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Norm of a vector problem.

  1. Jul 20, 2011 #1
    I'm trying to do a problem concerning converging sequences in normed linear spaces. Can anyone help me prove that if x=(x1,x2........,xn) is a vector in an n dimensional vector space then |xi| where i=1,2.....,n; is always less than or equal to ||x|| (norm of x). Maybe start out by writing x as a sum of n multiples of the basis vectors?
     
    Last edited: Jul 20, 2011
  2. jcsd
  3. Jul 20, 2011 #2

    lanedance

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    its not quite true, should be less than or equal..

    is this the usual norm or just a norm in general?
     
  4. Jul 20, 2011 #3
    yeah, i forgot to put in the "or equal to". It is any norm in general.
     
  5. Jul 20, 2011 #4

    lanedance

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    so what do you know about a norm that may help?

    in particular, i would look at the triangle inequality
     
  6. Jul 21, 2011 #5

    lanedance

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    how about considering
    x1 = (x1,0,0,..) and
    x = x1 + u
    or maybe even better
    x1 = x+(-u)
     
  7. Jul 21, 2011 #6
    uhhh
    So I have ||x1|| = |x1| <= ||x|| + ||u||
     
  8. Jul 22, 2011 #7
    I still don't see how the triangle inequality implies this =/. I think I proved it using orthogonality and inner product properties in the finite dimensional case.

    Assume the negation is true. There exists a non-zero vector x=(x1,x2.....xn) such that
    |x1| > ||x|| writing out with an orthonormal basis, we get ||x1e1|| > ||x1e1.......+xnen||.
    On squaring, we'd get x1^2 > x1^2 +...... + xn^2 whiich is a contradiction?
     
  9. Jul 23, 2011 #8

    lanedance

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    so this assumes the standard norm - is that ok?
     
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