Suppose I have an object in $\mathbb{R}^5$ described by:
$$x_1+x_2+x_3+x_4+x_5=1$$ $$x_1+2x_2+3x_3+4x_4+5x_5=6$$ $$x_1+7x_2+8x_3+9x_4+10x_5=11$$ $$x_1,x_2,x_3,x_4,x_5 \geq 0$$
Is there a way that I could map this object onto the k-dimensional unit-simplex, described by (I don't really know what dimension is proper for having a bijection, but as long as it is a mapping to an unit-simplex that is fine):
$${x_1}\prime + {x_2}\prime + ... +{x_k}\prime = 1$$ $${x_1}\prime, {x_2}\prime, ..., {x_k}\prime \geq 0$$
Best case scenario, I want a mapping that is invertible and linear. Moreover, the inverse image of the mapping should be equal to the domain of the map.
I am interested in this because: I want to sample from the object in $\mathbb{R}^5$, but it is difficult to do so. I can, however, sample easily from the lower-dimensional simplex. Thus, if I could have an invertible mapping with nice properties, this might solve my problem!
EDIT: turns out the region is empty. Go Figure.
the object in question is a convex pentagon in a 2-plane; The plane can be described by finding a favorite point $P$ in it in $\mathbb R^5;$ then find, say, an orthonormal basis $\vec{u}, \vec{v}$ for the plane given when all three right-hand sides are changed to $0.$ Your pentagon is then given as $$ P + s \vec{u} + t \vec{v}, $$ with five linear inequalities on the pair $(s,t)$ in order to satisfy all the $x_i \geq 0.$
$$ \left( \begin{array}{ccccc} 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 & 5 \\ 1 & 7 & 8 & 9 & 10 \end{array} \right) $$ $$ \left( \begin{array}{ccccc} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 & 4 \\ 0 & 6 & 7 & 8 & 9 \end{array} \right) $$ $$ \left( \begin{array}{ccccc} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 & 4 \\ 0 & 0 & -5 & -10 & -15 \end{array} \right) $$ $$ \left( \begin{array}{ccccc} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 & 4 \\ 0 & 0 & 1 & 2 & 3 \end{array} \right) $$ $$ \left( \begin{array}{ccccc} 1 & 0 & -1 & -2 & -3 \\ 0 & 1 & 2 & 3 & 4 \\ 0 & 0 & 1 & 2 & 3 \end{array} \right) $$ $$ \left( \begin{array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 & -2 \\ 0 & 0 & 1 & 2 & 3 \end{array} \right) $$
I get $$ \vec{u} = \frac{1}{\sqrt 6} \left(0,1,-2,1,0 \right) $$ $$ \vec{v} = \frac{1}{\sqrt {30}} \left(0,2,-1,-4,3 \right) $$ which you can check yourself, easier to ignore the square roots.
Next you need the one favorite point $P,$ which amounts to repeating this whole matrix business, but augmented with the original $1,6,11$ as another column, finding row echelon form, and picking a solution with all $x_i \geq 0.$
Alright, well, did it myself, I no longer believe that the region is feasible. Indeed, I get the augmented matrix $$ \left( \begin{array}{ccccc|c} 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & -1 & -2 & -3\\ 0 & 0 & 1 & 2 & 3 & 4 \end{array} \right) $$ which requires, for all nonegative $x_i,$ that
$$ x_4 + 2 x_5 \geq 3, $$ $$ 2 x_4 + 3 x_5 \leq 4. $$ So, if you copied the problem correctly, they are just screwing with you. since $$ - x_4 - 2 x_5 \leq -3, $$ $$ -3 x_4 - 6 x_5 \leq -9, $$ $$ 4 x_4 + 6 x_5 \leq 8, $$ $$ x_4 \leq -1. $$
Alright, let me change the problem to force a solution with all entries equal to $1,$ so the new system is $$ \color{magenta}{x_1+x_2+x_3+x_4+x_5=5,}$$ $$ \color{magenta}{x_1+2x_2+3x_3+4x_4+5x_5=15,}$$ $$ \color{magenta}{x_1+7x_2+8x_3+9x_4+10x_5=35,}$$ $$x_1,x_2,x_3,x_4,x_5 \geq 0.$$ As a result, we have parametrized the set as $$ (1,1,1,1,1) + \frac{s}{\sqrt 6} \left(0,1,-2,1,0 \right) + \frac{t}{\sqrt {30}} \left(0,2,-1,-4,3 \right) $$ or $$ \left(1, \; 1 + \frac{s}{\sqrt 6} + \frac{2t}{\sqrt {30}}, \; 1 - \frac{2s}{\sqrt 6} - \frac{t}{\sqrt {30}}, \; 1 + \frac{s}{\sqrt 6} - \frac{4t}{\sqrt {30}}, \; 1 + \frac{3t}{\sqrt {30}} \right). $$
In this case, the polygon is $st$ space is a non-empty convex set, as $(s=0,t=0)$ is in it, $$ \color{magenta}{ 1 + \frac{s}{\sqrt 6} + \frac{2t}{\sqrt {30}} \geq 0,} $$ $$ \color{magenta}{ 1 - \frac{2s}{\sqrt 6} - \frac{t}{\sqrt {30}} \geq 0,} $$ $$ \color{magenta}{1 + \frac{s}{\sqrt 6} - \frac{4t}{\sqrt {30}} \geq 0,} $$ $$ \color{magenta}{ 1 + \frac{3t}{\sqrt {30}} \geq 0.} $$