What are differences between affine space and vector space?












63












$begingroup$


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 $ABbb{R}^2$ 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 $ABbb{R}^2$.




Please explain.



Edit:



How come $ABbb{R}^2$ has point (0,0) without special significance? and why the addition of two vectors in $ABbb{R}^2$ 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.










share|cite|improve this question











$endgroup$












  • $begingroup$
    What is it precisely that you need explained?
    $endgroup$
    – xyzzyz
    Aug 1 '14 at 13:59










  • $begingroup$
    @xyzzyz I have made the edit in response to your comment.
    $endgroup$
    – user41451
    Aug 1 '14 at 14:04










  • $begingroup$
    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?
    $endgroup$
    – Zhen Lin
    Aug 1 '14 at 14:22
















63












$begingroup$


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 $ABbb{R}^2$ 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 $ABbb{R}^2$.




Please explain.



Edit:



How come $ABbb{R}^2$ has point (0,0) without special significance? and why the addition of two vectors in $ABbb{R}^2$ 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.










share|cite|improve this question











$endgroup$












  • $begingroup$
    What is it precisely that you need explained?
    $endgroup$
    – xyzzyz
    Aug 1 '14 at 13:59










  • $begingroup$
    @xyzzyz I have made the edit in response to your comment.
    $endgroup$
    – user41451
    Aug 1 '14 at 14:04










  • $begingroup$
    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?
    $endgroup$
    – Zhen Lin
    Aug 1 '14 at 14:22














63












63








63


53



$begingroup$


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 $ABbb{R}^2$ 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 $ABbb{R}^2$.




Please explain.



Edit:



How come $ABbb{R}^2$ has point (0,0) without special significance? and why the addition of two vectors in $ABbb{R}^2$ 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.










share|cite|improve this question











$endgroup$




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 $ABbb{R}^2$ 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 $ABbb{R}^2$.




Please explain.



Edit:



How come $ABbb{R}^2$ has point (0,0) without special significance? and why the addition of two vectors in $ABbb{R}^2$ 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.







linear-algebra affine-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 1 '14 at 14:30









mle

1,62711228




1,62711228










asked Aug 1 '14 at 13:56









user41451user41451

420155




420155












  • $begingroup$
    What is it precisely that you need explained?
    $endgroup$
    – xyzzyz
    Aug 1 '14 at 13:59










  • $begingroup$
    @xyzzyz I have made the edit in response to your comment.
    $endgroup$
    – user41451
    Aug 1 '14 at 14:04










  • $begingroup$
    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?
    $endgroup$
    – Zhen Lin
    Aug 1 '14 at 14:22


















  • $begingroup$
    What is it precisely that you need explained?
    $endgroup$
    – xyzzyz
    Aug 1 '14 at 13:59










  • $begingroup$
    @xyzzyz I have made the edit in response to your comment.
    $endgroup$
    – user41451
    Aug 1 '14 at 14:04










  • $begingroup$
    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?
    $endgroup$
    – Zhen Lin
    Aug 1 '14 at 14:22
















$begingroup$
What is it precisely that you need explained?
$endgroup$
– xyzzyz
Aug 1 '14 at 13:59




$begingroup$
What is it precisely that you need explained?
$endgroup$
– xyzzyz
Aug 1 '14 at 13:59












$begingroup$
@xyzzyz I have made the edit in response to your comment.
$endgroup$
– user41451
Aug 1 '14 at 14:04




$begingroup$
@xyzzyz I have made the edit in response to your comment.
$endgroup$
– user41451
Aug 1 '14 at 14:04












$begingroup$
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?
$endgroup$
– Zhen Lin
Aug 1 '14 at 14:22




$begingroup$
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?
$endgroup$
– Zhen Lin
Aug 1 '14 at 14:22










7 Answers
7






active

oldest

votes


















62












$begingroup$

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.



$,,,,,,,,,$enter image description here



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.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    is there a version for dummies to understand affine space?
    $endgroup$
    – SuperUberDuper
    May 11 '16 at 11:40






  • 3




    $begingroup$
    I don't think so. This is the only "simple" (completely impossible to understand) explanation I found on the internet.
    $endgroup$
    – Marcus Johansson
    May 11 '16 at 21:21










  • $begingroup$
    Imagine a sheet of paper with two dots drawn on it. It doesn't make sense to add the dots, unless I give you another dot called the "origin" (in which case you draw lines from the origin to p1 and p2, and form their parallelogram). But even without an origin you can do things like find the point that is 75% of the way from p1 to p2. Without an origin you have an affine space; with it you have a vector space. The things you can meaningfully do are different in both cases.
    $endgroup$
    – A_P
    Oct 2 '18 at 14:05



















