One-sided inverses of a stochastic matrix Suppose I have a matrix $A \in [0;1]^{n \times m}$ which is broad ($n < m$), full-rank, and row-stochastic, i.e.,
$$
A \mathbf{1}_m = \mathbf{1}_n
$$
where $\mathbf{1}_k$ denotes the all-ones vector of size $k$.
You can easily convince yourself that the Moore-Penrose inverse
$A^\mathsf{T}\bigl(AA^\mathsf{T}\bigr)^{-1}$ is not generally row-stochastic.
But if I consider the class of all right-inverses
$$
A^+=BA^\mathsf{T}\bigl(ABA^\mathsf{T}\bigr)^{-1}
$$
with arbitrary square full-rank $B$, can I at least ensure by an appropriate choice of $B$ that the rows sum to $1$, i.e.,
$$
BA^\mathsf{T}\bigl(ABA^\mathsf{T}\bigr)^{-1} \mathbf{1}_n = \mathbf{1}_m
$$
?
NB: I am not requiring that the entries of $A^+$ be in $[0;1]$, but only that the row entries sum up to one.
 A: No, because $A$ may not have any row-stochastic right-inverse in the first place. E.g. suppose
$$
A=\pmatrix{\frac13&\frac13&\frac13\\ \frac13&\frac{1+\epsilon}3&\frac{1-\epsilon}3}.
$$
Then every right-inverse of $A$ must be of the form $\pmatrix{a&d\\ b&e\\ c&f}$ with $a+b+c=d+e+f+\epsilon(e-f)=3$. If this is a stochastic matrix, its entries must be bounded between $0$ and $1$. Therefore, $e-f$ is bounded and when $\epsilon>0$ is sufficiently small, the sum of all entries of this right-inverse is close to $6$. Yet, the sum of all entries of a $3\times2$ row-stochastic matrix must be equal to $3$. So, $A$ cannot possibly possess a right-inverse that is also row-stochastic.
Edit. In the new setting where the desired right inverse is not required to be a nonnegative matrix, the answer is affirmative. This follows from the following two facts:

*

*When $A$ is a row-stochastic matrix of full row rank, it has a right inverse $R$ such that $R\mathbf1_n=\mathbf1_m$. Pick any right inverse $R_0$ of $A$ and define
$R=R_0+\frac1n(\mathbf1_m-R_0\mathbf1_n)\mathbf1_n^\top$. Then
\begin{aligned}
AR&=AR_0+\frac1n(A\mathbf1_m-AR_0\mathbf1_n)\mathbf1_n^\top\\
&=I_n+\frac1n(\mathbf1_n-I_n\mathbf1_n)\mathbf1_n^\top\\
&=I_n,\\
R\mathbf1_n
&=R_0\mathbf1_n+\frac1n(\mathbf1_m-R_0\mathbf1_n)\mathbf1_n^\top\mathbf1_n\\
&=R_0\mathbf1_n+(\mathbf1_m-R_0\mathbf1_n)\\
&=\mathbf1_m.\\
\end{aligned}

*When $A$ is a real matrix of full row rank, every right inverse $R$ of $A$ can be written in the form of $BA^\top(ABA^\top)^{-1}$. In fact, if we define $B=RA+I_m-A^\top(AA^\top)^{-1}A$, then
\begin{aligned}
ABA^\top
&=ARAA^\top+AA^\top-AA^\top(AA^\top)^{-1}AA^\top\\
&=AA^\top+AA^\top-AA^\top\\
&=AA^\top.
\end{aligned}
Hence
\begin{aligned}
BA^\top(ABA^\top)^{-1}
&=\left[RA+I_m-A^\top(AA^\top)^{-1}A\right]A^\top(AA^\top)^{-1}\\
&=R+A^\top(AA^\top)^{-1}-A^\top(AA^\top)^{-1}\\
&=R.
\end{aligned}
