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Does the basis of an Euclidean space have to be ordered by definition? Or can be left unordered?

I was also wondering about what is the morphism (i.e. the mapping that can preserve all the structures) on Euclidean spaces? Is it Euclidean transformation (rigid transformation), consisting of rotation, translation and reflection? Or the reflection is not part of it, because the basis of a Euclidean space must be ordered.

Thanks and regards!

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From the answers below it appears the answer to your question is yes and no.:) A basis can be ordered or unordered, it depends on what you're using it for as to whether or not you want it ordered. For example, for representing linear maps I use the phrase "ordered basis" rather than the term "basis" to make it clear what conditions I need on the basis. –  Ryan Budney Oct 3 '10 at 16:01
    
@Ryan: If what you're using it for intrinsically requires a basis order --- as opposed to it being merely convenient for presentation to label the basis vectors with an index set, which has a conventional ordering --- then I'm not sure that what you're studying is just "geometry" or "linear algebra" any more. Similarly, someone studying group theory, who finds that their problem simply requires that some notion of "distance" be continuously deformed, is not just working on "group theory", but algebraic topology. Do group actions have to preserve open sets? –  Niel de Beaudrap Oct 8 '10 at 6:08
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4 Answers 4

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Bases must be ordered, otherwise you couldn't speak of coordinates as ordered tuples. And this could make things pretty messy. For instance, if the standard basis for $\mathbb{R}^2$ could be either $e_1, e_2$ or $e_2, e_1$ simultaneously, then it would be difficult to speak of "the point of coordinates $(2,1)$".

Reflexions send bases to bases, because they are isomorphisms. But this doesn't mean they necessarily send a particular basis to the same particular basis. So they can change the order of the vectors of the basis. They don't preserve orientation, but this is an extra piece of structure, not included in the definition of Euclidean space.

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Yes, a basis should be thought of as an ordered list of vectors. However, one should be aware that some textbooks define a basis (incorrectly, IMO) as a set of vectors; as far as I can see, there is no advantage to this. –  Darsh Ranjan Oct 3 '10 at 4:31
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@Darsh. It is in fact standard to define a basis as a set of vectors rather than as an ordered tuple. Most of the time we are interested in linear combination of basis vectors where order does not matter because vector addition is commutative. –  Jyotirmoy Bhattacharya Oct 3 '10 at 5:41
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On bases

Bases do not have to be ordered. I respectfully present the following as a counterpoint to Agustí Roig's response.

First of all, a basis is simply a set of vectors, by definition; and sets are unordered. If you wish to enumerate the vectors in a basis, you will of course do so in some order, but this order is arbitrary.

This arbitrary order of enumeration is the reason why we describe vectors as tuples. The order of the coefficients in a tuple matters only in as much as it must be in agreement with the arbitrary order which was selected for the basis vectors. To wit: let x, y be two arbitrary linearly independent vectors. The tuple [ 7  5 ] with respect to the (enumeration of the) basis v1 = x, v2 = y represents the same vector as the tuple [ 5  7 ] with respect to the (different enumeration of the same) basis v2 = x, v1 = y. The tuple is just a representation of a vector, relative to an arbitrarily chosen order for the basis.

Even more foundationally: "tuples" can be regarded as functions from the integers (e.g. the indices '1', '2', etc.) to the reals, complex numbers, or whichever set you draw your coefficients from. So [ 7  5 ] can be thought of as 'really being' the function mapping

'1' $\mapsto$ 7,
'2' $\mapsto$ 5.

Our enumeration of the basis has a similar role: choosing the enumeration v1 = x, v2 = y is equivalent to defining a mapping

'1' $\mapsto$ x,
'2' $\mapsto$ y.

Why do I have '1' and '2' in quotes? Because they're just labels; any other labels would do just as well. For instance, I could replace '1' and '2' with the vectors x and y themselves. Then we could define vectors by coefficient functions such as

x $\mapsto$ 7,
y $\mapsto$ 5;

that is, the coefficient 7 is associated to the vector x, and the coefficient 5 is associated to y, to represent the vector 7x + 5y. This is what we really mean anyway; never mind any ordering of the vectors in your basis. With this, the arbitrary ordering of the basis disappears, leaving nothing but what it is we really mean by it all.

So: when we order our bases, its only to make it easier to present things — it is not part of the definition, or part of the structure of the objects we really care about.

On morphisms

A morphism on a Euclidean space ought to preserve all of the structures of Euclidean geometry. In particular, it should map circles to other circles, to preserve the structures guaranteed by the third postulate; so it must be an isothety (a rigid transformation up to scaling). This includes, but is not restricted to, the rigid transformations.

Can you make any argument why the morphisms should consist only of rigid transformations (excluding isothetic maps such as v $\mapsto$ 2v)?

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This began as a reply to Jyotirmoy's comment on Agustí's answer, but it became way too long.