34












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @Elchanan Solomon Can you suggest any book that contains materials about the difference in linear and Affine spaces!
    $endgroup$
    – Noor Aslam
    Jul 30 '18 at 3:26



















22












$begingroup$

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 $x,yin X$, 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 $X$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can you give concrete example of affine space where (0,0) does not have special significance?
    $endgroup$
    – user41451
    Aug 1 '14 at 14:09






  • 1




    $begingroup$
    @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.
    $endgroup$
    – Adam Hughes
    Aug 1 '14 at 14:10








  • 3




    $begingroup$
    @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.)
    $endgroup$
    – rschwieb
    Aug 1 '14 at 14:11








  • 31




    $begingroup$
    @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".
    $endgroup$
    – Marcin Łoś
    Aug 1 '14 at 19:06






  • 2




    $begingroup$
    A nice physical analogy just come into my mind: Potentials; there is no meaning for electric potential by it's own, only a difference between two of them has sense and measurable, similarly for gravity, when we calculating work done due to a fallen apple, we don't care how high it was relative to earth level, only the traveled distance (difference between two positions) is relevant, this is the affine space, no special origin, and only difference makes sense.
    $endgroup$
    – TMS
    Jan 14 '16 at 15:04



















15












$begingroup$

An example: Consider an $(mtimes n)$ system of linear equations:
$$sum_{k=1}^n b_{ik}>x_k=c_iqquad(1leq ileq m) ,tag{1}$$
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 $Asubset{mathbb R}^n$. 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 ${bf x}_pin A$ by whichever means you can declare this point as "origin" of $A$ and then introduce in $A$ coordinates as follows: The homogeneous system
$$sum_{k=1}^n b_{ik}>x_k=0qquad(1leq ileq m)$$
associated to $(1)$ has $d$ linearly independent solutions ${bf f}_jin{mathbb R}^n$ $>(1leq jleq d)$, and the set $A$ can then be written as
$$A=left{{bf x}_p+sum_{j=1}^d y_j{bf f}_j Biggm| y_jin{mathbb R} quad (1leq jleq d)right} .$$
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 ${mathbb R}^n$.






share|cite|improve this answer









$endgroup$





















    9












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$





















      6












      $begingroup$

      I have read some intuitions on the difference, although they are fine to me, I suggest this probably simpler explanation, which is the one I would like to read. I hope it can help someone:



      Assuming you understand what a vector space is, we can define an affine space as follows:




      An affine space $A$ is a set of elements (called points) with a difference
      function. This difference is a binary function, which takes two points $p$ and
      $q$ (both in $A$) and yields an element (a vector) $v$ of a vector space $V$
      (for each unique $A$, there is an unique $V$, which is the vector space
      associated to $A$). We write $v=p-q$. Additionally, this difference function
      must ensure that, for any point $p$ in $V$, it holds $p-p=0$, where $0$ is the
      null vector of $V$.




      The first difference (which arises to me) between affine and vector space is that this affine space definition does not mention any origin point for the affine space (the affine space has no one), while each vector space has an origin (the null vector). In geometric terms, this implies that, if $v$ is a vector, we can define a line as the set of all vectors parallel to $v$, that is the set $tv$ for all reals $t$. This line always passes through the origin (for $t=0$).



      However, a single point in an affine space does not have an associated line as in a vector space. To build a line in an affine space, we need two points, $p$ and $q$. Then the line going through $p$ and $q$ is the set of points $p+t(q-p)$, for every real $t$. This need for two points comes from the lack of an origin in the affine space.






      share|cite|improve this answer











      $endgroup$





















        2












        $begingroup$

        A subset $A$ of a vector space $X$ is affine if for any two points $x,yin A$, the line $ell$ through $x$ and $y$ is contained in $A$. That is,
        $A$ is affine if
        $$alpha x+(1-alpha)yin A$$
        for all $x,yin A$ and $alphainmathbb{R}$.
        This definition is equivalent to the axiomatic ones which may obscure the idea at 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 $ntimes m$ 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 $Csubset X$, 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
        $$operatorname{aff}( C )=Big{sum^n_{k=1}alpha_k x_k:
        ninmathbb{N},x_kin C,alpha_kinmathbb{R},sum^n_{k=1}alpha_k=1Big}.$$






        share|cite|improve this answer











        $endgroup$












          protected by Saad Oct 2 '18 at 14:30



          Thank you for your interest in this question.
          Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).



