Solving the LP $\min\{c^Tx\; :\; Ax=b\}$ Let $A\in\Bbb{R}^{m\times n}$, with rank$(A)=m, b\in\Bbb{R}^m, c\in\Bbb{R}^n$. Solve the LP$$\min\{c^Tx\; :\; Ax=b\}$$ and determine when it is unbounded:
$\min\{c^Tx\; :\; Ax=b\}=\max\{-b^Ty\; :\; A^Ty=-c\}$
If there exists no $y$, such that $A^Ty=-c$ then our original LP is unbounded.
We can rewrite $y$ as $y=-A^{-T}c$, because the pseudoinverse of $A$ exists, as $A$ is given to have full rank. Further, this means $y$ is unique.
So assuming that there exists such a $y=-A^{-T}c$, then we have an optimal $x$ with which the solution to the LP is
$c^Tx=-b^Ty\iff c^Tx=b^TA^{-T}c$
Does this work? Any feedback is appreciated. Thank you

$m < n$ and $rank(A)=m \implies$ $n - m$  free parameters exist. I suggest to consider $x \leq 0$ in order to show easily $c^Tx$ is unbounded below.
On the other hand, if $\mathbf x \in R^n$, because $rank(A)=m$ then $n - m$ free parameters exist, so you could prove that $c^Tx$ is unbounded below meaning that for each $k \in R$ at least one $x_0$ exists so that $c^Tx_0 < k$.
Let $B$ be the submatrix of $A$ having $m$ linearly independent columns (rows). Because $ \det(B) \ne 0 \implies B^{-1}$. We can write $A=(B|D)$ and consider $\mathbf x=( \mathbf x_B, \mathbf x_D)= (x_{B,1}, \cdots, x_{B,m}, x_{D,m+1}, \cdots,  x_{D,n})$ putting in evidence the $m$ free parameters. Now, we can observe that $A \mathbf x=B \mathbf x_B + D \mathbf x_D = \mathbf  b$, because $B^{-1}$ exists we get:
$ \mathbf x_B = B^{-1}b - B^{-1}D \mathbf x_D$.
So, we have $<c,x> = <c_B , x_B> + <c_D , x_D> = c_B (B^{-1}b - B^{-1}D x_D) + C_D x_D $ and objective function can be written as
$ \min <\mathbf c, \mathbf x> = c_B B^{-1}b + \min < C_D - B^{-1}D , \mathbf x_D > $ giving an explicit dependency from $x_D$.
As a result, $ <\mathbf c, \mathbf x>$ is unbounded if and only if $< C_D - B^{-1}D , \mathbf x_D > $ is unbounded.
At this point, it is sufficient to take $ (C_D - B^{-1}D) \mathbf x_D < k $
