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102
Advanced/Linear Algebra 101/main.tex
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@ -9,7 +9,16 @@
%\usepackage{lua-visual-debug} %\usepackage{lua-visual-debug}
\usepackage{tikz-3dplot} \usepackage{tikz-3dplot}
\usetikzlibrary{quotes,angles} \usetikzlibrary{
quotes,
angles,
matrix,
decorations.pathreplacing,
calc,
positioning,
fit
}
\input{tikzset}
\begin{document} \begin{document}
@ -24,97 +33,10 @@
\input{parts/0 notation} \input{parts/0 notation}
\input{parts/1 vectors} \input{parts/1 vectors}
\input{parts/2 dotprod}
\section{Dot Products} \input{parts/3 matrices}
\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} \section{Bonus}

90
Advanced/Linear Algebra 101/parts/2 dotprod.tex Normal file
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@ -0,0 +1,90 @@
\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

219
Advanced/Linear Algebra 101/parts/3 matrices.tex Normal file
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@ -0,0 +1,219 @@
\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.
\definition{}<matvec>
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 $v$:
$$
Av =
\begin{bmatrix}
\text{---} a_1 \text{---} \\
\text{---} a_2 \text{---}
\end{bmatrix}
\begin{bmatrix}
| \\
v \\
| \\
\end{bmatrix}
=
\begin{bmatrix}
r_1v \\
r_2v
\end{bmatrix}
$$
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{}
Say you multiply a size-$m$ vector by an $m \times n$ matrix. What is the size of your result?
\vfill
\problem{}
Compute the following:
$$
\begin{bmatrix}
1 & 2 \\
3 & 4 \\
5 & 6
\end{bmatrix}
\begin{bmatrix}
5 \\ 3
\end{bmatrix}
$$
\vfill
\pagebreak
\definition{}
We 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 $B$:
$$
AB =
\begin{bmatrix}
\text{---} a_1 \text{---} \\
\text{---} a_2 \text{---}
\end{bmatrix}
\begin{bmatrix}
| & | \\
v_1 & v_2 \\
| & | \\
\end{bmatrix}
=
\begin{bmatrix}
r_1v_1 & r_1v_2 \\
r_2v_1 & r_2vm_2 \\
\end{bmatrix}
$$
\begin{center}
\begin{tikzpicture}[>=stealth,thick,baseline]
\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=3mm]A-1-1) rectangle ([xshift=2mm,yshift=-3mm]A-2-1) {};
\draw[rounded corners,fill=black!30!white,draw=none] ([xshift=-3mm,yshift=2mm]B-1-1) rectangle ([xshift=3mm,yshift=-2mm]B-1-2) {};
\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=3mm]A-1-2) rectangle ([xshift=2mm,yshift=-3mm]A-2-2) {};
\draw[rounded corners] ([xshift=-3mm,yshift=2mm]B-2-1) 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{}
Consider the following matrix product. \\
Compute it 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{}
Look back to \ref{matvec}. \\
Convince yourself that vectors are matrices. \\
Can you multiply a matrix by a vector, as in $vA$? \\
How does the dot prouduct relate to matrix multiplication? (transpose)
\vfill
\pagebreak

36
Advanced/Linear Algebra 101/tikzset.tex Normal file
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@ -0,0 +1,36 @@
\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
}
}