          Would you like to answer one of these unanswered questions instead?














          7 Answers
          7






          active

          oldest

          votes








          7 Answers
          7






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          62












          $begingroup$

          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.



          $,,,,,,,,,$enter image description here



          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.






          share|cite|improve this answer











          $endgroup$









          • 2




            $begingroup$
            is there a version for dummies to understand affine space?
            $endgroup$
            – SuperUberDuper
            May 11 '16 at 11:40






          • 3




            $begingroup$
            I don't think so. This is the only "simple" (completely impossible to understand) explanation I found on the internet.
            $endgroup$
            – Marcus Johansson
            May 11 '16 at 21:21










          • $begingroup$
            Imagine a sheet of paper with two dots drawn on it. It doesn't make sense to add the dots, unless I give you another dot called the "origin" (in which case you draw lines from the origin to p1 and p2, and form their parallelogram). But even without an origin you can do things like find the point that is 75% of the way from p1 to p2. Without an origin you have an affine space; with it you have a vector space. The things you can meaningfully do are different in both cases.
            $endgroup$
            – A_P
            Oct 2 '18 at 14:05
















          62












          $begingroup$

          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.



          $,,,,,,,,,$enter image description here



          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.






          share|cite|improve this answer











          $endgroup$









          • 2




            $begingroup$
            is there a version for dummies to understand affine space?
            $endgroup$
            – SuperUberDuper
            May 11 '16 at 11:40






          • 3




            $begingroup$
            I don't think so. This is the only "simple" (completely impossible to understand) explanation I found on the internet.
            $endgroup$
            – Marcus Johansson
            May 11 '16 at 21:21










          • $begingroup$
            Imagine a sheet of paper with two dots drawn on it. It doesn't make sense to add the dots, unless I give you another dot called the "origin" (in which case you draw lines from the origin to p1 and p2, and form their parallelogram). But even without an origin you can do things like find the point that is 75% of the way from p1 to p2. Without an origin you have an affine space; with it you have a vector space. The things you can meaningfully do are different in both cases.
            $endgroup$
            – A_P
            Oct 2 '18 at 14:05














          62












          62








          62





          $begingroup$

          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.



          $,,,,,,,,,$enter image description here



          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.






          share|cite|improve this answer











          $endgroup$



          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.



          $,,,,,,,,,$enter image description here



          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 1 '14 at 14:17

























          answered Aug 1 '14 at 14:12









          Elchanan SolomonElchanan Solomon

          21.8k54277




          21.8k54277








          • 2




            $begingroup$
            is there a version for dummies to understand affine space?
            $endgroup$
            – SuperUberDuper
            May 11 '16 at 11:40






          • 3




            $begingroup$
            I don't think so. This is the only "simple" (completely impossible to understand) explanation I found on the internet.
            $endgroup$
            – Marcus Johansson
            May 11 '16 at 21:21










          • $begingroup$
            Imagine a sheet of paper with two dots drawn on it. It doesn't make sense to add the dots, unless I give you another dot called the "origin" (in which case you draw lines from the origin to p1 and p2, and form their parallelogram). But even without an origin you can do things like find the point that is 75% of the way from p1 to p2. Without an origin you have an affine space; with it you have a vector space. The things you can meaningfully do are different in both cases.
            $endgroup$
            – A_P
            Oct 2 '18 at 14:05














          • 2




            $begingroup$
            is there a version for dummies to understand affine space?
            $endgroup$
            – SuperUberDuper
            May 11 '16 at 11:40






          • 3




            $begingroup$
            I don't think so. This is the only "simple" (completely impossible to understand) explanation I found on the internet.
            $endgroup$
            – Marcus Johansson
            May 11 '16 at 21:21










          • $begingroup$
            Imagine a sheet of paper with two dots drawn on it. It doesn't make sense to add the dots, unless I give you another dot called the "origin" (in which case you draw lines from the origin to p1 and p2, and form their parallelogram). But even without an origin you can do things like find the point that is 75% of the way from p1 to p2. Without an origin you have an affine space; with it you have a vector space. The things you can meaningfully do are different in both cases.
            $endgroup$
            – A_P
            Oct 2 '18 at 14:05








          2




          2




          $begingroup$
          is there a version for dummies to understand affine space?
          $endgroup$
          – SuperUberDuper
          May 11 '16 at 11:40




          $begingroup$
          is there a version for dummies to understand affine space?
          $endgroup$
          – SuperUberDuper
          May 11 '16 at 11:40




