Added content
This commit is contained in:
parent
9c3b88827f
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@ -9,7 +9,16 @@
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%\usepackage{lua-visual-debug}
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\usepackage{tikz-3dplot}
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\usetikzlibrary{quotes,angles}
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\usetikzlibrary{
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quotes,
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angles,
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matrix,
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decorations.pathreplacing,
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calc,
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positioning,
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fit
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}
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\input{tikzset}
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\begin{document}
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@ -24,97 +33,10 @@
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\input{parts/0 notation}
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\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{} \\
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The dot product maps two elements of $\mathbb{R}^n$ to one element of $\mathbb{R}$:
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\footnotetext{
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\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.
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}
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$$
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a \cdot b = \sum_{i = 1}^n a_ib_i = a_1b_1 + a_2b_2 + ... + a_nb_n
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$$
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\problem{}
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Compute $[2, 3, 4, 1] \cdot [2, 4, 10, 12]$
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\vfill
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\problem{}
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Show that the dot product is
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\begin{itemize}
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\item Commutative
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\item Distributive
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\item Homogeneic: $x(a \cdot b) = xa \cdot b = a \cdot xb$
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\item Positive definite: $a \cdot a \geq 0$, with equality iff $a = 0$
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\end{itemize}
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\input{parts/2 dotprod}
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\input{parts/3 matrices}
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\vfill
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\pagebreak
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\problem{}
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Say you have two vectors, $a$ and $b$. Show that $\langle a, b \rangle$ = $||a||~||b||\cos(\alpha)$ \\
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\hint{What is $c$ in terms of $a$ and $b$?}
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\hint{The law of cosines is $a^2 + b^2 - 2ab\cos(\alpha) = c^2$}
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\hint{The length of $a$ is $||a||$}
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\draw[->]
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(0,0) coordinate (o) -- node[above left] {$a$}
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(1,2) coordinate (a)
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;
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\draw[->]
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(o) -- node[below] {$b$}
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(3,0.5) coordinate (b)
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;
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\draw[
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draw = gray,
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text = gray,
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-
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] (a) -- node[above] {$c$} (b);
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\draw
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pic[
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"$\alpha$",
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draw = orange,
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text = orange,
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<->,
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angle eccentricity = 1.2,
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angle radius = 1cm
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]
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{ angle = b--o--a }
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;
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\end{tikzpicture}
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\end{center}
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\vfill
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\problem{}
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If $a$ and $b$ are perpendicular, what must $\langle a, b \rangle$ be? Is the converse true?
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\vfill
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\pagebreak
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\section{Bonus}
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90
Advanced/Linear Algebra 101/parts/2 dotprod.tex
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90
Advanced/Linear Algebra 101/parts/2 dotprod.tex
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@ -0,0 +1,90 @@
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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{} \\
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The dot product maps two elements of $\mathbb{R}^n$ to one element of $\mathbb{R}$:
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\footnotetext{
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\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.
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}
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$$
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a \cdot b = \sum_{i = 1}^n a_ib_i = a_1b_1 + a_2b_2 + ... + a_nb_n
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$$
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\problem{}
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Compute $[2, 3, 4, 1] \cdot [2, 4, 10, 12]$
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\vfill
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\problem{}
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Show that the dot product is
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\begin{itemize}
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\item Commutative
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\item Distributive
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\item Homogeneic: $x(a \cdot b) = xa \cdot b = a \cdot xb$
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\item Positive definite: $a \cdot a \geq 0$, with equality iff $a = 0$
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\end{itemize}
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\vfill
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\pagebreak
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\problem{}
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Say you have two vectors, $a$ and $b$. Show that $\langle a, b \rangle$ = $||a||~||b||\cos(\alpha)$ \\
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\hint{What is $c$ in terms of $a$ and $b$?}
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\hint{The law of cosines is $a^2 + b^2 - 2ab\cos(\alpha) = c^2$}
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\hint{The length of $a$ is $||a||$}
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\draw[->]
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(0,0) coordinate (o) -- node[above left] {$a$}
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(1,2) coordinate (a)
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;
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\draw[->]
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(o) -- node[below] {$b$}
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(3,0.5) coordinate (b)
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;
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\draw[
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draw = gray,
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text = gray,
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-
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] (a) -- node[above] {$c$} (b);
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\draw
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pic[
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"$\alpha$",
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draw = orange,
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text = orange,
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<->,
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angle eccentricity = 1.2,
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angle radius = 1cm
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]
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{ angle = b--o--a }
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;
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\end{tikzpicture}
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\end{center}
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\vfill
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\problem{}
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If $a$ and $b$ are perpendicular, what must $\langle a, b \rangle$ be? Is the converse true?
