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Pointwise-product (meaning?)

  1. Mar 8, 2009 #1
    Given two real vectors in [tex](u_1,\ldots,u_n)[/tex] and [tex](v_1,\ldots,v_n)[/tex], does the pointwise-product (elementwise-product) [tex](u_1v_1,\ldots,u_nv_n)[/tex] have a geometrical meaning? or can it at least be expressed as a combination of other known products (e.g. outer-product, contraction, and so on...)?
  2. jcsd
  3. Mar 8, 2009 #2
    The problem is that this product depends on the choice of orthonormal basis, so it does not have a well defined geometric meaning. To illustrate:

    The vectors (1,0) and (0,1) have pointwise product (0,0), but rotate them 45° and you get [tex](1/\sqrt{2},1/\sqrt{2})[/tex] and [tex](-1/\sqrt{2},1/\sqrt{2})[/tex] which have pointwise product (-1/2,1/2) and this is not the vector (0,0) rotated 45°.

    The scalar product and cross product on the other hand are preserved under rotations.
  4. Mar 8, 2009 #3
    I'll make one up :). You could consider it as a scaling of the axis. Of course that would be equivalent to taking one vector and forming a matrix with it's elements on the diagonal and then right multiplying the other element by this matrix.

    To make this invariant under rotation, remember that If

    A is our transformation which is equivalent to the element wise product.

    and U is a rotation.

    Then if:
    A'=U A(U^-1)


    AX is equivalent to A'X'
  5. Mar 9, 2009 #4
    what about this...?

    I found a simple way to make the pointwise-product equivalent to a sum of 2-blades (a concept from geometric algebra), but unfortunately it only works when n=2 :(
    Let's pretend that the two vectors [tex]\textbf{a}=(a_1,a_2)[/tex] and [tex]\textbf{b}=(b_1,b_2)[/tex] are actually:

    [tex]\textbf{a}=a_{1} \textbf{e}_1 +a_{2}\textbf{e}_{2}[/tex]

    [tex]\textbf{b}=b_{1} \textbf{e}_3 +b_{2}\textbf{e}_{4}[/tex]

    Let's define these two other auxiliary vectors:

    [tex]\widehat{\textbf{a}}=a_{1} \textbf{e}_1 - a_{2}\textbf{e}_{2}[/tex]

    [tex]\widehat{\textbf{b}}=b_{1} \textbf{e}_3 - b_{2}\textbf{e}_{4}[/tex]

    (it actually looks like complex conjugation). Now if we take the following sum of outer-products (wedge-product) we get what we wanted:

    [tex]\frac{1}{2}(\textbf{a} \wedge \textbf{b} + \widehat{\textbf{a}} \wedge \widehat{\textbf{b}}) = a_1b_1(\textbf{e1}\wedge \textbf{e3}) + a_2b_2(\textbf{e2}\wedge \textbf{e4})[/tex]

    So we got that in two dimensions, the pointwise-product can be expressed as a sum of two certain 2-blades, which IS indeed a 2-blade representing a plane in 4 dimensions.

    The big question is: can this process be generalized to n-dimensions?
  6. Mar 9, 2009 #5

    matt grime

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    In order for that to be a genuine geometric notion, you need to show that it is invariant under change of basis, for example. Then there is the issue of what it means. Your original idea was as a map

    [tex]V\times V \to V[/tex]

    now you have a map

    [tex]V\times V ---> \wedge^2 (V)[/tex]

    which in general is not isomorphic to V.

    What you describe originally, is a perfectly well defined algebraic construction: it is the direct product of your underlying field F (be it R, C, Q etc) with itself n times (n the dimension of the vector space) viewed as a ring.
  7. Mar 9, 2009 #6
    First of all, thanks to all the persons who showed interest in this post.

    for matt grime:
    I guess (at least, I hope) I understand your remarks. Thanks for pointing out that what I did was a direct product.

    Unfortunately I don't fully get why the fact of having a map [tex]V\times V \rightarrow \wedge^2 (V)[/tex] should prevent the product from having a geometric meaning.

    Moreover, could you be more specific when you say that one must check if the product is invariant under a change of basis? (Note: the vectors [tex]\textbf{e}_i[/tex] can be anything as long as they are not parallel).

    And finally, I'm not yet able to figure out whether it's possible or not to extend that operation to n-dimensions.

    Thanks in advance!
    Last edited: Mar 9, 2009
  8. Mar 9, 2009 #7

    matt grime

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    I didn't say it didn't have a geometric meaning (you might want to look up algebraic geometry).