          3




          3




          $begingroup$
          I don't think so. This is the only "simple" (completely impossible to understand) explanation I found on the internet.
          $endgroup$
          – Marcus Johansson
          May 11 '16 at 21:21




          $begingroup$
          I don't think so. This is the only "simple" (completely impossible to understand) explanation I found on the internet.
          $endgroup$
          – Marcus Johansson
          May 11 '16 at 21:21












          $begingroup$
          Imagine a sheet of paper with two dots drawn on it. It doesn't make sense to add the dots, unless I give you another dot called the "origin" (in which case you draw lines from the origin to p1 and p2, and form their parallelogram). But even without an origin you can do things like find the point that is 75% of the way from p1 to p2. Without an origin you have an affine space; with it you have a vector space. The things you can meaningfully do are different in both cases.
          $endgroup$
          – A_P
          Oct 2 '18 at 14:05




          $begingroup$
          Imagine a sheet of paper with two dots drawn on it. It doesn't make sense to add the dots, unless I give you another dot called the "origin" (in which case you draw lines from the origin to p1 and p2, and form their parallelogram). But even without an origin you can do things like find the point that is 75% of the way from p1 to p2. Without an origin you have an affine space; with it you have a vector space. The things you can meaningfully do are different in both cases.
          $endgroup$
          – A_P
          Oct 2 '18 at 14:05











          34












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            @Elchanan Solomon Can you suggest any book that contains materials about the difference in linear and Affine spaces!
            $endgroup$
            – Noor Aslam
            Jul 30 '18 at 3:26
















          34












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            @Elchanan Solomon Can you suggest any book that contains materials about the difference in linear and Affine spaces!
            $endgroup$
            – Noor Aslam
            Jul 30 '18 at 3:26














          34












          34








          34





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 1 '14 at 14:35









          Zhen LinZhen Lin

          60.2k4108220




          60.2k4108220












          • $begingroup$
            @Elchanan Solomon Can you suggest any book that contains materials about the difference in linear and Affine spaces!
            $endgroup$
            – Noor Aslam
            Jul 30 '18 at 3:26


















          • $begingroup$
            @Elchanan Solomon Can you suggest any book that contains materials about the difference in linear and Affine spaces!
            $endgroup$
            – Noor Aslam
            Jul 30 '18 at 3:26
















          $begingroup$
          @Elchanan Solomon Can you suggest any book that contains materials about the difference in linear and Affine spaces!
          $endgroup$
          – Noor Aslam
          Jul 30 '18 at 3:26




          $begingroup$
          @Elchanan Solomon Can you suggest any book that contains materials about the difference in linear and Affine spaces!
          $endgroup$
          – Noor Aslam
          Jul 30 '18 at 3:26











          22












          $begingroup$

          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 $x,yin X$, 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 $X$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you give concrete example of affine space where (0,0) does not have special significance?
            $endgroup$
            – user41451
            Aug 1 '14 at 14:09






          • 1




            $begingroup$
            @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.
            $endgroup$
            – Adam Hughes
            Aug 1 '14 at 14:10








          • 3




            $begingroup$
            @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.)
            $endgroup$
            – rschwieb
            Aug 1 '14 at 14:11








          • 31




            $begingroup$
            @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".
            $endgroup$
            – Marcin Łoś
            Aug 1 '14 at 19:06






          • 2




            $begingroup$
            A nice physical analogy just come into my mind: Potentials; there is no meaning for electric potential by it's own, only a difference between two of them has sense and measurable, similarly for gravity, when we calculating work done due to a fallen apple, we don't care how high it was relative to earth level, only the traveled distance (difference between two positions) is relevant, this is the affine space, no special origin, and only difference makes sense.
            $endgroup$
            – TMS
            Jan 14 '16 at 15:04
















          22












          $begingroup$

          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 $x,yin X$, 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 $X$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you give concrete example of affine space where (0,0) does not have special significance?
            $endgroup$
            – user41451
            Aug 1 '14 at 14:09






          • 1




            $begingroup$
            @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.
            $endgroup$
            – Adam Hughes
            Aug 1 '14 at 14:10








          • 3




            $begingroup$
            @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.)
            $endgroup$
            – rschwieb
            Aug 1 '14 at 14:11








          • 31




            $begingroup$
            @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".
            $endgroup$
            – Marcin Łoś
            Aug 1 '14 at 19:06






          • 2




            $begingroup$
            A nice physical analogy just come into my mind: Potentials; there is no meaning for electric potential by it's own, only a difference between two of them has sense and measurable, similarly for gravity, when we calculating work done due to a fallen apple, we don't care how high it was relative to earth level, only the traveled distance (difference between two positions) is relevant, this is the affine space, no special origin, and only difference makes sense.
            $endgroup$
            – TMS
            Jan 14 '16 at 15:04














