# What are differences between affine space and vector space?

I know smilar questions have been asked and I have looked at them but none of them seems to have satisfactory answer. I am reading the book a course in mathematics for student of physics vol. 1 by Paul Bamberg and Shlomo Sternberg. In Chapter 1 authors define affine space and writes:

The space $\Bbb{R}^2$ is an example of a vector space. The distinction between vector space $\Bbb{R}^2$ and affine space $A\Bbb{R}^2$ lies in the fact that in $\Bbb{R}^2$ the point (0,0) has a special significance ( it is the additive identity) and the addition of two vectors in $\Bbb{R}^2$ makes sense. These do not hold for $A\Bbb{R}^2$.

Edit:

How come $A\Bbb{R}^2$ has point (0,0) without special significance? and why the addition of two vectors in $A\Bbb{R}^2$ does not make sense? Please give concrete examples instead of abstract answers . I am a physics major and have done courses in Calculus, Linear Algebra and Complex Analysis.

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What is it precisely that you need explained? –  xyzzyz Aug 1 '14 at 13:59
@xyzzyz I have made the edit in response to your comment. –  user41451 Aug 1 '14 at 14:04
First, do you understand the definition of affine space that the authors have given? If so, can you distinguish between the notion of a vector space and the notion of an affine space? –  Zhen Lin Aug 1 '14 at 14:22

Consider the vector space $\mathbb{R}^3$. Inside $\mathbb{R}^3$ we can choose two planes, $P_1$ and $P_2$. The plane $P_1$ passes through the origin but the plane $P_2$ does not. It is a standard homework exercise in linear algebra to show that the $P_1$ is a sub-vector space of $\mathbb{R}^3$ but the plane $P_2$ is not. However, the plane $P_2$ resembles a $2$-dimensional vector space in many ways, primarily in that it exhibits a linear structure. In fact, $P_2$ is a classical example of an affine space.

$\,\,\,\,\,\,\,\,\,$

One defect of the plane $P_2$ is that it has no distinguished origin. One can artificially choose a point and redefine the algebraic operations in such a way to give it an origin, but that is not inherent to $P_2$. Another problem is that the sum of two vectors in $P_2$ is no longer in $P_2$. One can think of $AR^{2}$ as being modeled on this situation.

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The easiest way for me to tell the two structures apart is their axioms.

A vector space is an algebraic object with its characteristic operations, and an affine space is a group action on a set, specifically a vector space acting on a set faithfully and transitively.

Why do we say that the origin is no longer special in the affine space? The issue is that both $V$ and $X$ are usually written as $\Bbb R^n$, although we are thinking of each of the two copies of this in different ways. The deal is that the set $X=\Bbb R^n$ really doesn't distinguish any of its elements... they're all the same. But in the vector space $\Bbb R^n$, you can spot the origin right away, called out in the axioms.

Why do we say affine points can be subtracted but not added? That makes it seem like there are indeed operations within the affine space just like there are in the vector space, blurring the picture.

The reason is precisely because of transitivity: if $V$ acts on $X$ so that $X$ is an affine space (written additively), then for any $x,y\in X$, there is a $v$ such that $v + x = y$. I've written the group action additively here, but it is suggestive to rewrite this as $y-x = v$ and confuse the element $v$ of the vector space with an element of $X$.

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Can you give concrete example of affine space where (0,0) does not have special significance? –  user41451 Aug 1 '14 at 14:09
@user41451 In any affine space it has no special significance. That's the point of affine space (no pun intended): the axioms don't mention it. In every example of affine space the point the point $(0,0,\ldots, 0)$ has no special significance and cannot be distinguished from the abstract structure. –  Adam Hughes Aug 1 '14 at 14:10
@user41451 : Perhaps if you just look at it from the abstract view you'll see. Here, I am handing you a set $X$ and a vector space $V$ which acts on $X$ faithfully and transitively. Now: tell me which element of $x$ is "$0$" . (In fact, the whole idea of finding it is absurd, since $0$ is notation for something in another set.) –  rschwieb Aug 1 '14 at 14:11
@user41451 You wake up one day on an infinite flat desert. You have a compass, so you know what does it mean "2 meters north", or "one kilometer east". These notions make sense as the movement to make, or difference between two points. But how do you determine where you are? It's just sand everywhere. Nothing you can do or observe allows you to conclude "yup, that's the origin". –  Marcin Łoś Aug 1 '14 at 19:06

Consider an infinite sheet (of idealised paper, if you like). If it is blank, then there is absolutely no way to distinguish between any two points on the sheet. Nonetheless, if you do have two points on the sheet, you can measure the distance between them. And if there is a uniform magnetic field parallel to the sheet, then you can even measure the bearing from one point to another. Thus, given any point $P$ on the sheet, you can uniquely describe every other point on the sheet by its distance and bearing from $P$; and conversely, given any distance and bearing, there is a point with that distance and bearing from $P$. This is the situation that the notion of a 2-dimensional affine space is an abstraction of.

