For $A \in \mathrm{Lin}(\mathbb{R}^n, \mathbb{R}^m)$ , $m=n$ and $A$ is invertible, I am able to prove the existence of a solution $x$ such that $A x \succeq 0$ where $\succeq$ denotes componentwise nonnegativity.
I prove existence by using the fact that the rows of $A$ form a basis. But if $m \neq n$, then can I say anything about the solvability of $A x \succeq 0$ ?
I am interested in developing a solution using only ideas from linear algebra (without invoking Farkas Lemma and linear programming)
Certainly $A \cdot 0 = 0$ is always componentwise nonnegative.
For a nonnegative, nonzero solution it's enough that the columns of $A$ span $\mathbb{R}^m$, since every element of $\mathbb{R}^m$ will have the form $Ax.$
Otherwise, there's no reason why the range of $A$ should contain any nonzero vectors whose entries all have the same sign, and you can't conclude anything based only on the size of the matrix.