          22












          22








          22





          $begingroup$

          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 $x,yin X$, 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 $X$.






          share|cite|improve this answer











          $endgroup$



          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 $x,yin X$, 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 $X$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 1 '14 at 14:15

























          answered Aug 1 '14 at 14:04









          rschwiebrschwieb

          105k12100245




          105k12100245












          • $begingroup$
            Can you give concrete example of affine space where (0,0) does not have special significance?
            $endgroup$
            – user41451
            Aug 1 '14 at 14:09






          • 1




            $begingroup$
            @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.
            $endgroup$
            – Adam Hughes
            Aug 1 '14 at 14:10








          • 3




            $begingroup$
            @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.)
            $endgroup$
            – rschwieb
            Aug 1 '14 at 14:11








          • 31




            $begingroup$
            @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".
            $endgroup$
            – Marcin Łoś
            Aug 1 '14 at 19:06






          • 2




            $begingroup$
            A nice physical analogy just come into my mind: Potentials; there is no meaning for electric potential by it's own, only a difference between two of them has sense and measurable, similarly for gravity, when we calculating work done due to a fallen apple, we don't care how high it was relative to earth level, only the traveled distance (difference between two positions) is relevant, this is the affine space, no special origin, and only difference makes sense.
            $endgroup$
            – TMS
            Jan 14 '16 at 15:04


















          • $begingroup$
            Can you give concrete example of affine space where (0,0) does not have special significance?
            $endgroup$
            – user41451
            Aug 1 '14 at 14:09






          • 1




            $begingroup$
            @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.
            $endgroup$
            – Adam Hughes
            Aug 1 '14 at 14:10








          • 3




            $begingroup$
            @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.)
            $endgroup$
            – rschwieb
            Aug 1 '14 at 14:11








          • 31




            $begingroup$
            @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".
            $endgroup$
            – Marcin Łoś
            Aug 1 '14 at 19:06






          • 2




            $begingroup$
            A nice physical analogy just come into my mind: Potentials; there is no meaning for electric potential by it's own, only a difference between two of them has sense and measurable, similarly for gravity, when we calculating work done due to a fallen apple, we don't care how high it was relative to earth level, only the traveled distance (difference between two positions) is relevant, this is the affine space, no special origin, and only difference makes sense.
            $endgroup$
            – TMS
            Jan 14 '16 at 15:04
















          $begingroup$
          Can you give concrete example of affine space where (0,0) does not have special significance?
          $endgroup$
          – user41451
          Aug 1 '14 at 14:09




          $begingroup$
          Can you give concrete example of affine space where (0,0) does not have special significance?
          $endgroup$
          – user41451
          Aug 1 '14 at 14:09




          1




          1




          $begingroup$
          @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.
          $endgroup$
          – Adam Hughes
          Aug 1 '14 at 14:10






          $begingroup$
          @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.
          $endgroup$
          – Adam Hughes
          Aug 1 '14 at 14:10






          3




          3




          $begingroup$
          @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.)
          $endgroup$
          – rschwieb
          Aug 1 '14 at 14:11






          $begingroup$
          @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.)
          $endgroup$
          – rschwieb
          Aug 1 '14 at 14:11






          31




          31




          $begingroup$
          @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".
          $endgroup$
          – Marcin Łoś
          Aug 1 '14 at 19:06




          $begingroup$
          @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".
          $endgroup$
          – Marcin Łoś
          Aug 1 '14 at 19:06




          2




          2




          $begingroup$
          A nice physical analogy just come into my mind: Potentials; there is no meaning for electric potential by it's own, only a difference between two of them has sense and measurable, similarly for gravity, when we calculating work done due to a fallen apple, we don't care how high it was relative to earth level, only the traveled distance (difference between two positions) is relevant, this is the affine space, no special origin, and only difference makes sense.
          $endgroup$
          – TMS
          Jan 14 '16 at 15:04




          $begingroup$
          A nice physical analogy just come into my mind: Potentials; there is no meaning for electric potential by it's own, only a difference between two of them has sense and measurable, similarly for gravity, when we calculating work done due to a fallen apple, we don't care how high it was relative to earth level, only the traveled distance (difference between two positions) is relevant, this is the affine space, no special origin, and only difference makes sense.
          $endgroup$
          – TMS
          Jan 14 '16 at 15:04











