatrix}+\begin{bmatrix}v_1+w_1,v_2+w_2,\hdots,v_n+w_n\end{bmatrix} \\

&=\begin{bmatrix}u_1,u_2,\hdots,u_n\end{bmatrix}+\(\begin{bmatrix}v_1,v_2,\hdots,v_n\end{bmatrix}+\begin{bmatrix}w_1,w_2,\hdots,w_n\end{bmatrix}\) \\

&=\vv{u}+\(\vv{v}+\vv{w}\)

\end{align*}

\end{itemize}

\end{proof}

\begin{definition}{Definition: Vector Space}

$\mathbb{R}^n$ is called a \textbf{vector space} (a set of vectors along with addition and scalar multiplication that satisfy the above 8 properties).

\end{definition}

\pagebreak

\subsection{Geometric Representation of Vectors in $\mathbb{R}^2$ and $\mathbb{R}^3$}

\begin{definition}{Definition: Standard Position}

A vector is in \textbf{standard position} if its initial point is the origin.

\end{definition}

\begin{tblr}{width=\textwidth,

cells = {halign = c, valign = m}, colspec={X X}}

{$\mathbb{R}^2=\left\{\begin{bmatrix}x \\ y\end{bmatrix}\Big|x,y\in\mathbb{R}\right\}$} &

{\raisebox{-.5\height}{\begin{tikzpicture}

\begin{axis} [

width = \linewidth,

axis x line=middle,

axis y line=middle,

ymin = -2, ymax = 2,

xmin = -2, xmax = 2,

xticklabels = {100},

yticklabels = {100},

ticks = none,

axis line style = {<->, ultra thick},

xlabel = {$x$},

ylabel = {$y$},

clip = false,

]

\addplot[mark=*] coordinates {(0.5,-0.5)} node[right] {$A$} node[below] {$\(a_1, a_2\)$};

\addplot[mark=*] coordinates {(0.25,1)} node[below right] {$B$} node[above right] {$\(b_1, b_2\)$};

\draw[->, red, ultra thick] (0.5,-0.5) -- (0.25,1);

\addplot[mark=*] coordinates {(0,0)};

\addplot[mark=*] coordinates {(-0.25,1.5)} node[above left] {$\(b_1-a_1, b_2-a_2\)$};

\draw[->, red, ultra thick] (0,0) -- (-0.25, 1.5);

\end{axis}

\end{tikzpicture}}

$\begin{aligned}

\vv{v}&=\vv{AB}=\begin{bmatrix}b_1 -a_1\\b_2-a_2\end{bmatrix}\text{Compact Form} \\

&= \vv{\(0,0\)\(b_1-a_1,b_2-a_2\)}

\end{aligned}$}

\end{tblr}

\begin{center}

\begin{tikzpicture}[scale = 2]

\draw[dotted, blue, ultra thick, <->] (-1.5, -1.5) -- (1.5, 1.5);

\draw[->, ultra thick] (0,0) -- (0.5, 0.5) node[right] {$\vv{u}$};

\fill (0,0) circle[radius=1pt] node[left] {$\vv{0}$};

\draw[->, ultra thick] (2,0) -- (1.75, -0.25) node[above, pos = 0] {$-\frac{1}{2}\vv{u}$};

\fill (2, 0) circle[radius = 1pt];

\draw[->, ultra thick] (4,0) -- (5, 1) node[right] {$2\vv{u}$};

\fill (4, 0) circle[radius = 1pt];

\fill (6, 0) circle[radius = 1pt] node[right] {$0\vv{u} = \vv{0}$};

\end{tikzpicture} \\

\begin{flushleft}

All scalar multiples of $\vv{u}$ (in standard position) lie on the dotted line. This is due to all non zero multiples of $\vv{u}$ being parallel to $\vv{u}$.\\

\end{flushleft}

\newline

\begin{tblr}{colspec = {X X X}, cells = {halign = c}, width = \textwidth}

{\begin{tikzpicture}[scale = 4]

\draw[->, ultra thick] (0,0) -- (0.86602540378443864676372317075294, 0.5) node[rotate = 30, above, pos = 0.5] {$\vv{u}$};

\fill (0,0) circle[radius = 1pt];

\end{tikzpicture}} &

{\begin{tikzpicture}[scale = 4]

\draw[->, ultra thick] (0,0) -- (0.5, -0.86602540378443864676372317075294) node[rotate = -60, above, pos = 0.5] {$\vv{v}$};

\fill (0,0) circle[radius = 1pt];

\end{tikzpicture}} &

{\begin{tikzpicture}[scale = 4]

\fill (1.3660254037844386467637231707529, -0.36602540378443864676372317075294) circle[radius = 1pt];

\draw[->, ultra thick] (0,0) -- (0.86602540378443864676372317075294, 0.5) node[rotate = 30, above, pos = 0.5] {$\vv{u}$};

\draw[->, ultra thick] (0.86602540378443864676372317075294, 0.5) -- (1.3660254037844386467637231707529, -0.36602540378443864676372317075294) node[rotate = -60, above, pos = 0.5] {$\vv{v}$};

\draw[->, ultra thick, red] (0, 0) -- (1.3660254037844386467637231707529, -0.36602540378443864676372317075294) node[rotate = -15, below, pos = 0.5] {$\vv{u} + \vv{v}$};

\fill (0,0) circle[radius = 1pt];

\end{tikzpicture}}

\end{tblr} \\

Head to Tail Rule \\

\newline

\begin{tikzpicture}[scale = 4]

\fill (1.3660254037844386467637231707529, -0.36602540378443864676372317075294) circle[radius = 1pt];

\draw[->, ultra thick] (0,0) -- (0.86602540378443864676372317075294, 0.5) node[rotate = 30, above, pos = 0.5] {$\vv{u}$};

\draw[->, ultra thick, dotted] (0.86602540378443864676372317075294, 0.5) -- (1.65469976687, -0.86602540378443864676372317075294);

\draw[->, ultra thick] (0,0) -- (0.5, -0.86602540378443864676372317075294) node[rotate = -60, below, pos = 0.5] {$\vv{v}$};

\draw[->, ultra thick, dotted] (0.5, -0.86602540378443864676372317075294) -- (1.75, -0.144337567297);

\draw[->, ultra thick, red] (0, 0) -- (1.3660254037844386467637231707529, -0.36602540378443864676372317075294) node[rotate = -15, below, pos = 0.5] {$\vv{u} + \vv{v}$};

\fill (0,0) circle[radius = 1pt];

\end{tikzpicture} \\

Paralleogram Rule

\end{center}

\begin{example}{Example: Is Every Vector In $\mathbb{R}^2$ A Linear Combination?}

Is every vector in $\mathbb{R}^2$ a linear combination of $\vv{a}=\begin{bsmallmatrix}1\\1\end{bsmallmatrix}$ and $\vv{b}=\begin{bsmallmatrix}2\\-1\end{bsmallmatrix}$? \\

$$\begin{bmatrix}x\\y\end{bmatrix}=c_1\begin{bmatrix}1\\1\end{bmatrix}+c_2\begin{bmatrix}2\\-1\end{bmatrix}$$

\begin{tblr}{width=\textwidth,

cells = {halign = c, valign = m}, colspec={X X}}

{$\begin{aligned}

\vv{a}-\vv{b} &= \vv{v} \\

\frac{5}{3}\vv{b}-\frac{3}{4}\vv{a}&=\vv{w}

\end{aligned}$} &

{\raisebox{-.5\height}{\begin{tikzpicture}[scale = 1.3]

\fill (1,1) circle[radius = 3pt];

\fill (-1,2) circle[radius = 3pt];

\draw[->, ultra thick] (0,0) -- (1,1) node[above, pos = 0.5, rotate = 45] {$\vv{a}$};

\draw[<->, ultra thick, dotted] (-3,-3) -- (2,2);

\draw[<->, ultra thick, dotted] (-2.5,1.25) -- (3.5,-1.75);

\draw[red, ->, ultra thick] (0, 0) -- (3.3333333333333333333333333333333, -1.6666666666666666666666666666667) node[above, pos = 0.8, rotate = 333.43494882292]{$\frac{5}{3}\vv{b}$};

\fill (2,-1) circle[radius = 3pt];

\draw[->, ultra thick] (0,0) -- (2,-1) node[above, pos = 0.5, rotate = 333.43494882292] {$\vv{b}$};

\draw[->, ultra thick, blue] (0,0) -- (-2,1) node[above, pos = 0.5, rotate = -26.56505117708] {$-\vv{b}$};

\draw[->, ultra thick, blue] (-2, 1) -- (-1, 2) node[pos = 0.5, rotate = 45, above] {\vv{a}};

\draw[->, ultra thick, red] (0,0) -- (-0.75,-0.75) node[above, pos = 0.5, rotate = 45] {$-\frac{3}{4}\vv{a}$};

\draw[->, ultra thick, purple] (0,0) -- (-1, 2) node[above, pos = 0.5, rotate = -63.43494882292] {$\vv{v}$};

\fill (2.5833333333333333333333333333333,-2.4166666666666666666666666666667) circle[radius = 3pt];

\draw[->, ultra thick, red] (3.3333333333333333333333333333333, -1.6666666666666666666666666666667) -- (2.5833333333333333333333333333333,-2.4166666666666666666666666666667) node[above, pos = 0.5, rotate = 45] {$-\frac{3}{4}\vv{a}$};

\draw[->, ultra thick, orange] (0,0) -- (2.5833333333333333333333333333333,-2.4166666666666666666666666666667) node[above, pos = 0.5, rotate = 316.909152433] {\vv{w}};

\fill (0,0) circle[radius = 3pt];

\end{tikzpicture}}}

\end{tblr}

\end{example}

\begin{definition}{Definition: Span $\mathbb{R}^n$}

We say the vectors $\vv{v_1},\vv{v_2},\hdots,\vv{v_k}$ \textbf{span or generate $\mathbb{R}^n$} if every vector in $\mathbb{R}^n$ is a linear combination of $\vv{v_1},\vv{v_2},\hdots,\vv{v_k}$.

\begin{example}{Example: Vectors That Span $\mathbb{R}^2$}

$\begin{bmatrix}1 \\ 1\end{bmatrix}$ and $\begin{bmatrix}2\\-1\end{bmatrix}$ span $\mathbb{R}^2$.

\end{example}

\end{definition}

\begin{example}{Example: Do They Span $\mathbb{R}^2$}

Do the following vectors span $\mathbb{R}^2$?

\begin{tasks}[style = enumerate](2)

\task{$\begin{bmatrix} 1 \\2\end{bmatrix}$

\raisebox{-.5\height}{\begin{tikzpicture}[scale = 0.25]

\fill (0,0) circle[radius = 10pt];

\draw[->, black, ultra thick] (0,0) -- (1,2);

\draw[<->, black, ultra thick, dotted] (-2,-4) -- (2,4);

\end{tikzpicture}} no}

\task{$\begin{bmatrix} 1 \\2\end{bmatrix}$, $\begin{bmatrix}-2\\-4\end{bmatrix}$

\raisebox{-.5\height}{\begin{tikzpicture}[scale = 0.25]

\fill (0,0) circle[radius = 10pt];

\draw[->, black, ultra thick] (0,0) -- (1,2);

\draw[->, black, ultra thick] (0,0) -- (-2,-4);

\draw[<->, black, ultra thick, dotted] (-3,-6) -- (3,6);

\end{tikzpicture}} no}

\task{$\begin{bmatrix} 1 \\2\end{bmatrix}$, $\begin{bmatrix}-2\\-4\end{bmatrix}$, $\begin{bmatrix}3\\-3\end{bmatrix}$

\raisebox{-.5\height}{\begin{tikzpicture}[scale = 0.25]

\fill (0,0) circle[radius = 10pt];

\draw[->, black, ultra thick] (0,0) -- (1,2);

\draw[->, black, ultra thick] (0,0) -- (-2,-4);

\draw[<->, black, ultra thick, dotted] (-3,-6) -- (3,6);

\draw[<->, black, ultra thick, dotted] (-3,3) -- (4,-4);

\draw[->, black, ultra thick] (0,0) -- (3,-3);

\end{tikzpicture}} yes}

\task{$\begin{bmatrix} 1 \\2\end{bmatrix}$, $\begin{bmatrix}-2\\-4\end{bmatrix}$, $\begin{bmatrix}3\\-3\end{bmatrix}$, $\begin{bmatrix}1\\0\end{bmatrix}$

\raisebox{-.5\height}{\begin{tikzpicture}[scale = 0.25]

\fill (0,0) circle[radius = 10pt];

\draw[->, black, ultra thick] (0,0) -- (1,2);

\draw[->, black, ultra thick] (0,0) -- (-2,-4);

\draw[<->, dotted, black, ultra thick] (-2,0) -- (2,0);

\draw[->, black, ultra thick] (0,0) -- (1,0);

\draw[<->, black, ultra thick, dotted] (-3,-6) -- (3,6);

\draw[<->, black, ultra thick, dotted] (-3,3) -- (4,-4);

\draw[->, black, ultra thick] (0,0) -- (3,-3);

\end{tikzpicture}} yes}

\end{tasks}

To span $\mathbb{R}^2$, we need at least 2 nonzero vectors in $\mathbb{R}^2$ that are not parallel (not scalar multiples of each other).

\end{example}

\newpage

\section{Section 1.2 Dot Product}

\subsection{Introduction}

\begin{definition}{Definition: Dot Product}

The \textbf{dot product} of vectors $\vv{u},\vv{v}\in\mathbb{R}^n$ is a scalar in $\mathbb{R}$ and defined by:

$$\vv{u}\cdot\vv{v}=\begin{bmatrix}u_1\\u_2\\\vdots\\u_n\end{bmatrix}\cdot\begin{bmatrix}v_1\\v_2\\\vdots\\v_n\end{bmatrix}=u_1v_1+u_2v_2+\hdots+u_nv_n$$

\begin{example}{Example: Dot Product}

$\begin{bmatrix}2\\1\\-3\\4\end{bmatrix}\cdot\begin{bmatrix}1\\3\\1\\0\end{bmatrix}=2+3+\(-3\)+0=2$

\end{example}

Note: $\vv{u}\cdot\vv{v}\ne\vv{u}\vv{v}$ as $\vv{u}\vv{v}=\text{undefined}$

\end{definition}

\newpage

\subsection{Properties of Dot Product}

Let $\vv{u},\vv{v},\vv{w}\in\mathbb{R}^n$ and $c\in\mathbb{R}$. Then

\begin{enumerate}

\item$\vv{u}\cdot\vv{v}=\vv{v}\cdot\vv{u}$ (Commutativity)

\item$\vv{0}\cdot\vv{u}=0\(\text{for } \vv{0}\in\mathbb{R}^n\)$

\item$\vv{u}\cdot\(\vv{v}+\vv{w}\)=\vv{u}\cdot\vv{v}+\vv{u}\cdot\vv{w}$ (Distributivity)

\item$\(c\vv{u}\)\cdot\vv{v}=\vv{u}\cdot\(c\vv{v}\)=c\(\vv{u}\cdot\vv{v}\)$

\item$\vv{u}\cdot\vv{u}\ge0\text{ and }\vv{u}\cdot\vv{u}=0\text{ if and only if }\vv{u}=\vv{0}$

\end{enumerate}

\begin{proof} \newline

\begin{itemize}

\item Commutativity \\ \newline

Applying the definition of dot product to $\vv{u}\cdot\vv{v}$ and $\vv{v}\cdot\vv{u}$ we obtain

\begin{align*}

\vv{u}\cdot\vv{v}&=u_1v_1+u_2v_2+\hdots+u_nv_n \\

&=v_1u_1+v_2u_2+\hdots+v_nu_n \text{ (Multiplication of Real Numbers is Commutative)}\\

&=\vv{v}\cdot\vv{u}

\end{align*}

\item $\(c\vv{u}\)\cdot\vv{v}=c\(\vv{u}\cdot\vv{v}\)$ \\ \newline

Using the definitions of scalar multiplication and dot product, we have

\begin{align*}

\(c\vv{u}\)\cdot\vv{v}&=\begin{bmatrix}cu_1,cu_2,\hdots,cu_n\end{bmatrix}\cdot\begin{bmatrix}v_1,v_2,\hdots,v_2\end{bmatrix} \\

&=cu_1v_1+cu_2v_2+\hdots+cu_nv_n \\

&=c\(u_1v_1+u_2v_2+\hdots+u_nv_n\) \\

&=c\(\vv{u}\cdot\vv{v}\)

\end{align*}

\end{itemize}

\end{proof}

Question: Is $\overbrace{\(\vv{u}\cdot\vv{v}\)}^{\in\mathbb{R}}\cdot\vv{w}$ defined for vectors $\vv{u},\vv{v},\vv{w}\in\mathbb{R}^n$? No, as $\vv{u}\cdot\vv{v}$ returns a scalar which also means $\(\vv{u}\cdot\vv{v}\)\vv{w}$ is defined.

\begin{definition}{Definition: Length/Magnitude/Norm}

The \textbf{length/magnitude/norm} of a vector $\vv{u}\in\mathbb{R}^n$ is defined to be $\norm{\vv{u}}=\sqrt{\vv{u}\cdot\vv{u}}$ which also means $\vv{u}\cdot\vv{u}=\norm{\vv{u}}^2$.

$$\text{If }\vv{u}=\begin{bmatrix}u_1\\u_2\\\vdots\\u_n\end{bmatrix}\text{ then }\norm{\vv{u}}=\sqrt{u_1^2+u_2^2+\hdots+u_n^2}$$

Notes:

\begin{tasks}[label = $\star$]

\task $\norm{c\vv{u}}=\abs{c}\times\norm{\vv{u}}$

\task $\norm{\vv{u}+\vv{v}}\ne\norm{\vv{u}}+\norm{\vv{v}}$

\end{tasks}

\end{definition}

\begin{definition}{Definition: Unit Vector}

A \textbf{unit vector} is a vector whose magnitude is one.

\begin{example}{Example: Unit Vector}

$\begin{bmatrix}-1/3\\2/3\\2/3\end{bmatrix}$ thus $\norm{\begin{bmatrix}-1/3\\2/3\\2/3\end{bmatrix}}=\sqrt{\frac{1}{9}+\frac{4}{9}+\frac{4}{9}}=1$

\end{example}

\end{definition}

\begin{definition}{Definition: Normalizing}

For any nonzero vector $\vv{u}$, we can create a unit vector in the same drection as $\vv{u}$ by \textbf{normalizing} $\vv{u}$.

$$\underbrace{\frac{\vv{u}}{\norm{\vv{u}}}=\frac{1}{\norm{\vv{u}}}\vv{u}}_{\text{normalization of }\vv{u}}$$

\begin{example}{Example: Normalizing}

$\vv{u}=\begin{bmatrix}1\\-1\\2\\3\end{bmatrix}$

\begin{align*}

\norm{\vv{u}}&=\sqrt{1+1+4+9} =\sqrt{15} \\

\vv{v}&=\frac{1}{\sqrt{15}}\begin{bmatrix}1\\-1\\2\\3\end{bmatrix}

\end{align*}

\end{example}

\end{definition}

\begin{definition}{Definition: Distance}

The \textbf{distance} or $d\(\vv{x},\vv{y}\)$ between the vectors $\vv{x},\vv{y}\in\mathbb{R}^n$ is defined by

$$d\(\vv{x},\vv{y}\)=\norm{\vv{x}-\vv{y}}.$$

\begin{center}

\begin{tikzpicture}[scale = 3]

\fill (0.5,0.86602540378443864676372317075294) circle[radius = 1pt];

\fill (0.86602540378443864676372317075294, 0.5) circle[radius = 1pt];

\draw[ultra thick, black] (0.5,0.86602540378443864676372317075294) -- (0.86602540378443864676372317075294, 0.5) node [right, pos = 0.5] {$d\(\vv{x},\vv{y}\)$};

\draw[ultra thick, -stealth, red] (0,0) -- (0.5,0.86602540378443864676372317075294) node[above, pos = 0.5, rotate = 60] {\vv{y}};

\draw[ultra thick, -stealth, blue] (0,0) -- (0.86602540378443864676372317075294, 0.5) node[above, pos = 0.5, rotate = 30] {\vv{x}};

\fill (0,0) circle[radius = 1pt];

\end{tikzpicture}

\end{center}

\begin{example}{Example: Distance}

$\vv{x}=\begin{bmatrix}1\\3\\2\end{bmatrix}, \vv{y}=\begin{bmatrix}-2\\4\\1\end{bmatrix}$

$$d\(\vv{x},\vv{y}\)=\norm{\vv{x}-\vv{y}}=\norm{\begin{bmatrix}1-\(-2\)\\3-4\\2-1\end{bmatrix}}=\norm{\begin{bmatrix}3\\-1\\1\end{bmatrix}}=\sqrt{9+1+1}=\sqrt{11}$$

\end{example}

\end{definition}

\begin{definition}{Definition: Orthogonal}

Two vectors $\vv{u},\vv{v}\in\mathbb{R}^n$ are \textbf{orthogonal} if $\vv{u}\cdot\vv{v}=0$.

\begin{example}{Example: Orthogonal}

$$\begin{bmatrix}1\\3\\1\\-1\end{bmatrix}\cdot\begin{bmatrix}2\\1\\-4\\1\end{bmatrix}=2+3-4-1=0$$

\end{example}

Notes:

\begin{tasks}[label = $\star$]

\task Nonzero vectors in $\mathbb{R}^2$ or $\mathbb{R}^3$ that are orthogonal are perpendicular.

\task {$\vv{u}\cdot\vv{0}=0$ so $\vv{0}\in\mathbb{R}^n$ is orthogonal to every vector in $\mathbb{R}^n$.

\begin{tikzpicture}[scale = 2]

\fill (0,0) circle[radius = 1.5pt];

\draw[ultra thick, -stealth, black] (0,0) -- (1,0);

\end{tikzpicture}}

\task $\vv{0}$ is the only vector orthogonal to itself.

\end{tasks}

\end{definition}

\begin{example}{Example: Find Orthogonal Vectors}

Find a vector orthogonal to both $\begin{bmatrix}1\\-1\\2\end{bmatrix}$ and $\begin{bmatrix}2\\1\\1\end{bmatrix}$.\\

\newline

Some plug and chug solutions are $\begin{bmatrix}-1\\1\\1\end{bmatrix}$ and $\begin{bmatrix}1\\-1\\-1\end{bmatrix}$ but below is a general solution.

$$\begin{bmatrix}x\\y\\z\end{bmatrix}\cdot\begin{bmatrix}1\\-1\\2\end{bmatrix}=0\text{ and }\begin{bmatrix}x\\y\\z\end{bmatrix}\cdot\begin{bmatrix}2\\1\\1\end{bmatrix}=0$$

\begin{align*}

x-y+2z&=0 & 2x+y+z&=0 \\

x&=y-2z & 2\(y-2z\)+y+z&=0 \\

x&=y-2y & 2y-4z+y+z&=0 \\

x&=-y & 3y-3z&=0 \\

& & y&=z

\end{align*}

$$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-a\\a\\a\end{bmatrix}$$

\end{example}

\newpage

\section{Section 3.1/3.2 Matrices}

\subsection{Introduction}

\begin{definition}{Definition: Matrix}

An $m\times n$ \textbf{matrix} $A$ is a rectangular array of $mn$ entries:

$$A=\begin{bmatrix}a_{11}&a_{12}&a_{13}&\hdots&a_{1n}\\a_{21}&a_{22}&a_{23}&\hdots&a_{2n}\\\vdots&\vdots&\vdots&\hdots&\vdots\\a_{m1}&a_{m2}&a_{m3}&\hdots&a_{mn}\end{bmatrix}$$

Notes:

\begin{itemize}

\item $m\times n$ is the \textbf{size} of the matrix $A$ with $m$ being the number of rows and $n$ being the number of columns.

\item $a_{ij}$ is the $\(i,j\)$ entry of $A$ (the entry in the $i^{\text{th}}$ row and $j^{\text{th}}$ column of $A$)

\item $\vv{a_j}$ is the $j^{\text{th}}$ column of $A$

\item $\vv{A_i}$ is the $i^{\text{th}}$ row of $A$

\end{itemize}

\begin{example}{Example: Matrix}

$$B=\underbrace{\begin{bmatrix}3&-2&1\\4&8&9\end{bmatrix}}_{2\times3}$$

\begin{itemize}

\item $b_{2,1}=4$, $b_{12}=-2$

\item $\vv{b_2}=\begin{bmatrix}-2\\8\end{bmatrix}$

\item $\vv{B_2}=\begin{bmatrix}4&8&9\end{bmatrix}$

\end{itemize}

\end{example}

\end{definition}

\begin{definition}{Definition: Main Diagonal}

The \textbf{main diagonal entries} of a matrix $A$ are entries of the form $a_{ii}$. \\

\newline

The \textbf{main diagonal} of $A$ consists of all main diagonal entries of $A$.

\begin{example}{Example: Main Diagonal}

\begin{center}

$C=$\begin{blockarray}{cccc}

\begin{block}{ [ ccc c ]}

\bigstrut[t]

\tikzmarknode[inner sep=2pt]{A11}{-8} & 9 & 4 \\

2 & 1 & 0 \\

-1 & 2 & \tikzmarknode[inner sep=2pt]{A33}{9} \bigstrut[b] \\

\end{block}

\end{blockarray}

\end{center}

\begin{tikzpicture}[remember picture,overlay]

\draw[red, thick] plot[smooth cycle] coordinates {(A11.north)

(A33.north east) (A33.south) (A11.south west)};

\end{tikzpicture}

$C$ is a $3\times3$ matrix with the red circle indicating the main diagonal of $C$.

\end{example}

\end{definition}

\begin{definition}{Definition: Square Matrix}

A \textbf{square matrix} is a matrix with size $n\times n$.

\end{definition}

\begin{definition}{Definition: Zero Matrix}

A \textbf{zero matrix} (deonoted $0$ or $0_{mn}$) is an $m\times n$ matrix whose entries are all zero.

\begin{example}{Example: Zero Matrix}

$$0_{4,2}=\begin{bmatrix}0&0\\0&0\\0&0\\0&0\end{bmatrix}$$

\end{example}\

\end{definition}

\begin{definition}{Definition: Identity Matrix}

The \textbf{identity matrix} $I_n$ is the $m\times n$ matrix whose main diagonal entries are all $1$ and all other entries are $0$.

\begin{example}{Example: Identity Matrix}

\begin{itemize}

\item $I_2=\begin{bmatrix}1&0\\0&1\end{bmatrix}$

\item $I_4=\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}$

\end{itemize}

\end{example}

\end{definition}

\begin{definition}{Definition: Equal Matrices}

We say \textbf{$A=B$} if $A$ and $B$ are both $m\times n$ and $a_{ij}=b_{ij}$ for all $i,j$.

