Geometric interpretation of $\mathbb R^n$

Question

Motivated by this question Basis of a basis I've been thinking when we say that a vector can have a geometrical interpretation, are we talking about the vectors themselves or the vectors coordinates in the usual basis $\{(1,0),(0,1)\}$ in for example $\mathbb R^2$?(see picture below to $\mathbb R^2$ case) I think vectors of $\mathbb R^n$ are just n-tuples of real numbers and then don't have a geometrical interpretation, am I right? I'm asking this silly question, because there are a lot of abuse of notation in this part of linear algebra and I want to know the correct concepts in order to not get lost in the future.

Thanks in advance

Answer 1

This is one case where it may be easier to consider a more complicated situation. Think of some reasonable surface in $\mathbb{R}^n$, a sphere say. If we're doing geometry we want to be to be able to refer to specific points on the surface, so we decide to assign each point an address, which will be a $k$-tuple of real numbers. We will want distinct points to get distinct addresses, but this turns out to be easy, so then we might add more conditions---for example, if two points are close, their coordinates should be close. But in the end, each point has a coordinate assigned to it.

The problem you're facing is that when we study the geometry of $\mathbb{R}^n$, we use $n$-tuples for addresses, and so our set of addresses looks the same as our set of points.

A short summary is that if you're doing geometry, it may help to think of a geometric point in $\mathbb{R}^n$ as being something distinct from a vector. It is determined by a vector, but which vector you use depends on the basis you choose.

Answer 2

Vectors are abstract elements of a vector space, in fact a vector space is a set of elements called "vectors" with some nice properties. Note that you can always do formal operations with vectors without using coordinates and only the vector space axioms: $ v,w \in V(K) $ vector space on $K$ field. $$ v+w = (v+w) \in V(K) $$ the parenthesis are here only to help you think about $v+w$ as an unique element of $V(K)$.

But if you want to represent this new vector and work with it in the usual manner you have to choose a basis of your space, and in some cases (the cases I think you are interested in, so finite dimensional spaces) use the isomorphism between your space with that basis and $\mathbb{R}^n$ for some $n \in \mathbb{N}$.

In this way we obtain a representation of a vector by it's coordinates and work with them like they are vectors themselves. Note that if you change the basis you change the isomorphism as well and therefore your coordinates.

Answer 3

In $\mathbb{R}^n$, under the strictest possible definitions, there is no difference between a vector and a coordinate vector in the standard basis. This is because $\mathbb{R}^n$ is defined to be $n$-tuples of real numbers, and the coordinates of a vector in an $n$-dimensional real vector space is defined to be an $n$-tuple of real numbers. In the standard basis, which uses the following real vectors (not coordinates): $$\begin{bmatrix}1\\0\\\vdots\\0\end{bmatrix} \qquad \begin{bmatrix}0\\1\\\vdots\\0\end{bmatrix} \qquad \cdots \qquad \begin{bmatrix}0\\0\\\vdots\\1\end{bmatrix} \qquad $$

Therefore, the $k^\text{th}$ element of a standard-basis coordinate vector for a real vector in $\mathbb{R}^n$ is precisely the $k^\text{th}$ element of the ($n$-tuple, which is a real) vector.

So on a purely set-theoretic level, these two objects are fundamentally, and precisely, the same object.

In general vector spaces the vectors may be weirder. I don't know how much experience you have with vector spaces, but they can be defined very abstractly. For example, polynomials with complex coefficients of degree less than $n$ form a real vector space, where addition and scalar multiplication act in the way you think they do. And this is one of the most normal vector spaces… they only get weirder from here.

But we know ('we' meaning the the mathematical community--I wouldn't expect you to personally be aware) that any finite-dimensional real vector space (FDRVS) is ''basically the same as'' some Euclidean space. Precisely, the two vector spaces are isomorphic, or with equal precision but with more linear-algebra-y words, there is a bijective linear transformation from any FDRVS to a Euclidean space. And as you might imagine, lists of numbers are easier to wrap our heads around than the crazy things inhabiting most vector spaces, so this is why we are interested in coordinates in the first place.

Therefore, because the distinction between the two ways of thinking is so useful in literally every single other vector space, we typically move those ideas over to $\mathbb{R}^n$. In the process, you realize that a change of basis can be described well in this way. So even in $\mathbb{R}^n$ there is a distinction between a real vector and a coordinate vector, where the coordinates are derived from any basis except the standard basis.

