Which vector to take first in Gram–Schmidt process? When trying to ortonormalize a basis using the Gram–Schmidt process often quite ugly results pop out. I know the basic algorithm, but have been told that the good choice of vectors would have an influence on how the result looks like. So in which way should I proceed so that calculations are as friendly as possible? I was thinking about selecting the $v_k$ in such a way that $proj_{v_k}(*)$ is not unnesessarily complicated - i.e. that $v_k$ itself does not contain to many irrational components. 
 A: This is a kinda trivial advice: If your given basis already contains an orthonormal (or at least -gonal) subset, take those vectors first. Then nothing will happen in the first steps.
For an example try to use Gram-Schmidt on $e_1,e_2,e_2+e_3$ and then on $e_2+e_3,e_2,e_1$.
A: There are no irrationals in Gram-Schmidt !
The basic step is normalization of a vector and substraction of its projection on another. This is achieved by
$$b\leftarrow b-\left(b\cdot\frac{a}{\|a\|}\right)\frac{a}{\|a\|}=b-\frac{(b\cdot a)\,a}{\|a\|^2}.$$
If the components of $a,b$ are rational, they remain so because the norm is squared. If they are integer, you can even stick to integers by moving the denominator
$$b\leftarrow b\,\|a\|^2-(b\cdot a)\,a.$$
If needed, you normalize all vectors in the end. (And at any time you can simplify factors common to all components.)

For example,
$$\left(\begin{matrix}1 &1 &1\\1&1&0\\1&0&0\end{matrix}\right)$$
gives
$$\left(\begin{matrix}1 &1 &1\\1&1&\bar2\\1&\bar1&0\end{matrix}\right),$$
by
$$(1,1,0)-\frac{2\,(1,1,1)}3=\frac{(1,1,\bar2)}3$$
and
$$(1,0,0)-\frac{1\,(1,1,1)}3-\frac{1\,(1,1,\bar2)}6=\frac{(1,\bar1,0)}6.$$
A: In many cases you can avoid having to use Gramm-Schmidt all together. Gramm-Schmidt is, for example, often used to orthogonally diagonalize symmetric matrices. For example, consider the matrix $$A=\begin{pmatrix}
0&2&2\\
2&0&2\\
2&2&0
\end{pmatrix}.$$
The characteristic polynomial is $P_A(X)=-(X+2)^2(X-4)$. Hence we have to calculate the eigenspaces for the eigenvalues $-2$ and $4$. The eigenspace corresponding to the value $-2$ can be found as the null space of 
$$\begin{pmatrix}
2&2&2\\2&2&2\\2&2&2
\end{pmatrix}\sim \begin{pmatrix}
1&1&1\\0&0&0\\0&0&0
\end{pmatrix}.$$
Hence a basis for this eigenspace is $v_1=(-1,1,0), v_2=(-1,0,1)$. In the same fashion one finds that $v_3=(1,1,1)$ yields a basis for the eigenspace corresponding to the eigenvalue $4$.
The next step is to orthonormalize this basis. (By basic linear algebra we already know that $v_3$ is orthogonal to $v_1$ and $v_2$, hence we have to concentrate on $v_1$ and $v_2$.) Now notice that $<v_1,v_2>=1$. So normally you would now use Gramm-Schmidt to orthonormalize this basis.
But, notice that $<v_2,v_2>=2$. Hence $$<xv_1+yv_2,v_2>=x<v_1,v_2>+y<v_2,v_2>=x+2y.$$
So take $x=2$ and $y=-1$. Call $w_1=xv_1+yv_2=(-1,2,-1)$. Then $w_1,v_2$ still forms a basis of the eigenspace corresponding to $-2$ and $<w_1,v_2>=0$. Thus the eigenbasis $w_1,v_2,v_3$ is already an orthogonal basis. We can then simply normalize each vector to obtain an orthonormalized basis.
So while this is not a direct answer to your question, it is often useful to remember that in many applications you have a choice in basis to start with. You can then introduce parameters (like the $x$ and $y$) to modify your basis. Often you will see immediately what $x$ and $y$ should be in order to simplify some calculations.
