Gram-Schmidt

**Mollier** · February 13th 2011, 11:30 PM

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

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

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

which in turn implies that

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

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

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

so

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

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

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

Thanks.

**Ackbeet** · February 15th 2011, 10:26 AM

The second-to-last equality is just a complex number written as its magnitude times a phase.

The last equality is just the expression

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

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

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