Dot product
$\require{mathtools}
\newcommand{\nc}{\newcommand}
%
%%% GENERIC MATH %%%
%
% Environments
\newcommand{\al}[1]{\begin{align}#1\end{align}} % need this for \tag{} to work
\renewcommand{\r}{\mathrm}
\renewcommand{\t}{\textrm}
%
% Delimiters
% (I needed to create my own because the MathJax version of \DeclarePairedDelimiter doesn't have \mathopen{} and that messes up the spacing)
% .. one-part
\newcommand{\p}[1]{\mathopen{}\left( #1 \right)}
\renewcommand{\P}[1]{^{\p{#1}}}
\renewcommand{\b}[1]{\mathopen{}\left[ #1 \right]}
\newcommand{\set}[1]{\mathopen{}\left\{ #1 \right\}}
\newcommand{\abs}[1]{\mathopen{}\left\lvert #1 \right\rvert}
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\newcommand{\norm}[1]{\mathopen{}\left\lVert #1 \strut \right\rVert}
\newcommand{\frob}[1]{\norm{#1}_\mathrm{F}}
\newcommand{\mix}[1]{\mathopen{}\left\lfloor #1 \right\rceil}
%% .. two-part
\newcommand{\inco}[2]{#1 \mathop{}\middle|\mathop{} #2}
\newcommand{\co}[2]{ {\left.\inco{#1}{#2}\right.}}
\newcommand{\cond}{\co} % deprecated
\newcommand{\pco}[2]{\p{\inco{#1}{#2}}}
\newcommand{\bco}[2]{\b{\inco{#1}{#2}}}
\newcommand{\setco}[2]{\set{\inco{#1}{#2}}}
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%
% Greek
\newcommand{\eps}{\epsilon}
\newcommand{\veps}{\varepsilon}
\newcommand{\vpi}{\varpi}
% the following cause issues with real LaTeX tho :/ maybe consider naming it \fhi instead?
\let\fi\phi % because it looks like an f
\let\phi\varphi % because it looks like a p
\renewcommand{\th}{\theta}
\newcommand{\Th}{\Theta}
\newcommand{\om}{\omega}
\newcommand{\Om}{\Omega}
%
% Miscellaneous
\newcommand{\LHS}{\mathrm{LHS}}
\newcommand{\RHS}{\mathrm{RHS}}
\DeclareMathOperator{\cst}{const}
% .. operators
\DeclareMathOperator{\poly}{poly}
\DeclareMathOperator{\polylog}{polylog}
\DeclareMathOperator{\quasipoly}{quasipoly}
\DeclareMathOperator{\negl}{negl}
\DeclareMathOperator*{\argmin}{arg\thinspace min}
\DeclareMathOperator*{\argmax}{arg\thinspace max}
% .. functions
\DeclareMathOperator{\id}{id}
\DeclareMathOperator{\sign}{sign}
\DeclareMathOperator{\err}{err}
\DeclareMathOperator{\ReLU}{ReLU}
% .. analysis
\let\d\undefined
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\newcommand{\df}[2]{ {\f{\d #1}{\d #2}}}
\newcommand{\ds}[2]{ {\sl{\d #1}{\d #2}}}
\newcommand{\ddf}[3]{ {\f{\dd{#1} #2}{\p{\d #3}^{#1}}}}
\newcommand{\dds}[3]{ {\sl{\dd{#1} #2}{\p{\d #3}^{#1}}}}
\renewcommand{\part}{\partial}
\newcommand{\partf}[2]{\f{\part #1}{\part #2}}
\newcommand{\parts}[2]{\sl{\part #1}{\part #2}}
\newcommand{\grad}[1]{\mathop{\nabla\!_{#1}}}
% .. sets
\newcommand{\es}{\emptyset}
\newcommand{\N}{\mathbb{N}}
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\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\F}{\mathbb{F}}
\newcommand{\zo}{\set{0,1}}
\newcommand{\pmo}{\set{\pm 1}}
\newcommand{\zpmo}{\set{0,\pm 1}}
% .... set operations
\newcommand{\sse}{\subseteq}
\newcommand{\out}{\not\in}
\newcommand{\minus}{\setminus}
\newcommand{\inc}[1]{\union \set{#1}} % "including"
\newcommand{\exc}[1]{\setminus \set{#1}} % "except"
% .. over and under
\renewcommand{\ss}[1]{_{\substack{#1}}}
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\newcommand{\ol}{\overline}
\newcommand{\tld}{\widetilde} % deprecated
\renewcommand{\~}{\widetilde}
\newcommand{\HAT}{\widehat} % deprecated
\renewcommand{\^}{\widehat}
\newcommand{\rt}[1]{ {\sqrt{#1}}}
\newcommand{\for}[2]{_{#1=1}^{#2}}
\newcommand{\sfor}{\sum\for}
\newcommand{\pfor}{\prod\for}
% .... two-part
\newcommand{\f}{\frac}
\renewcommand{\sl}[2]{#1 /\mathopen{}#2}
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% .. arrows
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% .. operators and relations
\renewcommand{\*}{\cdot}
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% .. punctuation and spacing
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%
% Levels of closeness
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\newcommand{#5}{\mathrel{#1\!\!#1\!\!#1}}
\newcommand{#6}{\mathrel{(\sdot{#1})}}
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}
\let\lt\undefined
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% .. vanilla versions (is it within a constant?)
