In my specific problem, I have a $(2\times 4)$ matrix $A$ of non-zero (complex) entries. The rows are linearly independent, hence $\text{rank}(A)=2$. Besides, the first column of $A$ is composed of the same non-zero number. Can I infer that, by removing the first column, the obtained $(2\times 3)$ matrix will still have (full) rank 2?
Does this expand to larger matrices of size $(n-2 \times n)$ where a single columns of same entries is removed?
No. Suppose that$$A=\begin{bmatrix}1&1&1&0\\1&2&2&0\end{bmatrix}.$$Then the rows are linearly independent. But $\operatorname{rank}\left(\left[\begin{smallmatrix}1&1&0\\2&2&0\end{smallmatrix}\right]\right)=1$.