Given a polyhedron $P$ specified by a set of linear constraints $P=\{x \in \mathbb{R}^n \mid Ax \le b \}$, what are the conditions on the matrix $A$ such that $P$ is bounded?
I have the following intuition but lack the proof: Rewrite (WLOG I believe) $P=\{x \in \mathbb{R}^n \mid \langle \alpha_i,x\rangle \le 1,i=1,2,\ldots,m \ge n+1\}$, then $P$ is bounded iff the cone formed by the vectors $\alpha_i$ equals $\mathbb{R}^n$. (Here $(\alpha_i)_j = A_{ij}/b_i$ .)
As shown in this answer a non-empty polyhedron ${\cal P}$ is unbounded if, and only if, for every $z\in\cal P$ there exists a ray with origin in $z$ included in $\cal P$.
If --in addition-- we express ${\cal P} = \{x\mid Ax \preceq b\}$, we obtain that the existence of such a ray is equivalent to the existence of $v\ne0$ such that $Av\preceq0$.