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Thread: independant proof

  1. #1
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    independant proof

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

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

    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 $\displaystyle ||{\sum_{i=1}^n{k_iy_i}}||^2=\sum_{i=1}^n|k_i|^2$ norm of one vector is 1 but here we have a sum
    If $\displaystyle \{u_i:i=1,\ldots,n\}$ is an orthonormal system then, $\displaystyle ||{\sum_{i=1}^n{k_iu_i}}||^2 =<\sum_{i=1}^n{k_iu_i},\sum_{i=1}^n{k_iu_i}>=...$

    Conclude.
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  5. #5
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    Re: independant proof

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

    Conclude.
    $\displaystyle <\sum_{i=1}^n{u_i},\sum_{i=1}^n{u_i}>$
    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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