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Gram-schmidts orthonormalization

  1. Nov 11, 2009 #1
    Apply Gram-Schmidts process to the sebust S of the inner product space V to obtain an orthogonal basis for span(S). Then normalize the vectors in this basis to obtain an orthognormal basis for span(S)

    V=span(S) where S={(1,i,0), (1-i,2,4i)} and x=(3+i,4i,-4).

    Isn't the length of (1,i,0) zero? I'm confused about finding orthonormal basis for a complex set. Can anyone solve this problem step by step for me?
    Thank you very much!
     
  2. jcsd
  3. Nov 11, 2009 #2

    lanedance

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    I won;t solve the problem for you, but am happy to have a look at your work

    the length of a= (1,i,0) is not zero, it is usually given by the complex innner product, which in matrix notation, if a is a complex column vector this becomes
    [tex] ||a||^2 = <a,a> = (a^*)^Ta[/[/tex]
    where * denotes the complex conjugate

    so just like a dot product, but with the complex conjugate of itself
     
  4. Nov 11, 2009 #3
    Thank you lanedance. I realized it after some searching on the internet. :)
    I have another dumb question...:P
    What's the length of (1-5i/2, 7-i/2,4i)? Is it the square root of (1-5i/2)squared+(7-i/2)squared+(4i)squared, where i2=-1?
     
  5. Nov 11, 2009 #4

    lanedance

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    the length squred will be the following dot product
    [tex] (1-5i/2, 7-i/2,4i)* \bullet (1-5i/2, 7-i/2,4i) [/tex]

    so taking the congujate
    [tex]= (1+5i/2, 7+i/2,-4i) \bullet (1-5i/2, 7-i/2,4i) [/tex]

    then the multiply out as a normal dot product

    if you're still confused, start with: what is the magnitude of the 1+5i/2 in the 1D complex space?
     
    Last edited: Nov 11, 2009
  6. Nov 11, 2009 #5
    I get it.
    Thank you very much, lanedance. :)
     
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