The duality in the definition actually has a relatively significant implication for your question. Both of the kinds of vectors have a geometric interpretation, but we think of them in very different ways.

The easier way to think of is the coordinate vectors. Here, you can choose an arbitrary origin, and then use the basis vectors to "grid" the space. From there you use the coordinate vectors to mark off the tick marks on that grid to find the location of the points corresponding to each of the vectors, and voila, geometry!

But you might notice something a little strange about that construction: you have to put the basis vectors on there, but if you don't have a grid already in place, how is that possible? The answer is that we mark the basis vectors using their real vectors. But the real vectors are lists of numbers. So in fact, $\mathbb{R}^n$ really does come with a basis that is in some sense ''correct'', which is why the standard basis has that name.

A real vector is therefore plotted in exactly the same way as a coordinate vector, except then afterwards you forget that the axes were there. In this sense, the study of real vectors as opposed to their more user-friendly coordinate counterparts is sometimes called coordinate-free.

To be clear, the standard basis is therefore fundamentally different than other bases. To be cute about it, you can't buy an $\mathbb{R}^n$ for your office desk without a standard basis preinstalled. But we only need the standard basis to help us define the actual object that is Euclidean space. Once the space is there, you can choose the elements which are the basis under consideration without knowing how they were created.

This is often said in the following way: Coordinate vectors require axes, and real vectors don't. But remember, since real vectors and coordinate vectors are actually the same in the standard basis, this truism is actually a bit off. Hopefully this explanation helps you to understand the finer points of this distinction.

(In the picture you've shown us, it can't be determined which kinds of vectors are being used because it uses the standard basis. But the dashed lines seem to suggest that the illustrator, at least, was thinking of the coordinate vectors. The biggest clue is the fact that the axes are still there in the picture, and you're not just given three arrows and told what vectors they represent. But this is more likely a pedagogical consideration than anything substantive)

Answer 4

There is no single (or perhaps I should say no fundamental) geometric interpretation of vectors in $\mathbb R^n$. Nevertheless, there is a common interpretation in which vectors are interpreted as weighted directions.

Take the arrows you've so carefully drawn, and now wipe out the coordinate axes. You no longer have well-defined coordinates for your vectors; you could draw new coordinate axes any way you like (provided they're not coincident lines or anything funny like that), and you would generate an entirely valid set of new tuples that describe your vectors in a new basis. But you've done absolutely nothing to your vectors. They stayed the same; you just chose a new basis. This is exactly what's called a passive transformation: the object has stayed the same; it's merely its expression in a new and different basis that you've obtained.

But there are other geometric interpretations of vectors, and other objects that can correspond to directions. There is, for example, a projective interpretation. To see this, imagine a 3d space, and construct a flat plane that is offset from the origin (say the plane $z=1$ for instance). In the projective interpretation, a vector tells us about position on the projective $z=1$ plane: that is, take any vector and extend it to an infinite line. Where that infinite line intersects $z=1$ is a "position". Take two vectors, and you define two positions on the projective plane, which in turn define a line in the projective plane. This is just the beginning of projective geometry (often presented using "homogeneous" coordinates, which you can google), and there are other types of geometry even more complicated in interpretation of the elements than this (see conformal geometry).

Nevertheless, the basic techniques of vector algebra underlie all these interpretations.

Answer 5

Yes, vectors can be given a geometric interpretation:

A vector is a line segment in $\mathbb R^2$ with a specific direction.

Depending on context operations on vectors can be given geometric interpretations. For example,

1. Multiplying a vector by a scalar increases the length of the line segment.

2. Multiplying the vector by a matrix of the right dimension rotates, scales the vector

and so on.

Answer 6

Vectors don't have to have a geometric interpretation. Like most objects in maths they are simply things that satisfy a bunch of axioms. However, as with most such objects, the "usual" interpretation of vectors is chosen such that there are clear geometric interpretations. You can clearly draw diagrams to see how addition of vectors, scalar multiplication of vectors, magnitudes of vectors, etc. all make intuitive sense in this case. However, don't forget that we can easily talk about differently defined vector spaces for which the meanings of these are not as clear geometrically.