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**:

- Set $A_{(0)} := A$ and $k:=0$
- Set $d_{(k)} := A_{(k)} \mathbf{1}$
- Set $A_{(k+1)} := B^{1/2}D_{(k)}^{-1/2} \cdot A_{(k)} \cdot D_{(k)}^{-1/2}B^{1/2}$
- 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.