\end{definition}

\newpage

\subsection{Matrix Addition and Scalar Multiplication}

If $A$ and $B$ are both $m\times n$ matrices, then $A+B$ is the $m\times n$ matrix where:

$$\(A+B\)_{ij}=a_{ij}+b_{ij}$$

\begin{example}{Example: Adding Matrices}

$$\begin{bmatrix}1&2\\-1&3\\0&8\\4&9\end{bmatrix}+\begin{bmatrix}-2&9\\1&4\\2&2\\0&8\end{bmatrix}=\begin{bmatrix}-1&11\\0&7\\2&10\\4&17\end{bmatrix}$$

\end{example}

For $c\in\mathbb{R}$ and $m\times n$ matrix $A$, $cA$ is the $m\times n$ matrix where:

$$\(cA\)_{ij}=ca_{ij}$$

\begin{example}{Example: Scalar Multiplication With Matrices}

$$2\begin{bmatrix}-1&3&8\\4&9&1\end{bmatrix}=\begin{bmatrix}-2&6&16\\8&18&2\end{bmatrix}$$

\end{example}

Note: A column vector $\vv{v}\in\mathbb{R}^n$ is an $n\times 1$ matrix.

\begin{tasks}[label = $\star$]

\task {The properties of Matrix Addition and Scalar Multiplication correspond directly to the properties discussed in Section 1.1. For vector addition and scalar multiplication.}

\end{tasks}

\begin{definition}{Definition: Transpose}

Let $A$ be an $m\times n$ matrix. The $n\times m$ matrix $A^T$ (\textbf{``$A$ Transpose''}) is the mattrix where:

$$\(A^T\)_{ij}=a_{ji}$$

\begin{example}{Example: Transpose}

\begin{itemize}

\item $A=\underbrace{\begin{bmatrix}2&-1&3\\0&4&5\end{bmatrix}}_{2\times3}$, $A^T=\underbrace{\begin{bmatrix}2&0\\-1&4\\3&5\end{bmatrix}}_{3\times2}$

\item $\vv{v}=\begin{bmatrix}1\\2\\3\\4\end{bmatrix}\text{Column Vector } 4\times1$, $\vv{v}^T=\underbrace{\begin{bmatrix}1&2&3&4\end{bmatrix}}_{\text{Row Vector } 1\times4}$

\end{itemize}

\end{example}

Notes:

\begin{itemize}

\item $\(A^T\)^T=A$

\item $\(A+B\)^T=A^T+B^T$

\end{itemize}

\end{definition}

\begin{definition}{Definition: Symmetric Matrix}

If $A=A^T$, we call $A$ a \textbf{symmetric matrix}.

\begin{example}{Example: Symmetric Matrix}

\begin{center}

$A=$\begin{blockarray}{cccc}

\begin{block}{ [ ccc c ]}

\bigstrut[t]

\tikzmarknode[inner sep=2pt]{A11}{1} & 3 & 8 \\

3 & 2 & 4 \\

8 & 4 & \tikzmarknode[inner sep=2pt]{A33}{9} \bigstrut[b] \\

\end{block}

\end{blockarray}

\end{center}

\begin{tikzpicture}[remember picture,overlay]

\draw[red, thick, dotted] plot[smooth cycle] coordinates {(A11.north)

(A33.north east) (A33.south) (A11.south west)};

\end{tikzpicture}

\end{example}

\end{definition}

\newpage

\subsection{Matrix-Vector Products}

Let $A$ be a $m\times n$ matrix and $\vv{x}\in\mathbb{R}^n$ then:

\begin{align*}

A\vv{x}&=\begin{bmatrix}\vv{a_1},\vv{a_2},\vv{a_3},\hdots,\vv{a_n}\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\\\vdots\\x_n\end{bmatrix} \\

&=x_1\vv{a_1}+x_2\vv{a_2}+x_3\vv{a_3}+\hdots+x_n\vv{a_n}

\end{align*}

Notes:

\begin{itemize}

\item $A\vv{x}$ is a linear combination of the columns of $A$.

\item {$A\vv{x}$ is only defined if the number of columns in $A$ equals the number of rows/components in $\vv{x}$. For $A\vv{x}=\vv{b}$ to be defined then $A$ needs to be $m\times n$ and $\vv{x}$ needs to be $n\times 1$.}

\end{itemize}

\begin{example}{Example: Matrix-Vector Multiplication Example 1}

\begin{align*}

\underbrace{\begin{bmatrix}1&3&-1\\2&4&5\end{bmatrix}}_{\textcolor{red}{2}\times\textcolor{violet}{3}}\underbrace{\begin{bmatrix}-3\\0\\6\end{bmatrix}}_{\textcolor{violet}{3}\times\textcolor{red}{1}}&=3\begin{bmatrix}1\\2\end{bmatrix}+0\begin{bmatrix}3\\4\end{bmatrix}+6\begin{bmatrix}-1\\5\end{bmatrix} \\

&=\begin{bmatrix}-3\\-6\end{bmatrix}+\vv{0}+\begin{bmatrix}-6\\30\end{bmatrix} \\

&=\underbrace{\begin{bmatrix}-9\\24\end{bmatrix}}_{\textcolor{red}{2}\times\textcolor{red}{1}}

\end{align*}

\end{example}

\begin{example}{Example: Matrix-Vector Multiplication Example 2}

\begin{align*}

\underbrace{\begin{bmatrix}1&3&8\\0&-1&1\\2&1&4\\1&2&1\end{bmatrix}}_{\textcolor{red}{4}\times\textcolor{violet}{3}}\underbrace{\begin{bmatrix}2\\-1\\1\end{bmatrix}}_{\textcolor{violet}{3}\times\textcolor{red}{1}}&=2\begin{bmatrix}1\\0\\2\\1\end{bmatrix}-1\begin{bmatrix}3\\-1\\1\\2\end{bmatrix}+1\begin{bmatrix}8\\1\\4\\1\end{bmatrix} \\

&=\begin{bmatrix}2\\0\\4\\2\end{bmatrix}-\begin{bmatrix}3\\-1\\1\\2\end{bmatrix}+\begin{bmatrix}8\\1\\4\\1\end{bmatrix}\\

&=\underbrace{\begin{bmatrix}7\\2\\7\\1\end{bmatrix}}_{\textcolor{red}{4}\times\textcolor{red}{1}}

\end{align*}

\end{example}

\begin{example}{Example: Matrix-Vector Multiplication Example 3}

\begin{center}

\begin{blockarray}{cccc}

\begin{block}{ [ ccc c ]}

\bigstrut[t]

\tikzmarknode[inner sep=2pt]{A11}{1} & \tikzmarknode[inner sep=2pt]{A12}{3} & \tikzmarknode[inner sep=2pt]{A13}{8} \\

0 & -1 & 1 \\

2 & 1 & 4 \\

1 & 2 & 1\bigstrut[b] \\

\end{block}

\end{blockarray}

\begin{blockarray}{cccc}

\begin{block}{ [ ccc c ]}

\bigstrut[t]

\tikzmarknode[inner sep=2pt]{B11}{2} \\ \tikzmarknode[inner sep=2pt]{B21}{-1} \\ \tikzmarknode[inner sep=2pt]{B31}{1} \bigstrut[b] \\

\end{block}

\end{blockarray}$=\begin{bmatrix}1\(2\)+3\(-1\)+8\(1\)\\0\(2\)-1\(-1\)+\(1\)\(1\)\\2\(2\)+1\(-1\)+4\(1\)\\1\(2\)+2\(-1\)+1\(1\)\end{bmatrix}=\begin{bmatrix}7\\2\\7\\1\end{bmatrix}$

\end{center}

\begin{tikzpicture}[remember picture,overlay]

\draw[red, thick] plot[smooth cycle] coordinates {(A11.north) (A12.north) (A13.north) (A13.east) (A13.south) (A12.south) (A11.south) (A11.west)};

\draw[red, thick] plot[smooth cycle] coordinates {(B11.north) (B11.east) (B21.east) (B31.east) (B31.south) (B31.west) (B21.west) (B11.west)};

\end{tikzpicture}

\end{example}

\newpage

\subsection{Matrix Multiplication}

Let $A$ be an $m\times \textcolor{red}{n}$ matrix and let $B$ be an $\textcolor{red}{n}\times q$ matrix. Then $AB$ is the $m\times q$ matrix where:

\begin{align*}

AB&=A\begin{bmatrix}\vv{b_1}&\vv{b_2}&\vv{b_3}\hdots \vv{b_q}\end{bmatrix} \\

&=\begin{bmatrix}A\vv{b_1}&A\vv{b_2}&A\vv{b_3}\hdots A\vv{b_q}\end{bmatrix}

\end{align*}

Notes:

\begin{itemize}

\item $AB=C$ where $A$ is $\textcolor{red}{m}\times\textcolor{violet}{n}$, $B$ is $\textcolor{violet}{n}\times\textcolor{red}{q}$ and $C$ is $\textcolor{red}{m}\times\textcolor{red}{q}$

\item {Question: If $A$ is $m\times n$ and $\vv{x}\in\mathbb{R}^n$, is $\vv{x}A$ defined? $\vv{x}A$ would be undefined as $\vv{x}$ is $\textcolor{red}{n}\times\textcolor{violet}{1}$ and $A$ is $\textcolor{violet}{m}\times \textcolor{red}{n}$ so unless $m=1$ then $\vv{x}A=\text{undefined}$. In conclusion $A\vv{x}\ne\vv{x}A$.}

\end{itemize}

\begin{example}{Example: Matrix Multiplication Example 1}

$$AB=\underbrace{\begin{bmatrix}1&3&-1\\2&0&4\end{bmatrix}}_{2\times\textcolor{red}{3}}\underbrace{\begin{bmatrix}-1&0&2&1\\1&1&2&-1\\0&1&-1&2\end{bmatrix}}_{\textcolor{red}{3}\times4}=\underbrace{\begin{bmatrix}2&2&9&-4\\-2&4&0&10\end{bmatrix}}_{2\times4}$$

\begin{align*}

\begin{bmatrix}1&3&-1\\2&0&4\end{bmatrix}\begin{bmatrix}-1\\1\\0\end{bmatrix}&=-1\begin{bmatrix}1\\2\end{bmatrix}+1\begin{bmatrix}3\\0\end{bmatrix}+0\begin{bmatrix}-1\\4\end{bmatrix}=\begin{bmatrix}2\\-2\end{bmatrix} \\

\begin{bmatrix}1&3&-1\\2&0&4\end{bmatrix}\begin{bmatrix}0\\1\\1\end{bmatrix}&= 0\begin{bmatrix}1\\2\end{bmatrix}+1\begin{bmatrix}3\\0\end{bmatrix}+1\begin{bmatrix}-1\\4\end{bmatrix}=\begin{bmatrix}2\\4\end{bmatrix}\\

\begin{bmatrix}1&3&-1\\2&0&4\end{bmatrix}\begin{bmatrix}2\\2\\-1\end{bmatrix}&= \begin{bmatrix}9\\0\end{bmatrix}\\

\begin{bmatrix}1&3&-1\\2&0&4\end{bmatrix}\begin{bmatrix}1\\-1\\2\end{bmatrix}&=\begin{bmatrix}-4\\10\end{bmatrix}

\end{align*}

\end{example}

Notes:

\begin{itemize}

\item $AB=C$ as $A$ is $2\times\textcolor{red}{3}$ and $B$ is $\textcolor{red}{3}\times4$ making $C$ being $2\times4$

\item $BA=\text{undefined}$ as $B$ is $3\times\textcolor{red}{4}$ and $A$ is $\textcolor{red}{2}\times3$, thus as $\textcolor{red}{4}\ne\textcolor{red}{2}$ then $BA$ is undefined

\item In general, $AB\ne BA$ thus \textbf{order matters} for matrix multiplication.

\end{itemize}

\begin{example}{Example: Matrix Multiplication Example 2}

\begin{align*}

\text{\begin{blockarray}{cccc}

\begin{block}{ [ ccc c ]}

\bigstrut[t]

\tikzmarknode[inner sep=3pt]{A11}{1} & \tikzmarknode[inner sep=3pt]{A12}{3} & \tikzmarknode[inner sep=3pt]{A13}{-1} \\

2 & 0 & 4 \\

\end{block}

\end{blockarray}

\begin{blockarray}{cccc}

\begin{block}{ [ ccc c ]}

\bigstrut[t]

\tikzmarknode[inner sep=3pt]{B11}{-1} & 0 & 2 & 1 \\

\tikzmarknode[inner sep=3pt]{B21}{1} & 1 & 2 & -1\\

\tikzmarknode[inner sep=3pt]{B31}{0} & 1 & -1 & 2\bigstrut[b] \\

\end{block}

\end{blockarray}}&=\begin{bmatrix}-1+3+0&0+3-1&2+6+1&1-3-2\\-2+0+0&0+0+4&4+0-4&2+0+8\end{bmatrix} \\

&=\begin{bmatrix}2&2&9&-4\\-2&4&0&10\end{bmatrix}

\end{align*}

\begin{tikzpicture}[remember picture,overlay]

\draw[red, thick] plot[smooth cycle] coordinates {(A11.north) (A12.north) (A13.north) (A13.east) (A13.south) (A12.south) (A11.south) (A11.west)};

\draw[red, thick] plot[smooth cycle] coordinates {(B11.north) (B11.east) (B21.east) (B31.east) (B31.south) (B31.west) (B21.west) (B11.west)};

\end{tikzpicture}

\end{example}

\begin{example}{Example: Matrix Multiplication Example 3}

$$\underbrace{\begin{bmatrix}1&-1&0&2\\3&-2&4&1\\0&1&-1&2\end{bmatrix}}_{\textcolor{violet}{3}\times\textcolor{red}{4}}\underbrace{\begin{bmatrix}2&3\\1&-1\\0&2\\4&1\end{bmatrix}}_{\textcolor{red}{4}\times\textcolor{violet}{2}}=\underbrace{\begin{bmatrix}9&6\\8&20\\9&-1\end{bmatrix}}_{\textcolor{violet}{3}\times\textcolor{violet}{2}}$$

\begin{align*}

\begin{bmatrix}1&-1&0&2\\3&-2&4&1\\0&1&-1&2\end{bmatrix}\begin{bmatrix}2\\1\\0\\4\end{bmatrix}&=2\begin{bmatrix}1\\3\\0\end{bmatrix}+1\begin{bmatrix}-1\\-2\\1\end{bmatrix}+0\begin{bmatrix}0\\4\\-1\end{bmatrix}+4\begin{bmatrix}2\\1\\2\end{bmatrix}=\begin{bmatrix}9\\8\\9\end{bmatrix} \\

\text{\begin{blockarray}{cccc}

\begin{block}{ [ ccc c ]}

\bigstrut[t]

\tikzmarknode[inner sep=3pt]{A11}{1} & \tikzmarknode[inner sep=3pt]{A12}{-1} & \tikzmarknode[inner sep=3pt]{A13}{0} & \tikzmarknode[inner sep=3pt]{A14}{2}\\

3 & -2 & 4 & 1\\

0 & 1 & -1 & 2\\

\end{block}

\end{blockarray}

\begin{blockarray}{cccc}

\begin{block}{ [ ccc c ]}

\bigstrut[t]

\tikzmarknode[inner sep=3pt]{B11}{3}\\

\tikzmarknode[inner sep=3pt]{B21}{-1}\\

\tikzmarknode[inner sep=3pt]{B31}{2}\\

\tikzmarknode[inner sep=3pt]{B41}{1}\\

\end{block}

\end{blockarray}}&=\begin{bmatrix}3\(1\)-1\(-1\)+2\(0\)+1\(2\)\\3\(3\)-1\(-2\)+2\(4\)+1\(1\)\\3\(0\)-1\(1\)+2\(-1\)+1\(2\)\end{bmatrix}=\begin{bmatrix}6\\20\\-1\end{bmatrix}

\begin{tikzpicture}[remember picture,overlay]

\draw[red, thick] plot[smooth cycle] coordinates {(A11.north) (A12.north) (A13.north)(A14.north) (A14.east) (A14.south)(A13.south) (A12.south) (A11.south) (A11.west)};

\draw[red, thick] plot[smooth cycle] coordinates {(B11.north) (B11.east) (B21.east) (B31.east)(B41.east) (B41.south) (B41.west) (B31.west)(B21.west) (B11.west)};

\end{tikzpicture} \\

\text{\begin{blockarray}{cccc}

\begin{block}{ [ ccc c ]}

\bigstrut[t]

\tikzmarknode[inner sep=3pt]{A11}{1} & \tikzmarknode[inner sep=3pt]{A12}{-1} & \tikzmarknode[inner sep=3pt]{A13}{0} & \tikzmarknode[inner sep=3pt]{A14}{2}\\

3 & -2 & 4 & 1\\

0 & 1 & -1 & 2\\

\end{block}

\end{blockarray}

\begin{blockarray}{cccc}

\begin{block}{ [ ccc c ]}

\bigstrut[t]

\tikzmarknode[inner sep=3pt]{B11}{2} & 3\\

\tikzmarknode[inner sep=3pt]{B21}{1} & -1\\

\tikzmarknode[inner sep=3pt]{B31}{0} & 2\\

\tikzmarknode[inner sep=3pt]{B41}{4} & 1\\

\end{block}

\end{blockarray}}&=\begin{bmatrix}9&6\\8&20\\9&-1\end{bmatrix}

\begin{tikzpicture}[remember picture,overlay]

\draw[red, thick] plot[smooth cycle] coordinates {(A11.north) (A12.north) (A13.north)(A14.north) (A14.east) (A14.south)(A13.south) (A12.south) (A11.south) (A11.west)};

\draw[red, thick] plot[smooth cycle] coordinates {(B11.north) (B11.east) (B21.east) (B31.east)(B41.east) (B41.south) (B41.west) (B31.west)(B21.west) (B11.west)};

\end{tikzpicture} \\

\end{align*}

\end{example}

\newpage

\subsection{Properties of Matrix Multiplication}

Let $A,B,C$ be matrices of appropriate size. Let $k\in\mathbb{R}$.

\begin{enumerate}

\item $\(AB\)C=A\(BC\)$ (Associativity)

\item $\(AB\)^T=B^TA^T$

\item $A\(B+C\)=AB+AC$ (Left Distributivity)

\item $\(B+C\)A=BA+CA$ (Right Distributivity)

\item $k\(AB\)=\(kA\)B=A\(kB\)$

\item {If $A$ is $m\times n$, then $AI_n=I_mA=A$. With $I_n$ and $I_m$ being identity matrices. (Multiplicative Identity)}

\end{enumerate}

In General:

\begin{itemize}

\item $AB\ne BA$

\item If $AB=AC$, we cannot assume $B=C$.

\item If $AB=0$ (where $0$ is a zero matrix), we cannot assume either $A=0$ or $B=0$.

\end{itemize}

\begin{proof} \newline

\begin{itemize}

\item To prove $A\(B+C\)=AB+AC$ (Left Distributivity), we let the rows of $A$ be denoted by $\vv{A_i}$ and the columns of $B$ and $C$ by $b_i$ and $c_i$. Then the $j$th column of $B+C$ is $b_j+c_j$ (since addition is defined componentwise), and thus

\begin{align*}

\[A\(B+C\)\]_{ij}&=\vv{A_i}\cdot\(\vv{b_j}+\vv{c_j}\) \\

&=\vv{A_i}\cdot\vv{b_j}+\vv{A_i}\cdot\vv{c_j} \\

&=\(AB\)_{ij}+\(AC\)_{ij} \\

&=\(AB+AC\)_{ij}

\end{align*}

Since this is true for all $i$ and $j$, we must have $A\(B+C\)=AB+AC$.

\item To prove $AI_n=A$ (Multiplicative Identity), we note that the identity matrix $I_n$ can be column-partitioned as

$$I_n=\begin{bmatrix}\vv{e_1}&\vv{e_2}&\hdots&\vv{e_n}\end{bmatrix}$$

where $\vv{e_i}$ is a standard unit vector. Therefore,

\begin{align*}

AI_n&=\begin{bmatrix}A\vv{e_1}&A\vv{e_2}&\hdots&Ae_n\end{bmatrix} \\

&=\begin{bmatrix}\vv{a_1}&\vv{a_2}&\hdots&\vv{a_n}\end{bmatrix} \\

&=A

\end{align*}

by the fact that $A\vv{e_j}$ is the $j$th column of $A$.

\end{itemize}

\end{proof}

\begin{example}{Example: $AB\ne BA$}

\begin{align*}

\underbrace{\begin{bmatrix}1&-1\\0&0\end{bmatrix}}_{2\times2}\underbrace{\begin{bmatrix}1&1\\1&1\end{bmatrix}}_{2\times2}&=\underbrace{\begin{bmatrix}0&0\\0&0\end{bmatrix}}_{2\times2} \\

\underbrace{\begin{bmatrix}1&1\\1&1\end{bmatrix}}_{2\times2}\underbrace{\begin{bmatrix}1&-1\\0&0\end{bmatrix}}_{2\times2}&=\underbrace{\begin{bmatrix}1&-1\\1&-1\end{bmatrix}}_{2\times2}

\end{align*}

\end{example}

\begin{example}{Example: \text{$AB=AC$} but $B\ne C$}

$$A=\begin{bmatrix}1&0\\0&0\end{bmatrix}\text{ }B=\begin{bmatrix}2&3\\4&5\end{bmatrix}\text{ }C=\begin{bmatrix}2&3\\0&8\end{bmatrix}$$

\begin{align*}

AC&=\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}2&3\\0&8\end{bmatrix}=\begin{bmatrix}2&3\\0&0\end{bmatrix} \\

AB&=\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}2&3\\4&5\end{bmatrix}=\begin{bmatrix}2&3\\0&0\end{bmatrix}

\end{align*}

$$AB=AC \text{ but } B\ne C$$

\end{example}

\newpage

\subsection{Properties of Matrix Powers}

If $A$ is $n\times n$, then:

$$\begin{aligned}

A^2&=AA \\

A^3&=AAA = A^2A\\

A^4&=AAAA=A^2A^2=A^3A \\

&=\hdots

\end{aligned}$$

Properties for $r$ and $s$ are positive integers:

\begin{itemize}

\item $A^rA^s = A^{r+a}$

\item $\(A^r\)^s=A^{rs}$

\end{itemize}

\begin{example}{Example: Matrix Squared}

$$\begin{bmatrix}-1&2\\3&1\end{bmatrix}^2=\begin{bmatrix}-1&2\\3&1\end{bmatrix}\begin{bmatrix}-1&2\\3&1\end{bmatrix}=\begin{bmatrix}7&0\\0&7\end{bmatrix}=7I_2$$

\end{example}

If $A$ wasn't a squared matrix ex. $3\times2$ then you can't multiply it by itself as their isn't as many columns as their are rows, $\(3\times\textcolor{red}{2}\)\(\textcolor{red}{3}\times2\)$.

\begin{example}{Example: Properties of Matrix Powers}

Simplify $\(A+B\)^2$. (where $A$ and $B$ are $n\times n$)

$$\(A+B\)\(A+B\)=A^2+AB+BA+B^2$$

Do note that it is not $A^2+2AB+B^2$ as order matters in matrix multiplication.

$$\(AB\)^2=ABAB\ne A^2B^2$$

\end{example}

Consider The Question: Is $\begin{bsmallmatrix}3\\1\\-4\end{bsmallmatrix}$ a linear combination of $\begin{bsmallmatrix}2\\-1\\2\end{bsmallmatrix}$ and $\begin{bsmallmatrix}1\\1\\3\end{bsmallmatrix}$?

\begin{itemize}

\item {Are there $x,y\in\mathbb{R}$ such that:

$$x\begin{bmatrix}2\\-1\\2\end{bmatrix}+y\begin{bmatrix}1\\1\\3\end{bmatrix}=\begin{bmatrix}3\\1\\-4\end{bmatrix}\text{? Vector Form}$$}

\item {Are there $x,y\in\mathbb{R}$ such that:

$$\begin{bmatrix}2&1\\-1&1\\2&3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3\\1\\-4\end{bmatrix}\text{? Matrix-Form or Matrix-Vector Form}$$}

\item {Are there $x,y\in\mathbb{R}$ such that:

$$\sysdelim..\systeme{

2x+y=3,

-x+y=1,

2x+3y=-4

}\text{Equation Form}$$}

\end{itemize}

\newpage

\section{Section 2.1 Intro to Linear Systems}

\begin{definition}{Definition: Linear Equation}

A \textbf{linear equation} in the $n$ \textbf{variables} $x_1,x_2,\hdots,x_n$ is an equation that can be written in the form:

$$a_1x_1+a_2x_2+\hdots+a_nx_n=b$$

where $a_1,a_2,\hdots,a_n$ are \textbf{coefficents} and $b$ is the \textbf{constant term}.

\begin{example}{Example: Linear Equation}

\begin{tblr}{width = \textwidth, colspec={XX}, cells = {halign = c, valign = m}}

{\raisebox{-0.5\height}{$\begin{aligned}

2x&=3y+1 \\

2y-3y&=1

\end{aligned}$}} & {$$xy=1$$ Non Linear} \\

{Linear}

\end{tblr}

\end{example}

\end{definition}

\begin{definition}{Definition: Solution}

A \textbf{solution} to a linear equation is an assignment of the variables that results in a true statement.

\begin{example}{Example: Solution}

$$x+y+z=1$$

\begin{tblr}{width = \textwidth, colspec={X|X|X}, cells = {halign = c, valign = m, mode = math}}

{\begin{aligned}x&=4\\y&=-2\\x&=-1\end{aligned}} & {\begin{aligned}x&=\frac{1}{3}\\y&=\frac{1}{3}\\z&=\frac{1}{3}\end{aligned}} & {\begin{aligned}x&=0\\y&=0\\z&=1\end{aligned}}

\end{tblr}

3 distinct solutions (but many others exist)

\end{example}

\end{definition}

\begin{definition}{Definition: Linear System/System of Linear Equations}

A set of $m$ linear equations in $n$ variables $x_1,x_2,\hdots,x_n$ is called an $m\times n$ \textbf{linear system/system of linear equations}. \\

\newline

A \textbf{solution} to an $m\times n$ linear system is an assignment of the variables that simultaneously satisfies \textbf{all} $m$ equations.

