diff --git a/Advanced/Linear Algebra 101/main.tex b/Advanced/Linear Algebra 101/main.tex deleted file mode 100755 index 2507b3e..0000000 --- a/Advanced/Linear Algebra 101/main.tex +++ /dev/null @@ -1,71 +0,0 @@ -% use [nosolutions] flag to hide solutions. -% use [solutions] flag to show solutions. -\documentclass[ - solutions, - hidewarning, - %singlenumbering -]{../../resources/ormc_handout} -\usepackage{../../resources/macros} - - -%\usepackage{lua-visual-debug} - -\usepackage{tikz-3dplot} -\usetikzlibrary{ - quotes, - angles, - matrix, - decorations.pathreplacing, - calc, - positioning, - fit -} -\input{tikzset} - - -\uptitlel{Advanced 2} -\uptitler{Spring 2023} -\title{Linear Algebra 101} -\subtitle{Prepared by \githref{Mark} on \today} - -\begin{document} - - \maketitle - - \input{parts/0 notation} - \input{parts/1 vectors} - \input{parts/2 dotprod} - \input{parts/3 matrices} - - - - \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: - - $$ - ||x \cdot y|| \leq ||x||~||y|| - $$ - - \vfill - - -\end{document} \ No newline at end of file diff --git a/Advanced/Linear Algebra 101/parts/0 notation.tex b/Advanced/Linear Algebra 101/parts/0 notation.tex deleted file mode 100755 index fc6a12b..0000000 --- a/Advanced/Linear Algebra 101/parts/0 notation.tex +++ /dev/null @@ -1,87 +0,0 @@ -\section{Notation and Terminology} - -\definition{} -\begin{itemize} - \item $\mathbb{R}$ is the set of all real numbers. - \item $\mathbb{R}^+$ is the set of positive real numbers. Zero is not positive. - \item $\mathbb{R}^+_0$ is the set of positive real numbers and zero. -\end{itemize} - -Mathematicians are often inconsistent with their notation. Depending on the author, their mood, and the phase of the moon, $\mathbb{R}^+$ may or may not include zero. We will use the definitions above. - - -\definition{} -Consider two sets $A$ and $B$. The set $A \times B$ consists of all tuples $(a, b)$ where $a \in A$ and $b \in B$. \\ -For example, $\{1, 2, 3\} \times \{\heartsuit, \star\} = \{(1,\heartsuit), (1, \star), (2,\heartsuit), (2, \star), (3,\heartsuit), (3, \star)\}$ \\ -This is called the \textit{cartesian product}. - -\vspace{4mm} - -You can think of this as placing the two sets \say{perpendicular} to one another: - -\begin{center} -\begin{tikzpicture}[ - scale=1, - bullet/.style={circle,inner sep=1.5pt,fill} -] - \draw[->] (-0.2,0) -- (4,0) node[right]{$A$}; - \draw[->] (0,-0.2) -- (0,3) node[above]{$B$}; - - \draw (1,0.1) -- ++ (0,-0.2) node[below]{$1$}; - \draw (2,0.1) -- ++ (0,-0.2) node[below]{$2$}; - \draw (3,0.1) -- ++ (0,-0.2) node[below]{$3$}; - - \draw (0.1, 1) -- ++ (-0.2, 0) node[left]{$\heartsuit$}; - \draw (0.1, 2) -- ++ (-0.2, 0) node[left]{$\star$}; - - \node[bullet] at (1, 1){}; - \node[bullet] at (2, 1) {}; - \node[bullet] at (3, 1) {}; - \node[bullet] at (1, 2) {}; - \node[bullet] at (2, 2) {}; - \node[bullet] at (3, 2) {}; - - - \draw[rounded corners] (0.5, 0.5) rectangle (3.5, 2.5) {}; - \node[above] at (2, 2.5) {$A \times B$}; - -\end{tikzpicture} -\end{center} - -\problem{} -Let $A = \{0, 1\} \times \{0, 1\}$ \\ -Let $B = \{ a, b\}$ \\ -What is $A \times B$? - -\vfill - -\problem{} -What is $\mathbb{R} \times \mathbb{R}$? \\ -\hint{Use the \say{perpendicular} analogy} - -\vfill -\pagebreak - -\definition{} -$\mathbb{R}^n$ is the set of $n$-tuples of real numbers. \\ -In English, this means that an element of $\mathbb{R}^n$ is a list of $n$ real numbers: \\ - -\vspace{4mm} - -Elements of $\mathbb{R}^2$ look like $(a, b)$, where $a, b \in \mathbb{R}$. \hfill \note{\textit{Note:} $\mathbb{R}^2$ is pronounced \say{arrgh-two.