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% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
\documentclass[
solutions,
nowarning,
%singlenumbering
]{../../resources/ormc_handout}
%\usepackage{lua-visual-debug}
\usepackage{tikz-3dplot}
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\usetikzlibrary{quotes,angles}
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\begin{document}
\maketitle
<Advanced 2>
<Spring 2023>
{Linear Algebra 101}
{
Prepared by Mark on \today \\
}
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\input{parts/0 notation}
\input{parts/1 vectors}
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\section{Dot Products}
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\definition{}
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We can also define the \textit{dot product} of two vectors.\footnotemark{} \\
The dot product maps two elements of $\mathbb{R}^n$ to one element of $\mathbb{R}$:
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\footnotetext{
\textbf{Bonus content. Feel free to skip.}
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Formally, we would say that the dot product is a map from $\mathbb{R}^n \times \mathbb{R}^n$ to $\mathbb{R}$. Why is this reasonable?
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\vspace{2mm}
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It's also worth noting that a function $f$ from $X$ to $Y$ can defined as a subset of $X \times Y$, where for all $x \in X$ there exists a unique $y \in Y$ so that $(x, y) \in f$. Try to make sense of this definition.
}
$$
a \cdot b = \sum_{i = 1}^n a_ib_i = a_1b_1 + a_2b_2 + ... + a_nb_n
$$
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\problem{}
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Compute $[2, 3, 4, 1] \cdot [2, 4, 10, 12]$
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\vfill
\problem{}
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Show that the dot product is
\begin{itemize}
\item Commutative
\item Distributive
\item Homogeneic: $x(a \cdot b) = xa \cdot b = a \cdot xb$
\item Positive definite: $a \cdot a \geq 0$, with equality iff $a = 0$
\end{itemize}
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\vfill
\pagebreak
\problem{}
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Say you have two vectors, $a$ and $b$. Show that $\langle a, b \rangle$ = $||a||~||b||\cos(\alpha)$ \\
\hint{What is $c$ in terms of $a$ and $b$?}
\hint{The law of cosines is $a^2 + b^2 - 2ab\cos(\alpha) = c^2$}
\hint{The length of $a$ is $||a||$}
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\begin{center}
\begin{tikzpicture}[scale=1]
\draw[->]
(0,0) coordinate (o) -- node[above left] {$a$}
(1,2) coordinate (a)
;
\draw[->]
(o) -- node[below] {$b$}
(3,0.5) coordinate (b)
;
\draw[
draw = gray,
text = gray,
-
] (a) -- node[above] {$c$} (b);
\draw
pic[
"$\alpha$",
draw = orange,
text = orange,
<->,
angle eccentricity = 1.2,
angle radius = 1cm
]
{ angle = b--o--a }
;
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\end{tikzpicture}
\end{center}
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\vfill
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\problem{}
If $a$ and $b$ are perpendicular, what must $\langle a, b \rangle$ be? Is the converse true?
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\vfill
\pagebreak
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\section{Bonus}
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\problem{}
Show that the euclidean norm satisfies the triangle inequalty:
$$
||x+y|| \leq ||x|| + ||y||
$$:
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\vfill
\problem{}
Show that the eucidean norm satisfies the reverse triangle inequality:
$$
||x - y|| \geq |~||x|| - ||y||~|
$$
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\vfill
\problem{}
Prove the Cauchy-Schwartz inequality:
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$$
||\langle x, y \rangle|| = ||x||~||y||
$$
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\vfill
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\end{document}