    I said that the second map you wrote down is fundamentally different from the first one as the image spaces is the second exterior power, not the original vector space, as you wrote originally. "Wedge two" of a two dimensional space is one dimensional, so I don't know why you think you have a plane in 4 dimensions. (I don't know what blades are, so it is entirely possible I'm misunderstanding you.)
  9. Mar 9, 2009 #8
    you are absolutely right when you say that the image space formed by that operation is not anymore the original vector space but something else; it was essentially a dirty trick to make things work as I wanted. However, I myself have still doubts on the correctness of that.
    In fact, in order to make it work I had to make use of two extra dimensions:

    [tex]\textbf{a}=a_{1} \textbf{e}_1 +a_{2}\textbf{e}_{2}=(a_1, a_2, 0, 0)[/tex]

    [tex]\textbf{b}=b_{1} \textbf{e}_3 +b_{2}\textbf{e}_{4}=(0, 0, b_1, b_2)[/tex]

    Taking a wedge product would yield a 2-blade in a 4-dimensional space [tex]\wedge^2 (R^4)[/tex]: that's why I was talking of a plane in a 4D space as a geometric interpretation. I agree that the space [tex]\wedge^2 (R^4)[/tex] has dimensionality 6.

    The term k-blade was introduced by Hestenes, a pioneer of geometric-algebra. You can create a k-blade by "wedging" k non parallel n-vectors, and it is possible to show that such a product can be interpreted geometrically as a subspace of dimensionality k with magnitude and orientation.


    I hope I clarified the misunderstandings and corrected the imprecise statements.
    Perhaps it is also good to formulate better my original question:

    given an "array" of real numbers, if we assume that it was created by pointwise multiplication of two other (unknown) "arrays", can we give a geometrical interpretation to such objects?
    In fact, I have a space whose elements are all created by pointwise multiplication of element-pairs of another space. What I wish to find is a correct metric to compute a "distance" between such elements.
  10. Mar 10, 2009 #9

    matt grime

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    Umm, you say correctly that [itex]wedge^2(R^4)[/itex] is 6 dimensional, but also talk of something living in R^4. You're confusing me.
  11. Mar 10, 2009 #10
    Ok...I admit I was wrong. I am sorry for confusing you. The truth is that I'm new to geometric algebra and my weak knowledge often leads me to such pesky mistakes. Let's forget about the four dimensions.

    I think the correct statement is that a 2-blade in [tex]\wedge^2 (R^4)[/tex] has a geometrical interpretation of a subspace of dimensionality 2, living in a 6D space.

    Similarly a 2-blade in [tex]\wedge^2 (R^3)[/tex] would represent an oriented plane (with some magnitude) in 3D

    Hopefully I am not wrong this time
  12. Mar 10, 2009 #11

    matt grime

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    I think I've got the basics of blades now.

    They are bases for something (what I might want to call Grassman space, perhaps).

    But when you say that u/\v is a 2 blade and spans a 2-dimensional subspace it is a 2-d subspace of what we've been calling V, the vector space you first thought of.

    Geometric algebra makes the set of k-dimensional subspaces of an n-dimensional space into an n choose k dimensional subspace (the wedge-k space).

    So, in fact

    inside R^4: the blade u/\v represents the subspace spanned by u and v.
  13. Mar 10, 2009 #12
    This is ironic, but now it's me the one who is confused :)
    I just thought that when you wedge two vectors [tex]u,v \in R^4[/tex] you get an entity [tex](u\wedge v) \in \wedge^2(R^4)[/tex] which lives in a space having [tex] \left( \begin{array}{c} 4 & 2 \\ \end{array} \right) [/tex] dimensions (6).

    I guess I am missing a small hint to figure out why it's actually a subspace of a 4D space.

    Btw, coming back to the original question do you think the approach I was following could lead anywhere?
  14. Mar 10, 2009 #13

    matt grime

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    Yes, that is true, but that 6-dimensional space parametrizes the (algebra of) two dimensional subspaces of R^4. The wedge of two (non-parallel) vectors is a (non-zero) vector in the exterior product, but it corresponds to a plane in R^4 - the plane spanned by the two vectors you wedged together.

    No, for the reason given in the very first reply by yatt: the product you describe depends on a choice of basis. I.e. it is not well defined: if you compose two elements and assign a meaning to it, and then I do the same we will get different answers depending on how we chose to represent the vectors.
  15. Mar 10, 2009 #14
    Re: what about this...?

    This is still not invariant under change of basis, which a slight variant of the argument I used before shows.

