Suppose that:
$$\{x\in\mathbb{R}^n\mid a_i^Tx\le b_i, i=1,2,\dots,m\}=\{x\in\mathbb{R}^n\mid g_j^Tx\le h_j, j=1,2,\dots,k\}$$
How can I show that if the vectors $a_1,\dots,a_m$ span $\mathbb{R}^n$ then $g_1,\dots,g_k$ span $\mathbb{R}^n$ too?
I actually don't know how to use the hypothesis that both sets are equal. This is what I did for now:
If $a_1,\dots,a_m$ span $\mathbb{R}^n$, then $m\ge n$ and (wlog) we can extract a basis $a_1,\dots,a_n$. Hence every vector in $\mathbb{R}^n$ (including all $g_j$, $b$ and $h$) can be written as a linear combination of $a_1,\dots,a_n$.
But I admit that from this point I don't know where to go and I am not sure how to proceed. Thanks in advance for the help.
This is only a partial answer.
Let $v_1, …, v_r ∈ ℝ^n$, $c_1, …, c_r ∈ ℝ$ and $A = \{x ∈ ℝ^n;\; \text{$v_i^T x ≤ c_i$, for $i = 1, …, r$}\}$. Observe that $0 ∈ A$ if and only if $\min \{c_1, …, c_r\} ≥ 0$.
Let’s assume $0 ∈ A$. Let $L$ be a line through the origin in $ℝ^n$. Then $L ⊂ A$ if and only if $L ⊂ 〈v_1, …, v_r〉^{\perp}$. So $v_1, …, v_r$ span $ℝ^n$ if and only if there is no line in $A$.
Therefore, in the case the null vector is contained in your sets, this already proves the assertion, as your sets are designed just like $A$.