\begin{example}{Example: Linear System/System of Linear Equations}

$$\sysdelim..\systeme{

x+y+z=1,

2x-y+z=2

}$$

Is $x=4$, $y=-2$, $z=-1$ a solution to this system? \textbf{No}, as the arguments don't satisfie both equations.

\begin{align*}

x+y+z&=4-2-1=1 \text{ }\checkmark \\

2x-y+z&=2\(4\)-\(-2\)+\(-1\)=9\ne2

\end{align*}

Is $x=1$, $y=0$, $z=0$ a solution? \textbf{Yes}

\begin{align*}

x+y+z&=1+0+0=1\text{ }\checkmark \\

2x-y+z&=2\(1\)-0+0=2\text{ }\checkmark

\end{align*}

\end{example}

\end{definition}

\begin{example}{Example: Solution to System of Equations}

Find all solutions to:

$$\begin{cases}

\sysdelim..\systeme{

3x+y=9,

x-2y=-4

}

\end{cases}$$

$$\begin{aligned}

3x+y&=9 & x-2y&=-4 \\

y&=9-3x\Rightarrow& x-2\(9-3x\)&=-4 \\

y&=9-3\(2\)& x-18+6x&=-4 \\

\Aboxed{y&=3} & 7x&=14 \\

& & \Leftarrow \Aboxed{x&=2}

\end{aligned}$$

One Unique Solution, which can also be seen graphically

\begin{center}

\begin{tikzpicture}

\begin{axis} [

width = 0.5\linewidth,

axis x line=middle,

axis y line=middle,

ymin = -10, ymax = 10,

xmin = -10, xmax = 10,

xlabel = {$x$},

ylabel = {$y$},

%xtick = {1, 2, 3, 4, 5, 6},

%ytick = {-1, 1, 2, 3, 4, 5, 6, 7, 8},

%xticklabels={},

every axis plot post/.append style={ultra thick},

clip = true,

axis line style = {Stealth-Stealth, thick},

legend style = {draw = none},

legend pos = outer north east

]

\addplot [

black,

samples = 200,

style = {Stealth-Stealth},

domain = -1/3:19/3,

]

{9-(3*x)};

\addplot [

black,

style = {Stealth-Stealth},

samples = 200,

domain = -10:10,

]

{(-4-x)/-2};

\addplot[mark=*] coordinates {(2,3)} node[below right] {$\(2,3\)$};

\end{axis}

\end{tikzpicture}

\end{center}

\end{example}

\begin{tblr}{width = \linewidth, colspec = {XXX}, cells = {halign = c, valign = m}}

{\begin{tikzpicture}

\begin{axis} [

width = \linewidth,

axis x line=middle,

axis y line=middle,

ymin = -5, ymax = 5,

xmin = -5, xmax = 5,

xlabel = {$x$},

ylabel = {$y$},

xtick = {100},

ytick = {100},

%xticklabels={},

every axis plot post/.append style={ultra thick},

clip = true,

axis line style = {Stealth-Stealth, thick},

legend style = {draw = none},

legend pos = outer north east

]

\addplot [

black,

style = {Stealth-Stealth},

samples = 200,

domain = -5:4,

] {x+1};

\addplot [

black,

style = {Stealth-Stealth},

samples = 200,

domain = -4:5,

] {x-1};

\end{axis}

\end{tikzpicture}} &

{\begin{tikzpicture}

\begin{axis} [

width = \linewidth,

axis x line=middle,

axis y line=middle,

ymin = -5, ymax = 5,

xmin = -5, xmax = 5,

xlabel = {$x$},

ylabel = {$y$},

xtick = {100},

ytick = {100},

%xticklabels={},

every axis plot post/.append style={ultra thick},

clip = true,

axis line style = {Stealth-Stealth, thick},

legend style = {draw = none},

legend pos = outer north east

]

\addplot [

black,

style = {Stealth-Stealth},

samples = 200,

domain = -5:5,

] {x};

\addplot [

black,

style = {Stealth-Stealth},

samples = 200,

domain = -4:5,

] {-x+1};

\addplot[mark=*] coordinates {(0.5,0.5)};

\end{axis}

\end{tikzpicture}} &

{\begin{tikzpicture}

\begin{axis} [

width = \linewidth,

axis x line=middle,

axis y line=middle,

ymin = -5, ymax = 5,

xmin = -5, xmax = 5,

xlabel = {$x$},

ylabel = {$y$},

xtick = {100},

ytick = {100},

%xticklabels={},

every axis plot post/.append style={ultra thick},

clip = true,

axis line style = {Stealth-Stealth, thick},

legend style = {draw = none},

legend pos = outer north east

]

\addplot [

black,

style = {Stealth-Stealth},

samples = 200,

domain = -5:5,

] {x};

\end{axis}

\end{tikzpicture}} \\

{No Solution} & {One Unique Solution} & {Infinitely Many Solutions}

\end{tblr} \newline

Any $m\times n$ linear system must have either no solution (inconsistent system), one unique solution, or infinitely many solutions. With one unique solution and infinitely many solutions being consistant systems.

\begin{example}{Example: Back and Forward Substitution}

Solve the following linear system: \\

\begin{tblr}{width = \linewidth, colspec = {X|X}, cells = {halign = l, valign = m}, row{2} = {halign = c}}

{$$\sysdelim..\systeme{

x+y-z=4,

2y+z=10,

3z=12

}$$ Upper Triangular Form so use Back-Substitution:} & {$$\sysdelim..\systeme{

x=3,

x-2y=9,

x+y-z=4

}$$Lower Triangular Form so use Forward-Substitution:} \\

{$\begin{aligned}

2y+4&=10 & x+3-4&=4 \\

2y&=6 & x&=5 \\

y&=3

\end{aligned}$ \\

\fbox{$\begin{aligned}x&=5\\y&=3\\z&=4\end{aligned}$}\\\newline One Unique Solution} &

{$\begin{aligned}

3-2y&=9 & 3-3-z&=4 \\

-2y&=6 & z&=-4 \\

y&=-3

\end{aligned}$ \\

\fbox{$\begin{aligned}x&=3\\y&=-3\\z&=-4\end{aligned}$}\\\newline One Unique Solution}

\end{tblr}

\end{example}

\begin{example}{Example: Getting a System of Equations into Back-Substitution}

Solve the following linear system:

$$\sysdelim..\sysautonum{\(*\)}\systeme{

-x-2y+z=1,

x+y-2z=1,

2x+6y+2z=0

}$$ First goal is to get the system into Upper Trangular Form, to do this first add equation $\(1\)$ and $\(2\)$ to get:

$$\(1\)+\(2\)=0-y-z=2$$

Next add $2\(1\)$ and $\(3\)$ to get:

$$2\(1\)+\(3\)=0+2y+4z=2$$

Combining everything into a new system of equations to get:

$$\sysdelim..\sysautonum{\(*\)}\systeme{

-x-2y+z=1,

-y-z=2,

2y+4z=3

}$$

Next add $2\(2\)$ and $\(3\)$ to get:

$$2\(2\)+\(3\)=0+2z=6$$

Combining everything into a new system of equations to get:

$$\sysdelim..\systeme{

-x-2y+z=1,

-y-z=2,

2z=6

}$$

And using Back-Substitution to solve gets

\fbox{$\begin{aligned}x&=12\\y&=-5\\z&=3\end{aligned}$}. This can also be written in different forms:

\begin{itemize}

\item {Equation Form:

$$\sysdelim..\systeme{

-x-2y+z=1,

x+y-2z=1,

2x+6y+2z=0

}$$}

\item {Vector Form:

$$x\begin{bmatrix}-1\\1\\2\end{bmatrix}+y\begin{bmatrix}-2\\1\\6\end{bmatrix}+z\begin{bmatrix}1\\-2\\2\end{bmatrix}=\begin{bmatrix}1\\1\\0\end{bmatrix}$$}

\item {Matrix Form/Matrix-Vector Form

$$\begin{bmatrix}-1&-2&1\\1&1&-2\\2&6&2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\1\\0\end{bmatrix}$$}

\item {Solution in Equation Form:$$\begin{aligned}x&=12\\y&=-5\\z&=3\end{aligned}$$}

\item {Solution in Vector Form:

\begin{align*}

\vv{x}&=\begin{bmatrix}12\\-5\\3\end{bmatrix} \\

\begin{bmatrix}-1&-2&1\\1&1&-2\\2&6&2\end{bmatrix}\begin{bmatrix}12\\-5\\3\end{bmatrix}&=\begin{bmatrix}1\\1\\0\end{bmatrix}

\end{align*}}

\end{itemize}

\end{example}

\begin{example}{Example: Bound and Free Variables}

Solve the following linear system:

$$\sysdelim..\systeme{

x_1+x_2-x_3+x_4=0,

x_2+x_3+2x_4=0

}$$

As you can't find $x_4$ and $x_3$ then you make them a variable say $t$ and $s$ with $t,s\in\mathbb{R}$ giving:

$$\begin{aligned}x_3&=s\\x_4&=t\end{aligned}$$

Now solving for $x_2$ gives:

\begin{align*}

x_2+s+2t&=0 \\

x_2&=-s-2t

\end{align*}

Now solving for $x_1$ gives:

\begin{align*}

x_1+\(-s-2t\)-s+t&=0\\

x_1-2s-t&=0\\

x_1&=2s+t

\end{align*}

\begin{itemize}

\item {Parametric Equation Form of Solution Set

\begin{center}\fbox{$\begin{aligned}

x_1&=2s+t\\

x_2&=-s-2t\\

x_3&=s\\

x_4&=t

\end{aligned}$}\end{center}}

\item {Parametric Vector Form of The Solution Set

\begin{align*}

\vv{x}&=\begin{bmatrix}2s+t\\-s-2t\\s\\t\end{bmatrix}\\

&=\begin{bmatrix}2s\\-s\\s\\0\end{bmatrix}+\begin{bmatrix}t\\-2t\\0\\t\end{bmatrix}\\

\Aboxed{\vv{x}&=s\begin{bmatrix}2\\-1\\1\\0\end{bmatrix}+t\begin{bmatrix}1\\-2\\0\\1\end{bmatrix}}

\end{align*}}

\end{itemize}

In this example $x_1$ and $x_2$ are called \textbf{bound variables}. \\

\newline

The non-sound variables are called \textbf{free variables}. We assign parameters to free variables. ($x_3$ and $x_4$ are free variables)

\end{example}

\begin{example}{Example: No Solution}

Solve the following linear system:

$$\sysdelim..\systeme{

4x-8y=6,

2x-4y=4

}$$

Try to get the equation into back substitution form by canceling the $2x$ in the second equation:

\begin{center}

\begin{tabular}{r}

{$-\frac{1}{2}\(4x-8y=6\)$} \\

{$+1\(2x-4y=4\)$} \\ \hline

{$0+0=1$}

\end{tabular}

\end{center}

This now makes the linear system:

\begin{align*}

4x-8y&=6\\

\Aboxed{0&=1}

\end{align*}

As $0\ne1$ then the system has no solution, which also makes it inconsistent.

\end{example}

\newpage

\section{Section 2.2 Solving Linear Systems}

\begin{definition}{Definition: Augmented Matrix}

Consider the linear system $A\vv{x}=\vv{b}$ where $A$ is an $m\times n$ matrix and $\vv{b}\in\mathbb{R}^m$. $A$ is called the \textbf{coefficent matrix} corresponding to the linear system. \\

\newline

$\[A\Big|\vv{b}\]$ is called the \textbf{augmented matrix} corresponding to the linear system.

\begin{example}{Example: Augmented Matrix}

$$\sysdelim..\systeme{

2x-3y+0z=4,

4x+y-5z=0

}$$

\begin{itemize}

\item {$\underbrace{\begin{bmatrix}2&-3&0\\4&1&-5\end{bmatrix}}_{\text{Coefficent Matrix}}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}4\\0\end{bmatrix}$}

\item {$\underbrace{\begin{bmatrix}2&-3&0&\aug&4\\4&1&-5&\aug&0\end{bmatrix}}_{\text{Augmented Matrix}}$}

\end{itemize}

\end{example}

\end{definition}

\begin{example}{Example: Augmented Matrix To System Of Equations With A Unique Solution}

Solve the linear system coresponding to the given augmented matrix:

\begin{center}

$\begin{bNiceMatrix}% don't forget the %

[first-row,]

x& y& z&&\text{constants} \\

1 & 1 & 1 & \aug&1 \\

0 & 2 & 1 &\aug& 0 \\

0 & 0 & -1 & \aug&2

\end{bNiceMatrix}$

$$\sysdelim..\systeme{

x+y+z=1,

2y+z=0,

-z=2

}$$Use Back-Substitution \\ \newline

\fbox{$\begin{aligned}x&=2\\y&=1\\z&=-2\end{aligned}$} $\vv{x}=\begin{bmatrix}2\\1\\-2\end{bmatrix}$ \\ \newline

One Unique Solution

\end{center}

\end{example}

\begin{example}{Example: Augmented Matrix to System of Equations With Infinite Solutions}

Solve the linear system coresponding to the given augmented matrix:

\begin{center}

$\begin{bNiceMatrix}% don't forget the %

[first-row,]

a& b& c&d&&\text{constants} \\

1 & 2 & 0 &0& \aug&5 \\

0 & 0 & 1 &0&\aug& 3 \\

0 & 0 & 0 &1& \aug&-2

\end{bNiceMatrix}$

$$\sysdelim..\systeme{

a+2b=5,

c=3,

d=-2

}$$

\fbox{$\begin{aligned}a&=5-2t\\b&=t\\c&=3\\d&=-2\end{aligned}$} ($a,c,d$ are \textbf{bound variables}. $b$ is a \textbf{free variable}.) \\ \newline

Infinetly Many Solutions \\ \newline

$\begin{aligned}

\vv{x}&=\begin{bmatrix}5-2t\\t\\3\\-2\end{bmatrix} \\

&=\begin{bmatrix}5\\0\\3\\-2\end{bmatrix}+\begin{bmatrix}-2t\\t\\0\\0\end{bmatrix} \\

\Aboxed{&=\begin{bmatrix}5\\0\\3\\-2\end{bmatrix}+t\begin{bmatrix}-2\\1\\0\\0\end{bmatrix}}

\end{aligned}$\\\newline Parametric Vector Form of Solution Set

\end{center}

\end{example}

\begin{definition}{Definition: Row Echelon Form (REF)}

A mattrix is said to be in \textbf{row echelon form (REF)} if:

\begin{enumerate}

\item Any zero rows, rows consisting entirely of zeros, in the matrix are at the bottom.

\item In each nonzero row, the first nonzero entry (called the \textbf{leading entry}) is in a column to the left of all leading entries below it.

\end{enumerate}

\begin{example}{Example: What is and isn't in REF}

\begin{multicols}{2}

\begin{itemize}

\item {$\underbrace{\begin{bNiceMatrix}

2&0&3&9\\

0&0&4&0\\

0&0&0&1

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-3) circle (2mm) ;

\tikz \draw (3-4) circle (2mm) ;

\end{bNiceMatrix}}_{\text{REF}}$}

\item {$\underbrace{\begin{bNiceMatrix}

0&2\\

0&0\\

0&0

\CodeAfter

\tikz \draw (1-2) circle (2mm) ;

\end{bNiceMatrix}}_{\text{REF}}$}

\item {$\underbrace{\begin{bNiceMatrix}

2&0&3&1\\

0&3&1&1\\

0&2&0&1

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (2mm) ;

\tikz \draw (3-2) circle (2mm) ;

\end{bNiceMatrix}}_{\text{\textbf{Not} REF}}$}

\item {$\underbrace{ \text{\begin{blockarray}{cccc}

\begin{block}{ [ ccc c ]}

\bigstrut[t]

0 & 1 & 0 & 0\\

\tikzmarknode[inner sep=3pt]{A11}{0} & \tikzmarknode[inner sep=3pt]{A12}{0} & \tikzmarknode[inner sep=3pt]{A13}{0} & \tikzmarknode[inner sep=3pt]{A14}{0}\\

0 & 0 & 3 & 1\\

\end{block}

\end{blockarray}}

\begin{tikzpicture}[remember picture,overlay]

\draw[red, thick] plot[smooth cycle] coordinates {(A11.north) (A12.north) (A13.north)(A14.north) (A14.east) (A14.south)(A13.south) (A12.south) (A11.south) (A11.west)};

\end{tikzpicture}}_{\text{\textbf{Not} REF}}$}

\end{itemize}

\end{multicols}

\end{example}

\end{definition}

\begin{definition}{Definition: Reduced Row Echelon Form (RREF)}

A matrix is in \textbf{reduced row echelon form (RREF)} if:

\begin{enumerate}

\item It is in REF

\item The leading entry in each nonzero row is one (called a \textbf{leading one})

\item Each leading one is the only nonzero entry in its column.

\end{enumerate}

\begin{example}{Example: Reduced Row Echelon Form}

$$\begin{bNiceMatrix}

1&0&2&3&0 \\

0&1&-1&0&0 \\

0&0&0&0&1

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (2mm) ;

\tikz \draw (3-5) circle (2mm) ;

\end{bNiceMatrix}$$

\end{example}

\end{definition}

Notes:

\begin{itemize}

\item The leading entries in a REF or RREF of a matrix are called \textbf{pivots}.

\item Given a nonzero matrix, there are infinitely many REF of the matrix, but only one \textbf{unique} RREF.

\end{itemize}

Given a matrix, we can get a REF of the matrix using \textbf{3 elementary row operations}:

\begin{enumerate}

\item {\textbf{Row Interchange:} $R_i\leftrightarrow R_j$

\begin{example}{Example: Row Interchange}

$$\begin{bmatrix}0&2\\1&3\end{bmatrix}\xrightarrow{R_1\leftrightarrow R_2}\begin{bmatrix}1&3\\0&2\end{bmatrix}$$

\end{example}}

\item {Scale a row by a \textbf{nonzero} scalar $k$: $kR_i$

\begin{example}{Example: Scale Row}

$$\begin{bmatrix}2&4\\0&1\end{bmatrix}\xrightarrow{\frac{1}{2}R_1}\begin{bmatrix}1&2\\0&1\end{bmatrix}$$

\end{example}}

\item {\textbf{Row Replacement: } $R_i+kR_j$ (replaces $R_i$)

\begin{example}{Example: Row Replacement}

$$\begin{bmatrix}1&3\\-2&4\end{bmatrix}\xrightarrow{R_2+2R_1}\begin{bmatrix}1&3\\0&10\end{bmatrix}$$

\end{example}It is also the equivalent of:

\begin{tabular}{r}

{$\begin{pmatrix}-2&4\end{pmatrix}$} \\

{$+2\begin{pmatrix}1&3\end{pmatrix}$} \\ \hline

{$\begin{matrix}0&10\end{matrix}$}

\end{tabular}}

\end{enumerate}

\begin{example}{Example: Find The REF And RREF Of A Matrix}

Find a REF of the following matrix:

$$\begin{bmatrix}1&2&-1&-1\\3&5&-2&-1\\2&2&1&1\end{bmatrix}$$

\begin{flalign*}

&\begin{bNiceMatrix}

1&2&-1&-1\\

\underline{3}&5&-2&-1\\

\underline{2}&2&1&1

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\end{bNiceMatrix}\noindent\xrightarrow{R_2-3R_1}\begin{bNiceMatrix}

1&2&-1&-1\\

0&\underline{-1}&1&2\\

\underline{2}&2&1&1

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\end{bNiceMatrix}\xrightarrow{R_3-2R_1}\begin{bNiceMatrix}

1&2&-1&-1\\

0&\underline{-1}&1&2\\

0&\underline{-2}&3&3

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (3mm) ;

\end{bNiceMatrix} \\

&\xrightarrow{R_3-2R_2}\underbrace{\begin{bNiceMatrix}

1&2&-1&-1\\

0&-1&1&2\\

0&0&1&-1

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (3mm) ;

\tikz \draw (3-3) circle (2mm) ;

\end{bNiceMatrix}}_{\text{REF}}\xrightarrow{-R_2}\begin{bNiceMatrix}

1&2&-1&-1\\

0&1&-1&-2\\

0&0&1&-1

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (2mm) ;

\tikz \draw (3-3) circle (2mm) ;

\end{bNiceMatrix}\xrightarrow{R_1-2R_2}\begin{bNiceMatrix}

1&0&1&3\\

0&1&-1&-2\\

0&0&1&-1

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (2mm) ;

\tikz \draw (3-3) circle (2mm) ;

\end{bNiceMatrix}\\

&\xrightarrow[R_1-R_3]{R_2+R_3}\underbrace{\begin{bNiceMatrix}

1&0&0&4\\

0&1&0&-3\\

0&0&1&-1

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (2mm) ;

\tikz \draw (3-3) circle (2mm) ;

\end{bNiceMatrix}}_{\text{RREF}}&&

\end{flalign*}

\end{example}

\begin{example}{Example: Find The RREF Of A Matrix}

Row reduce the matrix below to its RREF:

$$\begin{bmatrix}1&-2&1&4&1\\-1&2&1&2&0\\2&-4&0&2&1\end{bmatrix}$$

\begin{align*}

&\begin{bNiceMatrix}

1&-2&1&4&1\\

-1&2&1&2&0\\

2&-4&0&2&1

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\end{bNiceMatrix}\xrightarrow[R_2+R_1]{R_3-2R_1}\begin{bNiceMatrix}

1&-2&1&4&1\\

0&0&2&6&1\\

0&0&\underline{-2}&-6&-1

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-3) circle (2mm) ;

\end{bNiceMatrix}\xrightarrow{\frac{1}{2}R_2}\begin{bNiceMatrix}

1&-2&1&4&1\\

0&0&1&3&\frac{1}{2}\\

0&0&\underline{-2}&-6&-1

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-3) circle (2mm) ;

\end{bNiceMatrix}\\

&\xrightarrow[R_1-R_2]{R_3+2R_2}\begin{bNiceMatrix}

1&-2&0&1&\frac{1}{2}\\

0&0&1&3&\frac{1}{2}\\

0&0&0&0&0

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-3) circle (2mm) ;

\end{bNiceMatrix}&&

\end{align*}

\end{example}

\begin{definition}{Definition: Row Equivalent}

The matrices $A$ and $B$ are \textbf{row equivalent} if there is a sequence of elementary row operations that transforms $A$ into $B$.

\end{definition}

Notes:

\begin{itemize}

\item If $\[A\Big|\vv{b}\]$ and $\[C\Big|\vv{d}\]$ are \textbf{row equivalent}, then the linear systems $A\vv{x}=\vv{b}$ and $c\vv{x}=\vv{d}$ have the \textbf{same solution set}.

\end{itemize}

\begin{example}{Example: Row Equivalent Matrices Sharing A Solution Set}

Provided:

$$\begin{bmatrix}1&2&-1&\aug&-1\\3&5&-2&\aug&-1\\2&2&1&\aug&1\end{bmatrix}\xrightarrow{\text{Row Equivalent}}\begin{bmatrix}1&0&0&\aug&4\\0&1&0&\aug&-3\\0&0&1&\aug&-1\end{bmatrix}$$

The matrix $\begin{bmatrix}1&0&0&\aug&4\\0&1&0&\aug&-3\\0&0&1&\aug&-1\end{bmatrix}$ has the solution set of \fbox{$\begin{aligned}x_1&=4\\x_2&=-3\\x_3&=-1\end{aligned}$} and as $\begin{bmatrix}1&2&-1&\aug&-1\\3&5&-2&\aug&-1\\2&2&1&\aug&1\end{bmatrix}$ is row equivalent then they have the same solution set thus:

\begin{align*}

\begin{bmatrix}1&2&-1\\3&5&-2\\2&2&1\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}&=\begin{bmatrix}-1\\-1\\1\end{bmatrix} \\

\begin{bmatrix}1&2&-1\\3&5&-2\\2&2&1\end{bmatrix}\fbox{$\begin{bmatrix}4\\-3\\-1\end{bmatrix}$}&=\begin{bmatrix}-1\\-1\\1\end{bmatrix}\checkmark

\end{align*}

\end{example}

\begin{example}{Example: Row Equivalent Matrices Sharing Infinitely Many Solutions}

Provided:

$$\begin{bmatrix}1&-2&1&4&\aug&1\\-1&2&1&2&\aug&0\\2&-4&0&2&\aug&1\end{bmatrix}\xrightarrow{\text{Row Equivalent}}\begin{bNiceMatrix}[first-row]

&\downarrow&&\downarrow&&\\

1&-2&0&1&\aug&\frac{1}{2}\\

0&0&1&3&\aug&\frac{1}{2}\\

0&0&0&0&\aug&0

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-3) circle (2mm) ;

\end{bNiceMatrix}$$

$x_1$ and $x_3$ are bound variables, are a leading coefficent \\

$x_2$ and $x_4$ are free variables, don't have a leading coefficent \\

This can also be seen in the system of equations below:

$$\sysdelim..\systeme{

x_1-2x_2+x_4=\frac{1}{2},

x_3+3x_4=\frac{1}{2},

0=0

}$$

This makes

\begin{align*}

x_1&=\frac{1}{2}+2x_2-x_4 \\

x_3&=\frac{1}{2}-3x_4

\end{align*}

Giving the solution set of \fbox{$\begin{aligned}x_1&=\frac{1}{2}+2s-t\\x_2&=s\\x_3&=\frac{1}{2}-3t\\x_4&=t\end{aligned}$} or Infinitely Many Solutions, which can also be written in Parametric Vector Form:

\begin{align*}

\vv{x}&=\begin{bmatrix}\frac{1}{2}+2s-t\\s\\\frac{1}{2}-3t\\t\end{bmatrix} \\

&=\begin{bmatrix}\frac{1}{2}\\0\\\frac{1}{2}\\0\end{bmatrix}+\begin{bmatrix}2s\\s\\0\\0\end{bmatrix}+\begin{bmatrix}-t\\0\\-3t\\t\end{bmatrix} \\

\Aboxed{&=\begin{bmatrix}\frac{1}{2}\\0\\\frac{1}{2}\\0\end{bmatrix}+s\begin{bmatrix}2\\1\\0\\0\end{bmatrix}+t\begin{bmatrix}-1\\0\\-3\\1\end{bmatrix}}

\end{align*}

\end{example}

\begin{example}{Example: Row Reducing With No Solution}

$$\begin{bNiceMatrix}

1&2&-1&1&\aug&5\\

\underline{2}&3&-1&4&\aug&8\\

\underline{1}&4&-3&-3&\aug&6

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\end{bNiceMatrix}\xrightarrow[R_2-2R_1]{R_3-R_1}\begin{bNiceMatrix}

1&2&-1&1&\aug&5\\

0&-1&1&2&\aug&-2\\

0&\underline{2}&-2&-4&\aug&1

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (3mm) ;

