# Least amount of structure for vector space.

1. Dec 10, 2007

### matheinste

Hello all.

Back to basics again.

When defining a set of geometric vectors for a vector space of n dimensions how can we define such a set without a certain amount of structure already defined upon the n dimensional space. We presumably need some concept of direction to determine linear dependence or independence. What is the least amount of structure required to define a set of basis vectors.

Matheinste

2. Dec 10, 2007

### trambolin

Let me add some flavor in it... How to make a concrete argument using quotient spaces?

3. Dec 10, 2007

### Ben Niehoff

A vector space must obey vector algebra: i.e., you can add vectors, and multiply them by scalars, and there are identity elements for both operations. To talk about linearly-independent bases, you can use determinants to define them abstractly.

You don't need any more structure than that, though. In particular, you don't need to have a metric, or a norm, or any concept of "angle" or "distance".

4. Dec 10, 2007

### HallsofIvy

You need a basis. That is the "least amount of structure" you can have for a vector space.

5. Dec 10, 2007

### matheinste

Hello again.

We need a basis and this basis must be some linear combination of the underlying set of vectors which we have selected from which to construct our space. In the case of geometric vectors over the reals how can we select a set of 'vectors' from which to construct an analogue of a real physical space without some idea of direction.

Matheinste.

6. Dec 11, 2007

I believe the idea of direction is attached to the "real physical space", that's the problem.

7. Dec 11, 2007

### HallsofIvy

No. The basis IS the "underlying set of vectors". It is ALL of the vectors in the space that are linear combinations of those.

A set of vectors {v1, v2, ..., vn} are "independent" if and only if the only way you can have a1v1+ a2v2+ ...+ anvn= 0 is to have a1= a2= ....= an= 0. If the set of vectors is independent then it is a basis for an n dimensional vector space. It is not necessary to have "some idea of direction" to do that. In fact, in addition to a vector space, you need an "inner product" to define "direction" at all. And, for any vector space, there are an infinite number of different ways to do that.

8. Dec 11, 2007

### matheinste

Thanks radou and HallsofIvy for your responses.

As radou says the problem with geometric vectors is probably my trying to mix the physical with the abstract.

Going back to a basis. Of course a basis spans the whole space and there may be many of them (bases) and in the case geometric vector spaces, inner products. The definition of a vector space, in general, is a set of elements which we call vectors ------------------etc---- I thought that our choice (if we have one) of the members of this set was the fundamental set from which a vector space is constructed. For instance in the case of the reals over themselves the chosen set could be the natural number one and of course this by scalar multiplication generates the whole space. We may, having constructed the space, wish to chose another number as a basis.

I may be being over fussy about about this point and if I the distinction s meaningless I will forget about it. It is probably of no practical importance anyway.

Thankyou for your input. Matheinste.

9. Dec 11, 2007

There is a distinction between "constructing a vector space" and "spanning it with a spanning set" (in your case, a basis). To construct a vector space would mean to find a set with two operations defined, addition and scalar multiplication, such that its elements satisfy certain properties and such that specific elements exists (identity, inverse etc.). You were talking about spanning the reals with the set {1}, which has nothing to do with the construction of a vector space (unless I misunderstood you, of course).

10. Dec 11, 2007