Matrix with some constraint on coefficients Let $A \in \mathbb{R}^{n\times m}$ be a matrix satisfying the following property:
$$
a_{ij} - a_{ik} = a_{hj} - a_{hk},
$$
for every $i,h = 1, \dots, n$ and every $j, k = 1, \dots, m$.
Question: what can I say about $A$? I know that it is quite a vague question, but I would like to better understand what kind of matrices satisfy the property above.
The only thing I notices so far is that the difference between any two rows $A_i = (a_{i1}, \dots, a_{im})$ and $A_j = (a_{j1}, \dots, a_{jm})$ is a constant vector $(k, \dots, k)$ for some constant $k$ that depends on $i$ and $j$.
Similarly, the difference between any two columns is a constant vector.
From this, it's easy to see that the rank of $A$ is at most 2.
Is there anything else that I can say and that I have overlooked?
 A: As you noted, $A$ has rank at most $2$. In fact, we can get a nice rank factorization of $A$.
I will focus on the case that $A$ has rank exactly $2$ (i.e. its rows are non-constant). Let $k_{i}$ denote the constant such that $A_i - A_1 = (k_i,\dots,k_i)$. Let $e$ denote the row-vector $e = (1,\dots,1)$. Then we can write $A = CF$, where
$$
C = \pmatrix{1&0\\1&k_2\\ \vdots & \vdots \\ 1 & k_n},\quad  F = \pmatrix{A_1\\ e} = \pmatrix{a_{11} & \cdots & a_{1m}\\1 & \cdots & 1}.
$$

This leads to some nice consequences. If we want the singular values of $A$, then we know from its rank that $A$ has zero as a singular value with multiplicity at least $n-2$. Moreover, from the fact that $A^TA = F^TC^TCF$, we can conclude its non-zero singular values are the square roots of the eigenvalues of the matrix
$$
(C^TC)(FF^T) = \pmatrix{n & \sum_i k_i \\ \sum_i k_i & \sum_i k_i^2} \pmatrix{\sum_i a_{1i}^2 & \sum_i a_{1i}\\ \sum_i a_{1i} & n}.
$$
If $A$ is square, then we know from its rank that it has $0$ as an eigenvalue with multiplicity at least $n-2$. Its non-zero eigenvalues must be equal to the eigenvalues of
$$
FC = \pmatrix{\sum_{i=1}^n a_{1i} & \sum_{i=2}^n a_{1i}k_i\\ n & \sum_{i=2}^n k_i}.
$$