          15












          $begingroup$

          An example: Consider an $(mtimes n)$ system of linear equations:
          $$sum_{k=1}^n b_{ik}>x_k=c_iqquad(1leq ileq m) ,tag{1}$$
          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 $Asubset{mathbb R}^n$. 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 ${bf x}_pin A$ by whichever means you can declare this point as "origin" of $A$ and then introduce in $A$ coordinates as follows: The homogeneous system
          $$sum_{k=1}^n b_{ik}>x_k=0qquad(1leq ileq m)$$
          associated to $(1)$ has $d$ linearly independent solutions ${bf f}_jin{mathbb R}^n$ $>(1leq jleq d)$, and the set $A$ can then be written as
          $$A=left{{bf x}_p+sum_{j=1}^d y_j{bf f}_j Biggm| y_jin{mathbb R} quad (1leq jleq d)right} .$$
          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 ${mathbb R}^n$.






          share|cite|improve this answer









          $endgroup$


















            15












            $begingroup$

            An example: Consider an $(mtimes n)$ system of linear equations:
            $$sum_{k=1}^n b_{ik}>x_k=c_iqquad(1leq ileq m) ,tag{1}$$
            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 $Asubset{mathbb R}^n$. 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 ${bf x}_pin A$ by whichever means you can declare this point as "origin" of $A$ and then introduce in $A$ coordinates as follows: The homogeneous system
            $$sum_{k=1}^n b_{ik}>x_k=0qquad(1leq ileq m)$$
            associated to $(1)$ has $d$ linearly independent solutions ${bf f}_jin{mathbb R}^n$ $>(1leq jleq d)$, and the set $A$ can then be written as
            $$A=left{{bf x}_p+sum_{j=1}^d y_j{bf f}_j Biggm| y_jin{mathbb R} quad (1leq jleq d)right} .$$
            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 ${mathbb R}^n$.






            share|cite|improve this answer









            $endgroup$
















              15












              15








              15





              $begingroup$

              An example: Consider an $(mtimes n)$ system of linear equations:
              $$sum_{k=1}^n b_{ik}>x_k=c_iqquad(1leq ileq m) ,tag{1}$$
              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 $Asubset{mathbb R}^n$. 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 ${bf x}_pin A$ by whichever means you can declare this point as "origin" of $A$ and then introduce in $A$ coordinates as follows: The homogeneous system
              $$sum_{k=1}^n b_{ik}>x_k=0qquad(1leq ileq m)$$
              associated to $(1)$ has $d$ linearly independent solutions ${bf f}_jin{mathbb R}^n$ $>(1leq jleq d)$, and the set $A$ can then be written as
              $$A=left{{bf x}_p+sum_{j=1}^d y_j{bf f}_j Biggm| y_jin{mathbb R} quad (1leq jleq d)right} .$$
              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 ${mathbb R}^n$.






              share|cite|improve this answer









              $endgroup$



              An example: Consider an $(mtimes n)$ system of linear equations:
              $$sum_{k=1}^n b_{ik}>x_k=c_iqquad(1leq ileq m) ,tag{1}$$
              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 $Asubset{mathbb R}^n$. 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 ${bf x}_pin A$ by whichever means you can declare this point as "origin" of $A$ and then introduce in $A$ coordinates as follows: The homogeneous system
              $$sum_{k=1}^n b_{ik}>x_k=0qquad(1leq ileq m)$$
              associated to $(1)$ has $d$ linearly independent solutions ${bf f}_jin{mathbb R}^n$ $>(1leq jleq d)$, and the set $A$ can then be written as
              $$A=left{{bf x}_p+sum_{j=1}^d y_j{bf f}_j Biggm| y_jin{mathbb R} quad (1leq jleq d)right} .$$
              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 ${mathbb R}^n$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Aug 1 '14 at 14:31









              Christian BlatterChristian Blatter

              172k7113326




              172k7113326























                  9












                  $begingroup$

                  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.






                  share|cite|improve this answer









                  $endgroup$


















                    9












                    $begingroup$

                    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.






                    share|cite|improve this answer









                    $endgroup$
















                      9












                      9








                      9





                      $begingroup$

                      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.






                      share|cite|improve this answer









                      $endgroup$



                      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.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Aug 5 '14 at 14:05









                      YakkYakk

                      81657




                      81657























                          6












                          $begingroup$

                          I have read some intuitions on the difference, although they are fine to me, I suggest this probably simpler explanation, which is the one I would like to read. I hope it can help someone:



                          Assuming you understand what a vector space is, we can define an affine space as follows:




                          An affine space $A$ is a set of elements (called points) with a difference
                          function. This difference is a binary function, which takes two points $p$ and
                          $q$ (both in $A$) and yields an element (a vector) $v$ of a vector space $V$
                          (for each unique $A$, there is an unique $V$, which is the vector space
                          associated to $A$). We write $v=p-q$. Additionally, this difference function
                          must ensure that, for any point $p$ in $V$, it holds $p-p=0$, where $0$ is the
                          null vector of $V$.




