
GramSchmidt
Hi,
I learned GramSchmidt 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.

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