Undergrad Difference between R^n and other vector spaces

[Mr Davis 97]

I feel like the vector space $\mathbb{R}^n$ differs from other vector spaces, like $\mathbb{P}$. For example, if we wrote down an element of $\mathbb{P}$, like $1+2t^2$, this is an object in its own right, with no reference to any coordinate system or basis. However, when I write down an element of $\mathbb{R}^2$, like $\begin{bmatrix}1 \\ 2 \end{bmatrix}$, I feel like there is an inherent assumption that these are coordinates in terms of some basis, like the standard basis, and not just objects in their own right that exist independently of any basis. Is this just a notational issue or is there actually something different about $\mathbb{R}^n$ compared to all other vector spaces?

 

[andrewkirk]

I haven't come across the notation $\mathbb P$ but, given the example, my guess is that you might be referring to the vector space $\mathbb R[x]$, which is the vector space of polynomials with real coefficients.

There is no need for the inherent assumption you refer to for $\mathbb R^2$. Indeed, it depends on what one means by $\mathbb R^2$. When working algebraically with objects like vector spaces, we usually abuse notation in a way that identifies the particular vector space we are talking about with the equivalence class of all vector spaces that are isomorphic to it. So when we refer to $\mathbb R^2$ we might be referring to the set of ordered pairs (2-tuples) of real numbers, or we might be referring to a 2D plane with a single point identified as the origin, with operations of addition and multiplication defined in the usual way. Set theoretically, those two are completely different, but as vector spaces they are isomorphic. For the 2-tuple version there is a 'natural basis' made of the tuples (1 0) and (0 1). But for the 'pointed plane' there is no such natural basis, as there are no axes defined. There is no 'preferred direction'. We need to arbitrarily select a direction in order to start constructing a basis, using that direction as one of the axes.

So it depends on whether you are referring to a specific instance of an isomorphism class of vector spaces (ie to a specific set, with attached algebraic structure), or to the isomorphism class itself.

Like the 2-tuple version of $\mathbb R^2$, the polynomial space has a natural basis if a polynomial is defined as any sequence of real numbers with only a finite number of nonzeroes, with the usual rules of polynomial multiplication and addition. That basis is (1, 0, 0,...), (0, 1, 0, 0, ...) etc

But just like with $\mathbb R^2$ we can construct a vector space that is isomorphic to the polynomial space, but has no obvious natural basis, as follows:

Define a vector space as the following set of infinitely differentiable functions:
$$P\triangleq \{f\in C^\infty :\ \exists n\in\mathbb N(f^(n)=0)\}$$
with the usual rules of function addition and scalar multiplication.

Although, thinking about it, it seems to me that an attempt to construct a basis for $P$ is most likely to lead to the basis I gave above. So in a sense there is a natural basis for $P$, but it's just not immediately obvious with that second definition. Looked at that way, it seems that the isomorphism class of polynomial vector spaces has a more natural basis than does the isomorphism class of $\mathbb R^2$.

 

[Stephen Tashi]

Is this just a notational issue or is there actually something different about $\mathbb{R}^n$ compared to all other vector spaces?

Try to phrase the question so it isn't self-answering.

If we compare $\mathbb{R}^n$ to some other vector space, there is something different about $\mathbb{R}^n$. Otherwise, the other vector space wouldn't be an "other" vector space, it would be the same as $\mathbb{R}^n$.

For example, if we wrote down an element of $\mathbb{P}$, like $1+2t^2$, this is an object in its own right, with no reference to any coordinate system or basis.

What shall we mean by "an object in its own right"? Perhaps you mean that the set of polynomials of degree two in one variable with real coefficients has properties in addition to the property of being (isomorphic to) the vector space $\mathbb{R}^3$. For example, it makes sense to speak of the maximum value of such a polynomial in the interval [0,1], but if we are only given that we have an element $v\in \mathbb{R}^3$, we have no definition for "the maximum value of $v$ in [0,1]".

It's true that one can define a vector space of polynomials without specifying a basis for it. However, if you write carefully, you can define the vector space $\mathbb{R}^n$ without specifying a basis for it.

However, when I write down an element of $\mathbb{R}^2$, like $\begin{bmatrix}1 \\ 2 \end{bmatrix}$, I feel like there is an inherent assumption that these are coordinates in terms of some basis

That's a correct feeling. The notation assumes that the mathematical structure being denoted (whatever it is) can be represented as n-tuples of numbers and the usual operations on the n-tuples as vectors. The vector space of N-tuples of numbers has the property that $\{( 1,0,0,.), (0,1,0,0,..)...\}$ is a basis. However, you could define the vector space $\mathbb{R}^n$ without mentioning any basis for it. The existence of the standard basis is then a theorem rather than part of a definition.