\end{bNiceMatrix}\xrightarrow{R_3+2R_2}\begin{bNiceMatrix}[first-row]

&&&&&\downarrow \\

1&2&-1&1&\aug&5\\

0&-1&1&2&\aug&-2\\

0&0&0&0&\aug&-3

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (3mm) ;

\tikz \draw (3-6) circle (3mm) ;

\end{bNiceMatrix}$$

Giving the system of equations:

$$\sysdelim..\systeme{

x_1+2x_2-x_3+x_4=5,

-x_2+x_3+2x_4=-2,

0=-3

}$$

No Solution as $0\ne3$

\end{example}

\begin{definition}{Definition: Homogeneous Linear System}

An $m\times n$ linear system of the form $A\vv{x}=\vv{0}$ is called a \textbf{homogeneous linear system}. (all constant terms are equal to zero)

\end{definition}

\begin{definition}{Definition: Non-Homogeneous Linear System}

If at least one constant term is non zero in the linear system, we say the linear system is \textbf{non-homogeneous}. $\(A\vv{x}=\vv{b}\text{ where }\vv{b}\ne\vv{0}\) $

\end{definition}

Note that $A\vv{x}=\vv{0}$ is \textbf{always} consistent because $\vv{x}=\vv{0}$ is a solution to $A\vv{x}=\vv{0}$ $\(A\vv{0}=\vv{0}\)$ with $\vv{x}=\vv{0}$ being called the \textbf{trivial solution}.

\begin{example}{Example: Solving A Homogeneous And Non Homogeneous Linear System}

Solve the linear systems corresponding to the given augmented matrices: \\ \newline

\begin{tblr}{width = \linewidth, colspec = {X|X}, cells = {valign = m, halign = c}}

{$\begin{aligned}

&\begin{bNiceMatrix}

1&2&3&\aug&0\\

\underline{-1}&4&3&\aug&0

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\end{bNiceMatrix} \\

\xrightarrow{R_2+R_1}&\begin{bNiceMatrix}

1&2&3&\aug&0\\

0&6&6&\aug&0

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (2mm) ;

\end{bNiceMatrix} \\

\xrightarrow{\frac{1}{6}R_2}&\begin{bNiceMatrix}

1&2&3&\aug&0\\

0&1&1&\aug&0

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (2mm) ;

\end{bNiceMatrix} \\

\xrightarrow{R_1-2R_2}&\begin{bNiceMatrix}

1&0&1&\aug&0\\

0&1&1&\aug&0

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (2mm) ;

\end{bNiceMatrix}

\end{aligned}$\\ \newline

$\begin{aligned}

x_1+x_3&=0\Longrightarrow x_1=-x_3 \\

x_2+x_3&=0\Longrightarrow x_2=-x_3

\end{aligned}$\\ \newline

\fbox{$\begin{aligned}x_1&=-t\\x_2&=-t\\x_3&=t\end{aligned}\hspace{1cm}\vv{x}=t\begin{bmatrix}-1\\-1\\1\end{bmatrix}$}} &

{$\begin{aligned}

&\begin{bNiceMatrix}

1&2&3&\aug&5\\

\underline{-1}&4&3&\aug&7

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\end{bNiceMatrix} \\

\xrightarrow{R_2+R_1}&\begin{bNiceMatrix}

1&2&3&\aug&5\\

0&6&6&\aug&12

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (2mm) ;

\end{bNiceMatrix} \\

\xrightarrow{\frac{1}{6}R_2}&\begin{bNiceMatrix}

1&2&3&\aug&5\\

0&1&1&\aug&2

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (2mm) ;

\end{bNiceMatrix} \\

\xrightarrow{R_1-2R_2}&\begin{bNiceMatrix}

1&0&1&\aug&1\\

0&1&1&\aug&2

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (2mm) ;

\end{bNiceMatrix}

\end{aligned}$\\ \newline

$\begin{aligned}

x_1+x_3&=1\Longrightarrow x_1=1-x_3 \\

x_2+x_3&=2\Longrightarrow x_2=2-x_3

\end{aligned}$\\ \newline

\fbox{$\begin{aligned}x_1&=1-t\\x_2&=2-t\\x_3&=t\end{aligned}\hspace{1cm}\vv{x}=\begin{bmatrix}1\\2\\0\end{bmatrix}+t\begin{bmatrix}-1\\-1\\1\end{bmatrix}$}}

\end{tblr} \\ \newline

Notice that the differences between them are only a constant matrix.

\end{example}

\begin{example}{\text{Example: Relating $A\vv{x}=\vv{b}$ and $A\vv{x}=\vv{0}$}}

Suppose $\textcolor{red}{A}$ is an $m\times n$ matrix. If $\textcolor{red}{A}\vv{x}=\vv{b}$ has one unique solution, what is the solution set for $\textcolor{red}{A}\vv{x}=\vv{0}$?

\begin{center}

\fbox{$\vv{x}=\vv{0}$}

\end{center}

\end{example}

\begin{definition}{Definition: Rank}

The \textbf{rank} of a matrix $A$, denoted $\text{rank}\(A\)$, is the number of nonzero rows in a REF of $A$.

\begin{example}{Example: What Is The Maximum Rank Of A Matrix? Example 1}

If $A$ is a $5\times7$ matrix, what is the maximum possible value of rank $A$? \fbox{5}

\end{example}

\end{definition}

\begin{example}{Example: What Is The Maximum Rank Of A Matrix? Example 2}

If $A$ is a $7\times5$ matrix, what is the maximum possible value of $\text{rank}\(A\)$?

$$\begin{bNiceMatrix}

1&0&0&0&0 \\

0&1&0&0&0 \\

0&0&1&0&0 \\

0&0&0&1&0 \\

0&0&0&0&1 \\

0&0&0&0&0 \\

0&0&0&0&0 \\

0&0&0&0&0 \\

0&0&0&0&0 \\

\CodeAfter

\SubMatrix{\{}{1-1}{5-1}{.}[left-xshift=0.8mm]

\end{bNiceMatrix}$$

\fbox{$\text{rank}\(A\)=5$} as there is a max of 5 rows and columns such that there is a maximum of one pivot per row and column in REF.

\end{example}

\newpage

\section{Section 2.3 Span and Linear Independence}

\begin{example}{Does \text{$A\vv{x}=\vv{b}$} Have A Solution For All \text{$\vv{b}\in\mathbb{R}^3$}?}

Does $A\vv{x}=\vv{b}$ have a solution for every $\vv{b}\in\mathbb{R}^3$? \\ \newline

\begin{tblr}{width = \linewidth, colspec = {X|X}, cells = {halign = c, valign = m}}

{$A=\begin{bmatrix}1&0&1\\0&1&0\\1&0&2\end{bmatrix}\hspace{1cm}\vv{b}=\begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix}$ \\ \vspace{2.5ex}

$\begin{aligned}

\[A\Big|\vv{b}\] = &\begin{bNiceMatrix}

1&0&1&\aug&b_1 \\

0&1&0&\aug&b_2 \\

1&0&2&\aug&b_3

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\end{bNiceMatrix} \\

\xrightarrow{R_3-R_1}&\underbrace{\begin{bNiceMatrix}

1&0&1&\aug&b_1 \\

0&1&0&\aug&b_2 \\

0&0&1&\aug&b_3-b_1

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (2mm) ;

\tikz \draw (3-3) circle (2mm) ;

\end{bNiceMatrix}}_{\text{REF}}

\end{aligned}$ \\ \vspace{2ex}

\parbox{\linewidth}{

\begin{flushleft}

One unique solution exists to $A\vv{x}=\vv{b}$ for every $\vv{b}\in\mathbb{R}^3$ (no leading entry/pivot in constants column and all variables are bound)

\end{flushleft}} \\ \newline

$\text{rank}\(A\)=3$} &

{$A=\begin{bmatrix}1&0&1\\0&1&0\\1&0&1\end{bmatrix}\hspace{1cm}\vv{b}=\begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix}$ \\ \vspace{2.5ex}

$\begin{aligned}

\[A\Big|\vv{b}\] = &\begin{bNiceMatrix}

1&0&1&\aug&b_1 \\

0&1&0&\aug&b_2 \\

1&0&1&\aug&b_3

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\end{bNiceMatrix} \\

\xrightarrow{R_3-R_1}&\underbrace{\begin{bNiceMatrix}

1&0&1&\aug&b_1 \\

0&1&0&\aug&b_2 \\

0&0&0&\aug&b_3-b_1

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (2mm) ;

\end{bNiceMatrix}}_{\text{REF}}

\end{aligned}$ \\ \vspace{2ex}

\parbox{\linewidth}{

\begin{flushleft}

2 Possibilities: If $b_3-b_1\ne0$, there is a leading entry in the constants column so \textbf{no solution} exists to $A\vv{x}=\vv{b}$. If $b_3-b_1=0$, infinitely many solutions exists ($x_3$ is a free variable) $A\vv{x}=\vv{b}$ is \textbf{not} always consistent.

\end{flushleft}} \\ \newline

$\text{rank}\(A\)=2$}

\end{tblr}

\end{example}

For an $\textcolor{red}{m}\times n$ matrix $A$, $A\vv{x}=\vv{b}$ is consistent (one or infinitely many solutions exists) for all $\vv{b}\in\mathbb{R}^m$ if and only if $\text{rank}A=\textcolor{red}{m}$. \\ \newline

If $A\vv{x}=\vv{b}$ is consistent for all $\vv{b}\in\mathbb{R}^m$, then every vector in $\mathbb{R}^m$ is a linear combination of the columns in $A$. In other words the columns of $A$ \textbf{span} $\mathbb{R}^m$.

\begin{definition}{Definition: Span}

The set of linear combinations of the vectors $\vv{v_1},\vv{v_2},\hdots,\vv{v_k}\in\mathbb{R}^n$ is the span of $\vv{v_1},\vv{v_2},\hdots,\vv{v_k}$, denoted $\text{span}\(\vv{v_1},\vv{v_2},\hdots,\vv{v_k}\)$.

$$\text{span}\(\vv{v_1},\vv{v_2},\hdots,\vv{v_k}\)=\left\{c_1\vv{v_1}+c_2\vv{v_2}+\hdots+c_k\vv{v_k}\big|c_1,c_2,\hdots,c_k\in\mathbb{R}\right\}$$

We say the vectors $\vv{v_1},\vv{v_2},\hdots,\vv{v_k}$ span $\mathbb{R}^n$ or generate $\mathbb{R}^n$ if:

$$\text{span}\(\vv{v_1},\vv{v_2},\hdots,\vv{v_k}\)=\mathbb{R}^n$$

\begin{example}{Example: Comparing A Set And A span}

$$\text{span}\(\left. \begin{bmatrix}1\\2\end{bmatrix}\)=\left\{c\begin{bmatrix}1\\2\end{bmatrix}\right|c\in\mathbb{R}\right\}$$

\begin{center}

\begin{tikzpicture}

\begin{axis} [

width = 0.5\linewidth,

axis x line=middle,

axis y line=middle,

ymin = -5, ymax = 5,

xmin = -5, xmax = 5,

xlabel = {$x$},

ylabel = {$y$},

%xtick = {100},

%ytick = {100},

%xticklabels={},

every axis plot post/.append style={ultra thick},

clip = true,

axis line style = {Stealth-Stealth, thick},

legend style = {draw = none},

legend pos = outer north east,

title = {$\left\{\begin{bsmallmatrix}1\\2\end{bsmallmatrix}\right\}$ vs. $\text{span}\(\begin{bsmallmatrix}1\\2\end{bsmallmatrix}\)$}

]

\addplot [

black,

style = {Stealth-Stealth, dotted},

samples = 200,

domain = -2.5:2.5,

] {2*x} node[above right, pos = 0, rotate =63.434948822922010648427806279547] {$\text{span}\(\begin{bsmallmatrix}1\\2\end{bsmallmatrix}\)$};

\draw[-Stealth, ultra thick] (0,0) -- (1,2) node[right] {$\left\{\begin{bsmallmatrix}1\\2\end{bsmallmatrix}\right\}$};

\end{axis}

\end{tikzpicture}

\end{center}

\end{example}

\end{definition}

\begin{example}{Example: span That Equals \text{$\mathbb{R}^n$}}

$$\text{span}\(\left. \begin{bmatrix}1\\2\end{bmatrix},\begin{bmatrix}2\\-2\end{bmatrix}\)=\left\{c\begin{bmatrix}1\\2\end{bmatrix}+d\begin{bmatrix}2\\-2\end{bmatrix}\right|c,d\in\mathbb{R}\right\}$$

\begin{center}

\begin{tikzpicture}

\begin{axis} [

width = 0.5\linewidth,

axis x line=middle,

axis y line=middle,

ymin = -5, ymax = 5,

xmin = -5, xmax = 5,

xlabel = {$x$},

ylabel = {$y$},

%xtick = {100},

%ytick = {100},

%xticklabels={},

every axis plot post/.append style={ultra thick},

clip = true,

axis line style = {Stealth-Stealth, thick},

legend style = {draw = none},

legend pos = outer north east,

]

\addplot [

black,

style = {Stealth-Stealth, dotted},

samples = 200,

domain = -2.5:2.5,

] {2*x};

\draw[-Stealth, ultra thick] (0,0) -- (1,2);

\addplot [

black,

style = {Stealth-Stealth, dotted},

samples = 200,

domain = -5:5,

] {-x};

\draw[-Stealth, ultra thick] (0,0) -- (2,-2);

\addplot[mark=*] coordinates {(4,3)};

\addplot[mark=*] coordinates {(3.5,-2)};

\end{axis}

\end{tikzpicture} \\ \newline

$\text{span}\(\begin{bmatrix}1\\2\end{bmatrix},\begin{bmatrix}2\\-2\end{bmatrix}\)=\mathbb{R}^2$ thus $\begin{bmatrix}1\\2\end{bmatrix}$ and $\begin{bmatrix}2\\-2\end{bmatrix}$ span $\mathbb{R}^2$.

\end{center}

\end{example}

\begin{example}{Example: Algebraic span}

\begin{align*}

\text{span}\( \begin{bmatrix}1\\1\\0\\1\end{bmatrix},\begin{bmatrix}0\\1\\0\\0\end{bmatrix},\begin{bmatrix}1\\0\\1\\1\end{bmatrix}\)&=\left\{\left. c_1\begin{bmatrix}1\\1\\0\\1\end{bmatrix}+c_2\begin{bmatrix}0\\1\\0\\0\end{bmatrix}+c_3\begin{bmatrix}1\\0\\1\\1\end{bmatrix}\right| c_1,c_2,c_3\in\mathbb{R}\right\} \\

&=\left\{\left. \begin{bmatrix}\underline{c_1+c_3}\\c_1+c_2\\c_3\\\underline{c_1+c_3}\end{bmatrix}\right|c_1,c_2,c_3\in\mathbb{R}\right\}

\end{align*}

Every vector in the set has its 1\textsuperscript{st} and 4\textsuperscript{th} component equal to each other.

\end{example}

\begin{example}{Example: If A Vector Is In A span}

Is $\vv{b}=\begin{bmatrix}1\\0\\0\end{bmatrix}$ in $\text{span}\(\vv{v_1},\vv{v_2},\vv{v_3}\)$ where $\vv{v_1}=\begin{bmatrix}1\\1\\-4\end{bmatrix}$, $\vv{v_2}=\begin{bmatrix}0\\-1\\3\end{bmatrix}$, $\vv{v_3}=\begin{bmatrix}3\\-1\\0\end{bmatrix}$? \\ \newline

This is the equivelent of asking the following questions:

\begin{itemize}

\item Is $\vv{b}\in\text{span}\(\vv{v_1},\vv{v_2},\vv{v_3}\)$? \textbf{No}

\item Is $\vv{b}$ a linear combination of $\vv{v_1},\vv{v_2}$ and $\vv{v_3}$? \textbf{No}

\item Does $x_1\vv{v_1}+x_2\vv{v_2}+x_3\vv{v_3}=\vv{b}$ have a solution? \textbf{No}

\item Is $\begin{bmatrix}\vv{v_1}&\vv{v_2}&\vv{v_3}\end{bmatrix}\vv{x}=\vv{b}$ consistent? \textbf{No}

\end{itemize}

$$\begin{bmatrix}1&0&3&\aug&1\\1&-1&-1&\aug&0\\-4&3&0&\aug&0\end{bmatrix}\xrightarrow[R_2-R_1]{R_3+4R_1}\begin{bmatrix}1&0&3&\aug&1\\0&-1&-4&\aug&-1\\0&3&12&\aug&4\end{bmatrix}\xrightarrow{R_3+3R_2}\begin{bNiceMatrix}

1&0&3&\aug&1\\

0&-1&-4&\aug&-1\\

0&0&0&\aug&1

\CodeAfter

\tikz \draw (3-5) circle (2mm) ;

\end{bNiceMatrix}$$

As $0\ne1$ then there is no solution such that $A\vv{x}=\vv{b}$ is consistent so $\vv{b}$ is \textbf{not} in $\text{span}\(\vv{v_1},\vv{v_2},\vv{v_3}\)$. This also means that all of the above questions are false.

\end{example}

Let $A$ be an $m\times n$ matrix with column vectors $\vv{a_1},\vv{a_2},\hdots,\vv{a_n}$. Let $\vv{b}\in\mathbb{R}^m$. The linear system $A\vv{x}=\vv{b}$ is consistant if and only if $\vv{b}\in\text{span}\(\vv{a_1},\vv{a_2},\hdots,\vv{a_n}\)$.

\begin{theorem}{Theorem: Relating Consistence And span}

Let $A$ be an $m\times n$ matrix with column vectors $\vv{a_1},\vv{a_2},\hdots,\vv{a_n}$. Then the following statements are equivalent (either all True of all are False):

\begin{enumerate}

\item $\text{rank}A=m$

\item $A\vv{x}=\vv{b}$ is consistant for all $\vv{b}\in\mathbb{R}^m$

\item $\text{span}\(\vv{a_1},\vv{a_2},\hdots,\vv{a_n}\)=\mathbb{R}^m$ (the column vectors of $A$ span $\mathbb{R}^m$) ($\vv{a_1},\vv{a_2},\hdots,\vv{a_n}$ span $\mathbb{R}^m$)

\end{enumerate}

\end{theorem}

\begin{example}{Example: Finding Rank To See If A span Spans $\mathbb{R}^m$}

Does $\text{span}\(\begin{bmatrix}1\\2\\3\end{bmatrix},\begin{bmatrix}-1\\-1\\2\end{bmatrix},\begin{bmatrix}2\\-1\\1\end{bmatrix},\begin{bmatrix}3\\4\\8\end{bmatrix}\)=\mathbb{R}^3$?

$$\begin{bmatrix}1&-1&2&3\\2&-1&-1&4\\3&-2&1&8\end{bmatrix}\xrightarrow[R_3-3R_1]{R_2-2R_1}\begin{bmatrix}1&-1&2&3\\0&1&-5&-2\\0&1&-5&-1\end{bmatrix}\xrightarrow{R_3-R_2}\begin{bmatrix}1&-1&2&3\\0&1&-5&-2\\0&0&0&-3\end{bmatrix}$$

$\text{rank}A=3=\text{number of rows in }A$ thus $\text{span}\(\begin{bmatrix}1\\2\\3\end{bmatrix},\begin{bmatrix}-1\\-1\\2\end{bmatrix},\begin{bmatrix}2\\-1\\1\end{bmatrix},\begin{bmatrix}3\\4\\8\end{bmatrix}\)=\mathbb{R}^3$.

\end{example}

\begin{example}{Example: Do The Columns span $\mathbb{R}^m$?}

Do the columns of $A=\begin{bmatrix}1&2&0\\-1&0&2\\0&1&1\end{bmatrix}$ span $\mathbb{R}^3$?

$$A=\begin{bmatrix}1&2&0\\-1&0&2\\0&1&1\end{bmatrix}\xrightarrow{R_2+R_1}\begin{bmatrix}1&2&0\\0&2&2\\0&1&1\end{bmatrix}\xrightarrow{R_3-\frac{1}{2}R_2}\begin{bmatrix}1&2&0\\0&2&2\\0&0&0\end{bmatrix}$$

$\text{rank}A=2\ne\text{number of rows in }A$ so the columns of $A$ do \textbf{not} span $\mathbb{R}^3$.

\end{example}

\begin{example}{Example: If More Vectors Are Added To A Span Do Past Elements Change?}

Suppose $\vv{x}\in\text{span}\(\vv{u},\vv{v}\)$. Is $\vv{x}\in\text{span}\(\vv{u},\vv{v},\vv{w}\)$? \\ \newline

\textbf{Yes}, as since $\vv{x}\in\text{span}\(\vv{u},\vv{v}\)$ then $\vv{x}=c_1\vv{u}+c_2\vv{v}$ (for some $c_1,c_2\in\mathbb{R}$) which also means that $\vv{x}=c_1\vv{u}+c_2\vv{v}+0\vv{w}$ so $\vv{x}\in\text{span}\(\vv{u},\vv{v},\vv{w}\)$ as it can be written as a linear combination of $\vv{u},\vv{v}$ and $\vv{w}$.

\end{example}

\begin{example}{Example: Do Spans Change When Elements Are Linear Combinations?}

Suppose $\vv{w}=3\vv{u}-2\vv{v}$ and $\vv{x}\in\text{span}\(\vv{u},\vv{v},\vv{w}\)$. Is $\vv{x}\in\text{span}\(\vv{u},\vv{v}\)$?

\begin{align*}

\vv{x}&=c_1\vv{u}+c_2\vv{v}+c_3\vv{w} \\

&=c_1\vv{u}+c_2\vv{v}+c_3\(3\vv{u}-2\vv{v}\) \\

&=c_1\vv{u}+c_2\vv{v}+3c_3\vv{u}-2c_3\vv{v} \\

&=\(c_1+3c_3\)\vv{u}+\(c_2-2c_3\)\vv{v}

\end{align*}

\textbf{Yes}, as you can write $\vv{x}$ as a linear combination of $\vv{u}$ and $\vv{v}$. This also means that

$$\text{span}\(\vv{u},\vv{v},\vv{w}\)=\text{span}\(\vv{u},\vv{v}\)$$

\end{example}

\begin{theorem}{Theorem: A Span Can Contain Linear Combinations}

Let $\vv{v_1},\vv{v_2},\hdots,\vv{v_k}\in\mathbb{R}^n$ if $\vv{v_i}\in\text{span}\(\vv{v_1},\vv{v_2},\hdots,\vv{v}_{i-1},\vv{v}_{i+1},\hdots,\vv{v_k}\)$ then

$$\text{span}\(\vv{v_1},\vv{v_2},\hdots,\vv{v_k}\)=\text{span}\(\vv{v_1},\vv{v_2},\hdots,\vv{v}_{i-1},\vv{v}_{i+1},\hdots,\vv{v_k}\)$$

\begin{example}{Example: Span Having Linear Combination}

If $\vv{b}\in\text{span}\(\vv{u},\vv{v},\vv{w}\)$ then $$\text{span}\(\vv{u},\vv{v},\vv{w},\vv{b}\)=\text{span}\(\vv{u},\vv{v},\vv{w}\)$$

\end{example}

\end{theorem}

\begin{example}{Example: Span And Homogeneous Linear Systems}

Suppose $\vv{b}\in\text{span}\(\vv{u},\vv{v}\)$. How many solutions exist to the homogeneous linear system \\ $x_1\vv{u}+x_2\vv{v}+x_3\vv{b}=\vv{0}$?

$$\begin{bmatrix}\vv{u}&\vv{v}&\vv{b}\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\vv{0}$$

One solution is $\underbrace{x_1=0,x_2=0,x_3=0}_{\text{Trivial Solution}}$: $0\vv{u}+0\vv{v}+0\vv{b}=0\checkmark$

\begin{align*}

\vv{b}\in\text{span}\(\vv{u},\vv{v}\)&\rightarrow\vv{b}\text{ is a linear combination of }\vv{u}\text{ and }\vv{v} \\

&\rightarrow\vv{b}=c\vv{u}+d\vv{v}\text{ for some }c,d\in\mathbb{R} \\

&\rightarrow\vv{0}=c\vv{u}+d\vv{v}-\vv{b} \\

&\rightarrow \fbox{$\begin{aligned}x_1&=c\\x_2&=d\\x_3&=-1\end{aligned}$} \text{ is also a solution}

\end{align*}

As there are at least 2 solutions to the homogeneous linear system then there are \textbf{Infinitely Many Solutions} to $x_1\vv{u}+x_2\vv{v}+x_3\vv{b}=\vv{0}$.

\end{example}

\begin{definition}{Definition: Linearly Independent And Linearly Dependent}

The vectors $\vv{v_1},\vv{v_2},\hdots,\vv{v_k}\in\mathbb{R}^n$ are \textbf{linearly independent} if the homogeneous linear system $x_1\vv{v_1}+x_2\vv{v_2}+\hdots+x_k\vv{v_k}=\vv{0}$ has only the trivial solution $x_1=x_2=\hdots=x_k=0$. If infinitely many solutions exist to $x_1\vv{v_1}+x_2\vv{v_2}+\hdots+x_k\vv{v_k}=\vv{0}$ then the vectors $\vv{v_1},\vv{v_2},\hdots,\vv{v_k}$ are \textbf{linearly dependent}.

\end{definition}

Notes:

\begin{itemize}

\item In previous example, $\vv{u},\vv{v}$ and $\vv{b}$ are linearly dependent vectors.

\item {We use linearly independent/linearly dependent to describe a set of vectors $\vv{v_1},\vv{v_2},\hdots,\vv{v_k}$ or to describe the columns of $A$:

\begin{itemize}

\item If \underline{$A\vv{x}=\vv{0}$} has only the trivial solution $\vv{x}=\vv{0}$, the columns of $A$ are linearly independent, If \underline{$A\vv{x}=\vv{0}$} has infinitely many solutions, the columns of $A$ are linearly dependent.

