How can someone prove that the $\{dim(H)=\frac{n(n-1)}{2} uncorrect\ see\ update\}$ where $H$ is the Hamilton path polytope of a complete graph $K_n$, that consists of vertices $x\in\{0,1\}^{\mid E\mid}$
Update: I really did a silly mistake it's $\frac{n(n-1)}{2}-1$
Update:
Definition: The polytope of all Hamilton paths in a graph G=(V,E) is the convex hull of $\{x_1,\ldots,x_n\}$ characteristic vectors of those paths.
Here's my understanding of the construction. We take the set of edges $E$ of some graph $G$, and index them $E = \{e_1, \dotsc, e_m\}$ in some arbitrary order, where $m = |E|$. Then, to each Hamiltonian path in $G$ (i.e. a path visiting every vertex exactly once), we associate a point in $\mathbb{R}^m$ whose $i$th coordinate is $1$ if $e_i$ is in the path, and $0$ otherwise (the "characteristic vector" of the path). The convex hull $P$ of all these points is the "Hamilton path polytope" of $G$.
When $G = K_n$, then $m = \binom{n}{2}$. The affine dimension of $P$ is therefore at most $\binom{n}{2}$. Since every Hamiltonian path has $n - 1$ edges, every characteristic vector contains exactly $n-1$ coordinates which are $1$. Thus the whole polytope $P$ lies in the hyperplane where the sum of all the coordinates is $n-1$; so the affine dimension of $P$ is at most $\binom{n}{2} - 1$.
On the other hand, pick an ordering of the edges so that the first $n$ edges $e_1, \dotsc, e_n$ form a cycle $C$ in $G$. Leaving out any one of these yields a Hamiltonian path, and the characteristic vectors of the $n$ Hamiltonian paths thus obtained are linearly independent. To see this, just check that the determinant of the $n \times n$ matrix $$ (1 - \delta_{ij})_{i,j} = \begin{bmatrix} 0 & 1 & 1 & \cdots & 1 \\ 1 & 0 & 1 & \cdots & 1 \\ 1 & 1 & 0 & & \vdots \\ \vdots & \vdots & & \ddots & 1 \\ 1 & 1 & \cdots & 1 & 0 \end{bmatrix} $$ is nonzero. In fact, it's $(-1)^{n-1}(n-1)$.
Now consider each edge from $e_{n+1}$ up to $e_m$. Let $e_j$ be such an edge. Because $G$ is a complete graph, we can find a Hamiltonian path using $e_j$ and otherwise only edges from $\{ e_i \mid i < j \}$. In fact, since $e_j$ is a chord in the cycle $C$, we can find such a path using just edges from $C$, as suggested by this picture:
It follows that for each $j$ there is a characteristic vector whose nonzero entries are either among the first $n$, or in the never-before-used $j$th coordinate; it is thus linearly independent from all the previously added vectors.
In this way, we build a set of $m = \binom{n}{2}$ linearly independent vectors among the points in $P$. If they are linearly independent, then they are certainly affinely independent. So there is a set of at least $\binom{n}{2}$ affinely independent vectors among our points; therefore the affine dimension is at least $\binom{n}{2} - 1$.
Remembering the upper bound from the beginning, we conclude that the affine dimension is exactly $\binom{n}{2} - 1$.