% 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}
\usetikzlibrary{quotes,angles}

\begin{document}

	\maketitle
		<Advanced 2>
		<Spring 2023>
		{Linear Algebra 101}
		{
			Prepared by Mark on \today \\
		}


	\input{parts/0 notation}
	\input{parts/1 vectors}

	\section{Dot Products}

	\definition{}
	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 $\mathbb{R}$:

	\footnotetext{
		\textbf{Bonus content. Feel free to skip.}

		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?

		\vspace{2mm}

		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.
	}

	$$
		a \cdot b = \sum_{i = 1}^n a_ib_i = a_1b_1 + a_2b_2 + ... + a_nb_n
	$$

	\problem{}
	Compute $[2, 3, 4, 1] \cdot [2, 4, 10, 12]$

	\vfill

	\problem{}
	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}


	\vfill
	\pagebreak




	\problem{}
	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 $||a||$}


	\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 }
		;

	\end{tikzpicture}
	\end{center}

	\vfill

	\problem{}
	If $a$ and $b$ are perpendicular, what must $\langle a, b \rangle$ be? Is the converse true?


	\vfill
	\pagebreak

	\section{Bonus}

	\problem{}
	Show that the euclidean norm satisfies the triangle inequalty:
	$$
	||x+y|| \leq ||x|| + ||y||
	$$:

	\vfill

	\problem{}
	Show that the eucidean norm satisfies the reverse triangle inequality:

	$$
		||x - y|| \geq |~||x|| - ||y||~|
	$$

	\vfill

	\problem{}
	Prove the Cauchy-Schwartz inequality:

	$$
		||\langle x, y \rangle|| = ||x||~||y||
	$$

	\vfill


\end{document}