Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Least amount of structure for vector space.

  1. Dec 10, 2007 #1
    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.

  2. jcsd
  3. Dec 10, 2007 #2
    Let me add some flavor in it... How to make a concrete argument using quotient spaces?
  4. Dec 10, 2007 #3

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    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".
  5. Dec 10, 2007 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    You need a basis. That is the "least amount of structure" you can have for a vector space.
  6. Dec 10, 2007 #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.

  7. Dec 11, 2007 #6


    User Avatar
    Homework Helper

    I believe the idea of direction is attached to the "real physical space", that's the problem.
  8. Dec 11, 2007 #7


    User Avatar
    Staff Emeritus
    Science Advisor

    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.
  9. Dec 11, 2007 #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.
  10. Dec 11, 2007 #9


    User Avatar
    Homework Helper

    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).
  11. Dec 11, 2007 #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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Least amount of structure for vector space.
  1. Vector spaces (Replies: 10)

  2. Vector space (Replies: 16)

  3. Vector space (Replies: 2)

  4. Vector spaces (Replies: 6)

  5. Vector space (Replies: 4)