\end{itemize}}

\end{itemize}

\begin{example}{Example: Are Vectors Linearly Independent Or Dependent?}

Determine if the following vectors are linearly independent or linearly dependent: \\ \newline

\begin{tblr}{width = \linewidth, colspec = {X|X}, cells = {halign = c, valign = m}}

{$\begin{bmatrix}1\\0\\1\end{bmatrix},\begin{bmatrix}0\\1\\0\end{bmatrix},\begin{bmatrix}1\\0\\-1\end{bmatrix}$ \\ \vspace{2.5ex}

Solve $x_1\begin{bmatrix}1\\0\\1\end{bmatrix}+x_2\begin{bmatrix}0\\1\\0\end{bmatrix}+x_3\begin{bmatrix}1\\0\\-1\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$ \\ \vspace{2.5ex}

$\begin{aligned}

\begin{bNiceMatrix}

1&0&1&\aug&0\\

0&1&0&\aug&0\\

1&0&-1&\aug&0

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\end{bNiceMatrix}\xrightarrow{R_3-R_1}&\begin{bmatrix}1&0&1&\aug&0\\0&1&0&\aug&0\\0&0&-2&\aug&0\end{bmatrix} \\

\xrightarrow{-\frac{1}{2}R_3}&\begin{bmatrix}1&0&1&\aug&0\\0&1&0&\aug&0\\0&0&1&\aug&0\end{bmatrix} \\

\xrightarrow{R_1-R_3}&\begin{bmatrix}1&0&0&\aug&0\\0&1&0&\aug&0\\0&0&1&\aug&0\end{bmatrix}

\end{aligned}$ \\ \newline

\parbox{\linewidth}{\begin{flushleft}

As all variables are bound then there is only one solution that exists so the vectors are linearly independent.\end{flushleft}} \\ \newline

\fbox{$\begin{aligned}x_1&=0\\x_2&=0\\x_3&=0\end{aligned}$}} &

{$\begin{bmatrix}1\\2\\1\end{bmatrix},\begin{bmatrix}1\\1\\1\end{bmatrix},\begin{bmatrix}1\\0\\1\end{bmatrix}$ \\ \vspace{2.5ex}

Solve $x_1\begin{bmatrix}1\\2\\1\end{bmatrix}+x_2\begin{bmatrix}1\\1\\1\end{bmatrix}+x_3\begin{bmatrix}1\\0\\1\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$ \\ \vspace{2.5ex}

$\begin{aligned}

\begin{bNiceMatrix}

1&1&1&\aug&0\\

2&1&0&\aug&0\\

1&1&1&\aug&0

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\end{bNiceMatrix}\xrightarrow{R_2-2R_1}&\begin{bmatrix}1&1&1&\aug&0\\0&-1&-2&\aug&0\\1&1&1&\aug&0\end{bmatrix} \\

\xrightarrow{R_3-R_1}&\begin{bmatrix}1&1&1&\aug&0\\0&-1&-2&\aug&0\\0&0&0&\aug&0\end{bmatrix} \\

\end{aligned}$ \\ \newline

\parbox{\linewidth}{\begin{flushleft}

There are infinitely many solutions as the vectors are linearly dependent as $x_3$ is a free variable ($x_1,x_2$ are bound).\end{flushleft}} \\ \newline

$\begin{aligned}

x_1-x_3&=0 \\

x_2+2x_3&=0 \\

0&=0

\end{aligned}\hspace{1cm}$\fbox{$\begin{aligned}x_1&=t\\x_2&=-2t\\x_3&=t\end{aligned}$}}

\end{tblr}

\end{example}

\begin{example}{Example: Are The Columns Of A Matrix Linearly Independent?}

Are the columns of $A=\begin{bmatrix}1&3&5\\2&4&6\end{bmatrix}$ linearly independent?

$$\text{Solve }A\vv{x}=\vv{0}\text{: }\begin{bmatrix}1&3&5&\aug&0\\2&4&6&\aug&0\end{bmatrix}\xrightarrow{R_2-2R_1}\begin{bNiceMatrix}[first-row]

&&\downarrow&&\\

1&3&5&\aug&0\\

0&-2&-4&\aug&0

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (3mm) ;

\end{bNiceMatrix}$$

A free variable exists in the system $A\vv{x}=\vv{0}$ so $A\vv{x}=\vv{0}$ has infinitely many solutions, thus the columns of $A$ are linearly dependent.

\end{example}

\begin{theorem}{Theorem: Linear Combinations In Sets And Linearly Dependent}

The vectors $\vv{v_1},\vv{v_2},\hdots,\vv{v_k}\in\mathbb{R}^n$ are linearly dependent if and only if one of the vectors in the set is a linear combination of the other vectors in the set.

\end{theorem}

\begin{proof}

If one of the vectors, say $\vv{v_1}$, is a linear combination of the others, then there are scalars $c_2,\hdots,c_m$ such that $\vv{v_1}=c_2\vv{v_2}+\hdots+c_m\vv{v_m}$. Rearranging, we obtain $\vv{v_1}-c_2\vv{v_2}-\hdots-c_m\vv{v_m}=\vv{0}$, which implies that $\vv{v_1},\vv{v_2},\hdots,\vv{v_m}$ are linearly dependent, since at least one of the scalars (namely, the coefficient $1$ of $\vv{v_1}$) is nonzero. \\ \newline

Conversely, suppose that $\vv{v_1},\vv{v_2},\hdots,\vv{v_m}$ are linearly dependent. Then there are scalars $c_1,c_2,\hdots,c_m$, not all zero, such that $c_1\vv{v_1}+c_2\vv{v_2}+\hdots+c_m\vv{v_m}=\vv{0}$. Suppose $c_1\ne0$. Then

$$c_1\vv{v_1}=-c_2\vv{v_2}-\hdots-c_m\vv{v_m}$$

and we may multiply both sides by $\dfrac{1}{c_1}$ to obtain $\vv{v_1}$ as a linear combination of the other vectors:

$$\vv{v_1}=-\(\frac{c_2}{c_1}\)\vv{v_2}-\hdots-\(\frac{c_m}{c_1}\)\vv{v_m}$$

\end{proof}

\begin{example}{Example: Linear Combination In Set Of Vectors}

$\begin{bmatrix}1\\2\\1\end{bmatrix},\begin{bmatrix}1\\1\\1\end{bmatrix},\begin{bmatrix}1\\0\\1\end{bmatrix}$ are linearly dependent.

$$x_1\begin{bmatrix}1\\2\\1\end{bmatrix}+x_2\begin{bmatrix}1\\1\\1\end{bmatrix}+x_3\begin{bmatrix}1\\0\\1\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$

Which has the solution set of $\begin{aligned}x_1&=t\\x_2&=-2t\\x_3&=t\end{aligned}\text{ }\(t\in\mathbb{R}\)$. If $t=2$ then $\begin{aligned}x_1&=2\\x_2&=-4\\x_3&=2\end{aligned}$ thus:

\begin{align*}

2\begin{bmatrix}1\\2\\1\end{bmatrix}-4\begin{bmatrix}1\\1\\1\end{bmatrix}+2\begin{bmatrix}1\\0\\1\end{bmatrix}&=\begin{bmatrix}0\\0\\0\end{bmatrix} \\

4\begin{bmatrix}1\\1\\1\end{bmatrix}-2\begin{bmatrix}1\\0\\1\end{bmatrix}&=2\begin{bmatrix}1\\2\\1\end{bmatrix}\\

2\begin{bmatrix}1\\1\\1\end{bmatrix}-1\begin{bmatrix}1\\0\\1\end{bmatrix}&=\begin{bmatrix}1\\2\\1\end{bmatrix}

\end{align*}

Making $\begin{bmatrix}1\\2\\1\end{bmatrix}$ a linear combination of $\begin{bmatrix}1\\1\\1\end{bmatrix}$ and $\begin{bmatrix}1\\0\\1\end{bmatrix}$.

\end{example}

\begin{theorem}{Theorem: Relating rank\text{,} \text{$A\vv{x}=\vv{0}$}\text{,} And Linearly Independent Columns}

Let $A$ be an $m\times n$ matrix with column vectors $\vv{a_1},\vv{a_2},\hdots,\vv{a_n}$. The following are equivalent (all are True or False):

\begin{enumerate}

\item $\text{rank}A=n$

\item $A\vv{x}=\vv{0}$ has only the trivial solution $\vv{x}=\vv{0}$.

\item The columns of $A$ are linearly independent ($\vv{a_1},\vv{a_2},\hdots,\vv{a_n}$ are linearly independent)

\end{enumerate}

\end{theorem}

\begin{example}{Example: If \text{$\text{rank}=n$} Then Linearly Independent}

$$A=\begin{bmatrix}1&1&-1\\0&-2&4\\2&0&2\\2&4&-5\end{bmatrix}\xrightarrow{\text{Row Equivalent}}\begin{bNiceMatrix}

1&0&0\\

0&1&0\\

0&0&1 \\

0&0&0

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (2mm) ;

\tikz \draw (3-3) circle (2mm) ;

\end{bNiceMatrix}$$

$\text{rank}A=3=\text{number of columns in } A$, thus the columns of $A$ are linearly independent.

\end{example}

\begin{example}{Example: Linearly Dependent Or Linearly Independent Based On Size}

If $B$ is $3\times5$, are the columns of $B$ linearly independent or linearly dependent? \\ \newline

$\text{rank}B$ is at most 3 so $\text{rank}B\ne5$ thus the columns of $B$ are linearly dependent.

\end{example}

Note:

\begin{itemize}

\item Any set of vectors from $\mathbb{R}^n$ containing \textbf{more then} $n$ vectors must be linearly dependent.

\begin{proof}

Let $\vv{v_1},\vv{v_2},\hdots,\vv{v_m}$ be (column) vectors in $\mathbb{R}^n$ and let $A$ be the $n\times m$ matrix $\begin{bmatrix}\vv{v_1}&\vv{v_2}&\hdots&\vv{v_m}\end{bmatrix}$ with these vectors as its columns. Then $\vv{v_1},\vv{v_2},\hdots,\vv{v_m}$ are linearly dependent if and only if the homogeneous linear system with augmented matrix $\begin{bmatrix}A&\aug&\vv{0}\end{bmatrix}$ has a nontrivial solution. But this will always be the case if $A$ has more columns than rows; it is the case here, since number of columns $m$ is greater than number of rows $n$.

\end{proof}

\end{itemize}

\begin{example}{Example: Is A Set Of Vectors Containg \text{$\vv{0}$} Linearly Independent}

$\vv{u},\vv{v},\vv{0}\in\mathbb{R}^n$ Are $\vv{u},\vv{v},\vv{0}$ linearly independent? \\ \newline

Can be rephrased as, how many solutions exists to:

$$x_1\vv{u}+x_2\vv{v}+x_3\vv{0}=\vv{0}$$

Of which there is the trivial solution of $\begin{aligned}x_1&=0\\x_2&=0\\x_3&=0\end{aligned}$ and the non trivial solution of $\begin{aligned}x_1&=0\\x_2&=0\\x_3&=2\end{aligned}$. As there are two possible solutions then there are infinitely many solutions so the set of vectors is linearly dependent.

\end{example}

Note:

\begin{itemize}

\item If $\vv{0}$ is in a set of vectors, the set must be linearly dependent.

\end{itemize}

The \textbf{rank} of a matrix is equal to the number of linearly independent columns in the matrix.

\begin{example}{Example: How Many Linearly independent Columns?}

$$A=\begin{bNiceMatrix}[first-row]

\downarrow&\downarrow&&&\downarrow\\

1&0&5&0&2\\

0&1&3&4&1\\

-1&1&-2&4&1\\

0&0&0&0&1

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (2mm) ;

\tikz \draw (3-5) circle (2mm) ;

\end{bNiceMatrix}\longrightarrow\begin{bNiceMatrix}[first-row]

\downarrow&\downarrow&&&\downarrow\\

1&0&5&0&0\\

0&1&3&4&0\\

0&0&0&0&1\\

0&0&0&0&0

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (2mm) ;

\tikz \draw (3-5) circle (2mm) ;

\end{bNiceMatrix}=U$$

$\text{rank}A=3\ne5$ so $\vv{a_1},\vv{a_2},\vv{a_3},\vv{a_4},\vv{a_5}$ are linearly dependent, however $\vv{a_1},\vv{a_2},\vv{a_5}$ are linearly independent as they have a leading coefficient, and aren't linear combinations of other columns.

\begin{tblr}{width = \linewidth,colspec = {X|X}, cells = {mode = dmath, valign = m, halign = c}}

{\begin{aligned}5\begin{bmatrix}1\\0\\0\\0\end{bmatrix}+3\begin{bmatrix}0\\1\\0\\0\end{bmatrix}&=\begin{bmatrix}5\\3\\0\\0\end{bmatrix} \\

5\begin{bmatrix}1\\0\\-1\\0\end{bmatrix}+3\begin{bmatrix}0\\1\\1\\0\end{bmatrix}&=\begin{bmatrix}5\\3\\-2\\0\end{bmatrix}\end{aligned}} &

{\begin{aligned}4\begin{bmatrix}0\\1\\0\\0\end{bmatrix}&=\begin{bmatrix}0\\4\\0\\0\end{bmatrix} \\

4\begin{bmatrix}0\\1\\1\\0\end{bmatrix}&=\begin{bmatrix}0\\4\\4\\0\end{bmatrix}\end{aligned}}

\end{tblr}

$$\text{span}\(\vv{a_1},\vv{a_2},\vv{a_3},\vv{a_4},\vv{a_5}\)=\text{span}\(\vv{a_1},\vv{a_2},\vv{a_5}\)$$

We need $n$ linearly independent vectors from $\mathbb{R}^n$ to span $\mathbb{R}^n$. \\ \newline

Also note that although row equivalent matrices share a lot of properties they aren't equal, $\vv{a_1}\ne\vv{u_1}$.

\end{example}

Notes:

\begin{itemize}

\item Any 2 linearly independent vectors from $\mathbb{R}^2$ spans $\mathbb{R}^2$.

\item Any 3 linearly independent vectors from $\mathbb{R}^3$ spans $\mathbb{R}^3$.

\item In general, any $n$ linearly independent vectors from $\mathbb{R}^n$ spans $\mathbb{R}^n$.

\end{itemize}

\begin{example}{Example: \text{$\vv{a_1}\ne\vv{u_1}$}}

Suppose $\begin{bmatrix}\vv{v_1}&\vv{v_2}&\vv{v_3}&\vv{v_4}&\vv{v_5}\end{bmatrix}=\begin{bmatrix}1&2&0&2&0\\2&1&3&2&4\\1&2&0&2&0\\0&2&-2&0&0\end{bmatrix}$ has RREF \\$\begin{bmatrix}\vv{u_1}&\vv{u_2}&\vv{u_3}&\vv{u_4}&\vv{u_5}\end{bmatrix}=\begin{bNiceMatrix}

1&0&2&0&4\\

0&1&-1&0&0\\

0&0&0&1&-2\\

0&0&0&0&0

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (2mm) ;

\tikz \draw (3-4) circle (2mm) ;

\end{bNiceMatrix}$.

\begin{center}$\vv{v_1},\vv{v_2},\vv{v_4}$ are linearly independent, leading coefficent\end{center}

\begin{align*}

2\vv{v_1}-\vv{v_2}&=\vv{v_3} & 2\vv{u_1}-\vv{u_2}&=\vv{u_3} \\

4\vv{v_1}-2\vv{v_4}&=\vv{v_5} & 4\vv{u_1}-2\vv{u_4}&=\vv{u_5}

\end{align*}

\end{example}

Notes:

\begin{itemize}

\item In general, $\vv{v_i}\ne\vv{u_i}$ in row equivalent matrices but relationships bewtween column vectors is preserved.

\end{itemize}

\newpage

\section{Section 3.3 Inverses of Matrices}

\begin{definition}{Definition: Invertable}

An $n\times n$ matrix $A$ is \textbf{invertible} if there is an $n\times n$ matrix $B$ such that $AB=I_n$ and $BA=I_n$ (where $I_n$ is the $n\times n$ identity matrix). We call $B$ the \textbf{inverse} of $A$, denoted $A^{-1}$.

\begin{example}{Example: Invertible Matrix}

Verify $\displaystyle{\frac{1}{16}\begin{bmatrix}2&-8\\1&4\end{bmatrix}}$ is the inverse of $\begin{bmatrix}4&8\\-1&2\end{bmatrix}$.

$$\frac{1}{16}\begin{bmatrix}2&-8\\1&4\end{bmatrix}\begin{bmatrix}4&8\\-1&2\end{bmatrix}=\frac{1}{16}\begin{bmatrix}16&0\\0&16\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}=I_2$$

$$\begin{bmatrix}4&8\\-1&2\end{bmatrix}\(\frac{1}{16}\begin{bmatrix}2&-8\\1&4\end{bmatrix}\)=I_2$$

\end{example}

\end{definition}

Notes:

\begin{itemize}

\item If $A$ is invertible then $A^{-1}$ is unique.

\begin{proof}

In mathematics, a standard way to show that there is just one of something is to show that there cannot be more than one. So, suppose that $A$ has two inverses, say $A'$ and $A''$. Then

$$AA'=I=A'A\qquad\text{and}\qquad AA''=I=A''A$$

Thus,

$$A'=A'I=A'\(AA''\)=\(A'A\)A''=IA''=A''$$

Hence, $A'=A''$, and the inverse is unique.

\end{proof}

\item {Not all square matrices are invertible. $n\times n$ matrices can be grouped onto 2 categories:

\begin{itemize}

\item Invertible/nonsingular

\item Noninvertible/singular

\end{itemize}}

\end{itemize}

\begin{theorem}{Theorem: You Don't Need \text{$AC=I_n$ and $CA=I_n$}}

Let $A$ be an $n\times n$ matrix. If there is a matrix $C$ such that $C$ is $n\times n$ and $AC=I_n$ or $CA=I_n$ then $C$ is the inverse of $A$. \\ \newline

Notes:

\begin{itemize}

\item If $C=A^{-1}$, then $A=C^{-1}$.

\end{itemize}

\end{theorem}

If $\begin{bmatrix}a\end{bmatrix}\ne\begin{bmatrix}0\end{bmatrix}$, then $\displaystyle{\begin{bmatrix}a\end{bmatrix}^{-1}=\begin{bmatrix}\frac{1}{a}\end{bmatrix}}$ \\ \newline

If $\begin{bmatrix}a\end{bmatrix}=\begin{bmatrix}0\end{bmatrix}$, $\begin{bmatrix}a\end{bmatrix}$ is not invertible.

\newpage

\subsection{Inverses of $2\times2$ Matrices}

\begin{theorem}{Theorem: Finding The Inverse Of A \text{$2\times2$} Matrix}

Let $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ then:

\begin{itemize}

\item If $ad-bc=0$, then $A$ is not invertible.

\item {If $ad-bc\ne0$, then $A$ is invertible and is:

$$A^{-1}=\frac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$}

\end{itemize}

\end{theorem}

\begin{proof}

Suppose that $\text{det}A=ad-bc\ne0$ Then

$$\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}=\begin{bmatrix}ad-bc&-ab+ba\\cd-dc&-cb+da\end{bmatrix}=\begin{bmatrix}ad-bc&0\\0&ad-bc\end{bmatrix}=\text{det}A\begin{bmatrix}1&0\\0&1\end{bmatrix}$$

Similarly,

$$\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\begin{bmatrix}a&b\\c&d\end{bmatrix}=\text{det}A\begin{bmatrix}1&0\\0&1\end{bmatrix}$$

Since $\text{det}A\ne0$, we can multiply both sides of each equation by $\dfrac{1}{\text{det}A}$ to obtain

$$\begin{bmatrix}a&b\\c&d\end{bmatrix}\(\frac{1}{\text{det}A}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\)=\begin{bmatrix}1&0\\0&1\end{bmatrix}\qquad\text{and}\qquad\(\frac{1}{\text{det}A}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\)\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$$

The the matrix $\dfrac{1}{\text{det}A}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$ satisfies the definition of an inverse, so $A$ is invertible. Since the inverse of $A$ is unique we must have

$$A^{-1}=\frac{1}{\text{det}A}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$

Conversely, assume that $ad-bc=0$. We will consider separately the cases where $a\ne0$ and where $a=0$. If $a\ne0$, then $d=\dfrac{bc}{a}$, so the matrix can be written as

$$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}a&b\\ac/a&bc/a\end{bmatrix}=\begin{bmatrix}a&b\\ka&kb\end{bmatrix}$$

where $k=\dfrac{c}{a}$. In other words, the second row of $A$ is a multiple of the first. If $A$ has an inverse $\begin{bmatrix}w&x\\y&z\end{bmatrix}$, then

$$\begin{bmatrix}a&b\\ka&kb\end{bmatrix}\begin{bmatrix}w&x\\y&z\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$$

and the corresponding system of linear equations

\begin{align*}

aw+by&=1 \\

ax+bz&=0 \\

kaw+kby&=0 \\

kax+kbz&=0

\end{align*}

has no solution. \\

If $a=0$, then $ad-bc=0$ implies that $bc=0$, and therfore either $b$ or $c$ is $0$. Thus, $A$ is of the form

$$\begin{bmatrix}0&0\\c&d\end{bmatrix}\qquad\text{or}\qquad\begin{bmatrix}0&b\\0&d\end{bmatrix}$$

In the first case, $\begin{bmatrix}0&0\\c&d\end{bmatrix}\begin{bmatrix}w&x\\y&z\end{bmatrix}=\begin{bmatrix}0&0\\ *&*\end{bmatrix}\ne\begin{bmatrix}1&0\\0&1\end{bmatrix}$ Similarly $\begin{bmatrix}0&b\\0&d\end{bmatrix}$ cannot have an inverse. \\

Consequently, if $ad-bc=0$, then $A$ is not invertible.

\end{proof}

\begin{definition}{Definition: Determinate}

Let $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$. The \textbf{determinate} of $A$, denoted $\text{det}A$ is:

$$\text{det}A=ad-bc=\abs{\begin{matrix}a&b\\c&d\end{matrix}}$$

\end{definition}

\begin{example}{Example: Finding The Inverse Of A \text{$2\times2$} Matrix}

Find $A^{-1}$ or determine $A$ is not invertible: \\ \newline

\begin{tblr}{width = \linewidth, colspec={Q|Q|X}, row{1} = {valign=m,halign=c}, row{2} = {valign=t,halign=l}, row{3} = {valign=m,halign=c}}

{$A=\begin{bmatrix}1&-1\\2&1\end{bmatrix}$ \\ \vspace{2.5ex}

$\begin{aligned}\text{det}A&=1\(1\)-\(-1\)\(2\)\\ &=3\end{aligned}$ \\ \vspace{2.5ex}

$\begin{aligned}

A^{-1}&=\frac{1}{1\(1\)-\(-1\)\(2\)}\begin{bmatrix}1&1\\-2&1\end{bmatrix} \\

&=\frac{1}{3}\begin{bmatrix}1&1\\-2&1\end{bmatrix}

\end{aligned}$\vspace{2.5ex}} &

{$A=\begin{bmatrix}-1&1\\1&-1\end{bmatrix}$ \\ \vspace{2.5ex}

$\begin{aligned}\text{det}A&=\(-1\)\(-1\)-\(1\)\(1\)\\ &=0\end{aligned}$ \\ \vspace{2.5ex}

$A$ is Not Invertible} &

{$A=\begin{bmatrix}x&kx\\y&ky\end{bmatrix}$ \\ \vspace{2.5ex}

$\begin{aligned}\text{det}A&=kxy-kxy\\ &=0\end{aligned}$\\ \vspace{2.5ex}

$A$ is Not Invertible} \\

{Verify: $A^{-1}A$\vspace{1ex}} & {} \\

{$\displaystyle{\frac{1}{3}\begin{bmatrix}1&1\\-2&1\end{bmatrix}\begin{bmatrix}1&-1\\2&1\end{bmatrix}=\frac{1}{3}\begin{bmatrix}3&0\\0&3\end{bmatrix}=I_2\checkmark}$}

\end{tblr} \\ \newline

\textbf{Conclusion:} If $A$ is a $n\times n$ matrix with linearly dependent columns, then $A$ is not invertible.

\end{example}

\begin{theorem}{Theorem: If A Matrix Is Invertible Then It Is Linearly Independent}

If $A$ is an invertible $n\times n$ matrix, then for each $\vv{b}\in\mathbb{R}^n$, the linear system $A\vv{x}=\vv{b}$ has the unique solution $\vv{x}=A^{-1}\vv{b}$.

\begin{align*}

A\vv{x}&=\vv{b} & A\(A^{-1}\vv{b}\)&=\(AA^{-1}\)\vv{b} \\

A^{-1}A\vv{x}&=A^{-1}\vv{b} & &=I_n\vv{b} \\

I_n\vv{x}&=A^{-1}\vv{b} & &=\vv{b}\checkmark \\

\vv{x}&=A^{-1}\vv{b}

\end{align*}

\end{theorem}

\begin{proof}

We are asked to prove two things: that $A\vv{x}=\vv{b}$ has a solution and that it has only one solution. (In mathematics, such a proof is called an “existence and uniqueness” proof.) \\ \newline

To show that a solution exists, we need only verify that $\vv{x}=A^{-1}\vv{b}$ works. We check that

$$A\(A^{-1}\vv{b}\)=\(AA^{-1}\)\vv{b}=I\vv{b}=\vv{b}$$

So $A^{-1}\vv{b}$ satisfies the equation $A\vv{x}=\vv{b}$, and hence there is at least this solution. \\ \newline

To show that this solution is unique, suppose $\vv{y}$ is another solution. Then $A\vv{y}=\vv{b}$, and multiplying both sides of the equation by $A^{-1}$ on the left, we obtain the chain of implications

\begin{align*}

A\vv{y}&=\vv{b} \\

A^{-1}\(A\vv{y}\)&=A^{-1}\vv{b} \\

\(A^{-1}A\)\vv{y}&=A^{-1}\vv{b} \\

I\vv{y}&=A^{-1}\vv{b} \\

\vv{y}&=A^{-1}\vv{b}

\end{align*}

Thus, $\vv{y}$ is the same solution as before, and therefore the solution is unique.

