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Linear spaces vs. vector spaces

  1. Feb 6, 2013 #1
    Does a vector in an absract vector space (without any further structure i.e. no inner product or norm) have the properties usually associated with vectors, that is, magnitude and direction? If not, isn't the name vector space a bit misleading and it would be more appropriate to call it a linear space?
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  3. Feb 6, 2013 #2


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    Hi TrickyDicky! :smile:
    Take eg ℝ2

    we can give it the usual norm, or we can give it the norm √(x2 + 4y2) …

    either way, it's a vector space, but one is squashed relative to the other, and so distances and directions are changed (and equal distances or angles become unequal)

    even without a norm at all, parallel vectors are still parallel, and can be compared with each other in magnitude (using the scalar multiplication of a vector space)

    and an angle which includes another angle will still do so …

    so a vector space without a norm does have distance and direction :wink:

    (though distances in two different directions cannot be directly compared)​
  4. Feb 6, 2013 #3
    Hi tiny-tim
    I'm referring to the case without any norm or inner product. I agree one can compare the collinear elements of the space and say that one is twice or half as long as the other, but that doesn't allow us to assign them a specific length and therefore measure any distance.
    With respect to parallelism, I gathered from another thread in the relativity forum that unless you add a connection (so you'd need more structure) you cannot properly talk about parallelism, between elements of the space, at most you can talk about collinearity. So certainly you cannot assign an angle to the elements of the space either, don't you agree?
  5. Feb 6, 2013 #4


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    ah, i meant parallel at the same point (in other words, yes, collinear)
    yes, without a norm, we can only compare collinear distances
    yes … no norm, no angle :smile:
  6. Feb 6, 2013 #5


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    You can always define an inner product and a norm if the vector space is finite-dimensional.

    Two vectors are parallel if and only if they're linearly dependent, i.e. if one of them is a scalar times the other. You must be thinking of manifolds, not vector spaces.

    Angles require something like an inner product. With an inner product on a vector space over ℝ, you can define the angle θ(x,y) between x and y by
    $$\cos\theta(x,y)=\frac{\langle x,y\rangle}{\|x\|\|y\|}.$$ By the way, some people do call vector spaces "linear spaces" or even "linear vector spaces".
  7. Feb 6, 2013 #6
    Certainly, my question is simply about the case with no additional structure.
    Yes, but note that a vector space over the reals is automatically a manifold. Also see above the discussion on collinearity vs. parallelism.
    I know, my question was triggered by a book on linear algebra that had a chapter called linear spaces that in its first page had a footnote about why they didn't call it vector spaces as most linear algebra books. Their explanation was something to the effect that the authors didn't consider the elements of these spaces as vectors because they didn't have the properties that geoemtrical vectors had. I found it odd.
    Last edited: Feb 6, 2013
  8. Feb 6, 2013 #7
    Of course defining vectors over the reals simply as ordered n-tuples of n real numbers should suffice to avoid misunderstandings.
    I guess the problem is the representation we make of them, which is usually a geometric interpretation that may lead to ambiguities as to how much structure is added tacitly to the basic abstract vector space.
  9. Feb 6, 2013 #8


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    AFAIK "vector space" and "linear space" are synonyms, in this context. However the name "linear space" is also used for a different mathmatical structure: http://en.wikipedia.org/wiki/Linear_space_(geometry [Broken])

    I think the motivation for changing the name "vector space" to "linear space" is that the elements of a "vector space" can be mathematical objects which don't have any obvious geometrical interpretation as "vectors". For example the elements of a linear space might be "matrices whose elements are functions", not "numbers".

    I don't think the issue of norms is relevant. As Fredrik said, you can define an infinite number of different norms on any linear space. Interpreting one particular norm as a "length" in Euclidean geometry is certainly interesting and useful in physics or engineering, but it does't have much mathematical significance.
    Last edited by a moderator: May 6, 2017
  10. Feb 7, 2013 #9
    In #6 where I wrote " a vector space over the reals is automatically a manifold" it should say "a topological vector space over the reals is automatically a manifold".
  11. Feb 7, 2013 #10
    That's not true. A topological vector space doesn't need to have a nice manifold structure. For example, if I take a vector space and equip it with the trivial topology, then I have a topological vector space which is not a manifold. So it all depends on the topology.