    Edit: as matt grime just remarked...
  16. Mar 10, 2009 #15
    Ok, so you say that it is not possible to assign a general geometric meaning to the pointwise product because it depends on the choice of the basis in which we express the two vectors.

    Well, let's put a restriction then: what if we agree that the vectors are expressed in the canonical basis? Is there any possibility to assign a meaning to this product at least for a chosen basis?
  17. Mar 10, 2009 #16

    matt grime

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    Trivially, yes. You're using the identification of V with the diagonal matrices in End(V) - which depends on a choice of basis - and then multiplying by a matrix which is a geometric operation.

    But, at the risk of being Rusmfeldian, is it a meaningful meaning? Find an application for it, and then you'll have an answer. I don't see one, really - the basis dependence is a complete killer.

    There is a meaning for the point-wise product of matrices, but I can't remember what it is. Something in functional analysis maybe.
  18. Mar 10, 2009 #17
    ok, thanks a lot for your help. It has been very useful and appreciated.
    Essentially my problem is that I have a a vector space of (discrete finite) functions, which are basically vectors in [tex]R^n[/tex].

    All these vectors (function) undergo an operation which contains a pointwise-product.
    The operation consists in taking a vector u, finding its own corresponding [tex]\hat{u}[/tex] and then taking the pointwise product [tex]u \bullet \hat{u}[/tex].

    (The operation ^ is essentially a mirroring of the function around the y-axis).

    My question was: how do I compare the distance of the element which I created in such a way? That's why I was trying to find a meaning for the pointwise product
  19. Mar 28, 2009 #18
    I have to resume this thread since a possible solution occurred to me yesterday.
    Let's define the following very simple admissible coordinate-transformation pair:

    [tex]\mathcal{T}: \bar{x_i} = ln(x_i)[/tex]

    [tex]\mathcal{T}^{-1}: x_i = e^{\bar{x_i}}[/tex]

    If we consider two vectors [tex]\mathbf{a} = (a_1,\ldots,a_n)[/tex] and [tex]\mathbf{b} = (b_1,\ldots,b_n)[/tex] we have that:

    [tex](a_{1}b_{1},\ldots,a_{n}b_{n}) =: \mathbf{a} \bullet \mathbf{b} = \mathcal{T}^{-1}(\mathcal{T}\mathbf{a} + \mathcal{T}\mathbf{b})[/tex]

    In this sense, pointwise product between two vectors whose components are in [tex]\mathcal{R}^{+} [/tex], is equivalent to performing vector-addition onto their transformed counterparts, and then transforming back the result.

    Does this sound correct?
    Could it be generalized in order to handle also vectors with zero or negative components (by using complex numbers)?
  20. Mar 28, 2009 #19

    matt grime

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    I can't see any way in which that is 'admissible' given that you can't apply T to the origin. Also, since T isn't remotely linear, then just because you've done something that makes use of component wise addition in some array doesn't imply anything useful about the original vector space.

    So, you've just made a quite natural construction (the product of rings) into a very unnatural one only defined on a subset of the original space (and, no it can never be extended to any vector with any component zero since 0 is an essential singularity of log).
  21. Mar 28, 2009 #20
    Ok, I admit I misused the term 'admissible'.
    However let's try to make things clearer:

    1) Let's restrict our discussion to only vectors with positive (nonzero) components.
    2) Let's put the same restriction to the functions which define the transformation [tex]\mathcal{T}[/tex]. In other words we consider only those transformation-functions in the domain [tex](0,+\infty)[/tex]
    3) With this restriction the Jacobian of [tex]\mathcal{T}[/tex] is never 0 in the domain [tex](0,+\infty)[/tex]. Moreover, the functions are smooth, so they do have an inverse, which are trivially given by the transformation [tex]\mathcal{T}^{-1}[/tex]. We have a one-to-one mapping.
    4) Performing pointwise multiplication between vectors with positive components, must be indeed equivalent to performing vector-addition between their mapped counterparts.

    Was I wrong in any of those step?

    I agree with you that the transformations are not linear, that the result is trivial, natural and maybe of little interest, but I still think it is correct.
    And this would answer (partly) my question, by which I essentially wanted to know if there was ANY kind of geometry behind the pointwise-multiplication. Well, apparently there is, but only under certain restrictions.

    I fully understand that things that do not generalize, and are just particular cases do not appeal the mathematician who is primarily interested in exploring abstract concepts.
    However I can say that vectors with strictly positive values are still very useful in practice. One example that comes to my mind are images: which are represented commonly by array of positive values.

    Thanks again for your interest in this thread. Your posts are always extremely helpful!
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