\end{proof}

\begin{example}{Example: Using The Inverse To Solve A Linear System}

Find $\begin{bmatrix}5&3\\3&2\end{bmatrix}^{-1}$ then use the inverse to solve. \\ \newline

Find determinate:

$$\text{det}\begin{bmatrix}5&3\\3&2\end{bmatrix}=5\(2\)-3\(3\)=1$$

Find Inverse:

$$\begin{bmatrix}5&3\\3&2\end{bmatrix}^{-1}=\frac{1}{1}\begin{bmatrix}2&-3\\-3&5\end{bmatrix}$$

Solve the Linear Systems:

\begin{align*}

\begin{bmatrix}5&3\\3&2\end{bmatrix}\vv{x}&=\begin{bmatrix}1\\2\end{bmatrix} & \begin{bmatrix}5&3\\3&2\end{bmatrix}\vv{x}&=\begin{bmatrix}0\\0\end{bmatrix} \\

\vv{x}&=\begin{bmatrix}2&-3\\-3&5\end{bmatrix}\begin{bmatrix}1\\2\end{bmatrix} & \vv{x}&=\begin{bmatrix}2&-3\\-3&5\end{bmatrix}\begin{bmatrix}0\\0\end{bmatrix} \\

\vv{x}&=\begin{bmatrix}-4\\7\end{bmatrix} & \vv{x}&=\begin{bmatrix}0\\0\end{bmatrix}

\end{align*}

\end{example}

Note:

\begin{itemize}

\item {If $A$ is $n\times n$ and invertible, then $A\vv{x}=\vv{0}$ has only the trivial solution of $\vv{x}=\vv{0}$ or in other words the columns of $A$ are linearly independent.}

\end{itemize}

\begin{theorem}{Theorem: Properties Of Invertible Matrices}

Let $A$ and $B$ be invertible $n\times n$ matrices. Let $c$ be a nonzero scalar. Let $k$ be a nonnegatve integer. Then:

\begin{enumerate}

\item $\(A^{-1}\)^{-1}=A$

\item $\(AB\)^{-1}=B^{-1}A^{-1}$

\item $\(A^T\)^{-1}=\(A^{-1}\)^T$

\item $\displaystyle{\(cA\)^{-1}=\frac{1}{c}A^{-1}}$

\item $\(A^k\)^{-1}=\(A^{-1}\)^k=A^{-k}$

\end{enumerate}

Remember:

$$AA^{-1}=A^{-1}A=I_n$$

\end{theorem}

\begin{proof} \newline

\begin{itemize}

\item $\(A^{-1}\)^{-1}=A$ \\

To show that $A^{-1}$ is invertible, we must argue that there is a matrix $X$ such that

$$A^{-1}X=I=XA^{-1}$$

But $A$ certainly satisfies these equations in place of $X$, so $A^{-1}$ is invertible and $A$ is an inverse of $A^{-1}$. Since inverses are unique, this means that $\(A^{-1}\)^{-1}=A$.

\item $\(AB\)^{-1}=B^{-1}A^{-1}$\\

Here we must show that there is a matrix $X$ such that

$$\(AB\)X=I=X\(AB\)$$

The claim is that substituting $B^{-1}A^{-1}$ for $X$ works. We chack that

$$\(AB\)\(B^{-1}A^{-1}\)=A\(BB^{-1}\)A^{-1}=AIA^{-1}=AA^{-1}=I$$

where we have used associativity to shift the parentheses. Similarly, $\(B^{-1}A^{-1}\)\(AB\)=I$, so $AB$ is invertible and its inverse is $B^{-1}A^{-1}$.

\item $\(A^n\)^{-1}=\(A^{-1}\)^n$ \\

The basis step is when $n=0$, in which case we are being asked to prove that $A^0$ is invertible and that

$$\(A^0\)^{-1}=\(A^{-1}\)^0$$

This is the same as showing that $I$ is invertible and that is $I^{-1}=I$, which is clearly true. \\ \newline

Now we assume that the result is true when $n=k$, where $k$ is a specific nonnegative integer. That is, the induction hypothesis is to assume that $A^k$ is invertible and that

$$\(A^k\)^{-1}=\(A^{-1}\)^k$$

The induction step requires that we prove that $A^{k+1}$ is invertible and that $\(A^{k+1}\)^{-1}=\(A^{-1}\)^{k+1}$. Now we know that $A^{k+1}=A^kA$ (from the proof directly above) is invertible, since $A$ and (by hypothesis) $A^k$ are both invertible. Moreover,

\begin{align*}

\(A^{-1}\)^{k+1}&=\(A^{-1}\)^kA^{-1} \\

&=\(A^k\)^{-1}A^{-1}\text{ (by the Induction Hypothesis)} \\

&=\(AA^k\)^{-1} \text{ (by the proof directly above)} \\

&=\(A^{k+1}\)^{-1}

\end{align*}

Therfore, $A^n$ is invertible for all nonnegative integers $n$, and $\(A^n\)^{-1}=\(A^{-1}\)^n$ by the principle of mathematical induction.

\end{itemize}

\end{proof}

\begin{theorem}{Theorem: A Matrix Is Invertible If It Is Row Equivalent To $I_n$}

An $n\times n$ matrix $A$ is invertible if and only if $\underbrace{A \text{ is row equivalant to } I_n}_{I_n \text{ is the RREF of } A}$.

\end{theorem}

\begin{example}{Example: How Can We Find $A^{-1}$ (if $A^{-1}$ exists)?}

Consider a $3\times 3$ matrix $A$. Suppose $A^{-1}$ exists. Then:

\begin{align*}

AA^{-1}&=I_n \\

A\begin{bmatrix}\vv{x_1} & \vv{x_2} & \vv{x_3}\end{bmatrix} &= \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix} \\

\begin{bmatrix}A\vv{x_1} & A\vv{x_2} & A\vv{x_3}\end{bmatrix} &=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}

\end{align*}

This results in 3 linear systems:

$$A\vv{x_1}=\begin{bmatrix}1\\0\\0\end{bmatrix}\hspace{1cm}A\vv{x_2}=\begin{bmatrix}0\\1\\0\end{bmatrix}\hspace{1cm}A\vv{x_3}=\begin{bmatrix}0\\0\\1\end{bmatrix}$$

We can solve these linear systems to find the columns of $A^{-1}$.

\end{example}

\begin{example}{Example: Finding $A^{-1}$}

Let $A=\begin{bmatrix}-2&5\\-1&2\end{bmatrix}$. Solve $A\vv{x_1}=\begin{bmatrix}1\\0\end{bmatrix}$ and $A\vv{x_2}=\begin{bmatrix}0\\1\end{bmatrix}$.

\begin{align*}

\begin{bmatrix}-2&5&\aug&1\\-1&2&\aug&0\end{bmatrix}\xrightarrow{R_1\leftrightarrow R_2}&\begin{bmatrix}-1&2&\aug&0\\-2&5&\aug&1\end{bmatrix} & \begin{bmatrix}-2&5&\aug&0\\-1&2&\aug&1\end{bmatrix}\xrightarrow{R_1\leftrightarrow R_2}&\begin{bmatrix}-1&2&\aug&1\\-2&5&\aug&0\end{bmatrix} \\

\xrightarrow{-R_1}&\begin{bmatrix*}[r]1&-2&\aug&0\\-2&5&\aug&1\end{bmatrix*} & \xrightarrow{-R_1}&\begin{bmatrix*}[r]1&-2&\aug&-1\\-2&5&\aug&0\end{bmatrix*} \\

\xrightarrow{R_2+2R_1}&\begin{bmatrix*}[r]1&-2&\aug&0\\0&1&\aug&1\end{bmatrix*} & \xrightarrow{R_2+2R_1}&\begin{bmatrix*}[r]1&-2&\aug&-1\\0&1&\aug&-2\end{bmatrix*} \\

\xrightarrow{R_1+2R_2}&\begin{bmatrix*}[r]1&0&\aug&2\\0&1&\aug&1\end{bmatrix*} & \xrightarrow{R_1+2R_2}&\begin{bmatrix*}[r]1&0&\aug&-5\\0&1&\aug&-2\end{bmatrix*}

\end{align*}

Notice that both of the row reductions are the same steps so instead we can solve both systems at once

$$\begin{bmatrix}A&\!\!\!\left|\begin{matrix}\phantom{}\\\phantom{}\end{matrix}\right.\!\!\!&\begin{matrix}1&0\\0&1\end{matrix}\end{bmatrix}\xrightarrow{Row\hspace{1ex}Reduce}\begin{bmatrix*}[r]1&0&\aug&2&-5\\0&1&\aug&1&-2\end{bmatrix*}$$

Verify $A^{-1}=\begin{bmatrix}2&-5\\1&-2\end{bmatrix}$: $AA^{-1}=\begin{bmatrix}-2&5\\-1&2\end{bmatrix}\begin{bmatrix}2&-5\\1&-2\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}=I_2\checkmark$

\end{example}

For an $n\times n$ matrix, to find $A^{-1}$ or determine $A$ is not invertible:

\begin{enumerate}

\item Augment $A$ with $I_n$ $\begin{bmatrix}A&\aug&I_n\end{bmatrix}$

\item {Row Reduce $\begin{bmatrix}A&\aug&I_n\end{bmatrix}$ until either:

\begin{enumerate}

\item We determine $A$ is not row equivalent to $I_n$ thus $A$ is not invertible.

\item We get RREF of $\begin{bmatrix}A&\aug&I_n\end{bmatrix}$ and have $\begin{bmatrix}I_n&\aug&A^{-1}\end{bmatrix}$.

\end{enumerate}}

\end{enumerate}

\begin{example}{Example: Finding $A^{-1}$ At Once}

Find $A^{-1}$ or determine $A$ is not invertible: \\ \newline

\begin{tblr}{width = \linewidth, cells = {halign = c, valign = t}, colspec = {X|X}}

{$A=\begin{bmatrix}1&0&2\\1&1&2\\0&1&1\end{bmatrix}$ \\ \vspace{2.5ex}

$\begin{aligned}

\begin{bmatrix}A&\aug&I_3\end{bmatrix}=&\begin{bmatrix}1&0&2&\aug&1&0&0\\1&1&2&\aug&0&1&0\\0&1&1&\aug&0&0&1\end{bmatrix} \\

\xrightarrow{R_2-R_1}&\begin{bmatrix*}[r]1&0&2&\aug&1&0&0\\0&1&0&\aug&-1&1&0\\0&1&1&\aug&0&0&1\end{bmatrix*} \\

\xrightarrow{R_3-R_2}&\begin{bmatrix*}[r]1&0&2&\aug&1&0&0\\0&1&0&\aug&-1&1&0\\0&0&1&\aug&1&-1&1\end{bmatrix*} \\

\xrightarrow{R_1-2R_3}&\begin{bmatrix*}[r]1&0&0&\aug&-1&2&-2\\0&1&0&\aug&-1&1&0\\0&0&1&\aug&1&-1&1\end{bmatrix*}

\end{aligned}$ \\ \vspace{2.5ex}

$A=\begin{bmatrix}1&0&2\\1&1&2\\0&1&1\end{bmatrix} \hspace{1cm} A^{-1}=\begin{bmatrix*}[r]-1&2&-2\\-1&1&0\\1&-1&1\end{bmatrix*}$

\parbox{\linewidth}{\begin{flushleft}

Verify: $A^{-1}A=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I_3\checkmark$

\end{flushleft}}} & {$A=\begin{bmatrix}1&0&3&4\\0&1&1&1\\0&0&1&0\\0&1&2&1\end{bmatrix}$ \\ \vspace{2.5ex}

$\begin{aligned}

\begin{bmatrix}A&\aug&I_4\end{bmatrix}=&\begin{bNiceArray}{rrrr | rrrr}

1&0&3&4&1&0&0&0\\

0&1&1&1&0&1&0&0\\

0&0&1&0&0&0&1&0\\

0&\underline{1}&2&1&0&0&0&1

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (2mm) ;

\end{bNiceArray} \\

\xrightarrow{R_4-R_2}&\begin{bNiceArray}{rrrr | rrrr}

1&0&3&4&1&0&0&0\\

0&1&1&1&0&1&0&0\\

0&0&1&0&0&0&1&0\\

0&0&1&0&0&-1&0&1

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (2mm) ;

\tikz \draw (3-3) circle (2mm) ;

\end{bNiceArray} \\

\xrightarrow{R_4-R_3}&\begin{bNiceArray}{rrrr | rrrr}

1&0&3&4&1&0&0&0\\

0&1&1&1&0&1&0&0\\

0&0&1&0&0&0&1&0\\

0&0&0&\underline{0}&0&-1&-1&1

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (2mm) ;

\tikz \draw (3-3) circle (2mm) ;

\tikz \draw (4-6) circle (3mm) ;

\end{bNiceArray}

\end{aligned}$

\parbox{\linewidth}{\begin{flushleft}

$A$ is not row equivalent to $I_4$ so $A^{-1}$ DNE.

\end{flushleft}}}

\end{tblr}

\end{example}

\begin{theorem}{Theorem: The Fundamental Theorem Of Invertible Matrices (FTIM)}

Let $A$ be an $n\times n$ matrix. The following statements are equivalent (either all are true or all are false):

\begin{enumerate}

\item $A$ is invertible, ($A^{-1}$ exists)

\item $A$ is row equivalent to $I_n$. ($I_n$ is the RREF of $A$)

\item rank$A=n$

\item $A\vv{x}=\vv{b}$ has a solution for every $\vv{b}\in\mathbb{R}^n$. (more specifically, there is one unique solution $\vv{x}=A^{-1}\vv{b}$.)

\item $A\vv{x}=\vv{0}$ has only the trivial solution $\vv{x}=\vv{0}$.

\item The columns of $A$ span $\mathbb{R}^n$

\item The columns of $A$ are linearly independent.

\end{enumerate}

\end{theorem}

Notes:

\begin{itemize}

\item FTIM only applies to $n\times n$ matrices.

\item The negation of the FTIM gives a list of equivalent statements about noninvertible matrices.

\end{itemize}

\begin{example}{Example: FTIM Negation}

Let $A$ and $B$ be $n\times n$ matrices. If $B$ is not invertible, is $AB$ invertible?

\begin{align*}

B \text{ is not invertible }\Leftrightarrow&\text{ By FTIM, } B\vv{x}=\vv{0} \text{ has infinitely many solutions} \\

\Leftrightarrow& AB\vv{x}=A\vv{0} \text{ has infinitely many solutions} \\

\Leftrightarrow& \(AB\)\vv{x}=\vv{0}\text{ has infinitely many solutions} \\

\Leftrightarrow& \text{ By FTIM, $AB$ is not invertible.}

\end{align*}

\end{example}

\newpage

\section{Section 3.5 Subspaces}

\begin{definition}{Definition: Subspace}

A set of vectors $W $ in $\mathbb{R}^n$ is a \textbf{subspace} of $\mathbb{R}^n$ if:

\begin{enumerate}

\item $\vv{0}$ is in $W$

\item $W$ is closed under addition: For vectors $\vv{u}$ and $\vv{v}$ that are in $W$, then $\vv{u}+\vv{v}$ is also in $W$.

\item $W$ is also closed under scalar multiplication: For all $\vv{v}$ in $W$ and $c\in\mathbb{R}$, then $c\vv{v}$ is in $W$.

\end{enumerate}

\begin{example}{Example: Subspaces}

\begin{enumerate}

\item $\mathbb{R}^n$ is a subspace of $\mathbb{R}^n$.

\item $\left\{\vv{0}\right\}$ is a subspace of $\mathbb{R}^n$ (called the \textbf{zero space} of $\mathbb{R}^n$)

\end{enumerate}

\end{example}

\end{definition}

\begin{example}{Example: Not A Subspace Through Counter Example}

Is the set $W=\left\{\begin{bmatrix}0\\0\\0\end{bmatrix},\begin{bmatrix}1\\1\\1\end{bmatrix},\begin{bmatrix}1\\0\\1\end{bmatrix}\right\}$ a subspace of $\mathbb{R}^3$? \\ \newline

$W$ is not closed under addition because $\begin{bsmallmatrix}1\\1\\1\end{bsmallmatrix}$ and $\begin{bsmallmatrix}1\\0\\1\end{bsmallmatrix}$ are in $W$ but $\begin{bsmallmatrix}1\\1\\1\end{bsmallmatrix}+\begin{bsmallmatrix}1\\0\\1\end{bsmallmatrix}=\begin{bsmallmatrix}2\\1\\2\end{bsmallmatrix}$ is not in $W$. \\ \newline

$W$ is also not closed under scalar multiplication as $\begin{bsmallmatrix}1\\1\\1\end{bsmallmatrix}$ is in $W$ but $2\begin{bsmallmatrix}1\\1\\1\end{bsmallmatrix}=\begin{bsmallmatrix}2\\2\\2\end{bsmallmatrix}$ is not in $W$. \\ \newline

Thus $W$ is not a subspace of $\mathbb{R}^3$.

\end{example}

Notes:

\begin{itemize}

\item $\left\{\vv{0}\right\}$ is the only finite subsace of $\mathbb{R}^n$. All other subspaces of $\mathbb{R}^n$ contain infinitely many vectors.

\end{itemize}

\begin{example}{Example: Determine If A Set Is A Subspace}

Sketch each set in $\mathbb{R}^2$. Determine if the set is a subspace of $\mathbb{R}^2$.

\begin{enumerate}

\item {$S_1=\left\{\begin{bmatrix}x\\y\end{bmatrix}\in\mathbb{R}^2\Big|x+y=1\right\}$ \\

\begin{center}

\begin{tikzpicture}

\begin{axis} [

width = 0.5\linewidth,

axis x line=middle,

axis y line=middle,

ymin = -2, ymax = 2,

xmin = -2, xmax = 2,

xtick = {1},

ytick = {1},

axis line style = {Stealth-Stealth, ultra thick},

xlabel = {$x$},

ylabel = {$y$},

clip = false,

]

\addplot[mark=*, ultra thick] coordinates {(0,1)};

\addplot[mark=*, ultra thick] coordinates {(1,0)};

\addplot [

black,

samples = 200,

style = {Stealth-Stealth, thick},

domain = -1:2,

]

{1-x};

\end{axis}

\end{tikzpicture}

\end{center}

$\vv{0}$ is not in $S_1$ $(0+0\ne1)$ so $S_1$ is not a subspace.}

\item{$S_2=\left\{\begin{bmatrix}x\\y\end{bmatrix}\in\mathbb{R}^2\Big|x^2-y^2=0\right\}$ \\

\begin{center}

\begin{tikzpicture}

\begin{axis} [

width = 0.5\linewidth,

axis x line=middle,

axis y line=middle,

ymin = -2.5, ymax = 2.5,

xmin = -2.5, xmax = 2.5,

axis line style = {Stealth-Stealth, ultra thick},

xlabel = {$x$},

ylabel = {$y$},

clip = false,

]

\addplot[mark=*, ultra thick] coordinates {(0,0)};

\addplot [

black,

samples = 200,

style = {Stealth-Stealth, thick},

domain = -2.5:2.5,

]

{x} node [right] {$y=x$};

\addplot [

black,

samples = 200,

style = {Stealth-Stealth, thick},

domain = -2.5:2.5,

]

{-x} node[left, pos = 0] {$y=-x$};

\draw[red, ultra thick, -Stealth] (0,0) -- (1,1);

\draw[red, ultra thick, -Stealth] (0,0) -- (-1,1);

\draw[blue, ultra thick, -Stealth] (0,0) -- (0,2);

\end{axis}

\end{tikzpicture}

\end{center}

$\vv{0}$ is in $S_2$ because $0^2-0^2=0$ \\ \newline $\begin{bsmallmatrix}1\\1\end{bsmallmatrix}$ and $\begin{bsmallmatrix}-1\\1\end{bsmallmatrix}$ are in $S_2$ but $\begin{bsmallmatrix}1\\1\end{bsmallmatrix}+\begin{bsmallmatrix}-1\\1\end{bsmallmatrix}=\begin{bsmallmatrix}0\\2\end{bsmallmatrix}$ is not in $S_2$ $\(0^2-2^2\ne0\)$ so $S_2$ is \textbf{not} closed under addition, thus $S_2$ is \textbf{not} a subspace of $\mathbb{R}^2$.}

\item{$S_3=\left\{\begin{bmatrix}x\\y\end{bmatrix}\in\mathbb{R}^2\Big|x\ge0\text{ and }y\ge0\right\}$ \\

\begin{center}

\begin{tikzpicture}

\begin{axis} [

width = 0.5\linewidth,

axis x line=middle,

axis y line=middle,

ymin = -2.5, ymax = 2.5,

xmin = -2.5, xmax = 2.5,

axis line style = {Stealth-Stealth, ultra thick},

xlabel = {$x$},

ylabel = {$y$},

clip = false,

]

\fill[red, fill=red!30, draw = red] (0,0) rectangle (2,2);

\draw[thick, -Stealth] (0,0) -- (1,1);

\draw[thick, -Stealth] (0,0) -- (-2,-2);

\end{axis}

\end{tikzpicture}

\end{center}

$\begin{bsmallmatrix}1\\1\end{bsmallmatrix}$ is in $S_3$ $\(1\ge0\)$, and $-2\in\mathbb{R}$, however $-2\begin{bsmallmatrix}1\\1\end{bsmallmatrix}=\begin{bsmallmatrix}-2\\-2\end{bsmallmatrix}$ is \textbf{not} in $S_3$. Therfore $S_3$ is not closed under scalar multiplication so $S_3$ is \textbf{not} a subspace.}

\item{$S_4=\left\{\begin{bmatrix}x\\y\end{bmatrix}\in\mathbb{R}^2\Big|x-y=0\right\}$ \\

\begin{center}

\begin{tikzpicture}

\begin{axis} [

width = 0.5\linewidth,

axis x line=middle,

axis y line=middle,

ymin = -2.5, ymax = 2.5,

xmin = -2.5, xmax = 2.5,

axis line style = {Stealth-Stealth, ultra thick},

xlabel = {$x$},

ylabel = {$y$},

clip = false,

]

\addplot[mark=*, ultra thick] coordinates {(0,0)};

\addplot [

black,

samples = 200,

style = {Stealth-Stealth, thick},

domain = -2.5:2.5,

]

{x};

\end{axis}

\end{tikzpicture}

\end{center}

\begin{enumerate}

\item {$\begin{bmatrix}0\\0\end{bmatrix}$ is in $S_4$ because $0-0=0$.}

\item {Is $S_4$ closed under addition? \\ \newline

Let $\begin{bmatrix}x_1\\y_1\end{bmatrix},\begin{bmatrix}x_2\\y_2\end{bmatrix}\in S_4$. Then $x_1-y_1=0$ and $x_2-y_2=0$. Is $\begin{bmatrix}x_1\\y_1\end{bmatrix}+\begin{bmatrix}x_2\\y_2\end{bmatrix}=\begin{bmatrix}x_1+x_2\\y_1+y_2\end{bmatrix}$ in $S_4$?

\begin{align*}

\(x_1+x_2\)-\(y_1+y_2\)&=x_1+x_2-y_1-y_2 \\

&=x_1-y_1+x_2-y_2 \\

&=0 + 0 \\

&= 0

\end{align*}

So $\begin{bmatrix}x_1+x_2\\y_1+y_2\end{bmatrix}$ is in $S_4$ so $S_4$ is closed under addition.}

\item {Is $S_4$ closed under scalar multiplication. Let $\begin{bmatrix}x\\y\end{bmatrix}\in S_4$ and $c\in\mathbb{R}$. Then $x-y=0$. Is $c\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}cx\\cy\end{bmatrix}$ in $S_4$?

\begin{align*}

cx-cy&=c\(x-y\) \\

&=c\(0\) \\

&=0

\end{align*}

$c\begin{bmatrix}x\\y\end{bmatrix}$ is in $S_4$ so $S_4$ is closed under scalar multiplication.}

\end{enumerate}

Therfore $S_4$ is a subspace of $\mathbb{R}^2$ as $\vv{0}$ is in $S_4$, $S_4$ is closed under addition and also closed under scalar multiplication.}

\end{enumerate}

\end{example}

\begin{example}{Example: Is \text{$A\vv{x}=\vv{0}$} A Subspace?}

Let $A$ be an $m\times n$ matrix. Let $G=\left\{\vv{x}\in\mathbb{R}^n\Big|A\vv{x}=\vv{0}\right\}$. Is $G$ a subspace of $\mathbb{R}^n$?

\begin{enumerate}

\item {$\vv{0}$ is in $G$ because, $A\vv{0}=\vv{0}\checkmark$.}

\item {Is $G$ closed under addition? Let $\vv{x},\vv{y}\in G$ then $A\vv{x}=\vv{0}$ and $A\vv{y}=\vv{0}$. Is $\vv{x}+\vv{y}$ in $G$?

\begin{align*}

A\(\vv{x}+\vv{y}\)&=A\vv{x}+A\vv{y} \\

&=\vv{0}+\vv{0}\\

&=\vv{0}\checkmark

\end{align*}

$\vv{x}+\vv{y}$ is in $G$ so $G$ is closed under addition.}

\item {Is $G$ closed under scalar multiplication? Let $\vv{x}\in G$ and $c\in\mathbb{R}$ then $A\vv{x}=\vv{0}$. Is $c\vv{x}$ in $G$?

\begin{align*}

A\(c\vv{x}\)&=cA\vv{x} \\

&=c\(\vv{0}\) \\

&=\vv{0}\checkmark

\end{align*}

$c\vv{x}$ is in $G$ so $G$ is closed under scalar multiplication.}

\end{enumerate}

Therfore $G$ is a subspace of $\mathbb{R}^n$.

\end{example}

Notes:

\begin{itemize}

\item {In the previous example $G$ is the solution set to the homogeneous linear system $A\vv{x}=\vv{0}$}

\item {The solution set to $A\vv{x}=\vv{0}$ is a subspace called the \textbf{null space} of $A$.}

\end{itemize}

\begin{example}{Example: Is span\text{$\(\vv{u},\vv{v}\)$} A Subspace Of $\mathbb{R}^6$?}

Let $\vv{u}$ and $\vv{v}$ be vectors in $\mathbb{R}^6$. Is span$\(\vv{u},\vv{v}\)$ a subspace of $\mathbb{R}^6$?

$$\text{span}\(\vv{u},\vv{v}\)=\left\{c\vv{u}+d\vv{v}\Big|c,d\in\mathbb{R}\right\}$$

\begin{enumerate}

\item {Is $\vv{0}$ in span$\(\vv{u},\vv{v}\)$? \begin{center}$\vv{0}=0\vv{u}+0\vv{v}$ so $\vv{0}$ is in span$\(\vv{u},\vv{v}\)\checkmark$.\end{center}}

\item {Is span$\(\vv{u},\vv{v}\)$ closed under addition? \\ \newline

Let $\vv{x},\vv{y}\in\text{span}\(\vv{u},\vv{v}\)$. Then:

$$\vv{x}=c_1\vv{u}+d_1\vv{v}\quad\text{ and }\quad\vv{y}=c_2\vv{u}+d_2\vv{v}$$

Is $\vv{x}+\vv{y}$ in span$\(\vv{u},\vv{v}\)$?

\begin{align*}

\vv{x}+\vv{y}&=c_1\vv{u}+d_1\vv{v}+c_2\vv{u}+d_2\vv{v} \\

&=c_1\vv{u}+c_2\vv{u}+d_1\vv{v}+d_2\vv{v} \\

&=\(c_1+c_2\)\vv{u}+\(d_1+d_2\)\vv{v}\checkmark \\

\end{align*}

$\vv{x}+\vv{y}$ is in span$\(\vv{u},\vv{v}\)$ so span$\(\vv{u},\vv{v}\)$ is closed under addition.}

\item {Is span$\(\vv{u},\vv{v}\)$ closed under scalar multiplication? \\ \newline

Let $\vv{x}\in\text{span}\(\vv{u},\vv{v}\)$ and $k\in\mathbb{R}$. Then:

$$\vv{x}=c\vv{u}+d\vv{v}$$

Is $k\vv{x}$ in span$\(\vv{u},\vv{v}\)$?

$$k\vv{x}=k\(c\vv{u}+d\vv{v}\)=\(kc\)\vv{u}+\(kd\)\vv{v}\checkmark$$

Thus $k\vv{x}$ is in span$\(\vv{u},\vv{v}\)$ so span$\(\vv{u},\vv{v}\)$ is closed under scalar multiplication.}

\end{enumerate}

Therfore, span$\(\vv{u},\vv{v}\)$ is a subspace of $\mathbb{R}^6$.

