1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Inner Product Space - Pythagorean?

  1. Feb 5, 2015 #1

    ElijahRockers

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    Let ##V## be an inner product space and let ##V_0## be a finite dimensional subspace of ##V##. Show that if ##v ∈ V## has ##v_0 = proj_{V_0}(v)##:

    ||v - vo||^2 = ||v||^2 - ||vo||^2

    2. Relevant equations
    General inner product space properties, I believe.

    3. The attempt at a solution
    So Vo is finite dimensional and has an orthonormal basis {v1, v2,...., vn} for finite n. I can say that right? If so,

    ##v_0 = \sum_{i=1}^{n}<v,v_i>v_i## (Not sure if that helps really)

    Also <v - vo , vo> = 0 since they are orthogonal.

    That's about where I get stuck. I see that the expression I'm supposed to arrive at is basically pythagorean theorem for inner product spaces, and I can see it in my head quite easily if n is 2 dimensional.

    So I tried working backwards:

    ||v - vo||^2 = ||v||^2 - ||vo||^2

    ||v - vo||^2 + ||vo||^2 = ||v||^2

    And using inner product properties I can get:

    <v - vo , v - vo> + <vo, vo> = <v , v>

    I don't know if that really helps though.

    Any advice is appreciated.
     
  2. jcsd
  3. Feb 5, 2015 #2

    pasmith

    User Avatar
    Homework Helper

    Expand [itex]\|v - v_0\|^2 = \langle v - v_0, v - v_0 \rangle[/itex]. You should get four terms. Why do the two terms you don't want vanish?
     
  4. Feb 5, 2015 #3

    ElijahRockers

    User Avatar
    Gold Member

    Didn't think I could...

    So would it be:
    ##<v,v> -2<v,v_0 > + 2<v_0 , v_0> = <v, v>##
    ##<v,v_0 > = <v_0, v_0>##
    ##v = v_0##
    ##v - v_0 = 0##

    so ##<v - v_0 , v_0> = 0##

    Is that right?
     
  5. Feb 5, 2015 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You say you can show <v - vo , v - vo> + <vo, vo> = <v , v>? Isn't that the same thing as ||v - vo||^2 + ||vo||^2 = ||v||^2?
     
  6. Feb 5, 2015 #5

    ElijahRockers

    User Avatar
    Gold Member

    Yea. but as far as I am able to tell I'm supposed to be able to show that given vo = proj_Vo (v). I was working backwards from ||v - vo||^2 etc
     
  7. Feb 5, 2015 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    How did you define projection? Have you shown <v-vo,vo>=0? That's all you really need. Expand ||v|| = ||(v-vo)+vo||.
     
  8. Feb 5, 2015 #7

    ElijahRockers

    User Avatar
    Gold Member

    Yes, I defined that in my original post. That's what I was thinking! Thanks!

    EDIT: Shown, not defined.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Inner Product Space - Pythagorean?
  1. Inner Product Space (Replies: 1)

  2. Inner product spaces (Replies: 1)

Loading...