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Inner product concept question.

  1. Apr 21, 2010 #1
    I am having difficulties with understanding some aspects of inner products. For example,

    ||u||² = <u,u>

    Where <u,u> denoted the inner product of "u" with itself.

    My problem here is that we can define any inner product we wish. For example, if I defined,

    <u,v> = u1v1 + 3(u2v2)
    Then the above equation for finding the magnitude of "u" doesn't agree with the other formula for finding the magnitude of the vector..

    ||u|| = sqrt(u1^2 + u2^2)

    Why does this happen? There are other equations where this same thing happens. Shouldnt everything agree?
     
  2. jcsd
  3. Apr 21, 2010 #2
    It's not clear to me why you feel you can define any inner product we wish. Assuming we are talking about 3D Euclidean space and normal vector calculus, there is only one correct definition of inner product based on the key geometrically intrinsic property, i.e. the magnitude or length of a vector.

    [tex] \vec U \bullet \vec V ={{\vert \vec U \vert ^2 +\vert \vec V \vert ^2 - \vert \vec V -\vec U \vert ^2}\over{2}}[/tex]

    This definition clearly results in the following:

    [tex]\vec U \bullet \vec U =\vert \vec U \vert ^2 [/tex]
     
  4. Apr 21, 2010 #3
    In my book it says that you can define a different inner product as long as it obeys all the same rules of an inner product. For example, one of the problems in my book says,

    "Find the inner product of u and v, if <u,v> = 3(u1v1) + (u2v2)"
     
  5. Apr 21, 2010 #4

    radou

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    On an inner product space, you can naturally define the norm of a vector x with ||x|| = √<x, x>, regardless of how the inner product is defined.
     
  6. Apr 21, 2010 #5
  7. Apr 21, 2010 #6
    OK, I guess this is a generalization of the concept I'm familiar with.

    Now I'm curious about your question. What are the stated rules for an inner product according to your book?
     
  8. Apr 21, 2010 #7
  9. Apr 21, 2010 #8
    So tell me if im wrong with this assumption.

    The norm of a vector is defined by its inner product. A vector could have a multiple amount of norms depending on which inner product is defined. The actual length of the vector in R^n is defined by the dot product definition of the inner product, and other inner products do not necessarily define the length of the vector.

    Would this be correct?
     
  10. Apr 21, 2010 #9
    Yes you are correct. The inner product is used to define the norm, and of course there are different norms you can use on a space that give you different properties. The norm on an arbitrary vector space is a friendly extension of the familiar concept of the length of a vector in R2.
     
  11. Apr 21, 2010 #10
    On a related note, I've seen Euclidean space defined as an affine space with an inner product (in Bowen & Wang: Introduction to Vectors and Tensors, Vol 2). Is this definition standard, and, if so, how do people refer unambiguously to the traditional ones, En? Traditional/canonical Euclidean n-space?
     
  12. Apr 22, 2010 #11

    HallsofIvy

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    Science Advisor

    In finite dimensional vector spaces, you define "norm" by "inner product". In infinite dimensional spaces, however, you can have a norm where there is no corresponding inner product.

    For example, if we define [itex]l_1[/itex] to be the set of all infinite sequences [itex]\{a_i\}[/itex] such that [itex]\sum |a_i|[/itex] is finite, then we can define the norm to be that sum, even though there is no inner product that will give that norm.

    If we define [itex]l_2[/itex] to be the set of all sequences [itex]\{a_i\}[/itex] such that [itex]\sum a_i^2[/itex] is finite, then we can show that [itex]\sum a_ib_i[/itex], where [itex]\{a_i\}[/itex] and [itex]\{b_i\}[/itex] are two such sequences, is also finite and we can define the inner product to be that sum. In that case we define the norm of [itex]\{a_i\}[/itex] to be the square root of its inner product with itself.

    The same things can be said of the set of "absolutely integrable functions" on an interval and the set of "square integrable" functions on an interval, using the integral rather than sum.
     
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