    What is true is that there is a very canonical way to make a finite-dimensional real vector space into a manifold. That doesn't mean that every topology will make the vector space into a manifold.
  12. Feb 7, 2013 #11
  13. Feb 7, 2013 #12
    No, it's right. It's exactly what I said. Every real vector space V has a canonical manifold structure as follows: take a basis [itex]\{E_1,...,E_n\}[/itex]. Then consider the map

    [tex]V\rightarrow \mathbb{R}^n:x^i E_i\rightarrow (x^1,...., x^n)[/tex]

    This induces a maximal atlas which defines a smooth structure on the vector space V. Furthermore, the maximal atlas turns out to be independent of the choice of the basis.

    What you said, is that every topological vector space is a manifold. This is wrong, as the trivial topology shows.
  14. Feb 7, 2013 #13
    I didn't write that anywhere. Check my posts and try to quote that statement. I wrote first that every vector space over the reals is a manifold, and then corrected it to every topological vector space over the reals is a manifold. Now I,ve managed to confuse myself and I'm not sure if you are sayin that both statements are wrong, that only one is wrong(wich one?). I guess I have difficulties sorting out the subtleties of the mathematical language, is not having a natural manifold structure equivalent to being a manifold?
    Thanks for your patience.
  15. Feb 7, 2013 #14
    You said that

    This is wrong. A topological manifold is, by definition, a topological space which satisfies some properties (usually locally Euclidean, Hausdorff and second countable).
    If you speak about a topological vector space, then you have some topology. Saying that a topological vector space is a manifold means exactly that the topology on your space satisfies the properties. Like I pointed out, this is not necessarily true.

    What is true is that every finite-dimensional real vector space induces naturally a topology which makes it into a manifold.

    There are many topologies which you can put on a vector space and which make it into a topological vector space. But there is only one topology which make it into a manifold.
  16. Feb 7, 2013 #15
    We are seemingly not understanding each other, let's use a definition quoted from Wikipedia: http://en.wikipedia.org/wiki/Manifold

    "The broadest common definition of manifold is a topological space locally homeomorphic to a topological vector space over the reals."

    Is it not a topological vector space over the reals also a topological vector space over the reals locally? In that case it seems to me being a topological vector space over the reals is enough to count as a manifold since such object is trivially locally homeomorphic to a topological vector space over the reals. Sorry about the annoying repetition of phrases.
  17. Feb 7, 2013 #16
    Doesn't this finite-dimensional real vector space need to be a topological real vector space, I thought you said in the other thread that the algebraic structure only was not enough to have a manifold ?
  18. Feb 7, 2013 #17
    Sure, if you follow that definition, then you are entirely correct. But that "broad definition" is not the common definition studied in differential geometry. If you are talking about differential geometry and you mention "topological manifold", then people will automatically assume that you're talking about a space locally homeomorphic to [itex]\mathbb{R}^n[/itex] with perhaps some other properties. People will not think of Banach manifolds or spaces with the trivial topology, unless you specifically say it.
  19. Feb 7, 2013 #18
    I am saying that if I am given a finite-dimensional vector space V, then there exists a canonical topology [itex]\mathcal{T}[/itex] such that [itex](V,\mathcal{T})[/itex] is a topological manifold (in the sense that it is locally Euclidean, Hausdorff and second countable). Of course, [itex](V,\mathcal{T})[/itex] will also be a topological vector space.
  20. Feb 7, 2013 #19
    Thanks for your replies.
  21. Feb 7, 2013 #20
    Out of curiosity, [itex]\mathbb{R}^n[/itex] in your post is referring to Euclidean space or to a topological real vector space?
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