The problem set up is as follows: Let $\omega^{(i)} \in \mathbb{R}^n$, for $i=1,2,...,k$, $k \le n$, be i.i.d. random vectors (whose distribution is irrelevant). Also, let $A \in \mathbb{R}^{m \times n}$.
The paper I'm reading then claims that (1) since $\omega^{(i)}$ are random vectors, they lie in general linear position, and therefore (2) no linear combination of $\omega^{(i)}$ lies in the null space of $A$.
I understand why (1) holds but can't see the reason for (2). The paper doesn't explain this result or give any citation for a proof, so I assume it's a classical result, but Google/Scholar searches didn't yield anything, and I don't know enough about algebraic geometry to even begin a proof.
Either posting a proof or simply a link to one would be sufficient for me. Thanks in advance!
According to the Omnomnomnom's post, assume that $dim(\ker(A))=p\leq n-k$. Then we randomly choose a basis $B=\{e_1,\cdots,e_{n}\}$ s.t. $span(\{e_1,\cdots,e_{p}\})=\ker(A)$ (firstly randomly choose the $p $ first vectors in $\ker(A)$). Then, in the basis $B$, $\det(e_1,\cdots e_{n-k},v_1,\cdots,v_k)=\det(\begin{pmatrix}I_{n-k}&U_{n-k,k}\\0_k&V_{k,k}\end{pmatrix})=\det(V)$ where $V$ is random (if the $(v_i)$ and $A$ are independent). Finally $\det(V)\not= 0$ almost everywhere and we are done.