9
$\begingroup$

I have been thinking recently about the transition from abstract mathematical objects and vector spaces to visual represenations of these objects. In particular, I was thinking about our typical representation of the vector space $\mathbb{R}^2$ consisting of perpendicular axes. Before reading further, please pretend that you do not have any visual intuition of what perpendicular or orthogonal means, but rather that all you have is the formal, abstract definition of the vector space $\mathbb{R}^2$ with its typical operations and metric.

With respect to the typical scalar dot product, the standard basis vectors $(1,0)$ and $(0,1)$ are obviously orthogonal, since their dot product gives zero.

Before we continue, consider the definition of a circle $C$ with radius $r$ and center $(a,b)$. We define this as the set of all points $(x,y)$ such that the distance with respect to our (Euclidian) metric between $(a,b)$ and $(x,y)$ is $r$. Symbolically,

$$\displaystyle C:=\{ \left(x,y\right)\in\mathbb{R}^2: \left(x-a\right)^2+\left(y-b\right)^2=r^2 \}$$

Suppose we are now thinking of a way to give objects (such as our circle) visual representations, and let us try to do this using two number lines that intersect at some point we shall call the origin. One way to do this would be use axes that are perpendicular to each other according to our typical, geometric notion of "perpendicular". Another way could be tilting the $y$-axis a little bit. Below is a graphic of the circle $C$ visualized in both these coordainte systems with $(0,1)$ in blue and $(1,0)$ in red:

A graphic showing the same set C graphed in different coordinate systems

As we can see in (a), no matter between which point of the circle and the origin you draw a vector, it will always look equally long. In (b), it seems that even though the distance between any point of the circle should be the same (and it is, abstractly), any vector you draw will not always look equally long. Indeed, a vector which looks longer than another could actually be shorter or longer or the same length according to our metric. So in this representation, it seems we can no longer compare lengths visually anymore, which is a huge downside.

Is this a good argument (and perhaps even the reason) for why the axes should be perpendicular in our typical, geometric understanding of perpendicular, even though the abstract vector space itself tells us nothing about how "perpendicular" looks?

$\endgroup$
11
  • 2
    $\begingroup$ seems to me that once you use "Euclidian" distance, you are committed to ortogonality. $\endgroup$ Commented Aug 31 at 15:34
  • 1
    $\begingroup$ @Anna: You might search (and/or ask History of Science and Math SE for help searching) for what source docs indicate René Descartes himself was thinking when he invented (what we call) Cartesian coordinates. See also Pierre de Fermat and Nicole Oresme. Also, as Wikipedia notes, "The concept of using a pair of axes was introduced later, after Descartes' La Géométrie was translated into Latin in 1649 by Frans van Schooten and his students", so contemporary(ish) commentary is likely also insightful. $\endgroup$ Commented Aug 31 at 16:43
  • 3
    $\begingroup$ This doesn't really address your concern (axes angle still involved), but it's worth mentioning that in older texts coordinate systems were usually studied more generally, where the axes were not necessarily perpendicular, although perpendicular axes received the greatest amount of attention. Google "oblique coordinates", and also look at older (roughly, anything published before the late 1920s) analytic geometry texts (most are freely available at Google Books and/or Internet Archive), some of which are listed in this Mathematics SE answer. $\endgroup$ Commented Aug 31 at 18:18
  • 5
    $\begingroup$ If we still have visual intuition of distance, we would like the intuitive distance in our representation to correspond to the actual distance, and that forces perpendicular axes. $\endgroup$ Commented Sep 1 at 9:05
  • 2
    $\begingroup$ The whole point of drawing things in mathematics is precisely to draw on our pre-existing spatial intuition. So, the question "if we forget about our pre-existing spatial intuition, why should we draw things this way and not that way" does not make a whole lot of sense to me. $\endgroup$ Commented Sep 1 at 10:01

4 Answers 4

Reset to default
15
$\begingroup$

Not an answer, but a frame change. You ask about

transition from abstract mathematical objects and vector spaces to visual representations of these objects.

That's not how mathematics works. It doesn't start with formal definitions. The definitions that turn out to be useful (like vector spaces with dot products) are invented to capture preexisting intuition - in this case the visual geometric idea of perpendicularity. So you should never need to invent the intuition in order to understand the abstraction.

All too often an abstract algebra text or class moves too quickly to formality, before the reader/student has a chance to absorb enough examples to appreciate the abstraction.

$\endgroup$
1
  • 2
    $\begingroup$ Thank you for your answer. I understand that in real life, we often move from intuition to formal abstractions instead of the other way around. I simply find it interesting to think about going the other way (my real life intuition of the physical world is HORRIBLE). $\endgroup$ Commented Sep 1 at 13:27
7
$\begingroup$

I, unfortunately, am unable to implement your instruction to "pretend that [I] do not have any visual intuition of what perpendicular or orthogonal means". That intuition was developed in my mind long before I learned the formalities of coordinate systems of vector spaces.

