# Find a vector in the column space whose entries are distinct

Let $$A$$ be a real matrix of size $$m\times n$$, i.e., $$A \in \mathbb{R}^{m\times n}$$. Suppose all rows of $$A$$ are distinct, i.e., $$A_i \ne A_j$$ for $$i\ne j$$. Note that $$A$$ might not be a full rank matrix as it allows $$A_i = 2A_j$$. I want to find a vector (at least existence) in the column space of $$A$$ whose entries are all distinct.

I know that if $$m \le n$$ and $$A$$ is a full rank matrix, for any vector $$b$$ whose entries are distinct, we can find a vector $$Aw \in Col(A)$$ where $$w = A^\dagger b$$ ($$A^\dagger$$ is the Moore-Penrose inverse of $$A$$). However, I am not sure how to prove or disprove this in other cases, i.e., $$A$$ is not full rank and/or $$m > n$$.

The required vector $$x$$ always exists.
To begin, as the first two rows of $$A$$ are different, $$a_{1j}\ne a_{2j}$$ for some $$j$$. If we take $$x=e_j$$, the first two entries of $$Ax=a_{\ast j}$$ will be distinct.
Now suppose we have found an $$x$$ such that the first $$r\,(\ge2)$$ entries of $$y=Ax$$ are distinct. Suppose $$y_{r+1}=y_i$$ for some $$i\le r$$. Pick an index $$j$$ such that $$a_{ij}\ne a_{r+1,j}$$. Replace $$x$$ by $$x+te_j$$ for a sufficiently small $$t>0$$. With this new $$x$$, the first $$r+1$$ entries of $$Ax$$ are distinct. Continue in the manner, we will finally obtain a desired vector $$x$$.