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I Non-negativity of the inner product

  1. Apr 6, 2017 #1
    The inner product axioms are the following:
    ##\text{(a)} \ \langle x+z,y \rangle = \langle x,y \rangle + \langle z,y \rangle##
    ##\text{(b)} \ \langle cx,y \rangle = c\langle x,y \rangle##
    ##\text{(c)} \ \overline{\langle x,y \rangle} = \langle y,x \rangle##
    ##\text{(d)} \ \langle x,x \rangle > 0 \ \text{if} \ x \ne 0##

    What about these axioms imply that ##\langle x,y \rangle \ge 0## for all ##x## and ##y##?
     
  2. jcsd
  3. Apr 6, 2017 #2

    fresh_42

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    What does ##\overline{\langle x,y \rangle}## mean?
     
  4. Apr 6, 2017 #3
    Complex conjugation, in the case that the underlying field is complex
     
  5. Apr 6, 2017 #4

    fresh_42

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    In that case: what does ##\langle x,y \rangle \geq 0## mean?
     
  6. Apr 6, 2017 #5
    Well I'm not sure. I though the inner product could never be negative
     
  7. Apr 7, 2017 #6

    fresh_42

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    As you see, you have to decide, what your underlying field is. Complex numbers cannot be (Archimedean) ordered, so ##\langle x,y \rangle \geq 0## usually makes no sense here (unless you define an order that doesn't have much to do with the corresponding arithmetic). It can make sense for ##\langle x,x \rangle## as, e.g. the ordinary dot product ##\langle x,x \rangle = x^\tau \cdot \bar{x} \in \mathbb{R}##.

    So for the real case, do you have any ideas?
     
  8. Apr 7, 2017 #7
    Well I just realize that the inner product can be negative... Well then I guess that leads me to another question. I know that the norm of a vector ##\sqrt{\langle x, x\rangle}## can intuitevly be thought of as length of a vector. However, what is the geometric intuition behind the inner product of any two distinct vectors, especially that in the case of the reals the inner product can be negative?
     
  9. Apr 7, 2017 #8

    fresh_42

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    In the usual sense, i.e. the one you mentioned ##\langle x, x\rangle = \sum_i x_i^2##, this is the ordinary Euclidean metric. It is simply the theorem of Pythagoras. If you change the coordinates in a way, that leaves you with the two you actually need, as ##x,y## always lie in a plane, you get the school version of it.
    One can always write ##\langle x, y\rangle = (x,y) B (x,y)^\tau ## with a matrix ##B##. In the case of the Euclidean metric we have ##B=I##. And as we're on a physics website: the Minkowski metric is given by ##B = \operatorname{diag}(1,-1,-1,-1)##, resp. ##B = \operatorname{diag}(-1,1,1,1)## depending on whether a mathematician or a physicist wrote the book.

    Normally one wants do define a distance, i.e. a metric by a inner product, so it makes sense to require that there will be no negative distances. But in principle every matrix ##B## is allowed, although one might not call the result a inner product, rather a bilinear form B instead (see the other thread).
     
  10. Apr 7, 2017 #9
    I think you may have asked the original question because you confused inner product with norm measure (it happens).
    For ordinary geometrical vectors:
    a•b = ||a|| ||b|| cos(a,b)
    (the angle (a,b) has orientation from a to b and + or - sign ...)
    Play with that and I think it can answer a lot.

    P.S. Also (perhaps part of the confusion), in abstract and linear algebra, a [vector] space with norm does not necessarily have inner product (it can just be a space with norm, and norm is by axiom always real and possitive, or zero). Whereas, on the other hand, a space with inner product always has a norm (at least the obvious one that has been mentioned).
     
    Last edited: Apr 7, 2017
  11. Apr 7, 2017 #10
    2 typos here: 1. (corr.) "wants to". 2. (corr.) "an inner".
     
  12. Apr 7, 2017 #11

    WWGD

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    Even when working in a Complex V.S, inner-products are Real-valued , e.g. http://mathworld.wolfram.com/InnerProduct.html

    EDIT: This is incorrect, As Stavros pointed out. A Complex inner-product can be Complex-valued.
     
    Last edited: Apr 7, 2017
  13. Apr 7, 2017 #12
    You mean for x=y? (by using axiom (c) we get that the inner product is real). For x≠y it can be complex (otherwise why the complex conjugate? - look also carefully at your quoted ref.).

    In the axioms it is forgotten to be mentioned (implied) that the inner product on a space V is a mapping from VxV to C ...
     
  14. Apr 7, 2017 #13

    WWGD

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    Ah, my bad, when I read scalar I assumed Real-valued for some reason.
     
  15. Apr 7, 2017 #14
    Ok. 'Scalar' as opposed to 'vector' ...

    Also in the ref., or in general, one has to realize the differences (and connections) between 'metric space', 'normed space' and 'inner product space'. (+ there are more useful types mentioned there, e.g. 'Hilbert space', Lorentzian inner product and metric etc. ...)
     
  16. Apr 7, 2017 #15

    WWGD

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    I was thinking of Real vector spaces and somehow did not read carefully-enough, my bad, wake-up time, I am having my double shot :).
     
  17. Apr 7, 2017 #16
    Lol. Can happen to anyone ...
     
  18. Apr 8, 2017 #17

    WWGD

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    My doubt is then how do we do geometry in a Complex V.S if we do not have the expression :

    ##\frac {a.b}{||a||||b||}=cos(\theta) ## , since a.b ( the dot product) is not a Real number?
     
  19. Apr 8, 2017 #18

    fresh_42

    Staff: Mentor

    We do geometry in the corresponding real V.S##^2##. Otherwise we get imaginary and negative volumes, I guess. Or we define a distance by ##d(a,b)=|\,||a||-||b||\,|##. Your question is in its kernel: How can a complex vector space be oriented, if we don't consider it as a real one of twice the dimension? I think it can't for the same reasons the complex numbers can't be Archimedean ordered, but I don't have the definitions or a proof in mind, i.e. a list of requirements.
     
  20. Apr 8, 2017 #19

    WWGD

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    Do you mean orientation in terms of a choice of basis (so that we have two orientations, + and -) and two bases B,B' are equivalent iff the change-of-basis matrix has + determinant and are inequivalent otherwise? EDIT: Is orientation/orientability a Topological or Geometric property? EDIT: So we define two possible orientations on a pair ( Vector Space, Basis) , but, as you said, we need to then consider "Realified" bases for the Complex spaces ( or viceversa)?
     
    Last edited: Apr 8, 2017
  21. Apr 8, 2017 #20

    fresh_42

    Staff: Mentor

    The usual dot product on a complex vector space is given as sesquiliniear form by ##v \cdot w = \sum_i \bar{v}_i w_i## or the other way around. This gives us the real valued Euclidean norm ##\langle v,v \rangle = \sum_i ||v_i||^2##. If we allow the inner product to be complex, we cannot use it to apply real arithmetic as the basis of geometry. Otherwise we will have find a way to get rid of the imaginary part, e.g. by simply ignoring the ##\mathcal{i}##, i.e. the substitution ##\mathcal{i} \mapsto 1##. I just don't know where this leads to or what will be left of any geometry in these cases. So as long as the requirements aren't defined, it's hard to tell what's possible or not. I cannot imagine how such a solution should look like. If we simply want to use the ordinary Euclidean geometry, then ##\mathbb{C}=\mathbb{R}^2## will do.
     
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