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Math Help - independant proof

  1. #1
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    independant proof

    there is \{y_{1}..y_{n}\} a group of vectors in R^{n} ,there is \{u_{1}..u_{n}\} orthonormal basis of R^{n}.
    the first part i have proved that for  \lambda_{1}..\lambda_{n}\ in R
    \parallel\sum_{i=1}^{n}\lambda_{i}y_{i}\parallel^{  2}\leq(\sum_{i=1}^{n}|\lambda_{i}|^{2})(\sum_{i=1}  ^{n}\parallel y_{i}\parallel^{2})
    the second part is my question
    prove that if \{y_{1}..y_{n}\} is a set for which \sum_{i=1}^{n}\parallel y_{i}\parallel^{2}<1 then \{y_{1}+u_{1,}..,y_{n}+u_{n}\} is independant?
    how i tried to solve it:
    a set is independant if for k_{1}(y_{1}+u_{1})+..+k_{n}(y_{n}+u_{n})=0the only way it to be zero is if k1=..=kn=0
    i opened it and got k_{1}y_{1}+..+k_{n}y_{n}=k_{1}u_{1}+..+k_{n}u_{n}
    so on both side we have equal vectors so is their norm
    (\sum_{i=1}^{n}|k_{i}|^{2})(\sum_{i=1}^{n}\paralle  l y_{i}\parallel^{2})=(\sum_{i=1}^{n}|k_{i}|^{2})(\s  um_{i=1}^{n}\parallel u_{i}\parallel^{2})<br />
what next how to prove that k_{1}=..=k_{n}=0
    ??
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: independant proof

    We have \sum_{i=1}^{n}k_iy_i=-\sum_{i=1}^{n}k_iu_i . Taking norms we obtain ||{\sum_{i=1}^n{k_iy_i}}||^2=\sum_{i=1}^n|k_i|^2 . But \sum_{i=1}^n|k_i|^2\leq (\sum_{i=1}^n|k_i|^2)(\sum_{i=1}^n||y_i||^2) . As \sum_{i=1}^n||y_i||^2\right< 1 , necessarily \sum_{i=1}^n|k_i|^2=0 which implies k_i=0 for all i=1,\ldots,n .
    Last edited by FernandoRevilla; July 26th 2011 at 04:50 AM.
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  3. #3
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    Re: independant proof

    Quote Originally Posted by FernandoRevilla View Post
    We have \sum_{i=1}^{n}k_iy_i=-\sum_{i=1}^{n}k_iu_i . Taking norms we obtain ||{\sum_{i=1}^n{k_iy_i}}||^2=\sum_{i=1}^n|k_i|^2 . But \sum_{i=1}^n|k_i|^2\leq (\sum_{i=1}^n|k_i|^2)(\sum_{i=1}^n||y_i||^2) . As \sum_{i=1}^n||y_i||^2\right< 1 , necessarily \sum_{i=1}^n|k_i|^2=0 which implies k_i=0 for all i=1,\ldots,n .
    why if orthonormal then ||{\sum_{i=1}^n{k_iy_i}}||^2=\sum_{i=1}^n|k_i|^2
    ?

    norm of one vector is 1
    but here we have a sum
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: independant proof

    Quote Originally Posted by transgalactic View Post
    why if orthonormal then ||{\sum_{i=1}^n{k_iy_i}}||^2=\sum_{i=1}^n|k_i|^2 norm of one vector is 1 but here we have a sum
    If \{u_i:i=1,\ldots,n\} is an orthonormal system then, ||{\sum_{i=1}^n{k_iu_i}}||^2 =<\sum_{i=1}^n{k_iu_i},\sum_{i=1}^n{k_iu_i}>=...

    Conclude.
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  5. #5
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    Re: independant proof

    Quote Originally Posted by FernandoRevilla View Post
    If \{u_i:i=1,\ldots,n\} is an orthonormal system then, ||{\sum_{i=1}^n{k_iu_i}}||^2 =<\sum_{i=1}^n{k_iu_i},\sum_{i=1}^n{k_iu_i}>=...

    Conclude.
    <\sum_{i=1}^n{u_i},\sum_{i=1}^n{u_i}>
    again we have here a sum of vectors not a single vector,
    the norm of the single vector is 1


    why the resolt is 1 too?
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