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# 数学代写|几何组合代写Geometric Combinatorics代考|MAT361 Identities in the Algebra of Polyhedra

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## 数学代写|几何组合代写Geometric Combinatorics代考|Identities in the Algebra of Polyhedra

What can we do with polyhedra? One important observation is that the image of a polyhedron under a linear transformation is a polyhedron.

Theorem 1. Let $P \subset \mathbb{R}^d$ be a polyhedron and let $T: \mathbb{R}^d \longrightarrow \mathbb{R}^k$ be a linear transformation. Then $T(P) \subset \mathbb{R}^k$ is a polyhedron. Furthermore, if $P$ is a rational polyhedron and $T$ is a rational linear transformation (that is, the matrix of $T$ is rational), then $T(P)$ is a rational polyhedron.

The crucial step in the proof. Let us consider the following particular case: $k=d-1$ and $T$ is the projection onto the first $(d-1)$ coordinates: $\left(x_1, \ldots, x_d\right) \longmapsto$ $\left(x_1, \ldots, x_{d-1}\right)$. Suppose that the polyhedron $P$ is defined by a system of linear inequalities:
$$\sum_{j=1}^d a_{i j} x_j \leq b_i \quad \text { for } \quad i=1, \ldots, m$$
Let us look at the coefficients of $x_d$.
Let $I_{+}=\left{i: a_{i d}>0\right}, I_{-}=\left{i: a_{i d}<0\right}$, and $I_0=\left{i: a_{i d}=0\right}$. Then a point $y=\left(x_1, \ldots, x_{d-1}\right)$ belongs to $T(P)$ if and only if
$$\sum_{j=1}^{d-1} a_{i j} x_j \leq b_j \quad \text { for } \quad i \in I_0$$
and there exists $x_d$ such that
(2)
\begin{aligned} &x_d \leq \frac{b_i}{a_{i d}}-\sum_{j=1}^{d-1} \frac{a_{i j}}{a_{i d}} x_j \quad \text { for } \quad i \in I_{+} \ &x_d \geq \frac{b_i}{a_{i d}}-\sum_{j=1}^{d-1} \frac{a_{i j}}{a_{i d}} x_j \quad \text { for } \quad i \in I_{-} \end{aligned}

## 数学代写|几何组合代写Geometric Combinatorics代考|A plausible argument

A plausible argument. We don’t really prove this important theorem, although we come very close. We start by showing that the theorem is not obviously false.
We notice that if $P$ is non-empty and does not contain vertices then $P$ contains a line and hence we can choose $g=[P]$.

Suppose we have been sloppy and included in the sum not only all vertices $v$ of $P$ but also some non-vertices $v \in P$. No harm done: if $v \in P$ is a non-vertex then $\operatorname{co}(P, v)$ contains a line and so we just have to adjust $g$. This shows that the formula is robust enough.

Suppose that the theorem holds for some polyhedron $P \subset \mathbb{R}^d$ and let $T: \mathbb{R}^d \longrightarrow$ $\mathbb{R}^k$ be a sufficiently generic linear transformation. We claim that the theorem holds for the image $T(P)$. Indeed, by Theorem 2 the transformation $T$ gives rise to the transformation $\mathcal{T}$ on the algebra of polyhedra. Let us apply $\mathcal{T}$ to both sides of the identity. We have $\mathcal{T}[P]=[T(P)]$ and $\mathcal{T}[\operatorname{co}(P, v)]=[T(\operatorname{co}(P, v))]=$ $[\operatorname{co}(T(P), T(v))]$, cf. Review Problem 10.

We have to be somewhat careful with $g$ : we know that $g$ is a linear combination of indicators of polyhedra with lines. If we are unlucky, the kernel of $T$ may “eat up” some of those lines and $\mathcal{T}(g)$ will not lie in $\mathcal{P}_0\left(\mathbb{R}^k\right)$. This is the reason why we chose $T$ to be “generic”. Thus if we prove the theorem for some “model” polyhedra $P$, we can extend it (with some care) to polyhedra obtained from $P$ by linear transformations.

## 数学代写|几何组合代写Geometric Combinatorics代考|Identities in the Algebra of Polyhedra

$$\sum_{j=1}^d a_{i j} x_j \leq b_i \quad \text { for } \quad i=1, \ldots, m$$

$$\sum_{j=1}^{d-1} a_{i j} x_j \leq b_j \quad \text { for } \quad i \in I_0$$

(2)
$$x_d \leq \frac{b_i}{a_{i d}}-\sum_{j=1}^{d-1} \frac{a_{i j}}{a_{i d}} x_j \quad \text { for } \quad i \in I_{+} \quad x_d \geq \frac{b_i}{a_{i d}}-\sum_{j=1}^{d-1} \frac{a_{i j}}{a_{i d}} x_j \quad \text { for } \quad i \in I_{-}$$

## MATLAB代写

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