Now suppose we have marked a point $O$ on the sheet. Then we can "add" points $P$ and $Q$ on the sheet by drawing the usual parallelogram diagram. The result $P + Q$ of the "addition" depends on the choice of $O$ (and, of course, $P$ and $Q$), but nothing else. This is what the notion of a 2-dimensional vector space is an abstraction of.

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An example: Consider an $(m\times n)$ system of linear equations: $$\sum_{k=1}^n b_{ik}\>x_k=c_i\qquad(1\leq i\leq m)\ ,\tag{1}$$ where $d:=n-{\rm rank}(B)\geq1$, and ${\bf c}\ne{\bf 0}\in{\mathbb R}^m$. When this system has at least one solution ${\bf x}_p$ ($p$ for "particular") then the full set of solutions is a $d$-dimensional affine space $A\subset{\mathbb R}^n$. Two points in $A$ cannot be added to produce a new point in $A$, nor can points be scaled in $A$, and there is no distinguished point in $A$ that may serve as origin.

However you can say the following: Having found a point ${\bf x}_p\in A$ by whichever means you can declare this point as "origin" of $A$ and then introduce in $A$ coordinates as follows: The homogeneous system $$\sum_{k=1}^n b_{ik}\>x_k=0\qquad(1\leq i\leq m)$$ associated to $(1)$ has $d$ linearly independent solutions ${\bf f}_j\in{\mathbb R}^n$ $\>(1\leq j\leq d)$, and the set $A$ can then be written as $$A=\left\{{\bf x}_p+\sum_{j=1}^d y_j{\bf f}_j\ \Biggm|\ y_j\in{\mathbb R} \quad (1\leq j\leq d)\right\}\ .$$ The $y_j$ can then serve as coordinates in $A$, so that $A$ looks as it were a $d$-dimensional coordinate space. But note that "addition" in this space refers to the chosen point ${\bf x}_p$, and not to the origin of the base space ${\mathbb R}^n$.

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Vector spaces and Affine spaces are abstractions of different properties of Euclidean space. Like many abstractions, once abstracted they become more general.

A Vector space abstracts linearity/linear combinations. This involves the concept of a zero, scaling things up and down, and adding them to each other.

An Affine space abstracts the affine combinations. You can think of an affine combination as a weighted average, or a convex hull (if you limit the coefficients to be between 0 and 1).

As it turns out, you do not need a zero, nor do you need the concept of "scaling", nor do you need full on addition, in order to have a concept of weighted average and convex hull within a space.

Now, you can take your affine space $\mathbb {A}$ , pick any point $o$ from it, and talk about ${\mathbb A}-o$ as a vector space.

Mapping your $n$ dimensional affine space over $\mathbb {R}$ to $\mathbb{R}^n$ is in effect picking a point, and mapping it to a space with more structure than your original affine space. So you end up with the origin $o$ appearing special, but that is an artifact of your mapping.

If you look at the Earth, the lines of longitude have a zero point, but that zero point is arbitrary -- it has no meaning. The lines of longitude are an affine space. We measure them in degrees (or radians), and we have picked a zero, but other than it being useful to agree where the zero is, it isn't a special line.

The space of rotations around a circle, on the other hand, have a zero that is meaningful -- zero means you don't rotate. We measure them as a vector space.

The lines of longitude are measured as rotations away from our arbitrary point we assigned zero. But what matters about them is the ability to say how far apart two longitude are from each other, not any one line's absolute value.

If we where doing some math and it would be useful to move the zero of longitude, we are free to do so. But if we want to move the zero in the space of rotation (to say bending things 90 degrees) we are not nearly as free.

In general, your location is an affine space, as there is no special place, and scaling your location by a factor of 3 makes no sense, and adding two locations makes no sense -- but taking the average of two locations makes sense.

The (directed) distance between locations is a vector space. Saying something is twice as far as another distance makes sense, the "same place" (distance zero) makes sense, and adding two directed distances together makes sense.

And you can pick a spot and describe locations as the directed distance from that particular spot, but the spot picked was arbitrary, and if it would be useful to pick a different spot, you are free to.

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A subset $A$ of a vector space $X$ is affine if for any two points $x,y\in A$, the line $\ell$ through $x$ and $y$ is contained in $A$. That is, $A$ is affine if $$\alpha x+(1-\alpha)y\in A$$ for all $x,y\in A$ and $\alpha\in\mathbb{R}$. This definition is equivalent to the axiomatic ones which may obscure the idea for a first reading.

Example 1. of an affine space in $\mathbb{C}^m$ is the set of solutions to the equation $Ax=b$, where $A$ is an complex $n\times m$ matrix.

Example 2. A line in any vector space is affine.

Example 3. The intersection of affine sets in a vector space $X$ is also affine. Given a set $C\subset X$, the smallest affine set containing $C$ -denoted by $\operatorname{aff}( C )$- is the intersection of all affine sets in $X$ that contained $C$. It is easy to check that $$\operatorname{aff}( C )=\Big\{\sum^n_{k=1}\alpha_k x_k: n\in\mathbb{N},x_k\in C,\alpha_k\in\mathbb{R},\sum^n_{k=1}\alpha_k=1\Big\}.$$

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