An arrangement of hyperplanes in $\mathbb{R}^d$ is simple if the hyperplanes are in general position (for every $1\leq k\leq d+1$, the intersection of $k$ hyperplanes is $(d-k)$-dimensional).
My question is that how can I show that if $n\geq d+1$, an arrangement of $n$ hyperplanes is simple if and only if every $d$ intersect at a single point and no $d+1$ have a point in common?
If you have an arbitrary system of linear equations and add to it one equation, the dimension of the space of solutions can decrease by at most 1. Do take any $d$ equations of your hyperplanes, order them arbitrarily, and denote by $n_k$ the dimension of the space of solutions of the first $k$ equations. Then $$d=n_0\geq n_1\geq\ldots\geq n_d=0,$$ and $n_j-n_{j+1}\leq 1$. It follows that all $n_{j}-n_{j+1}=1$, for all $j=0,\ldots,d$, and thus $n_k=d-k$.
Remark: only one condition was used: that any $d$ equations have one solution.
Remark 2. If you define $\mathrm{dim}(\emptyset)=-1$, then the same argument goes through and shows that your last condition alone is equivalent to the rest.