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\vfill
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\pagebreak
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219
Advanced/Linear Algebra 101/parts/3 matrices.tex
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219
Advanced/Linear Algebra 101/parts/3 matrices.tex
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\section{Matrices}
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\definition{}
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A \textit{matrix} is a two-dimensional array of numbers: \\
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$$
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A =
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\begin{bmatrix}
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1 & 2 & 3 \\
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4 & 5 & 6
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\end{bmatrix}
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$$
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The above matrix has two rows and three columns. It is thus a $2 \times 3$ matrix.
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\definition{}<matvec>
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We can define the product of a matrix $A$ and a vector $v$:
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$$
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Av =
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\begin{bmatrix}
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1 & 2 & 3 \\
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4 & 5 & 6
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\end{bmatrix}
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\begin{bmatrix}
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a \\ b \\ c
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\end{bmatrix}
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=
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\begin{bmatrix}
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1a + 2b + 3c \\
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4a + 5b + 6c
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\end{bmatrix}
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$$
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Note that each element of the resulting $2 \times 1$ matrix is the dot product of a row of $A$ with $v$:
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$$
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Av =
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\begin{bmatrix}
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\text{---} a_1 \text{---} \\
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\text{---} a_2 \text{---}
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\end{bmatrix}
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\begin{bmatrix}
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| \\
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v \\
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| \\
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\end{bmatrix}
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=
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\begin{bmatrix}
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r_1v \\
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r_2v
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\end{bmatrix}
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$$
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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.
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\problem{}
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Say you multiply a size-$m$ vector by an $m \times n$ matrix. What is the size of your result?
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\vfill
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\problem{}
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Compute the following:
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$$
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\begin{bmatrix}
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1 & 2 \\
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3 & 4 \\
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5 & 6
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\end{bmatrix}
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\begin{bmatrix}
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5 \\ 3
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\end{bmatrix}
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$$
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\vfill
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\pagebreak
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\definition{}
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We also multiply a matrix by a matrix:
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$$
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AB =
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\begin{bmatrix}
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1 & 2 \\
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3 & 4
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\end{bmatrix}
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\begin{bmatrix}
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10 & 20 \\
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100 & 200
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\end{bmatrix}
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=
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\begin{bmatrix}
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210 & 420 \\
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430 & 860
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\end{bmatrix}
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$$
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Note each element of the resulting matrix is dot product of a row of $A$ and a column of $B$:
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$$
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AB =
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\begin{bmatrix}
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\text{---} a_1 \text{---} \\
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\text{---} a_2 \text{---}
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\end{bmatrix}
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\begin{bmatrix}
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| & | \\
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v_1 & v_2 \\
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| & | \\
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\end{bmatrix}
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=
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\begin{bmatrix}
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r_1v_1 & r_1v_2 \\
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r_2v_1 & r_2vm_2 \\
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\end{bmatrix}
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$$
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\begin{center}
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\begin{tikzpicture}[>=stealth,thick,baseline]
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\begin{scope}[layer = nodes]
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\matrix[
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matrix of math nodes,
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left delimiter={[},
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right delimiter={]}
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] (A) at (0, 0){
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1 & 2 \\
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3 & 4 \\