                          The first difference (which arises to me) between affine and vector space is that this affine space definition does not mention any origin point for the affine space (the affine space has no one), while each vector space has an origin (the null vector). In geometric terms, this implies that, if $v$ is a vector, we can define a line as the set of all vectors parallel to $v$, that is the set $tv$ for all reals $t$. This line always passes through the origin (for $t=0$).



                          However, a single point in an affine space does not have an associated line as in a vector space. To build a line in an affine space, we need two points, $p$ and $q$. Then the line going through $p$ and $q$ is the set of points $p+t(q-p)$, for every real $t$. This need for two points comes from the lack of an origin in the affine space.






                          share|cite|improve this answer











                          $endgroup$


















                            6












                            $begingroup$

                            I have read some intuitions on the difference, although they are fine to me, I suggest this probably simpler explanation, which is the one I would like to read. I hope it can help someone:



                            Assuming you understand what a vector space is, we can define an affine space as follows:




                            An affine space $A$ is a set of elements (called points) with a difference
                            function. This difference is a binary function, which takes two points $p$ and
                            $q$ (both in $A$) and yields an element (a vector) $v$ of a vector space $V$
                            (for each unique $A$, there is an unique $V$, which is the vector space
                            associated to $A$). We write $v=p-q$. Additionally, this difference function
                            must ensure that, for any point $p$ in $V$, it holds $p-p=0$, where $0$ is the
                            null vector of $V$.




                            The first difference (which arises to me) between affine and vector space is that this affine space definition does not mention any origin point for the affine space (the affine space has no one), while each vector space has an origin (the null vector). In geometric terms, this implies that, if $v$ is a vector, we can define a line as the set of all vectors parallel to $v$, that is the set $tv$ for all reals $t$. This line always passes through the origin (for $t=0$).



                            However, a single point in an affine space does not have an associated line as in a vector space. To build a line in an affine space, we need two points, $p$ and $q$. Then the line going through $p$ and $q$ is the set of points $p+t(q-p)$, for every real $t$. This need for two points comes from the lack of an origin in the affine space.






                            share|cite|improve this answer











                            $endgroup$
















                              6












                              6








                              6





                              $begingroup$

                              I have read some intuitions on the difference, although they are fine to me, I suggest this probably simpler explanation, which is the one I would like to read. I hope it can help someone:



                              Assuming you understand what a vector space is, we can define an affine space as follows:




                              An affine space $A$ is a set of elements (called points) with a difference
                              function. This difference is a binary function, which takes two points $p$ and
                              $q$ (both in $A$) and yields an element (a vector) $v$ of a vector space $V$
                              (for each unique $A$, there is an unique $V$, which is the vector space
                              associated to $A$). We write $v=p-q$. Additionally, this difference function
                              must ensure that, for any point $p$ in $V$, it holds $p-p=0$, where $0$ is the
                              null vector of $V$.




                              The first difference (which arises to me) between affine and vector space is that this affine space definition does not mention any origin point for the affine space (the affine space has no one), while each vector space has an origin (the null vector). In geometric terms, this implies that, if $v$ is a vector, we can define a line as the set of all vectors parallel to $v$, that is the set $tv$ for all reals $t$. This line always passes through the origin (for $t=0$).



                              However, a single point in an affine space does not have an associated line as in a vector space. To build a line in an affine space, we need two points, $p$ and $q$. Then the line going through $p$ and $q$ is the set of points $p+t(q-p)$, for every real $t$. This need for two points comes from the lack of an origin in the affine space.






                              share|cite|improve this answer











                              $endgroup$



                              I have read some intuitions on the difference, although they are fine to me, I suggest this probably simpler explanation, which is the one I would like to read. I hope it can help someone:



                              Assuming you understand what a vector space is, we can define an affine space as follows:




                              An affine space $A$ is a set of elements (called points) with a difference
                              function. This difference is a binary function, which takes two points $p$ and
                              $q$ (both in $A$) and yields an element (a vector) $v$ of a vector space $V$
                              (for each unique $A$, there is an unique $V$, which is the vector space
                              associated to $A$). We write $v=p-q$. Additionally, this difference function
                              must ensure that, for any point $p$ in $V$, it holds $p-p=0$, where $0$ is the
                              null vector of $V$.




