1. ## 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.

2. 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

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