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](https://mathoverflow.net/questions/36174/multinomial-transformation-for-matrices/156583) 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.



Injectivity of matrix "fingerprint"
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 $T(X)\neq T(X')$.