\end{example}

\begin{theorem}{Theorem: A span Is A Subspace}

Let $\vv{v_1},\vv{v_2},\hdots,\vv{v_k'}$ be vectors in $\mathbb{R}^n$. Then the set span$\(\vv{v_1},\vv{v_2},\hdots,\vv{v_k}\)$ is a subspace of $\mathbb{R}^n$.

\end{theorem}

\begin{example}{Example: Spans Are A Subspace}

\begin{itemize}

\item span$\(\begin{bmatrix}1\\3\\-1\\2\\0\end{bmatrix},\begin{bmatrix}2\\-1\\1\\0\\1\end{bmatrix}\)$ is a subspace of $\mathbb{R}^5$.

\item span$\(\begin{bmatrix}1\\2\end{bmatrix},\begin{bmatrix}3\\1\end{bmatrix},\begin{bmatrix}4\\6\end{bmatrix}\)$ is a subspace of $\mathbb{R}^2$.

\end{itemize}

\end{example}

\begin{definition}{Definition: Spanning Set}

Let $W$ be a subspace of $\mathbb{R}^n$. We say the vectors $\vv{v_1},\vv{v_2},\hdots,\vv{v_k}$ in $W$ span/generate $W$ if $$\text{span}\(\vv{v_1},\vv{v_2},\hdots,\vv{v_k}\)=W$$. We call $\{\vv{v_1},\vv{v_2},\hdots,\vv{v_k}\}$ a \textbf{spanning set} for $W$.

\begin{example}{Example: Spanning Set}

\begin{tblr}{width = \linewidth, colspec = {XX}}

{$\begin{aligned}

S_4&=\left\{\begin{bmatrix}x\\y\end{bmatrix}\in\mathbb{R}^2\Big|x-y=0\right\} \\

S_4&=\text{span}\(\begin{bmatrix}1\\1\end{bmatrix}\)

\end{aligned}$} &

{\raisebox{-.5\height}{\begin{tikzpicture}

\begin{axis} [

width = \linewidth,

axis x line=middle,

axis y line=middle,

ymin = -2.5, ymax = 2.5,

xmin = -2.5, xmax = 2.5,

axis line style = {Stealth-Stealth, ultra thick},

xlabel = {$x$},

ylabel = {$y$},

clip = false,

]

\addplot[mark=*, ultra thick] coordinates {(0,0)};

\addplot [

black,

samples = 200,

style = {Stealth-Stealth, thick},

domain = -2.5:2.5,

]

{x};

\draw[ultra thick, -Stealth, red] (0,0) -- (1,1);

\end{axis}

\end{tikzpicture}}}

\end{tblr}

$\left\{\begin{bmatrix}1\\1\end{bmatrix}\right\}$ is a spanning set for $S_4$, as $\begin{bmatrix}1\\1\end{bmatrix}$ spans $S_4$. With that said $$\text{span}\(\begin{bmatrix}0\\0\end{bmatrix}\)=\left\{\begin{bmatrix}0\\0\end{bmatrix}\right\}\ne S_4$$ and $\left\{\begin{bmatrix}1\\1\end{bmatrix},\begin{bmatrix}2\\2\end{bmatrix}\right\}$ is also a spanning set for $S_4$.

\end{example}

$\star$ Note: Spanning sets are not unique. The number of vectors in a spanning set is not unique.

\end{definition}

\begin{example}{Example: Finding A Spanning Set Of A Matrix}

$A=\begin{bNiceMatrix}[first-row]

\downarrow&&\downarrow\\

1&0&2&0\\

2&0&1&3\\

3&0&-1&7

\end{bNiceMatrix}\xrightarrow{Row\hspace{1ex}Reduce}\begin{bNiceMatrix}[first-row]

\downarrow&&\downarrow\\

1&0&0&2\\

0&0&1&-1\\

0&0&0&0

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-3) circle (2mm) ;

\end{bNiceMatrix}=U$

\begin{align*}

W&=\text{span}\(\begin{bmatrix}1\\2\\3\end{bmatrix},\begin{bmatrix}0\\0\\0\end{bmatrix},\begin{bmatrix}2\\1\\-1\end{bmatrix},\begin{bmatrix}0\\3\\7\end{bmatrix}\) \\

&=\text{span}\(\begin{bmatrix}1\\2\\3\end{bmatrix},\begin{bmatrix}2\\1\\-1\end{bmatrix}\)

\end{align*}

$\underbrace{\left\{\begin{bmatrix}1\\2\\3\end{bmatrix},\begin{bmatrix}0\\0\\0\end{bmatrix},\begin{bmatrix}2\\1\\-1\end{bmatrix},\begin{bmatrix}0\\3\\7\end{bmatrix}\right\}}_{\text{Linearly Dependent}}$ is a spanning set for $W$. \\ \newline

$\underbrace{\left\{\begin{bmatrix}1\\2\\3\end{bmatrix},\begin{bmatrix}2\\1\\-1\end{bmatrix}\right\}}_{\text{Linearly Independent}}$ is also a spanning set for $W$. \\

$\star$ Note: span$\(\begin{bmatrix}1\\2\\3\end{bmatrix},\begin{bmatrix}2\\1\\-1\end{bmatrix}\)\ne\text{span}\(\begin{bmatrix}1\\0\\0\end{bmatrix},\begin{bmatrix}0\\1\\0\end{bmatrix}\)$

\end{example}

To get a spanning set for the subspace $W$ that contains the fewest number of vectors, we want to find the \textbf{linearly independent spanning set (basis)} for $W$.

\begin{definition}{Definition: Basis}

Let $W$ be a subspace of $\mathbb{R}^n$ and assume $W\ne\left\{\vv{0}\right\}$. The ordered set of vectors \\ $\mathcal{B}=\left\{\vv{v_1},\vv{v_2},\hdots,\vv{v_k}\right\}$ in $W$ is called a \textbf{basis} for $W$ if:

\begin{enumerate}

\item $W=\text{span}\(\vv{v_1},\vv{v_2},\hdots,\vv{v_k}\)$

\item and $\left\{\vv{v_1},\vv{v_2},\hdots,\vv{v_k}\right\}$ is linearly independent

\end{enumerate}

\textbf{Special Case:} The empty set denoted $\emptyset$, is a basis for $\left\{\vv{0}\right\}$.

\begin{example}{Example: Different Basis Of $\mathbb{R}^2$}

Find 3 distinct bases for $\mathbb{R}^2$.

$$\left\{\begin{bmatrix}1\\1\end{bmatrix},\begin{bmatrix}1\\-1\end{bmatrix}\right\}\quad\left\{\begin{bmatrix}0\\1\end{bmatrix},\begin{bmatrix}1\\0\end{bmatrix}\right\}\quad\left\{\begin{bmatrix}1\\0\end{bmatrix},\begin{bmatrix}0\\1\end{bmatrix}\right\}$$

$\star$ Note: $\left\{\begin{bmatrix}0\\1\end{bmatrix},\begin{bmatrix}1\\0\end{bmatrix}\right\}$ and $\left\{\begin{bmatrix}1\\0\end{bmatrix},\begin{bmatrix}0\\1\end{bmatrix}\right\}$ are two different bases.

\end{example}

\end{definition}

\begin{definition}{Definition: Standard Basis}

The \textbf{standard basis} for $\mathbb{R}^n$ is the ordered set $\mathcal{B}=\left\{\vv{e_1},\vv{e_2},\hdots,\vv{e_n}\right\}$ where $\vv{e_i}$ is the $i^{\text{th}}$ column vector of $I_n$.

\begin{example}{Example: Standard Basis}

\begin{itemize}

\item {$\left\{\begin{bmatrix}1\\0\\0\end{bmatrix},\begin{bmatrix}0\\1\\0\end{bmatrix},\begin{bmatrix}0\\0\\1\end{bmatrix}\right\}$ is the standard basis for $\mathbb{R}^3$.}

\item {$\left\{\begin{bmatrix}1\\0\\0\\0\end{bmatrix},\begin{bmatrix}0\\1\\0\\0\end{bmatrix},\begin{bmatrix}0\\0\\1\\0\end{bmatrix},\begin{bmatrix}0\\0\\0\\1\end{bmatrix}\right\}$ is the standard basis for $\mathbb{R}^4$.}

\end{itemize}

\end{example}

\end{definition}

$\star$ Note: When $W$ is a nonzero subspace of $\mathbb{R}^n$, there are infinitely many bases for $W$.

\begin{example}{Example: How Many Ways Are There To Write A Vector In A Subspace?}

Consider the subspace $W$ of $\mathbb{R}^3$ where $\underbrace{\mathcal{B}=\left\{\vv{v_1},\vv{v_2}\right\}}_{W=\text{span}\(\vv{v_1},\vv{v_2}\)}$ is a basis for $W$. If $\vv{b}$ is a vector in $W$, how many solutions exist to the linear system $x_1\vv{v_1}+x_2\vv{v_2}=\vv{b}$?

\begin{itemize}

\item {A solution exists because $\vv{v_1}$ and $\vv{v_2}$ span $W$ (every vector in $W$ is a linear combination of $\vv{v_1}$ and $\vv{v_2}$).}

\item {One unique solution exists because $\vv{v_1}$ and $\vv{v_2}$ are linearly independent (there are no free variables in the linear system).}

\end{itemize}

\end{example}

\begin{theorem}{Theorem: There Is One Way To Write A Vector In A Subspace}

Let $W$ be a subspace of $\mathbb{R}^n$ where $W\ne\left\{\vv{0}\right\}$. Let $\mathcal{B}=\left\{\vv{v_1},\vv{v_2},\hdots,\vv{v_k}\right\}$ be a basis for $W$. Let $\vv{v}$ be a vector in $W$ Then there is exactly one way to write $\vv{v}$ as a linear combination of $\vv{v_1},\vv{v_2},\hdots,\vv{v_k}$.

$$c_1\vv{v_1}+c_2\vv{v_2}+\hdots+c_k\vv{v_k}=\vv{v}$$

$c_1,c_2,\hdots,c_k$ are the \textbf{coordinates of $\vv{v}$ with respect to $\mathcal{B}$.} \\

$\begin{bmatrix}\vv{v}\end{bmatrix}\mathcal{B}=\begin{bmatrix}c_1\\c_2\\\vdots\\c_k\end{bmatrix}$ is the \textbf{coordinate vector of $\vv{v}$ with respect to $\mathcal{B}$}.

\end{theorem}

\begin{example}{Example: Finding The Coordinate Vector Graphically}

In $\mathbb{R}^2$: $\mathcal{B}=\left\{\vv{u},\vv{v}\right\}$ is a basis for $\mathbb{R}^2$.

\begin{center}

\begin{tikzpicture}[scale = 0.5]

\fill (0,0) circle[radius=7pt];

\draw[ultra thick, Stealth-Stealth, dotted] (-3,-6) -- (3,6);

\draw[ultra thick, -Stealth] (0,0) -- (1,2) node[right] {\vv{u}};

\draw[ultra thick, Stealth-Stealth, dotted] (-3,3) -- (3,-3);

\draw[ultra thick, -Stealth] (0,0) -- (2,-2) node[right] {\vv{v}};

\draw[dotted, ultra thick] (-1,-5) -- (-2,-4);

\draw[dotted, ultra thick] (-1,-5) -- (1,-1);

\draw[ultra thick, -Stealth] (0,0) -- (-1,-5) node[below right] {$\vv{w}=-2\vv{u}+\frac{1}{2}\vv{v}$};

\end{tikzpicture}

\end{center}

$\begin{bmatrix}\vv{w}\end{bmatrix}\mathcal{B}=\begin{bmatrix}-2\\1/2\end{bmatrix}$ and for $\mathcal{B}_2=\left\{\vv{v},\vv{u}\right\}$ then $\begin{bmatrix}\vv{w}\end{bmatrix}\mathcal{B}_2=\begin{bmatrix}1/2\\-2\end{bmatrix}$.

\end{example}

\begin{example}{Example: Finding The Coordinate Vector}

The set $\mathcal{B}=\left\{\begin{bmatrix}1\\0\\1\end{bmatrix},\begin{bmatrix}0\\-1\\1\end{bmatrix},\begin{bmatrix}1\\1\\1\end{bmatrix}\right\}$ is a basis for $\mathbb{R}^3$.

\begin{enumerate}

\item {Let $\vv{v}=\begin{bmatrix}10\\11\\6\end{bmatrix}$. Find $\begin{bmatrix}\vv{v}\end{bmatrix}\mathcal{B}$. \\ \newline

Firstly Solve:

$$x_1\begin{bmatrix}1\\0\\1\end{bmatrix}+x_2\begin{bmatrix}0\\-1\\1\end{bmatrix}+x_3\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}10\\11\\6\end{bmatrix}$$

Next row reduce to RREF:

$$\begin{bmatrix}1&0&1&\aug&10\\0&-1&1&\aug&11\\1&1&1&\aug&6\end{bmatrix}\xrightarrow{\text{Row Reduce}}\begin{bmatrix}1&0&0&\aug&3\\0&1&0&\aug&-4\\0&0&1&\aug&4\end{bmatrix}$$

Therefore $\[\vv{v}\]\mathcal{B}=\begin{bmatrix}3\\-4\\7\end{bmatrix}$.}

\item {Let $\[\vv{u}\]\mathcal{B}=\begin{bmatrix}2\\-1\\1\end{bmatrix}$. Find $\vv{u}$.

$$\vv{u}=2\begin{bmatrix}1\\0\\1\end{bmatrix}-1\begin{bmatrix}0\\-1\\1\end{bmatrix}+1\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}3\\2\\2\end{bmatrix}$$}

\end{enumerate}

\end{example}

$\star$ Note: Every subspace of $\mathbb{R}^n$ has a basis and bases for nonzero subspaces are not unique. However, every basis for a subspace $W$ contains the same number of vectors.

\begin{example}{Example: Every Bases Having The Same Number Of Vectors}

If $W$ is a subspace of $\mathbb{R}^5$ and $\mathcal{B}=\left\{\vv{u},\vv{v},\vv{w}\right\}$ is a basis for $W$, then every basis for $W$ contains exactly 3 vectors. In other words dim$W=3$.

\end{example}

\begin{definition}{Definition: Dimension}

Let $W$ be a subspace of $\mathbb{R}^n$. The number of vectors in a basis for $W$ is called the \textbf{dimension} of $W$, denoted dim$W$.

\begin{example}{Example: Dimension}

\begin{itemize}

\item dim$\left\{\vv{0}\right\}=0$

\item dim$\(\mathbb{R}^2\)=2$

\item dim$\(\mathbb{R}^n\)=n$

\item {\begin{tblr}{width = \linewidth, colspec = {XX}}

{$\begin{aligned}

W&=\text{span}\(\begin{bmatrix}1\\1\end{bmatrix}\) \\

\mathcal{B}_W&=\left\{\begin{bmatrix}1\\1\end{bmatrix}\right\}

\end{aligned}$ \\

$\text{dim}W=1$ \\

$W$ is a one-diminsional subspace of $\mathbb{R}^2$.} & {\raisebox{-.5\height}{\begin{tikzpicture}

\begin{axis} [

width = \linewidth,

axis x line=middle,

axis y line=middle,

ymin = -2.5, ymax = 2.5,

xmin = -2.5, xmax = 2.5,

axis line style = {Stealth-Stealth, ultra thick},

xlabel = {$x$},

ylabel = {$y$},

xtick = {-2,-1,1,2},

ytick = {-2,-1,1,2},

clip = false,

]

\addplot[mark=*, ultra thick] coordinates {(0,0)};

\addplot [

black,

samples = 200,

style = {Stealth-Stealth, thick},

domain = -2.5:2.5,

]

{x} node[right] {$W$};

\draw[ultra thick, -Stealth, red] (0,0) -- (1,1);

\end{axis}

\end{tikzpicture}}}

\end{tblr}}

\end{itemize}

\end{example}

\end{definition}

\begin{example}{Example: Finding The Basis Of A Matrix}

$A=\begin{bNiceMatrix}[first-row]

&&\xmark&\xmark&\\

3&-1&3&2&2 \\

2&1&2&3&3 \\

1&2&1&3&3 \\

4&1&4&5&6

\end{bNiceMatrix}\xrightarrow{Row\hspace{1ex}Reduce}\begin{bNiceMatrix}[first-row]

&&\downarrow&\downarrow&\\

1&0&1&1&0 \\

0&1&0&1&0 \\

0&0&0&0&1 \\

0&0&0&0&0

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (2mm) ;

\tikz \draw (3-5) circle (2mm) ;

\end{bNiceMatrix}=U$

\begin{align*}

W&=\text{span}\(\begin{bmatrix}3\\2\\1\\4\end{bmatrix},\begin{bmatrix}-1\\1\\2\\1\end{bmatrix},\begin{bmatrix}3\\2\\1\\4\end{bmatrix},\begin{bmatrix}2\\3\\3\\5\end{bmatrix},\begin{bmatrix}2\\3\\3\\6\end{bmatrix}\) \\

&=\text{span}\(\begin{bmatrix}3\\2\\1\\4\end{bmatrix},\begin{bmatrix}-1\\1\\2\\1\end{bmatrix},\begin{bmatrix}2\\3\\3\\6\end{bmatrix}\) \\

\mathcal{B}_W&=\left\{\begin{bmatrix}3\\2\\1\\4\end{bmatrix},\begin{bmatrix}-1\\1\\2\\1\end{bmatrix},\begin{bmatrix}2\\3\\3\\6\end{bmatrix}\right\}

\end{align*}

dim$W$=3, thus $W$ is a 3-dimensional subspace of $\mathbb{R}^4$.

\end{example}

\begin{theorem}{Theorem: If Vectors Span $W$}

Let $W$ be a subspace of $\mathbb{R}^n$ and suppose $\vv{v_1},\vv{v_2},\hdots,\vv{v_k}$ are in $W$ as well as span $W$. Then:

\begin{enumerate}

\item dim$W\le k$

\item {and If dim$W=k$, then $\vv{v_1},\vv{v_2},\hdots,\vv{v_k}$ are linearly independent.}

\end{enumerate}

\end{theorem}

\begin{theorem}{Theorem: If Vectors In $W$ Are Linearly Independnet}

Let $W$ be a subspace of $\mathbb{R}^n$ and suppose $\vv{v_1},\vv{v_2},\hdots,\vv{v_k}$ are linearly independent vectors in $W$. Then:

\begin{enumerate}

\item dim$W\ge k$

\item {and If dim$W= k$, then $\vv{v_1},\vv{v_2},\hdots,\vv{v_k}$ span$W$.}

\end{enumerate}

\end{theorem}

$\star$ Note: If $W$ is a subspace of $\mathbb{R}^n$, then $0\le\text{dim}W\le n$.

\begin{itemize}

\item If dim$W=0$, then $W=\left\{\vv{0}\right\}$.

\item If dim$W=n$, then $W=\mathbb{R}^n$.

\end{itemize}

\newpage

\subsection{Subspaces in $\mathbb{R}^3$ (Geometrically)}

\begin{example}{Example: Subspaces In $\mathbb{R}^3$ Geometrically}

\begin{tblr}{width = \linewidth, colspec={QQXQ}, cells = {halign = c, valign = m}, hlines, vlines}

{Set} & {Dimension} & {Plot} & {Description} \\

{$S_0=\left\{\vv{0}\right\}$} & {dim$\(S_0\)=0$} & {\raisebox{-.5\height}{\begin{tikzpicture}

\begin{axis}[

width = \linewidth,

ymin = -2.5, ymax = 2.5,

xmin = -2.5, xmax = 2.5,

zmin = -2.5, zmax = 2.5,

axis line style = {ultra thick},

xlabel = {$x$},

ylabel = {$y$},

zlabel = {$z$},

clip = false,

ticks = none,

3d box=background,

grid=major,

]

\addplot3[mark=*, ultra thick] coordinates {(0,0,0)};

\end{axis}

\end{tikzpicture}}} & {Point/Origin}\\

{$S_1=\text{span}\(\vv{u}\)$ \\

$\(\vv{u}\ne0\)$} & {dim$\(S_1\)=1$} & {\raisebox{-.5\height}{\begin{tikzpicture}

\begin{axis}[

width = \linewidth,

ymin = -2.5, ymax = 2.5,

xmin = -2.5, xmax = 2.5,

zmin = -2.5, zmax = 2.5,

axis line style = {ultra thick},

xlabel = {$x$},

ylabel = {$y$},

zlabel = {$z$},

clip = false,

ticks = none,

3d box=background,

grid=major,

]

\addplot3 [red,domain=-2.5:2.5,samples y=1, Stealth-Stealth, ultra thick] (0,x,x);

\addplot3[mark=*, ultra thick] coordinates {(0,0,0)};

\end{axis}

\end{tikzpicture}}} & {Line through Origin} \\

{$S_2=\text{span}\(\vv{u},\vv{v}\)$ \\

$\(\vv{u},\vv{v}\text{ are lin. indep}\)$} & {dim$\(S_2\)=2$} & {\raisebox{-.5\height}{\begin{tikzpicture}

\begin{axis}[

width = \linewidth,

ymin = -2.5, ymax = 2.5,

xmin = -2.5, xmax = 2.5,

zmin = -2.5, zmax = 2.5,

axis line style = {ultra thick},

xlabel = {$x$},

ylabel = {$y$},

zlabel = {$z$},

clip = false,

ticks = none,

3d box=background,

grid=major,

]

\fill[red, fill=red, draw = red, fill opacity=0.2, ultra thick] (2.5,-2.5,-2.5) -- (2.5,2.5,-2.5) -- (-2.5,2.5,2.5) -- (-2.5,-2.5,2.5) -- cycle;

\addplot3[mark=*, ultra thick] coordinates {(0,0,0)};

\draw[black, ultra thick, -Stealth] (0,0,0) -- (0,2.5,0);

\draw[black, ultra thick, -Stealth] (0,0,0) -- (-1.25, 0,1.25);

\end{axis}

\end{tikzpicture}}} & {Plane through Origin} \\

{$S_3=\text{span}\(\vv{u},\vv{v},\vv{w}\)$ \\

$\(\vv{u},\vv{v},\vv{w}\text{ are lin. indep}\)$} & {dim$\(S_3\)=3$} & {\raisebox{-.5\height}{\begin{tikzpicture}

\begin{axis}[

width = \linewidth,

ymin = -2.5, ymax = 2.5,

xmin = -2.5, xmax = 2.5,

zmin = -2.5, zmax = 2.5,

axis line style = {ultra thick},

xlabel = {$x$},

ylabel = {$y$},

zlabel = {$z$},

clip = false,

ticks = none,

3d box=background,

grid=major,

]

\addplot3[mark=*, ultra thick] coordinates {(0,0,0)};

\draw[draw =red,fill=red, fill opacity=0.2, ultra thick] (-2.5,-2.5,-2.5) -- (2.5,-2.5,-2.5) -- (2.5,-2.5,2.5) -- (-2.5,-2.5,2.5) -- cycle;

\draw[draw =red,fill=red, fill opacity=0.2, ultra thick] (2.5,-2.5,-2.5) -- (2.5,2.5,-2.5) -- (2.5,2.5,2.5) -- (2.5,-2.5,2.5) -- cycle;

\draw[draw =red,fill=red, fill opacity=0.2, ultra thick] (-2.5,-2.5,2.5) -- (-2.5,2.5,2.5) -- (2.5,2.5,2.5) -- (2.5,-2.5,2.5) -- cycle;

\end{axis}

\end{tikzpicture}}} & {$S_3=\mathbb{R}^3$}

\end{tblr}

\end{example}

\newpage

\subsection{Column Space and Null Space of a Matrix}

\begin{definition}{Definition: Column Space}

Suppose $A$ is an $m\times n$ matrix with columns $\vv{a_1},\vv{a_2},\hdots,\vv{a_n}$. Then the \textbf{column space} of $A$ is the set:

\begin{align*}

\text{col}A&=\text{span}\(\vv{a_1},\vv{a_2},\hdots,\vv{a_n}\) \\

&=\left\{c_1\vv{a_1}+c_2\vv{a_2}+\hdots+c_n\vv{a_n}\Big|c_1,c_2,\hdots,c_n\in\mathbb{R}\right\} \\

&=\left\{\vv{b}\in\mathbb{R}^m\Big|A\vv{x}=\vv{b}\text{ is consistant}\right\}

\end{align*}

$\star$ Note: col$A$ is a subspace of $\mathbb{R}^m$.

