[m]\setminus\{i\} \right\}$. In this way, we divide the whole space $\operatorname{\mathbb{R}}^n$ into parts such that any two of them have only trivial intersection $\{ \pmb{0} \}$, each part resulted is called a [*region*]{} of $\operatorname{\mathbb{R}}^n$ with respect to $\{ \pmb{b}_i : i \in [m] \}$.

Because $X$ is a subspace of $\mathbb{R}^n$, it is naturally divided into a number of parts by those dividers $\{ \pmb{b}_i^{\perp} : i \in [m] \}$, each of which is called a [*region*]{} of $X$. More precisely, a region of $X$ is obtained by intersecting $X$ with some region of $\operatorname{\mathbb{R}}^n$. A moment’s reflection shows that any region of $X$ is convex.

Now let us try to carve up $X$ inside, that shows another way of defining region. First, we figure out those orthogonal projections of the group $\pmb{b}_1,\ldots,\pmb{b}_m$ onto $X$, which are denoted by $\pmb{x}_1,\ldots,\pmb{x}_m$. Because ${\mathrm{span} \hspace{0.4mm} \{ \pmb{b}_i : i \in [m] \}} = \operatorname{\mathbb{R}}^n$, ${\mathrm{span} \hspace{0.4mm} \{ \pmb{x}_i : i \in [m] \}} = X$. It is clear that each vector $\pmb{x}_i$ in the group such that $\| \pmb{x}_i \| \neq 0$ possesses uniquely one orthogonal complement in $X$, which is denoted by $\pmb{x}_i^{\perp}$ and called the [*divider of $X$ associated 