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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.

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    $\begingroup$ For any matrix $A$ won't $T(A)=T(\lambda A)$ be the same $\endgroup$
    – meh
    Feb 7, 2014 at 18:46
  • $\begingroup$ Yes, but $\lambda A$ won't be in $S$. $\endgroup$
    – Kevin P. Costello
    Feb 7, 2014 at 21:18
  • $\begingroup$ @Costello . Ah, my bad. $\endgroup$
    – meh
    Feb 8, 2014 at 3:36
  • $\begingroup$ Nonnegative entries... Can't the denominators in $D_A$ vanish? $\endgroup$
    – darij grinberg
    Feb 9, 2014 at 20:39
  • $\begingroup$ @darijgrinberg Thank you, I meant "strictly positive" (and corrected the question above). $\endgroup$
    – Bill Bradley
    Feb 10, 2014 at 20:21

1 Answer 1

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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 $T(X)\neq T(X')$.

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