TITLE: Injectivity of matrix "fingerprint"
QUESTION [7 upvotes]: Consider $S$, the set of all $n\times m$ real matrices with specified row sums $(r_1,...,r_n)$, column sums $(c_1,...,c_m)$, and strictly positive entries.
For any matrix $A$, define
$$ D_A(i,j)=\frac{A_{i,j}A_{i+1,j+1}}{A_{i+1,j}A_{i,j+1}} $$
Let $T:S\rightarrow (R^+)^{(n-1)\times (m-1)}$ be the map where the $(i,j)$-th entry of $T(A)$ is $D_A(i,j)$.
Question: Given $T(A)$, can we recover $A\in S$ (i.e. is $T$ injective)?
Motivation: This question arose in the context of a MathOverflow question about a matrix version of the log-likelihood ratio.
Comment #1: For any $1\leq i<i'\leq n$ and $1\leq j<j'\leq m$, the $D(i,j)$ are sufficient to compute
$$ \frac{A_{i,j}A_{i',j'}}{A_{i',j}A_{i,j'}} $$
Comment #2: $T$ is surjective (where "$R^+$" means "strictly positive reals").  So the answer to this question will determine if $T$ is a bijection.
Comment #3: If there is a standard name for the $D_A(i,j)$ quantity, please let me know.

REPLY [4 votes]: Yes, the map $T$ is injective. Suppose you have matrices $X,X'$ with equal row and column sums and positive entries, so that $X'=X+E$ where $E=(e_{ij})$ is not identically zero. It is a nice combinatorial exercise to show that there must exist indices $p,q,r,s$ so that $e_{pq}$ and $e_{rs}$ are both nonnegative and $e_{ps},e_{rq}$ are both non positive and they are not all zero. Then we have $$\frac{x'_{pq}x'_{rs}}{x'_{ps}x'_{rq}} > \frac{x_{pq}x_{rs}}{x_{ps}x_{rq}}$$ which shows $T(X)\neq T(X')$.