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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

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    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

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    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.
     
  22. Apr 8, 2017 #21

    WWGD

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    Just throwing this in , have not fully thought through: for a,b in ## \mathbb C ## if <a,b>=c ; c in ##\mathbb C ## can't we talk about ## cos \theta ## in terms of the argument of c (having chosen a branch) EDIT: dividing by the product of scalars ||a||||b|| will not change the angle?
     
  23. Apr 13, 2017 #22
    There is a simple counterexample. Consider the inner product of two scalars:
    ##\langle x,y \rangle = x \overline{y}##
    It is easy to show that that satisfies all four of these axioms. Now take <x,-x>. It equals - <x,x>, and from the axioms, it is negative unless x = 0.
     
  24. Apr 14, 2017 #23

    Mark Harder

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    What do you mean by this? Are a and b meant to be scalars ∈ ℂ? If so, what does <a,b> mean? Or do you mean that a,b ∈ℂn?
     
  25. Apr 14, 2017 #24

    WWGD

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    I was thinking of an inner-product in the Complexes but it could also be in ## \mathbb C^n ## , so a,b would be in ## \mathbb C^n ## itself. EDIT: for ##\mathbb C ## we could have , e.g., <z,w>:=zw^ , where b^ is the conjugate of w , i.e., if w=a+ib , then w^:=a-ib.
     
  26. Apr 15, 2017 #25

    mathwonk

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    This is an interesting point I just noticed the other day in re- writing my linear algebra notes. the usual dot product v.w obtained by letting v amd w be coordinate column vectors and taking the transpose of v, and matrix - multiplying that row vector by the column vector w, is complex valued and can be negative. The result is that when one defines the "orthogonal complement", of a subspace, it may share some non zero vectors with the original subsopace. e.g. over the complex field C we have the vector <1,i > in the plane C^2, which is orthogonal to itself. Thus to do spectral theory and complex geometry, one uses the hermitian inner product whereby one first conjugates all the entries of w and then does what we said before. The axioms originally given by the OP suggest this meaning, and in this setting the hermitian dot product of v with itelf is always real and indeed also non -negative. I.e. even though the underlying field is C, the hermitian inner product is real valued and non negative. This does follow from the axioms given above using the conjugation symmetry property, it seems. oops, i didn't see the typo there, he probably meant that <x,x> ≥ 0 for all x. I.e. <x,y> isn't even non negative for real dot products.

    Interestingly, even with the other more primitive inner product, the orthogonal complement of a subspace still has the expected complementary dimension, but it is perhaps better viewed as a subspace of the dual space.
     
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