\end{definition}

\begin{example}{Example: Is $\vv{b}$ In col$A$}

$A=\begin{bmatrix}1&2&-1\\2&3&-1\\1&4&-3\end{bmatrix}$, $\vv{b}=\begin{bmatrix}3\\2\\11\end{bmatrix}$

\begin{enumerate}

\item {Is $\vv{b}$ in col$A$? \\ \newline

Firstly solve $A\vv{x}=\vv{b}$:

$$\begin{bmatrix}A&\aug&\vv{b}\end{bmatrix}=\begin{bmatrix}1&2&-1&\aug&3\\2&3&-1&\aug&2\\1&4&-3&\aug&11\end{bmatrix}\xrightarrow{Row\hspace{1ex}Reduce}\begin{bNiceMatrix}

1&0&1&\aug&-5\\

0&1&-1&\aug&4 \\

0&0&0&\aug&0

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (2mm) ;

\end{bNiceMatrix}$$

Infinitely Many solutions to $A\vv{x}=\vv{b}$, thus a solution exists so $\vv{b}$ is in col$A$.}

\item {Find a basis for col$A$.

\begin{align*}

\text{col}A&=\text{span}\(\begin{bmatrix}1\\2\\1\end{bmatrix},\begin{bmatrix}2\\3\\4\end{bmatrix},\begin{bmatrix}-1\\-1\\3\end{bmatrix}\)=\text{span}\(\begin{bmatrix}1\\2\\1\end{bmatrix},\begin{bmatrix}2\\3\\4\end{bmatrix}\) \\

A&=\begin{bNiceMatrix}[first-row]

\downarrow&\downarrow&\xmark\\

1&2&-1\\

2&3&-1\\

1&4&-3

\end{bNiceMatrix}\xrightarrow{Row\hspace{1ex}Reduce}\begin{bNiceMatrix}[first-row]

\downarrow&\downarrow&\xmark\\

1&0&1\\

0&1&-4 \\

0&0&0

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (2mm) ;

\end{bNiceMatrix} \\

\mathcal{B}_{colA}&=\left\{\begin{bmatrix}1\\2\\1\end{bmatrix},\begin{bmatrix}2\\3\\4\end{bmatrix}\right\}

\end{align*}

As dim$\(\text{col}A\)=2$ then col$A$ is a 2 dimensional subspace of $\mathbb{R}^3$.}

\end{enumerate}

\end{example}

\begin{theorem}{\text{Theorem: Relating Rank, Dimension, And Column Space}}

Let $A$ be an $m\times n$ matrix. Then:

$$\text{rank}A=\text{dim}\(\text{col}A\)$$

$\star$ Note: If the columns of $A$ span $\mathbb{R}^m$, then $\text{col}A=\mathbb{R}^m$.

\end{theorem}

Elementary Row operations can affect the column space of a matrix.

\begin{example}{Example: Row Operations Affect Column Space}

$A=\begin{bmatrix}1&1\\1&1\end{bmatrix}\xrightarrow{R_2-R_1}\begin{bmatrix}1&1\\0&0\end{bmatrix}=U$ \\

\begin{tblr}{width = \linewidth, colspec={XX}, cells = {halign = c, valign=m}}

{$\begin{aligned}

\text{col}A&=\text{span}\(\begin{bmatrix}1\\1\end{bmatrix}\) \\

\text{col}U&=\text{span}\(\begin{bmatrix}1\\0\end{bmatrix}\) \\

\text{col}A&\ne\text{col}U

\end{aligned}$} &

{\raisebox{-.5\height}{\begin{tikzpicture}

\begin{axis} [

width = \linewidth,

axis x line=middle,

axis y line=middle,

ymin = -2.5, ymax = 2.5,

xmin = -2.5, xmax = 2.5,

axis line style = {Stealth-Stealth, ultra thick},

xlabel = {$x$},

ylabel = {$y$},

clip = false,

]

\addplot[mark=*, ultra thick] coordinates {(0,0)};

\addplot [

right,

ultra thick,

samples = 200,

style = {Stealth-Stealth, thick},

domain = -2.5:2.5,

]

{x} node[right] {col$A$} ;

\draw[ultra thick, Stealth-Stealth, red] (-1.5,0) -- (1.5,0) node[pos = 0, above]{col$U$};

\end{axis}

\end{tikzpicture}}}

\end{tblr}

\end{example}

\begin{definition}{Definition: Null Space}

We definie the \textbf{null space} of an $m\times n$ matrix $A$ to be the set:

$$\text{null}A=\left\{\vv{x}\in\mathbb{R}^n\Big|A\vv{x}=\vv{0}\right\}$$

$\star$ Note: null$A$ is the subspace of $\mathbb{R}^n$.

\end{definition}

\begin{example}{Example: Finding null$A$}

$A=\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}\xrightarrow{Row\hspace{1ex}Reduce}\begin{bmatrix}1&0&-1\\0&1&2\end{bmatrix}$

Finding a basis for null$A$. \\ \newline

Solve $A\vv{x}=\vv{0}$:

\begin{align*}

x_1-x_3&=0 & x_1&=t \\

x_2+2x_3&=0 & x_2&=-2t \\

&&x_3&=t

\end{align*}

Get Parmetric Vector Form of solution set:

$$\vv{x}=\begin{bmatrix}t\\-2t\\t\end{bmatrix}=t\begin{bmatrix}1\\-2\\1\end{bmatrix}$$

The vectors multiplied by parametric form a basis for null$A$:

$$\mathcal{B}_{\text{null}A}=\left\{\begin{bmatrix}1\\-2\\1\end{bmatrix}\right\}$$

$\text{dim}\(\text{null}A\)=1$, thus null$A$ is a one dimensional subspace of $\mathbb{R}^3$.

\end{example}

\begin{definition}{Definition: Nullity}

The nullity of a matrix $A$ is the dimension of null$A$:

$$\text{nullity}\(A\)=\text{dim}\(\text{null}A\)$$

\end{definition}

\begin{example}{Example: Finding Basis For col$A$ And null$A$ And nullity$A$}

Suppose $A=\begin{bNiceMatrix}[first-row]

\downarrow&\downarrow&&&\downarrow \\

1&0&1&1&0 \\

0&-1&0&-1&0 \\

-1&2&-1&1&1 \\

0&1&0&1&1

\end{bNiceMatrix}$ has RREF $\begin{bNiceMatrix}

1&0&1&1&0 \\

0&1&0&1&0 \\

0&0&0&0&1 \\

0&0&0&0&0

\CodeAfter

\tikz \draw (1-1) circle (2mm) ;

\tikz \draw (2-2) circle (2mm) ;

\tikz \draw (3-5) circle (2mm) ;

\end{bNiceMatrix}$

Find a basis for col$A$ and a basis for null$A$. Then find rank$A$ and nullity$A$.

$$\mathcal{B}_{\text{col}A}=\left\{\begin{bmatrix}1\\0\\-1\\0\end{bmatrix},\begin{bmatrix}0\\-1\\2\\1\end{bmatrix},\begin{bmatrix}0\\0\\1\\1\end{bmatrix}\right\}$$

Has leading entry's in columns 1, 2 and 5. Which also means rank$A=3$. \\

\newline

Find null$A$ (solve $A\vv{x}=\vv{0}$):

\begin{align*}

x_1+x_3+x_4&=0 & x_1&=-s-t \\

x_2+x_4&=0 & x_2&=-t \\

x_5&=0 & x_3&=s \\

&& x_4&=t \\

&& x_5&=0

\end{align*}

$$\vv{x}=\begin{bmatrix}-s-t\\-t\\s\\t\\0\end{bmatrix}=s\begin{bmatrix}-1\\0\\1\\0\\0\end{bmatrix}+t\begin{bmatrix}-1\\-1\\0\\1\\0\end{bmatrix}$$

$$\mathcal{B}_{\text{null}A}\left\{\begin{bmatrix}-1\\0\\1\\0\\0\end{bmatrix},\begin{bmatrix}-1\\-1\\0\\1\\0\end{bmatrix}\right\}$$

nullity$A=2$

\end{example}

$\star$ Note: If null$A=\left\{\vv{0}\right\}$, then $\mathcal{B}_{\text{null}A}=\emptyset$.

\begin{theorem}{Theorem: Rank-Nullity Theorem}

If $A$ is an $m\times n$ matrix, then:

$$\text{rank}A+\text{nullity}A=n$$

\end{theorem}

\begin{theorem}{Theorem: FTIM Countined}

Let $A$ be an $n\times n$ matrix. The following are equivalent: \\ \newline

Original FTIM:

\begin{enumerate}

\item {$A$ is invertible, ($A^{-1}$ exists)}

\item $A$ is row equivalent to $I_n$. ($I_n$ is the RREF of $A$)

\item rank$A=n$

\item $A\vv{x}=\vv{b}$ has a solution for every $\vv{b}\in\mathbb{R}^n$. (more specifically, there is one unique solution $\vv{x}=A^{-1}\vv{b}$.)

\item $A\vv{x}=\vv{0}$ has only the trivial solution $\vv{x}=\vv{0}$.

\item The columns of $A$ span $\mathbb{R}^n$

\item {The columns of $A$ are linearly independent.}

\end{enumerate}

Added:

\begin{enumerate}[resume]

\item {nullity$A=0$}

\item col$A=\mathbb{R}^n$

\item null$A=\left\{\vv{0}\right\}$

\item {The columns of $A$ form a basis for $\mathbb{R}^n$.}

\end{enumerate}

\end{theorem}

\newpage

\section{Section 3.6 Transformations}

\begin{definition}{Definition: Transformation/Function/Mapping}

A \textbf{transformation/function/mapping} $T$ from $\mathbb{R}^n$ to $\mathbb{R}^m$ is a rule that assigns to each vector $\vv{x}\in\mathbb{R}^n$ a vector $T\(\vv{x}\)\in\mathbb{R}^m$.

$$T:\mathbb{R}^n\rightarrow\mathbb{R}^m\qquad\vv{x}\mapsto\underbracket{T\(\vv{x}\)}_{\text{`` $T$ of $\vv{x}$ ''}}$$

$\mathbb{R}^n$ is the \textbf{domain} of $T$. \\ \newline

$\mathbb{R}^m$ is the \textbf{codomain} of $T$. \\ \newline

For $\vv{x}\in\mathbb{R}^n$, $T(x)\in\mathbb{R}^m$ is called the \textbf{image} of $\vv{x}$.\\ \newline

The set of all images is called the \textbf{range} of $T$.

$$\text{range}\(T\)=\left\{T\(\vv{x}\)\Big|\vv{x}\in\text{domain}\(T\)\right\}$$

\end{definition}

\begin{example}{Example: The Domain And Codomain Of $T$}

\begin{align*}

&T_1\(\vv{x}\)=3\vv{x}& &T_1:\text{ }\mathbb{R}^n\rightarrow \mathbb{R}^n \\

&T_2\(\begin{bmatrix}x\\y\\z\end{bmatrix}\)=\begin{bmatrix}x\\y\end{bmatrix}& &T_2:\text{ }\mathbb{R}^3\rightarrow \mathbb{R}^2

\end{align*}

\end{example}

\begin{example}{Example: Transformation}

\begin{align*}

S\(x_1,\text{ }x_2\)&=\(x_2-3x_1,\text{ }4+x_2,\text{ }9,\text{ }3-x_1\) & &S:\text{ }\mathbb{R}^2\rightarrow \mathbb{R}^4 \\

S\(2,\text{ }1\)&=\(1-3\(2\),\text{ }4+1,\text{ }9,\text{ }3-2\) \\

&=\(-5,\text{ }5,\text{ }9,\text{ }1\)

\end{align*}

\end{example}

\begin{definition}{Definition: Matrix Transformation}

Let $A$ be an $m\times n$ matrix. $A$ transformation $T:\mathbb{R}^n\rightarrow\mathbb{R}^m$ defined by $T\(\vv{x}\)=A\vv{x}$ is called a \textbf{matrix transformation} and $A$ is called the \textbf{standared matrix} of $T$.

\begin{example}{Example: Matrix Transformation}

$T\(\vv{x}\)=\underbrace{\begin{bmatrix}1&3\\2&1\\1&4\end{bmatrix}}_{3\times2}\underbrace{\vv{x}}_{2\times1}$ \\

Domain: $\mathbb{R}^2$ \\ \newline

Codomain: $\mathbb{R}^3$

\end{example}

\end{definition}

\begin{example}{Example: Doing Stuff With Transformations}

Let $A=\begin{bmatrix}1&2&3\\0&1&0\end{bmatrix}$ and $\vv{u}=\begin{bmatrix}0\\1\\-1\end{bmatrix}$. Define $T\(\vv{x}\)=A\vv{x}$.

\begin{enumerate}

\item{Find $T\(\vv{u}\)$

$$T\(\vv{u}\)=A\vv{u}=\begin{bmatrix}1&2&3\\0&1&0\end{bmatrix}\begin{bmatrix}0\\1\\-1\end{bmatrix}=\begin{bmatrix}-1\\1\end{bmatrix}$$}

\item{If $T\(\vv{x}\)=\vv{0}$, Find $\vv{x}$? \\ \newline

Firstly what are we meant to do:

$$T\(\vv{x}\)=A\vv{x}=\begin{bmatrix}1&2&3\\0&1&0\end{bmatrix}\vv{x}=\vv{0}$$

Next Row Reduce:

$$\begin{bmatrix}1&2&3&\aug&0\\0&1&0&\aug&0\end{bmatrix}\xrightarrow{R_1-2R_2}\begin{bmatrix}1&0&3&\aug&0\\0&1&0&\aug&0\end{bmatrix}$$

Solve System of Equations Using Back Substitution:

\begin{align*}

x_1+3x_3&=0 & x_1&=-3t \\

x_2&=0 & x_2&=0 \\

&&x_3&=t

\end{align*}

$\vv{x}=t\begin{bmatrix}-3\\0\\1\end{bmatrix}$

Check: $A\vv{x}=\begin{bmatrix}1&2&3\\0&1&0\end{bmatrix}\begin{bmatrix}-3t\\0\\t\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\checkmark$}

\item{What is the range of $T$?

\begin{align*}

\text{range}\(T\)&=\left\{T\(\vv{x}\)\Big|\vv{x}\in\text{domain}\(T\)\right\} \\

&=\left\{A\vv{x}\Big|\vv{x}\in\text{domain}\(T\)\right\} \\

&=\left\{\vv{b}\in\mathbb{R}^2\Big|A\vv{x}=\vv{b}\text{ is consistant}\right\} \\

&=\text{col}A \\

&=\text{span}\(\begin{bmatrix}1\\0\end{bmatrix},\begin{bmatrix}2\\1\end{bmatrix},\begin{bmatrix}3\\0\end{bmatrix}\) \\

&=\text{span}\(\begin{bmatrix}1\\0\end{bmatrix},\begin{bmatrix}2\\1\end{bmatrix}\) \\

\Aboxed{\text{range}\(T\)&=\mathbb{R}^2}

\end{align*}

\begin{center}

\begin{tikzpicture}

\fill (0,0) circle[radius=4pt];

\draw[dotted, ultra thick, Stealth-Stealth] (-3,-1.5) -- (3,1.5);

\draw[ultra thick, -Stealth] (0,0) -- (2,1);

\draw[dotted, ultra thick, Stealth-Stealth] (-3,0) -- (3,0);

\draw[ultra thick, -Stealth] (0,0) -- (1,0);

\end{tikzpicture}

\end{center}}

\end{enumerate}

\end{example}

\begin{definition}{Definition: Kernal}

The set of all vectors $\vv{x}$ that $T$ maps to $\vv{0}$ is called the \textbf{kernal} of $T$.

$$\text{ker}T=\left\{\vv{x}\in\text{domain}\(T\)\Big|T\(\vv{x}\)=\vv{0}\right\}$$

\end{definition}

Let $A$ be an $m\times n$ matrix and define $T:\mathbb{R}^n\rightarrow\mathbb{R}^m$ by $T\(\vv{x}\)=A\vv{x}$. Then:

\begin{enumerate}

\item {range$\(T\)=\text{col}A$}

\item {and ker$\(T\)=\text{null}A$}

\end{enumerate}

\subsection{Linear Transformations}

\begin{definition}{Definition: Linear Transformation}

A transformation $T$ is \textbf{linear} if for all $\vv{u}$ and $\vv{v}$ in the domain of $T$ and all $c\in\mathbb{R}$, we have:

\begin{enumerate}

\item $T\(\vv{u}+\vv{v}\)=T\(\vv{u}\)+T\(\vv{v}\)$

\item $T\(c\vv{u}\)=cT\(\vv{u}\)$

\end{enumerate}

1 and 2 can be combined as followes: $T$ is linear provided

$$T\(c\vv{u}+d\vv{v}\)=cT\(\vv{u}\)+dT\(\vv{v}\)$$

for all $\vv{u},\vv{v}\in\text{domain}\(T\)$ and $c,d\in\mathbb{R}$.

\end{definition}

\begin{theorem}{Theorem: All Matrix Transformations Are Linear Transformations}

All Matrix Transformations Are Linear Transformations

\begin{proof}

Let $T$ be a matrix transformation with standared matrix $A$. Then $T\(\vv{x}\)=A\vv{x}$. Let $\vv{u}$ and $\vv{v}$ be vectors in the domain of $T$ and let $c,d\in\mathbb{R}$. Then:

\begin{align*}

T\(c\vv{u}+d\vv{v}\)&=A\(c\vv{u}+d\vv{v}\) \\

&=A\(c\vv{u}\)+A\(d\vv{v}\) \\

&=cA\vv{u}+dA\vv{v} \\

&=cT\(\vv{u}\)+dT\(\vv{v}\)

\end{align*}

Therfore $T$ is linear.

\end{proof}

\end{theorem}

$\star$ Note: If $T$ is linear, then $T\(\vv{0}\)=\vv{0}$. If $T\(\vv{0}\)\ne\vv{0}$, then $T$ is \textbf{not} linear.

\begin{example}{Example: $T$ Is Not Linear As \text{$T\(\vv{0}\)\ne\vv{0}$}}

$T\(x,y,z\)=\(4+x,\hspace{1ex}y+x\)$

$$T\(0,0,0\)=\(4+0,0+0\)=\(4,0\)\ne\(0,0\)$$

Thus as $T\(\vv{0}\)\ne\vv{0}$ then $T$ is not linear.

\end{example}

\begin{example}{Example: $T$ May Or May Not Be Linear}

$T\(x,y\)=\(x^2,y^2\)$

$$T\(0,0\)=\(0^2,0^2\)=\(0,0\)$$

If $T\(\vv{0}\)=\vv{0}$, $T$ may or may not be linear. We need to work with the definition.\\ \newline

Is $T\(c\(x,y\)\)=cT\(x,y\)$ for all $c\in\mathbb{R}$, $\(x,y\)\in\mathbb{R}^2$?

$$T\(c\(x,y\)\)=T\(cx,cy\)=\(c^2x^2,c^2y^2\)$$

$$cT\(x,y\)=c\(x^2,y^2\)=\(cx^2,cy^2\)$$

As $\(c^2x^2,c^2y^2\)\ne\(cx^2,cy^2\)$ then there is a \textbf{Counter Example:}

$$T\(-1\(2,3\)\)=T\(-2,-3\)=\(4,9\)$$

$$-1T\(2,3\)=-1\(4,9\)=\(-4,-9\)$$

Which as $\(4,9\)\ne\(-4,-9\)$ then $T$ is not linear.

\end{example}

\begin{example}{Example: Is $T$ Linear?}

Suppose $T:\mathbb{R}^2\rightarrow\mathbb{R}^3$ is defined by:

$$T\(x,y\)=\(3x,0,x+y\)$$

Is $T$ linear? \\

$\star$ Note: Does $T\(c\vv{u},d\vv{v}\)=cT\(\vv{u}\)+dT\(\vv{v}\)$? \\ \newline

Let $\vv{u},\vv{v}\in\mathbb{R}^2$ and $c,d\in\mathbb{R}$, then:

$$\vv{u}=\(u_1,u_2\)\qquad\vv{v}=\(v_1,v_2\)$$

that also means that:

$$c\vv{u}+d\vv{v}=\(cu_1+dv_1,cu_2+dv_2\)$$

Now Calculating $T\(c\vv{u}+d\vv{v}\)$:

\begin{align*}

T\(c\vv{u}+d\vv{v}\)&=T\(c\(u_1,u_2\)+d\(v_1,v_2\)\) \\

&=T\(\(cu_1,cu_2\)+\(dv_1,dv_2\)\) \\

&=T\(cu_1+dv_1,cu_2+dv_2\) \\

&=\(3\(cu_1+dv_1\),0,cu_1+dv_1+cu_2+dv_2\) \\

&=\(3cu_1+3dv_1,0,cu_1+dv_1+cu_2+dv_2\)

\end{align*}

Now Calculating $cT\(\vv{u}\)+dT\(\vv{v}\)$:

\begin{align*}

cT\(\vv{u}\)+dT\(\vv{v}\)&=cT\(u_1,u_2\)+dT\(v_1,v_2\) \\

&=c\(3u_1,0,u_1+u_2\)+d\(3v_1,0,v_1+v_2\) \\

&=\(3cu_1,0,cu_1+cu_2\)+\(3dv_1,0,dv_1+dv_2\) \\

&=\(3cu_1+3dv_1,0,cu_1+cu_2+du_1+dv_2\)

\end{align*}

As $T\(c\vv{u}+d\vv{v}\)=cT\(\vv{u}\)+dT\(\vv{v}\)$ then $T$ is linear.

\end{example}

All linear transformations are matrix transformations: if $T$ is linear, there exists a matrix $A$ such that $T\(\vv{x}\)=A\vv{x}$. ($A$ is the \textbf{standard matrix} of $T$) \\ \newline

How can we find the standard matrix of a linear transformation? \\ \newline

Suppose $T:\mathbb{R}^2\rightarrow\mathbb{R}^4$ is a linear transformation, then for $\vv{x}=\begin{bmatrix}x_1\\x_2\end{bmatrix}\in\mathbb{R}^2$, we have:

\begin{align*}

T\(\vv{x}\)&=T\(\begin{bmatrix}x_1\\x_2\end{bmatrix}\) \\

&=\underbrace{T\(x_1\begin{bmatrix}1\\0\end{bmatrix}+x_2\begin{bmatrix}0\\1\end{bmatrix}\)}_{\text{$T$ is linear}} \\

&=x_1T\(\begin{bmatrix}1\\0\end{bmatrix}\)+x_2T\(\begin{bmatrix}0\\1\end{bmatrix}\) \\

&=x_1T\(\vv{e_1}\)+x_2T\(\vv{e_2}\) \\

&=\fbox{$\begin{bmatrix}T\(\vv{e_1}\)&T\(\vv{e_2}\)\end{bmatrix}$}\begin{bmatrix}x_1\\x_2\end{bmatrix} \\

&=A\vv{x}

\end{align*}

\begin{definition}{Definition: Standard Matrix}

The \textbf{standard matrix} $A$ of a linear transformation $T:\mathbb{R}^n\rightarrow\mathbb{R}^m$ is the $m\times n$ matrix

$$A=\begin{bNiceMatrix}T\(\vv{e_1}\)&T\(\vv{e_2}\)&\Hdotsfor{1}&T\(\vv{e_n}\)\end{bNiceMatrix}$$

where $\left\{\vv{e_1},\vv{e_2},\hdots,\vv{e_n}\right\}$ is the standard basis for $\mathbb{R}^n$. \\ \newline

$\star$ Note: $A$ is unique.

\end{definition}

\begin{example}{Example: Finding The Standard Matrix Of The Previous Example}

$T\(x,y\)=\(3x,0,x+y\)\qquad T:\mathbb{R}^2\rightarrow\mathbb{R}^3$ \\ \newline

We've shown $T$ is linear in the previous example, now find the standard matrix for $T$. \\ \newline

As $A=\begin{bNiceMatrix}T\(\vv{e_1}\)&T\(\vv{e_2}\)&\Hdotsfor{1}&T\(\vv{e_n}\)\end{bNiceMatrix}$ and:

\begin{align*}

T\(1,0\)=\(3,0,1\) \\

T\(0,1\)=\(0,0,1\)

\end{align*}

then $A=\begin{bmatrix}3&0\\0&0\\1&1\end{bmatrix}$ this can also be seen below:

$$T\(\begin{bmatrix}x\\y\end{bmatrix}\)=A\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3&0\\0&0\\1&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3x\\0\\x+y\end{bmatrix}$$

\end{example}

\begin{example}{Example: Finding The Standard Matrix}

$T\(x_1,x_2,x_3\)=\(4x_2,-x_1,x_2-x_1,x_3\)\qquad T:\mathbb{R}^3\rightarrow\mathbb{R}^4$ is linear. Find the standard matrix of $T$

\begin{align*}

T\(1,0,0\)&=\(0,-1,-1,0\) \\

T\(0,1,0\)&=\(4,0,1,0\) \\

T\(0,0,1\)&=\(0,0,0,1\)

\end{align*}

Thus $A=\begin{bmatrix}0&4&0\\-1&0&0\\-1&1&0\\0&0&1\end{bmatrix}$

$$T\(\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\)=\begin{bmatrix}0&4&0\\-1&0&0\\-1&1&0\\0&0&1\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}4x_2\\-x_1\\-x_1+x_2\\x_3\end{bmatrix}$$

\end{example}

\subsection{Onto and One-to-One}

\begin{definition}{Example: Onto}

A linear transformation $T:\mathbb{R}^n\rightarrow\textcolor{blue}{\mathbb{R}^m}$ is \textbf{onto} if range$\(T\)=\textcolor{blue}{\mathbb{R}^m}$. \\ \newline

$\star$ Note:

\begin{itemize}

\item If $T\(\vv{x}\)=A\vv{x}$, then \ul{range$\(T\)=\text{col}\(A\)$}, so $T$ is onto if \ul{col$A=\textcolor{blue}{\mathbb{R}^m}$}.

\item $T$ is onto if and only if $T\(\vv{x}\)=$\ul{$A\vv{x}=\vv{b}$} is consistant for every $\vv{b}\in\mathbb{R}^m$.

\end{itemize}

\end{definition}

\begin{definition}{Definition: One-To-One}

A linear transformation $T:\mathbb{R}^n\rightarrow\mathbb{R}^m$ is \textbf{one-to-one} (1-1) if and only if $T\(\vv{u}\)=T\(\vv{v}\)$ implies $\vv{u}=\vv{v}$. (If $\vv{u}\ne\vv{v}$, then $T\(\vv{u}\)\ne T\(\vv{v}\)$ when $T$ is 1-1) \\ \newline

$\star$ Note: If $T\(\vv{x}\)=A\vv{x}$, $T$ is 1-1 if and only if $T\(\vv{x}\)=$\ul{$A\vv{x}=\vv{b}$} has at most one solution for each $\vv{b}$ in $\mathbb{R}^m$.

\end{definition}

\begin{theorem}{Theorem: One-To-One And Kernal Are Related}

$T$ is 1-1 if and only if ker$T=\left\{\vv{0}\right\}$. \\ \newline

$\star$ Note: If $T\(\vv{x}\)=A\vv{x}$, then ker$T=\text{null}A$ so $T$ is $\text{1-1}$ if and only if null$A=\left\{\vv{0}\right\}$.

\end{theorem}

\begin{example}{Example: One-to-One And Onto Using Real-Valued Functions From $\mathbb{R}$ To $\mathbb{R}$}

Notes:

\begin{itemize}

\item $f:\mathbb{R}\rightarrow\mathbb{R}$ is onto if range of $f=\text{ codomain of }f=\mathbb{R}$

\item $f$ is one-to-one if graph of $f$ passes the horizontal line test (no 2 distinct inputs have the same output)

\end{itemize}

\begin{enumerate}