Limit of alternated row and column normalizations Let $E_0$ be a matrix with non-negative entries.
Given $E_n$, we apply the following two operations in sequence to produce $E_{n+1}$.
A. Divide every entry by the sum of all entries in its column (to make the matrix column-stochastic).
B. Divide every entry by the sum of all entries in its row (to make the matrix row-stochastic).
For example:
$E_0=\begin{pmatrix}
\frac{2}{5} & \frac{1}{5} & \frac{2}{5} & 0 & 0\\ 
\frac{1}{5} & 0 & \frac{7}{10} & \frac{1}{10} & 0\\ 
0 & 0 & 0 & \frac{3}{10} & \frac{7}{10}
\end{pmatrix}\overset{A}{\rightarrow}\begin{pmatrix}
\frac{2}{3} & 1 & \frac{4}{11} & 0 & 0\\ 
\frac{1}{3} & 0 & \frac{7}{11} & \frac{1}{4} & 0\\ 
0 & 0 & 0 & \frac{3}{4} & 1
\end{pmatrix}\overset{B}{\rightarrow}\begin{pmatrix}
\frac{22}{67} & \frac{33}{67} & \frac{12}{67} & 0 & 0\\ 
\frac{44}{161} & 0 & \frac{12}{23} & \frac{33}{161} & 0\\ 
0 & 0 & 0 & \frac{3}{7} & \frac{4}{7}
\end{pmatrix}=E_1$
What is the limit of $E_n$ as $n \to \infty$?

Additional remarks.
In my problem, the matrix has $c\in \{1,2,\dots,5\}$ rows and $r=5$ columns (note that the two letters are reversed, but in the original context of this problem these letters $r$ and $c$ do not actually stand for rows and columns). So $E_0$ can be $1\times 5$, $2\times 5$, ... or $5\times 5$.
We denote with $(e_n)_{ij}$ the entries of $E_{n}$; hence $(e_n)_{ij}\in[0;1]$ and $\forall i \sum_{j=1}^{r}(e_n)_{ij}=1$ for $n>0$.
I managed to express $(e_{n+1})_{ij}$ as a function of $(e_{n})_{ij}$ :
$$(e_{n+1})_{ij}=\frac{\frac{(e_{n})_{ij}}{\sum_{k=1}^{c}(e_n)_{kj}}}{\sum_{l=1}^{r}\frac{(e_n)_{il}}{\sum_{k=1}^{c}(e_n)_{kl}}}$$
What I can't seem to find now is an expression $(e_{n})_{ij}$ as a function of $(e_{0})_{ij}$, to be able to calculate $\underset{n \to +\infty }{lim}(e_n)_{ij}$
I wrote code to compute this iteration; when I ran it with the previous example $E_0$, I found out that:
$E_0=\begin{pmatrix}
\frac{2}{5} & \frac{1}{5} & \frac{2}{5} & 0 & 0\\ 
\frac{1}{5} & 0 & \frac{7}{10} & \frac{1}{10} & 0\\ 
0 & 0 & 0 & \frac{3}{10} & \frac{7}{10}
\end{pmatrix}\overset{n \rightarrow+\infty}{\rightarrow}E_n=\begin{pmatrix}
\frac{7}{25} & \frac{3}{5} & \frac{3}{25} & 0 & 0\\ 
\frac{8}{25} & 0 & \frac{12}{25} & \frac{1}{5} & 0\\ 
0 & 0 & 0 & \frac{2}{5} & \frac{3}{5}
\end{pmatrix}$
Not only do the row sums equal $1$, but the column sums equal $\frac{3}{5}$: it seems that in this process column sums converge to $\frac{c}{r}$.
I'm not a mathematician so I was looking for a simple inductive proof. I tried to express $E_2$ (and so on) as a function of $E_0$, but it quickly gets overwhelming, starting from $E_2$...
 A: When $E_0$ is square (i.e., $r = c$) this procedure is called Sinkhorn iteration or the Sinkhorn-Knopp algorithm (see this Wikipedia page). You can find a wealth of results by Googling those terms, the most well-known of which is that if $E_0$ has strictly positive entries (and again, is square) then the limit of $E_n$ indeed exists and is doubly stochastic.
A: The paragraph below applies to a different problem, where row normalization is split out from column normalization, so I have an $"E_{n+1/2}"$ which will be sometimes different from both $E_n$ and $E_{n+1}$. (If the starting matrix "looks like" an order k square stochastic matrix, then it will be invariant under both normalizations.) 
Note that some cases will not converge to a single limit. Given a c by r matrix of all ones, normalizing by column (of r entries) results in all entries being 1/r, while normalizing by rows gives all entries being 1/c, so the sequence will fluctuate between these two. Except when the entries are all zero, I would expect a similar oscillation with any other starting nonzero binary matrix.  You might be able to establish oscillation for matrices with more distinct values.
Getting back to the posted problem, the transformations have an invariance under permutations of rows, similarly of columns. Thus if the input looks like an upper two by two diagonal block matrix and a lower two by three nonzero matrix, the upper block may converge on a stochastic two by two block, while the lower will be influenced (if it converges at all) by a ratio of 3/2. So the block structure will influence the results, and the ratio c/r might not apply.
Gerhard "Goes This Way And That" Paseman, 2019.12.28.
