4

Let $A$ be an entry-wise strictly positive matrix. Sinkhorn's theorem assures that there exist strictly positive diagonal matrices (unique up to scaling) $D_1, D_2$ such that $D_1 A D_2$ is doubly stochastic.

If I am given with two probability vectors $p,q$. Then, do there exist two strictly positive diagonal matrix $D_1, D_2$ such that $D_1 A D_2$ is stochastic (i.e., the sums of the rows are $1$) and $(D_1 A D_2)^\top p=q$?

I think such modification is possible. However, I am not able to find any proof of this. Can anyone please help me in this regard?

Rodrigo de Azevedo
  • 23,223
DeltaEpsilon
  • 1,222
  • Does the $t$ perform matrix transposition? – Rodrigo de Azevedo Feb 06 '25 at 14:46
  • Yes, it is the usual transport. Like for doubly stochastic matrix we have the conditions $A$1$=$1 and $A^t$1$=$1. – DeltaEpsilon Feb 06 '25 at 16:40
  • So, you have $2n$ unknowns, $2n$ non-negativity constraints and $n + 2n = 3n$ equations, right? – Rodrigo de Azevedo Feb 06 '25 at 16:58
  • You may need to require that $q>0$, otherwise, if $q$ has zero entries, then $Sp\ne q$ whenever $S>0$, $p\ge0$ and $p\ne0$. – user1551 Feb 06 '25 at 17:21
  • Okay, so let us assume $q>0$. For instance, take $q=(1,\ldots,1)$. Then if $p>0$, then I can do that. But what happens if some entries of p are $0$? – DeltaEpsilon Feb 06 '25 at 20:56
  • 1
    The Iterative Proportional Fitting algorithm is closely related to Sinkhorn, but it is not limited to generating doubly stochastic matrices. It allows you to specify the row & column sums independently. – greg Feb 08 '25 at 15:01
  • As the sums of the columns of $(D_1 A D_2)^\top$ are all 1, the entries of the vectors $p$ and $q$ must sum up to the same number. – Helmut Feb 08 '25 at 15:35
  • 2
    This is only possible if $p$ and $q$ satisfy additional assumptions. Let $\delta = \min_i q_i$ and let $e$ be the vector with all entries $=1$. Then $(D_1AD_2)^T(p-\delta e) = q - \delta e$. This is zero for some coordinate, which is only possible if some entries of $p - \delta e$ are negative, i.e. $\min_i p_i < \min_i q_i$. Similarly, $\max q_i < \max_i p_i$. – Hans Engler Feb 14 '25 at 00:44

0 Answers0