As I stated in my comment on Agustí's answer, some textbooks define a basis as a set, and I think this is wrong. Obviously, some people disagree (just read some of the other answers and comments). I'll try to explain why a basis should never be thought of as a set.

Do the columns of the following matrix form a basis for $\mathbb{R}^2$? $$\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 1\end{bmatrix}$$ The answer is "yes" if a basis is a set, but obviously the right answer is "no." This makes it clear that whatever sort of collection to which we apply concepts like "linear [in]dependence" or "basis", it must allow repeated vectors; otherwise, a lot of basic theorems from linear algebra would require extra clauses to deal with matrices with repeated rows or columns. This rules out sets as our collections of choice.

To find out what our collection of choice should be, let's revisit the definition of a basic concept like linear dependence (in somewhat imprecise "collection-agnostic" language (whose meaning should nevertheless be clear)):

The vectors $v_1,\ldots,v_n$ are linearly dependent if there are scalars $c_1,\ldots,c_n$, not all zero, such that $c_1v_1+\cdots+c_nv_n=0$.

It's obvious here that the natural object to apply such a concept to is an indexed tuple of vectors (indexed by $\{1,\ldots,n\}$ above, but we could formulate the concept equally well for an arbitrary finite index set): we begin with an indexed tuple of vectors and then consider a tuple of scalars indexed by the same set, and then we apply a function to these two tuples that is now completely specified (taking the linear combination). We could have started with a completely unindexed, unordered collection of vectors with repetition allowed (i. e., a "multiset"), but then to consider a linear combination, we would have had to choose some indexing of the multiset first and then to consider some scalars indexed by our arbitrary index set in order to form a linear combination. The core point is that the natural input to the "linear combination" operation is a tuple of vectors and a tuple of scalars indexed by the same finite set. This is the fundamental operation on a vector space, so to me, it makes sense to formulate everything in terms of indexed tuples of vectors whenever possible.

I talked about indexing lists of vectors in the previous paragraph, but indexing isn't the same as ordering. When I say "order," though, I'm actually being a bit imprecise. It's usually not the order relation itself that matters, but the choice of index set does matter. For example, if you want to write down the coordinate matrix of a vector or the matrix of a linear transformation relative to a basis, then your index set pretty much has to be $\{1,...,n\}$ (or $\{0,...,n-1\}$ if you've ever programmed in a language other than Matlab or Fortran) because whatever you're using to index the elements of the basis, you're also using to index the rows or columns of your matrices. Since one of the main uses of bases is to reduce things to matrices, the sets $\{1,...,n\}$ are very convenient for indexing. In this very common context, all the basic concepts like "basis," "linear independence," "spanning," etc. would apply to finite sequences of vectors. In no context, however, do they naturally apply to finite sets of vectors.

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Your complaint is not so much that we should be using sequences instead of sets, as it is that we should be using bags (i.e. multi-sets, sets where membership has multiplicity) rather than sets. This obviously applies for the columns of the matrix you describe; and the case of the definition of linear independence hilights this, where even in the sum the particular order does not matter, nor does the choice of indexing scheme, so long as it is a bijection between some index-set and the vector-set (or vector-bag) under consideration. –  Niel de Beaudrap Oct 3 '10 at 11:58
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Anyway: I would readily concede that bags are a better choice than sets for describing collections of vectors for the purpose of describing linear independence, etc. However, I think that details such as a specific choice of indexing into the definitions only cloud details, and introduce distinctions which are on the whole totally uninteresting. –  Niel de Beaudrap Oct 3 '10 at 12:02
    
@Niel, I think the definitions should be chosen to make the surrounding theory as natural and simple as possible, and as such, sequences are still better than multisets. Again, all of the definitions are about linear combinations, and "linear combination" is an operation on a tuple of vectors and a tuple of scalars. Moreover, while the property of "being a basis" is indeed independent of the indexing, it's still useful to talk about permuting a basis, e. g., when discussing orientation. –  Darsh Ranjan Oct 3 '10 at 19:32
    
well, it's also useful to talk about permuting a basis when talking about permutation matrices, or anything where permutations play a role. (For instance, the sign of the cross product can be seen as taking the 'sign' of a permutation, rather than saying anything about the geometry per se.) — I respectfully disagree that 'sequences' make the theory more natural, or simple; and if it makes it pedagogically easier, this can just as easily be an accident of the fact that we teach sequences, but not multisets, in grade-school. –  Niel de Beaudrap Oct 4 '10 at 6:32
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Orientation does not play any role in Euclidean geometry. So it is most natural to define a Euclidean space as a finite-dimensional vector space with an inner product. Lengths, distances and angles are then defined in terms of this inner product. We are interested in transformations which preserve lengths and angles, which are precisely the orthogonal transformations including reflections. There is no need to explicitly choose a basis at any point.

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