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Normed linear space vs inner product space and more

  1. Oct 15, 2013 #1
    Correct me if I'm wrong here but it is my understanding that vector spaces are given structure such as inner products, because it allows us to use these structured vector spaces to describe and analyse physical things with them.

    So physical properties such as 'distance' cannot be analysed in a general vector space but have analogies in a vector space [itex]\mathbb{R}^2[/itex] and [itex]\mathbb{R}^3[/itex] over the Real Numbers in an inner product space.

    Assuming my understanding is correct I'm a bit confused at the connection/relationship between Metric spaces, Inner product spaces and Normed linear spaces and which are more general and which are subsets of which.

    My understanding is that Metric spaces are subsets of Inner product spaces which are subsets of Normed linear spaces.

    An Inner product space is a normed linear space with a specific definition of the norm, namely [itex]||a|| = \sqrt{\langle a|a\rangle}[/itex] whereas a normed linear space is any vector space together with a function that obeys the properties of a norm.

    Is this correct?
     
  2. jcsd
  3. Oct 15, 2013 #2

    Office_Shredder

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    Inner product spaces are all normed linear spaces as you describe, but all normed linear spaces are metric spaces - you had that one on the wrong end.
     
  4. Oct 15, 2013 #3
    Ah so the metric space are the most general, i thought it was the least general and that misunderstanding had me going around in circles!

    Many thanks.
     
  5. Oct 15, 2013 #4

    D H

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    Not quite. A vector space is not necessarily a metric space, and a metric space is not necessarily a vector space. You can't say either one is more general than the other.

    On the other hand, the intersection of the two concepts -- that's where the normed vector spaces lie.
     
    Last edited: Oct 15, 2013
  6. Oct 15, 2013 #5

    Office_Shredder

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    This isn't right.... every normed vector space is a metric space, but there are vector spaces that have metrics on them which do not come from norms.
     
  7. Oct 15, 2013 #6

    D H

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    Fixed that. The normed vector spaces are a subset of the intersection of the vector spaces and the metric spaces. A metric on a vector space that is not induced by a norm is the distance ##d(\vec a,\vec b)## is 0 if ##\vec a - \vec b = 0##, 42 otherwise.
     
  8. Oct 15, 2013 #7
    Hmm, so i've had a bit of a think about where i was going wrong and the comments posted here have helped me get on the right track I believe.

    Can i just clarify what you mean by "every normed vector space is a metric space"

    Do you mean every normed vector space is intrinsically a metric space?
    That doesn't make sense to me, it seems to me that you can have a normed vector space that is not a metric space.

    Or do you meant that once you have a normed vector space , one can always compose some function of this norm that is a metric?
    This makes sense to me... as far as my current understanding goes at least.
     
  9. Oct 15, 2013 #8

    WannabeNewton

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    You cannot have a normed vector space which is not a metric space: a norm on a vector space immediately determines a metric structure for the vector space.
     
  10. Oct 15, 2013 #9

    Office_Shredder

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    Yes, the metric is d(x,y) = ||x-y||
     
  11. Oct 15, 2013 #10
    I'm sure once again there is some underlying misconception i have of the topic so far, so I will lay out my reasoning and hopefully we can spot where I'm going wrong.

    The definition i've seen of a metric space is: an ordered pair (M,d) where M is a set and d is a metric, that is a function d: M x M -> R and the function d must have certain properties.

    A norm however is just a function f: M -> R
    nothing in the norm itself has any of the properties that a metric requires.

    Thus you can have a normed vector space that is not a metric space.
    Ie the combination of the normed vector space and its metric d(x,y) is a metric space. But the normed vector space on its own is not.

    Or is it that any vector space that has a structure imposed on it such that one can define a function that acts as a metric is called a metric space.

    Sorry if it comes off as being pedantic, but i'm keen to avoid as few false conceptions as possible in my understanding as this is as fundamental as it gets and misconceptions here will spiral out of control in no time!
     
  12. Oct 15, 2013 #11

    Office_Shredder

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    I just said it in my last post, given a norm || || on a vector space V, then (V,d) is a metric space where d(x,y) = ||x-y||. Given any normed space you can trivially turn it into a metric space, and the whole point of defining a norm is that you get access to that metric which is why a normed vector space is in practice always assumed to have the metric.
     
  13. Oct 15, 2013 #12
    Why the number 42 may I ask?

    BiP
     
  14. Oct 15, 2013 #13

    D H

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    Google the phrase "the answer to life, the universe, and everything".
     
  15. Oct 15, 2013 #14

    Office_Shredder

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    Any number will work, 42 was picked either randomly or because D H is a fan of Hitchhiker's Guide to the Galaxy
     
  16. Oct 15, 2013 #15

    D H

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    There are many similarities:
    [tex]
    \begin{matrix}
    ||\vec a||\ge 0 & d(a,b) \ge 0 \\
    ||\vec a||=0 \Leftrightarrow \vec a = 0 & d(a,b)=0 \Leftrightarrow a=b \\
    ||\vec a + \vec b|| \le ||\vec a|| + ||\vec b|| \quad & d(a,c) \le d(a,b) + d(b,c)
    \end{matrix}[/tex]
     
  17. Oct 15, 2013 #16
    If [itex]X[/itex] is a vector space, we can think of a norm as being a special type of metric on [itex]X[/itex]. As mentioned above, any norm [itex]||\cdot||:X\to\mathbb R_+[/itex] induces a metric [itex]d_{||\cdot||}: X\times X\to \mathbb R[/itex] via [itex]d_{||\cdot||}(x,y):=||x-y||[/itex]. It's easy to check that [itex]d_{||\cdot||}[/itex] is indeed a metric, and it satisfies a couple other nice properties:
    - It's translation-invariant, i.e. [itex]d_{||\cdot||}(x+z,y+z)=d_{||\cdot||}(x,y)[/itex] for any [itex]x,y,z\in X[/itex].
    - It's (positively) homogeneous, i.e. [itex]d_{||\cdot||}(\alpha x,0)=|\alpha| d_{||\cdot||}(x,0)[/itex] for any [itex]x\in X, \alpha\in \mathbb R[/itex].

    A nice exercise is the converse, that if [itex]\rho:X\times X\to \mathbb R_+ [/itex] is a metric which is translation-invariant and homogeneous, then the map [itex]||\cdot||_{\rho}: X\to \mathbb R_+[/itex] given by [itex]||x||_{\rho}:= \rho(x,0)[/itex] is a norm on [itex]X[/itex].
     
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