                              The first difference (which arises to me) between affine and vector space is that this affine space definition does not mention any origin point for the affine space (the affine space has no one), while each vector space has an origin (the null vector). In geometric terms, this implies that, if $v$ is a vector, we can define a line as the set of all vectors parallel to $v$, that is the set $tv$ for all reals $t$. This line always passes through the origin (for $t=0$).



                              However, a single point in an affine space does not have an associated line as in a vector space. To build a line in an affine space, we need two points, $p$ and $q$. Then the line going through $p$ and $q$ is the set of points $p+t(q-p)$, for every real $t$. This need for two points comes from the lack of an origin in the affine space.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Dec 23 '15 at 10:39









                              Carlos U.A.

                              31




                              31










                              answered Dec 23 '15 at 9:57









                              Carlos UACarlos UA

                              6111




                              6111























                                  2












                                  $begingroup$

                                  A subset $A$ of a vector space $X$ is affine if for any two points $x,yin A$, the line $ell$ through $x$ and $y$ is contained in $A$. That is,
                                  $A$ is affine if
                                  $$alpha x+(1-alpha)yin A$$
                                  for all $x,yin A$ and $alphainmathbb{R}$.
                                  This definition is equivalent to the axiomatic ones which may obscure the idea at 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 $ntimes m$ 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 $Csubset X$, 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
                                  $$operatorname{aff}( C )=Big{sum^n_{k=1}alpha_k x_k:
                                  ninmathbb{N},x_kin C,alpha_kinmathbb{R},sum^n_{k=1}alpha_k=1Big}.$$






                                  share|cite|improve this answer











                                  $endgroup$


















                                    2












                                    $begingroup$

                                    A subset $A$ of a vector space $X$ is affine if for any two points $x,yin A$, the line $ell$ through $x$ and $y$ is contained in $A$. That is,
                                    $A$ is affine if
                                    $$alpha x+(1-alpha)yin A$$
                                    for all $x,yin A$ and $alphainmathbb{R}$.
                                    This definition is equivalent to the axiomatic ones which may obscure the idea at 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 $ntimes m$ 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 $Csubset X$, 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
                                    $$operatorname{aff}( C )=Big{sum^n_{k=1}alpha_k x_k:
                                    ninmathbb{N},x_kin C,alpha_kinmathbb{R},sum^n_{k=1}alpha_k=1Big}.$$






                                    share|cite|improve this answer











                                    $endgroup$
















                                      2












                                      2








                                      2





                                      $begingroup$

                                      A subset $A$ of a vector space $X$ is affine if for any two points $x,yin A$, the line $ell$ through $x$ and $y$ is contained in $A$. That is,
                                      $A$ is affine if
                                      $$alpha x+(1-alpha)yin A$$
                                      for all $x,yin A$ and $alphainmathbb{R}$.
                                      This definition is equivalent to the axiomatic ones which may obscure the idea at 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 $ntimes m$ 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 $Csubset X$, 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
                                      $$operatorname{aff}( C )=Big{sum^n_{k=1}alpha_k x_k:
                                      ninmathbb{N},x_kin C,alpha_kinmathbb{R},sum^n_{k=1}alpha_k=1Big}.$$






                                      share|cite|improve this answer











                                      $endgroup$



                                      A subset $A$ of a vector space $X$ is affine if for any two points $x,yin A$, the line $ell$ through $x$ and $y$ is contained in $A$. That is,
                                      $A$ is affine if
                                      $$alpha x+(1-alpha)yin A$$
                                      for all $x,yin A$ and $alphainmathbb{R}$.
                                      This definition is equivalent to the axiomatic ones which may obscure the idea at 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 $ntimes m$ 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 $Csubset X$, 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
                                      $$operatorname{aff}( C )=Big{sum^n_{k=1}alpha_k x_k:
                                      ninmathbb{N},x_kin C,alpha_kinmathbb{R},sum^n_{k=1}alpha_k=1Big}.$$







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Nov 30 '18 at 18:35

























                                      answered Jun 3 '15 at 3:24









                                      Oliver DiazOliver Diaz

                                      30216




                                      30216

















                                          protected by Saad Oct 2 '18 at 14:30



                                          Thank you for your interest in this question.
                                          Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).



                                          Would you like to answer one of these unanswered questions instead?



                                          Popular posts from this blog

                                          Ellipse (mathématiques)

                                          Quarter-circle Tiles

                                          Mont Emei