#math/linear-algebra #review
An affine hull is the smallest flat geometric object that contains a set of points.
- One point $v_{1}$, its affine hull is that single point.
- Two points $v_{1},v_{2}$, its affine hull is a line through the two points.
- Three non-colinear points $v_{1},v_{2},v_{3}$ is a plane containing the three.
The affine hull of a set of points consists of all linear combinations of those points whose coefficients sum to $1$. Precisely, given vectors $v_{1}, \ldots{}, v_{n}$ the affine hull of these points is given by
$
\operatorname{aff}(v_1,\ldots,v_n)
=
\left\{
\sum_{i=1}^{n}\lambda_i v_i
:
\lambda_i\in\mathbb{R},\
\sum_{i=1}^{n}\lambda_i=1
\right\}.
$
**Case 1:**
Take the standard basis vectors $e_{1}=(1,0),e_{2}=(0,1)$. Their affine hull is the line $x+y=1$^[1]. Choose
$
-w=\left(\frac12,\frac12\right)\in\operatorname{aff}(e_1,e_2),
\qquad
w=\left(-\frac12,-\frac12\right).
$
Adding $w$ translates the entire affine line so that the point $-w$ moves to the origin:
$
(-w)+w=0,
\qquad
e_1+w=\left(\frac12,-\frac12\right),
\qquad
e_2+w=\left(-\frac12,\frac12\right).
$
```tikz
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}[
scale=1.8,
>=stealth,
line cap=round,
line join=round,
every node/.style={font=\small}
]
% Original affine line through the standard basis vectors
\begin{scope}
\node at (0.5,1.52) {$\mathrm{aff}(e_1,e_2):\ x+y=1$};
\draw[->,black!30] (-0.35,0) -- (1.45,0) node[right] {$x$};
\draw[->,black!30] (0,-0.35) -- (0,1.35) node[above] {$y$};
\draw[very thick,blue,<->] (-0.25,1.25) -- (1.25,-0.25);
\draw[->,thick,teal!80!black] (0,0) -- (1,0)
node[midway,below] {$e_1$};
\draw[->,thick,teal!80!black] (0,0) -- (0,1)
node[midway,left] {$e_2$};
\fill[blue] (1,0) circle (1.4pt);
\fill[blue] (0,1) circle (1.4pt);
\fill[orange!90!black] (0.5,0.5) circle (1.8pt)
node[above right] {$-w$};
\draw[->,thick,dashed,orange!90!black] (0.5,0.5) -- (0,0)
node[midway,above left] {$w$};
\fill (0,0) circle (1.3pt) node[below left] {$0$};
\end{scope}
% Translation arrow
\draw[->,ultra thick,orange!90!black] (1.55,0.55) -- (2.45,0.55)
node[midway,above] {$+w$};
% The translated line, now a linear subspace through the origin
\begin{scope}[shift={(3.55,0)}]
\node at (0,1.52) {$\mathrm{aff}(e_1+w,e_2+w):\ x+y=0$};
\draw[->,black!30] (-1,0) -- (1,0) node[right] {$x$};
\draw[->,black!30] (0,-0.9) -- (0,1.1) node[above] {$y$};
\draw[very thick,blue,<->] (-0.82,0.82) -- (0.82,-0.82);
\fill[teal!80!black] (0.5,-0.5) circle (1.8pt)
node[below right] {$e_1+w$};
\fill[teal!80!black] (-0.5,0.5) circle (1.8pt)
node[above left] {$e_2+w$};
\fill[orange!90!black] (0,0) circle (1.9pt)
node[above right] {$(-w)+w=0$};
\end{scope}
\end{tikzpicture}
\end{document}
```
So, $\text{dim span}(e_{1}+w,e_{2}+w)=1$.
**Case 2:**
This makes an interesting appearance in [[2C3 - Problem Set|Linear Algebra Done Right - 2C3 Problem 9]], where the $\text{dim }\text{span}(v_{1}+w, \ldots,v_{m}+w)=m-1$ when $-w\in \text{aff}(v_{1}, \ldots{}, v_{m})$.
---
[1]: Given vectors $v_{1},v_{2}\in \mathbb{R}^{n}$ the line that passes through them is $\{av_{1}+bv_{2}:a,b\in \mathbb{R} \text{ and }a+b=1\}$. Alternatively, you can write $\{ (1-t)v_{1}+tv_{2} : t\in \mathbb{R}\}$.