}} -Elements of $\mathbb{R}^5$ look like $(a_1, a_2, a_3, a_4, a_5)$, where $a_n \in \mathbb{R}$. \\ - -$\mathbb{R}^1$ and $\mathbb{R}$ are identical. - -\vspace{4mm} - -Intuitively, $\mathbb{R}^2$ forms a two-dimensional plane, and $\mathbb{R}^3$ forms a three-dimensional space. \\ -$\mathbb{R}^n$ is hard to visualize when $n \geq 4$, but you are welcome to try. - -\problem{} -Convince yourself that $\mathbb{R} \times \mathbb{R}$ is $\mathbb{R}^2$. \\ -What is $\mathbb{R}^2 \times \mathbb{R}$? - - -\vfill -\pagebreak \ No newline at end of file diff --git a/Advanced/Linear Algebra 101/parts/1 vectors.tex b/Advanced/Linear Algebra 101/parts/1 vectors.tex deleted file mode 100755 index 97c20f9..0000000 --- a/Advanced/Linear Algebra 101/parts/1 vectors.tex +++ /dev/null @@ -1,88 +0,0 @@ -\section{Vectors} - -\definition{} -Elements of $\mathbb{R}^n$ are often called \textit{vectors}. \\ -As you may already know, we have a few operations on vectors: -\begin{itemize} - \item Vector addition: $[a_1, a_2] + [b_1, b_2] = [a_1+b_1, a_2+b_2]$ - \item Scalar multiplication: $x \times [a_1, a_2] = [xa_1, xa_2]$. -\end{itemize} -\note{ - The above examples are for $\mathbb{R}^2$, and each vector thus has two components. \\ - These operations are similar for all other $n$. -} - -\problem{} -Compute the following or explain why you can't: -\begin{itemize} - \item $[1, 2, 3] - [1, 3, 4]$ \note{Subtraction works just like addition.} - \item $4 \times [5, 2, 4]$ - \item $a + b$, where $a \in \mathbb{R} -^5$ and $b \in \mathbb{R}^7$ -\end{itemize} - -\vfill - -\problem{} -Consider $(2, -1)$ and $(3, 1)$ in $\mathbb{R}^2$. \\ -Can you develop geometric intuition for their sum and difference? - -\begin{center} - \begin{tikzpicture}[scale=1] - - \draw[->] - (0,0) coordinate (o) -- node[below left] {$(2, -1)$} - (2, -1) coordinate (a) - ; - - \draw[->] - (a) -- node[below right] {$(3, 1)$} - (5, 0) coordinate (b) - ; - - \draw[ - draw = gray, - text = gray, - -> - ] - (o) -- node[above] {$??$} - (b) coordinate (s) - ; - - \end{tikzpicture} - \end{center} - - -\vfill -\pagebreak - -\definition{Euclidean Norm} -A \textit{norm} on $\mathbb{R}^n$ is a map from $\mathbb{R}^n$ to $\mathbb{R}^+_0$ \\ -Usually, one thinks of a norm as a way of measuring \say{length} in a vector space. \\ -The norm of a vector $v$ is written $||v||$. \\ - -\vspace{2mm} - -We usually use the \textit{Euclidean norm} when we work in $\mathbb{R}^n$. \\ -If $v \in \mathbb{R}^n$, the Euclidean norm is defined as follows: \\ -If $v = [v_1, v_2, ..., v_n]$, -$$ - ||v|| = \sqrt{v_1^2 + v_2^2 + ... + v_n^2} -$$ -This is simply an application of the Pythagorean theorem. - -\problem{} -Compute the euclidean norm of -\begin{itemize} - \item $[2, 3]$ - \item $[-2, 1, -4, 2]$ -\end{itemize} - -\vfill - -\problem{} -Show that $a \cdot a$ is $||a||^2$. - -\vfill - -\pagebreak \ No newline at end of file diff --git a/Advanced/Linear Algebra 101/parts/2 dotprod.tex b/Advanced/Linear Algebra 101/parts/2 dotprod.tex deleted file mode 100644 index 9f52b2d..0000000 --- a/Advanced/Linear Algebra 101/parts/2 dotprod.tex +++ /dev/null @@ -1,93 +0,0 @@ -\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 be 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 $a \cdot (b + c) = a \cdot b + a \cdot c$ - \item Homogeneous: $x(a \cdot b) = xa \cdot b = a \cdot xb$ \\ - \note{$x \in \mathbb{R}$, and $a, b$ are vectors.