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 $m > n$.
Any comments/suggestions/answers will be very appreciated.
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 $x$.