Given some hyperplane arrangement $\mathcal{A}$, we call any subset $\mathcal{B}\subseteq \mathcal{A}$ $\textit{central}$ if $$\displaystyle \bigcap_{H\in \mathcal{B}}H\neq \emptyset.$$
There is a proposition (3.11.3 in Stanley's Enumerative Comb. vol. I, ed. 2) that says:
Let $\mathcal{A}$ be an arrangement in an $n$-dimensional vector space. Then the characteristic polynomial is $$\chi_\mathcal{A} (x)=\displaystyle \sum_{\substack{\mathcal{B}\subseteq \mathcal{A}\\ \mathcal{B}\, \text{central}}} (-1)^{|\mathcal{B}|}x^{n-\text{rank}(\mathcal{B})}.$$
I'm working on the following problem:
For the arrangement $\mathcal{A}$ (all in $\mathbb{R}^n$), show that the characteristic polynomial is as indicated:
$x_i=x_j$ for $1\leq i < j \leq n$ and $x_i=0$ for $1\leq i \leq n$. Then
$$\chi_\mathcal{A} (x)=(x-1)^2(x-2)(x-3)\cdots (x-(n-1)). $$
I tried an example (did this in Paint):
Could someone help me understand what I did wrong? Thank you!
You misunderstood how the arrangement should look. For $n=2$, you have the lines (hyperplanes) $x_1=0$, $x_2=0$, and $x_1=x_2$. So your hyperplane arrangement looks like
Remember, the rank of the central arrangement $\mathcal{B}$ is the dimension of the space spanned by the normal vectors of the hyperplanes in $\mathcal{B}$. Now it should make sense why the characteristic polynomial is $(x-1)^3=x^3-3x^2+3x-1$.
For $n=3$, this is harder to draw (I don't have the software for it), but your hyperplanes would be $x_1=0$, $x_2=0$, $x_3=0$, $x_1=x_2$, $x_1=x_3$, and $x_2=x_3$. I actually can't understand why the characteristic polynomial is $(x-1)^2(x-2)$.
Maybe someone could explain how to prove that the characteristic polynomial is $(x-1)^2(x-2)(x-3)\cdots (x-(n-1))$ for any $n\in \mathbb{N}$??