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Thread: Gram-Schmidt

  1. #1
    Member Mollier's Avatar
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    Gram-Schmidt

    Hi,

    I learned Gram-Schmidt a few years ago, and now that I'm reading about it from another source I've having some problems with it.

    Suppose that $\displaystyle O_k=\{u_1,u_2,\dots,u_k\}$ is an orthonormal basis for $\displaystyle S_k=span\{x_1,x_2,\dots,x_k\}$, and consider the problem of finding one additional vector $\displaystyle u_{k+1}$ such that $\displaystyle O_{k+1}=\{u_1,u_2,\dots,u_k,u_{k+1}\}$ is an orthonormal basis for $\displaystyle S_{k+1}=span\{x_1,x_2,\dots,x_k,x_{k+1}\}$.

    For this to hold, the Fourier expansion of $\displaystyle x_{k+1}$ with respect to $\displaystyle O_{k+1}$ must be,

    $\displaystyle x_{k+1}=\sum^{k+1}_{i=1}(u^*_ix_{k+1})u_i$,

    which in turn implies that

    $\displaystyle u_{k+1}=\frac{x_{k+1}-\sum^{k}_{i=1}(u^*_ix_{k+1})u_i}{u^*_{k+1}x_{k+1}}$.

    Since |$\displaystyle |u_{k+1}||=1$, we have that

    $\displaystyle |u^*_{k+1}x_{k+1}|=|| x_{k+1} - \sum^{k}_{i=1}(u^*_ix_{k+1})u_i||$,

    so

    $\displaystyle u^*_{k+1}x_{k+1}=e^{i\theta}||x_{k+1}-\sum^{k}_{i=1}(u^*_ix_{k+1})u_i||$ for some $\displaystyle 0\leq\theta\2\pi$, and

    $\displaystyle u_{k+1}=\frac{x_{k+1}-\sum^{k}_{i=1}(u^*_ix_{k+1})u_i}{e^{i\theta}||x_{k +1}-\sum^{k}_{i=1}(u^*_ix_{k+1})u_i||}$

    I do not understand the last two equalities. Would be great if someone could help me out a bit.

    Thanks.
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  2. #2
    A Plied Mathematician
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    The second-to-last equality is just a complex number written as its magnitude times a phase.

    The last equality is just the expression

    $\displaystyle \displaystyle u^*_{k+1}x_{k+1}=e^{i\theta}\left\|x_{k+1}-\sum^{k}_{i=1}(u^*_ix_{k+1})u_i\right\|$ substituted into the denominator of

    $\displaystyle \displaystyle u_{k+1}=\dfrac{x_{k+1}-\sum^{k}_{i=1}(u^*_ix_{k+1})u_i}{u^*_{k+1}x_{k+1}} .$
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