The same is true historically: Euclid's Definition 10 in Book 1 already has a crystal clear notion of perpendicular lines, one which incorporated a good mathematical model of real-life notions of perpendicularity:

Book 1, Definition 10: A line crossing another line that makes adjacent angles that are equal to each other ... is then called perpendicular to the line it crosses.

I would venture to say that the formalities of Cartesian coordinate systems, and the later developments of vector spaces and inner products, were developed (in part) to incorporate still better mathematical models of perpendicularity.

$\endgroup$
0
$\begingroup$

The abstract definition of the circle centered at $(a,b)$ is $v\cdot v=r^2$ where $v=(x-a,y-a)$. If you use orthogonal coordinates, this abstract equation reduces to your equation $\left(x-a\right)^2+\left(y-b\right)^2=r^2$. If you use polar $(R,\theta)$ coordinates it reduces to $R=r$. If you use another coordinate system where the dot product is given by $u\cdot v=u'Av$ where $A$ is a positive definite matrix then it reduces to $v'Av=r^2$. If $A$ is the identity, then this simplifies to $v\cdot v=v'v=r^2$ which is your equation $\left(x-a\right)^2+\left(y-b\right)^2=r^2$. The abstract equation is the same in all cases.

The abstract dot product also defines a right angle: the vectors $u$ and $v$ are orthogonal if $u\cdot v=0$. In fact it defines all lengths and angles. $||v||^2=v\cdot v$ and $u\cdot v=||u||~ ||v||~ \cos\theta$.

For the Pythagoras theorem, observe that

$$||u+v||^2= (u+v)\cdot (u+v)=||u||^2+||v||^2 +2 u\cdot v=||u||^2+||v||^2 +2 ||u||~ ||v||~ \cos\theta$$

and the last term disappears for non zero $u$ and $v$ if and only if $u$ and $v$ are orthogonal. This means that you cannot use the Pythagoras theorem unless the sides are orthogonal.

In short the abstract dot product not only defines the metric but also completely defines the geometry. The mistake is to assume that for any coordinate system, the abstract dot product is represented by $u'v$ instead of $u'Av$.

On the other hand if the metric does not come from a dot product, then the circle has a very different shape. For example, if you use the $L^1$ metric $||(a,b)||=|a|+|b|$, then the "circle" is actually a square.

$\endgroup$
0
$\begingroup$

I am unsure if it is possible to fully grasp any of the notations or what a coordinate system even is without having intuition about how mathematics relates to visual spaces to begin with. Understanding coordinate systems in vector spaces benefits from geometric intuition in visual space in $\mathbb{R}^2$

However, I will propose an example that may play into your having this intuition.

In $\mathbb{R}^2$, geometric shapes transform under linear transformations. Say the line segment joining $ u, v \in \mathbb{R}^2 $ is defined as S := {(1 - λ)u + λv : 0 ≤ λ ≤ 1}. If $ T: \mathbb{R}^2 \to \mathbb{R}^2 $ is a linear transformation, then T(S) = {(1 - λ)T(u) + λT(v) : 0 ≤ λ ≤ 1} is a line segment joining T(u) and T(v).

An invertible linear transformation maps a triangle to another triangle, preserving convexity but possibly altering side lengths or angles (an isosceles triangle may become scalene for example). Non-invertible transformations may collapse a triangle to a line or point (degenerate case). Other convex shapes transform similarly: a square may become a parallelogram, and a circle may become an ellipse, due to stretching or shearing. Changing the basis of $ \mathbb{R}^2 $ alters the matrix of $ T $, but geometric properties like line segments and convexity remain the same under such changes.

To connect the abstract definition of $\mathbb{R}^2$, as the set of ordered pairs (x, y) with standard addition and scalar multiplication, to its common visual representation, consider its structure without assuming geometric notions like perpendicularity. Formally, $\mathbb{R}^2$ is equipped with the Euclidean metric $d((x_1, y_1), (x_2, y_2)) = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$, and we choose the standard basis { (1, 0), (0, 1) }. A vector (x, y) = x(1, 0) + y(0, 1) is represented by coordinates x and y, which we can plot as a point on a plane. The basis vectors (1, 0) and (0, 1) are linearly independent and satisfy $\langle (1, 0), (0, 1) \rangle = 0$, where the inner product is $\langle (x_1, y_1), (x_2, y_2) \rangle = x_1 x_2 + y_1 y_2$. This orthogonality leads to the visual interpretation of perpendicular axes, as the angle between vectors, defined via the inner product $(\cos \theta = \frac{\langle u, v \rangle}{\|u\| \|v\|})$, is 90 degrees when the inner product is zero. Thus, the coordinate plane with perpendicular axes emerges naturally from the algebraic structure, providing a visual framework for transformations like those above.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.