I am not a mathematician but I do use some tools from geometry in robotics. So, I apologize if what I am writing here is not mathematically consistent but I really do need your help.
I have a linear system $y = Ax$ with $x \in \mathbb{R}^n$, $y \in \mathbb{R}^m$ and $n>m$. I want to find a set $\mathcal{Y} = \{y \in \mathbb{R}^m \, | \, y = Ax,\, x \in \mathcal{X}\}$, where $\mathcal{X}$ is an $n$-dimensional hypercube that may be expressed as $\mathcal{X} = \{ x \in \mathbb{R}^n \, | \, C^{T}x \leq d \}$, where $C = \begin{bmatrix} \,I_n \,| \, -I_n\, \end{bmatrix}$ with $I_n$ being an $n$ identity matrix and $d = \begin{bmatrix} \,d_M^T \,| \, -d_m^T\, \end{bmatrix}^T$ with $d_M$ and $d_m$ define the upper and lower bounds of $\mathcal{X}$. The fact that $x = A^+y+N\lambda$, then finding $\mathcal{Y}$ amounts to computing the orthogonal projection of the hyper-parallelpiped defined by $\{\, (y,\, \lambda) \in \mathbb{R}^m \times \mathbb{R}^{n-m}\, | \, P y + Q \lambda \leq d \}$ onto $y$ (i.e. the elimination of $\lambda$), where $ P = \begin{bmatrix} \,A^+ \, \\ -A^+ \end{bmatrix}, \, Q = \begin{bmatrix} \, N \, \\ -N \, \end{bmatrix} $
My question is how eliminate $\lambda$ without using classical Fourier-Motzkin elimination?
I would like to share with you what I did for this problem. Let us denote $p_i^T \in \mathbb{R}^m$ and $q_i^T \in \mathbb{R}^{n-m}$ as the $i$-th row of $P$ and $Q$, respectively, $\forall \, i = 1, \, 2, \,, \ldots, \, 2n$. The hyper-parallelepiped is defined by the intersection of $2n$ halfspaces. In principle, in order to eliminate $l$ variables from a system of linear inequalities, $l+1$ inequalities are needed. So to eliminate $\lambda \in \mathbb{R}^{n-m}$ we need $n-m+1$ lines from the $2n$ inequalities. Let $P_j$, $Q_j$ and $d_j$ denote the $j$ combination representing an $n-m+1$ lines from $P$, $Q$, and $d$. If there exist a positive vector in the null space $Q_j^T$ i.e. $\mathcal{N}(Q_j^T)>0$ then we eliminate $\lambda$ to get $\mathcal{N}^T(Q_j^T)P_j y \leq \mathcal{N}^T(Q_j^T)d_j$. If $\mathcal{N}(Q_j^T)$ is not positive then we can neglect this combination. Taking all combinations into account we get all facets of projection.
One key note about the method described above is that I always get a $2C_{n-m+1}^{n}$ that give me $\mathcal{N}(Q_j^T)>0$ which is actually the exact number of facets that I got from other different methods. Any interpretation for that and a proof? How can we prove in math terms that facets we got belongs to the projection?