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};
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\matrix[
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matrix of math nodes,
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left delimiter={[},
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right delimiter={]}
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] (B) at (2, 0) {
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10 & 20 \\
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100 & 200 \\
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};
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\node at (3.25, 0) {$=$};
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\matrix[
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matrix of math nodes,
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left delimiter={[},
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right delimiter={]}
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] (C) at (4.5, 0) {
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210 & 420 \\
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430 & 860 \\
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};
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\end{scope}
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\draw[rounded corners,fill=black!30!white,draw=none] ([xshift=-2mm,yshift=3mm]A-1-1) rectangle ([xshift=2mm,yshift=-3mm]A-2-1) {};
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\draw[rounded corners,fill=black!30!white,draw=none] ([xshift=-3mm,yshift=2mm]B-1-1) rectangle ([xshift=3mm,yshift=-2mm]B-1-2) {};
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\draw[rounded corners,fill=black!30!white,draw=none] ([xshift=-4mm,yshift=2mm]C-1-1) rectangle ([xshift=4mm,yshift=-2mm]C-1-1) {};
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\draw[rounded corners] ([xshift=-2mm,yshift=3mm]A-1-2) rectangle ([xshift=2mm,yshift=-3mm]A-2-2) {};
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\draw[rounded corners] ([xshift=-3mm,yshift=2mm]B-2-1) rectangle ([xshift=3mm,yshift=-2mm]B-2-2) {};
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\draw[rounded corners] ([xshift=-4mm,yshift=2mm]C-2-2) rectangle ([xshift=4mm,yshift=-2mm]C-2-2) {};
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\end{tikzpicture}
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\end{center}
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\problem{}
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Compute the following matrix product. \\
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$$
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\begin{bmatrix}
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1 & 2 \\
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3 & 4 \\
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5 & 6
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\end{bmatrix}
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\begin{bmatrix}
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9 & 8 & 7 \\
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6 & 5 & 4
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\end{bmatrix}
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$$
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\vfill
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\problem{}
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Consider the following matrix product. \\
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Compute it or explain why you can't.
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$$
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\begin{bmatrix}
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1 & 2 & 3 \\
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4 & 5 & 6
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\end{bmatrix}
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\begin{bmatrix}
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10 & 20 \\
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30 & 40
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\end{bmatrix}
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$$
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\vfill
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\problem{}
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If $A$ is an $m \times n$ matrix and $B$ is a $p \times q$ matrix, when does the product $AB$ exist?
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\vfill
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\pagebreak
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\problem{}
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Look back to \ref{matvec}. \\
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Convince yourself that vectors are matrices. \\
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Can you multiply a matrix by a vector, as in $vA$? \\
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How does the dot prouduct relate to matrix multiplication? (transpose)
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\vfill
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\pagebreak
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36
Advanced/Linear Algebra 101/tikzset.tex
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36
Advanced/Linear Algebra 101/tikzset.tex
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@ -0,0 +1,36 @@
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\usetikzlibrary{arrows.meta}
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\usetikzlibrary{shapes.geometric}
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\usetikzlibrary{patterns}
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% We put nodes in a separate layer, so we can
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% slightly overlap with paths for a perfect fit
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\pgfdeclarelayer{nodes}
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\pgfdeclarelayer{path}
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\pgfsetlayers{main,nodes}
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% Layer settings
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\tikzset{
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% Layer hack, lets us write
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% later = * in scopes.
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layer/.style = {
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execute at begin scope={\pgfonlayer{#1}},
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execute at end scope={\endpgfonlayer}
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},
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%
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% Nodes
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main/.style = {
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draw,
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circle,
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fill = white
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},
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%
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% Paths
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path/.style = {
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line width = 4mm,
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draw = black,
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% Lengthen paths so they're
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% completely under nodes.
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line cap = rect,
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opacity = 0.3
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}
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}
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