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B Does the L2 norm of a vector destroy all directional info?

  1. May 24, 2016 #1
    Sorry I'm a little rusty with my math and proof logic, and this feels like a dumb question, but oh well! The Euclidian norm of a vector in ℝ3 is [tex] \|{v}\| = \sqrt{x^2 + y^2 + z^2}[/tex] where [tex] \|{v}\| \geq 0. [/tex] I'm trying to show that there is always an infinite number of solutions for arbitrary positive v, in other words that there are an infinite number of vectors of any fixed length. I intuitively know this to be true by visually imagining that fixing v gives a sphere of possible solutions.

    I can disprove through counterexample easily that [tex] \{ \forall v \mid \|{v}\| = \sqrt{{x_1}^2 + {y_1}^2 + {z_1}^2} > 0 \} \Rightarrow

    \\ \{\nexists (x_2, y_2, z_2) \neq (x_1, y_1, z_1) \mid \|{v}\| = \sqrt{{x_2}^2 + {y_2}^2 + {z_2}^2} = \sqrt{{x_1}^2 + {y_1}^2 + {z_1}^2} \}.[/tex] I'm thinking I could just convert or invert this somehow logically to show what I want. I know the way to show that only one solution exists for an equation, but not how to show an infinite number of solutions exists. I'm recalling free variables in linear systems but I don't see how I might apply that here in a general case.

    Assuming this to be true, this means that if I want to destroy the directional information of a trajectory I can just take it's norm at every point? The reason I want to do this is to train a classifier to classify two types of trajectories but not by using the starting and ending points or any other directional information, because the two types of trajectories end at different positions in space and that's too "obvious." Instead I'd take the magnitude of the position, velocity, acceleration, and jerk, hopefully still retaining some other kinds of qualities of the trajectory. Another way to do this might be to warp every trajectory to the same starting and end points in space, but I still worry the trajectories are still discernible solely by using location/directional information in some way. I think I'd have to do this anyway, too, when using position magnitude. To put this all another way, I'm not really interested in the shape of the trajectory, per se, but more in it's qualities and dynamics. Like, two trajectories can look exactly the same in space, but not in time. Hope this is clear enough. Happy to provide more info if necessary.
     
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  3. May 24, 2016 #2

    mathman

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    In n dimensions (for your example, n = 3), all vectors (v) from the origin to the surface of a sphere of radius r have ||v||=r.
     
  4. May 25, 2016 #3
    Is there any way to prove this?
     
  5. May 25, 2016 #4

    mathman

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    By the definition of a sphere.
     
  6. May 26, 2016 #5

    lavinia

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    You can parameterize a portion of the sphere and then observe that there are infinitely many parameters.

    For the sphere in three space one could use

    ##(cos(α)cos(θ),cos(α)sin(θ),sin(α))##
     
  7. May 27, 2016 #6

    micromass

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    And this generalizes in a rather obvious way to multiple dimensions: https://en.wikipedia.org/wiki/N-sphere#Spherical_coordinates
     
  8. May 27, 2016 #7

    mathman

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    In n dimensions, a spherical surface of radius r centered at the origin consists of all points [itex](x_1,x_2,....,x_n)\ where\ x_1^2+x_2^2+....x_n^2=r^2[/itex]. Therefore [itex]||(x_1,x_2,....,x_n)||=r[/itex] .
     
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