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When showing a super simple property about norms, I have a doubt about how much I can assume

  1. Sep 25, 2014 #1
    Sorry, I wasn't sure of the best way to phrase this. This is a common problem I keep having.

    Here's the definition of a norm:

    Let E be a vector space V defined over a field F. A norm on V is a function p: [itex]V \rightarrow \mathbb{R}[/itex] such that:

    [itex]\forall a \in F[/itex] and [itex]\forall u,b \in V[/itex]:

    (i) p(av) = |a|p(v)
    (ii) p(u + v) [itex]\leq[/itex] p(u) + p(v)
    (iii) p(v) = 0 [itex] \iff [/itex] v = 0
    Now, an obvious property from these axioms is:

    (iv) p(v) [itex] \geq [/itex] 0 ([itex]\forall v \in V[/itex])
    We can see this as follows:

    Select some vector v [itex]\in V[/itex]. Then:

    p(v + (-v)) [itex]\leq[/itex] p(v) + p(-v) (by ii)
    p(v + (-v)) [itex]\leq[/itex] p(v) + |-1|p(v) = 2p(v) (by i)
    p(0) [itex]\leq[/itex] 2p(v)
    0 [itex]\leq[/itex] 2p(v) (by iii)
    0 [itex]\leq[/itex] p(v)


    =======

    Here's the part I get confused on.... How do I know I can actually do all of these steps? For example - I do know that for some arbitrary vector v [itex]\in V[/itex], I can get a -v to use in this little proof. I know I can do this because V is a vector space, and the vectors in a vector space make an abelian group which means that every vector has an inverse. But for example - how do I "know" that I can do division by 2 in that last step? (Going from 0 [itex]\leq[/itex] 2p(v) to 0 [itex]\leq[/itex] p(v)). I think the reason I can do it in this case is because the scalar field F that V is defined over is a division ring (by definition of a vector space)... but I had to go back and look that up before I would allow myself to do this. In fact, is this the reason why I can do it, or is it some other reason? After learning about abstract algebra, I'm just starting to be weary of making any move because I feel like I'm assuming too much. For example - even just doing the step p(v) + |-1|p(v) = 2p(v), I feel uneasy after learning about abstract algebra, because I feel I'm thinking of 1 as I would in the real numbers... or the 0/2 = 0 step... I feel like every move I take, I have to question if I'm doing something because I'm thinking only in terms of number systems I'm familiar with (like the real numbers), and I worry maybe I'm assuming too much.
     
    Last edited: Sep 25, 2014
  2. jcsd
  3. Sep 25, 2014 #2

    mathman

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    The coefficients outside of p(v) are real numbers. The norms are also non-negative real numbers. The list of properties all involve real numbers.
     
  4. Sep 25, 2014 #3
    Right, you may consider a norm to be a mapping from the algebraic structure V, a vector space, to the algebraic structure (|R, +, x), a ring (though we may also consider it to be the additive group, I believe). That is, we aren't just mapping to the set of real numbers, we are mapping to the ring of reals equipped with two binary operations, which are the standard multiplication and addition, it is implicitly so for these mappings. I think this is a good question, and it shows a good quality to have for studying mathematics.
     
    Last edited: Sep 25, 2014
  5. Sep 26, 2014 #4
    Just wanted to say thanks for the responses. Just out of curiosity - is vector space definition ever generalized so that the set its mapped to is any general ring? Why is it limited to R?
     
  6. Sep 26, 2014 #5
    (If I understand your question) a norm always maps to |R because it is a notion of distance, or length.

    This is dependent on the definition of a norm, not that of the vector space.

    If you look at the properties of a norm closely, you will see that it is an abstraction of the concept of length.
     
    Last edited: Sep 26, 2014
  7. Sep 26, 2014 #6
    Oh yeah, I see now I guess it wouldn't make sense for it to be mapped to anything other than R, since it's trying to capture the idea of length. Thanks for the response!

    Can I ask a sort of unrelated topic - can you suggest any more of these abstractions that I could go and study about? For example - I know about metrics (for distance), norms (for length), topologies (not really sure what it's trying to abstract... the idea of the "shape"?) What else is there? I mean - cool abstractions of every day concepts that we just think about in a Euclidean way. Not requesting that you take your time to explain these things, just if you have any terms you could throw out, then I could go read about them.
     
  8. Sep 26, 2014 #7
    Yes, it is only natural that we think of length using the real numbers.

    I think that no level of abstraction is lost here. The properties of the norm and vector space itself seem to imply that the norm must map into, at the very least, some ordered (triangle inequality) field satisfying the archimedean property (because the complex do), and that would make it isomorphic to a subfield of the reals, anyway. But, I'm starting to not know what I'm talking about, so maybe someone else can comment on that further.

    I think metrics and norms are the main abstractions of ideas we think about in a Euclidean way. There are objects such as semi-norms and pseudo-metrics, which are norm and metric like objects that do not satisfy the full definition. I find these "almost metrics" to be particularly interesting, because some of them generate topologies that are not metrizable.
     
  9. Sep 27, 2014 #8

    Fredrik

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    The set of scalars can be a ring that isn't a field, but in those cases, the term "module" is used instead of "vector space". I haven't studied norms on modules, but I think that they would still be defined as functions into ##\mathbb R##, because we want to be able to interpret ##\|x-y\|## as the distance between x and y.
     
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