Explicit Representation of the SU(N) Simple Roots in with redundant coefficents? Commonly the simple roots for $SU(n)$ groups are given as $n$ dimensional vectors, although root-space is $n-1$ dimensional.
The $SU(n)$ Wikipedia article explains:

Here, we use n redundant coordinates instead of n − 1 to emphasize the symmetries of the root system (the n coordinates have to add up to zero). In other words, we are embedding this n − 1 dimensional vector space in an n-dimensional one.

I want to know how can I transform back from these redundant coordinates to the non-redundant ones.
A concrete example: 
I need an explicit representation of the simple roots of $SU(5)$ in four dimensions. Unfortunately these are usually given in five-dimensions, because the coefficients are then all integers. 
$$\{\text{(1, -1, 0, 0, 0)},\text{(0, 1, -1, 0, 0)},\text{(0, 0, 1, -1, 0)},\text{(0, 0, 0, 1, -1)}\} $$
How can I compute the corresponding coefficents in four-dimension? The result should be sth. like 
$$\{1,0,0,0\} , \left\{-\frac{1}{2},\frac{\sqrt{3}}{2},0,0\right\}, \left\{0,0,-\frac{1}{\sqrt{3}},\sqrt{\frac{2}{3}}\right\} \left\{0,0,-\frac{\sqrt{\frac{3}{2}}}{2},\frac{\sqrt{\frac{5}{2}}}{2}\right\}, $$
 A: You just need to find a set of points in $\mathbb{R}^4$ which have the same inner products with each other as the original points.  This is equivalent to finding an orthogonal transformation of $\mathbb{R}^5$ that sends the original hyperplane $H : x_1 + x_2 + \ldots + x_5 = 0$ to one of the coordinate hyperplanes, say $x_5=0$.  
Written as a matrix in the original coordinates, the transformation will have $(1,1,1,1,1)/\sqrt{5}$ in the bottom row (assuming your vectors are column vectors). Then complete to an orthogonal matrix in an arbitrary way: the other rows must be linearly independent unit vectors in $\mathbb{R}^5$ orthogonal to $(1,1,1,1,1)/\sqrt{5}$.  There are lots of ways to do this.  Now when you apply this matrix to your original vectors, the result will always have a 0 in the last coordinate, which you can drop, leaving a point in $\mathbb{R}^4$.
Your question suggests that you are expecting an answer of a particular form, but in fact, the answer isn't unique and there are lots of different answers you could get which are all correct, depending on how you complete the orthogonal matrix.  Maybe there is a particularly "nice" completion that gives you the answer you suggest, but I don't see it right now.