} - \item Positive definite: $a \cdot a \geq 0$, with equality iff $a = 0$ \\ - \note{$a \in \mathbb{R}^n$, and $0$ is the zero vector.} -\end{itemize} - - -\vfill -\pagebreak - - - - -\problem{} -Say you have two vectors, $a$ and $b$. Show that $a \cdot b$ = $||a||~||b||\cos(\alpha)$, \\ -where $\alpha$ is the angle between $a$ and $b$. \\ -\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 $a \cdot b$ be? Is the converse true? - - -\vfill -\pagebreak \ No newline at end of file diff --git a/Advanced/Linear Algebra 101/parts/3 matrices.tex b/Advanced/Linear Algebra 101/parts/3 matrices.tex deleted file mode 100644 index 60683d2..0000000 --- a/Advanced/Linear Algebra 101/parts/3 matrices.tex +++ /dev/null @@ -1,284 +0,0 @@ -\section{Matrices} - -\definition{} -A \textit{matrix} is a two-dimensional array of numbers: \\ -$$ -A = -\begin{bmatrix} - 1 & 2 & 3 \\ - 4 & 5 & 6 -\end{bmatrix} -$$ -The above matrix has two rows and three columns. It is thus a $2 \times 3$ matrix. \\ -\vspace{1mm} -The order \say{first rows, then columns} is usually consistent in linear algebra. \\ -If you look closely, you may also find it in the next definition. - -\definition{} -We can define the product of a matrix $A$ and a vector $v$: - -$$ -Av = -\begin{bmatrix} - 1 & 2 & 3 \\ - 4 & 5 & 6 -\end{bmatrix} -\begin{bmatrix} - a \\ b \\ c -\end{bmatrix} -= -\begin{bmatrix} - 1a + 2b + 3c \\ - 4a + 5b + 6c -\end{bmatrix} -$$ -Note that each element of the resulting $2 \times 1$ matrix is the dot product of a row of $A$ with $v$: - -$$ -Av = -\begin{bmatrix} - \text{---} r_1 \text{---} \\ - \text{---} r_2 \text{---} -\end{bmatrix} -\begin{bmatrix} - | \\ - v \\ - | \\ -\end{bmatrix} -= -\begin{bmatrix} - r_1 \cdot v \\ - r_2 \cdot v -\end{bmatrix} -$$ - -Naturally, a vector can only be multiplied by a matrix if the number of rows in the vector equals the number of columns in the matrix. - -\problem{} -Compute the following: - -$$ -\begin{bmatrix} - 1 & 2 \\ - 3 & 4 \\ - 5 & 6 -\end{bmatrix} -\begin{bmatrix} - 5 \\ 3 -\end{bmatrix} -$$ - -\vfill - - -\problem{} -Say you multiply a size-$m$ vector $v$ by an $m \times n$ matrix $A$. \\ -What is the size of your result $Av$? - -\vfill -\pagebreak - -\definition{} -We can also multiply a matrix by a matrix: - -$$ -AB = -\begin{bmatrix} - 1 & 2 \\ - 3 & 4 -\end{bmatrix} -\begin{bmatrix} - 10 & 20 \\ - 100 & 200 -\end{bmatrix} -= -\begin{bmatrix} - 210 & 420 \\ - 430 & 860 -\end{bmatrix} -$$ -Note each element of the resulting matrix is dot product of a row of $A$ and a column of $B$: - -$$ -AB = -\begin{bmatrix} - \text{---} r_1 \text{---} \\ - \text{---} r_2 \text{---} -\end{bmatrix} -\begin{bmatrix} - | & | \\ - v_1 & v_2 \\ - | & | \\ -\end{bmatrix} -= -\begin{bmatrix} - r_1 \cdot v_1 & r_1 \cdot v_2 \\ - r_2 \cdot v_1 & r_2 \cdot v_2 \\ -\end{bmatrix} -$$ - -\begin{center} -\begin{tikzpicture} - - \begin{scope}[layer = nodes] - \matrix[ - matrix of math nodes, - left delimiter={[}, - right delimiter={]} - ] (A) at (0, 0){ - 1 & 2 \\ - 3 & 4 \\ - }; - - \matrix[ - matrix of math nodes, - left delimiter={[}, - right delimiter={]} - ] (B) at (2, 0) { - 10 & 20 \\ - 100 & 200 \\ - }; - - \node at (3.25, 0) {$=$}; - - \matrix[ - matrix of math nodes, - left delimiter={[}, - right delimiter={]} - ] (C) at (4.5, 0) { - 210 & 420 \\ - 430 & 860 \\ - }; - \end{scope} - - \draw[rounded corners,fill=black!30!white,draw=none] ([xshift=-2mm,yshift=2mm]A-1-1) rectangle ([xshift=2mm,yshift=-2mm]A-1-2) {}; - - \draw[rounded corners,fill=black!30!white,draw=none] ([xshift=-3mm,yshift=2mm]B-1-1) rectangle ([xshift=3mm,yshift=-2mm]B-2-1) {}; - - \draw[rounded corners,fill=black!30!white,draw=none] ([xshift=-4mm,yshift=2mm]C-1-1) rectangle ([xshift=4mm,yshift=-2mm]C-1-1) {}; - - - \draw[rounded corners] ([xshift=-2mm,yshift=2mm]A-2-1) rectangle ([xshift=2mm,yshift=-2mm]A-2-2) {}; - - \draw[rounded corners] ([xshift=-3mm,yshift=2mm]B-1-2) rectangle ([xshift=3mm,yshift=-2mm]B-2-2) {}; - - \draw[rounded corners] ([xshift=-4mm,yshift=2mm]C-2-2) rectangle ([xshift=4mm,yshift=-2mm]C-2-2) {}; -\end{tikzpicture} -\end{center} - - -\problem{} -Compute the following matrix product. \\ - -$$ -\begin{bmatrix} - 1 & 2 \\ - 3 & 4 \\ - 5 & 6 -\end{bmatrix} -\begin{bmatrix} - 9 & 8 & 7 \\ - 6 & 5 & 4 -\end{bmatrix} -$$ - -\vfill - - -\problem{} -Compute the following matrix product or explain why you can't. - -$$ -\begin{bmatrix} - 1 & 2 & 3 \\ - 4 & 5 & 6 -\end{bmatrix} -\begin{bmatrix} - 10 & 20 \\ - 30 & 40 -\end{bmatrix} -$$ - -\vfill - - -\problem{} -If $A$ is an $m \times n$ matrix and $B$ is a $p \times q$ matrix, when does the product $AB$ exist? - - -\vfill -\pagebreak - -\problem{} -Is matrix multiplication commutative? \\ -\note{Does $AB = BA$ for all $A, B$? \\ You only need one counterexample to show this is false.} - -\vfill - - -\definition{} -Say we have a matrix $A$. The matrix $A^T$, pronounced \say{A-transpose}, is created by turning rows of $A$ into columns, and columns into rows: - -$$ -\begin{bmatrix} - 1 & 2 & 3 \\ - 4 & 5 & 6 -\end{bmatrix} ^ T -= -\begin{bmatrix} - 1 & 4 \\ - 2 & 5 \\ - 3 & 6 -\end{bmatrix} -$$ - -\problem{} -Compute the following: - -\hfill -$ -\begin{bmatrix} - a & b \\ - c & d -\end{bmatrix} ^ T -$\hfill -$ -\begin{bmatrix} - 1 \\ - 3 \\ - 3 \\ - 7 \\ -\end{bmatrix} ^ T -$\hfill -$ -\begin{bmatrix} - 1 & 2 & 4 & 8 \\ -\end{bmatrix} ^ T -$ -\hfill~ - -\vfill -\pagebreak - -The \say{transpose} operator is often used to write column vectors in a compact way. \\ -Vertical arrays don't look good in horizontal text. - -\problem{} -Consider the vectors $a = [1, 4, 3]^T$ and $b = [9, 1, 4]^T$ \\ -\begin{itemize} - \item Compute the dot product $a \cdot b$. - \item Can you redefine the dot product using matrix multiplication? -\end{itemize} -\note{As you may have noticed, a vector is a special case of a matrix.} - -\vfill - -\problem{} -A \textit{column vector} is an $m \times 1$ matrix. \\ -A \textit{row vector} is a $1 \times m$ matrix. \\ -We usually use column vectors. Why? \\ -\hint{How does vector-matrix multiplication work?} - - -\vfill -\pagebreak \ No newline at end of file diff --git a/Advanced/Linear Algebra 101/tikzset.tex b/Advanced/Linear Algebra 101/tikzset.tex deleted file mode 100644 index ffbdd51..0000000 --- a/Advanced/Linear Algebra 101/tikzset.tex +++ /dev/null @@ -1,36 +0,0 @@ -\usetikzlibrary{arrows.meta} -\usetikzlibrary{shapes.geometric} -\usetikzlibrary{patterns} - -% We put nodes in a separate layer, so we can -% slightly overlap with paths for a perfect fit -\pgfdeclarelayer{nodes} -\pgfdeclarelayer{path} -\pgfsetlayers{main,nodes} - -% Layer settings -\tikzset{ - % Layer hack, lets us write - % later = * in scopes. - layer/.style = { - execute at begin scope={\pgfonlayer{#1}}, - execute at end scope={\endpgfonlayer} - }, - % - % Nodes - main/.style = { - draw, - circle, - fill = white - }, - % - % Paths - path/.style = { - line width = 4mm, - draw = black, - % Lengthen paths so they're - % completely under nodes. - line cap = rect, - opacity = 0.3 - } -} \ No newline at end of file