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Inner products

  1. Oct 4, 2007 #1
    We define: <u,v>=u2v2+u3v3

    For vectors u=(u1,u2,u3) and v=(v1,v2,v3) in R3.

    Explain the reasons why this is not an inner product on R3.



    I have completed the 4 axioms as below:

    1. <u,v>= u2v3 + u3v3

    =v2u2 + v3u3

    =<v,u>



    2.<cu,v> = cu1v2+cu2v2

    = c(u2v2+u3v3)

    = c<u,v>

    3.<u,v+w>=u2(v2+w2)+u3(v3,w3)

    = u2v2+u2w2+u3v3+u3w3

    = <u,v>+<u,w>

    4.a) <u,u>=u22 + u32

    greater than or equal to zero as u2^2 greater than or equal to zero and u3^2 is greater than or equal to zero

    b) <u,u>=0 then u2=0 and u3=0.



    Somehow ive wrongly proved all the axioms :S im assuming this does not define an inner product as it does not include u1 and v1 in the inner product, therefore cannot be an inner product in R^3. I would greatly appreciate anyone looking over my work to help me! Thanks
     
  2. jcsd
  3. Oct 4, 2007 #2

    mjsd

    User Avatar
    Homework Helper

    your last step is wrong

    proving u2 = u3=0 leaves u1 arbitrary and so the vector u = (u1, 0, 0) which is non-zero violates axiom 4 b).
     
  4. Oct 4, 2007 #3
    Thanks mjsd!! So it is correct that none of the other axioms apart from 4.b) depend on u1 or v1?
     
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