Least amount of structure for vector space.

In summary, in order to define a set of geometric vectors for a vector space of n dimensions, the minimum amount of structure required is for the set to obey vector algebra and for there to be a basis of linearly-independent vectors. This does not require any additional structures such as a metric or norm, and the choice of basis does not necessarily have to have any concept of direction. The underlying set of vectors is the basis for the vector space and its construction does not involve spanning the space with a specific set of elements.
  • #1
matheinste
1,068
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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
 
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  • #2
Let me add some flavor in it... How to make a concrete argument using quotient spaces?
 
  • #3
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
You need a basis. That is the "least amount of structure" you can have for a vector space.
 
  • #5
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
matheinste said:
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.

I believe the idea of direction is attached to the "real physical space", that's the problem.
 
  • #7
matheinste said:
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.
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
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
matheinste said:
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

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
Hello radou.

Thanks for pointing that out. I had of course overlooked an essential part of the definition that the set of vectors has the structure of a group etc. I fully understand what you are saying.

I think I see the answer to my original question. Its pretty obvious that the least required structure for a vector space is that required by the definition. Geometric "vectors" require additional structure depending upon what we want to do with them.

Thanks Matheinste.
 

1. What is the definition of "Least amount of structure" for vector space?

The least amount of structure for a vector space refers to the minimum set of axioms or rules that are required for the set to be considered a vector space. This includes the existence of a zero vector, closure under addition and scalar multiplication, and the existence of additive and multiplicative identities.

2. How is the "Least amount of structure" related to the dimension of a vector space?

The dimension of a vector space is directly related to the least amount of structure required for the set to be a vector space. A vector space with a higher dimension will have a greater number of required axioms or rules, while a lower dimensional vector space will have a smaller set of required axioms.

3. Can a vector space have too much structure?

Yes, a vector space can have too much structure. This means that the set will have more axioms or rules than necessary to be considered a vector space. Having too much structure can lead to unnecessary complexity and can make it more difficult to perform operations on the vectors in the space.

4. What are some examples of vector spaces with the least amount of structure?

Some examples of vector spaces with the least amount of structure include the real numbers, complex numbers, and n-dimensional Euclidean space. These sets have the minimum set of axioms required for them to be considered vector spaces.

5. Why is understanding the "Least amount of structure" important in mathematics and science?

Understanding the least amount of structure required for a set to be considered a vector space is important because it allows for the development of a foundational understanding of vector spaces. It also helps to simplify the study of vector spaces and allows for the application of vector spaces in various fields of science, such as physics and engineering.

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