# Convergence of Symmetric Iterative Proportional Fitting

Let $A$ denote a symmetric matrix of non-negative entries, whose rows (and columns) sum up to the all-positive vector $d := A\mathbf{1}$ with $\mathbf{1}$ the all-one-vector. Denote $D := diag(d)$ by a capital letter, similar for any other vector. Let $b$ denote any all-positive vector.

Assume there exists a positive diagonal matrix $S$ such that $A^*:=SAS$ provides $b$ as its row (and column) sums, that is $A^*\mathbf{1} = b$.

It is well-known that iterated proportional fitting provides a sequence $(A=:\tilde A_{(0)}, \tilde A_{(1)}, \tilde A_{(2)}, \ldots)$ that converges to $A^*$. It alternately scales rows appropriately, then columns, then rows, and so forth. This implies that no $\tilde A_{(k)}$ for $k>0$ is symmetric.

However, I am interested in the following symmetrization:

1. Set $A_{(0)} := A$ and $k:=0$
2. Set $d_{(k)} := A_{(k)} \mathbf{1}$
3. Set $A_{(k+1)} := B^{1/2}D_{(k)}^{-1/2} \cdot A_{(k)} \cdot D_{(k)}^{-1/2}B^{1/2}$
4. Increase $k \gets k+1$ and continue at Step 2

This provides a sequence of symmetric matrices $(A=:\tilde A_{(0)}, \tilde A_{(1)}, \tilde A_{(2)}, \ldots)$. It is known that it converges to $A^*$ in the case that $b = \mathbf{1}$, thus, if the limit is doubly stochastic.

But does this also hold for $b \ne \mathbf{1}$?

I strongly believe this, but I cannot find any results on this, and the proof of the case $b = \mathbf{1}$ seems not to generalize.

Indeed he proves that SIPF converges to the same limit as IPF, for any all-positive vector $b$.
This further implies that the limit matrix $A^*$ is precisely the Bregman projection of $A$ onto the set $\mathcal{S}(b,A)$ defined by all symmetric non-negative matrices of row/column sums $b$, where the projection is with respect to relative entropy error (generalized Kullback-Leibler divergence) $$D(X\|Y) := \sum_{i,j} x_{ij} \log(x_{ij}/y_{ij}) - x_{ij} + y_{ij}.$$
$$A^* \ = \ \text{arg}\,\text{min}_{X \in S(b,A)} \ D(X\|A)$$
(whenever existing, i.e., whenever $\mathcal{S}(b,A) \ne \emptyset$).