Polyhedron's Representations and spanning the Euclidian space Let's say you have to different representations of the same polyhedron $P\neq \emptyset$:

$$P=\{x\in \mathbb{R}^n\;|\;h_i^Tx\leq c_i, i=1,...,k \} =\{x\in \mathbb{R}^n\;|\;g_j^Tx\leq d_i, j=1,...,l \}. $$

I'd like to show that given those representations, if I know that the vectors $h_1, . . . ,h_k$ are spanning $\mathbb{R}^n$, then so do also the vectors $g_1, ..., g_l$.

How would you do that ?

I was thinking about using the properties of the polyhedron, seen as geometric objects, but I ended up showing nothing.
Alright, I've found a way to do it :

Let's consider the matrix $A$ whose lines are the vectors $a_i^T$. It has to have full column-rank, since the $a_i$ are spanning $\mathbb{R}^n$ (we necessarily have $m\geq n$).

Since it has full column-rank, it means our polyhedron can't contain a line. 
(I let you prove the lemma saying a polyhedron does not contain a line if and only if its matrix has full column-rank).

But now, since it doesn't contain a line, it has to have one extreme point (with $n$ active constraints), because if you take a point in P, choose a direction in which every active constraints remain active and move along this direction, you'll end up meeting another constraint, which will become active. (You have to met such a constraint, since $P$ can't contain a line.) Just repeat the procedure until you've got $n$ linearly independent active constraints. Actually, what you'll get is a basic feasible solution, in terms of Linear Programming.

Now you've got a point with $n$ linearly independent active constraints and you know you can represent it using the $h_i$ too. Let's do it. And that's it, you've got $n$ linearly independent vectors $h_j$, since $\mathbb{R}^n$ is of dimension $n$, you know that those $h_j$ are spanning the whole space.
Alright, I've found a way to do it :

Let's consider the matrix $A$ whose lines are the vectors $a_i^T$. It has to have full column-rank, since the $a_i$ are spanning $\mathbb{R}^n$ (we necessarily have $m\geq n$).

Since it has full column-rank, it means our polyhedron can't contain a line. 
(I let you prove the lemma saying a polyhedron does not contain a line if and only if its matrix has full column-rank).

But now, since it doesn't contain a line, it has to have one extreme point (with $n$ active constraints), because if you take a point in P, choose a direction in which every active constraints remain active and move along this direction, you'll end up meeting another constraint, which will become active. (You have to met such a constraint, since $P$ can't contain a line.) Just repeat the procedure until you've got $n$ linearly independent active constraints. Actually, what you'll get is a basic feasible solution, in terms of Linear Programming.

Now you've got a point with $n$ linearly independent active constraints and you know you can represent it using the $h_i$ too. Let's do it. And that's it, you've got $n$ linearly independent vectors $h_j$, since $\mathbb{R}^n$ is of dimension $n$, you know that those $h_j$ are spanning the whole space.