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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.

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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.

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  • $\begingroup$ Thanks for pointing this out. In fact I only care about row sums being equal to one, I don't care about entries of the right-inverse being within $[0;1]$. I have corrected the problem statement accordingly. $\endgroup$ – jens Sep 30 '15 at 15:46

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