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% .. dotted versions (is it equal in the limit?)
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% .. log versions (is it equal up to log?)
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%
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% Logic and bit operations
\newcommand{\fa}{\forall}
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% Linear algebra
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\DeclareMathOperator{\rank}{rank}
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\newcommand{\transp}{\mathsf{T}}
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% .. named tensors
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%
% Probability
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% .. operators
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\DeclareMathOperator*{\K}{K}
\DeclareMathOperator*{\corr}{corr}
\DeclareMathOperator*{\median}{median}
\DeclareMathOperator*{\maj}{maj}
% ... information theory
\let\H\undefined
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\DeclareMathOperator*{\KL}{KL}
% .. other divergences
\newcommand{\dTV}{d_{\mathrm{TV}}}
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%
%%% SPECIALIZED COMPUTER SCIENCE %%%
%
% Complexity classes
% .. classical
\newcommand{\Poly}{\mathsf{P}}
\newcommand{\NP}{\mathsf{NP}}
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\renewcommand{\L}{\mathsf{L}}
% .. probabilistic
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% .. circuits
\newcommand{\NC}{\mathsf{NC}}
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\newcommand{\Ppoly}{\mathsf{P}/\poly}
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% .. resources
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% .. keywords
\newcommand{\coclass}{\mathsf{co}}
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% Boolean analysis
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% .. functions (small caps sadly doesn't work)
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% Alignment
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% In "text"
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% remove these last two if using real LaTeX
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%
% Fonts
% .. bold
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% .. calligraphic
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\newcommand{\TY}{\mathtt{Y}}
\newcommand{\TZ}{\mathtt{Z}}$
Adapted from this 3b1b video.
The dot product $u\cdot v \ce u_1v_1+\cdots+u_nv_n$
- is bilinear,
- gives a norm $\norm{u}^2_2 = \inner{u,u}$,
- is related to orthogonal projection.
See also: Geometric properties of linear algebra
Why in the world is it related to orthogonal projection?
Claim. If $u$ is a unit vector, then $u\cdot v$ is the (signed) length of the projection of $v$ on the line containing $u$.
Proof. This projection operation is linear in $v$, so it’s enough to prove that the projection of each basis vector $e_i$ is indeed $u \cdot e_i =u_i$. And since $u$ and $e_i$ are both unit vectors, by symmetry we can equivalently project $u$ onto $e_i$, which amounts to taking its $i\nth$ coordinate $u_i$.
General expression
More generally, let’s define the dot product $u \cdot v$ to be the product of the length of $u$ and of $v$’s orthogonal projection onto $u$ (with a minus sign if they point in opposite directions), and try to figure out how to compute it.
We can geometrically verify that it is symmetric
and linear in the second component
so it’s a bilinear operation.
This means that we can decompose it as
\[
u \cdot v = \sum_{i,j} u_iv_i (e_i \cdot e_j)
\]
where $e_i$ is the $i\nth$ basis vector. And it’s clear that $e_i \cdot e_j$ should be $1$ iff $i=j$ and $0$ otherwise, so we get
\[
u \cdot v = u_1v_1 + \cdots+ u_n v_n.
\]
TODO:
- I’m still somehow deeply unsatisfied with both these proofs.
- Why does the dot product not change if your rotate both vectors? It must be because the transpose of a rotation matrix is its own inverse, but why is that the case?